tag:blogger.com,1999:blog-15936079425057947052024-03-20T02:00:17.026-07:00Quantum GazetteAn exploration into quantum physics and science writingGelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.comBlogger31125tag:blogger.com,1999:blog-1593607942505794705.post-89215200982491625862020-03-13T17:48:00.001-07:002020-03-26T06:36:18.770-07:00On the security of quantum key distribution protocolsIn an earlier post, we described the <a href="https://quantumgazette.blogspot.com/2016/09/the-bb84-protocol-for-quantum-key.html" target="_blank">BB84 protocol</a>, which allows two parties, say Alice and Bob, to share a secret bit string (called a key) using quantum states.<br />
<br />
The protocol works because it uses a set of quantum states that cannot be fully distinguished from each other. Still, you may ask why would this be enough? If we run the protocol, how do we know that the keys the Alice and Bob get are indeed identical and private?<br />
<br />
This means we must show that Alice's key is the same as Bob's key, and that any outside party, say Eve, cannot have too much knowledge about either key. That is exactly what we would need to do to prove the security of a quantum key distribution (QKD) protocol.<br />
<br class="Apple-interchange-newline" />
We say that a QKD protocol is <b>secure</b> if it produces a correct and secret key. A key is <b>correct</b> if Alice and Bob obtain precisely the same bit string. A key is <b>secret</b> if given a set of $N$ possible keys, the probability that Alice and Bob get any particular key is $\frac{1}{N}$. In other words, the key is uniformly distributed.<br />
<br />
<a name='more'></a><br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiH8WX93gqAIzTfxeIzHnfm6GVag6-d-jQuIGJxzNjJt2gnqU-IanE0zDQp6fkDxziK3uYY9dFP3vq-TjK8Dh0REwRFMX7zImd2-y5FVpE5IinsakKtk0VANnCCsu0MvNELWz46_coVS_87/s1600/qkdCorrSec.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="610" data-original-width="1111" height="350" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiH8WX93gqAIzTfxeIzHnfm6GVag6-d-jQuIGJxzNjJt2gnqU-IanE0zDQp6fkDxziK3uYY9dFP3vq-TjK8Dh0REwRFMX7zImd2-y5FVpE5IinsakKtk0VANnCCsu0MvNELWz46_coVS_87/s640/qkdCorrSec.png" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">A standard QKD protocol involves a quantum channel where Alice can send quantum states to Bob, and a classical channel where they can discuss publicly. An eavesdropper Eve has access to both channels but she cannot alter messages in the classical channel. The output of the protocol is a pair of random strings called keys, one each for Alice and Bob. The protocol is secure if (i) Alice and Bob obtain the same key and (ii) the key is uniformly distributed.</td></tr>
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<br />
It is easy to understand the correctness of a key since obviously Alice and Bob cannot share a key if they have different bit strings. It maybe a bit less clear why the secrecy of the key depends on the probability distribution of the set of possible keys.<br />
<br />
Roughly speaking, if some of the bit strings are more likely than others, this tells Eve something about the key. This can be seen by considering a somewhat extreme case. Suppose a certain QKD protocol generates a 2-bit key $k$ such that<br />
<br />
$\Pr[k = 00] = \Pr[k = 01] = 0.45, \quad \Pr[k =10] = \Pr[k = 11] = 0.05.$<br />
<br />
In this case, Eve can just conclude that the first bit of the key is probably 0, and she has a 90% chance of being right. In some sense, this describes how Eve tries to learn about the key: she attacks the protocol so it produces key that is much less uniformly distributed.<br />
<br />
We see that the worst case for Eve is when even after her attack, all the possible keys are equally likely. Here the best she can do is guess purely at random. So now it should be clear why a uniformly random key implies a secret key.<br />
<br />
Of course, in practice we can slightly relax our security demands. We might allow a small chance that Alice and Bob do not get the same key, or we might be satisfied with a key that is nearly uniformly distributed. As long as the probability of obtaining a correct and secret key is very high, that is certainly good enough.<br />
<br />
<br />
<b><u>How to achieve a correct key</u></b><br />
<b><br /></b>
In the BB84 protocol, when Eve is absent then Alice and Bob can generate a secure key from those qubits that Alice sent and Bob measured in the same basis. If Eve attempts to identify the states sent by Alice to Bob, she inevitably introduces errors by making Alice and Bob's keys not match. Alice and Bob can find this out if they use some of their data to check that their keys are the same.<br />
<br />
Note that if the quantum channel is noisy, it also can create errors in the key. So how do we actually get a correct and secret key in the presence of noise, whether due to Eve or to imperfections in the channel? What we can show is that if the noise is low enough, then it is possible to correct the errors and reduce Eve's knowledge about the key. In BB84, once Alice and Bob determine which bits were obtained from the same basis, then Alice and Bob will each have a sifted key.<br />
<br />
The first step is to perform error correction on the sifted keys over the public classical channel, which is called <b>information reconciliation</b>. The general approach is based on the idea of error-correcting codes. Here we will consider an example using the Brassard-Salvail Cascade method [1].<br />
<br />
In Cascade, the sifted keys are split into a number of blocks and the parity of each block are compared. If a difference in parity is found, Alice and Bob perform a binary search to locate and correct the error. That is, they divide the block into two halves, compare the sub-block parities, and continue dividing until the error is found. Once they compare all blocks and correct for all mismatch parities, the round is over.<br />
<br />
In the next round, Alice and Bob first randomly reorder their bit strings then divide them into blocks with twice the size of the previous round. They repeat the process of comparing blocks and finding errors. After four rounds, the corrected strings will be identical with high probability.<br />
<br />
Note that each time Alice and Bob compare parities, they are essentially revealing partial information about the key to Eve. Thus, it is crucial to optimize the initial size $w$ of the blocks relative to the tolerated error rate $e$. In the original Cascade, they recommended an intial block size of $w \approx \lceil \frac{0.73}{e} \rceil$ bits.<br />
<br />
To demonstrate Cascade, suppose Alice and Bob start with the following sifted keys:<br />
<br />
Alice's string: 1010 0110 0100 0010<br />
Bob's string: 1010 0100 0100 0010<br />
<br />
In the first round, they choose a block size of 4. Computing the parities, they find that<br />
<br />
Alice's parities: 0 0 1 1<br />
Bob's parities: 0 1 1 1<br />
<br />
They find a difference in the second block so they perform binary search. Splitting the second block into two halves:<br />
<br />
Alice's 2nd block: 01 10<br />
Bob's 2nd block: 01 00<br />
<br />
Computing the parity of the first half of the second block:<br />
<br />
Alice's 2nd block 1st half parity: 1<br />
Bob's 2nd block 1st half parity: 1<br />
<br />
The first half has same parity so the error must be in the second half. Comparing the first bits of the second half:<br />
<br />
Alice's 2nd block, 1st bit of 2nd half: 1<br />
Bob's 2nd block, 1st bit of 2nd half: 0<br />
<br />
They found the mismatched bits so they can make them equal say if Bob flips his bit. There are no more blocks with different parities so the round is over. Since our example only had a single bit error, the string are already matching.<br />
<br />
Alice's string: 1010 0110 0100 0010<br />
Bob's string: 1010 0110 0100 0010<br />
<br />
However, Alice and Bob would not know this so they will proceed to the second round. To begin, Alice and Bob first have to randomly rearrange their bit strings. This is necessary because if there are two errors in a block, this will not be detected by comparing block parities. So in each round we rearrange the strings in order to redistribute the errors.<br />
<br />
In our example, suppose Alice and Bob perform the following reordering:<br />
<br />
$(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15) \mapsto (1,9,2,10,3,4,15,16,5,6,11,12,7,13,14,8)$<br />
<br />
Afterwards, we have<br />
<br />
Alice's string: 10011010 01001000<br />
Bob's string: 10011010 01001000<br />
<br />
since now the block size is 8. Again Alice and Bob should compare the block parities:<br />
<br />
Alice's parities: 0 0<br />
Bob's parities: 0 0<br />
<br />
They find they are the same so no corrections are needed for this round. Since our example used only 16 bits we can stop here but normally Alice and Bob will proceed for two more rounds before halting.<br />
<br />
<br />
<u><b>How to achieve a secret key</b></u><br />
<br />
Our example shows that for low error rate (like 1 bit error in 16 bits), we can find and correct errors to make Alice and Bob's keys identical. However, this comes at the price of revealing a number of parity bits. In the example, we had to reveal a total of 8 parity bits over two rounds. This means at least half of the key is already known by Eve.<br />
<br />
Clearly, we need to reduce Eve's partial information, whether it was gained during eavesdropping on the quantum channel, or on the classical channel during information reconciliation. This step is known as <b>privacy amplification</b>. It is generally done using a universal hash function, chosen randomly from a publicly known family of such functions. It maps a bit string into a shorter string, where how much the length is reduced is based on a bound on how much Eve knows about the key.<br />
<br />
The bound can calculated based on the expected error rate in the quantum channel and the bit revealed during information reconciliation. A precise calculation of the best bound can be quite complicated. However, there is a relative simple result we can use if we are willing to sacrifice more bits than is absolutely necessary to achieve secrecy.<br />
<br />
Bennett, Brassard, Crépeau and Maurer derived the following theorem [2]: Suppose that for some $n$-bit string, eavesdropping reveals at most $t$ bits to Eve. Let $s < n - t$ be a security parameter and $r = n - t - s$. Then the Eve's expected information about the key can be bounded by $\frac{2^{-s}}{\ln(2)} \approx (1.443)2^{-s}$.<br />
<br />
To see how this works, in our example above with $n=16$ suppose that Alice and Bob chose the error rate threshold to be 12.5%. This means that if all the errors are due to Eve, then Eve can potential know 2 of 16 bits after eavesdropping on the quantum channel. Since 8 parity bits were revealed, then in the worst case Eve might know $t = 10$ of the bits of the corrected keys.<br />
<br />
Suppose we choose $s = 4$. Then $r = n - t- s = 2$. Now Alice and Bob can choose a random full-rank $r \times n$ binary matrix (all entries 0 or 1) to apply to their $n$-bit key when written as a $n \times 1$ vector. Thus, we compute each bit of the key as the parity of several bits. The idea then is if Eve knows bit $x$ but not bit $y$, then she does not know $x \oplus y$.<br />
<br />
In our example, suppose Alice and Bob picked the matrix<br />
<br />
$H = \begin{pmatrix}<br />
0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 \\<br />
1 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 1<br />
\end{pmatrix}. $<br />
<br />
Let $v$ be the key written as a column vector, i..e,<br />
<br />
$v = \begin{pmatrix}<br />
1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0<br />
\end{pmatrix}^{T}.$<br />
<br />
Then<br />
$ K = Hv = \begin{pmatrix}<br />
1 \\ 0<br />
\end{pmatrix}.$<br />
<br />
The theorem above tells us that on average, Eve will know $\frac{2^{-4}}{\ln(2)} \approx 0.0902$ bits of this 2-bit key.<br />
<br />
Of course, in our example, Eve still knows a bit too much since on average she still knows roughly 4.5% of the key. But if we had a longer string where we can choose a larger $s$ we can reduce Eve's knowledge more significantly.<br />
<br />
<br />
<br />
<br />
<b>References:</b><br />
<br />
[1] G. Brassard and L. Salvail "Secret key reconciliation by public discussion", Advances in Cryptology: Eurocrypt 93 Proc. (1993) 410-23.<br />
<br />
[2] C. Bennett, G. Brassard, C. Crépeau, and U. M. Maurer, "Generalized privacy amplification", IEEE Trans. Inform. Tehory 41 (1995) 1915-23.<br />
<br />Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-87189194690107508482018-05-15T00:23:00.000-07:002018-05-15T00:25:18.092-07:00Quantum machine learning algorithm for supervised cluster assignmentMachine learning refers to various methods for deriving patterns from data that can be used to interpret new inputs. This is what is meant by computers that can "learn" without being explicitly programmed. Machine learning techniques are often associated with the development of artificial intelligence, but they are also prevalent in applications that involve large amounts of data, such as in image and speech recognition, and in risk assessment for business and finance.<br />
<br />
Broadly speaking, there are three forms of machine learning: (i) supervised, in which a machine infers a function from a sample of input-output relations (called training data); (ii) unsupervised, in which a machine tries to uncover a hidden structure from assorted data; and (iii) reinforced, in which a machine attempts to develop a strategy for winning a game given a set of rules and objectives.<br />
<br />
Formally, machine learning techniques involve data represented as vectors in a high-dimensional space. As it turns out, quantum information has taught us that quantum computers are good at manipulating high-dimensional vectors for certain tasks. The general aim of quantum machine learning is to develop quantum algorithms that exhibit significant speedup over classical machine learning techniques.<br />
<br />
In many cases, the quantum learning approach seems to amount to running a classical learning problem on a quantum computer, hoping to find a speedup. But in other cases such as neural networks or Bayesian decision theory, the approach is to translate stochastic methods into the language of quantum information, hoping to develop purely quantum methods with no real classical counterpart.<br />
<br />
<a name='more'></a><br />
Here we will describe a quantum algorithm by Seth Lloyd, Masoud Mohseni, and Patrick Rebentrost for evaluating distances and inner products between vectors that has exponential speedup over known classical methods. This was chosen because it is one of the earliest known applications of quantum information to machine learning, and mostly because it is simple enough to be described in detail.<br />
<br />
In the problem of supervised cluster assignment, we are given an unassigned input $\vec{u}$ and a sample of $M$ vectors $\vec{v}_{j}$ and $\vec{w}_{k}$ for two clusters $V$ and $W$. The task is to assign $u$ to either $V$ or $W$. In this case, the typical approach is to compute the mean or centroid of the samples for $V$ and $W$ and find which of these two centroids is closer to $\vec{u}$. If we consider vectors of real numbers then it is sufficient to use the Euclidean distance<br />
<br />
$D(\vec{a},\vec{b}) = | \vec{a} - \vec{b}| = \sqrt{| \vec{a} |^2 + | \vec{b} |^2 - 2 \vec{a} \cdot \vec{b}}$<br />
<br />
as a measure of distance between any two vectors $\vec{a}$ and $\vec{b}$.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5tND8Bx1Lln53t6v9xx1ii7ui2pPB2ylgH_5VkM2g0e4zbkbYumamCdrzAl4B9zMz9HCdQCtKhgXDcYWMU5kktwrlRScDJLm4PTFDmOoUZrCb4DVojj8wdXKbfUHUGnPsXFzMdhV3OwbV/s1600/Slide1.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="590" data-original-width="917" height="410" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5tND8Bx1Lln53t6v9xx1ii7ui2pPB2ylgH_5VkM2g0e4zbkbYumamCdrzAl4B9zMz9HCdQCtKhgXDcYWMU5kktwrlRScDJLm4PTFDmOoUZrCb4DVojj8wdXKbfUHUGnPsXFzMdhV3OwbV/s640/Slide1.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">A diagram for the supervised cluster assignment. Given a sample of vectors from $V$ (squares) and $W$ (triangles) from previously cataloged data and a new input vector $\vec{u}$ (blue dot), the task is to assign $\vec{u}$ to one of the clusters. The standard approach here is to compute the mean or centroid of each sample (green and red crosses), and find which centroid is nearer to $\vec{u}$.</td></tr>
</tbody></table>
<br />
<br />
We will use a quantum algorithm to compute the distance between any pair of vectors. To do this efficiently, we assume that our data for $\vec{u}$, $\vec{v}_{k}$, and $\vec{w}_k$ are stored in some quantum memory, which a quantum computer can access in order to create quantum states of the form<br />
<br />
$|a\rangle = \dfrac{ \vec{a} }{ \sqrt{| \vec{a}|} }$,<br />
<br />
that is, the data vectors are directly encoded into quantum states where the data vector lengths are stored in the norms, i.e., $\langle a | a \rangle = | \vec{a} |$.<br />
<br />
The quantum algorithm computes the distance between vectors $\vec{a}$ and $\vec{b}$ by estimating the inner product between the quantum states<br />
<br />
$| \psi \rangle = \dfrac{1}{\sqrt{2}} \left( \dfrac{|0\rangle |a\rangle}{\sqrt{|\vec{a}|}}<br />
+ \dfrac{|1\rangle |b\rangle}{\sqrt{ |\vec{b}|}} \right) $ and $|\phi \rangle = \dfrac{1}{\sqrt{Z}} \left( |\vec{a}| |0\rangle - |\vec{b}| |1\rangle \right)$<br />
<br />
where $Z = |\vec{a}|^{2} + |\vec{b}|^{2}$. These two states look complicated but they can easy enough to construct if the data for $\vec{a}$ and $\vec{b}$ are available in a quantum memory. We construct $|\psi\rangle$ and $|\phi\rangle$ because it can be shown that<br />
<br />
$\langle \phi | \psi \rangle = \dfrac{1}{\sqrt{2Z}} ( |\vec{a}| \langle 0| - |\vec{b}| \langle 1|) \left( \dfrac{|0\rangle |a\rangle}{\sqrt{|\vec{a}|}} + \dfrac{|1\rangle |b\rangle}{\sqrt{ |\vec{b}|}} \right)<br />
= \dfrac{1}{\sqrt{2Z}} \left( \sqrt{|\vec{a}|} |a\rangle - \sqrt{|\vec{b}|} |b\rangle \right)$,<br />
<br />
and so<br />
<br />
$|\langle \phi | \psi \rangle|^2 = \dfrac{1}{2Z} \left\lVert \sqrt{|\vec{a}|} |a\rangle - \sqrt{|\vec{b}|} |b\rangle \right\rVert^2 = \dfrac{1}{2Z} \left( |\vec{a}| \langle a | a \rangle + |\vec{ab}| \langle b | b \rangle - \sqrt{ |\vec{a}| |\vec{b}| } \langle a | b \rangle - \sqrt{ |\vec{b}| |\vec{a}|} \langle b | a \rangle \right) = \dfrac{1}{2Z} \left( |\vec{a}|^2 + |\vec{b}|^2 - 2 \sqrt{ |\vec{a}| |\vec{b}|} \langle a | b \rangle \right) = \dfrac{D(\vec{a},\vec{b})^2}{2Z}.$<br />
<br />
To estimate $| \langle \phi | \psi \rangle |^2$, we will use a swap test between $|\phi\rangle$ and the qubit part of $|\psi\rangle$.<br />
<br />
A quantum swap test is often used to test if two states $|\alpha\rangle$ and $|\beta\rangle$ are the same or not. It involves the following circuit:<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdwuWZA4bGV-ZFkrIGhf6gWeiCyo0LWb306X6OypXmuz_44RaAJY-_EGgpSvIeWjpmE_3SWER1V2SDH6fGy97NSITFykxzGwA-ApCCpehYUnnE9oEWgIzy5DTtBwR_VpjCsKSE6gRFuAgG/s1600/Slide2.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="720" data-original-width="1280" height="360" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdwuWZA4bGV-ZFkrIGhf6gWeiCyo0LWb306X6OypXmuz_44RaAJY-_EGgpSvIeWjpmE_3SWER1V2SDH6fGy97NSITFykxzGwA-ApCCpehYUnnE9oEWgIzy5DTtBwR_VpjCsKSE6gRFuAgG/s640/Slide2.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Quantum swap test for states $|\alpha\rangle$ and $|\beta\rangle$. We apply a Hadamard gate to an ancillary qubit and use this as the control qubit of a controlled-swap operation between $|\alpha\rangle$ and $|\beta\rangle$. After that, we again apply a Hadamard gate to the ancillary qubit and finally measure that qubit in the standard qubit. If we run this circuit many times, we can estimate the probability of measuring $|0\rangle$ for the ancillary qubit.</td></tr>
</tbody></table>
<br />
<br />
Going over how the state changes in the circuit shown above:<br />
<br />
$|0\rangle |\alpha\rangle |\beta\rangle \mapsto \dfrac{1}{\sqrt{2}} \left (|0\rangle + |1\rangle) |\alpha\rangle |\beta\rangle \right) \mapsto \dfrac{1}{\sqrt{2}} \left (|0\rangle |\alpha\rangle |\beta\rangle + |1\rangle |\beta\rangle |\alpha\rangle \right) \mapsto \dfrac{1}{2} \left [ |0\rangle \left( |\alpha\rangle |\beta\rangle + |\beta\rangle |\alpha\rangle \right) + |1\rangle \left( |\alpha\rangle |\beta\rangle - |\beta\rangle |\alpha\rangle \right) \right] $<br />
<br />
From the last step, it can be seen that if $|\alpha\rangle = |\beta\rangle$ then measuring the ancillary qubit yields $|0\rangle$ with probability 1. More generally,<br />
<br />
$\Pr[ \text{obtain } |0\rangle] = \dfrac{1}{4} \lVert |\alpha\rangle |\beta\rangle + |\beta\rangle |\alpha\rangle \rVert^{2} = \dfrac{1 + | \langle \alpha | \beta \rangle |^2}{2}$.<br />
<br />
When we run the swap test for $|\phi \rangle $ and the ancillary qubit of $|\psi \rangle$, the state changes in the circuit as follows:<br />
<br />
$|0\rangle |\phi\rangle |\psi \rangle \mapsto \left( \dfrac{ |0\rangle + | 1\rangle}{2} \right) |\phi\rangle \left( \dfrac{|0\rangle |a\rangle}{\sqrt{|\vec{a}|}} + \dfrac{|1\rangle |b\rangle}{\sqrt{ |\vec{b}|}} \right)<br />
\mapsto \dfrac{1}{2} \left[ |0\rangle \left( \dfrac{|\phi\rangle |0\rangle |a\rangle}{\sqrt{|\vec{a}|}} + \dfrac{ |\phi\rangle |1\rangle |b\rangle}{\sqrt{ |\vec{b}|}} \right) + |1\rangle \left( \dfrac{|0\rangle |\phi\rangle |a\rangle}{\sqrt{|\vec{a}|}} + \dfrac{|1\rangle |\phi\rangle |b\rangle}{\sqrt{ |\vec{b}|}} \right)\right]<br />
\mapsto \dfrac{ |0\rangle}{2\sqrt{2}} \left[ \left( |\phi\rangle |0\rangle + |0\rangle |\phi\rangle \right) \dfrac{|a\rangle}{\sqrt{|\vec{a}|}} + \left( |\phi\rangle |1\rangle + |1\rangle |\phi\rangle \right) \dfrac{|b\rangle}{\sqrt{|\vec{b}|}} \right] +<br />
\dfrac{ |1\rangle}{2\sqrt{2}} \left[ \left( |\phi\rangle |0\rangle - |0\rangle |\phi\rangle \right) \dfrac{|a\rangle}{\sqrt{|\vec{a}|}} + \left( |\phi\rangle |1\rangle - |1\rangle |\phi\rangle \right) \dfrac{|b\rangle}{\sqrt{|\vec{b}|}} \right] $.<br />
<br />
The probability of obtaining $|0\rangle$ when the first qubit is measured is given by the norm of the post-measurement state.<br />
<br />
Let $| \Omega_{j}\rangle = |\phi\rangle |j\rangle + |j\rangle |\phi\rangle$ for $j = 0,1$. Using the definition of $\phi$, it is a bit tedious but straightforward to verify that<br />
<br />
$ \langle \Omega_0 |\Omega_0\rangle = \dfrac{4 |\vec{a}|^2 + 2 |\vec{b}|^2}{Z}, \quad<br />
\langle \Omega_1 |\Omega_1\rangle = \dfrac{4 |\vec{b}|^2 + 2 |\vec{a}|^2}{Z}, \quad<br />
\langle \Omega_0 |\Omega_1\rangle = \langle \Omega_0 |\Omega_0\rangle = \dfrac{-2 |\vec{a}| |\vec{b}|}{Z} $<br />
<br />
Then we can easily compute<br />
<br />
$\Pr[ \text{obtain } |0\rangle] = \dfrac{1}{8} \left\lVert<br />
\dfrac{ |\Omega_{0}\rangle |a\rangle}{\sqrt{|\vec{a}|}} + \dfrac{ |\Omega_{1}\rangle |b\rangle}{\sqrt{|\vec{b}|}} \right\rVert^2 = \dfrac{1}{8} \left(<br />
\dfrac{ \langle \Omega_0 |\Omega_0\rangle \langle a |a \rangle}{|\vec{a}|} +<br />
\dfrac{ \langle \Omega_1 |\Omega_1\rangle \langle b | b \rangle}{|\vec{b}|} +<br />
\dfrac{ \langle \Omega_0 |\Omega_1\rangle \langle a |b \rangle}{\sqrt{|\vec{a}| |\vec{b}|}} +<br />
\dfrac{ \langle \Omega_1 |\Omega_0\rangle \langle b |a \rangle}{\sqrt{|\vec{b}| |\vec{a}|}} \right)<br />
= \dfrac{1}{8Z} \left[ 4 ( |\vec{a}|^2 + |\vec{b}|^2) + 2( |\vec{a}|^2 + |\vec{b}|^2 - 2 \vec{a}\cdot \vec{b} \right] = \dfrac{1}{8Z} \left( 4Z + 2 D(\vec{a},\vec{b})^2 \right) $<br />
<br />
so an estimate for this probability yields an estimate for the distance between $\vec{a}$ and $\vec{b}$. We just need to make several copies of the states and run the swap test a few times, and estimate the probability from the fraction of outcomes where $|0\rangle$ was obtained.<br />
<br />
Finally, we apply this to the problem of cluster assignment by setting $\vec{a} = \vec{u}$ and computing the distance for $\vec{b}_0 = \frac{1}{M} \sum_j \vec{v}_j$ and $\vec{b}_1 = \frac{1}{M} \sum_k \vec{w}_k$, and assign $\vec{u}$ to $V$ if it is closer to $\vec{b}_0$, or to $W$ if it is closer to $\vec{b}_1$.<br />
<br />
What we have seen above is just one example of a problem that involves taking the vector dot product in a high-dimensional real vector space, where typical classical methods take $O(N)$ steps to compute, while the quantum version takes $O(\log N)$ steps. However, it is important to note that this speedup comes with some caveats, such as the need for a quantum memory in order to prepare the necessary states in the above method, and such requirements are not necessarily easy to achieve in practice.<br />
<br />
In the same paper, Lloyd, Mohseni, and Rebentrost used the quantum algorithm above as a subroutine in an adiabatic quantum algorithm for solving the $k$-means problem. The $k$-means problem is an unsupervised learning problem of assigning $M$ vectors to $k$ clusters in such a way that the average distance to the centroid of the cluster is minimized. The adiabatic algorithm they describe is based on the standard classical method used for solving the $k$-means problem, but with the optimization being run on an adiabatic quantum computer.<br />
<br />
<br />
<b>Reference:</b><br />
<br />
S. Lloyd, M. Mohseni, P. Rebentrost, "Quantum algorithms for supervised and unsupervised machine learning", arXiv:1307.0411v2.Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-87732316667881785232018-02-27T13:11:00.000-08:002018-02-27T16:44:30.463-08:00Nonlocality, steering, and entanglementThe nonlocality of quantum theory was first pointed out by Albert Einstein, Boris Podolsky, and Nathan Rosen, collectively known as EPR, in a 1935 paper that argued for the incompleteness of quantum mechanics.<br />
<br />
The gist of their argument can be seen through an example due to David Bohm, who considered the state<br />
<br />
$|\Psi\rangle = \dfrac{|00\rangle + |11\rangle }{\sqrt{2}} = \dfrac{|++\rangle + |--\rangle }{\sqrt{2}}, $<br />
<br />
also known as an EPR pair, which here we expressed in both the standard basis $\{ |0\rangle, |1\rangle \}$ and Hadamard basis $\{ |+\rangle, |-\rangle \}$.<br />
<br />
Suppose Alice gets the first qubit and Bob gets the second qubit. If Alice measures her qubit in the standard basis, she immediately projects Bob's qubit onto either the state $|0\rangle$ or $|1\rangle$. She produces a similar kind of projection of Bob's system if she measures in the Hadamard basis.<br />
<br />
This seems problematic because it suggests that Alice can affect Bob's qubit even though their qubits no longer interact, in which case you would expect that nothing that Alice does on her qubit should influence Bob's.<br />
<br />
Einstein, Podolsky, and Rosen thought (incorrectly as it turns out) that this nonlocal effect implies that the quantum state provides an incomplete description of Alice and Bob's physical systems. They attempt to support this intuition by arguing that one might be able to explain what Bob obtains if he measures his qubit through local hidden variables (LHV) that are missing from quantum theory.<br />
<br />
Like EPR, Erwin Schrödinger was bothered by the idea of nonlocality. However, unlike EPR, Schrodinger believed that a quantum state provided a complete description of a local, isolated system.<br />
<br />
<a name='more'></a>Thus, his response, Schrödinger introduced the notions of an entangled state for describing states like $|\Psi\rangle$, and steering for describing Alice's ability to influence Bob's system with her measurement choice. Schrödinger rejected the EPR argument for LHVs, instead, he argued (wrongly) that where quantum theory failed was in describing delocalized entangled systems, i.e., Bob's system is in some definite state, a local hidden state (LHS), which renders steering an unobservable effect in practice.<br />
<br />
It turns out that steering is a form of quantum correlation that is distinct from entanglement and nonlocality. In fact, it lies in between the two, since not all entangled states can be used to demonstrate steering, but not all steerable states can be used to violate a Bell inequality, which would imply that a quantum correlation can not be explained by a LHV model (i.e. it is nonlocal).<br />
<br />
Some intuition for distinguishing the three quantum-correlation-types can be obtained from the quantum information perspective if we think of the following tasks for three parties Alice, Bob, and Charlie.<br />
<br />
Suppose that Alice and Bob share an entangled state and their goal is to convince Charlie of this fact. In this scenario, Alice and Bob are not allowed to communicate but they are allowed to share some common random bits. They each can communicate directly to Charlie.<br />
<br />
(i) If Charlie trusts both Alice and Bob, then all they have to do is perform a tomographic measurement on their state. (A tomographic measurement is one whose results contain enough information to reconstruct the quantum state if it were unknown.) They give their results to Charlie, who can use the data to verify that the state is entangled.<br />
<br />
(ii) If Charlie trusts Bob but not Alice, then Alice and Bob have to demonstrate steering, since it would be enough to convince Charlie if there is no LHS model for the trusted party.<br />
<br />
(iii) If Charlie does not trust both, then Alice and Bob have to show that they can violate a Bell inequality, since this is the way Charlie can be guaranteed that the correlations that Alice and Bob can produce with their state is non-classical.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEipDd1qgNpBTgwhwmVex3aeAiZLElWqyIeIu0Fd8ieYuwokBsB31kIWM6CdX7djOwzU_gvngFFeYQVNjUQu7vJclZI6PE3QcFm8a-l01NxZGOi7SgjKL7luXTYqUSRud8bSI3akupYpr0hY/s1600/steering.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="1600" data-original-width="852" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEipDd1qgNpBTgwhwmVex3aeAiZLElWqyIeIu0Fd8ieYuwokBsB31kIWM6CdX7djOwzU_gvngFFeYQVNjUQu7vJclZI6PE3QcFm8a-l01NxZGOi7SgjKL7luXTYqUSRud8bSI3akupYpr0hY/s640/steering.png" width="339" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Nonlocality, steering, and entanglement from a task. Alice and Bob aim to convince Charlie that they share entangled qubits. Alice and Bob cannot communicate with each other but they are allowed to have some shared classical random bits. Here the white and black boxes are seen from Charlie's perspective. The white boxes denote a qubit basis measurement, labelled by a Bloch vector $\vec{n}$, while the black boxes refer to some unknown operation with input $x (y)$ and output $a (b)$. Describing from the top to bottom: If Charlie trusts both Alice and Bob then they can just perform some tomographically complete measurement to verify entanglement. If Charlie trusts only Bob, then Alice and Bob have to show that the state they share can be steered. If Charlie trusts neither, then they have to exhibit a Bell-inequality violation, by say, playing the CHSH game and winning sufficiently many games.</td></tr>
</tbody></table>
<br />
<br />
Like Bell nonlocality, steering can be demonstrated through simple tests, e.g. for qubits it is sufficient to consider two measurements for Alice, which prepares a collection of four states for Bob. This means steering can be certified using violation of steering inequalities, analogous to Bell inequality violations for exhibiting nonlocality.<br />
<br />
Note that if Bob's system can be described by a LHS, Bob's probabilities for measurement outcomes must be compatible with a quantum state. This suggests the use of uncertainty relations for deriving some steering criteria.<br />
<br />
One key difference between steering inequalities and Bell inequalities is that the latter is constructed without defining any particular measurement for Alice or Bob. For steering, the inequality does depend on Bob's measurement choice since we require outcome probabilities consistent with a quantum state.<br />
<br />
AS EPR and Schrödinger have noted, steering can be demonstrated using pure entangled states. And from what we know about the EPR pair $|\Psi\rangle$, it can be used to show nonlocality through violation of the CHSH inequality, a Bell inequality for qubits. So to again show that the ideas are distinct, we need to consider mixtures, quantum states that are necessarily described using density operators.<br />
<br />
The example that we will take here is the isotropic state for qubits<br />
<br />
$\rho = p|\Psi\rangle\langle\Psi| + (1-p)\dfrac{I}{4}$,<br />
<br />
which can be seen as a mixture of $|\Psi\rangle$ and the maximally mixed state for two qubits. Note that this is actually a family of quantum states with $0 \le p \le 1$. In standard matrix form, we would write it as<br />
<br />
$\rho = \dfrac{p}{2}<br />
\begin{pmatrix}<br />
1 \\ 0 \\ 0 \\ 1<br />
\end{pmatrix}<br />
\begin{pmatrix}<br />
1 & 0 & 0 & 1<br />
\end{pmatrix}<br />
+ \dfrac{1-p}{4}<br />
\begin{pmatrix}<br />
1 & 0 & 0 & 0 \\<br />
0 & 1 & 0 & 0 \\<br />
0 & 0 & 1 & 0 \\<br />
0 & 0 & 0 & 1<br />
\end{pmatrix}<br />
=<br />
\dfrac{1}{4}<br />
\begin{pmatrix}<br />
1+p & 0 & 0 & 2p \\<br />
0 & 1-p & 0 & 0 \\<br />
0 & 0 & 1-p & 0 \\<br />
2p & 0 & 0 & 1+p<br />
\end{pmatrix}$.<br />
<br />
We can determine for which values $p$ the state $\rho$ is entangled by looking at the eigenvalues of the partial transpose:<br />
<br />
$\rho^{T_B} =<br />
\begin{pmatrix}<br />
1+p & 0 & 0 & 0 \\<br />
0 & 1-p & 2p & 0 \\<br />
0 & 2p & 1-p & 0 \\<br />
0 & 0 & 0 & 1+p<br />
\end{pmatrix}<br />
$<br />
<br />
The Peres-Horodecki criterion says that when $\rho^{T_B}$ has a negative eigenvalue then $\rho$ is entangled. Thus, solving for the eigenvalues of $\rho^{T_B}$<br />
<br />
$\mathrm{Det}(\rho^{T_B} - \lambda I)<br />
= (1 + p - 4\lambda) ^2 [ (1- p - 4\lambda)^2 - 4p^2 ] = 0$<br />
<br />
so the four eigenvalues are $\lambda = (1+p)/4$ counted thrice, and $\lambda = (1-3p)/4$. Thus, the last one gives the smallest eigenvalue and we get $\lambda < 0$ when $ p > 1/3$.<br />
<br />
To establish Bell nonlocality for $\rho$, we consider the CHSH inequality given by<br />
<br />
$ | \langle Z X' \rangle + \langle Z Z' \rangle + \langle X X' \rangle - \langle X Z' \rangle |\le 2$,<br />
<br />
where $\langle A B \rangle = \mathrm{Tr}(\sigma A\otimes B)$ is the expectation value of observable $A \otimes B$ given the quantum state $\sigma$. In this case, the relevant observables (i.e. they give rise to the optimal violation) are<br />
<br />
$<br />
X =<br />
\begin{pmatrix}<br />
0 & 1 \\<br />
1 & 0<br />
\end{pmatrix},<br />
\quad<br />
Z =<br />
\begin{pmatrix}<br />
1 &0 \\<br />
0 & -1<br />
\end{pmatrix},<br />
\quad<br />
X' = \dfrac{1}{\sqrt{2}}<br />
\begin{pmatrix}<br />
1 & 1 \\<br />
1 & -1<br />
\end{pmatrix},<br />
\quad<br />
Z' = \dfrac{1}{\sqrt{2}}<br />
\begin{pmatrix}<br />
-1 & -1 \\<br />
-1 & 1<br />
\end{pmatrix}<br />
$.<br />
<br />
That is, Alice's observables are just the Pauli operators $X$ and $Z$, while Bob's operators are rotated versions $X'$ and $Z'$. As we have seen before, these observables correspond to Alice measuring her qubit in the standard or Hadamard basis, with Bob measuring his qubit in the basis $\{ \cos\theta |0\rangle + \sin\theta |1\rangle, \sin\theta |0\rangle - \cos\theta |1\rangle \}$, for $\theta = \pm\pi/8$. More specifically, a measurement in the qubit basis<br />
<br />
$\left\{<br />
\begin{pmatrix}<br />
a \\<br />
b<br />
\end{pmatrix},<br />
\begin{pmatrix}<br />
b^* \\<br />
-a^*<br />
\end{pmatrix}<br />
\right\}$<br />
<br />
yields observable<br />
<br />
$O =<br />
\begin{pmatrix}<br />
a \\<br />
b<br />
\end{pmatrix}<br />
\begin{pmatrix}<br />
a^* & b^*<br />
\end{pmatrix}<br />
-<br />
\begin{pmatrix}<br />
b^* \\<br />
-a^*<br />
\end{pmatrix}<br />
\begin{pmatrix}<br />
b & -a<br />
\end{pmatrix}<br />
=<br />
\begin{pmatrix}<br />
|a|^2 - |b|^2 & 2 a b^* \\<br />
2 b a^* & |b|^2 - |a|^2<br />
\end{pmatrix}<br />
$.<br />
<br />
Let $\beta = Z\otimes Z' + Z \otimes X' + X \otimes X' - X \otimes Z'$ be the CHSH operator. Then<br />
<br />
$\beta = \sqrt{2}<br />
\begin{pmatrix}<br />
1 & 0 & 0 & 1 \\<br />
0 & -1 & 1 & 0 \\<br />
1 & 1 & -1& 0 \\<br />
1 & 0 & 0 & 1<br />
\end{pmatrix}$,<br />
<br />
and so we can easily compute $\mathrm{Tr}(\rho \beta) = \dfrac{1}{2\sqrt{2}} (2+ 6p +2p - 2) =2\sqrt{2} p$. Hence, we observe a CHSH inequality violation when $ 2 - |\mathrm{Tr}(\rho \beta)| < 0$, which for the isotropic state happens when $ p > 1/\sqrt{2}$.<br />
<br />
Finally, we consider the following steering inequality:<br />
<br />
$\langle X X \rangle + \langle Y Y \rangle + \langle Z Z \rangle \le \sqrt{3}$,<br />
<br />
where $Y$ is the other Pauli operator $ Y =<br />
\begin{pmatrix}<br />
0 & -i \\<br />
i & 0<br />
\end{pmatrix}$.<br />
<br />
Here we have the steering operator $\Sigma = X \otimes X + Y \otimes Y + Z \otimes Z$. A direct calculation gives<br />
<br />
$\Sigma = \begin{pmatrix}<br />
1 & 0 & 0 & 2 \\<br />
0 &-1 & 0 & 0 \\<br />
0 & 0 & - 1 & 0 \\<br />
2 & 0 & 0 & 1<br />
\end{pmatrix}$,<br />
<br />
which means that $\mathrm{Tr}(\rho \Sigma) = \dfrac{1}{4} (2 + 10p + 2p - 2) = 3p$.<br />
<br />
Thus, the steering is observed when $\sqrt{3} - |\mathrm{Tr}(\rho \Sigma) | < 0$, which happens when $p > 1\sqrt{3}$. (Actually this result is conservative since the steering limit is $p > 1/2$ if we consider an infinite number of measurements, but that is obviously impractical to test.)<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj-FankBENvumNAjCwDVzG3sy2o9ExZEu4J_Sxpu_bM7GczmhHSEEi0xgR3UAbHQy3nubxJvAylrh3zvhuIdDaMHpdqH-plC62jqC9Q0PT9IYS5WZH_6PqKp2hogqJRSPOYzrcmhFCL-dZw/s1600/steeringPlot.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="592" data-original-width="1000" height="378" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj-FankBENvumNAjCwDVzG3sy2o9ExZEu4J_Sxpu_bM7GczmhHSEEi0xgR3UAbHQy3nubxJvAylrh3zvhuIdDaMHpdqH-plC62jqC9Q0PT9IYS5WZH_6PqKp2hogqJRSPOYzrcmhFCL-dZw/s640/steeringPlot.png" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Nonlocality, steering, and entanglement for the family of qubit isotropic states $\rho(p)$. This red line plots the smallest eigenvalue of the partial transpose of $\rho$. The green line plots $\sqrt{3} - |\mathrm{Tr}(\rho \Sigma)|$, which comes from a steering inequality. The blue line plots $2 - \mathrm{Tr}(\rho \beta) |$, which comes from the CHSH inequality. The value of $p$ where the red, green, and blue lines become negative indicate the respective bounds for entangled, steerable, and Bell nonlocal isotropic states.</td></tr>
</tbody></table>
<br />
<br />
So to recap, for the family of qubit isotropic states, we have seen that a particular $\rho$ is entangled only if $p > 1/3$, it is steerable for Pauli measurements only if $p > 1/\sqrt{3}$, and it exhibits Bell nonlocality only if $ p> 1/\sqrt{2}$.<br />
<br />
This simple example at least illustrates that these notions are indeed different for mixed states. It continues to be a pretty challenging task to quantify these various kinds of quantum correlations in higher dimensions.<br />
<br />
<b><br /></b>
<b>References:</b><br />
<br />
H. M. Wiseman, S. J. Jones, and A. C. Doherty, <a href="https://arxiv.org/pdf/quant-ph/0612147.pdf" target="_blank">Steering, Entanglement, Nonlocality, and the Einstein-Podolsky-Rosen paradox</a>, Physical Review Letters, 98 (2007) 140402.<br />
<br />
E.G. Cavalcanti, S.J. Jones, H.M. Wiseman and M.D. Reid, <span class="fontstyle0"><a href="https://arxiv.org/pdf/0907.1109v2.pdf" target="_blank">Experimental criteria for steering and the Einstein-Podolsky-Rosen paradox</a>, </span>Physical Review A 80 (2009) 032112.<br />
<br />Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-70164842386068949532018-02-20T00:02:00.000-08:002018-04-06T00:25:21.096-07:00Quantum security from an uncertainty gameThe uncertainty principle of quantum theory limits how much Eve can predict about the outcomes of two incompatible measurements on Adam's state.<br />
<br />
We will consider a guessing game that shows security against an eavesdropper Eve who can prepare quantum states but otherwise store and process classical information only. For example, Eve is allowed to make measurements on a qubit during its transmission but she does not possess any quantum memory for storing quantum states that she might entangle with the qubit.<br />
<br />
In such a setting, Eve's actions effectively chooses the state of the qubit sent to Adam, since she would know as much as she can about the qubit if she had prepared it herself. This situation can be analyzed in terms of the following guessing game, to be played by Adam and Eve:<br />
<br />
1. Eve prepares a qubit $| \psi \rangle$ and sends it to Adam.<br />
2. Adam chooses a uniformly random bit $\theta$. If $\theta = 0$, Adam measures $| \psi \rangle$ in the standard basis $\{ |0\rangle, |1\rangle \}$. If $\theta = 1$, Adam measures $| \psi \rangle$ in the Hadamard basis $\{ |+\rangle, |-\rangle \}$, where $|\pm\rangle = (|0\rangle \pm |1\rangle )/ \sqrt{2}$.<br />
3. Adam records a bit value $x$ as his measurement outcome. (By convention, $|+\rangle$ corresponds to bit value 0.)<br />
4. Adam reveals the measurement basis $\theta$ to Eve.<br />
<br />
Eve wins the game if she correctly guesses the value of $x$.<br />
<br />
<a name='more'></a><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgHUyYSCOR95yFscfyEWKFMnk2VsVHqgLAVoGIGLPNVEbwW_yg2ElZGA7iGKHGMsFg0AEWgJbWu7jVxNZoHpn82rI6dWZGsoaXDHTUmlvqUt3jP0NeVTZF3mdGebHlLc1_TFChyxwTWqTFN/s1600/Slide1.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="400" data-original-width="766" height="334" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgHUyYSCOR95yFscfyEWKFMnk2VsVHqgLAVoGIGLPNVEbwW_yg2ElZGA7iGKHGMsFg0AEWgJbWu7jVxNZoHpn82rI6dWZGsoaXDHTUmlvqUt3jP0NeVTZF3mdGebHlLc1_TFChyxwTWqTFN/s640/Slide1.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">The uncertainty guessing game, Eve prepares a qubit in state $|\psi \rangle$ and sends it to Adam. Adama randonly chooses a bit $\theta$, which he uses to decide whther to measure the qubit in the standard basis $\theta = 0$ or the Hadamard basis $\theta = 1$. In either case, Adam obtains a bit value $x$. Eve's goal is to guess Adam's result.</td></tr>
</tbody></table>
<br />
This game embodies the uncertainty principle because for incompatible measurements such as the standard and Hadamard basis, there is no quantum state that Eve can prepare that allows her to perfectly guess the outcome for both choices of measurements. In this case, uncertainty can be quantified in terms of Eve's average winning probability:<br />
<br />
$ \Pr [\text{Eve wins}] = ( \Pr [ x \text{ is guessed correctly} | \theta = 0] + \Pr[ x \text{ is guessed correctly} | \theta = 0] )/2 \le c $.<br />
<br />
The above inequality holds since the basis is chosen uniformly at random. Now we will show that if Eve has no additional information except for the basis announced by Adam, then the bound $c$ is strictly less than 1.<br />
<br />
We want to find the largest possible $c$ that Eve can achieve. To do this, Eve needs to find a state $ | \psi \rangle$ that maximizes her guessing probability in both bases because she does not know which basis Adam will measure.<br />
<br />
Intuitively she has to prepare a state where she know one of the outcomes is significantly more likely for both bases, so she can make that her guess. It should not matter which outcome is more likely, only that Eve knows which one it is for a given basis $\theta$. So we do not lose generality if we look for the state such that $x = 0$ is more likely for either basis.<br />
<br />
We also know that there is no observable physical difference between $| \psi \rangle$ and $e^{i \phi} | \psi \rangle$, which lets us consider<br />
<br />
$|\psi \rangle = a | 0 \rangle + (b+ i c) | 1 \rangle $ with $ a^2 + b^2 + c^2 = 1$.<br />
<br />
Now we have<br />
<br />
$\Pr[x = 0 | \theta = 0] = | \langle 0 | \psi \rangle |^2 = a^2 $, and<br />
<br />
$\Pr[x = 0 | \theta = 1] = | \langle + | \psi \rangle |^2 = \left| \dfrac{(a + b + ic) }{\sqrt{2}} \right|^2 = \dfrac{(a + b)^2 + c^2}{2}$.<br />
<br />
Since each $\theta$ is equally likely,<br />
<br />
$\Pr[x = 0 ] = \dfrac{a^2}{2} + \dfrac{(a + b)^2 + c^2}{4}$.<br />
<br />
Let $f(a,b,c) = \Pr[x = 0]$. We want to maximize $f(a,b,c)$ subject to the constraint $g(a,b,c) = a^2 + b^2 + c^2 - 1 = 0$. There are several ways to do this; here we will present two.<br />
<br />
The conventional approach here is to use the method of Lagrange multipliers. For this method, we define $ \mathcal{L} = f(a,b,c) + \lambda g(a,b,c)$, where $\lambda$ is a Lagrange multiplier.<br />
<br />
We take partial derivatives and set them to zero, and we would get candidate solutions for minimum and maximum values:<br />
<br />
$<br />
\begin{align}<br />
\nonumber<br />
\dfrac{\partial \mathcal{L}}{\partial \lambda} &= a^2 + b^2 + c^2 - 1 = 0,<br />
&<br />
\dfrac{\partial \mathcal{L}}{\partial a} &= a + \dfrac{a+b}{2} + 2 \lambda a = 0, \\<br />
\nonumber<br />
\\<br />
\nonumber<br />
\dfrac{\partial \mathcal{L}}{\partial b} &= \dfrac{a+b}{2} + 2 \lambda b = 0,<br />
&<br />
\dfrac{\partial \mathcal{L}}{\partial c} &= \dfrac{c}{2} + 2 \lambda c = 0.<br />
\end{align}<br />
$<br />
<br />
The first equation is just the constraint. From the last equation, either $\lambda = -1/4 $ or $c = 0$. If we take the first one and plug into the third equation, we get<br />
<br />
$ \dfrac{a+b}{2} - \dfrac{b}{2} = 0 \quad \implies \quad a = 0 $,<br />
<br />
which if we substitute into the second equation gives $b = 0$. Finally, the first equation implies $c = \pm 1$ and thus $|\psi\rangle = \pm i |1\rangle$. So this solution actually minimizes the outcome $x = 0$ since for this state $ \Pr[x = 0 | \theta = 0] = 0$ and $ \Pr[x = 0 | \theta = 1] = 1/2$.<br />
<br />
We get the maximum from taking $c = 0$, which means $a^2 + b^2 = 1$, and from the second and third equations:<br />
<br />
$ -2 \lambda = 1 + \dfrac{a+b}{2a} = \dfrac{a+b}{2b}$.<br />
<br />
It follows from the middle and right-hand-sides above that $a^2 - b^2 = 2 a b$. Using the constraint with $c = 0$ here leads to<br />
<br />
$a^2 - b^2 = a^2 - ( 1 - a^2) = 2 a b = 2 a \sqrt{1 - a^2}$.<br />
<br />
Squaring both sides gives<br />
<br />
$(2 a^2 -1)^2 = 4 a^2 ( 1- a^2) \quad \implies \quad 8 a^4 - 8 a^2 + 1 = 0$.<br />
<br />
Therefore by the quadratic formula, we obtain<br />
<br />
$a^2 = \dfrac{8 \pm \sqrt{64 - 4(8)}}{2(8)} = \dfrac{2 + \sqrt{2}}{8}$.<br />
<br />
The larger value has a plus sign so<br />
<br />
$a = \sqrt{\dfrac{2 + \sqrt{2}}{8}} = \sqrt{\dfrac{1}{2} \left( 1 + \dfrac{1}{\sqrt{2}} \right)} = \cos \left(\frac{\pi}{8}\right)$, and so $b = \sin\left(\frac{\pi}{8}\right)$.<br />
<div>
<br /></div>
With this solution, we find that<br />
<br />
$\Pr[x = 0] = \dfrac{1}{2}\cos^2 \left( \frac{\pi}{8}\right) + \dfrac{1}{4} + \dfrac{1}{2} \cos\left(\frac{\pi}{8}\right) \sin\left( \frac{\pi}{8}\right) = \dfrac{1}{2}\cos^2 \left( \frac{\pi}{8}\right) + \dfrac{1}{4} [1 + \sin\left( \frac{\pi}{4} \right) ] $<br />
$ \qquad \qquad = \dfrac{1}{2} \left( 1 + \dfrac{1}{\sqrt{2}} \right) = \cos^2 \left(\frac{\pi}{8}\right)$.<br />
<br />
A different way get to the result is to note that if we define<br />
<br />
$M = \dfrac{1}{2} \left( | 0 \rangle \langle 0 | + | + \rangle \langle + |\right) $<br />
<br />
then $\Pr[x = 0] = \langle \psi | M | \psi \rangle$. We can maximize this by looking for an eigenstate $|m\rangle$ of M, which corresponds to its maximum eigenvalue $\omega$. In that case,<br />
<br />
$\Pr[x = 0] = \langle m | M | m \rangle = \langle m | \omega | m \rangle = \omega$.<br />
<br />
We can express $M$ as a $2 \times 2$-matrix by defining the column vectors $|0\rangle = (1,0)$ and $|+\rangle = (1, 1)/ \sqrt{2}$,<br />
<br />
$M = \dfrac{1}{4}<br />
\begin{pmatrix}<br />
3 & 1 \\<br />
1 & 1<br />
\end{pmatrix}, \qquad \mathrm{Tr}(M) = 1, \mathrm{Det}(M) = \dfrac{1}{8}$,<br />
<br />
and solving for its eigenvalues:<br />
<br />
$\mathrm{eig}(M) = \dfrac{1}{2} \left( \mathrm{Tr}(M) \pm \sqrt{\mathrm{Tr}(M)^2 - 4 \mathrm{Det}(M)}\right)<br />
= \dfrac{1}{2} \left( 1 \pm \dfrac{1}{\sqrt{2}} \right) = \cos^2 \left(\frac{\pi}{8}\right)$.<br />
<br />
The larger eigenvalues has the plus sign so<br />
<br />
$\omega = \dfrac{1}{2} \left( 1 + \dfrac{1}{\sqrt{2}} \right) = \cos^2 \left(\frac{\pi}{8}\right)$.<br />
<div>
<br /></div>
The corresponding eigenvector can be solved from<br />
<br />
$\dfrac{1}{4}<br />
\begin{pmatrix}<br />
3 - \omega & 1 \\<br />
1 & 1 - \omega<br />
\end{pmatrix}<br />
\begin{pmatrix}<br />
x \\<br />
y<br />
\end{pmatrix} =<br />
\begin{pmatrix}<br />
0\\<br />
0<br />
\end{pmatrix}$.<br />
<br />
Working out the algebra, we obtain $y = (\sqrt{2}- 1)x$. Since we want the eigenvector to be normalized, we have<br />
<br />
$x^2 + (\sqrt{2} - 1)^2 x^2 = 1 , \qquad \implies \qquad<br />
x^2 = \dfrac{1}{4- 2\sqrt{2}} = \dfrac{4 + 2\sqrt{2}}{8} = \dfrac{1}{2} \left( 1 + \dfrac{1}{\sqrt{2}} \right) = \cos^2 \left(\frac{\pi}{8}\right) $.<br />
<br />
Thus, the eigenstate with maximum eigenvalue $\omega = \cos^2 \left(\frac{\pi}{8}\right) $ is<br />
<br />
$|m \rangle =<br />
\begin{pmatrix}<br />
x \\<br />
y<br />
\end{pmatrix} =<br />
\begin{pmatrix}<br />
\cos \left(\frac{\pi}{8}\right) \\<br />
\sin \left(\frac{\pi}{8}\right)<br />
\end{pmatrix}$<br />
<br />
where $\omega$ is also Eve's optimal probability for guessing $x$.<br />
<br />
So we have seen that for Eve who can prepare quantum states but has no quantum memory, her best strategy wins the guessing game with probability $\omega = \cos^2 \left(\frac{\pi}{8}\right) $ so the upper bound is $c = \cos^2 \left(\frac{\pi}{8}\right) $ for such an adversary.<br />
<br />
That Eve is unable to always guess correctly means that there is some secret information that we may extract from $x$. Typically, this involves running the game independently many times until you can extract a sufficient number of secret bits.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhIBzJ7KaV6vJDb5PlYxBbHXUaCa48DsVeTIz9XnGozNf7ON9Qh32Y05Pvlr_hx67xdfTL2GzUw7y6Sv8gwxlmub2SRoQV9IMxnOvqLVJsk52da5kLIRuaMVcUN5AdDI0vQPdmE8C9sJ293/s1600/Slide2.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="512" data-original-width="883" height="370" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhIBzJ7KaV6vJDb5PlYxBbHXUaCa48DsVeTIz9XnGozNf7ON9Qh32Y05Pvlr_hx67xdfTL2GzUw7y6Sv8gwxlmub2SRoQV9IMxnOvqLVJsk52da5kLIRuaMVcUN5AdDI0vQPdmE8C9sJ293/s640/Slide2.png" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">If Eve plays the guessing game with a quantum memory then she can always win the game. Her winning strategy is to prepare an EPR pair, and send one qubit to Adam while keeping one to herself. Thus, whether Adam measures in the standard or Hadamard basis, Eve just needs to wait for Adam to announce his basis $\theta$ so she can measure her share of the EPR pair in the same basis and obtain the same bit value $x$.</td></tr>
</tbody></table>
<br />
What happens if Eve is allowed to store some qubits? Then if you think about it for a bit, you will find that Eve can perfectly win the game. This is because all Eve needs to do is to prepare an EPR pair and keep one system for herself.<br />
<br />
Because an EPR pair can be written as<br />
<br />
$|\Phi\rangle = \dfrac{|00\rangle + |11\rangle }{\sqrt{2}}<br />
= \dfrac{ |++\rangle + |--\rangle}{\sqrt{2}}$,<br />
<br />
when Adam makes a measurement in either the standard or Hadamard basis, he gets a random outcome but he "steers" Eve's qubit to the same basis state as his qubit after the measurement. Eve just waits for Adam to reveal the basis $\theta$ and then she can measure her qubit in the same basis to get the same outcome as Adam.<br />
<br />
Are we out of luck then, since we can not be sure if an adversary Eve has quantum memory or not? No, not yet since we know that entanglement is monogamous.<br />
<br />
To restrict Eve's information about Adam's measurement results in the guessing game, we have to exploit both the uncertainty principle and the monogamy of entanglement. To make sure Eve does not gain any advantage from being able to store qubits, Adam needs to play with another honest party, Bob.<br />
<br />
What happens is that both Adam and Bob receive qubits. Adam still picks the basis $\theta$ to measure his qubit to get an outcome $x$. Afterwards, Adam announces $\theta$ to both Bob and Eve. Bob will then measure his qubit in the same basis $\theta$ to get an outcome $y$. Eve will make some measurement on whatever quantum state she kept for herself to get a bit outcome $z$. The goal of the game is then for Eve and Bob to both correctly guess Adam's bit $x$, i.e., Eve and Bob win the game if $x = y = z$.<br />
<br />
A full analysis of this monogamy-of-entanglement (ME) guessing game is a lot more involved so we will not go through it here. But what we can intuitively see is that if Bob tries to win the game all the time by being entangled to Adam with an EPR pair, then Eve will be completely disentangled with Adam and therefore the quantum state she stored will not have perfect information about $x$.<br />
<br />
In fact, the analysis of the three-player ME guessing game shows that the optimal winning probability is the same as in the two-player uncertainty game with no quantum memory: in fact, Eve can play the ME game using her optimal strategy with classical resources only.<br />
<br />
The analysis of the ME game has been used to show that BB84 is secure even for a receiving party who has untrusted measurement devices, and for one-round single-qubit position verification scheme.<br />
<br />
<b><br /></b>
<b>References:</b><br />
<br />
M. Tomamichel, S. Fehr, J. Kaniewski, S. Wehner, <a href="https://arxiv.org/abs/1210.4359" target="_blank">A Monogamy-of-Entanglement Game With Applications to Device-Independent Quantum Cryptography</a>, New Journal of Physics 15 (2013) 103002.<br />
<br />
<br />
<br />
<br />Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-59191685941048607132018-02-10T00:24:00.000-08:002018-02-16T20:52:43.829-08:00Delegated quantum computation Suppose Alice wants to perform a quantum computation. She can implement some basic quantum gates on the state $|0\rangle$ but she does not own a fully-functional quantum computer. Bob runs a company that sells time on their quantum computer and thus offers his services to Alice. But Alice does not trust Bob. In particular, she does not want Bob to learn her quantum state at any point during the calculation. Is there a way for Bob to help Alice securely? Yes, this is a problem that can be solved by delegated computation.<br />
<br />
In the task of delegated computation, Alice has an input $|q\rangle$ and a classical description $\mathsf{C}$ of the quantum circuit and she wants to obtain the output $\mathsf{C}(|q\rangle)$. Alice and Bob exchange a few messages, at the end of which Bob obtains a result $|r\rangle$, which he sends to Alice. Ideally, we want the protocol to be correct, verifiable, and blind.<br />
<br />
The protocol is said to be correct if Alice accepts Bob's result and $|r\rangle = \mathsf{C}(|q\rangle)$ when both parties are honest. The protocol is said to be verifiable if for any cheating Bob, Alice either aborts or finds $|r\rangle = \mathsf{C}(|q\rangle)$. The protocol is said to blind if for any cheating Bob, at the end of the protocol, Bob learns nothing about $|q\rangle$ or $\mathsf{C}$.<br />
<br />
<a name='more'></a>To fully describe the problem, we need to specify what resources Alice is allowed to have. For the protocol that we will describe here, Alice is allowed to (i) prepare qubits in $|0\rangle$, (ii) store and exchange two qubits (quantum swap gate), (iii) perform Pauli gates $X, Z$ where $X|0\rangle = |1\rangle, X|1\rangle = |0\rangle$ and $Z|0\rangle = |0\rangle, |Z1\rangle = (-1)|1\rangle$, (iv) generate random classical bits, and (v) measure qubits in the basis $|0\rangle, |1\rangle$.<br />
<br />
She can talk to Bob through a quantum channel where she can send single qubits one at a time to Bob, as well as classical channel where she can tell Bob about the quantum operations Bob has to perform on the qubits.<br />
<br />
Our approach here is based on the idea of computing on encrypted data. Suppose Alice has an input qubit in state $|q\rangle$. To encrypt this state using a quantum one-time pad (QOTP), we choose two random bits $j$ and $k$ and apply $Z^{k}X^{j}$ on $|q\rangle$ to get $|\tilde{q}\rangle = Z^{k}X^{j}|q\rangle$. If Alice keeps $j$ and $k$ to herself and sends only $|\tilde{q}\rangle$ to Bob, then she keeps her input private.<br />
<br />
To see how the one-time pad works, suppose that Alice has $|q\rangle = a |0\rangle + b|1\rangle$. Then $\mathrm{Enc}_{j,k}|q\rangle = Z^k x^j |q\rangle = (-1)^{jk} (a |j\rangle + b|j \oplus 1\rangle)$, where $\oplus$ denotes addition mod 2.<br />
<br />
The figure below shows how a qubit is encrypted using the quantum one-time pad. We also note that if the encrypted qubit is in the state $Z^{k}X^{j}|q\rangle$, this can be decrypted using the same classical bits $j$ and $k$, and by applying the $Z$ and $X$ in reverse.<br />
<br />
<div class="separator" style="clear: both; text-align: center;">
</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgenqMXB8PT7FFzi8vhhPoHZMAxemLv8mNhOl8PLqQCoY1eiL6JVeZDSDjB9yJ9NVXRtcBZz3g1N5d99w5kzdnPpqkGNEE1V9zHV4e73s9LU41Jq5owNopX6rWd-MLbZ4LWlGWkm8E-leMS/s1600/Slide1.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="693" data-original-width="1152" height="384" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgenqMXB8PT7FFzi8vhhPoHZMAxemLv8mNhOl8PLqQCoY1eiL6JVeZDSDjB9yJ9NVXRtcBZz3g1N5d99w5kzdnPpqkGNEE1V9zHV4e73s9LU41Jq5owNopX6rWd-MLbZ4LWlGWkm8E-leMS/s640/Slide1.jpg" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Encrypting a qubit with a quantum one-time pad. Two random classical bits $(j,k)$ are chosen for the key. The operator $Z^k X^j$ is applied to generate an encrypted qubit. The encrypted qubit can be decoded by applying the operator in reverse order with the same key bits used. Note that the one-time pad can be used to encrypt many qubits just by performing this single-qubit procedure on each qubit.</td></tr>
</tbody></table>
<div class="separator" style="clear: both; text-align: center;">
</div>
<br />
<br />
Because the one-time pad encrypts and decypts using only $X$ and $Z$ gates, we do not change the resources available to Alice if she is provided an encryption and decryption QOTP device instead of the Pauli $X$ and $Z$ gates. If Alice has the red Enc box, she can perform $X$ by choosing the bits $j=1, k=0$ and she can perform $Z$ by choosing $j=0,k=1$.<br />
<br />
Now in order for Alice to run a circuit $C$ on $|q\rangle$, she needs to find another circuit $\tilde{C}$ such that $\tilde{C}(\tilde{|q\rangle} = \tilde{C(|q\rangle)}$, that is, applying this new circuit to the encrypted input gives the encrypted version of the desired outcome $C(|q\rangle)$.<br />
<br />
Any quantum circuit can be expressed in terms of a universal set of gates: a set of gates is universal if any circuit can be implemented efficiently using gates from that set. Here we shall consider the universal gate set $G = {H, \mathrm{CNOT}, T}$, where<br />
<br />
$H$ is the Hadamard gate: $H|0\rangle = |+\rangle, H|1\rangle = |-\rangle$;<br />
<br />
$\mathrm{CNOT}$ is the controlled-NOT gate on two qubits: $\mathrm{CNOT} |0,x\rangle = |0,x\rangle, \mathrm{CNOT}|1,x\rangle = |1,x \oplus 1\rangle$, where $\oplus$ is addition mod 2;<br />
<br />
and $T$ is often called the $\pi/8$-phase gate: $T|0\rangle = |0\rangle, T|1\rangle = \sqrt{i}|1\rangle$.<br />
<br />
Our goal is to construct the equivalent circuits for implementing these quantum gates on an input encrypted using the quantum one-time pad. If we can show that Bob can help Alice implement gates from a universal set then we have effectively shown that he can help Alice perform any quantum computation.<br />
<br />
Let us consider the Hadamard gate first. Note that if $|q\rangle = a |0\rangle + |1\rangle$, then we expect $H|q\rangle = a |+\rangle + b |-\rangle$. But Alice will be sending a qubit encrypted using the one-time pad so what Bob gets would be $|\tilde{q}\rangle = (-1)^{jk} (a |j\rangle + b|j \oplus 1\rangle)$.<br />
<br />
It is easier to see what happens if we look at it case-by-case, and this is summarized in the table below. What the table tells us is after Alice applies $Z^k X^j$ on the qubit she sends to Bob, and Bob performs the Hadamard gate on the qubit and sends back the result, Alice can "undo" the encryption on the qubit by applying $X^k Z^j$.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhDrAUhAJ1VwtIGwPavONcX3I2bM9yHiSvGvGGdb4yeHccUsDjp5sRY_jOSOxJcjsU-9jyJFeJdkNTpSLtnGjbEsFBo8UjOjeIzTsxeUODwBiQ5HWy036KZ9EEnaHiLIWiDz1HoIOCAD10l/s1600/Slide3.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="425" data-original-width="1600" height="168" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhDrAUhAJ1VwtIGwPavONcX3I2bM9yHiSvGvGGdb4yeHccUsDjp5sRY_jOSOxJcjsU-9jyJFeJdkNTpSLtnGjbEsFBo8UjOjeIzTsxeUODwBiQ5HWy036KZ9EEnaHiLIWiDz1HoIOCAD10l/s640/Slide3.jpg" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">The table above shows the effect of the Hadamard gate applied to the encrypted qubit, for different values of the key bits $j,k$. It shows that we can get desired result, possibly up to an overall phase, if we apply $X^k Z^j$ on the qubit that Bob returns to Alice.</td></tr>
</tbody></table>
<br />
<br />
This means that if Alice encrypts using the bits $(j,k)$ and gives the encrypted qubit to Bob, after Bob performs the Hadamard gate and returns the result to Alice, she only has to decrypt with the bits exchanged, i.e. using the bits $(k,j)$. This circuit is shown below, where the blue box denotes an operation that honest Bob will perform. Later we will see what happens if Bob tries to cheat and does something different.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2zTUwlA8Te4mL8mkZjl8JPczsbmFbJBH4Fk9nxaBqKo6HrYSDtaTnOtFk7ASvdAMwhKxh1BGDo9drm7G0gORUn_ZSTe0JqLMX6pTCyPD14OAdLQJ4P2JjrzDP12PP3ASxFMDrsCck01y9/s1600/Slide2.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="202" data-original-width="1084" height="118" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2zTUwlA8Te4mL8mkZjl8JPczsbmFbJBH4Fk9nxaBqKo6HrYSDtaTnOtFk7ASvdAMwhKxh1BGDo9drm7G0gORUn_ZSTe0JqLMX6pTCyPD14OAdLQJ4P2JjrzDP12PP3ASxFMDrsCck01y9/s640/Slide2.jpg" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Delegated computation of the Hadamard gate. The red and green boxes are QOTP boxes that allow her to perform Pauli operations on qubits. Here we see that if Alice encrypts a qubit with key $(j,k)$ and Bob honestly performs $H$, then Alice can recover her desired output by interchanging the roles of the key bits during decryption. Note that the outcome might get an overall phase that might be relevant if this circuit is part of a larger one.</td></tr>
</tbody></table>
<br />
<br />
<div class="separator" style="clear: both; text-align: center;">
</div>
<div>
Next let us consider the CNOT gate. In this case, Alice has two qubits in state $|q_1\rangle$ and $|q_2\rangle$. If we performed the CNOT gate directly on the two qubits, we expect that for<br />
<br />
$|q_1\rangle = a |0\rangle + b |1\rangle$ and<br />
<br />
$|q_2\rangle = c |0\rangle + d |1\rangle$, then<br />
<br />
$\mathrm{CNOT} |q_1 q_2 \rangle = ac |00\rangle + ad |01\rangle + bd |10\rangle + bc |11\rangle$.<br />
<br />
Alice will encrypt her input qubits using the QOTP with keys $(j,k)$ and $(l,m)$, respectively. She sends the qubits to Bob and asks him to perform $CNOT|q_1,q_2\rangle$. If he does this honestly and returns two qubits $|\tilde{q}_1\rangle, |\tilde{q}_2\rangle$ to Alice. The state of these two qubits for an honest Bob is shown in the table below.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEholrYUmkEAzmKL0iQgG4qQ_o0Sp3YijlJNjB_U5Zcyp7l6c-OlSZnorFGONStT8__yyE2TGt2jKCwlxG1Ki6Se05WMnaT-m1Fvoc2vOXCMULzIlBw2leL0SgjNn1w7db9B1IGJwFfTsUQf/s1600/Slide6.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="720" data-original-width="1139" height="404" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEholrYUmkEAzmKL0iQgG4qQ_o0Sp3YijlJNjB_U5Zcyp7l6c-OlSZnorFGONStT8__yyE2TGt2jKCwlxG1Ki6Se05WMnaT-m1Fvoc2vOXCMULzIlBw2leL0SgjNn1w7db9B1IGJwFfTsUQf/s640/Slide6.jpg" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">The effect of the CNOT gates on the encrypted pair of qubits. It is still possible to recover the desired output just by decrypting the qubits using the same keys as before, plus some Pauli gates for corrections.</td></tr>
</tbody></table>
<br />
<br />
Once more, we can carefully work out what Alice needs to do to recover the desired outcome by studying individual cases. This is straightforward but a bit tedious since we have 16 cases to consider. However, you may check that what Alice needs to do is apply $Z^k X^j Z^m$ to the first qubit $|\tilde{q}_1\rangle$ and $X^j Z^m X^l$ to the second qubit $|\tilde{q}_2\rangle$. This gives the desired outcome up to possibly an overall minus sign that can be corrected later if needed. </div>
<div>
<br /></div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjHQnp_UlBPOjo-LA1fb0sNmixkWPFMHIpCtLSvLDfa3R2m1KN9IT2-uBQRTIkfI67Uo7hAbGnBrNTFGdmXD4Y3vW2wyZ4Mb_NUcJkjw7GwA8Z-0fldCCUwdhH9ExmedzVmTr8oI9dxntSu/s1600/Slide4.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="505" data-original-width="1476" height="218" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjHQnp_UlBPOjo-LA1fb0sNmixkWPFMHIpCtLSvLDfa3R2m1KN9IT2-uBQRTIkfI67Uo7hAbGnBrNTFGdmXD4Y3vW2wyZ4Mb_NUcJkjw7GwA8Z-0fldCCUwdhH9ExmedzVmTr8oI9dxntSu/s640/Slide4.jpg" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Delegated computation of the CNOT gate. Here Alice encrypts her two qubits $|q_1\rangle$ and $|q_2\rangle$ using keys $(j,k)$ and $(l,m)$, resepctively for the QOTP. Once the result of Bob's computation is received, Alice can recover the correct result as follows: for the first qubit, she decrypts using the key $(j,k)$ but afterwards she applies a correction $Z^m$ (which can be implemented with the red box using key $(0,m)$). For the second qubit, she first applies the correction $X^j$ (which can be implemented with the red box using key $(j,0)$) before decrypting using the same key $(l,m)$. </td></tr>
</tbody></table>
<div class="separator" style="clear: both; text-align: center;">
</div>
<br />
<br />
Finally let us consider the $T$ gate. This one gets a bit more complicated because Alice is limited to what she can recover using just Pauli gates. If we take as input a qubit in state $|q\rangle = a |0\rangle + b|1\rangle$, then applying $T$ gives us $T|q\rangle = a|0\rangle + b\sqrt{i}|1\rangle$.<br />
<br />
Again, Alice will encrypt the qubit using a pair of key bits $(j,k)$ then send the encrypted qubit to Bob. As we see in the table below, depending on the value of $j$, the state that an honest Bob returns will either be already be the one we want after decryption ($j=0$) or will require some correction by $T^2$.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiP0RdbQ_sgSuiinRAQJ3c9CR4AUgl9ySvMnYnnuEPKGKUS3MkP5FA9W8WgSXiMTzmZU8byMjQ_prwlOIRf2XXhxFna1_sDEw9mRmXBF23JHSj9sJfniMC9L-HEUwkRXstG58sxb4Sb3TsM/s1600/Slide7.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="508" data-original-width="1600" height="202" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiP0RdbQ_sgSuiinRAQJ3c9CR4AUgl9ySvMnYnnuEPKGKUS3MkP5FA9W8WgSXiMTzmZU8byMjQ_prwlOIRf2XXhxFna1_sDEw9mRmXBF23JHSj9sJfniMC9L-HEUwkRXstG58sxb4Sb3TsM/s640/Slide7.jpg" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">The $T$ gate acting on an encrypted qubit. Upon decryption, we already have the desired output when $j=0$, but non-Pauli gate corrections are needed when $j = 1$.</td></tr>
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The problem is that Alice can not perform $T^2$ since this can be implemented from $X$ and $Z$ only. The simplest solution is for Alice to ask Bob again. However, even if she does another encryption, Bob will notice that Alice asked for $T^2$ after $T$, which would reveal to him that $j =1$.<br />
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To keep $j$ secret from Bob, Alice should ask Bob to perform $T^2$ for both values of $j$. This can be achieved by preparing an extra qubit in state $|0\rangle$, and sending this qubit for the gate $T^2$ when $j = 0$. If we draw the quantum circuit, this corresponds to adding a controlled-swap gate after the qubit returned by Bob is decrypted. This is shown in the figure below.<br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJUZyz4bMIemi5qOd3fVhQapJT7ys45IMV79G7sPUBkXSHlKts8Qd_Me0ZQ3Yx8ZQO0Z1lZUloNTfkmRMGvoaKpJLgM0R576pMDYFYwCvrd3ZXOnxktY3UZgYCf8V2OR9IZRT3Yq9R095s/s1600/Slide5.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="641" data-original-width="1600" height="256" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJUZyz4bMIemi5qOd3fVhQapJT7ys45IMV79G7sPUBkXSHlKts8Qd_Me0ZQ3Yx8ZQO0Z1lZUloNTfkmRMGvoaKpJLgM0R576pMDYFYwCvrd3ZXOnxktY3UZgYCf8V2OR9IZRT3Yq9R095s/s640/Slide5.jpg" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Delegated computation of the $T$ gate. This calculation involves two stages because there is a slight complication when $j=1$ during the QOTP encryption. To get the correct result, we have to re-encrypt and resend the qubit to Bob and ask him to perform the $T^2$ gate. Aftewards, Alice can peform the final correction $Z^l$. Observe that when $j=0$, we already get the correct result just by decrypting the qubit received from an honest Bob. Thus, in order to keep the value of $j$ private, we send an different encrypted qubit to Bob, which does not affect the state of the other qubit.</td></tr>
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The circuit shows that when $j=0$, we already have the desired out so when we ask Bob to perform $T^2$, we swap in the encrypted $|0\rangle$. Afterwards we swap back the two qubits so we can later discard the extra qubit.<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiB09keTJ34IRuOyyOP1AFla-fjFaCzwUycBJ90-XWOqGGggP-cnu_lc92i1Vs8xwLR_AUciuYrUqN4JCkDB1YM9W_p01rfcpFZcW7782xw2GimqgR3yWHtN9sRz32ug9-X3AUsMM12N_ZS/s1600/Slide8.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="478" data-original-width="1600" height="190" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiB09keTJ34IRuOyyOP1AFla-fjFaCzwUycBJ90-XWOqGGggP-cnu_lc92i1Vs8xwLR_AUciuYrUqN4JCkDB1YM9W_p01rfcpFZcW7782xw2GimqgR3yWHtN9sRz32ug9-X3AUsMM12N_ZS/s640/Slide8.jpg" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Resending the qubit to Bob for the $T^2$ correction. We see that after decrypting, we need an additional $Z$ correction when $l = 1$, which Alice can do herself.</td></tr>
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The interesting case is when $j=1$. Here after swapping, we will be encrypting the qubit $|q'\rangle = a\sqrt{i}|0\rangle + b|1\rangle$ using the key $(l,m)$ and sending it to Bob. The table above shows what happens next for various $(l,m)$. If he is honest, he performs $T^2$ and sends back the result. Alice decrypts normally and applies a $Z$ correction when $l = 1$. Note that the result here comes with an extra phase factors but since these only contribute to an overall phase, they do not yield any physically observable effects.<br />
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We have shown a protocol for delegated computation of a universal set of gates. Thus, we have presented a way for Bob to assist Alice in her quantum computation, without revealing any of Alice's private data. The protocol works as intended when Bob is honest so it is correct.<br />
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Unfortunately, you may have noticed already that Alice has to tell Bob the operations he has to perform so the circuit is revealed to Bob. A simple way to make the computation blind is to make Alice run a fixed circuit that cycles through an $H, T$, and CNOT gate. If any of the gates is not needed, Alice sends a junk qubit to Bob. Then from Bob's point of view, he merely observes Alice asking him to perform a repeating sequence of $H, T$, and CNOT gates. Thus, we can make the protocol blind. (The drawback of doing it this way is that it becomes quite impractical if most of the gates are not needed.)<br />
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Finally we want to be able to verify if Bob indeed performed the operation Alice asked of him. In order for Alice to detect if Bob is cheating, she can perform some test runs on a random subset of her inputs. She can do two types of test runs. One is called the $X$-test and it uses as input a string of qubits all in state $0\rangle$. The other is called the $Z$-test and it uses as input a string of qubits all in state $|+\rangle$.<br />
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The trick here is for Alice to run some computation on these test qubits such that net effect is equivalent to the identity operator on these inputs. This requires a bit of fiddling around to make sure the computation is complicated enough that Bob will not notice that he is doing an identity operation overall. The idea is that Alice knows which qubits are involved in the test runs so she just checks that she recovers the input states in those computations.<br />
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Thus, we have described a protocol for delegated computation based on the QOTP. The main drawback of the protocol is that, in order to make it blind and verifiable, Alice has to perform additional computations on junk qubits or test runs, which makes the protocol quite impractical when Alice's delegated circuit is large.<br />
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There are more efficient ways to do delegated computation if we used a different quantum computing model (one based on so-called cluster states) or if we can talk to two quantum servers instead of just one; the latter one even allows Alice to be entirely classical, i.e., she is unable to do any quantum operations.<br />
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<b>Reference:</b><br />
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Andrew Childs, <a href="https://arxiv.org/abs/quant-ph/0111046v2" target="_blank">Secure Assisted Quantum Computation</a>, Quantum Information and Computation 5 (2005) 456.Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-65912671253042711212018-02-01T22:05:00.001-08:002018-02-16T10:00:24.120-08:00A simple protocol for quantum oblivious transferOne of the goals in cryptography is to allow parties who do not trust each other to solve problems together. One important problem that can be solved using such a secure multiparty computation is called secure function evaluation (SFE).<br />
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In SFE, we have two parties Alice and Bob who each hold some private data. Alice wants to use Bob's data to compute some function on her side, while Bob wants to use Alice's data to compute some function his side. The goal is to allow Alice and Bob to compute their functions without having to reveal their secrets.<br />
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We can describe a SEF protocol as follows: Denote Alice's secret input by $x$ and Bob's secret input by $y$. Alice and Bob can send messages to each other. At the end of the protocol, Alice outputs $a$ and Bob outputs $b$. Alice wants to compute the function $A(x,y)$ while Bob wants to compute $B(x,y)$.<br />
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If the protocol is correct, then $a = A(x,y)$ and $b = B(x,y)$ for honest Alice and Bob. If the protocol is secure, then Alice cannot know anything about Bob's input $y$ except what she might deduce from $A(x,y)$. Similarly, Bob cannot know anything about Alice's input $x$ except what he might deduce from $B(x,y)$.<br />
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Let us consider an example of SFE called oblivious transfer (OT). In an OT protocol, Alice has two input bit strings $s_0$ and $s_1$, and Bob has input bit $t$. Alice does not need to compute anything but Bob wants to retrieve $s_t$. An OT protocol is secure if Alice does not find $t$ and also if Bob only gets one of Alice's strings.<br />
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Suppose Alice has a database with two records $s_0$ and $s_1$. Bob wants to retrieve one of the records but he does not want Alice to find out which one. Meanwhile, Alice is willing to grant Bob access to one record, but not to the entire database. This is a practical problem that can be solved with OT.<br />
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<a name='more'></a>The main reason OT is interesting is because it has been shown that we can solve any SFE problem just by using a secure OT protocol as many times as needed. Unfortunately, it is also known that you cannot perform classical OT protocols perfectly: you will need to assume that a cheating party has some computational or physical limitations. But as long as these assumptions are reasonable, it allows us to construct OT protocols that are good enough in practice.<br />
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Here we describe a simple quantum protocol for OT. Informally, the protocol works as follows: Alice will encrypt her two secret bit strings using a one-time pad, which encrypts a message by adding it to a random secret key. Alice will send qubits to Bob in order to generate the keys to be used for encryption, in such a way that Alice has two keys, but Bob only has one key. Alice sends the two encrypted messages to Bob and Bob will be able retrieve one of the secrets using his key.<br />
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Now we will define the steps in our protocol more precisely. For concreteness, we will take Alice's input strings $s_0$ and $s_1$ to be three-bit messages in our example, but generalizing to $n$ bits is straightforward.<br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgBnG-5ZCHeRR296P-buzKCZUAGV1dP-xDJh5ZwRgx0DT8ZplY6GuJIhsp3sn4cuM3q3R7sPqsA9EgOwXZrnITiJyh1-i4P90RfmEow-PMCJkPRluEEBgleiO4UcugdcK-6xObtD9jRwIEG/s1600/Slide1.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="298" data-original-width="846" height="224" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgBnG-5ZCHeRR296P-buzKCZUAGV1dP-xDJh5ZwRgx0DT8ZplY6GuJIhsp3sn4cuM3q3R7sPqsA9EgOwXZrnITiJyh1-i4P90RfmEow-PMCJkPRluEEBgleiO4UcugdcK-6xObtD9jRwIEG/s640/Slide1.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">In oblivious transfer, Alice has 2 input strings $s_0$ and $s_1$ and Bob has an input bit $t$, and Bob has to output $s_t$. For our example, we will consider 3-bit messages for $s_0$ and $s_1$</td></tr>
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We will take our qubits to be in the form of photon polarization, which can be prepared in two different bases using polarization filters, namely the $+$ filter for vertical and horizontal polarizations, and the $\times$ filter for diagonal polarizations. We have previously seen these qubit states and measurements in the BB84 protocol for quantum key distribution.<br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgg4YxTl0DAp1yewYeYX-jAdiPW5djTzDhnQlTm-y0Ef5JNX5kdQXrHyKoKJwOS2Wpm9ypo_sAzs8Ff4lhaggldxsz-a1dI4EJGNVg9MAPT7gTU_6_y8ClENBisbwIoRfUZjJjgd78FNeIJ/s1600/Slide2.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="330" data-original-width="825" height="256" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgg4YxTl0DAp1yewYeYX-jAdiPW5djTzDhnQlTm-y0Ef5JNX5kdQXrHyKoKJwOS2Wpm9ypo_sAzs8Ff4lhaggldxsz-a1dI4EJGNVg9MAPT7gTU_6_y8ClENBisbwIoRfUZjJjgd78FNeIJ/s640/Slide2.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Alice and Bob have to decide on how to assign bit values to qubits encoded in different bases. Here we consider a qubit in the form of the polarization of a photon, where a qubit basis corresponds to filters oriented along different angles. We will use the $+$ filter and the $\times$ filter, which correspond to the computational basis and the Hadamard basis, respectively.</td></tr>
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For Alice to generate two random keys, she need to choose a random $2n$-bit string $k$, and a random $2n$-bit string $\theta$ that has $n$ 0s and $n$ 1s. In this case, $k$ denotes the bit values and $\theta$ denotes the basis or filter.<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilG9GoScdc0kKAAXSW4eHvJ-C__O9ABWpbR4x3xmwTnutSVh0pFPvj6iY2TxkETss_e7uxNno7POE17iECcns-eyGhb46XzWZMowzc3RT52cUCFv5ZfS2mEQ37lvgVUFuqT7B03bjh6VmF/s1600/Slide3.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="374" data-original-width="651" height="366" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilG9GoScdc0kKAAXSW4eHvJ-C__O9ABWpbR4x3xmwTnutSVh0pFPvj6iY2TxkETss_e7uxNno7POE17iECcns-eyGhb46XzWZMowzc3RT52cUCFv5ZfS2mEQ37lvgVUFuqT7B03bjh6VmF/s640/Slide3.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Alice picks a random string for each filter. She also picks a random sequence of filters that determines the order in which the bits of the random strings are sent to Bob. In practice, Alice would pick two random bit strings, one for the bit values and the other for the bases, which needs equal numbers of 0s and 1s.</td></tr>
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What happens is that Alice prepares photon $i$ in the state $| k_i \rangle_{\theta_i}$, where $\theta_i = 0$ indicates a qubit for the $+$ filter (computational basis states $|0\rangle$ and $|1\rangle$), while $\theta_i = 1$ indicates a qubit for the $\times$ filter (Hadamard basis states $|0\rangle_1 = |+\rangle$ and $|1\rangle_1 = |-\rangle$, where $|\pm\rangle = ( |0\rangle \pm |1\rangle )/\sqrt{2}$).<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh18clALUAKkja9lVeuj3NR__0jGoHNSyJIkuvfwlkfc9GlOx5AnIFzRQFgL_-tpw-3LuF4xQRAXri6deJkexiesed1aP1-jOA2rYW-XmZXiCjlLP_ccc6rBVrksi_swqecCDctULOwGUV-/s1600/Slide4.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="445" data-original-width="914" height="310" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh18clALUAKkja9lVeuj3NR__0jGoHNSyJIkuvfwlkfc9GlOx5AnIFzRQFgL_-tpw-3LuF4xQRAXri6deJkexiesed1aP1-jOA2rYW-XmZXiCjlLP_ccc6rBVrksi_swqecCDctULOwGUV-/s640/Slide4.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">The bit strings $k_0$ and $k_1$ and the sequence of filters help determine the photon polarization that Alice should send to Bob, and in what order.</td></tr>
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If we match the bits of $\theta$ with $k$, then Alice can denote by $k_0$ the bits in $k$ that align with 0s in $\theta$, and by $k_1$, the bits that align with the 1s in $\theta$.<br />
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We needed $k_0$ and $k_1$ to be of the same length as the input strings, which is why we needed $\theta$ to have the same number of 0s and 1s. In our example, the bits in $\theta$ corresponds to the sequence of $+$ and $\times$ filters that Alice used to prepare the photons that she will send to Bob.<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZLdq1u8124enlrJKeu15b-ZIzSY4wtQ-pzXXAt8UXhWRvqDTCL45AMKRaVW7jykHvur-fZ6i__CyPK7U-FfDzQDBofvfudI7jL1L15BcfKIktQumRNukzHGf5uZ8HkFWbSilC1Ddgkthb/s1600/Slide5.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="444" data-original-width="914" height="310" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZLdq1u8124enlrJKeu15b-ZIzSY4wtQ-pzXXAt8UXhWRvqDTCL45AMKRaVW7jykHvur-fZ6i__CyPK7U-FfDzQDBofvfudI7jL1L15BcfKIktQumRNukzHGf5uZ8HkFWbSilC1Ddgkthb/s640/Slide5.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Bob picks a filter $\theta$ and measure each qubit he gets from Alice using this filter. He records the outcomes of his measurements in the string $y$.</td></tr>
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Alice sends the qubits one by one to Bob. If $t = 0$. Bob measures each qubit with the $+$ filter and if $t = 1$ he measures each qubit with the $\times$ filter. He records the outcomes into the string $\tilde{k}$.<br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiVihYILspkarueoLY_ib0qRHGv-xQLQ49JClmeNm8XhpL_AzU_fkqXzn80rx7sZt4FMSihwX-q2IeFZTdH66GwU27b4BnNr4iVjdX5TCKNhWj1MbvYCN2ajENaw_FPgi1q0mQxf8jfTKRB/s1600/Slide6.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="445" data-original-width="891" height="318" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiVihYILspkarueoLY_ib0qRHGv-xQLQ49JClmeNm8XhpL_AzU_fkqXzn80rx7sZt4FMSihwX-q2IeFZTdH66GwU27b4BnNr4iVjdX5TCKNhWj1MbvYCN2ajENaw_FPgi1q0mQxf8jfTKRB/s640/Slide6.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Waiting some time after Bob receives all the qubits sent by Alice, Alice will send the random sequence of filters described by the string $\theta$. Bob can then check which qubits were measured with the correct filter, which means Alice and Bob have the same bit value in those positions.</td></tr>
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Bob confirms that he has received and measured each qubit. After some time has passed, Alice sends the bit string $\theta$ so Bob can identify the positions in $\tilde{y}$ where $\theta_i = \tilde{\theta}$, that is, he finds the qubits where they used the same filter. Let $\tilde{k}_t$ denote the bit string obtained from those positions.<br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgCzG-yLc0tVdL0br7U-Ml7p824tPH1bI2SWfHbz_7vCsZnARuuKpZJR6AAPCe912rhUjqiORMWhbI_yd3fyf199tC0URSuerZ0XYLXi3NDtuYkN8Civfnj0Y_kBceFY-Y5MMrTspKwc6gu/s1600/Slide7.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="310" data-original-width="669" height="296" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgCzG-yLc0tVdL0br7U-Ml7p824tPH1bI2SWfHbz_7vCsZnARuuKpZJR6AAPCe912rhUjqiORMWhbI_yd3fyf199tC0URSuerZ0XYLXi3NDtuYkN8Civfnj0Y_kBceFY-Y5MMrTspKwc6gu/s640/Slide7.png" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Alice can encode her inputs $s_0$ and $s_1$ with a one-time pad using her random keys $k_:0$ and $k_1$. Alice computes the strings $m_j = k_j \oplus s_j$ for $j=0,1$ and sends $m_0$ and $m_1$ to Bob. </td></tr>
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Alice computes $m_0 = k_0 \oplus s_0$ and $m_1 = k_1 \oplus s_1$ ($\oplus$ is the bitwise XOR operation) and sends $m_0$ and $m_1$ to Bob. Finally, Bob can compute his desired string using $s_t = m_t \oplus \tilde{k}_t$.<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhfVkg0yPzo2YY9vvTcMbkHFkP_95YawPXunZCCuzeUn0lvSc6-JeGsu4jqQGr4ufu3tWsgSFG_cYa5robSZa4n_U0rCXN26Vr8Em2_QnM4HOM6JyLG0naZhyAcn9UxqsPoN-zjUkRT3GDx/s1600/Slide8.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="308" data-original-width="617" height="318" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhfVkg0yPzo2YY9vvTcMbkHFkP_95YawPXunZCCuzeUn0lvSc6-JeGsu4jqQGr4ufu3tWsgSFG_cYa5robSZa4n_U0rCXN26Vr8Em2_QnM4HOM6JyLG0naZhyAcn9UxqsPoN-zjUkRT3GDx/s640/Slide8.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Bob can compute the desired input string $s_t$ by using his key string to decrypt $m_t$. He has only one key so he can only open one of the two encrypted messages.</td></tr>
</tbody></table>
<br />
<br />
It is easy to check that the protocol works for honest Alice and Bob. To see that it is secure, first let us consider a cheating Alice, who wants to find $t$. Observe that $t$ is not communicated in the protocol; in fact, Bob does not send messages to Alice. So there is no way for Alice to learn $t$.<br />
<br />
How about a cheating Bob? If you think about it for a little while, you might find out that the protocol is not secure if Bob has a means for storing qubits reliably. Such a quantum memory allows him to postpone measuring his qubits until after Alice reveals the string $\theta$. Bob can then measure each photon in his quantum memory using the correct filter, allowing him to recover both $k_0$ and $k_1$.<br />
<br />
So unfortunately, it turns out that you cannot perform OT perfectly: you will need to impose some limitations on at least one of the cheating parties. Here we can use a physical assumption for limiting Bob.<br />
<br />
Since much of the present-day technology for quantum memory is still quite new and untested, it is reasonable to assume that there is a limit to how many qubits Bob can reliably store in his quantum computer. By making sure there is a sufficient delay between Alice sending the qubits and revealing the bases $\theta$, Bob will be forced to measure some of the qubits that he cannot store in memory.<br />
<br />
To quantify security, we need to know how much Bob can learn about $k_0$ and $k_1$ given his limited capacity for storing qubits. This involves a computation involving Bob's min-entropy of the bit strings $k_0$ and $k_1$, which goes beyond our scope here.<br />
<br />
Still it follows from such an analysis that we can be secure against cheating Bob if $q \le m - \log^2 m$, where $q$ is the number of qubits Bob can store in his quantum memory, and $m$ is total number of qubits Alice sends to Bob in our OT protocol. Since $\log^2 n$ grows more slowly than $n$, we see that security is possible just by sending several more qubits than Bob can store.<br />
<br />
<br />
<b>Reference:</b><br />
<br />
Charles Bennett, Gilles Brassard, and Claude Crépeau, <a href="https://link.springer.com/content/pdf/10.1007%2F3-540-46766-1_29.pdf" target="_blank">Practical Quantum Oblivious Transfer</a>, in CRYPTO '91 LNCS vol. 576 (1992) 351-366.<br />
<br />
<br />Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-33478364608015589632017-12-31T07:04:00.004-08:002018-03-23T07:54:46.351-07:00Grover's algorithm for the unstructured search problemAfter Peter Shor showed how quantum computers can factor numbers efficiently, interest surged in finding other problems which quantum methods solved faster than the best known classical techniques. In 1996 Lov Grover developed such a quantum algorithm for solving the unstructured search problem.<br />
<br />
In a generic search problem, we are given a set $S$ of $N$ items. Some of the items in the set are marked. Our task is to find a marked item in the set $S$.<br />
<br />
You can think of a search problem like looking up a name in a list. In most cases, the list will be arranged in a way that makes finding names easier. For example, a list of names is often sorted alphabetically. Here we are talking about a list with no discernible ordering. In this case, the best we could do is to check each entry for the desired name.<br />
<br />
<a name='more'></a>In mathematical terms, a search problem involves a function $f$ which takes as input $x$ from the set $S$ of $N$ items, and outputs a bit, 0 or 1. The bit value tells us whether the input $x$ is a marked item or not. Thus, the goal in a search problem is to find a marked item in $S$, i.e an item $x$ such that $f(x)=1$. In this case, we say that $x$ is a solution to the search problem on $S$.<br />
<br />
When we talk about the complexity of a search algorithm, then typically the only cost we count is the queries made to the function $f$. This is the number we will compare between classical and quantum methods.<br />
<br />
Consider again the problem of searching a list of names. The function $f$ in this setting corresponds to a means of checking if an entry $x$ is the desired name. In practice, that might just mean reading the entry yourself, or maybe querying some machine that does this check for you. Whichever the case, the best you could do for an unstructured list is to query $f$ for each entry until you get to a solution, which might take around $N$ queries. Grover's algorithm solves the unstructured search problem using just around $\sqrt{N}$ queries, so the speedup we get is not as spectacular as in Shor's algorithm, but is no less significant.<br />
<br />
Here I will describe Grover's algorithm. In order to be explicit with some of the calculations, I will be using a specific example: we are going to search for a unique solution in a set with $8$ items. This means we will consider a function $f$ that takes 3-bit strings as an input, i.e., the items in the set are labeled with bit strings, and outputs a bit $f(x)$, which tells us if $x$ is marked or not.<br />
<br />
We will need some background material in order to discuss the details of the algorithm so we will go through them here briefly. First I will describe a few elementary quantum gates for building quantum circuits. In the figure below, the gates we consider are the NOT gate, the CNOT gate, and the Toffoli gate (these gates operate the same way on bits if we have classical Boolean circuits).<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjN3ef8QUnd7GjUh1RKMCOxr2a-TjwvNCRgCLhOO7yiFWNtH7iKj_hv-235XJLxzsXgSfALLniP4huNxCSmEtltkjii_Naci3iqOF8EhwtQSSqO0BPK4SFpoOi5ePPb_lz-uELir6hFEdPN/s1600/Slide1.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="794" data-original-width="1280" height="396" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjN3ef8QUnd7GjUh1RKMCOxr2a-TjwvNCRgCLhOO7yiFWNtH7iKj_hv-235XJLxzsXgSfALLniP4huNxCSmEtltkjii_Naci3iqOF8EhwtQSSqO0BPK4SFpoOi5ePPb_lz-uELir6hFEdPN/s640/Slide1.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Some elementary quantum gates that we will use to construct larger quantum gates.</td></tr>
</tbody></table>
<br />
<br />
The NOT gate is a single-qubit gate flips the computational basis states $|0\rangle$ into $|1\rangle$, and vice-versa. This is also known as the Pauli-$X$ gate.<br />
<br />
The CNOT gate is a two-qubit gate that flips computational basis states in the second register (target qubit) if the first register has a bit value 1 (control qubit). It is known as an entangling gate for two qubits since if we prepare the control qubit in the state $|+\rangle = ( |0\rangle + |1\rangle )/\sqrt{2}$ and the target qubit in the state $|0\rangle$, then the output after applying a CNOT gate is<br />
<br />
$ \text{CNOT} ( |0\rangle + |1\rangle )|0\rangle /\sqrt{2}<br />
= \text{CNOT} ( |00 \rangle + |10 \rangle ) /\sqrt{2}<br />
= ( |00\rangle + |11\rangle )/\sqrt{2}$,<br />
<br />
which is an EPR pair.<br />
<br />
The Toffoli gate, also known as the controlled-CNOT gate and was invented by Tommaso Toffoli, is a three-qubit gate that flips the computational basis states of the third qubit if and only if the first two qubits both have a bit value of 1. This gate is like a reversible version of the AND gate, which outputs 1 if and only if the two input bits are both 1. (A gate is reversible is we can determine what the input was if we are given only the output.)<br />
<br />
In quantum theory, a quantum operation must be unitary if they are to preserve the total probabilities. This means that if we want to make a quantum gate $U_{f}$ that computes a function $f$, this gate must be reversible. Fortunately, in most cases, this just means that when we build $U_{f}$, we should have a register or wire that holds the input.<br />
<br />
The figure below shows an oracle or black box for computing a function $f$ with a three-bit input and one-bit output. You will often see similar-looking oracles in texts about quantum algorithms. This is because for some computational problems, we are not so concerned with the details of how the function $f$ is computed, in which case, we think of how simple or complex an algorithm is only in terms of how many times we have to query the function $f$. For many problems, this already gives us a good idea if a problem is hard or easy to solve.<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiWZ0cgZcLInd6EZyvn4iS-zDp9e0GOFdFaZCj5UcZVDJ_k3vCdHQNRGFm-oU9wgU8wD-qU-x-gYRHLaX5tplSiBYB3LwSbGjrYK4rvTrRm1m6LqqmTa-GRk32Fhe0KA2c6iNcmiIgPG_P6/s1600/Slide2.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="794" data-original-width="1280" height="396" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiWZ0cgZcLInd6EZyvn4iS-zDp9e0GOFdFaZCj5UcZVDJ_k3vCdHQNRGFm-oU9wgU8wD-qU-x-gYRHLaX5tplSiBYB3LwSbGjrYK4rvTrRm1m6LqqmTa-GRk32Fhe0KA2c6iNcmiIgPG_P6/s640/Slide2.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">A standard quantum black box for the function $f$. It involves two sets of wires: a top set (here the top 3 wires) for the input to $f$ and a bottom set (here just the last wire since the output is 1 bit) for the output for $f$.</td></tr>
</tbody></table>
<br />
<br />
Still you might be curious how a function $f$ might be implemented in practice. Later when we describe Grover's algorithm, we will consider an unstructured search problem on a set of 8 items with a unique marked item. In terms of the function $f$, this means we can label the items with 3 bits (000, 001, all the way to 111), and we are looking for which 3-bit string x gives $f(x) = 1$.<br />
<br />
Let's suppose that $f(110) = 1$ so 110 is the unique solution to our search problem. Then one possible circuit that computes $f$ through a quantum gate $U_{f}$ is shown below. Observe that the circuit below uses two additional wires (the 4th and 5th wires from the top) that hold intermediate values of a computation. However, we use the trick of un-computation to reset these wires to their initial values (they start and end in state $|0\rangle$ ) so these wires can be discarded without affecting the qubits in the other wires. If you are thinking in terms of a black box $U_f$ then the 4th and 5th wires can be placed completely inside the black box.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh2K83flzO4iJ1xxw_h_dG2zcwjSJD8T9V1aqpJSi5yOkLsfqgrCOQhhmM3Q1V0ExtlejAGoXkO7fAdHmDJHFs20ZqqMvjg0l3H0XVyGSEsFDN1r1HTIzq3H9mNXJbVpieeRT3Q6ygC5Tgi/s1600/Slide4.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="794" data-original-width="1280" height="396" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh2K83flzO4iJ1xxw_h_dG2zcwjSJD8T9V1aqpJSi5yOkLsfqgrCOQhhmM3Q1V0ExtlejAGoXkO7fAdHmDJHFs20ZqqMvjg0l3H0XVyGSEsFDN1r1HTIzq3H9mNXJbVpieeRT3Q6ygC5Tgi/s640/Slide4.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">While quantum algorithms are often described using quantum black boxes when the function $f$ is queried, in practice the black boxes have to be implemented using quantum gates. Here we present a quantum circuit for out example function $f$ which has a marked item $x = 110$. </td></tr>
</tbody></table>
<br />
<br />
Another trick that will be useful to us is called the phase-kickback trick. Suppose we have a function $f$ with an input of $N$ bits and outputs a bit. We can construct a corresponding quantum gate $U_f$ that computes this $f$, which has a top wire of $N$ bits, and a bottom wire of 1 bit, as illustrated in the figure below. Let us examine what happens if we use an input of $|x\rangle$ on the top wire and $|-\rangle = ( |0\rangle - |1\rangle )/\sqrt{2}$ on the bottom wire:<br />
<br />
$U_{f} |x\rangle |-\rangle = U_{f} (|x\rangle |0\rangle - |x \rangle |1\rangle )/\sqrt{2}<br />
= (|x\rangle |f(x) \rangle - |x \rangle |\overline{f}(x) \rangle )/\sqrt{2} $<br />
$ = |x\rangle ( |f(x) \rangle -|\overline{f}(x)\rangle )/\sqrt{2}<br />
= |x\rangle (-1)^{f(x)} ( |0\rangle - |1\rangle )/\sqrt{2}<br />
= (-1)^{f(x)} |x\rangle |-\rangle$<br />
<br />
where $\overline{f}(x) = 1 \oplus f(x)$. So what we see is that the bottom wire receives a sign that depends on whether $f(x)$ is 0 or 1. However, since this phase depends only on what is on the top wire, we can think of this phase being "kicked back" onto the top wire (thus, the name phase kickback).<br />
<br />
If you have read other references for Grover's algorithm, they will sometimes talk about an oracle or black box for the function $f$ that marks $x$ as a solution by multiplying it by -1. Now you know how such a black box can be built for $f$ using phase kickback.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVK5JAZEhngyJXy8UERccMTF3cG7I2Wddhf0lwa17473wAjm2yAA8rsYFFbqaom0ueA9oQTswY0kedA3IO36ZlyZWA5L_PyfQnzMThehwLF_F_HQfts9Vz_IlcjlR7rdxRi2uaU85WYXVJ/s1600/Slide3.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="794" data-original-width="1280" height="396" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVK5JAZEhngyJXy8UERccMTF3cG7I2Wddhf0lwa17473wAjm2yAA8rsYFFbqaom0ueA9oQTswY0kedA3IO36ZlyZWA5L_PyfQnzMThehwLF_F_HQfts9Vz_IlcjlR7rdxRi2uaU85WYXVJ/s640/Slide3.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">The phase-kickback trick is often used to build a quantum gates that flips the phase of solution states given a quantum black box for computing $f$. Since the state $|-\rangle$ remains the same after, this wire can actually be considered as part of the black box for $f$.</td></tr>
</tbody></table>
<br />
<br />
The last thing we we require is a quantum gate that flips the phase (maps $|x\rangle to -|x\rangle$) except when $x$ is the all-zeroes string. As we have already seen, this can be achieved using the phase-kickback trick on a quantum gate $U_{g}$ that computes the function $g$ where $g(000) = 0$ and $g(x) = 1$ otherwise. This is shown as a black box in the figure below. Note that because this operation flips a known state $|0\rangle$ and is independent of $f$, then we consider the operation to be cost-free (we don't count queries to $g$ when comparing the number of queries between classical and quantum algorithms).<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiG4eCeXtpj2-rGOUxjxfsSNHaCh9iIDSag27doSxQTuWIu96z4_OVIsq2rAyotzen16ef7HjsyBMdjA74-bCAG0A1-GB1rH6fOpoRlwaAgGt6pHKcaTefEA2GaHmNSfvS6Q33gaJdPB4C2/s1600/Slide5.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="794" data-original-width="1280" height="396" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiG4eCeXtpj2-rGOUxjxfsSNHaCh9iIDSag27doSxQTuWIu96z4_OVIsq2rAyotzen16ef7HjsyBMdjA74-bCAG0A1-GB1rH6fOpoRlwaAgGt6pHKcaTefEA2GaHmNSfvS6Q33gaJdPB4C2/s640/Slide5.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">A quantum gate for the function $g$ that flips the phase of all computational basis states except the all-zeroes string.<br />
Since this operation does not involve any queries to $f$, then we do not count the queries made to it.</td></tr>
</tbody></table>
<br />
<br />
We do not need to know exactly how $g$ is computed in $U_{g}$ but for completeness I provided a sample quantum circuit for implementing $U_{g}$ below. Similar to what we had for $f$, the 4th and 5th wires from the top are additional wires for intermediate results, which would be fully contained within a black box for $U_g$.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3bNJa9z4SjkLaL10dDSkMcbinjXQudF9x6fKg5cGG9wIpx0-rDo0H1FXs0nMqwABu8ipylrl7xnrkcrj97UiiXq2Prar1zqw0-LtORnFjhKpGGQu7LkDIdtVTdv7XqlRjaDXNZn5Y-Cq1/s1600/Slide6.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="794" data-original-width="1280" height="396" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3bNJa9z4SjkLaL10dDSkMcbinjXQudF9x6fKg5cGG9wIpx0-rDo0H1FXs0nMqwABu8ipylrl7xnrkcrj97UiiXq2Prar1zqw0-LtORnFjhKpGGQu7LkDIdtVTdv7XqlRjaDXNZn5Y-Cq1/s640/Slide6.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">A possible quantum circuit for implementing the black box $U_g$ in practice.</td></tr>
</tbody></table>
<br />
<br />
Now we have all the ingredients to describe the steps in Grover's algorithm. Again, here we are solving specifically a search problem on a set of 8 items so we will use 4 qubits: 3 qubits are used for the input to $f$ and one qubit for the computed value of $f$.<br />
<br />
In the preparation phase, we start with 3 qubits in state $|0 \rangle$ and one qubit in the state $|-\rangle$ in the last wire, which you have seen will be used for implementing phase kickback. Then we apply the Hadamard gate to each wire with a qubit in $|0 \rangle$. Recall that the Hadamard gate is the quantum gate that maps $|0\rangle$ to $|+\rangle$ and $|1\rangle$ to $|-\rangle$. The goal here is to prepare an equal superposition state for all possible inputs to $f$.<br />
<br />
$ |0\rangle |0\rangle |0\rangle |-\rangle \mapsto |+\rangle |+\rangle |+\rangle |-\rangle $<br />
<br />
Next is the iteration phase. Here we will apply an operation a certain number of times before measuring the qubits. We shall call the repeating operation the Grover operator $G$ and it involves the following steps: (i) apply $U_f$, (ii) apply Hadamard gates to all wires except the bottom wire with the $|- \rangle$, (iii) apply $U_g$, and (iii) apply Hadamard gates again to all wires except the bottom wire. The quantum circuit for $G$ is shown below.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjhbRr5d_VCvdPKNkTms8oH3lfHEotqmwUHXWXsI00Ng2v_7Btp9qnHc7-cwy3MQelPNFc2uHUR2kB447J7WG4LYfK4kkOHNOEyUzbtk6cKWhc-83iB-I10yuMgynwlTL_JW_6irv-ST3F4/s1600/Slide7.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="794" data-original-width="1280" height="396" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjhbRr5d_VCvdPKNkTms8oH3lfHEotqmwUHXWXsI00Ng2v_7Btp9qnHc7-cwy3MQelPNFc2uHUR2kB447J7WG4LYfK4kkOHNOEyUzbtk6cKWhc-83iB-I10yuMgynwlTL_JW_6irv-ST3F4/s640/Slide7.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">The Grover operator $G$ that is repeated a number around $\sqrt{N}$ times during the iteration phase.</td></tr>
</tbody></table>
<br />
<br />
In the iteration phase, we apply the Grover operator $m = \pi\sqrt{N}/4 $ times, rounded down to the nearest number. In our example $N = 8$ so $m = 2.221$, which is rounded down to 2.<br />
<br />
We can also show how the quantum state in the wires change with each application of $G$. The calculation is straightforward but it can be rather tedious to illustrate step-by-step so we will only state the result after each application of $G$:<br />
<br />
$ G G |+\rangle |+\rangle |+\rangle |-\rangle $<br />
$= G (|0\rangle |0\rangle |0\rangle + |0\rangle |0\rangle |1\rangle +|0\rangle |1\rangle |0\rangle +|0\rangle |1\rangle |1\rangle<br />
+|1\rangle |0\rangle |0\rangle + |1\rangle |0\rangle |1\rangle + 5 |1\rangle |1\rangle |0\rangle +|1\rangle |1\rangle |1\rangle ) |-\rangle /(4\sqrt{2}) $<br />
$= (-|0\rangle |0\rangle |0\rangle - |0\rangle |0\rangle |1\rangle -|0\rangle |1\rangle |0\rangle -|0\rangle |1\rangle |1\rangle<br />
-|1\rangle |0\rangle |0\rangle - |1\rangle |0\rangle |1\rangle + 11 |1\rangle |1\rangle |0\rangle -|1\rangle |1\rangle |1\rangle ) |-\rangle /(8\sqrt{2}) $<br />
<br />
We end our quantum computation by measuring all the wires except the bottom one (with $|-\rangle$) in the computational basis to obtain some bit-string value, which we call $y$. From the calculation above, we see that we would get $y = 110$ with probability $(11/(8\sqrt{2}))^{2} = 0.945$ so with high probability, we will obtain the solution to our search problem. The quantum circuit for the Grover search in our example is illustrated in the figure below.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdWerrSQh7u4nnLgzCGQcPUzZHOz9_xpqUgzMPxhXNZSavbMIEM6v7iGMt3HvnNGKE8md771rT8k-i4qdbtWTLPdUdRH1IaXOUdXuuPl38zxjZ8ArX6-idMGu0gGEPiX5GIj0k1aJnRzGB/s1600/Slide8.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="794" data-original-width="1280" height="396" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdWerrSQh7u4nnLgzCGQcPUzZHOz9_xpqUgzMPxhXNZSavbMIEM6v7iGMt3HvnNGKE8md771rT8k-i4qdbtWTLPdUdRH1IaXOUdXuuPl38zxjZ8ArX6-idMGu0gGEPiX5GIj0k1aJnRzGB/s640/Slide8.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">A quantum circuit for Grover's algorithm for our example function $f$ with marked item $y = 110$. Since $N = 8$ in our problem, $\pi \sqrt{N}/4 \approx 2.21$ which is 2 when rounded down, so the Grover operator $G$ is applied twice.</td></tr>
</tbody></table>
<br />
To gain some intuition for why this quantum search algorithm works, it may be useful to understand what the operator $G$ does to the quantum state in the wires. Since an equal superposition state is prepared in the preparation phase, we can represent this quantum state in terms of a bar graph of the probability amplitude for each bit-string value. An equal superposition state means the bars in our graph will all be of the same height, illustrated as the blue bars in the diagram below.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi1juiHBi046onQYYzOHzxbnpv9HE2DqPGjY5H1D4TE8oXRABbyBJy6V9TvWXwTfNhqM0Y6YSRvVqxa72KYWvZ4fRz91YTr-dkGFonD4LksM09Feob8TPNjmA1bFSg9tWWvXpxMEKTy1anh/s1600/Slide9.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="794" data-original-width="1280" height="396" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi1juiHBi046onQYYzOHzxbnpv9HE2DqPGjY5H1D4TE8oXRABbyBJy6V9TvWXwTfNhqM0Y6YSRvVqxa72KYWvZ4fRz91YTr-dkGFonD4LksM09Feob8TPNjmA1bFSg9tWWvXpxMEKTy1anh/s640/Slide9.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">What the Grover operator does on the probability amplitudes: Initially we have the same amplitude for each basis state, shown as the blue bars of same height. Next the gate $U_f$ flips the phase of the solution state, which can be seen in the orange bars. Finally, the rest of the Grover operator performs an inversion about the mean. Here the mean is indicated by dashed line. You can see this inversion about this line by comparing the orange bars to the gray bars.</td></tr>
</tbody></table>
<br />
<br />
Now when we apply $U_f$, what happens is it flips the sign of the marked item. The rest of the operators in $G$ is actually a phase flip operator that flips the sign of the equal superposition state. Its net effect on the current quantum state is to flip the bar heights about the average bar height (which is the dashed line in the diagram). This is why the combination of $U_g$ sandwiched by Hadamard gates is sometimes called inversion about the mean.<br />
<br />
Thus, in the iteration phase, we repeatedly apply a phase flip on the solution state and a mean inversion until the amplitude of the solution is large enough. In example, we saw that the amplitude for $y = 110$ increased by roughly $1/\sqrt{N}$ when $G$ was used so this is why we expect $\sqrt{N}$ application of $G$ gives the solution with high probability.<br />
<br />
To summarize, Grover's algorithm for searching a set of $N$ items is run as follows:<br />
(i) Start with $N$ qubits in $|0\rangle$ on the top register and a qubit in state $|-\rangle$ on the bottom register.<br />
(ii) Apply Hadamard gates to the $N$ qubits on the top register.<br />
(ii) Apply the Grover operator $G$ a total $\pi \sqrt{N}/4$ times, rounded down if needed.<br />
(iv) Measure the $N$ qubits on the top register in the computational basis to obtain a bit string $y$.<br />
(v) Verify that $f(y) =1$ so $y$ is a solution to the search problem.<br />
<br />
We have seen how Grover's algorithm works and given some general idea of why it works but since the speed-up is not to large, is the result really that useful? While we often think of the algorithm in terms of searching a list, it is more accurate to think of it in terms of inverting a function. If we let $f(x) = z$ then the search problem is like trying to compute $x$ when you are given $z$. When thinking in these terms, then the possible applications become more apparent. Also, it turns out that you can show that for unstructured search, this is the best you can do, i.e., the number of queries used by Grover's algorithm is optimal.<br />
<br />
One particularly useful application is in the collision problem in cryptography. In the simplest setting, we are asked in a given function $f$ is one-to-one (each possible input has a unique output) or two-to-one. Thus, solving the problem involves finding a pair of distinct values $x$ and $x'$ such that $f(x) = f(x')$. Knowing how to solve this problem is useful in cryptography because for many schemes employ functions whose security relies on the condition that collisions are hard to find. Grover's algorithm can be used in this instance to show that a scheme can be broken by a quantum computer.<br />
<br />
And finally, even for unstructured search, a quadratic speedup may be significant in practice: it means a classical method that runs in 100 s might only take 10 s for a quantum computer.<br />
<br />
<br />
<b>References:</b><br />
<br />
Ryan O'Donnell, <a href="http://ryan%20o%27donnell/" target="_blank">Grover's Algorithm</a> (2015).<br />
<br />
Philip Kaye, Raymond Laflamme, and Michele Mosca, An Introduction to Quantum Computing (Oxford, 2007) chap. 8.<br />
<br />Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-78007434502870916092017-08-28T08:32:00.000-07:002018-02-16T09:59:40.913-08:00Quantum speed-up in Simon's problemIt has been said that quantum computers will allow us to perform certain computations faster than the regular PCs of today. While this sounds like a remarkable technological advance, it is necessary to clarify what is meant by this:<br />
<br />
1. A recipe for solving a computational problem is called an algorithm. In order to compare the speed of classical and quantum algorithms, we need a way of counting computational steps for classical and quantum algorithms that is fair to compare.<br />
<br />
2. How many steps an algorithm needs depends on whether it receives an easy or hard instance of the problem. Typically we are interested in algorithms that work the best on average. Another issue is that for the algorithms that we do know, we don't usually know if they are the best that we can do (you can sometimes prove a classical algorithm is optimal by using lower-bound arguments). So here we are really comparing quantum methods with best as-of-yet classical methods. Put differently, it maybe the case (though many believe it unlikely) that classical and quantum computers have the same computational power; we just have not found the best classical algorithms that match the performance of quantum ones.<br />
<br />
Here I will describe a problem for which quantum computers have a distinct advantage over classical ones. It is an example of what is called a black-box problem in computer science.<br />
<br />
<a name='more'></a>To explain what a black-box problem is like, consider a game where you have to guess what animal I am thinking about. To help you guess, you are allowed to ask me yes-no questions. In this analogy, the game is like a computational problem and the animal represents the answer to this problem. Your questions correspond to inputs of a function and I serve as a black box that computes this function for you. The number of times you ask me a question is the number of queries made to the black box, which counts as the number of steps in a computation.<br />
<br />
Black-box problems seem rather artificial since we are given access to functions that are computed "magically in one step" but they are convenient theoretical constructs that allow us to compare the performance of classical and quantum algorithms. Here we will consider a problem formulated by Daniel Simon in 1994, which resembles a quite natural real-world problem.<br />
<br />
In Simon's problem, you are given access to a black box which computes a function f whose input and output is n bits. There is a promise that whenever $f(x) = f(y)$ for some pair of inputs $x$ and $y$, then either $x = y$ or $x \oplus y$ = p (recall that the XOR of $x$ and $y$ or $x \oplus y$ means we take each corresponding bit and add them, but with $1+1=0$, e.g., $0101 \oplus 1100 = 1001$.) Here $p$ is called the period of $f$. The goal is to find $p$.<br />
<br />
Intuitively this is a very hard problem classically since you would need to find two different inputs $x$ and $y$ for which $f(x) = f(y)$, but there is nothing we know about $f$ that will help us find such a pair more easily than just randomly picking $x$ and $y$.<br />
<br />
From what is called the birthday attack, if a function has $m$ possible values that are equally likely, then we can expect to find two inputs $x$ and $y$ such that $f(x) =f(y)$ if we evaluate $f$ for random inputs about $1.25\sqrt{m}$ times. This means that any classical algorithm for solving Simon's problem for an $n$-bit function, which has $2^{n}$ possible output values,<br />
will require checking around $\sqrt{2^{n}}$ inputs.<br />
<br />
It is not too difficult to show how to solve the problem quantumly for an arbitrary function on n bits but to be more explicit in our calculations, we consider the example of a function $f$ on 3 bits where $f(x) = f(y)$ for $x \oplus y = 001 = p$. For completeness, let us suppose that $f(000) = 101, f(010) = 111, (f100) = 011$, and $f(110) = 010$. Actually it does not matter what the values for $f$ we get for these inputs; only the period $p$ matters for the algorithm.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgEhImAw4NRWrCRrN0Jb3r8grBN3w7G6U6E1Pwvk6a3ds1cqnlfrq-lUwzHUzKHZAbi_9jh2bc8-lyP8WvIOMEyTRjOESTDsIrdXnbjTKgtWZUqP0lkWxhiv_frx2groZF9cWnEA20_w-CX/s1600/simonalg.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="794" data-original-width="1280" height="396" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgEhImAw4NRWrCRrN0Jb3r8grBN3w7G6U6E1Pwvk6a3ds1cqnlfrq-lUwzHUzKHZAbi_9jh2bc8-lyP8WvIOMEyTRjOESTDsIrdXnbjTKgtWZUqP0lkWxhiv_frx2groZF9cWnEA20_w-CX/s640/simonalg.png" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><div style="text-align: start;">
An efficient quantum algorithm for solving Simon's problem. The top 3 qubits form the input register and the bottom 3 form the output register. The box labeled f is a quantum black box for the 3-qubit function. The box with a metered scale denotes a measurement of a qubit in the <span style="font-size: small;">$\{|0\rangle , |1\rangle \}$</span>-basis.</div>
</td></tr>
</tbody></table>
<br />
<br />
The figure above shows the circuit for Simon's algorithm on our 3-bit function $f$. Recall that the Hadmard gate is the qubit unitary gate which takes $|0 \rangle$ to $|+\rangle = ( |0\rangle + |1\rangle) / \sqrt{2}$ and $|1\rangle$ to $|-\rangle = ( |0\rangle - |1\rangle) / \sqrt{2}$. Notice also that a quantum black box works by having separate registers for the input (say its value is $x$) and output qubits (pre-assigned to some value $v$) and the black box acts on both registers simultaneously before returning the input $x$ and final output $v \oplus f(x)$, i.e., it takes<br />
$|x\rangle |v\rangle$ to $|x\rangle |v \oplus f(x)\rangle$<br />
<br />
To see how the quantum state evolves from left to right in the quantum circuit, we will write the qubits for the input and output registers separately. The initial state is<br />
<br />
$|000 \rangle |000 \rangle$.<br />
<br />
After applying Hadamard gates to the input qubits, we get<br />
<br />
$|+++\rangle |000\rangle $.<br />
<br />
After querying the quantum black box,<br />
<br />
$[ |00+\rangle |101 \rangle + |01+ \rangle |111\rangle + |10+ \rangle |011\rangle + |11+\rangle |010\rangle ]/2$,<br />
<br />
where the output values follow from how we defined $f$ in the above.<br />
<br />
After the second application of Hadamard gates on the input qubits, we have<br />
<br />
$[ |++0\rangle |101\rangle + |+-0 \rangle|111\rangle + |-+0\rangle |011\rangle + |--0\rangle |010\rangle ]/2$.<br />
<br />
After measuring the output qubits in the $\{|0\rangle , |1\rangle \}$-basis,<br />
<br />
$|++0 \rangle |101 \rangle$ or $|+-0 \rangle|111\rangle$ or $|-+0\rangle|011\rangle$ or $|--0\rangle|010\rangle$,<br />
<br />
each with probability 25%.<br />
<br />
After measuring the input qubits in the $\{|0\rangle , |1\rangle \}$-basis, we get<br />
<br />
$|000\rangle$ or $|010\rangle$ or |$100\rangle$ or $|110\rangle$<br />
<br />
from the input qubits, each with probability 25%.<br />
<br />
This means that if we run the algorithm a few times, it returns 3 bits such that the first 2 bits could be anything but the third bit is always 0. More precisely, if the algorithm returns the value z then we always get some z such that<br />
<br />
$z \cdot p = 0$,<br />
<br />
that is, the product of each bit of $z$ with each bit of the period $b$ is 0.<br />
<br />
This gives us a simple way of finding $f$: run the algorithm a few times until we get enough $z$ values to solve for all the bits of the period $p$.<br />
<br />
We showed the calculation for a particular function $f$ but the algorithm works in a similar fashion for any arbitrary function $f$ on bits with period $p$:<br />
<br />
(i) apply Hadamard gates to the input qubits<br />
<br />
(ii) query the quantum black box using all the qubits<br />
<br />
(iii) apply Hadamard gates to the output qubits<br />
<br />
(iv) measure the input qubits in the $\{|0\rangle , |1\rangle \}$-basis.<br />
<br />
The input register qubits yields a bit-string value whose bit-wise product with the period is 0 for all corresponding pairs of bits.<br />
<br />
We get random values for $z$ so there is no fixed number of times we need to repeat the circuit to get $p$. However, it can be shown that if $f$ is a function on $n$ bits, then the probability we can solve for $p$ exactly after $n$ repetitions of the circuit is bigger than 25%. This means we can almost always find the period just by always repeating the circuit $4n$ times for any function on $n$ bits.<br />
<br />
Therefore comparing the best classical algorithm, which would need $\sqrt{2^{n}}$ queries to a black box for the $n$-bit function $f$, and this quantum algorithm, which requires a small multiple of $n$, we see a big computational speed-up.<br />
<br />
For comparison, the classical and quantum algorithms require about the same number of steps for $n=10$ but for $n = 20$, we have $1.25\sqrt{2^{20}} = 1280$, so the classical algorithm needs more than 60 times the number of quantum queries.<br />
<br />
We have seen a quantum speed-up but it works for an artificial problem so how significant is it really in real-world terms? Well, it turns out that the problem of prime number factorization, which is believed to be a hard problem for classical computers to solve, can be solved if one can find the period of a function for integers. This is a slightly more complicated problem that one involving bit-strings but it can be solved by a quantum circuit with a structure similar to Simon's algorithm. We know this as Shor's algorithm.<br />
<br />
Many of today's practical crypto-systems are secure if we are correct in assuming that certain computational problems are hard even for the fastest PCs. But Shor's algorithm has shown us that some hard problems might not be so hard at all, given the right quantum technology. <br />
<br />
<br />
<b>References:</b><br />
<br />
Dave Bacon, <a href="https://courses.cs.washington.edu/courses/cse599d/06wi/lecturenotes8.pdf" target="_blank">Simon's Algorithm</a> (2006).<br />
<br />
John Watrous, <a href="https://cs.uwaterloo.ca/~watrous/LectureNotes/CPSC519.Winter2006/06.pdf" target="_blank">Simon's algorithm</a> (2006).<br />
<br />
<br />
<br />
<br />Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-26466382408531731002016-09-22T17:54:00.001-07:002018-02-16T09:59:28.427-08:00The BB84 protocol for quantum key distributionThe most successful application of quantum theory to cryptography is quantum key distribution (QKD). The goal of QKD is to generate an identical string of bits that is privately shared between two parties, which we shall call Alice and Bob.<br />
<br />
The particular QKD scheme that we will describe was proposed by Charles Bennett and Gilles Brassard in 1984, and is often referred to as BB84.<br />
<br />
The protocol has two main parts, a quantum and classical phase. In the quantum phase, Alice sends single photons to Bob over some public quantum channel. In the classical phase, Alice and Bob need to talk to each other over an authenticated classical channel, that is, it can be public but they need to verify that they are talking to the correct person.<br />
<br />
<a name='more'></a>Recall that light is an EM wave, that is a wave of paired electric fields and magnetic fields that are perpendicular to each other. If the electric field component of a beam of light vibrates along a single direction (like vertical in the figure below), we say that the beam of light is polarized in that direction.<br />
<br />
<div class="separator" style="clear: both; text-align: center;">
</div>
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVA7DUomeSb4UyKMw7II_PZaQkpIXlIyDUdogP7HxJ14rr7sxZWhadzVhNoHWd9sDqL4VNZToN_W1KyN3CHILZRWlqIzPGnyJJm3y9wWZtbzv0K96-N49o6S0Vn1iKVc-_NejL6OrUCeGL/s1600/qkd01.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="402" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVA7DUomeSb4UyKMw7II_PZaQkpIXlIyDUdogP7HxJ14rr7sxZWhadzVhNoHWd9sDqL4VNZToN_W1KyN3CHILZRWlqIzPGnyJJm3y9wWZtbzv0K96-N49o6S0Vn1iKVc-_NejL6OrUCeGL/s640/qkd01.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">An electromagnetic (EM) wave can be visualized as a pair of electric and magnetic fields that vibrate along directions that are perpendicular to the direction in which the wave is moving. Observe that the electric and magnetic field themselves are perpendicular to each other.</td></tr>
</tbody></table>
<br />
<br />
The polarization can be measured using polarizing filters, which is made of a special material that blocks one of two perpendicular directions.<br />
<br />
For BB84, it is enough to consider two kinds of polarizing filters. The first one we shall call the $+$ filter, since it involves the horizontal and vertical directions. The other one we shall call the $\times$ filter, since it involves the left-diagonal (\) and right-diagonal (/) directions.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgEYNr1K14f4aiHgNi3Cpii8LJsHJgeSX1-_HEwhxDye6RKDIKJSBDddvByo9gHq5OdxKH4bMWGeC9tm8ToD0cAaSh_z3PN4U5iz3A-vGxCoCdi5GOzd3WvzwIfLHqKowTe37izWav0Q9Ej/s1600/qkd02.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="308" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgEYNr1K14f4aiHgNi3Cpii8LJsHJgeSX1-_HEwhxDye6RKDIKJSBDddvByo9gHq5OdxKH4bMWGeC9tm8ToD0cAaSh_z3PN4U5iz3A-vGxCoCdi5GOzd3WvzwIfLHqKowTe37izWav0Q9Ej/s640/qkd02.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Alice and Bob agree on a bit table for encoding 0 and 1 in both the $+$ filter and $\times$ filter.</td></tr>
</tbody></table>
<br />
<br />
To perform BB84, Alice and Bob have to first agree on how bits will be encoded in the polarization directions for each filter. This means they should form a bit table like the one shown above.<br />
<br />
Once they have setup the table, the scheme begins with Alice producing a random string of bits and a random sequence of filters. She then prepares photons which are polarized in the directions according to the agreed-upon bit table, as illustrated below. She sends the photons one by one to Bob.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhv0Er_WSybZJVwJiIQtkae6kS7d2e9lVNP9A4ICkRGs4PfuhGJhQE1raZ7iEeVNn_EDjNGHksqzfqlWA9HSxCLWJzUZ6T5zRZjF-sh-y5Nzk764J6HKDcyhO2_YKSLpmlnhiXahUQ-1WK5/s1600/qkd03.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="380" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhv0Er_WSybZJVwJiIQtkae6kS7d2e9lVNP9A4ICkRGs4PfuhGJhQE1raZ7iEeVNn_EDjNGHksqzfqlWA9HSxCLWJzUZ6T5zRZjF-sh-y5Nzk764J6HKDcyhO2_YKSLpmlnhiXahUQ-1WK5/s640/qkd03.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: 12.8px;">Alice produces a random string of bits (first column) and a random sequence of filters (second column). She then uses the bit table to translate the bits into polarization directions (third column). She then sends the photons one by one to Bob.</span></td></tr>
</tbody></table>
<br />
<div class="separator" style="clear: both; text-align: center;">
</div>
<br />
<br />
On his end, Bob created his own random sequence of filters that he will use to measure the photons he gets from Alice (figure below). He translates the outcome of each measurement into bits using the bit table. Observe here that if Bob measures the polarized photon using a filter that does not match the filter chosen by Alice, then he gets a random outcome, that is, both 0 and 1 are equally likely for that filter. In quantum terms, this is because each polarization of the $+$ filter can be thought of as an equal superposition of polarization directions of the $\times$ filter, and vice-versa.<br />
<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjjWGV3eCiAOQEZ0MUcjMTMAnBsL6d4OcN0YD4RBGIrOg8Icz-qsMclsMz7vTJDfLBpddyTk_JOveXhKTDyBfIUtodEiW9cvjGxUkyY5SgOBio3MbOAhCHSGH5Fhryg3EoMFJ_bFW6ToUDK/s1600/qkd04.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="390" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjjWGV3eCiAOQEZ0MUcjMTMAnBsL6d4OcN0YD4RBGIrOg8Icz-qsMclsMz7vTJDfLBpddyTk_JOveXhKTDyBfIUtodEiW9cvjGxUkyY5SgOBio3MbOAhCHSGH5Fhryg3EoMFJ_bFW6ToUDK/s640/qkd04.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Bob creates a random sequence of filters for measuring each incoming photon. He gets a polarization direction as an outcome, which he can translate into a 0 or 1 according to the bit table. What follows is the classical phase of the scheme, where Alice and Bob compare filters to determine which bits can be used in the secret key.</td></tr>
</tbody></table>
<br />
<br />
At this point, the quantum phase of the scheme is done. For the classical phase, Alice and Bob can talk over the phone to perform the remaining step. Here Alice tells Bob what filters she used for sending each photon (but not the direction) and Bob tells Alice in which positions they used the same filter. The bits produced from the positions with matching filters will be identical and the resulting bit string is called the raw key.<br />
<br />
It is important to note that it is okay for Alice and Bob to reveal the filters because even if an eavesdropper named Eve listened in, knowing just the filters after the quantum phase gives no information about the bits.<br />
<br />
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<tr><td class="tr-caption" style="text-align: center;">Alice and Bob compare the filters they used over some authenticated classical channel. The positions where the filters match correspond to identical bits, so these together form their raw key.</td></tr>
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Of course, you might think that Eve would try something clever, like copy the photon polarization into some quantum memory so she can measure it later when the filters are revealed. However, we know from quantum theory that you can not copy arbitrary quantum states so the only way this copying tactic will work is if she already knew what polarization she was trying to copy.<br />
<br />
But what if Eve instead uses her own filter to measure photons and prepares new photons for Bob based on the outcome of her measurements? This sounds fine except that when Eve measures with the wrong filter and Bob measures with the right one,there is a 50-50 chance that Alice and Bob record different bits even though they used the same filter. Alice and Bob can actually use some of the raw key bits to detect such a discrepancy, especially if Eve tried to measure too many of the photons.<br />
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So far what we have described is how BB84 generates a shared key between Alice and Bob when there are no errors in the quantum channel. In this ideal setting, the raw key can be used as the shared secret key between Alice and Bob.<br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjb40hF1RmPeT8F4kPJtBL97Xyy9MzTRl4VGOFc2Avb23IUOgb41U9gwl4Gd-rQVwSbU9IIE7csBE095rqczGmH4Gabg9XMVOStjC-vkUIH8mXwCdcgRG5u7sx1mMrvc4tg5Wg3v8HonR_r/s1600/qkd06.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="444" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjb40hF1RmPeT8F4kPJtBL97Xyy9MzTRl4VGOFc2Avb23IUOgb41U9gwl4Gd-rQVwSbU9IIE7csBE095rqczGmH4Gabg9XMVOStjC-vkUIH8mXwCdcgRG5u7sx1mMrvc4tg5Wg3v8HonR_r/s640/qkd06.PNG" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: 12.8px;">A summary of the BB84 scheme. Alice sends a random sequence of photon polarizations to Bob. Bob measures them with a random sequence of filters. Afterwards, they talk to each other to compare the filters they used for each position. The ones where the filters match will correspond to identical bits, and form the shared key.</span></td></tr>
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<br />
In a more realistic setting, when Alice sends a polarized photon to Bob, the photon Bob gets might already have a different polarization direction. This would lead to a few cases where the bits don't match for Alice and Bob even though they use the same filter.<br />
<br />
In this case, Alice and Bob have to perform additional steps to correct for errors. This requires sacrificing some of the bits of the raw key for testing and correcting. Note that error correction corrects errors regardless of its source, so this also corrects for errors that are caused by Eve.<br />
<br />
This testing itself will reveal some information to Eve about the raw key, which might be useful to her. Therefore, Alice and Bob have to do privacy amplification next to reduce whatever useful information Eve might have gained from the error correction step. The bit string that they obtain after error correction and privacy amplification is the final secret key.<br />
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When we say that the key is secret, what we really mean is that if Eve that was trying to figure out the key, Alice and Bob can always detect Eve's presence by using some of the bits in the raw key for checking the error rate. When the error rate is above an acceptable level, Alice and Bob conclude that somebody has been eavesdropping and they discard the raw key.<br />
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<br />
<b>References:</b><br />
<br />
Charles Bennett and Gilles Brassard, <a href="http://researcher.watson.ibm.com/researcher/files/us-bennetc/BB84highest.pdf" target="_blank">Quantum cryptography: Public key distribution and coin tossing</a> in Proc. IEEE ICCSSP vol. 1 (1984) 175-179.<br />
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<br />Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-20106330382448359732016-07-15T07:29:00.003-07:002022-09-14T03:46:59.589-07:00The CHSH game and quantum entanglementA simple setting for demonstrating the usefulness of entanglement involves a two-player game known as the CHSH game. The game is a variant of an experimental setup (by Clauser, Horne, Shimony and Holt) that is often used to illustrate Bell's theorem.<br />
<br />
We shall call the two players 2 players Alice and Bob. We will also have Charlie as a referee that decides if Alice and Bob wins the game. They can decide on any strategy before the game commences but they cannot communicate with each other once the game starts.<br />
<br />
To begin, Charlie picks two uniformly random bits $x$ and $y$, and gives $x$ to Alice and $y$ to Bob. Alice answers the referee with bit $a$, while Bob replies with bit $b$. After getting $a$ and $b$, Charlie checks whether<br />
<br />
$a \oplus b = xy \mod{2}$,<br />
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that is, that the XOR of the output bits $a$ and $b$ is equal to the AND of input bits $x$ and $y$. If so, then Alice and Bob win the game.<br />
<br />
<a name='more'></a>So for Alice and Bob to win, their output bits need to satisfy the following conditions:<br />
<br />
$x$ $y$ Alice and Bob win if<br />
0 0 $a = b$<br />
0 1 $a = b$<br />
1 0 $a = b$<br />
1 1 $a = b \oplus 1$<br />
<br />
Now let us compare classical and quantum strategies for playing the game. A classical strategy is any method for Alice to output $a$ given $x$, and similarly for Bob to output $b$ given $y$ that uses only classical information and no communication between them (which is why Bob does not know $x$ and Alice does not know $y$). A quantum strategy allows Alice and Bob to use a shared quantum state and perform measurements on it to decide their output bits.<br />
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The interesting question to ask here is whether a quantum strategy allows Alice and Bob to win the CHSH game with a higher success rate than any classical strategy. For this, we need to find and compare the optimal winning strategies in each situation.<br />
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For classical strategies, Alice and Bob may try to output 0 or 1 with some probability, depending on the value of $x$ or $y$, respectively. However, it can be shown that these probabilistic strategies can always be achieved by some deterministic strategy that outputs a a specific value. That is, Alice can always do better by just mapping $x$ to a particular value for $a$ and same for Bob mapping $y$ to a particular value for $b$.<br />
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If we know that deterministic strategies are best, then it is straightforward to check that the best winning probability is $\frac{3}{4} = 75 \%$, which can be achieved by the trivial strategy of outputting $a = b = 0$, whatever $x$ and $y$ may be.<br />
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Is there a quantum strategy that does better? Indeed there is and we will describe it here. First, Alice and Bob need to share the entangled two-qubit state<br />
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$|S\rangle = \left( |0,0\rangle + |1,1\rangle \right) / \sqrt{2}$,<br />
<br />
where the first qubit belongs to Alice and the second qubit belongs to Bob.<br />
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To know what $a$ and $b$ values to output, Alice and Bob perform measurements on their share of $|S\rangle$. Here we mention the measurement bases they need to use, where the first state in each basis corresponds to outputting 0, and the second state corresponds to outputting 1.<br />
<br />
Alice $x$ measure in basis<br />
0 <i> $|0 \rangle, |1 \rangle$</i><br />
1 <i> $|+\rangle, |-\rangle$</i><br />
Bob $y$ measure in basis<br />
0 $|u_{0} \rangle , |u_{1} \rangle$<br />
1 $|v_{0} \rangle , |v_{1} \rangle$<br />
<br />
where we have<br />
<br />
$|+\rangle = \left( |0\rangle + |1\rangle \right) / \sqrt{2}, \quad |-\rangle = \left( |0\rangle - |1\rangle \right) / \sqrt{2}$,<br />
$|u_{0}\rangle = \cos{u} |0\rangle + \sin{u} |1\rangle, \quad |u_{1}\rangle = -\sin{u} |0\rangle + \cos{u} |1\rangle $,<br />
$|v_{0}\rangle = \cos{v} |0\rangle + \sin{v} |1\rangle, \quad |v_{1}\rangle = -\sin{v} |0\rangle + \cos{v} |1\rangle $<br />
<i><br /></i>
for $u = \pi/8$ and $v = -u$.<br />
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The bases are shown in the figure below, where $|0\rangle$ and $|1\rangle$ correspond to vectors on the horizontal and vertical axis of the plane, respectively.<br />
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<tr><td class="tr-caption" style="font-size: 12.8px;">Measurement bases for the optimal quantum strategy in the CHSH game. The label on the arrows indicates the value of <i>a</i> for Alice and the value of <i>b</i> for Bob.</td></tr>
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To compute the winning probability for this quantum strategy, we compute the probability for getting $a = b$ when $x$ and $y$ are not both 1, and the probability of getting $a \oplus b = 1$ when $x = y = 1$. For this, it is useful to know that<br />
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<i>$|S\rangle = \left( |+,+\rangle + |-,-\rangle \right) / \sqrt{2}$.</i><br />
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The calculation is tedious but straightforward so we will just show a few details:<br />
<br />
$xy$ Pr[win | $xy$ ]<br />
00 $ | \langle 0, u_0 | S \rangle |^2 + | \langle 1, u_1 | S \rangle |^2 = 2 | \cos u / \sqrt{2} |^2 = \cos^2{u}$<br />
01 $ | \langle 0, v_0 | S \rangle |^2 + | \langle 1, v_1 | S \rangle |^2 = 2 | \cos v / \sqrt{2} |^2 = \cos^2{v} = \cos^2{u}$<br />
10 $| \langle +, u_0 | S \rangle |^2 + | \langle -, u_1 | S \rangle |^2 = 2 | ( \cos u + \sin u) / 2 |^2 = (1 + \sin{2u})/2 = cos^2{u} $<br />
11 $ | \langle +, v_1 | S \rangle |^2 + | \langle -, v_0 | S \rangle |^2 = 2 | ( \cos v - \sin v) / 2 |^2 = (1 - \sin{2v})/2 = \cos^2{u} $<br />
<br />
where we used $\cos(-u) = \cos u, \sin(-u) = - \sin u$, and $\cos{2u} = \sin{2u}$ for $u = \pi/8$. Because each combination of $x$ and $y$ is equally likely, we obtain $\mathrm{Pr[win]} = \cos^2{\pi/8}$.<br />
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Since $\cos^2{\pi/8}= 0.8536 > 0.75$, this optimal quantum strategy is better than the optimal classical strategy by over 10%.<br />
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<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: 12.8px;">A summary of the CHSH game. Alice and Bob receive input bits $x$ and $y$ from Charlie. Their goal is to answer with output bits $a$ and $b$ such that $a \oplus b = x \cdot y$. Alice and Bob cannot communicate during the game but they can decide on a strategy beforehand. In the quantum version of the game, Alice and Bob are allowed to share a quantum state and make measurements on it. Shown above is the optimal quantum strategy, where $|\psi\rangle$ is a maximally entangled two-qubit state, and the circles below Alice and Bob depict what measurement each should perform for input bit 0 or 1.</span></td></tr>
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<b>References</b><br />
<br />
John Watrous, <a href="https://cs.uwaterloo.ca/~watrous/CPSC519/LectureNotes/20.pdf" target="_blank">Bell inequalities and nonlocality</a> (2016).<br />
<br />
Mark Wilde, <a href="http://www.markwilde.com/teaching/2015-fall-qit/lectures/lecture-06.pdf" target="_blank">CHSH game, Bell's theorem, and Tsirelson's bound</a> (2015).<br />
<br />Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com2tag:blogger.com,1999:blog-1593607942505794705.post-8591929151690832462015-03-22T00:27:00.000-07:002018-02-16T09:53:30.841-08:00The uncertainty principle and wave-particle duality are equivalent<span style="font-family: "georgia" , "times new roman" , serif;">In the 17th century, physicists debated over the true nature of light. When he observed that light is split into different colors by a prism of glass, Isaac Newton hypothesized that light is composed of particles he called corpuscles, which undergo refraction when they accelerate into a denser medium.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">At around the same time, </span><span style="font-family: "georgia" , "times new roman" , serif;">Christian Huygens proposed that light is made up of waves. In his treatise published in 1690, he described how light propagated by means of spherical waves, and explained how reflection and refraction occurs.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">In 1704, Newton published Opticks, expounding on his corpuscular theory of light. The debate raged over whether light was a particle or a wave for almost a century, and was not settled until Thomas Young's interference experiments with double slits, which could only be explained if light was a wave.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">The story did not end there. In 1901, Max Planck was able to explain the energy curve of blackbody radiation by supposing that light was emitted in small packets of energy. Planck thought of this light particles, or quanta, as a convenient mathematical device and did not believe them to be real. However, when the photoelectric effect was discovered in 1905, Albert Einstein showed that it could be explained in terms of wave packets of light we now call photons. In 1927, Louis de Broglie constructed a pilot wave theory that attempted to explain how particle and wave aspects of light can coexist.</span><br />
<a name='more'></a><span style="font-family: "georgia" , "times new roman" , serif;">We now know that this ability to exhibit wave and particle behavior is a property of any quantum system, so it is also true of things like electrons or, atoms. But there is a trade-off between observing particle aspects and wave aspects . For example, in the double slit experiment, trying to determine which slit a photon goes through results in interference fringes that are less visible. Mathematically, this is expressed as the Englert-Greenberger-Yasin duality relation</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">$D^2 + V^2 \le 1$,</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">which relates the visibility $V$ of interference fringes to the distinguishability $D$ of the photon's path in a double slit (or any other interference) experiment.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">In 1927, Niels Bohr formulated the principle of complementarity, which asserts that certain pairs of properties of quantum systems are complementary, in the sense, that they can not be both measured accurately in a single experiment. </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">Bohr conceived of the idea of complementarity from discussions with Werner Heisenberg regarding his then newly-discovered uncertainty principle. The uncertainty principle expressed a fundamental limit to how much certain pairs of properties such position and momentum, can be jointly measured. The formal inequality relating the standard deviation of position and momentum was first derived by Earle Hesse Kennard and Hermann Weyl. </span><span style="font-family: "georgia" , "times new roman" , serif;">The more general version commonly used today is due to Howard P. Roberston:</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">$\mathrm{stdev}(X) \ \mathrm{stdev}(Z) \ge \dfrac{| \langle [X,Z] \rangle |}{2}$,</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">where $X$ and $Z$ are complementary operators,<i> </i>$[X,Z] = XZ- ZX$ is called the commutator, $\mathrm{stdev}(A)$ denotes the standard deviation of $A$, and $\langle A \rangle$</span><span style="font-family: "georgia" , "times new roman" , serif;"> denotes the expectation value of </span><span style="font-family: "georgia" , "times new roman" , serif;">$A$</span><span style="font-family: "georgia" , "times new roman" , serif;">. When </span><span style="font-family: Georgia, 'Times New Roman', serif;">$</span><span style="font-family: "georgia" , "times new roman" , serif;">X</span><span style="font-family: Georgia, 'Times New Roman', serif;">$</span><span style="font-family: "georgia" , "times new roman" , serif;"> denotes position and $Z$</span><span style="font-family: "georgia" , "times new roman" , serif;"> denotes momentum, the right-hand-side is equal to $1/2$.</span><br />
<br />
<span style="font-family: "georgia" , "times new roman" , serif;">From Bohr's persepective, wave-particle duality is an expression of quantum complementarity, qualitatively no different from how the position and momentum of photons cannot be simultaneously determined in a double slit apparatus. Yet Berge Englert showed in 1996 that the duality relation can be derived without using any uncertainty relations, which seems to imply that they are logically independent concepts.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">It was only recently, in 2014, that </span><span style="font-family: "georgia" , "times new roman" , serif;">Patrick Coles, Jędrzej Kaniewski, and Stephanie Wehner showed</span><span style="font-family: "georgia" , "times new roman" , serif;"> how wave-particle duality and the uncertainty principle are formally equivalent if one uses modern entropic versions of uncertainty relations. This interesting result, which unifies two fundamental concepts in quantum mechanics is explored here.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<b style="font-family: Georgia, 'Times New Roman', serif;">Entropy as measure of uncertainty</b><br />
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<br />
<span style="font-family: "georgia" , "times new roman" , serif;">As we have seen above, Heisenberg uncertainty relations are typically expressed in terms of standard deviation. However, information theory has taught us that uncertainty is best measured in terms of entropy. Before we state the entropic version of the uncertainty principle, it maybe helpful to first go into a short digression into how entropy is defined.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br />
Consider the experiment of rolling a dice. It has 6 possible outcomes, which we denote by $x$</span><span style="font-family: "georgia" , "times new roman" , serif;">. If the probability of getting $x$</span><span style="font-family: "georgia" , "times new roman" , serif;"> is $p(x)$</span><span style="font-family: "georgia" , "times new roman" , serif;">, the amount of information in it can be measured by</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">$I(x) = \log[1/ p(x)]$</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">where here log is a base-2 logarithm, i.e., $\log(2^y) = y$. The formula $I(x)$ is called the surprisal of $x$, since observing an improbable outcome ($p(x)$ is small) is quite surprising (its corresponding $I(x)$ is big).</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">Entropy (denoted by $H$) is formally defined as the expectation value of the surprisal:</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">$H(X) = \sum_x p(x) \log [1 /p(x)]$,</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">where $X$ is the random variable that takes values $x$. It maybe helpful to think of $X$ as representing the outcome of some type of measurement.</span><br />
<br />
<span style="font-family: "georgia" , "times new roman" , serif;">For example, if $X$ describes the outcome of a dice roll, the entropy of $X$ is given by</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">$H(X) = p(1) \log (1 / p(1) ) + ... + p(6) \log (1 / p(6) ).$ </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">If we believe the dice to be fair, then all possible outcomes are equally probable, i.e., $p(1) = p(2) = ... = p(6) = 1/6$. The entropy in this case becomes</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">$H(X) = 6 (1/6) \log [1/(1/6)] = \log(6).$</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">If the probabilities are different, the entropy will become smaller than $\log(6)$. Suppose we have a loaded die that always turns up 1, then $p(1) = 1$, while the rest has probability 0. In that case, $H(X) = \log(1) = 0$, i.e., the entropy vanishes when the outcome of a roll is certain.<i> </i></span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">This is why we say entropy measures uncertainty: its value is the largest when we think that all possibilities are equally likely to happen, i.e, when we are completely unsure of the outcome, and zero when we know what the outcome will be.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;"><br /><br /><b>Entropic uncertainty relations</b></span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><b><br /></b></span>
<span style="font-family: "georgia" , "times new roman" , serif;">Entropic versions of the Heisenberg uncertainty relation were first considered by Isidore Hirschmann in the context of Fourier analysis in the 1950s., with the connection to quantum mechanics mostly established in the 1970s. The most well-known version today was proven by to Hans Maassen and Jos Uffink in 1988:</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">$H(X) + H(Z) \ge - 2 \ln c$</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">where $X$ and $Z$ refer to measurements with outcomes $x$ and $z$, respectively, and</span><br />
<br />
<span style="font-family: "georgia" , "times new roman" , serif;">$c = \max_{i,j} | \langle x | z \rangle|$</span><br />
<br />
<span style="font-family: "georgia" , "times new roman" , serif;">where $|x \rangle $ is the quantum state corresponding to outcome $x$ of measurement $X$, $|z \rangle $ is the quantum state corresponding to outcome $z $ of measurement $Z$, and $\langle x | z\rangle $ is the inner product between the two vectors. </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">For instance, if $X = Z$, then $c = 1, \ln c = 0$, and we get $H(X) \ge 0$. Of course, all this tells us is that it is possible to be certain of the outcome of measurement $X$.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">Conceptually what the Maassen-Uffink relation means is that the combined uncertainty of the outcomes of any pair of measurements is always larger than some quantity that depends only a particular pair of outcomes from each measurement that are hardest to distinguish from each other ($c$ is largest when $|x \rangle = |z \rangle$ ).</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">Actually, what Maassen and Uffink derived was a more general relation that involved a more general definition of entropy called Renyi entropy:</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">$H_a = (1/(1-a)) \log [ \sum_x p(x)^a ].$</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">The more general Maassen-Uffink relation reads</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">$H_a (X) + H_b (Z) \ge= -2 \ln c,$<i> </i> </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">where we must have $1/a + 1/b = 2$. The special case for the usual (Shannon) entropy is obtained when<i> </i>$a = b = 1$<i>.</i></span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"> </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;">The entropic uncertainty relation that is equivalent to the duality relation is </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">$H_\mathrm{min} (X) + H_\mathrm{max} (Z) \ge 1$,</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">which can be obtained from the general Maassen-Uffink relation when $a = \infty, b = 1/2$.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">$H_\infty$ is called the min-entropy or $H_\mathrm{min}$ because it is always the smallest one among all Renyi entropies. $H_{1/2}$ is called $H_\mathrm{max}$ to contrast it with $H_\mathrm{min}$, and also because $b=1/2$ is the largest Renyi entropy allowed by the relation.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">Although $H_\mathrm{max}$ and $H_\mathrm{min}$ sound rather technical, they have relatively simple operational meanings in cryptography. $H_\mathrm{min} (X)$ is related to the probability of guessing $X$ correctly using some optimal strategy, where the harder it is to guess the value of $X$, the larger its min-entropy gets. $H_\mathrm{max} (Z)$ describes the amount of secret information contained in $Z$. </span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">Thus, in cryptographic terms, the entropic uncertainty relation describes a certainty-confidentiality trade-off between $X$ and $Z$: if $X$ is easy to guess then $Z$ is more concealed but if $X$ is hard to guess then $Z$ is more exposed. Intuitively, this tells us we can only determine $X$ at the expense of $Z$, and vice-versa. </span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">Now we are ready to see that the same intuition holds if $X$ refers to some particle behavior of a quantum system and $Z$ refers to its wave behavior.</span><br />
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<b style="font-family: Georgia, 'Times New Roman', serif;">The complementary guessing game</b><br />
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<tr><td class="tr-caption" style="text-align: center;">Complementary guessing game. For each photon detected on the screen in the double-slit experiment, Alice's goal is either to determine which slit the photon passed through or which position was used for the source. Because the questions involve the complementary wave and particle behavior of light, trying to one behavior limits our ability to observe the other.</td></tr>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">Consider the double-slit experiment depicted above. We have a light source on the left that emits photons onto 2 narrow slits and interference fringes can be seen at a screen placed far beyond the slits. The idea here is to describe an individual photon as a qubit with 2 possible states depending on whether it is "made" to behave as a particle or as a wave.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">Let $P$ describe the path a photon traverses, where $|1 \rangle$ refers to the top slit and $|2 \rangle$ refers to the bottom slit. $P$ exhibits the particle behavior of light (since the two possible states tells us if light passes through the top or bottom slit).</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">To get two different wave behaviors, the source can be moved along the up-down direction, between two positions that we have labelled $|t\rangle $ and $|b \rangle$. If we write them in terms of the path states we would get</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">$|t) = [ |1 \rangle + e^{i g} |2\rangle ] / \sqrt{2}$, </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;">$|b) = [ |1 \rangle - e^{i g} |2 \rangle ] / \sqrt{2}$.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">where $g$ is some angle called the phase. Let $W$ take the values $|t \rangle$ and $|b \rangle$. $W$ exhibits the wave behavior of light (since its two possible states refers to light that passes partially through both slits).</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">(The basic idea is that particles have well-defined locations while waves have well-defined phases.)</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">Consider a guessing game where Alice chooses whether to run a particle experiment or wave experiment by deciding where the source is placed. Her task is to decide what is the state of each photon, that is, she needs to determine the value of $P$ if the source is kept at the middle, and the value of $W$ if the source is moved up or down. </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">The following entropic uncertainty relation restricts Alice's ability to guess </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">$H_\mathrm{min} (P) + \min\{ H_\mathrm{max} (W) \} \ge 1.$</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">Here, it is necessary to take the minimum of $H_\mathrm{max}$ because when Alice runs a wave experiment to guess $W$, there are many possible choices for $g$. However, she can always determine $g$ when measuring the visibility of the interference pattern. So mathematically, finding $g$ corresponds to finding that minimum value of $H_\mathrm{max}$.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">Now we can go back to the wave-particle duality relations. There are actually many versions of duality relation, which vary according to the specific details of the particle or wave experiments used in the guessing game. </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">In the simplest version, t</span><span style="font-family: "georgia" , "times new roman" , serif;">he distinguishability $D$ refers to Alice's ability to predict the path in a particle experiment, while the visibility $V$ is related to the probability the photon is detected at a particular spot on the screen.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">Coles, Kaniewski, and Wehner showed that in terms of the variables $P$ and $W$ in our guessing game above, we would have</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">$H_\mathrm{min} (P) = - \log [ (1 + D) / 2]$,</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;">$\min \{ H_\mathrm{max} (W) \} = \log [ 1 + \sqrt{1 - V^2 } ].$</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">(The above formulas are derived using many tools of quantum information theory, so unfortunately that goes well beyond what we can explain here concisely.)</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">From the formulas we observe that when $D$ is big, $H_\mathrm{min}$ is small, which is what we expect since high $D$ means the photon path is easy to guess. </span><span style="font-family: "georgia" , "times new roman" , serif;">Meanwhile, when $V$ is big, $H_\mathrm{max}$ is small, which again makes sense since when the interference fringes have low visibility, then the value of the phase becomes well-hidden.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">From the entropic uncertainty relation, we obtain</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">$- \log [ (1 + D)/2] + \log [1 + \sqrt{1 - V^2}] \ge 1.$</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">Using both sides as the exponent for 2 (since $2^{\log(x)} = x$) gives</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">$\left(\dfrac{2}{1+D}\right) [ 1 + \sqrt{1 - V^2}] \ge 2.$</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">Working out the algebra, we find that</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">$1 + \sqrt{1 - V^2} \ge 1 + D \quad \implies \quad 1 \ge D^2 + V^2$,</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;"></span>
<span style="font-family: "georgia" , "times new roman" , serif;">which is indeed (one of) the wave-particle duality relation(s). </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">This illustrates that duality relations are in fact, entropic uncertainty relations in disguise. </span><span style="font-family: "georgia" , "times new roman" , serif;">I suspect Niels Bohr would have been quite happy with such a result.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;"><b>Reference:</b></span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">P. Coles, J. Kaniewski, and S. Wehner, "</span><span style="font-family: "georgia" , "times new roman" , serif;">Equivalence of wave-particle duality to entropic uncertainty," </span><span style="font-family: "georgia" , "times new roman" , serif;"><i>Nature Communications</i> <b>5</b>, 5814 (2014).</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-6827843596482678732015-03-13T07:23:00.001-07:002018-02-16T09:58:37.766-08:00The monogamy of entanglement<span style="font-family: "georgia" , "times new roman" , serif;">Quantum information theory has taught us that entanglement is a useful resource for communication and information processing. As with any other resource, we would like to describe the properties of entanglement quantitatively, to help us determine the various ways in which it can be manipulated. One particular question that might come to mind is this: to what extent can entanglement be shared among different objects?</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">The answer to this question leads us to an important fundamental property of entanglement called monogamy: if two objects are maximally entangled to each other, then neither object is entangled to a third one. More generally, it says that the stronger the entanglement between two objects is, the weaker their entanglements are with other objects.</span><br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh5zroAK3PYVnkr2Z2QDAqtLL6M8NhPtzGQf75qNgRkMhDtmN3ZpXDTqAOAKawqHpuxvxRvrIRVntuzL8xcW7LSWkbjd8RtymTPUXI6y-31-KGjhTr0jxIa1Vv8cQ-7ZXuRe9b-4iq7Md3N/s1600/monogamyPic.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="360" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh5zroAK3PYVnkr2Z2QDAqtLL6M8NhPtzGQf75qNgRkMhDtmN3ZpXDTqAOAKawqHpuxvxRvrIRVntuzL8xcW7LSWkbjd8RtymTPUXI6y-31-KGjhTr0jxIa1Vv8cQ-7ZXuRe9b-4iq7Md3N/s1600/monogamyPic.png" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Monogamy of entanglement: Let the red, blue and yellow balls represent qubits and the lines connecting them represent the amount of entanglement between each pair. The left image shows some entanglement shared among the 3 qubits. However, if red and yellow are maximally entangled as shown on the right, then neither is entangled with blue. In general, the more entangled a pair of qubits are, the less entangled each one is with any other qubit.</td></tr>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">To really understand what monogamy means, we first need to describe how entanglement between pairs of objects are measured. That way, saying 2 objects are maximally entangled has a precise meaning. </span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">Measuring entanglement in general is a not so easy thing: it turns out you can get different numbers depending on context. However, to just understand the basic idea, it will be sufficient to consider 2 qubits.</span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">Suppose we have qubits A and B that are in a pure state which we can write as</span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;"><i>|A,B) = a |0,0) + b |0,1) + c |1,0) + d|1,1),</i></span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">where |0) and |1) represent a pair of distinguishable states for a qubit, and the notation |0,1) means that qubit A is in the state |0) and qubit B is in the state |1), and so on.</span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">In 1997, Hill and Wootters described a general formula for measuring the entanglement of 2 qubits called concurrence. For the state |A,B), the concurrence<i> C(A,B)</i> is given by</span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;"><i>C(A,B) = 2 sqrt{ det(M) }</i> </span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">where </span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">Here M is the matrix for the reduced state of A, that is, it describes the state of qubit A alone given that the state of the qubits A and B is |A,B). det(M) refers to the determinant of the matrix M. For our purposes, it is sufficient to just know the formula for the concurrence, in terms of the numbers a,b,c, and d.</span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">The only other thing we need to know about concurrence is the following. If <i>C(A,B) = 0</i>, this means qubits A and B have no entanglement between them. If <i>C(A,B) = 1</i>, qubits A and B are maximally entangled. </span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">For example, when we consider the state</span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>|Bell) = 1/sqrt(2) |0,0) + 1/sqrt(2) |1,1)</i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">then <i>det(M) = 1/4</i> and so <i>C(A,B) = 2 sqrt(1/4) = 1</i>. This means we have a maximally entangled state for 2 qubits. </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">Intuitively, this is nice because for |Bell), whenever we measure the two qubits in the same way, we find them in the same state, e.g., they are both in |0) or |1), and this is kind of what you expect for a state that supposedly has the maximum amount of entanglement.</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">Now that we know how to measure entanglement between pairs of qubits, we can ask what happens if we have three qubits A,B,C in a pure state and how the concurrence between different pairs of qubits are related. </span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">Again, the situation is quite complicated to analyze in general so we consider a simple example. Suppose we have the 3 qubits in the state</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>|A,B,C) = x |1,0,0) + y|0,1,0) + z |0,0,1).</i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">We can calculate the concurrences <i>C(A,B)</i> and <i>C(A,C)</i> and find that</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>C(A,B) = 2|xy|, </i></span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><i>C(A,C) = 2|yz|.</i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">If we compute the reduced state of A to get the matrix <i>M</i> in this case, we find that</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_bpvcxtUSJHwVkdbutfvhvHKqpcSf3x-rhHgcRQu7hU3UUkbuOuAN1taeIJAUYGTRKRMjY5byQr4PEYJkbUQPavjAjXEj3vfvpBwgv9kNuFDwpPvMbo4jXSaErwyW1-3g6MP31MGIQiVN/s1600/detM.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="76" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_bpvcxtUSJHwVkdbutfvhvHKqpcSf3x-rhHgcRQu7hU3UUkbuOuAN1taeIJAUYGTRKRMjY5byQr4PEYJkbUQPavjAjXEj3vfvpBwgv9kNuFDwpPvMbo4jXSaErwyW1-3g6MP31MGIQiVN/s1600/detM.png" width="640" /></a></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">Comparing with the concurrences, we find that</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>C(A,B)^2 + C(A,C)^2 = 4 det(M).</i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">This is a quantitative statement of the monogamy of entanglement. If we treat qubits B and C as one object, the right-hand side of the formula above is equal to the concurrence <i>C(A,BC)^2</i>. </span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">Roughly speaking, the formula says that the entanglement between qubits A and B and the entanglement between A and C is bounded by the entanglement between A and the joint system formed by qubits B and C. </span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">Of course, the above formula was obtained for a special example. Coffman, Kondu, and Wootters showed that if we consider a general state for 3 qubits, the formula becomes an inequality:</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>C(A,B)^2 + C(A,C)^2 <= 4 det(M).</i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><b>References:</b></span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">Scott Hill and William K. Wootters, "Entanglement of a Pair of Qubits," Phys. Rev. Lett. <b>78</b>, 5022 (1997).</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">Valerie Coffman, Joydip Kundu, and William K. Wootters, "Distributed Entanglement," Phys. Rev. A, <b>61</b>, 052306</span><span style="font-family: "georgia" , "times new roman" , serif;"> (2000).</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
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Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-23932036553869498862014-07-04T08:28:00.001-07:002018-02-16T09:58:09.543-08:00An arbitrary quantum cannot be cloned<div>
<span style="font-family: "georgia" , "times new roman" , serif;">One of the early important results in the study of quantum information is the no-cloning theorem, which tells us that there is no quantum operation that allows us to create multiple copies of an arbitrary quantum state.</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">This property is very different from what we expect from classical information, which you may reproduce as many times as you wish. For example, you can send a PDF file by email to many recipients while keeping a copy to yourself. The important point is that whatever the contents may be, you can make a duplicate of it.</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">Now consider a cloning machine<i> M</i> for qubits that can produce identical copies of the states <i>|u)</i> and<i> |d)</i>:</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>M |u) |0) = M |u) |u),</i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>M |d) |0) = M |d) |d) ,</i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">where <i>|0)</i> denotes any fixed initial state for <i>M</i>. This is necessary in pretty much the same way you would need a blank piece of paper before you can photocopy a printed document.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
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<div>
<span style="font-family: "georgia" , "times new roman" , serif;">Suppose we want to produce a copy of the state</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>|q) = a |u) + b|d).</i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">If we feed <i>|q)</i> as an input to machine <i>M</i>, we would get</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>M |q) |0) = M [ a |u) |0) + b |d) |0) ] </i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i> = a [M |u) |0)] + b [M|d) |0)]</i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i> = a |u) |u) + b |d) |d).</i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">If M could make a clone of |q), we expect the output <i>|q) |q)</i>,</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>|q) |q) = [a |u) + b|d)] [a |u) + b|d)] </i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i> = a^2 |u) |u) + ab |u) |d) + ba |d) |u) +b^2 |d) |d),</i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">which is not the same as </span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>M |q) |0) = a |u) |u) + b |d) |d).</i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">unless <i>a = 0</i> or <i>b = 0</i>, which just means the input state to <i>M</i> must be either <i>|u) </i>or<i> |d)</i>.</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">What this tells us is that a quantum cloning machine works only for states that are fully distinguishable by measurement, since we know that <i>|u) </i>and<i> |d)</i> are orthogonal quantum states, and we can discriminate one from the other by a measurement in the basis <i>{|u), |d)}</i>. </span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgGg1L4KzuxpHv3fVA8a9qcA7yBBIo_xJWPdiKKtx8C3Z03pHAcyyk9CD8FglvRfmRtfFPq2cvawAbzMQFjpnxrkjnt-fns8pAYm4pHt2HQuJ0wZch0_C-muYgbguKGUNADH2-C_cUf4owO/s1600/nocloningthm.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="360" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgGg1L4KzuxpHv3fVA8a9qcA7yBBIo_xJWPdiKKtx8C3Z03pHAcyyk9CD8FglvRfmRtfFPq2cvawAbzMQFjpnxrkjnt-fns8pAYm4pHt2HQuJ0wZch0_C-muYgbguKGUNADH2-C_cUf4owO/s1600/nocloningthm.png" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Attempt to construct a quantum cloning machine M for qubits. Assuming the machine produces exact copies of the states <i>|u)</i> and <i>|d)</i>, if we give M the arbitrary input state |<i>q)</i>, it outputs a state that is generally entangled and is not the same as having two copies of <i>|q)</i>, except when <i>|q) = |u)</i> or <i>|q) = |d)</i>.</td></tr>
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<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">You may wonder if we can even have a cloning machine like <i>M</i>, since we actually assume at the start that we have such a machine. There is no problem in doing this because if all our states are guaranteed to be either <i>|u) </i>and<i> |d)</i>, then we are effectively dealing with classical information. You can think of <i>|u)</i> as something like bit "0" and <i>|d)</i> as something like bit "1", and we know we can tell <i>|u)</i> and <i>|d)</i> apart by measurement, so it will be just like differentiating between the bits 0 and 1.</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">This means, for example, that it is possible to construct a cloning machine <i>M'</i> such that </span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>M' |+) |0) = M' |+) |+),</i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>M' |-) |0) = M' |-) |-) ,</i></span></div>
</div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">where</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>|+) = [ |0) + |1) ] /sqrt(2), </i></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>|-) = [ |0) - |1) ] /sqrt(2),</i></span></div>
</div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">since this is another basis of orthogonal states for qubits.</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">It is worth noting that the generalization actually holds true: quantum information that is encoded exclusively in orthogonal (hence, fully distinguishable) states behave like classical information.</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">What has been shown is that duplicating an arbitrary quantum state with a fixed machine is not possible. However, it is possible to construct a quantum cloning machine if we are willing to make some concessions. One way is to have a machine that sometimes produces exact copies but fails on other times. Another way is to have a machine that does not quite replicate the input but always produces a close approximation of the input state. </span></div>
<div>
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Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-11489827436347818412014-06-13T06:12:00.001-07:002018-02-16T09:57:37.343-08:00How to teleport a qubit<span style="font-family: "georgia" , "times new roman" , serif;">One of the fascinating things we can do with quantum entanglement is a scheme called quantum teleportation. In the original proposal by Charlie Bennett, Gilles Brassar, Claude Crepeau, Richard Jozsa, Asher Peres and Bill Wootters, it describes a way to transmit an arbitrary quantum state between two parties who may be far apart, using only a Bell state shared between the two parties, a few qubit operations that each party can perform independently, and two bits of information that can be communicated by one party to the other.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">Suppose Alice and Bob are in separate locations but they share a pair of electrons that are in the entangled state</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;"><i>|E) = (|u,u) + |d,d)) / sqrt(2)</i></span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">where as usual <i>|u)</i> denotes the state of an electron having its spin pointing in the up-direction, <i>|d)</i> denotes that with spin in the down-direction, and <i>|u,u)</i> refer to the state of the first and second electrons, respectively. Let's say that Bob has the first electron on his side and Alice has the second electron on her side. </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span><span style="font-family: "georgia" , "times new roman" , serif;">Alice also possesses a third electron in the state</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>|q) = a |u) + b |d)</i></span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">and she wants Bob to obtain this state. If Alice does not know what the value of <i>a</i> and <i>b</i> precisely, she can not clone the state and send a copy to Bob. However, since Alice and Bob have shared entanglement, it is possible to transfer the state of this electron into Bob's electron using teleportation, which is shown in the figure below. </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
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<tr><td class="tr-caption" style="text-align: center;">A sketch of how to teleport a qubit from Alice to Bob using an entangled state. Alice applies CNOT to the 2 electrons on her side and then proceeds to measure one of them in the <i> |+),|-) </i>basis while the other one in the<i> |u),|d) </i>basis. She tells Bob the 2 outcomes she gets. If Alice got a<i> |d)</i>, Bob applies an up-down flipper to his electron. If Alice got a<i> |-),</i> Bob applies a plus-minus flipper to his electron. After performing the required operations, the state of Bob's electron will be <i>|q)</i>.</td></tr>
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
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<span style="font-family: "georgia" , "times new roman" , serif;">We will now describe the method in detail. </span><span style="font-family: "georgia" , "times new roman" , serif;">First, we note that the combined state of the three electrons can be written as</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;"><i>|E, q) = a |u,u,u) + a |d,d,u) + b |u,u,d) + b |d,d,d).</i></span><br />
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;">Written this way, Bob holds the first qubit while Alice holds the second and third qubits. The first thing she does is perform a 2-qubit operation known as CNOT, or controlled NOT gate. The figure below illustrates how CNOT operates. </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh4oC3NOIYjexxeuM8zmaUh3RJXZBBrI2iC7kfO2BLjVbuDvJN90m42Pn1YlTXHi_O8shPct2g9LwoezSJk1GD_jdXFSecZS1JZffS8ih2fv-Gh640SidN-esK0yHrW2LFbLUNeQENI122Z/s1600/quantumTeleportation2.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="480" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh4oC3NOIYjexxeuM8zmaUh3RJXZBBrI2iC7kfO2BLjVbuDvJN90m42Pn1YlTXHi_O8shPct2g9LwoezSJk1GD_jdXFSecZS1JZffS8ih2fv-Gh640SidN-esK0yHrW2LFbLUNeQENI122Z/s1600/quantumTeleportation2.png" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">The controlled-NOT gate. For each box shown above, the top qubit controls whether the spin of the bottom qubit is flipped or not. More specifically, spin up is flipped to down and vice-versa only when the top qubit is spin down, as seen in the two boxes on the right.</td></tr>
</tbody></table>
<span style="font-family: "georgia" , "times new roman" , serif;"></span>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">For the three electrons we have, it will flip the third qubit from <i>|u) </i>to<i> |d)</i>, or vice-versa, if the second qubit is in the <i>|d)</i> state, and does nothing to the third qubit if the second qubit is in the <i>|u)</i> state. This means that after applying CNOT(2,3) to the 2nd and 3rd electrons, the combined state of the electrons becomes</span></div>
<div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span></div>
<div>
<i><span style="font-family: "georgia" , "times new roman" , serif;">CNOT(2,3) |E,q) = </span><span style="font-family: "georgia" , "times new roman" , serif;">a |u,u,u) + a |d,d,d) + b |u,u,d) + b |d,d,u).</span></i><br />
<br />
<span style="font-family: "georgia" , "times new roman" , serif;">Now Alice is ready to perform the following steps: </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">(a) She measures the spin of the 3rd electron in the up-down basis. Because she starts with</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<i><span style="font-family: "georgia" , "times new roman" , serif;">CNOT(2,3) |E,q) = </span><span style="font-family: "georgia" , "times new roman" , serif;">[a |u,u) + b |d,d)] |u) + [a |d,d) + b |u,u) ] |d),</span></i><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">we see that if she gets outcome <i>|u)</i>, the state of the first 2 qubits will be<i> </i></span><span style="font-family: "georgia" , "times new roman" , serif;"><i>a|u,u) + b|d,d)</i>.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;">On the other hand, if she gets outcome <i>|d)</i>, </span><span style="font-family: "georgia" , "times new roman" , serif;"> the state of the first 2 qubits will be </span><span style="font-family: "georgia" , "times new roman" , serif;"><i>a|d,d) + b|u,u)</i>.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">(b) She measures the spin of the 2nd electron in the plus-minus basis. In this case, it is useful to note that </span><br />
<br />
<span style="font-family: "georgia" , "times new roman" , serif;"><i>a |u,u) + b |d,d) = [ a |u,+) + b|d, +) + a |u,-) - b|d,-) ] / sqrt(2),</i></span><br />
<i style="font-family: Georgia, 'Times New Roman', serif;">a |d,d) + b |u,u) = [ a |d,+) + b|u, +) + a |d,-) - b|u,-) ] / sqrt(2),</i><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">where we used <i>|u) = [|+) + |-)]/sqrt(2) </i>and<i> </i></span><span style="font-family: "georgia" , "times new roman" , serif;"><i>|d) = [|+) - |-)]/sqrt(2)</i>. Thus, state the first qubit ends up is one of four possibilities: </span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<i><span style="font-family: "georgia" , "times new roman" , serif;">a|u) + b|d), </span><span style="font-family: "georgia" , "times new roman" , serif;">a|u) - b|d), </span><span style="font-family: "georgia" , "times new roman" , serif;">a|d) + b|u), </span><span style="font-family: "georgia" , "times new roman" , serif;">a|d) - b|u).</span></i><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">(c) The last thing Alce needs to do is tell Bob the 2 outcomes she obtained from the measurements. So we need her to have access to a phone or email or a fax machine. Bob will not need to say anything to Alice so her means of communication can be something that works just one way, from Alice to Bob.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">All this time,Bob just waits to hear from Alice about her measurement results. The results tell him what sort of corrections he needs to perform so that his electron ends up in the state |q). In particular, there are 2 kinds of corrections he might need, which uses 2 <a href="http://quantumgazette.blogspot.com/2014/03/spins-magnets-and-quantum-mechanics.html#more" target="_blank">different types of flippers</a>:</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">(i) If Alice reports an outcome |d), Bob applies an up-down flipper to his qubit.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">(ii) </span><span style="font-family: "georgia" , "times new roman" , serif;">If Alice reports an outcome |-), Bob applies a plus-minus flipper to his qubit, that is, a flipper that flips state |+) to |-), and vice-versa.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">If he gets both of those outcomes from Alice, he applies both flippers, the order of which is not important. If he gets neither of those outcomes, he does nothing.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">Finally, after doing corrections, Bob's qubit will be in the state</span><br />
<br />
<span style="font-family: "georgia" , "times new roman" , serif;"><i>a|u) + b|d) = |q)</i>,</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">which originally was the state of the third qubit.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">A summary of what happens when teleporting a qubit (after CNOT is used) is summarized below:</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhjSaFT_6YpKlZZtwW38v0jKQti0MDSBuDMlqqvdnQDUuaBt1WB2N4ID9NFZJSCtbSZK6ZFCHzDRAj_ERPa-R3QeNoM-vpTVj06E2ZOMviAf2HRk0GmT1A6slNN-Nb9qiCNRlPdp2lKdFly/s1600/quantumTeleportation3.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="243" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhjSaFT_6YpKlZZtwW38v0jKQti0MDSBuDMlqqvdnQDUuaBt1WB2N4ID9NFZJSCtbSZK6ZFCHzDRAj_ERPa-R3QeNoM-vpTVj06E2ZOMviAf2HRk0GmT1A6slNN-Nb9qiCNRlPdp2lKdFly/s1600/quantumTeleportation3.png" width="640" /></a></div>
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">There are a few important remarks that need to be said:</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">1. The state of the qubit is arbitrary since the operations Alice and Bob need to perform do not depend on the value of <i>a</i> and <i>b</i>.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">2. Teleportation does not make a second copy of the state. Because Alice has to measure the third electron in the up-down basis, at the end of the scheme, the original qubit would now be in either state <i>|u) </i>or<i> |d)</i>.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">3. Teleportation does not allow us to transmit information faster than the speed of light. Note that before Alice tells Bob about the measurement outcomes, there are 4 possible states for Bob's electron. As far as he is concerned, these four states are equally likely, without the extra information provided by Alice. But Alice has to send this information by some other method whose speed is most certainly limited by laws of relativity.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">4. Teleportation does not work in the way depicted in science fiction accounts like Star Trek, where actual material is transferred instantaneously from one location to another. Here, only the quantum state is transferred to Bob, the electron that originally carried that state stays on Alice side.</span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>
<span style="font-family: "georgia" , "times new roman" , serif;">5. Still, quantum teleportation seems to suggest that it may be possible in principle to transfer the quantum states of all the particles that make up an object and "reconstruct" it in another location. This has been achieved with states of a few photons, or more recently, <a href="http://www.nytimes.com/2014/05/30/science/scientists-report-finding-reliable-way-to-teleport-data.html" target="_blank">with electrons trapped in diamonds</a>. The problem with bigger objects is the operations you need to perform become increasingly complex as more and more particles are involved, not to mention you need in complete detail the quantum state of all particles if you want an accurate "reproduction".</span></div>
<div>
</div>
Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-36103799774638791632014-03-13T02:41:00.001-07:002014-03-13T02:45:40.130-07:00A quantum version of Zeno's paradox<!--[if gte mso 9]><xml>
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</xml><![endif]--><span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">The quantum Zeno effect describes the situation
where an unstable particle, say a radioactive atom, won’t decay if it is
observed continuously. More generally, it says that if you repeatedly interact
with a quantum system through measurement then you can effectively freeze its
quantum state.</span><br />
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<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">The phenomenon is named after a Greek philosopher of
ancient times, Zeno of Elea. Zeno is known best for a set of paradoxes (we know
of 9 of them) that he posed as arguments against Aristotle’s concept of motion. </span><span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">Here we are interested in the arrow or fletcher’s
paradox. </span><br />
<span style="font-size: small;"><br /></span>
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">If you observe an arrow flying through the air at some particular instant in time, then it would
have a definite position, meaning it isn’t moving at that specific moment. However,
you can think of the arrow’s motion as happening one moment at a time. This says that motion must be impossible since it is made up of this long
sequence of motionless moments. </span></div>
<div class="MsoNormal">
<span style="font-size: small;"><br /></span>
<span style="font-family: Georgia; font-size: small;">Of course, as far as we can tell, the world is not static and objects in it are not forever motionless. What’s lacking with Zeno’s assertion is the mathematical notion of continuity. Motion is possible because time doesn’t flow like a series of separate frames in a film but more like the seamless current of a steady stream.</span></div>
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<span style="font-size: small;"><br /></span></div>
<a name='more'></a><span style="font-size: small;"><br /></span>
<br />
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<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;"><span style="font-family: "Georgia","serif"; line-height: 115%;">(You might be more familiar with the paradox involving a
race between a tortoise and Achilles. The idea is if Achilles gives the
tortoise, he will never catch up because when you consider each time he halves
the distance between him and the tortoise, it’s always at a location behind
where the tortoise. Of course, the paradox is solved by the concept of a
convergent series.) </span></span><br />
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;"><span style="font-family: "Georgia","serif"; line-height: 115%;"> </span> </span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">Roughly speaking, the quantum Zeno effect says that the
state of a quantum system does not change if you keep checking whether the
state has changed or not. </span></div>
<div class="MsoNormal">
<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">They say it’s like when you’re heating some water and
waiting for it to boil. If you keep staring at the water, it never seems to
boil but if you look away long enough then it finally does boil. Of course, in
this case it’s merely the perception: when you want something to happen
immediately, it feels like it takes forever to happen. Your gaze does not
prevent the water from boiling, and you will see heated water boil if you wait long enough.</span></div>
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<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">However, if you repeatedly measure a quantum system often enough,
you can keep it from evolving into a different state.</span></div>
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<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">To describe the effect, we employ the qubit
represented by the spin of the valence electron of a silver atom in the
Stern-Gerlach apparatus described in <a href="http://quantumgazette.blogspot.ca/2014/03/spins-magnets-and-quantum-mechanics.html">the previous post</a>. There we learned that
if you applied a uniform magnetic field to a silver atom, it exerts a torque to
the spin of the atom’s valence electron, causing it to rotate about an axis
that’s parallel to the direction of the magnetic field. We used this
to construct a flipper, a device that flips the spin from pointing up to
pointing down as shown below.</span></div>
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<span style="font-size: small;"><br /></span></div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><span style="font-size: small;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg1_KZ44vTMA8OQFJfEzHJEXjiE4sYmf3lnA-FwLNXS3XfkDAstWs6nkPMprMK7pSijfbQDhTomLDCsP6HQ8Ox6rN0vwwvQOQyWpJeo_EtrFMH7_s5Xn9CWz-rx4ti-F1NPXUd-ddfTMoR7/s1600/Slide1.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg1_KZ44vTMA8OQFJfEzHJEXjiE4sYmf3lnA-FwLNXS3XfkDAstWs6nkPMprMK7pSijfbQDhTomLDCsP6HQ8Ox6rN0vwwvQOQyWpJeo_EtrFMH7_s5Xn9CWz-rx4ti-F1NPXUd-ddfTMoR7/s1600/Slide1.PNG" height="480" width="640" /></a></span></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: x-small;">An up-down flipper, one of the Stern-Gerlach devices discussed in the previous blog post. It flips the spin state of an atom from <i>|u)</i> to <i>|d)</i> and vice-versa.</span></td></tr>
</tbody></table>
<br />
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<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">We can change the duration of the magnetic field if we want
to control how much the spin rotates. For example, if we apply the uniform
field half as long as we would for a flipper, then the spin is rotated only
halfway towards the opposite direction. If our electron was initially at state <i>|u)</i> and the uniform field is pointed
away from your monitor, then it would end up at state <i>|+) = 1/sqrt(2) |u) + 1/sqrt(2) |d)</i>.</span></div>
<div class="MsoNormal">
<span style="font-size: small;"><br /></span></div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><span style="font-size: small;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgcg4CCN615lvwDstzjaoaaRU7AsIkCkUcJs6Vb-0vxqAEFRKXCn-h4OqAcnf6AmtrTku5crAKb6lv7S4ihty18F0eGUpjPVgCcAFvu7nZm5JMTJRHbsf-NhSWQz8maNZl8elyQxVsOw4Nq/s1600/Slide2.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgcg4CCN615lvwDstzjaoaaRU7AsIkCkUcJs6Vb-0vxqAEFRKXCn-h4OqAcnf6AmtrTku5crAKb6lv7S4ihty18F0eGUpjPVgCcAFvu7nZm5JMTJRHbsf-NhSWQz8maNZl8elyQxVsOw4Nq/s1600/Slide2.PNG" height="480" width="640" /></a></span></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: x-small;">Schematic diagram for the up x-selector, a device that rotates a spin pointing up an angle x about an axis parallel to the uniform magnetic field.</span></td></tr>
</tbody></table>
<div class="MsoNormal">
<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">In general, we want to imagine a rotator that rotates a spin
pointing up <i>x</i></span><span style="font-family: "Times New Roman","serif"; font-size: small; line-height: 115%;">°</span><span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;"> along an axis pointing to the right of
this page. We shall call this an up <i>x</i>-rotator
for short. The Stern-Gerlach apparatus for an up <i>x</i>-rotator is depicted above. For example, the device which
takes a spin from state <i>|u)</i> to state <i>|+)</i> is an up 90</span><span style="font-family: "Times New Roman","serif"; font-size: small; line-height: 115%;">°</span><span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">-rotator.</span></div>
<div class="MsoNormal">
<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">Suppose we have a beam of silver atoms
prepared in the state <i>|u).</i> What we want to do is flip the spin from up to down
using two 90</span><span style="font-family: "Times New Roman","serif"; font-size: small; line-height: 115%;">°</span><span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">-rotators (since a flip requires an angle
of 180</span><span style="font-family: "Times New Roman","serif"; font-size: small; line-height: 115%;">°</span><span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">) but
in between the rotators, let’s check if the spin is still pointing up. </span><span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">The setup for that experiment is shown in the figure below:
the beam is sent through a sequence of three devices, an up-selector sandwiched
by two 90</span><span style="font-family: "Times New Roman","serif"; font-size: small; line-height: 115%;">°</span><span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">-selectors.</span></div>
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<span style="font-size: small;"><br /></span>
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><span style="font-size: small;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgpzJ8yNRk6eqc6q29aqLKYD3mvpRiMvhDS6pXagQpXuIzIKva7KTcbM76LqV0pX5_PwG1lNrefGlPau1Ylir8fCKkYiqoGwNKprua457K8OIEweU1MhKv_XgUoe08RuiCuMCE2ezQyKiyy/s1600/Slide6.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgpzJ8yNRk6eqc6q29aqLKYD3mvpRiMvhDS6pXagQpXuIzIKva7KTcbM76LqV0pX5_PwG1lNrefGlPau1Ylir8fCKkYiqoGwNKprua457K8OIEweU1MhKv_XgUoe08RuiCuMCE2ezQyKiyy/s1600/Slide6.PNG" height="220" width="640" /></a></span></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: x-small;">In this Stern-Gerlach experiment, we have a beam of atoms with spin
pointing up, written in shorthand form as (1,0) on the left of the
figure. We have two 90<span style="font-family: "Times New Roman","serif"; line-height: 115%;">°</span>-rotators
that will attempt to flip the spin in two steps but in between we want
to "look" at the atoms and check if they are still pointing up. Because
about half the beam gets through the up-selector and the final state of
the atoms is |+), you could say that 1/4 of the original atoms
"survived" in the initial state.</span></td></tr>
</tbody></table>
</div>
<span style="font-size: small;"><br /></span>
<span style="font-size: small;"><br /></span>
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">The scenario is simple enough that we can just follow step
by step how the state of the atoms changes with each device: </span><br />
<span style="font-size: small;"><br /></span>
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">1. The first rotator
essentially turns the spin from state <i>|u)</i>
into state <i>|+)</i>. </span><br />
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">2. Next, the up-selector
checks every atom if it is <i>“|u) </i>or<i> |d)</i>” and transmits it only if the spin
is pointing up. </span><br />
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">3. About half the beam will pass the selector and another rotator
converts the state from <i>|u) </i>to<i> |+).</i> </span><br />
<span style="font-size: small;"><br /></span>
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">Of the remaining atoms, roughly
half will be measured in the <i>|u)</i>
state. All in all, ¼ of the atoms in the original beam gets through. That's what happens if we check once.</span><br />
<div class="MsoNormal">
<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">Now what we really want is to repeatedly check if the atoms are still in the <i>|u)</i> state. It means we want to have as
many selectors as possible. It also means that we can only apply a uniform magnetic
field for a brief period between adjacent selectors, so the angle <i>x</i> gets smaller as we add more selectors.</span></div>
<div class="MsoNormal">
<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">In this case, it is helpful to know how our rotator works
for any angle. Since we always start with a beam of atoms in the state <i>|u),</i> it
is simple to describe what happens to the state: it goes from <i>|u) </i>to cos<i>(x/2) |u) + </i>sin<i>(x/2)|d)</i>.
Using the shorthand notation for up-down qubits, we would write these states as <i>(1,0) </i>and<i> (</i>cos<i>(x/2), </i>sin<i>(x/2</i>)), respectively.</span></div>
<div class="MsoNormal">
<span style="font-size: small;"><br /></span></div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><span style="font-size: small;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjiSMuoNWkEg4Ahx28tuvAKUk8_tKVL6TQiHgj-0uEsuMFZJSH57AiStG4ZZndlvYByx8e3ZlpiULSt0Pppx9YCOKntpWnlNM3y-f1dHYSeDhLZfqnEoj5zJbYr_MdVtf4waNk3cUITVH7_/s1600/Slide7.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjiSMuoNWkEg4Ahx28tuvAKUk8_tKVL6TQiHgj-0uEsuMFZJSH57AiStG4ZZndlvYByx8e3ZlpiULSt0Pppx9YCOKntpWnlNM3y-f1dHYSeDhLZfqnEoj5zJbYr_MdVtf4waNk3cUITVH7_/s1600/Slide7.PNG" height="480" width="640" /></a></span></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: x-small;">A Stern-Gerlach experiment for demonstrating the quantum Zeno effect. We have <i>n</i> x-rotators that attempt to flip the spin of the initial beam of atoms in the state |u) (that is, <i>nx = 180<span style="font-family: "Times New Roman","serif"; line-height: 115%;">°</span></i>). In between every pair of adjacent x-rotators, an up-selector checks whether the atoms are still in their initial state. Because a fraction |cos(x/2)|^2 of the atoms get transmitted each time an up-selector is used and there are n selectors if we also measure the final beam, the probability any particular atom survives the experiment is [|cos(x/2))^2]^n, which is very nearly 100% when n is large enough. Roughly speaking, the act of repeatedly observing if the state of the atoms changed causes it not to change.</span></td></tr>
</tbody></table>
<div class="MsoNormal">
<br />
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">Let's see what happens in an experiment where we
check the state of the atoms <i>n</i> times.
Again, since we are trying to flip the spins while also repeatedly measuring if
they are still pointing up<i>,</i> we have
sequence of <i>n</i> rotators with <i>x = 180</i></span><span style="font-size: small;"><i><span style="font-family: "Times New Roman","serif"; line-height: 115%;">°</span></i><i><span style="font-family: "Georgia","serif"; line-height: 115%;">/n</span></i><span style="font-family: "Georgia","serif"; line-height: 115%;"> with an up-selector between
every pair of rotators, as depicted in the diagram above. </span></span></div>
<div class="MsoNormal">
<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">We are interested in large values of n so suppose we have <i>n = 100.</i> Then <i>x = 1.8</i></span><span style="font-family: "Times New Roman","serif"; font-size: small; line-height: 115%;"><i>°</i>
</span><span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">and so each rotator changes the state
from <i>|u) := (1,0)</i> to </span></div>
<div class="MsoNormal">
<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-size: small;"><i><span style="font-family: "Georgia","serif"; line-height: 115%;"> |x) := (</span></i><span style="font-family: "Georgia","serif"; line-height: 115%;">cos<i>(x/2), </i>sin<i>(x/2))</i> <i>=
(0.9999, 0.0157)</i>.</span></span></div>
<div class="MsoNormal">
<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">That is, once the beam hits the first selector, we
expect 99.97% of the atoms to be measured in the state |u) and transmitted. </span><span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">This process is repeated <i>n = 100</i> times, which tells us that the probability
any single atom makes it to the end is</span></div>
<div class="MsoNormal">
<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;"><i>prob("atom is intact"</i></span><span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;"><i>)</i> <i>= [|cos(x/2)|^2]^100 = 0.9756</i></span><span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">.</span></div>
<div class="MsoNormal">
<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">Therefore, we expect most of the atoms in the original beam to get
through.</span></div>
<div class="MsoNormal">
<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">Despite our attempts to flip the spin state from <i>|u) </i>to<i> |d)</i>, almost all of the atoms we get at the end of the experiment
will be found in state <i>|u) </i>just because we kept observing them. If you
take a larger value of <i>n</i> (which means <i>x</i> gets smaller at the same time), you get even
closer to 100%. Because we have effectively frozen the state of the atoms. in some sense</span><span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;"> we have achieved something similar to
Zeno’s arrow.</span></div>
<div class="MsoNormal">
<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">We generally expect the state to be altered by measurement, since a measurement "forces" an atom to choose one of the outcome states. What happens here, however, is that as we include more and more selectors, there is less time for each rotator to try and rotate the spins away from pointing up. By trying to do more (more rotators and selectors) we end up doing less (the state is more likely to remain the same).</span></div>
<div class="MsoNormal">
<span style="font-size: small;"><br /></span></div>
<div class="MsoNormal">
<span style="font-family: "Georgia","serif"; font-size: small; line-height: 115%;">The quantum Zeno effect might seem surprising at first, especially since it suggests that we can prevent a radioactive element such as uranium from decaying if we keep “looking” at it (with the right measurement) often enough. But all you need is a simple experiment with atoms and magnets to illustrate how it reasonably follows from the rules of quantum theory. By examining details more carefully, we find it’s not so surprising after all. </span></div>
Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-11110517665051832602014-03-08T03:30:00.001-08:002014-03-08T03:32:22.242-08:00Spins, magnets, and quantum mechanics<div class="separator" style="clear: both; text-align: center;">
</div>
<span style="font-family: Georgia, "Times New Roman", serif;">Quantum mechanics is often described as an area of physics that deal with energy and matter at the atomic scales, where different weird, unusual stuff happen. To some extent, it is true that quantum objects behave in ways that seem counter to our everyday, common-sense intuition. Unfortunately, focusing on these particular aspects of quantum theory might give the impression that it is mysterious, mystical, and difficult to understand. And that is simply not true. Things like superposition require a little getting used but, for the most part, quantum mechanics works in ways you expect and that naturally make sense. </span><br />
<span style="font-family: Georgia, "Times New Roman", serif;"><br /></span>
<span style="font-family: Georgia, "Times New Roman", serif;">I hope to demonstrate this by discussing an experiment with atoms and magnets that is explained in fairly simple terms using quantum mechanics. This is largely how I was introduced to the subject many years ago. But to start, we will need to go over some basic ideas regarding magnets and magnetic fields.</span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">Most of us are familiar with magnets from their ability to attract iron and similar metals but in physics, a magnet is just any material that produces a magnetic field. A magnet influences its surroundings through the magnetic field it creates and reacts to the magnetic fields it experiences from other magnetic objects. </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-family: Georgia, "Times New Roman", serif;">A nice thing
about magnets is we can understand how they work without having to be very precise. I'm sure many of you have played around with a bar magnet before, which is often found bent into a horseshoe shape, to create a region of particularly strong magnetic field in between the ends labeled north pole and south pole. You probably also know about and maybe experienced first-hand how opposite poles attract and similar ones repel, and how they attract
or repel more when two magnets are brought closer together. </span></span><br />
<span style="font-family: Georgia, "Times New Roman", serif;"> </span><br />
<span style="font-family: Georgia, "Times New Roman", serif;">Something less familiar is what determines the force at which a magnet attracts and repels objects. The strength of a magnet is measured by a property called magnetic moment, which is responsible for a magnet's tendency to align with magnetic fields.</span><br />
<br />
<a name='more'></a><br />
<span style="font-family: Georgia, "Times New Roman", serif;">According to Maxwell's equations describing electricity and magnetism, a moving electric charge creates a magnetic field, so in terms of physics, they behave like magnets in many situations. If we consider a loop carrying an electric current as shown in the figure below, we find that the magnetic moment depends on the
current flow and the area enclosed by the loop, specifically their
product. </span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjJ2XgiHUS556E7k1OxjK3Vw_dpH4Ua2MEfxLmqbuz2eyhG7Ug1fgzbOZ-wz7viZvxpw4NU69ZtGKt2SBTym6it6qDL6UlT6qjkx1DZvA7I0dodnHz4iPkmr3E5c_9KcftubbQ4zayDPdq6/s1600/Slide1.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjJ2XgiHUS556E7k1OxjK3Vw_dpH4Ua2MEfxLmqbuz2eyhG7Ug1fgzbOZ-wz7viZvxpw4NU69ZtGKt2SBTym6it6qDL6UlT6qjkx1DZvA7I0dodnHz4iPkmr3E5c_9KcftubbQ4zayDPdq6/s1600/Slide1.PNG" height="165" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">(a) For a loop forming a circle with area <i>A</i> and carrying an electric current <i>I</i> that flows counter-clockwise, the magnetic moment is given by <i>m = I x A</i> and directed upwards. (b) The magnetic moment of a bar of magnet is the combined effect of the magnetic properties of all its electrons, and points from the south pole towards the north pole of the magnet.</td></tr>
</tbody></table>
<br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">The magnetic moment is a vector quantity like velocity so it comes with a direction. Its direction is given by what's called the right-hand rule, which means you twist your fingers on your right hand in the same direction as the current flow and your thumb will point to the direction of the magnetic moment. </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">An electron possesses a magnetic moment like current-carrying loops. It will react to magnetic fields in the same way except that an electron's magnetic moment is a built-in property: because an electron is considered a pointlike object, there is no loop or current that gives rise to the magnetic moment. </span><br />
<span style="font-family: Georgia, "Times New Roman", serif;"><br /></span>
<span style="font-family: Georgia, "Times New Roman", serif;">Sometimes it is useful to imagine electrons as small round objects rotating like tops. It is exactly this picture of a spinning electron that gave rise to the term "electron spin," and an electron's magnetic moment is due to its spin. It should be emphasized that while this is a nice way to visualize an electron, it's not meant to be taken literally: if the electron did actually spin, the measured value of the magnetic moment suggests that it spins at a rate much faster than the speed of light, which is inconsistent with Einstein's relativity.</span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">The magnetism of an ordinary bar magnet is produced by the accumulated effect of the magnetic moment of its electrons. In this case, the magnetic moments are mostly aligned along the direction from the south pole to the north pole, which also sets the direction of the bar's magnetic moment.</span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgNkCSuQyLGZzgveT4aqqLBHKgDw94B0OvAuBU7-Eb7jRif7qKLr8IGDbPrI_CFwvudUxtelUiY8cQdCSRenSngJrUHitcgwW0ToTFt-wbq_7-vC5t0dluY7TOKmzHuDKdiniPobC1ZZQrS/s1600/Slide3.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgNkCSuQyLGZzgveT4aqqLBHKgDw94B0OvAuBU7-Eb7jRif7qKLr8IGDbPrI_CFwvudUxtelUiY8cQdCSRenSngJrUHitcgwW0ToTFt-wbq_7-vC5t0dluY7TOKmzHuDKdiniPobC1ZZQrS/s1600/Slide3.PNG" height="300" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Bar magnet in a uniform magnetic field. If its magnetic moment is at an angle compared to the direction of the field as shown, we see that the bar does not move up or down because it experiences equal forces on both poles. However, the bar will start to rotate around an axis that is parallel to the magnetic field due to the torque acting on the bar.</td></tr>
</tbody></table>
<br />
<br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">Consider a bar of magnet placed in a uniform magnetic field as shown above. If the source of uniform magnetic field (that is, the north and south poles of the magnet producing it) is located far away from the bar of magnet, then the same amount of force pushes or pulls at it along the direction of the field. The bar magnet remains in place. </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">But if the bar is placed at a certain angle to the direction of the field, it will spin around an axis oriented in the same direction as the field. The rotating motion is due to the torque acting on the bar magnet by virtue of its magnetic moment. That pretty much explains all that can happen in a uniform field.</span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">[For completeness, the torque is given by <i>T = m B sin x</i>, where <i>m</i> is the magnetic moment of the bar magnet, <i>B</i> is the strength of the magnetic field, and <i>x</i> is the angle between the directions of the magnetic moment and uniform field. The direction of the torque is perpendicular to both the magnetic moment and uniform field.] </span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhUZfUuvPH4FWDayryOIo6P_LegO-IJ382XUDTsISbc9pjrZ6su5wZC5o5lBty1Zpp-DiBj64krni6578wczhoWEnxHPzPd1rh5dxbNa2YJhaNEuk5bRDe2XqHBx4qHLleG6Li2iEu12HI3/s1600/Slide4.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhUZfUuvPH4FWDayryOIo6P_LegO-IJ382XUDTsISbc9pjrZ6su5wZC5o5lBty1Zpp-DiBj64krni6578wczhoWEnxHPzPd1rh5dxbNa2YJhaNEuk5bRDe2XqHBx4qHLleG6Li2iEu12HI3/s1600/Slide4.PNG" height="300" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Bar magnet in a non-uniform magnetic field. The left side shows the bar magnet placed at an angle to the magnetic field, which is generally pointing upwards. Because the south pole of the bar magnet meets the south pole side of the field, it is repelled. The same goes for the north pole in the bottom. However, the repulsion is stronger at the top because the field is stronger there. So the net effect is to slowly drag the bar magnet down, which is indicated by the green arrow. The right side shows the relative positions of different bar magnets after passing the non-uniform field. The amount and direction in which the bar is displaced depends on the orientation of their magnetic moment. </td></tr>
</tbody></table>
<br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">Now suppose we place the bar magnet in a non-uniform magnetic field, which can be produced by bending a huge bar magnet such that one of the pointy ends of the south pole is aligned to a flat edge of the north pole, like in the figure above. In this case, the magnetic field is stronger near the pointy end side than it is at the flat side. </span><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">If the bar's magnetic moment points in the same direction as the field, its north pole experiences a stronger attraction towards the upper part of the field (since the field pointing that way means it points towards a south pole) then it is pulled upwards as it moves across the field.</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">If
the bar's magnetic moment points opposite to the direction as the field, its south pole experiences a stronger repulsion away from the upper part
of the field then it is pulled downwards as it move across the field. This arrows on the screen in the figure indicate the relative position that the bar magnet lands on according to how it is oriented. If the electron were a classical object, the magnetism it acquires from its spin would cause it to behave in a similar fashion.</span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">We make our connection with quantum mechanics by using what we've learned about magnets to describe what's known as the Stern-Gerlach experiment. </span><br />
<br />
<div class="separator" style="clear: both; text-align: center;">
</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhi6FOGD3peC7s2xGUnxTSE9JYaasLJNx-HAc6csa84v-7BmVax00MpVEmMorAdYerxkPVntliUXtIZ4Y46XdD5TlMBCa93vbuM1McRkF3Aw0BHTAQ7W3H8NEeqLYPz5ReuHSYiLYpVOLfJ/s1600/Slide7.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhi6FOGD3peC7s2xGUnxTSE9JYaasLJNx-HAc6csa84v-7BmVax00MpVEmMorAdYerxkPVntliUXtIZ4Y46XdD5TlMBCa93vbuM1McRkF3Aw0BHTAQ7W3H8NEeqLYPz5ReuHSYiLYpVOLfJ/s1600/Slide7.png" height="480" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Schematic for the Stern-Gerlach experiment. A beam of silver atoms
produced from a furnace is sent through a non-uniform magnetic field and
collected on a detecting screen.If the magnetic properties of atoms
behave like bar magnets, we expect a continuous line of spots on the
screen since the magnetic moment of the silver atoms should not have any
preferred orientation. </td></tr>
</tbody></table>
<br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">In the original experiment done by Otto Stern and Walther Gerlach, pieces of silver were heated in a furnace until it boils. The resulting metallic vapor is sent through a collimating device, which produces a narrow beam of atoms. The beam of silver atoms then pass through a strong, non-uniform magnetic field. Finally, the atoms are collected on a photographic plate or some other type of detecting screen.</span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">If the magnetic properties of silver atoms are like ordinary bar magnets, we expect the atoms on the screen to spread out on a continuous line because if we pick a few atoms at random, there is no preferred direction for its magnetic moment.</span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgQMsrSMODR3K7-veJvuR8BwVay4Z90n8oqZb8S90ze8yHjVTom3mmtZcnjzIUV5YoQofOjCIKbh1zIuS3ZsD_VP9GEt1vbkrPQsMchema9wBxdtA9LmEkzk48bSElVcAf8VGRo3DuZSlb7/s1600/Slide5.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgQMsrSMODR3K7-veJvuR8BwVay4Z90n8oqZb8S90ze8yHjVTom3mmtZcnjzIUV5YoQofOjCIKbh1zIuS3ZsD_VP9GEt1vbkrPQsMchema9wBxdtA9LmEkzk48bSElVcAf8VGRo3DuZSlb7/s1600/Slide5.PNG" height="300" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">A silver atom with its valence electron with its magnetic moment pointing up, the atom get displaced to the top spot on the screen. If it were pointing down, it would be deflected to the bottom spot indicated.</td></tr>
</tbody></table>
<br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">When they did the experiment, Stern and Gerlach observed that the atoms mostly went into one of two spots on the screen. If we did think of the silver atoms as magnets, it seems that their magnetic moment, for whatever reason, points up or down only.</span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">The fact that there are two beam of atoms is split into two parts is already unexpected but what's really surprising is that you can change the orientation of the non-uniform magnetic field and in general you will get the same kind of splitting into two spots. Somehow the atoms have a magnetic property that has only two possible values whatever the alignment of the field is. </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">Of course, nowadays we understand why this happens: it's because almost all of the contribution to the magnetic moment of a silver atom comes from the single electron in its valence or outermost shell. In effect, a silver atom (or any atom with a single valence electron) responds to the magnetic field basically the same way an electron would. The Stern-Gerlach experiment actually proved that a quantum property called spin exists for electrons although this was not immediately realized at the time.</span><br />
<span style="font-family: Georgia, "Times New Roman", serif;"><br /></span>
<span style="font-family: Georgia, "Times New Roman", serif;">For our purposes, we will identify the magnetic moment of the electron in the Stern-Gerlach experiment with the spin of the electron. (Indeed, they are directly related although strictly speaking, because the charge of an electron is negative, the spin actually points opposite the direction of the magnetic moment.)</span><br />
<span style="font-family: Georgia, "Times New Roman", serif;"><br /></span>
<span style="font-family: Georgia, "Times New Roman", serif;">Through the Stern-Gerlach apparatus, we can prepare, control, and measure the spin of electrons in various ways and see a description in terms of qubits aids our understanding of what's happening.</span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">As we have doing in some of the last few entries, we consider an electron spin can point up or down, which we write as states <i>|u) </i>and <i>|d)</i> for the electron. In the Stern Gerlach experiment, the two spots we get on the screen correspond to measuring the spin of the valence electron in the silver atoms in</span><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-family: Georgia, "Times New Roman", serif;"> the states <i>|u)</i> and <i>|d)</i></span>.</span><br />
<span style="font-family: Georgia, "Times New Roman", serif;"><br /></span>
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEif2rp-m_uPp02tBcr2f5HcEZ_CwLll8fE6JK2_3KWOatjd0jQb7zkG_As6z_FkZc5MCFZZTLo328KMbutbGHSBcanl_5WilcjvvOJsq6QyImfqmlwd6wVwVUun7n5yelJyYu-ZoHSXRBSE/s1600/slide10.png" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEif2rp-m_uPp02tBcr2f5HcEZ_CwLll8fE6JK2_3KWOatjd0jQb7zkG_As6z_FkZc5MCFZZTLo328KMbutbGHSBcanl_5WilcjvvOJsq6QyImfqmlwd6wVwVUun7n5yelJyYu-ZoHSXRBSE/s1600/slide10.png" height="300" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">It
is possible to change the alignment of the non-uniform magnetic field.
Each orientation essentially corresponds to a measurement of the the
electron spin pointing up or down in that direction. The left side shows
one way to label the orientations using quantum states and the right
side shows how one might implement a "|+) or |-)" measurement using a
Stern-Gerlach apparatus.</td></tr>
</tbody></table>
<span style="font-family: Georgia, "Times New Roman", serif;"><br /></span>
<span style="font-family: Georgia, "Times New Roman", serif;">If atoms that are deflected up are labelled <i>|u) </i>and atoms that are deflected down are labeled <i>|d)</i>, we can also have superpositions of <i>|u)</i> and <i>|d)</i> if we rotate the orientation of the non-uniform magnetic field. For instance, in the figure above, we can put the field sideways so that it splits the beam of atoms into "<i>|+) </i>or<i> |-)</i>". </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">Any other orientation in the circle splits the beam of atoms into two groups whose electrons are in one of two possible spin states, <i><br /></i></span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;"><i>a |u) + b|d)</i> or<i> a|u) - b|d)</i>, </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">for some numbers <i>a</i> and <i>b</i>. It might be worth recalling here that<i> a = b = 1/sqrt(2)</i> for the<i>|+) and |-) </i>states. Also, the numbers <i>a</i> and <i>b</i> correspond to probabilities for measuring the electron with spin pointing up and down, where the specific probabilities are given by <i>|a|^2</i> and <i>|b|^2</i>, respectively.</span><br />
<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5Mlz7i7U5P3wGEYWhI67madT-ttaI3ZfPxbqFwDURnZJH7Zqv95XUmaCqmgGWwWpQ5rUu4KFaKW2F3MNgKjBk8avPMl_Wy3OqaJYq8qOuC719CoLh_GarXTtkxN8m3E1sgC5oPfoIDtTK/s1600/Slide8.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5Mlz7i7U5P3wGEYWhI67madT-ttaI3ZfPxbqFwDURnZJH7Zqv95XUmaCqmgGWwWpQ5rUu4KFaKW2F3MNgKjBk8avPMl_Wy3OqaJYq8qOuC719CoLh_GarXTtkxN8m3E1sgC5oPfoIDtTK/s1600/Slide8.png" height="300" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Schematic for |u) and |d) selectors. The top part shows how the selectors are implemented with the Stern-Gerlach apparatus and the bottom part shows a box notation for each selector. </td></tr>
</tbody></table>
<br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">There are two main types of devices we can utilize in the experiment. The first one is called a selector. A selector choose one of the two possible spin states for any particular orientation. The selectors for <i>|u) </i>and <i>|d)</i> are shown in the figure above. </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">As you can see, since the atoms coming out of a selector are guaranteed to be in the state |u) or |d) (or whatever other state is chosen), it allows us to prepare a beam of atoms in a known, definite state. </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">In the figure we've also introduced a box notation that describes what state a selector selects. The box notation will be useful when we describe situations where a number of devices are used in sequence.</span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6Y3ay2zIqFbz9xgUO7QbONzwQscgXcGF5CllGoz_2lbv2E_xhQfdL3e3LecMy-USLM9knUqSe6SKdtGanmJdiWoHij_egXjYhPm3Cy0IdzC1Mr5B1Sw5rdScXpWjVP2A7KSyTt-yq3T8z/s1600/slide11.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6Y3ay2zIqFbz9xgUO7QbONzwQscgXcGF5CllGoz_2lbv2E_xhQfdL3e3LecMy-USLM9knUqSe6SKdtGanmJdiWoHij_egXjYhPm3Cy0IdzC1Mr5B1Sw5rdScXpWjVP2A7KSyTt-yq3T8z/s1600/slide11.png" height="300" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">A flipper reverses the direction of the spin using a uniform magnetic field. As shown on the left box, if you have a spin pointing up and uniform magnetic
field pointing out towards you, then spin rotates in a
clockwise manner. Applying the field for a sufficient amount of time and the spin's direction is reversed. The box notation on the bottom right shows that our flipper here flips <i>|u) </i>and<i> |d)</i>. </td></tr>
</tbody></table>
<br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">Another
device is called a flipper and what it does is to flip the spin value,
so <i>|u)</i> is turned into <i>|d)</i> and vice-versa. With atoms, this can be
achieved by applying a uniform magnetic field of the right strength and
duration, as depicted in the figure above. </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">Using combinations of flippers and selectors, we can explore different ways to prepare, manipulate, and measure beams of atoms. One of the simplest situations to look at is using a flipper and a selector in sequence, which can be ordered in two ways. </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">If you have a
selector followed by a flipper, then the selector picks out the atoms
in the |u) state and the flipper flips those spins down into the |d)
state. So the atoms that come out of this particular arrangement will be
all have spins pointing down. </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">If
you have a flipper followed by a selector, then the flipper switches
the spins of the atoms first, so those pointing up will flip down and
those pointing down will flip up, and then the selector picks out all
the atoms that ended up in state |u). So the atoms resulting from this
arrangement will all have spins pointing up.</span><br />
<span style="font-family: Georgia, "Times New Roman", serif;"> </span><br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjN2sZOnzl-kNfN5wVqNm2Y59YO8MOmMQyK0TYHThvPQaA51gGQMoRumlqTFG4yFWw_FdNORl3uaa_GEpVHsz_Lzy0Cl_wNfIsYAFYdItVpNf6_0IiXzBkXLlVxFG-7YGn8K4EvxLPj9Nhy/s1600/slide13.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjN2sZOnzl-kNfN5wVqNm2Y59YO8MOmMQyK0TYHThvPQaA51gGQMoRumlqTFG4yFWw_FdNORl3uaa_GEpVHsz_Lzy0Cl_wNfIsYAFYdItVpNf6_0IiXzBkXLlVxFG-7YGn8K4EvxLPj9Nhy/s1600/slide13.png" height="300" width="400" /></a></div>
<br />
<br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">This leads us to our first lesson regarding operations we perform on quantum objects: the order in which they are applied is important. </span><br />
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEijENT92qx5_YIYYf3RFEDyxPwFXZhS0klicActiqdNzfmWFNflhtEXXn9FpVa2EQPf5lFDrQXehnL5ypqnk6QOIoTkiNyKgWRXuyFCFw9QU7RMWVJlSmk5SJ75XVSVUwb-YxZpncWVA_Al/s1600/slide14.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEijENT92qx5_YIYYf3RFEDyxPwFXZhS0klicActiqdNzfmWFNflhtEXXn9FpVa2EQPf5lFDrQXehnL5ypqnk6QOIoTkiNyKgWRXuyFCFw9QU7RMWVJlSmk5SJ75XVSVUwb-YxZpncWVA_Al/s1600/slide14.png" height="300" width="400" /></a></div>
<br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">Consider a slightly more advanced situation involving two flippers and one selector, shown in the figure above. What happens to a beam of atoms when we let them pass through an up-selector, sandwiched by two up-down flippers? Here is helpful to consider an incoming beam of atoms in the state <i>a|u) + b|d).</i> We choose this state only because we want to be able to say something about any arbitrary state. </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">Now, the beam first encounters the up-down flipper, which turns the state of the atoms into<i> b |u) + a|d)</i>, since it swaps the states <i>|u) </i>and<i> |d).</i> </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">Next, we have the up-selector, which picks up atoms pointing up. Since the state at this point is </span><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-family: Georgia, "Times New Roman", serif;"><i> b |u) + a|d), </i>the probability the spin is measured as pointing up is <i>|b|^2.</i> If we only care about the state of the atoms that go through and not how many, then all that matters is that the atoms that get transmitted will be in the <i>|u)</i> state.</span></span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-family: Georgia, "Times New Roman", serif;">Finally, the spins are flipped and since all the atoms at this point are in the state |u), they get flipped to state <i>|d)</i>. If we interpret the probability as the ratio of atoms that come out of the devices from those that went in, then we can also say that <i>|b|^2</i> of the original atoms in the incoming beam are left.</span></span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-family: Georgia, "Times New Roman", serif;">Compare this with what happens when we have a down-selector. If we again start with an incoming beam of atoms in the state </span></span><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-family: Georgia, "Times New Roman", serif;"><i>a|u) + b|d), </i>we expect a fraction <i>|b|^2</i> of them to be transmitted by the down-selector, and of course, the selected atoms will be in the <i>|d)</i> state.</span> </span></span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-family: Georgia, "Times New Roman", serif;">Since we made our observation for an arbitrary spin state, we learn the following: an up-selector sandwiched by two up-down flippers works in the exact same way as a down-selector. For now, the most general thing we can say about it is this: flippers and selectors can be combined in different ways to achieve the same result. </span> </span><br />
<br />
<br />
<div class="separator" style="clear: both; text-align: center;">
<span style="font-family: Georgia, "Times New Roman", serif;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg59lLj62p7g8qlZXvzZuspTAvkLwcmu6tOiDYdTHHR0GSM37_vnhJK_ypj3G8yNSomZ11xZsAadc0ju6EZdkXSpviFxCInKJtH2bzQPgs2hLANtSlPgRW__mMt-WPvxTDXbdImorIfH05G/s1600/slide15.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg59lLj62p7g8qlZXvzZuspTAvkLwcmu6tOiDYdTHHR0GSM37_vnhJK_ypj3G8yNSomZ11xZsAadc0ju6EZdkXSpviFxCInKJtH2bzQPgs2hLANtSlPgRW__mMt-WPvxTDXbdImorIfH05G/s1600/slide15.png" height="300" width="400" /></a></span></div>
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">The last example we wish to consider are sequences of selectors, like the two shown above. The top sequence is an up-selector followed by a down-selector. It's pretty easy to determine what happens here since the first selector produces a beam of |u) atoms and the second selector transmits only |d) atoms. Clearly, none of the atoms come out of this sequence.</span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">In the bottom sequence, we insert a plus-selector between the two selectors. Recall that a plus-selector transmits only atoms with spin in the <i>|+)</i> state. To follow what happens in this case, it is helpful to suppose an incoming beam of atoms in the state <i>a|u) + b|d)</i>. </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">First, the up-selector transmits the fraction |a|^2 of atoms that are measured in the <i>|u)</i> state. </span><span style="font-family: Georgia, "Times New Roman", serif;">Next, the plus-selector transmits atoms in state <i>|+)</i>. Because <i> </i></span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;"><i>|u) = 1/sqrt(2) |+) + 1/sqrt(2) |-),</i> </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">this means that <i>|1/sqrt(2)|^2 = 1/2 </i>of the atoms that pass through the up-selector get through the plus-selector as well. Finally, the down-selector transmits half of the beam transmitted by the first two selectors, since <i> </i></span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;"><i>|+) = </i></span><span style="font-family: Georgia, "Times New Roman", serif;"><i>1/sqrt(2) |u) + 1/sqrt(2) |d)</i>. </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">All in all, if we start with an beam of atoms in state </span><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-family: Georgia, "Times New Roman", serif;"><i>a|u) + b|d)</i>, we expect the fraction (|<i>a|^2)/4</i> of them to come out of the bottom sequence in the |d) state.</span></span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-family: Georgia, "Times New Roman", serif;">This is just an illustration of how a selector "disturbs" the quantum state of our beam of atoms. It's already apparent in how a single selector works, since you could say that any incoming beam forgets its previous state and is converted into the state being selected. The above example just shows that the amount of disturbance can be rather significant. </span></span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-family: Georgia, "Times New Roman", serif;">After playing around with beams of atoms, flippers, and selectors, what now? Perhaps it has not been obvious since the only quantum stuff we've really used are quantum states for describing the beam of atoms as they pass through devices but we've actually pretty much covered all the essential features of the quantum mechanics of qubits. All that I've really left out of the discussion is the mathematical ingredients that you would need if you wanted to be more rigorous and systematic. </span></span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh3ul_Mq8PP2YypoXwUXWTX7v2RymnwTzAeTS2ckhuE65NMk9IRM4GJ37dox-RpfUNuGE1c_qg3WWk1rWteXyxjAEL_39JvnvBNx7bx2uukLVtfbDtm5a6hAGNmI8io_2rbx9AdVDQFK-Zp/s1600/slide16.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh3ul_Mq8PP2YypoXwUXWTX7v2RymnwTzAeTS2ckhuE65NMk9IRM4GJ37dox-RpfUNuGE1c_qg3WWk1rWteXyxjAEL_39JvnvBNx7bx2uukLVtfbDtm5a6hAGNmI8io_2rbx9AdVDQFK-Zp/s1600/slide16.png" height="300" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Some of the mathematics used for describing qubit states and operations.</td></tr>
</tbody></table>
<br />
<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-family: Georgia, "Times New Roman", serif;">A few of these ingredients are shown above. Those familiar with the technical details know that states are represented by column vectors, the flippers are represented by matrices known as Pauli operators, and the selectors are represented by projection operators. </span></span><br />
<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-family: Georgia, "Times New Roman", serif;"><br /></span></span>
<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-family: Georgia, "Times New Roman", serif;">It's been a long journey but the main point we're trying to make in this post is this: sure, quantum mechanics involves plenty of stuff that are quite unexpected, bizarre even but despite these elements, it is sound, logical, and easy to understand once you become acquainted with its rules. It's quite cumbersome without the math, but if you just wanted a feel for what the essence of quantum physics is, you can go a long way with spins, magnets, flippers and selectors.</span></span><br />
<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-family: Georgia, "Times New Roman", serif;"> </span></span>Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-40006292282575403202014-03-03T06:25:00.000-08:002014-03-03T12:19:42.779-08:00Orthogonal states and quantum contextuality<span style="font-family: Georgia,"Times New Roman",serif;">In this post, we use the idea of orthogonal quantum states to describe a fascinating result in quantum mechanics known as the Kochen-Specker theorem. To begin, we review a bit of necessary mathematics.</span><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">Recall our usual example of a qubit represented by the spin of an electron. Typically, write the state <i>|E)</i> of such an electron as a superposition of the spin pointing up, which we write as the state <i>|u),</i> and spin pointing down, which we write as the state <i>|d)</i>: </span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;"><i>|E) = a |u) + b |d)</i>,</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">where a and b are numbers such that <i>|a|^2</i> is the probability of measuring the spin as up and <i>|b|^2</i> is the probability of measuring the spin as down.</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">We've seen before that the set of all possible states for a qubit corresponds to all possible directions the spin can point to, which can be described by using points on a sphere. However, we may choose to write our qubit states using any pair of polar opposite points, and normally we would choose the directions corresponding to north and south pole, which are labeled as the spin states <i>|u)</i> and |<i>d)</i>, respectively.</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"></span><br />
<a name='more'></a><span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">If we choose to always write states in terms of <i>|u)</i> and |<i>d),</i> then we may write the state <i>|E) </i>in the shorthand form</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<i><span style="font-family: Georgia,"Times New Roman",serif;">|E) := (a, b)</span></i><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">where the symbol := means that it is understood that the first number refers to spin up and the second number to spin down.</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">In general, you can have quantum systems that have properties that have more than two values that can be perfectly distinguished. The maximum number of states that can be fully distinguished from each other by a single measurement is what's called the dimension of a quantum system. </span><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">For example, a qutrit is a 3-dimensional quantum system, with distinguishable states commonly labeled as <i>|0), |1), </i>and<i> |2).</i> Like with qubits, there are many possible choices for triplets of states for expressing qutrits; for us what matters is that we may select any one of these possibilities.</span><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;"> If we know in advance the particular set of distinguishable states we want to use for writing down states, then we can use the shorthand form. For a qutrit, we would then have</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<i><span style="font-family: Georgia,"Times New Roman",serif;">a |0) + b|1) + c|2) := (a, b, c).</span></i><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">What's convenient with the shorthand form is that it allows us to describe an operation called an inner product in a simple way. Suppose we have two qutrits <i>Q</i> and <i>R</i>, with states</span><br />
<i><br /></i>
<i><span style="font-family: Georgia,"Times New Roman",serif;">|Q) := (a, b, c), </span></i><br />
<i><span style="font-family: Georgia,"Times New Roman",serif;">|R) := (x, y, z).</span></i><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">The inner product between states <i>|Q) </i>and<i> |R)</i>, is computed by multiplying corresponding numbers to each other and adding the products:</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /><i>(Q|R) = ax + by + cz.</i></span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">[If we have complex numbers, we have to take the complex conjugate of <i>|Q)</i> before multiplying to <i>|R)</i>. In this case <i>(R|Q)</i> is the complex conjugate of <i>(Q|R)</i>. If we only have real numbers, then <i>(Q|R)</i> and <i>(R|Q)</i> are equal.]</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">The significance of the inner product <i>(Q|R)</i> is that it can be used to find the probability of measuring the state <i>(Q|</i> when the state is initially <i>|R)</i>. More accurately, the probability is given by <i>| (Q|R) |^2</i>.</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">The inner product is actually something we have implicitly used with our discussion with qubits before. Recall the state <i>|+)</i>, which is an equal superposition state of <i>|u) </i>and<i> |d)</i>. In terms of <i>|u) </i>and<i> |d)</i>, we would write it in shorthand form as</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<i><span style="font-family: Georgia,"Times New Roman",serif;">|+) = (1/sqrt(2) , 1/sqrt(2) ).</span></i><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">Now the probability of measuring an electron with spin in state <i>|u) := (1, 0),</i> given that it was prepared in state <i>|+),</i> is given by the absolute-value square of the inner product</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<i><span style="font-family: Georgia,"Times New Roman",serif;">(u|+) = [1 x 1/sqrt(2)] + [0 x 1/sqrt(2)] = 1/sqrt(2),</span></i><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">so that <i>| (u|+)|^2 = 1/2</i>. </span><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">The main reason we introduced the inner product here is to be able to make the following statement: </span><span style="font-family: Georgia,"Times New Roman",serif;">If <i>|Q) </i>and <i>|R)</i> represent two quantum states that are perfectly distinguishable, then </span><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-family: Georgia,"Times New Roman",serif;"><i>|Q) </i>and <i>|R)</i></span> are orthogonal, that is, <i>(Q|R) = 0</i>.</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">Orthogonal quantum states are important because they correspond to properties that can be discriminated by measurement. If I present you with an electron whose spin I guarantee is either in state <i>|u) </i>or<i> |d)</i>, then you can perform a "</span><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-family: Georgia,"Times New Roman",serif;"><i>|u) </i>or<i> |d)</i></span>" measurement to determine in which state I prepared it. This fact is not terribly interesting with qubits, since there are only 2 possible outcomes in each measurement, but quantum systems of higher dimension, the orthogonality between states has a surprising consequence.</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">To get to that result, first we describe a particular game you can play with quantum states, which you might call the orthogonality game. To describe the rules of the game, let's first consider the following set of 9 qutrit states, in shorthand form, written in a 3 by 3 table:</span><br />
<br />
<table border="1" cellpadding="0" cellspacing="0" class="MsoNormalTable" style="border-collapse: collapse; border: none; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0in 5.4pt 0in 5.4pt; mso-yfti-tbllook: 1184;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"><td style="border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 54.9pt;" width="73"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(1, 0, 0)</span></i></div>
</td>
<td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 63.0pt;" width="84"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(0, 1, 1)</span></i></div>
</td>
<td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 67.5pt;" width="90"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(0, 1, -1)</span></i></div>
</td>
</tr>
<tr style="mso-yfti-irow: 1;">
<td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 54.9pt;" width="73"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(0, 1, 0)</span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 63.0pt;" width="84"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(1, 0, 1)</span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 67.5pt;" width="90"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(1, 0, -1)</span></i></div>
</td>
</tr>
<tr style="mso-yfti-irow: 2; mso-yfti-lastrow: yes;">
<td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 54.9pt;" width="73"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(0, 0, 1)</span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 63.0pt;" width="84"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(1, 1, 0)</span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 67.5pt;" width="90"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(1, -1, 0) </span></i></div>
</td>
</tr>
</tbody></table>
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">Because we only need to know whether pairs of qutrits are orthogonal or not, then we can just try to compute the inner product for every pair and check if it's zero or not. That's a bit of work but let me just tell you that the states in the first column and in every row are orthogonal to each other. This is easy to check by taking inner products of rows and columns.</span><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">Since orthogonal states can be discriminated from each other completely, we should be able to assign definite values to the properties they represent. Going back to qubits, if we had an electron spin pointing up, we could say something like <i>|u)</i> is ON and <i>|d)</i> is OFF since <i>|u)</i> and <i>|d)</i> are orthogonal.</span><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">We could do something similar with the states in the table above. If we use a color scheme of green and red for representing ON and OFF, respectively, then for sets of 3 orthogonal states, there should be one green color and two red colors. Again, this follows from the fact the states represent properties that can be distinguished by measurement: if a qutrit is measured in state <i>|0)</i>, then it is definitely not in state <i>|1) </i>or state <i>|2) </i>because these are states that are orthogonal to each other.</span><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">For our example, we should have one green color in the first column and one green color in each row. An example of a color scheme that obeys the orthogonality rules is</span><br />
<br />
<table border="1" cellpadding="0" cellspacing="0" class="MsoNormalTable" style="border-collapse: collapse; border: none; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0in 5.4pt 0in 5.4pt; mso-yfti-tbllook: 1184;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"><td style="background: #00B050; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 54.9pt;" width="73"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(1, 0, 0)</span></i></div>
</td>
<td style="background: red; border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 63.0pt;" width="84"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(0, 1, 1)</span></i></div>
</td>
<td style="background: red; border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 67.5pt;" width="90"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(0, 1, -1)</span></i></div>
</td>
</tr>
<tr style="mso-yfti-irow: 1;">
<td style="background: red; border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 54.9pt;" width="73"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(0, 1, 0)</span></i></div>
</td>
<td style="background: #00B050; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 63.0pt;" width="84"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(1, 0, 1)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 67.5pt;" width="90"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(1, 0, -1)</span></i></div>
</td>
</tr>
<tr style="mso-yfti-irow: 2; mso-yfti-lastrow: yes;">
<td style="background: red; border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 54.9pt;" width="73"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(0, 0, 1)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 63.0pt;" width="84"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(1, 1, 0)</span></i></div>
</td>
<td style="background: #00B050; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 67.5pt;" width="90"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(1, -1, 0)</span></i></div>
</td>
</tr>
</tbody></table>
<div class="MsoNormal">
</div>
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">As you can see, states like (1,0,0) and (1,0,1) are not orthogonal so they can both be green but states like (1,-1,0) and (0,0,1) are orthogonal so they have to be of different colors, according to our little game. </span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">The way we would like to interpret the colors is, for example, if you had a qutrit system with these "properties", then if you performed the measurement " (0,1,0) or (1,0,1) or (1,0,-1)" then you would certainly get the result for (1,0,1).</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">Another valid color scheme in our game would be</span><br />
<br />
<table border="1" cellpadding="0" cellspacing="0" class="MsoNormalTable" style="background: red; border-collapse: collapse; border: none; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0in 5.4pt 0in 5.4pt; mso-yfti-tbllook: 1184;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"><td style="border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 54.9pt;" width="73"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(1, 0, 0)</span></i></div>
</td>
<td style="background: #00B050; border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 63.0pt;" width="84"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(0, 1, 1)</span></i></div>
</td>
<td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 67.5pt;" width="90"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(0, 1, -1)</span></i></div>
</td>
</tr>
<tr style="mso-yfti-irow: 1;">
<td style="background: #00B050; border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 54.9pt;" width="73"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(0, 1, 0)</span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 63.0pt;" width="84"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(1, 0, 1)</span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 67.5pt;" width="90"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(1, 0, -1)</span></i></div>
</td>
</tr>
<tr style="mso-yfti-irow: 2; mso-yfti-lastrow: yes;">
<td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 54.9pt;" width="73"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(0, 0, 1)</span></i></div>
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<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 63.0pt;" width="84"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(1, 1, 0)</span></i></div>
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<td style="background: #00B050; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt; width: 67.5pt;" width="90"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-margin-top-alt: auto; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">(1, -1, 0)</span></i></div>
</td>
</tr>
</tbody></table>
<br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">Since there is at least one consistent way to assign green and red colors to the quantum states in our table, we say that these states are noncontextual. What we mean by that is for instance, you can fix the color of state (1,0,0) and the color will be the same whether you measure the first column or the first row (they form sets of orthogonal states and so either set can be fully distingushed by measurement). That is to say, the color assigned to each state is independent of the "measurement context".</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">Our little game leads us to ask the following question: is it always possible to find a consistent coloring for any group of quantum states where some states are orthogonal to each other? The answer is no and this is the main content of an important result in quantum theory known as the Kochen-Specker theorem.</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">Roughly speaking, the Kochen-Specker theorem tells us that if you tried to describe quantum systems such that properties have definite values that are always different for orthogonal states, and these values do not depend on the context in which they are measured, then there are certain sets of quantum states for which you will fail. </span><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">This should surprise you if only a little because what this means is that even though the possible values of a quantum property can be fully distinguished from each other (a set of orthogonal states represent the values of such a property), the exact value obtained for the property depends on how it is measured (because a quantum state can belong to different sets of orthogonal states, each set describing a different kind of measurement). This is clearly different from what we would expect from a particle described by classical physics, since we could say a ball, for example, has this position and that momentum, without having to describe how to measure these properties.</span><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">As it turns out, the Kochen-Specker theorem does not apply to qubits so it is always possible to play our orthogonality game with states of a single qubit. The inconsistency between coloring orthogonal states only arises when you have qutrits or systems of higher dimension.</span><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">In principle, the Kochen-Specker theorem is easy to prove, since all you have to do is give one example where you can't color all the states consistently, according to the rules of our game. In practice, it's not so easy to come up with a enough sets of orthogonal states where there is a contradiction (usually because the proof attempt immediately fails if you find even just one way of coloring the states consistently). </span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
<span style="font-family: Georgia,"Times New Roman",serif;">A simple example for proving the theorem involves the following 4-dimensional quantum states:</span><br />
<br />
<table border="0" cellpadding="0" cellspacing="0" class="MsoNormalTable" style="border-collapse: collapse; mso-padding-alt: 0in 5.4pt 0in 5.4pt; mso-yfti-tbllook: 1184;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"><td style="padding: 0in 5.4pt 0in 5.4pt; width: 149.4pt;" valign="top" width="199"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|A)
:= (0, 0, 0, 1)</span></i></div>
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<td style="padding: 0in 5.4pt 0in 5.4pt; width: 157.5pt;" valign="top" width="210"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|G)
:= (0, 1, 0, 0)</span></i></div>
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<td style="padding: 0in 5.4pt 0in 5.4pt; width: 135.0pt;" valign="top" width="180"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|M)
:= (1, 0, 0, 1)</span></i></div>
</td>
</tr>
<tr style="mso-yfti-irow: 1;">
<td style="padding: 0in 5.4pt 0in 5.4pt; width: 149.4pt;" valign="top" width="199"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|B)
:= (1, -1, 1, -1)</span></i></div>
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<td style="padding: 0in 5.4pt 0in 5.4pt; width: 157.5pt;" valign="top" width="210"><div class="MsoNormal" style="line-height: normal; mso-margin-bottom-alt: auto; mso-margin-top-alt: auto;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|H)
:= (1, 1, 1, 1)</span></i></div>
</td>
<td style="padding: 0in 5.4pt 0in 5.4pt; width: 135.0pt;" valign="top" width="180"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|N)
:= (1, 0, 0, -1)</span></i></div>
</td>
</tr>
<tr style="mso-yfti-irow: 2;">
<td style="padding: 0in 5.4pt 0in 5.4pt; width: 149.4pt;" valign="top" width="199"><div class="MsoNormal" style="line-height: normal; mso-margin-bottom-alt: auto; mso-margin-top-alt: auto;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|C)
:= (0, 0, 1, 0)</span></i></div>
</td>
<td style="padding: 0in 5.4pt 0in 5.4pt; width: 157.5pt;" valign="top" width="210"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|I)
:= (-1, 1, 1, 1)</span></i></div>
</td>
<td style="padding: 0in 5.4pt 0in 5.4pt; width: 135.0pt;" valign="top" width="180"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|O)
:= (1, -1, 0 ,0)</span></i></div>
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</tr>
<tr style="mso-yfti-irow: 3;">
<td style="padding: 0in 5.4pt 0in 5.4pt; width: 149.4pt;" valign="top" width="199"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|D)
:= (1, -1, -1, 1)</span></i></div>
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<td style="padding: 0in 5.4pt 0in 5.4pt; width: 157.5pt;" valign="top" width="210"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|J)
:= (1, 1, 0, 0)</span></i></div>
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<td style="padding: 0in 5.4pt 0in 5.4pt; width: 135.0pt;" valign="top" width="180"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|P)
:= (0, 0, 1, 1)</span></i></div>
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<tr style="mso-yfti-irow: 4;">
<td style="padding: 0in 5.4pt 0in 5.4pt; width: 149.4pt;" valign="top" width="199"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|E)
:= (1, 1, -1, 1)</span></i></div>
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<td style="padding: 0in 5.4pt 0in 5.4pt; width: 157.5pt;" valign="top" width="210"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|K)
:= (1, 0, 1, 0)</span></i></div>
</td>
<td style="padding: 0in 5.4pt 0in 5.4pt; width: 135.0pt;" valign="top" width="180"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|Q)
:= (0, 1, 0, -1)</span></i></div>
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</tr>
<tr style="mso-yfti-irow: 5; mso-yfti-lastrow: yes;">
<td style="padding: 0in 5.4pt 0in 5.4pt; width: 149.4pt;" valign="top" width="199"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|F)
:= (1, 1, 1, -1)</span></i></div>
</td>
<td style="padding: 0in 5.4pt 0in 5.4pt; width: 157.5pt;" valign="top" width="210"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|L)
:= (1, 0, -1, 0)</span></i></div>
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<td style="padding: 0in 5.4pt 0in 5.4pt; width: 135.0pt;" valign="top" width="180"><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";">|R)
:= (0, 1, -1, 0)</span></i></div>
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</tbody></table>
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">which can be grouped into columns of orthogonal states like this:</span><br />
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/* Style Definitions */
table.MsoNormalTable
{mso-style-name:"Table Normal";
mso-tstyle-rowband-size:0;
mso-tstyle-colband-size:0;
mso-style-noshow:yes;
mso-style-priority:99;
mso-style-qformat:yes;
mso-style-parent:"";
mso-padding-alt:0in 5.4pt 0in 5.4pt;
mso-para-margin-top:0in;
mso-para-margin-right:0in;
mso-para-margin-bottom:10.0pt;
mso-para-margin-left:0in;
line-height:115%;
mso-pagination:widow-orphan;
font-size:11.0pt;
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mso-ascii-font-family:Calibri;
mso-ascii-theme-font:minor-latin;
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mso-fareast-theme-font:minor-fareast;
mso-hansi-font-family:Calibri;
mso-hansi-theme-font:minor-latin;}
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{mso-style-name:"Table Grid";
mso-tstyle-rowband-size:0;
mso-tstyle-colband-size:0;
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border:solid windowtext 1.0pt;
mso-border-alt:solid windowtext .5pt;
mso-padding-alt:0in 5.4pt 0in 5.4pt;
mso-border-insideh:.5pt solid windowtext;
mso-border-insidev:.5pt solid windowtext;
mso-para-margin:0in;
mso-para-margin-bottom:.0001pt;
mso-pagination:widow-orphan;
font-size:11.0pt;
font-family:"Calibri","sans-serif";
mso-ascii-font-family:Calibri;
mso-ascii-theme-font:minor-latin;
mso-hansi-font-family:Calibri;
mso-hansi-theme-font:minor-latin;}
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<![endif]-->
<br />
<table border="1" cellpadding="0" cellspacing="0" class="MsoTableGrid" style="border-collapse: collapse; border: none; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0in 5.4pt 0in 5.4pt; mso-yfti-tbllook: 1184;">
<tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;">
<td style="border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|A)</span></span></i></div>
</td>
<td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|A)</span></span></i></div>
</td>
<td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|B)</span></span></i></div>
</td>
<td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|B)</span></span></i></div>
</td>
<td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|C)</span></span></i></div>
</td>
<td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|D)</span></span></i></div>
</td>
<td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|E)</span></span></i></div>
</td>
<td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|E)</span></span></i></div>
</td>
<td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|F)</span></span></i></div>
</td>
</tr>
<tr style="mso-yfti-irow: 1;">
<td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|C)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|G)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|D)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|H)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|G)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|H)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|F)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|I)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|I)</span></span></i></div>
</td>
</tr>
<tr style="mso-yfti-irow: 2;">
<td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|J)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|K)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|J)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|L)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|M)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|N)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|O)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|K)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|M)</span></span></i></div>
</td>
</tr>
<tr style="mso-yfti-irow: 3; mso-yfti-lastrow: yes;">
<td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|O)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|L)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|P)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|Q)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|N)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|R)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|P)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|Q)</span></span></i></div>
</td>
<td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-size: 12pt;">|R)</span></span></i></div>
</td>
</tr>
</tbody></table>
<br />
<span style="font-family: Georgia,"Times New Roman",serif;"></span>
<span style="font-family: Georgia,"Times New Roman",serif;">It would be a lot of work to check inner products for all columns but it is a worthwhile exercise to verify that any pair of states in some columns are indeed orthogonal. </span><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">According to the rules, we have to color the boxes such that there is exactly one green box in every column. For example, we can try to color <i>|A), |B), </i>and<i> |E) </i>green:</span><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;"><table border="1" cellpadding="0" cellspacing="0" class="MsoTableGrid" style="border-collapse: collapse; border: none; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0in 5.4pt 0in 5.4pt; mso-yfti-tbllook: 1184;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"><td style="background: #00B050; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|A)</span></i></div>
</td>
<td style="background: #00B050; border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|A)</span></i></div>
</td>
<td style="background: #00B050; border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|B)</span></i></div>
</td>
<td style="background: #00B050; border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|B)</span></i></div>
</td>
<td style="background: red; border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|C)</span></i></div>
</td>
<td style="background: red; border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|D)</span></i></div>
</td>
<td style="background: #00B050; border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|E)</span></i></div>
</td>
<td style="background: #00B050; border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|E)</span></i></div>
</td>
<td style="background: red; border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|F)</span></i></div>
</td>
</tr>
<tr style="mso-yfti-irow: 1;">
<td style="background: red; border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|C)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|G)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|D)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|H)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|G)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|H)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|F)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|I)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|I)</span></i></div>
</td>
</tr>
<tr style="mso-yfti-irow: 2;">
<td style="background: red; border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|J)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|K)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|J)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|L)</span></i></div>
</td>
<td style="background: #00B050; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|M)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|N)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|O)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|K)</span></i></div>
</td>
<td style="background: #00B050; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|M)</span></i></div>
</td>
</tr>
<tr style="mso-yfti-irow: 3; mso-yfti-lastrow: yes;">
<td style="background: red; border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|O)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|L)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|P)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|Q)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|N)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|R)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|P)</span></i></div>
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<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|Q)</span></i></div>
</td>
<td style="background: red; border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0in 5.4pt 0in 5.4pt;"><div align="center" class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; text-align: center;">
<i><span style="font-family: "Georgia","serif"; font-size: 12.0pt;">|R)</span></i></div>
</td>
</tr>
</tbody></table>
</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;">We see that we are left with the 6th column without any green because all four states in that column already show up in the other columns as red. </span><span style="font-family: Georgia,"Times New Roman",serif;">A few more attempts at coloring should convince you it can not be done.</span><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">In fact there is a rather straightforward way to see why the boxes can't be colored consistenly. There are 9 columns of orthogonal sets of states, so we expect 9 green boxes from the rules. However, each box contains one of 18 states, each of which appears twice. Therefore, any way we assign the green colors, we expect an even number of them, which contradicts the fact that we have an odd number of columns.</span>
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<span style="font-family: Georgia,"Times New Roman",serif;">Although our simple proof works with 4-dimensional quantum states, the theorem can be proven with qutrits. In fact, Simon Kochen and Ernst Specker originally proved the result for qutrits using 117 states, which are represented by this intricate but confusing diagram:</span><span style="font-family: Georgia,"Times New Roman",serif;"></span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhcf6sMSI-5PNPvAHQm6-19Z7Aserkqkh3pcsCoWp7Gf5bgQeWlClq9ThSv69GGuZ7j0yyJnyGd3TSI2YMy7TShrsPwKrMw1e1PJXVCmGFat2UZGf_zvMA3kIePov7cFEaOMna-iuIZ5tlU/s1600/fig2.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhcf6sMSI-5PNPvAHQm6-19Z7Aserkqkh3pcsCoWp7Gf5bgQeWlClq9ThSv69GGuZ7j0yyJnyGd3TSI2YMy7TShrsPwKrMw1e1PJXVCmGFat2UZGf_zvMA3kIePov7cFEaOMna-iuIZ5tlU/s1600/fig2.gif" height="375" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Kochen and Speckers 117 states for proving their theorem with qutrits</td></tr>
</tbody></table>
<div class="separator" style="clear: both; text-align: center;">
</div>
<span style="font-family: Georgia,"Times New Roman",serif;"></span>
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
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<span style="font-family: Georgia,"Times New Roman",serif;">We will not describe their proof here. Suffice to say, it involves the fact that if you look above, it is made up of a bunch of "hexagonal cones" like the one below</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
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<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjSiQJ3CNqb7D1lF2lWClXVEI-IK7aC5XZ0FUOMVtR3JRUhtvUAv4VNzrSd2hCj3kQI4d0NHUUMxljJ2Dno0YnB5aFEUhzfFmKEu9WhuGu6citsEkCqu4O9X1OxKzD5Wgo-WSLpq7Jo-Clc/s1600/fig1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjSiQJ3CNqb7D1lF2lWClXVEI-IK7aC5XZ0FUOMVtR3JRUhtvUAv4VNzrSd2hCj3kQI4d0NHUUMxljJ2Dno0YnB5aFEUhzfFmKEu9WhuGu6citsEkCqu4O9X1OxKzD5Wgo-WSLpq7Jo-Clc/s1600/fig1.gif" height="228" width="320" /></a></div>
<span style="font-family: Georgia,"Times New Roman",serif;"></span>
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
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<span style="font-family: Georgia,"Times New Roman",serif;">Kochen and Specker showed that this "hexagonal cone" of 10 points can not be colored in a consistent way if the angle between two differently colored points (black and white in this figure) is too small. And apparently, the angle is indeed too small in the 117-state diagram.</span><br />
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<span style="font-family: Georgia,"Times New Roman",serif;"></span>
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">Asher Peres has a slightly more accessible proof with qutrits using 33 states, which can be represented by points on a cube shown below. The complete
proof involves considering 16 triples and 24 pairs of states that are
orthogonal to each other from these 33 states. It is not so simple to describe how to arrive at a contradiction though (relative to the proof for the 18 states above) so the reader is free to consult the original reference listed below at his or her own peril.</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"></span>
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzBFRFDVy9MHFKs5zwwSR0_wMjyeq80gxl300YOgQIvAWEPNKM2Ub-cCAr068eJ43e1Ck4RWR2IUs-ytWsMolkrbgpaxHDQ3Zkdf1pWxTn6Xif2w8MhJ9df_htBckXIuIfhQeArHuhfUGG/s1600/peres33Rays.JPG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzBFRFDVy9MHFKs5zwwSR0_wMjyeq80gxl300YOgQIvAWEPNKM2Ub-cCAr068eJ43e1Ck4RWR2IUs-ytWsMolkrbgpaxHDQ3Zkdf1pWxTn6Xif2w8MhJ9df_htBckXIuIfhQeArHuhfUGG/s1600/peres33Rays.JPG" height="285" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">33 points on a cube used in Peres' proof of the Kochen-Specker theorem. If the center of the cube is (0,0,0), then the coordinates of the points represent the shorthand form for states of a qutrit. </td></tr>
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<span style="font-family: Georgia,"Times New Roman",serif;"></span>
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
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<span style="font-family: Georgia,"Times New Roman",serif;">We can summarize what we've learned here in a way similar to the lesson of Bell's theorem: There is no hidden mechanism that assigns definite values to quantum properties that are independent of the measurement used to determine them. </span><br />
<br />
<span style="font-family: Georgia,"Times New Roman",serif;">Asher Peres was fond of saying "unperformed experiments have no results." Echoing those words, we could say "unmeasured (quantum) properties have no (definite) values."</span><br />
<span style="font-family: Georgia,"Times New Roman",serif;"><br /></span>
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<span style="font-family: Georgia,"Times New Roman",serif;"></span>
<b><span style="font-family: Georgia,"Times New Roman",serif;">References:</span></b><br />
<br />
<span style="font-size: small;"><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-weight: normal;"><span style="color: black;">A. Cabello,
J. M. Estebaranz, and
G. G. Alcaine, </span>"<a href="http://arxiv.org/abs/quant-ph/9706009v1">Bell-Kochen-Specker theorem: A proof with 18 vectors</a>"<i> </i></span><span style="font-weight: normal;"><i>Physics Letters A</i> <b>212</b> (1996) 183.</span></span></span><br />
<span style="font-size: small;"><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-weight: normal;"><br /></span></span></span>
<span style="font-size: small;"><span style="font-family: Georgia,"Times New Roman",serif;">A. Peres, "<a href="http://citeseerx.ist.psu.edu/viewdoc/download?rep=rep1&type=pdf&doi=10.1.1.205.5481">Two simple proofs of the Kochen-Specker theorem</a>" <i>Journal of Physics A: Math. and Gen</i>. <b>24</b> (1991) L175.</span></span><br />
<br />
<span style="font-size: small;"><span style="font-family: Georgia,"Times New Roman",serif;"><span style="font-weight: normal;"><a href="http://plato.stanford.edu/entries/kochen-specker/">The Kochen-Specker Theorem</a>. Stanford Encyclopedia of Philosophy. </span></span></span>Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-5093251038012745082014-02-27T03:00:00.000-08:002014-03-03T12:25:12.429-08:00That remarkable theorem by Bell<span style="font-family: Georgia, Times New Roman, serif;">Before quantum physics came along, assigning a state to a physical system meant that you can uniquely determine the value of any measurable property using that state. In quantum mechanics, however, we've learned this is not the case: even if you know the quantum state exactly, for most measurements the best you can hope for is to estimate the probabilities for the various possible outcomes.</span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">You might wonder, is the inability to uniquely determine measurement outcomes just a lack of complete knowledge about the system? Or put differently, could you gain some additional information beyond the quantum state that will help you determine outcomes exactly? The answer is "no", quantum phenomena are fundamentally probabilistic in nature and one way to show this is using an argument first made by John Bell in the 1960s.</span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">Bell's theorem, as the result is called, says that there is no hidden mechanism that decides the outcome of a quantum measurement. The way the argument works is that if you assume that there is some unknown factor that manufactures the statistical results of a measurement, then you can establish a limit on the amount of correlation among certain properties you can measure, which you can state as a Bell inequality. You can then show that there are always some quantum states that can violate the inequality. In this post, we attempt to describe a simple example of Bell's theorem at work.</span><br />
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<a name='more'></a><br />
<span style="font-family: Georgia, Times New Roman, serif;">We will have to be a little bit more mathematical than usual in some parts but nothing too advanced is need: some arithmetic, absolute values, inequalities, a little trigonometry (really, just knowing what sine and cosine are), simple algebra such as if <i>x = a + b</i> and <i>y = a - b</i> then <i>a = (x + y)/2</i>, </span>
<span style="font-family: Georgia, Times New Roman, serif;">and taking "products" of states, that is <i>|a) x |b) = |a)|b) = |a,b)</i>.</span><br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiw1tsGKxPwf3CBLCp1xRL5H7bDHnOS1xppDi8Gl7BJHKIC_Kc825jFNLK-v4BWNwu4_Fh0JVe8uVasweSPuUV6AE61M64E1aY0DmENRw_NnecTl0-wmkvxSVz3mvp55zbZfPLhrZB0p9k6/s1600/bellExpt.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiw1tsGKxPwf3CBLCp1xRL5H7bDHnOS1xppDi8Gl7BJHKIC_Kc825jFNLK-v4BWNwu4_Fh0JVe8uVasweSPuUV6AE61M64E1aY0DmENRw_NnecTl0-wmkvxSVz3mvp55zbZfPLhrZB0p9k6/s1600/bellExpt.jpg" height="480" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">A typical experiment for testing a Bell inequality. A source emits many entangled pairs of electrons to detectors held by Alice on the left and Bob on the right. Each detector has 2 outcomes +1 or -1 for measuring the spin. There is a switch that lets Alice choose one of two possible directions for the spin measurement, labeled <i>a</i> and <i>a</i>'. Bob has a similar switch with settings <i>b</i> and <i>b'</i>.</td></tr>
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<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">Consider the spin of an electron as a qubit, with two states <i>|u)</i> for up and <i>|d)</i> for down that can be distinguished perfectly. Qubits can be in superposition and following what we've done previously, we can have states like</span><br />
<i><span style="font-family: Georgia, Times New Roman, serif;"><br /></span></i>
<i><span style="font-family: Georgia, Times New Roman, serif;">|+) = 1/sqrt(2) |u) + 1/sqrt(2) |d)</span></i><br />
<i><span style="font-family: Georgia, Times New Roman, serif;">|-) = 1/sqrt(2) |u) - 1/sqrt(2) |d).</span></i><br />
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<span style="font-family: Georgia, Times New Roman, serif;">Bell inequalities are often mentioned in the context of an experiment performed with entangled qubits, an example of which is shown above. In the setup, we have a source of pair of electrons in an entangled state </span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<i><span style="font-family: Georgia, Times New Roman, serif;">|Bell) = 1/sqrt(2) |u,u) + 1/sqrt(2) |d,d). </span></i><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">There are two detectors for measuring the spin of the electrons. We have Alice operating the left detector and Bob handling the right one. Each detector has two possible measurement settings, which we label a or a' on Alice's side, and b or b' on Bob's side. For now, it does not matter what the settings refer to exactly, only that Alice and Bob have 2 options for measuring the spin. </span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">Whichever setting they use, they get one of two possible outcomes, which we will call +1 or -1. Again it does not matter which outcome you call +1 or -1, but to be consistent, we will usually call the nearest state when you move clockwise from |u) the +1 outcome. A short sample of the results of this experiment is also provided above. </span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">Using Alice's and Bob's measurement results for electrons belonging to the same entangled pair, we can calculate the correlation</span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;"><i>C = [ (number of +1 pairs) - (number of -1 pairs) ] / (total number of detected pairs).</i></span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">We have a minus for the -1 pairs because they refer to outcomes that are anti-correlated. Observe that if Alice and Bob always both measure "<i>|u)</i> or <i>|d)</i>", then all detected pairs will be +1 pairs and <i>C = +1</i>. similarly if Alice and Both always both measures "<i>|u)</i> or <i>|d)</i>" but Bob has his detector inverted (so that <i>|u)</i> is the "+1" outcome for Alice while it is the -1 outcome for Bob), then all detected pairs will be -1 pairs and <i>C = -1</i> in that case. </span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">Alice and Bob will choose randomly between the measurement settings available to them so the only thing we can tell for certain is that C is a number between -1 and 1. </span><span style="font-family: Georgia, 'Times New Roman', serif;">For example, suppose that Alice measures "|u) or |d)" while Bob measures "<i>|+) </i>or <i>|-)</i>". If we write<i> |u) </i>and <i>|d)</i> in terms of <i>|+) </i>and <i>|-)</i>, we find that</span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<i><span style="font-family: Georgia, Times New Roman, serif;">|u,u) = 1/sqrt(2) |u,+) + 1/sqrt(2) |u,-), </span></i><br />
<i><span style="font-family: Georgia, Times New Roman, serif;">|d,d) = 1/sqrt(2) |d,+) - 1/sqrt(2) |d,-).</span></i><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">then</span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<i><span style="font-family: Georgia, Times New Roman, serif;">|Bell) = 1/2 |u,+) + 1/2 |d,+) + 1/2 |u,-) - 1/2 |d,-).</span></i><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">This means that there is 1/4 probability that you get any one of the four possible outcomes when Alice and Bob's measurements are taken together. This translates to getting, on average, the same number of +1 pairs as -1 pairs, which means <i>C = 0</i> for this example.</span><br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiyyuhw8l8gdLRHrFJ9OeNHs7OzfnR8mfadN3yS4BsINrk1TC-Gl38haRZMmHpwSveat4qpLtoXc7XwSgRpH_MiuxmhmD6p2aQEuOA5OD_ZdDtA_YkaPehQ0aK7SgGCyB1gkvsdPuZ6PxNI/s1600/bellQubit.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiyyuhw8l8gdLRHrFJ9OeNHs7OzfnR8mfadN3yS4BsINrk1TC-Gl38haRZMmHpwSveat4qpLtoXc7XwSgRpH_MiuxmhmD6p2aQEuOA5OD_ZdDtA_YkaPehQ0aK7SgGCyB1gkvsdPuZ6PxNI/s1600/bellQubit.jpg" height="313" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Circular slice of the sphere of qubit states that contains the states |u) and |+). We measure the spin along the direction indicated by a pair of points |x) and |x') on this circle, where x is the angle between |x) and |u).</td></tr>
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<br />
<span style="font-family: Georgia, Times New Roman, serif;">Generally, we let Alice or Bob choose a measurement setting that is a certain angle away from |u) in that circle on the set of qubit states that contains |u) and |+) (automatically |d) and |-), too, since these are oppositely directed). The setting can be identified with an orientation in this circle, described by pair of opposite points, which we label |x) and |x'), and are given by</span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<i><span style="font-family: Georgia, Times New Roman, serif;">|x) = cos(x/2) |u) + sin(x/2) |d),</span></i><br />
<i><span style="font-family: Georgia, Times New Roman, serif;">|x') = -sin(x/2) |u) + cos(x/2) |d). </span></i><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">Note that when <i>x = 0</i></span><span style="font-family: "Georgia","serif";">°</span><span style="font-family: Georgia, Times New Roman, serif;">, we get the pair |<i>u), |d)</i> and when <i>x = 90</i></span><span style="font-family: "Georgia","serif";">°</span><span style="font-family: Georgia, Times New Roman, serif;">, we get the pair <i>|+) , |-)</i>.</span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">We now have everything we need to "derive" a Bell inequality. The simplest version requires two settings for each detector, as we have in the experiment displayed earlier. Let <i>A(a)</i> be Alice's result (+1 or -1) on the left detector when it is set at <i>a</i>, and <i>B(b)</i> be Bob's result in the right detector when it is set at <i>b</i>, and same goes for <i>a'</i> and <i>b'</i>. </span><span style="font-family: Georgia, 'Times New Roman', serif;">Let <i>C(a,b)</i> be the correlation of measurement outcomes when Alice's detector setting is <i>a</i> and Bob's setting is <i>b</i>, and so on.</span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">Suppose there was some hidden variable <i>H</i> that determines the outcome for each entangled pair of electrons based on the setting of the detector. For simplicity, we will assume that <i>H</i> takes only 2 values, 0 or 1. (We can make the argument for an H that takes many more values but the calculation is simpler when we only have two.) If <i>H </i>determines the correlation, then</span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<i><span style="font-family: Georgia, Times New Roman, serif;">C(a,b) = p(0) A(a|0) B(b|0) + p(1) A(a|1) B(b|1)</span></i><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">where <i>p(0)</i> is the fraction of measured pairs with <i>H = 0</i>, <i>p(1) </i>is the same with <i>H = 1</i>, <i>A(a|0)</i> is Alice's result given that the detector setting is <i>a</i> and <i>H = 0</i>, and the same goes for the other possibilities. Since we assume <i>H</i> must be 0 or 1, we also have<i> p(0) + p(1) = 1.</i></span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">Consider <i>C(a,b) - C(a,b')</i>, which is equal to</span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<i><span style="font-family: Georgia, Times New Roman, serif;">p(0) A(a|0) B(b|0) + p(1) A(a|1) B(b|1) - p</span><span style="font-family: Georgia, 'Times New Roman', serif;">(0) A(a|0) B(b'|0) - p(1) A(a|1) B(b'|1).</span></i><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<span style="font-family: Georgia, 'Times New Roman', serif;">Now we add the following terms to it:</span><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<i><span style="font-family: Georgia, 'Times New Roman', serif;">p(0) A(a|0) B(b|0) A(a'|0) B(b'|0) + p</span><span style="font-family: Georgia, 'Times New Roman', serif;">(1) A(a|1) B(b|1) A(a'|1) B(b'|1) </span></i><br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;"> - p</span><span style="font-family: Georgia, 'Times New Roman', serif;">(0) A(a|0) B(b'|0) A(a'|0) B(b|0) - p</span><span style="font-family: Georgia, 'Times New Roman', serif;">(1) A(a|1) B(b'|1) A(a'|1) B(b|1).</span></i><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<span style="font-family: Georgia, 'Times New Roman', serif;">That might look like a mess but if you look carefully, all that was is a fancy way of adding zero, because we are adding then subtracting the same stuff. If we group similar terms, we find that</span><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<i><span style="font-family: Georgia, 'Times New Roman', serif;">C(a,b) - C(a,b') = p(0) A(a|0)B(b|0) [ 1 + A(a'|0)B(b'|0) ] </span></i><br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;"> + </span><span style="font-family: Georgia, 'Times New Roman', serif;"> </span><span style="font-family: Georgia, 'Times New Roman', serif;">p(1) A(a|1)B(b|1) [ 1 + A(a'|1)B(b'|1) ] </span></i><br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;"> -</span><span style="font-family: Georgia, 'Times New Roman', serif;"> </span><span style="font-family: Georgia, 'Times New Roman', serif;">p(0) A(a|0)B(b'|0) [ 1 + A(a'|0)B(b|0) ]</span></i><br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;"> -</span><span style="font-family: Georgia, 'Times New Roman', serif;"> </span><span style="font-family: Georgia, 'Times New Roman', serif;">p(1) A(a|1)B(b'|1) [ 1 + A(a'|1)B(b|1) ].</span></i><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<span style="font-family: Georgia, 'Times New Roman', serif;">Because the product <i>A(a|H)B(b|H)</i> is +1 or -1<i>,</i> its absolute value is +1, whatever the value of <i>H</i> is and even if we had setting <i>a'</i> or <i>b'. </i>Also, we know that <i>p</i></span><span style="font-family: Georgia, 'Times New Roman', serif;"><i>(0)</i> and <i>p(1)</i> are bigger than or equal to zero. </span><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<span style="font-family: Georgia, 'Times New Roman', serif;">An important result concerning absolute values is the so-called triangle inequality, which says </span><span style="font-family: Georgia, 'Times New Roman', serif;">that if you had any triangle with sides of lengths <i>x, y</i>, and <i>z</i>, then it must be true that</span><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<i><span style="font-family: Georgia, 'Times New Roman', serif;">|x - y| </span><span lang="EN-CA" style="font-family: Georgia; font-size: 12.0pt; mso-ansi-language: EN-CA; mso-bidi-font-family: "Times New Roman"; mso-bidi-language: AR-SA; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-US;">≤</span><span style="font-family: Georgia, 'Times New Roman', serif;"> z </span><span lang="EN-CA" style="font-family: Georgia; font-size: 12.0pt; mso-ansi-language: EN-CA; mso-bidi-font-family: "Times New Roman"; mso-bidi-language: AR-SA; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-US;">≤</span><span style="font-family: Georgia, 'Times New Roman', serif;"> x + y.</span></i><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<span style="font-family: Georgia, 'Times New Roman', serif;">You can check with examples to convince yourself that it's true. Like for a right triangle with sides of lengths 3, 4, and 5, you have |3-4| = 1 </span><span style="font-family: Georgia;"><</span><span style="font-family: Georgia, 'Times New Roman', serif;"> 5 <</span><span style="font-family: Georgia, 'Times New Roman', serif;"> 3+4 = 7.</span><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<span style="font-family: Georgia, 'Times New Roman', serif;">If we take the absolute value of <i>C(a,b) - C(a,b')</i>, we can use the triangle inequality and |A(a|H)B(b|H)| = 1, etc. to get </span><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<i><span style="font-family: Georgia, 'Times New Roman', serif;">|C(a,b) - C(a,b')| </span><span lang="EN-CA" style="font-family: Georgia; font-size: 12.0pt; mso-ansi-language: EN-CA; mso-bidi-font-family: "Times New Roman"; mso-bidi-language: AR-SA; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-US;">≤</span><span style="font-family: Georgia, 'Times New Roman', serif;"> </span><span style="font-family: Georgia, 'Times New Roman', serif;">p(0) [ 1 + A(a'|0)B(b'|0) ] + </span><span style="font-family: Georgia, 'Times New Roman', serif;"> </span><span style="font-family: Georgia, 'Times New Roman', serif;">p(1) [ 1 + A(a'|1)B(b'|1) ] </span></i><br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;"> +</span><span style="font-family: Georgia, 'Times New Roman', serif;"> </span><span style="font-family: Georgia, 'Times New Roman', serif;"> </span><span style="font-family: Georgia, 'Times New Roman', serif;">p(0) [ 1 + A(a'|0)B(b|0) ] + </span><span style="font-family: Georgia, 'Times New Roman', serif;"> </span><span style="font-family: Georgia, 'Times New Roman', serif;"> </span><span style="font-family: Georgia, 'Times New Roman', serif;">p(1) [ 1 + A(a'|1)B(b|1) ].</span></i><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<span style="font-family: Georgia, 'Times New Roman', serif;">From the way we define correlations, and since p(0) + p(1) = 1, we can write this as</span><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<i><span style="font-family: Georgia, 'Times New Roman', serif;">|C(a,b) - C(a,b')| </span><span lang="EN-CA" style="font-family: Georgia; font-size: 12.0pt; mso-ansi-language: EN-CA; mso-bidi-font-family: "Times New Roman"; mso-bidi-language: AR-SA; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-US;">≤</span><span style="font-family: Georgia, 'Times New Roman', serif;"> 2 + C(a'b') + C(a',b).</span></i><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<span style="font-family: Georgia, 'Times New Roman', serif;">Now this is what we get if we had added the messy stuff that was equal to zero. But we could have subtracted it instead, in which case we would end up with</span><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<i><span style="font-family: Georgia, 'Times New Roman', serif;">|C(a,b) - C(a,b')| </span><span lang="EN-CA" style="font-family: Georgia; font-size: 12.0pt; mso-ansi-language: EN-CA; mso-bidi-font-family: "Times New Roman"; mso-bidi-language: AR-SA; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-US;">≤</span><span style="font-family: Georgia, 'Times New Roman', serif;"> 2 - C(a'b') - C(a',b).</span></i><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<span style="font-family: Georgia, 'Times New Roman', serif;">Because it should not matter whether we add or subtract zero, the inequality must be true when the terms on the right side achieve the smaller value, which we can write as</span><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<i><span style="font-family: Georgia, 'Times New Roman', serif;">|C(a,b) - C(a,b')| </span><span lang="EN-CA" style="font-family: Georgia; font-size: 12.0pt; mso-ansi-language: EN-CA; mso-bidi-font-family: "Times New Roman"; mso-bidi-language: AR-SA; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-US;">≤</span><span style="font-family: Georgia, 'Times New Roman', serif;"> 2 - |C(a'b') + C(a',b)|.</span></i><br />
<div>
<br /></div>
<div>
<span style="font-family: Georgia, Times New Roman, serif;">Finally, writing the inequality in its usual form,</span></div>
<div>
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span></div>
<div>
<i><span style="font-family: Georgia, 'Times New Roman', serif;">|C(a,b) - C(a,b')| + |C(a'b') + C(a',b)| </span><span lang="EN-CA" style="font-family: Georgia; font-size: 12.0pt; mso-ansi-language: EN-CA; mso-bidi-font-family: "Times New Roman"; mso-bidi-language: AR-SA; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-US;">≤</span><span style="font-family: Georgia, 'Times New Roman', serif;"> 2.</span></i></div>
<div>
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span></div>
<span style="font-family: Georgia, Times New Roman, serif;">This is our Bell inequality. The same expression can be obtained if<i> H</i> were allowed to take a wide range of values. </span><span style="font-family: Georgia, 'Times New Roman', serif;">What it says is that if the results we get from measuring the spins of entangled pairs of electrons is determined by some hidden variable <i>H</i>, then even if we don't know what H is, we know that the correlations it produces must obey the inequality. This is a remarkable claim because it means that no matter how elaborate you design a local hidden mechanism to be, it won't break the inequality.</span><br />
<br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
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<div class="separator" style="clear: both; text-align: center;">
</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhhr9ZLInnF-15AAXa4SbzY7OEX-1DSw9ESsbg5l2dzWGm7T6XbOd9uWbhSGo-ucuSxiGtY7A3jron70PG2uoaU0k7vyBerJpnh3CKq1mTxTpabCr5tKCOlaxa0h0Zg1JvG3uPafbJuObr8/s1600/bellSettings.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhhr9ZLInnF-15AAXa4SbzY7OEX-1DSw9ESsbg5l2dzWGm7T6XbOd9uWbhSGo-ucuSxiGtY7A3jron70PG2uoaU0k7vyBerJpnh3CKq1mTxTpabCr5tKCOlaxa0h0Zg1JvG3uPafbJuObr8/s1600/bellSettings.jpg" height="480" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Measurement settings for Alice and Bob that achieve the maximum amount of violation of the Bell inequality we derived.</td></tr>
</tbody></table>
<span style="font-family: Georgia, 'Times New Roman', serif;"></span>
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<span style="font-family: Georgia, 'Times New Roman', serif;">So the question now is, do the correlations obtained from an actual experiment with electrons violate the inequality? The answer is yes, but as we have seen earlier, it is necessary to choose the detector settings for Alice and Bob to actually observe the violation. The simplest choice of settings is shown above, and ideally it would produce the maximum violation possible.</span><br />
<br />
<span style="font-family: Georgia, 'Times New Roman', serif;">The calculation needed to show the violation is straightforward but a bit tedious so we will describe it in detail in the following.</span><br />
<br />
<span style="font-family: Georgia, 'Times New Roman', serif;">Recall how the states <i>|x)</i> and <i>|x')</i> are related to <i>|u)</i> and<i> |d)</i>. From that, we can write</span><br />
<br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;">|u) = cos(x/2) |x) - sin(x/2) |x') </span></i><br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;">|d) = sin(x/2) |x) + cos(x/2) |x').</span></i><br />
<br />
<span style="font-family: Georgia, 'Times New Roman', serif;">To compute the correlations for the setting chosen in the picture, we need to write the state |Bell) in terms of the measurements that Alice and Bob perform. </span><br />
<br />
<span style="font-family: Georgia, 'Times New Roman', serif;">When the settings are a and b:</span><br />
<i><br /></i>
<i><span style="font-family: Georgia, 'Times New Roman', serif;">|u,u) = cos(x/2) |u,x) - sin(x/2) |u,x')</span></i><br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;">|d,d) = sin(x/2) |d,x) + cos(x/2) |d,x').</span></i><br />
<br />
<span style="font-family: Georgia, 'Times New Roman', serif;">If we label <i>|u)</i> and <i>|x)</i> by +1 and <i>|d)</i> and <i>|x') </i>by -1, the correlation <i>C(a,b)</i> can be computed by multiplying each value of a measurement with the probabilities of states that give that value. So we get</span><br />
<br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;">C(a,b) = (+1) ( p[|u,x)] + p[|d,x')] ) + (-1) </span><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">( p[|u,x')] + p[|d,x)] )</span></span></i><br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;"> = cos^2(x/2) - sin^2(x/2)</span></span></i><br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;"> = cos(x).</span></span></i><br />
<br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">For setting b, we have <i>x = 45</i></span></span><span style="font-family: "Georgia","serif"; mso-bidi-font-family: "Times New Roman";"><i>°</i></span> <span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">so <i>C(a,b) = 1/sqrt(2)</i>. For setting b', we have <i>x = 135</i></span></span><span style="font-family: "Georgia","serif"; mso-bidi-font-family: "Times New Roman";"><i>°</i></span><span style="font-family: "Georgia","serif";"></span><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">
so <i>C(a,b') = -1/sqrt(2).</i></span></span><br />
<br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">To calculate the correlations <i>C(a',b) </i>and <i>C(a',b')</i>, recall that</span></span><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span></span>
<i><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">|u,u) + |d,d) = |+,+) + |-,-)</span></span></i><br />
<br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">so for Bob, we can write <i>|+) </i>and <i>|-)</i> in terms of <i>|x) </i>and <i>|x')</i>:</span></span><br />
<br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">|+, +) = [ cos(x/2) + sin(x/2) ]/sqrt(2) ] |+, x) + </span></span><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;"> [ cos(x/2) - sin(x/2) ]/sqrt(2) |+, x'),</span></span></i><br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">|-, -) = [ cos(x/2) - sin(x/2) ]/sqrt(2) ] |-, x) - </span></span><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;"> [ cos(x/2) + sin(x/2) ]/sqrt(2) |-, x').</span></span></i><br />
<br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">Using <i>cos(</i></span></span><i><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">45</span></span><span style="font-family: "Georgia","serif"; mso-bidi-font-family: "Times New Roman";">°) = sin(</span></span></span><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">45</span></span></i><span style="font-family: "Georgia","serif"; mso-bidi-font-family: "Times New Roman";"><i>°) = 1/sqrt(2)</i> and the trigonometric identity <i>sin(x + y) = sin(x) cos(y) + cos(x) sin(y)</i>, we find that</span><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"></span><br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">C(a',b) = </span></span><span style="font-family: Georgia, 'Times New Roman', serif;">(+1) ( p[|+, x)] + p[|-, x')] ) + (-1) </span><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">( p[|+, x')] + p[|-, x)] )</span></span></i><br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;"> = sin^2(</span></span><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">45</span></span><span style="font-family: "Georgia","serif"; mso-bidi-font-family: "Times New Roman";">° + x/2) - </span><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">sin^2(</span></span><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">45</span></span><span style="font-family: "Georgia","serif"; mso-bidi-font-family: "Times New Roman";">° - x/2).</span></i><br />
<span style="font-family: "Georgia","serif"; mso-bidi-font-family: "Times New Roman";"><br /></span>
<span style="font-family: "Georgia","serif"; mso-bidi-font-family: "Times New Roman";">Using your handy calculator, you can check that<i> C(a'b) = 1/sqrt(2)</i> for <i>x = </i></span><span style="font-family: "Georgia","serif"; mso-bidi-font-family: "Times New Roman";"><i><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">45</span></span></i><span style="font-family: "Georgia","serif"; mso-bidi-font-family: "Times New Roman";"><i>° </i></span> and <i>C(a',b') = 1/sqrt(2)</i> for </span><span style="font-family: "Georgia","serif"; mso-bidi-font-family: "Times New Roman";"><i>x = </i></span><span style="font-family: "Georgia","serif"; mso-bidi-font-family: "Times New Roman";"><i><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">135</span></span></i><span style="font-family: "Georgia","serif"; mso-bidi-font-family: "Times New Roman";"><i>°</i>.</span></span><br />
<span style="font-family: "Georgia","serif"; mso-bidi-font-family: "Times New Roman";"><br /></span>
<span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;"> Putting all the correlations together to check the inequality, we get</span></span><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span></span>
<i><span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;"> </span></span></i><span style="font-family: Georgia, 'Times New Roman', serif;"><i>|C(a,b) - C(a,b')| + |C(a'b') + C(a',b)| = 4/sqrt(2) = 2 sqrt(2) > 2</i>.</span><br />
<br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;">The main assumption used in the Bell inequality is that any hidden mechanism that produces correlated results can only affect its immediate vicinity, since any sort of influence it makes is limited by the speed of light, according to special relativity. Since we should not be so eager to give up on Einstein's work, because entangled states can produce violations of a Bell inequality, it means no such local mechanism exists. The behavior of quantum objects represents a truly probabilistic phenomenon.</span></span><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span></span>
<span style="font-family: Georgia, Times New Roman, serif;">As my old advisor Berge Englert would say: </span><span style="font-family: Georgia, 'Times New Roman', serif;">"T</span><span style="font-family: Georgia, 'Times New Roman', serif;">his is a brutal fact of life. In a very profound sense, quantum mechanics is about learning to live with it. "</span><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<span style="font-family: Georgia, 'Times New Roman', serif;"><b>Reference:</b></span><br />
<span style="font-family: Georgia, 'Times New Roman', serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">B.-G. Englert, Lectures on Quantum Mechanics: Basic Matters, pp. 4-8 (World Scientific, Singapore, 2006).</span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-7786253974914814722014-02-24T04:16:00.002-08:002014-03-03T12:24:55.975-08:00Understanding quantum entanglement<span style="font-family: Georgia, Times New Roman, serif;">As mentioned before, a quantum computer exploits the rules of quantum theory for accomplishing feats that are considered impossible with conventional computers. Two features of quantum mechanics are often involved in achieving such feats: superposition, which was discussed in our <a href="http://quantumgazette.blogspot.ca/2014/02/a-note-on-quantum-superposition.html">last post</a>, and entanglement, which is the topic here.</span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, 'Times New Roman', serif;">Quantum entanglement is considered to be one of the counterintuitive aspects of quantum mechanics. However it is not a particularly difficult concept to grasp if we start with the notion of correlation. Roughly speaking, entangled quantum systems are objects whose properties so strongly correlated that using a state to describe all of them as a single unit describes them better than assigning a state to individual parts.</span><br />
<br />
<span style="font-family: Georgia, Times New Roman, serif;"></span><br />
<a name='more'></a>
<span style="font-family: Georgia, Times New Roman, serif;">To describe quantum entanglement in more detail, we need to introduce the idea of correlation between 2 or more objects. Suppose we have two coins. Each coin has two possible states: heads (H) or tails (T). Taken together, there are four possible outcomes when both coins are tossed: <i>HH, HT, TH, </i>and<i> TT</i>. </span><span style="font-family: Georgia, 'Times New Roman', serif;">In general, a correlation measures the probability we get the same outcome when we measure the same property for two or more systems.</span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">We say that the coins are perfectly correlated if we can determine the state of one coin with certainty when are told about the state of the other coin. One somewhat contrived way to achieve this would be to join a pair of coins like in the diagram below, so that they end of heads or tails together. In this case, if the chance of landing heads or tails is the same, then the state of the two coins is described by <i>1/2 HH + 1/2 TT</i>.</span><br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh8MdbPItfbfmLx4_aZgxyJFXbYe-H8wrIdtbp4rCmuScZC73P5Z5ZCn1pJvCidhwiPkPMtrQM2XS0xTzsGiJ1av2mDQEuLp04npWOAuc-04cm_d-3GYajxEhFTAokCEt9e2i5hdOOORBdk/s1600/joinedCoins.JPG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh8MdbPItfbfmLx4_aZgxyJFXbYe-H8wrIdtbp4rCmuScZC73P5Z5ZCn1pJvCidhwiPkPMtrQM2XS0xTzsGiJ1av2mDQEuLp04npWOAuc-04cm_d-3GYajxEhFTAokCEt9e2i5hdOOORBdk/s1600/joinedCoins.JPG" height="329" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Two coins attached to each other by a stick so that when they are flipped they also land heads or tails together. We say that the outcomes of the flip is perfectly correlated.</td></tr>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Perfect correlation is usually better understood when two states involved are always opposite each other (in this case we usually say they are anti-correlated), like if I promise that two closed boxes contain a pair of socks and you open one to reveal a left sock, then you know that the other box held a right sock (if I was telling the truth). The point remains that when objects are perfectly correlated or anti-correlated, information about some property of one can be used to determine the same property for the other. </span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">Let's consider the perfectly correlated qubits. Suppose we have two electrons, each with spin pointing up, which we write as the state |u), or down, which we write as state |d). Taken together, we have four states that we can perfectly distinguished by measurement: |u,u), |u,d), |d,u), and |d,d), where |u,d) means the first electron has spin pointing up and the second electron has spin pointing down, and so on.</span><br />
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<span style="font-family: Georgia, Times New Roman, serif;">An example of a perfectly correlated state for this pair of qubits is <i>1/sqrt(2) |u,u) + 1/sqrt(2) |d,d)</i>. This is an example of an entangled quantum state for two qubits. Observe that it is a superposition state and we know from the <a href="http://quantumgazette.blogspot.ca/2014/02/a-note-on-quantum-superposition.html">previous post</a> that this state leads to probabilities of a very different quality than what we get from a pair of random coins. </span><span style="font-family: Georgia, 'Times New Roman', serif;">However, that is still a somewhat mathematical way of looking at it. Is there an experimental way to distinguish between this entangled state, and a correlated mixture of bits?</span><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEihPvnNV7RABDUBN5t_-5dTufPlAbRatQnzzUzZirSsC0zn9B2JE_JGZ0Y0Z0Upp-MzAntENw-kbTglKll6XEzQMe6-rhEcYE-T9TYxQjP6j3AUEmrMKgtJJF5ZI5aYFHM5qkgfyWv9rmgv/s1600/qubitStates.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEihPvnNV7RABDUBN5t_-5dTufPlAbRatQnzzUzZirSsC0zn9B2JE_JGZ0Y0Z0Upp-MzAntENw-kbTglKll6XEzQMe6-rhEcYE-T9TYxQjP6j3AUEmrMKgtJJF5ZI5aYFHM5qkgfyWv9rmgv/s1600/qubitStates.png" height="300" width="400" /></a></div>
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span><span style="font-family: Georgia, Times New Roman, serif;">To find out, recall</span><span style="font-family: Georgia, 'Times New Roman', serif;"> that for a qubit such as the spin of an electron, up and down can be measured along any direction in 3-dimensional space. Using the coordinates on a globe to label directions, with "up or down" going along the line between north and south pole, we can try to measure the spin along "|+) or |-)" direction on the equator. With </span><br />
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<span style="font-family: Georgia, Times New Roman, serif;"><i>|+) = 1/sqrt(2) |u) + 1/sqrt(2) |d),</i> </span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><i>|-) = 1/sqrt(2) |u) - 1/sqrt(2) |d)</i>, </span><br />
<span style="font-family: Georgia, Times New Roman, serif;"><br /></span>
<span style="font-family: Georgia, Times New Roman, serif;">we can write |u) </span><span style="font-family: Georgia, 'Times New Roman', serif;">and |d) in terms of |+) and |-). Carrying out the calculation step-by-step, we have</span><br />
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<i><span style="font-family: Georgia, Times New Roman, serif;">|u,u) = |u)|u) = [ 1/sqrt(2) |+) + 1/sqrt(2) |-) ] </span><span style="font-family: Georgia, 'Times New Roman', serif;">[ 1/sqrt(2) |+) + 1/sqrt(2) |-) ] </span></i><br />
<i><span style="font-family: Georgia, 'Times New Roman', serif;"> = 1/2 |+,+) + 1/2 |+,-) + 1/2 |-,+) + 1/2 |-,-)</span></i><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">where roughly speaking, what we did was multiply numbers with numbers and states with states. Similarly, we get</span><br />
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<i><span style="font-family: Georgia, 'Times New Roman', serif;">|d,d) = |d)|d) = </span><span style="font-family: Georgia, 'Times New Roman', serif;">1/2 |+,+) - 1/2 |+,-) - 1/2 |-,+) + 1/2 |-,-).</span></i><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">Adding the two expressions, we see that </span><br />
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<i style="font-family: Georgia, 'Times New Roman', serif;">|u,u) + |d,d) =|+,+) + |-,-), </i><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">meaning that if we had measured the spins in "<i>|+)</i> or <i>|-)</i>" direction, we would still get correlated results. That is, if the first spin is in state <i>|+)</i>, the second one will also be found in state <i>|+)</i>, and similarly for |-).</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">If you don't find this to be surprising, remember how the "<i>|u)</i> or <i>|d)</i>" and "<i>|+)</i> or <i>|-)</i>" refer to two properties of an electron spin that obey an uncertainty principle, like position and momentum. Because we have so many different possible directions available, this means that we have an assortment of properties, any one of which can be used to find correlated results on spins measured in the same way. It is often in this sense that we say that a pair of entangled spins are not really two separate quantum systems but just a single one.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">Also, if we can keep the entangled electrons sufficiently isolated from the environment, we can separate them as far apart as galaxies before measuring their spins and the outcomes will still be correlated. This means that quantum entanglement is not due to some force or influence being exchanged by the electrons on each other to keep them synchronized, since any such signal must ultimately be limited by the speed of light according to Einstein's relativity.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">If each spin is measured along a different direction, there is some randomness to the outcomes so we wouldn't be able to guess accurately the result for one spin just from information of the other spin. This is why we cannot use entangled quantum states to send information faster than light, since any two people using entangled spins for example would have to measure them along the exact same direction, and guaranteeing that requires some for of communication between them. </span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">In those cases where the spins are measured along different orientations, there is still a way of showing through calculations that in general the correlations are stronger than one can achieve with classical bits of information, through violation of so-called Bell inequalities. </span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">A Bell inequality establishes some bound on the correlation between two systems if we assume that each system is local, which just means one system cannot suddenly affect the other if they are sufficiently far apart, and has properties that have well-defined values (that exist even if we don't measure them.) It can be shown that quantum states with some degree of entanglement can violate such a bound for some choices of measurement. </span><br />
<br />Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-44582979096233202592014-02-19T21:59:00.001-08:002016-07-17T11:00:11.903-07:00A note on quantum superposition <span style="font-family: "georgia" , "times new roman" , serif;">Quantum computing is commonly described as the means of harnessing the laws of quantum mechanics to process information, usually for the purpose of doing certain calculations faster than what you can achieve with conventional computers. </span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">A typical computer operates on binary digits, or bits, of information, which is</span><span style="font-family: "georgia" , "times new roman" , serif;"> a sequence of zeros and ones encoded by electrical signals. </span><span style="font-family: "georgia" , "times new roman" , serif;">A calculation is performed using a circuit of transistors designed to switch the signals on and off in a particular way, so that the final bit values determine the result. </span><span style="font-family: "georgia" , "times new roman" , serif;">The list of specific steps needed for handling bits in any such calculation is called an algorithm.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">In contrast, a quantum computer encodes information in quantum bits, or qubits. </span><span style="font-family: "georgia" , "times new roman" , serif;">Computation with qubits is different from computation with bits because quantum systems can be prepared and controlled in ways that can not be achieved with signals that represent bits. </span><span style="font-family: "georgia" , "times new roman" , serif;">Quantum algorithms describe ways in which qubits can be manipulated so that the correct measurement yields the desired outcome of a calculation with high probability.</span><br />
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<a name='more'></a><span style="font-family: "georgia" , "times new roman" , serif;">It is a characteristic feature of quantum theory that measurements in general do not have definite outcomes. Rather, what we have is a list of probabilities for each possible outcome. This begs the question, can't you just use randomized bits to generate the same kind of probabilities? You can't and part of the reason lies in the differences between a probabilistic bit and a quantum superposition.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">Consider a coin. When flipped it can come up heads or tails. If we represent heads by $H$ and tails by $T$, then we can represent the state of a coin by its probability to land heads or tails. Suppose we believe we have a fair coin with the same chance of landing heads or tails. The state $C$ that we assign to the coin is then $C = \frac{1}{2} H + \frac{1}{2} T$.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">Now consider an electron. You might recall from your chemistry classes that electrons have a property called spin, which can be up and down, and pair of such spins are required to fill in orbitals in an atom. In any case, spin is a quantum property of an electron that is an example of a qubit. </span><span style="font-family: "georgia" , "times new roman" , serif;">We can assign a quantum state to the electron that describes the probability of measuring it in a up or down orientation. </span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">Let $|u\rangle$<i> </i>denote the state of an electron with spin pointing up and $|d\rangle$ for spin pointing down. An example of a state the has equal chance of being measured with the electron spin pointing up or down is $|+\rangle = \frac{1}{\sqrt{2}} \left( |u\rangle + |d \rangle \right) $.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">If we think of $|u\rangle$ and $|d\rangle$ as the quantum versions of $H$ and $T$, then we see that one main difference between bit states and qubit states is the fact that the numbers that appear in quantum states are not probabilities but rather probability amplitudes. </span><span style="font-family: "georgia" , "times new roman" , serif;">In general, these amplitudes are complex numbers whose absolute values when squared correspond to probabilities.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">Since quantum states use complex numbers, there are many states that describe an electron that has equal chance of being found with the spin up or down. Another such example would be the state </span><span style="font-family: "georgia" , "times new roman" , serif;">$|-\rangle = \frac{1}{\sqrt{2}} \left( |u\rangle - |d \rangle \right) $. </span><span style="font-family: "georgia" , "times new roman" , serif;">We can easily check this since $\left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}$ </span><span style="font-family: "georgia" , "times new roman" , serif;">.</span><span style="font-family: "georgia" , "times new roman" , serif;"> </span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">In experiments, we verify the probabilities by first preparing many electrons in the same state and then performing identical spin measurements on each electron. The relative frequency of the each outcome provides an estimate of the probabilities.</span><br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgu6SDpyVkngdXoOP9zjK5zfXhRLizvmTO5EO_Yp486-o8AltbcNtCYpW25eexDthWn7n7Jhq4wAtdUeMUyUBEo7LyO9kCCNn-X22ATwKyAnIw32UyGnGzwX383himsEo2TIWVe9a-50qM6/s1600/qubitStates.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="300" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgu6SDpyVkngdXoOP9zjK5zfXhRLizvmTO5EO_Yp486-o8AltbcNtCYpW25eexDthWn7n7Jhq4wAtdUeMUyUBEo7LyO9kCCNn-X22ATwKyAnIw32UyGnGzwX383himsEo2TIWVe9a-50qM6/s1600/qubitStates.png" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Qubit states as arrows directed toward points on a sphere starting from its center.</td></tr>
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<span style="font-family: "georgia" , "times new roman" , serif;">The picture of a qubit provided by the spin of an electron is actually quite useful because just as up and down can point to any orientation in 3-dimensional space, the same holds for a qubit. In geometric terms, we would say that the set of all possible states for a qubit corresponds to points on a sphere, or since we are considering the direction of electron spin, to arrows that go from the center of the sphere. </span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">Imagine that sphere of qubit states to be labeled geographically like a globe. We can call state $|u\rangle$ the arrow pointing towards the north pole and $|d\rangle$ will be the one pointing towards the south pole. Furthermore, we can choose the state $|u\rangle$ to be the direction along the equator pointing towards the prime meridian, and the state $|-\rangle$ would be the pointing directly opposite that. Any direction lying on the equator would be a superposition state with the same probability of being measured in $|u\rangle$ or $|d\rangle$ but distinct directions will involve different complex numbers.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">One other thing to note from picturing qubit states as forming a sphere is that oppositely directed points, such as $|u\rangle$ and $|d\rangle$, or $|+\rangle$ and $|-\rangle$ , correspond to states that can be perfectly distinguished by measurement. What this means is that if we are promised that the electron has been prepared with spin state $|u\rangle$ or $|d\rangle$, then we can perform the measurement "$|u\rangle$ or $|d\rangle$" to determine which state it is. Formally, we would say the 2 states are orthogonal and any such pair of states define a projective measurement.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">What happens if we have an electron in state $|u\rangle$ and perform a "$|+\rangle$ or $|-\rangle$" measurement on it? To see what we get, observe that we can write </span><span style="font-family: "georgia" , "times new roman" , serif;">$ |u\rangle = \frac{1}{\sqrt{2}} \left( |+\rangle + |-\rangle \right) $ </span><span style="font-family: "georgia" , "times new roman" , serif;">so a </span><span style="font-family: "georgia" , "times new roman" , serif;">"$|+\rangle$ or $|-\rangle$" measurement</span><span style="font-family: "georgia" , "times new roman" , serif;"> will yield $|+\rangle$ or $|-\rangle$ with the same probability. We see that the 2 pairs of states $A = |u\rangle,|d\rangle$ and $B = |+\rangle,|-\rangle$ lead to a mirror situation of equal probabilities if we prepare a state from one pair and perform a measurement from the other pair. </span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">This means that we have an uncertainty principle between the properties </span><span style="font-family: "georgia" , "times new roman" , serif;">"$|u\rangle$ or $|d\rangle$" and </span><span style="font-family: "georgia" , "times new roman" , serif;">"$|+\rangle$ or $|-\rangle$", since knowing we have a state that belongs to pair $A$ means are totally uncertain of the outcome when we measure using the pair $B$, and vice-versa.</span><span style="font-family: "georgia" , "times new roman" , serif;"> This is in fact a qubit version of the well-known uncertainty principle often mentioned with regard to position and momentum of a quantum particle. </span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">Is it possible to prepare electron spins that behave like a fair coin? The answer is yes and this requires we prepare a probabilistic mixture instead of a superposition. For instance, we can have a device that randomly prepares electrons with spin up or down with equal probability, in which case the state is given by $M = \frac{1}{2} \left( |u\rangle\langle u| + |d \rangle \langle d| \right)$, using a notation similar to what is often used by physicists. </span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">At this point, it is worth mentioning that</span><span style="font-family: "georgia" , "times new roman" , serif;"> $|u\rangle\langle u|$ describes the same state as $|u\rangle$, but mathematically $|u\rangle$ would be expressed by a column vector while $|u\rangle\langle u|$ would be the corresponding matrix given by "multiplying" the column vector $|u\rangle$ with the row vector $\langle u|$. Also, it might be useful to note that terms that involve vectors for the same state, such as $|u\rangle\langle u|$ or $|d\rangle\langle d|$ correspond to diagonal entries of a matrix.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">The state $M$ is different from the state $|+\rangle $ and one simple way to show this is to compare matrices. Working it out, we get $ |+\rangle\langle +| = \frac{1}{2} \left( |u \rangle \langle u| + |d \rangle \langle d| + |u \rangle \langle d| + |d \rangle \langle u| \right) $</span><span style="font-family: "georgia" , "times new roman" , serif;">. </span><span style="font-family: "georgia" , "times new roman" , serif;">So we get cross-terms or off-diagonal entries for the matrix representing state $|+\rangle$ that we don't have with the state $M$. </span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">Going back to the picture of qubit states in terms of a sphere, mixtures such as $M$ would be shown as arrow directed toward points in the interior of the sphere. The state $M$ itself is actually just be just a point, the center of the sphere. In fact, it is the unique state representing an equal mixture of any pair of orthogonal states, that is, oppositely directed points on the sphere. For example, we could have written $M = \frac{1}{2} \left( |+\rangle\langle +| + |-\rangle \langle -| \right)$.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">This also tells us how we can differentiate between the states $M$ and $|+\rangle $. For any measurement we choose, $M$ will always give equal probabilities for the two states for that measurement. On the other hand, $|+\rangle$ will give different probabilities for different measurements, like for </span><span style="font-family: "georgia" , "times new roman" , serif;">"$</span><span style="font-family: "georgia" , "times new roman" , serif;">|u\rangle$</span><span style="font-family: "georgia" , "times new roman" , serif;"> or $</span><span style="font-family: "georgia" , "times new roman" , serif;">|d\rangle$</span><span style="font-family: "georgia" , "times new roman" , serif;">" we get probability 1/2 for either state but for </span><span style="font-family: "georgia" , "times new roman" , serif;">"</span><span style="font-family: "georgia" , "times new roman" , serif;">$|+\rangle$ or $|-\rangle$" it will always come out $|+\rangle$.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">The main lesson here is that quantum superpositions allow us to generate probabilities with qubits that would not be possible with bits, and this follows closely from the fact for a qubit, the set of all states form a ball (if mixtures are included) while for a random bit, we only have a number that tells you the probability of a zero (since if $\Pr[0] = x$, then $\Pr[1] = 1-x$).</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;"><br /></span>Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-10721132453960486332014-02-01T06:14:00.000-08:002014-03-03T12:24:27.512-08:00Energy transfer in light-harvesting pigments exhibits non-classical effects<div class="separator" style="clear: both; text-align: center;">
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<span style="font-family: Georgia, Times New Roman, serif;">In <a href="http://quantumgazette.blogspot.ca/2012/04/photosynthesis-is-partly-quantum.html">a previous post</a>, we described the coherent transfer of energy from sunlight in photosynthetic systems. What it shows there is some indication of quantum effects playing a useful role in organic processes. What it does not show is whether there exists a different mechanism for explaining the same effect without using quantum mechanics.</span><br />
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<span style="font-family: Georgia, Times New Roman, serif;">To convince anyone that quantum effects play an important function in biological processes such as photosynthesis, we must show that a classical explanation is not sufficient to account for the efficient transport of energy. In a paper by O'Reilly and Olaya-Castro, they demonstrate this using methods in the quantum theory of light. </span><br />
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<span style="font-family: Georgia, Times New Roman, serif;">Light is a form of electromagnetic radiation, where the term is reserved mainly for radiation that is visible to our eyes. It is composed of electromagnetic waves that vibrate at different frequencies, which we perceive as different shades of colors. Much of how light waves behave can be explained using a classical theory of waves, which describes how waves can be combined to produce various patterns of interference.</span><br />
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<span style="font-family: Georgia, Times New Roman, serif;">Quantum mechanics refines the classical description of light mainly by saying that the light waves are composed of particles called photons. In this way, energy lost or gained through electromagnetic radiation must come in lumps, like money comes in units of pennies.</span><br />
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<span style="font-family: Georgia, Times New Roman, serif;">This lends itself to a way of describing the state of light in terms of modes that can be populated by different numbers of photons, quite similar to having several boxes that hold different number of balls. (However, photons are quantum particles, the superposition principle allows them to be found in many different modes at the same time, something impossible with classical balls in boxes.)</span><br />
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<span style="font-family: Georgia, Times New Roman, serif;">Even if we think of light as particles, it should be possible to recover the simpler classical light wave when we consider a large number of photons. This happens for a single mode of light when the number of photons measured in it follows a Poisson distribution (the average number is the same as its variance), it means light in that mode is in what is called a coherent state. Lasers are a good example of photons in a coherent state. (The Poisson distribution for the number of photons describes the probability of finding a certain number of photons in a given light mode when the average photon count is known and the counts are independent of each other.)</span><br />
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<span style="font-family: Georgia, Times New Roman, serif;">Light is said to be non-classical if its state can not be written as a proper mixture of coherent states. Mathematically, if A, B, C represent coherent states and suppose that S = a A + b B + c C where S represents the state of light. Then S is a proper mixture of A, B, and C if the numbers a, b, and c are bigger than or equal to zero.</span><br />
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<span style="font-family: Georgia, Times New Roman, serif;">In pretty much the same way we describe light in terms of photons, we can describe vibrational motion in terms of quantum particles called phonons. In photosynthesis, phonons play a significant role in the coherent transfer of energy between pigments in a light harvesting system.</span><br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgGyMkekAtp6CmvSiExrt7Owxua0HvvhTeWR2DoVB3Vg7Kk9Vpk7TvuvkMkhe2oCHYxqz8QpsGBEwVqsYWOP9V8mOYPtfDRB4T3o92bzJAYTLk27lwpOjVarxVUlFlfVNx9GIfyVcusf4mE/s1600/vibrationAssistedTransfer.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgGyMkekAtp6CmvSiExrt7Owxua0HvvhTeWR2DoVB3Vg7Kk9Vpk7TvuvkMkhe2oCHYxqz8QpsGBEwVqsYWOP9V8mOYPtfDRB4T3o92bzJAYTLk27lwpOjVarxVUlFlfVNx9GIfyVcusf4mE/s1600/vibrationAssistedTransfer.gif" height="480" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">A simple model of energy transfer between photosynthetic pigments (blue ovals) mediated by a particle of vibration or phonon (yellow ball) in a low-energy thermal bath of proteins.</td></tr>
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<span style="font-family: Georgia, Times New Roman, serif;">In what is called vibration-assisted electronic energy transfer, the energy received from sunlight is transmitted from one pigment to another through modes of vibrations with the right amount of energy. As shown above, consider two neighboring chlorophyll pigments (blue ovals) with compatible energy levels, with vibrational motions that weakly interact in a protein environment acting as a room-temperature thermal bath. We start with the top left pigment carrying an excitation, represented by the red ball on the top energy level. This excitation is transferred to the bottom right pigment through a phonon, which helps the excitation bridge the gap between pigments. For the transfer to work, the phonon must have energy levels that match that of the pigments. We actually know of phycoerythrin pigments in algae and chlorophyll pigments in spinach that have roughly this kind of energy levels.</span><br />
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<span style="font-family: Georgia, Times New Roman, serif;">The Mandel Q-parameter is a measure of non-classical behavior and is given by Q = V/M - M - 1, where </span><span style="font-family: Georgia, 'Times New Roman', serif;">M is the mean number of phonons and V is the variance. If Q = 0, then the state of the phonons can be expressed as a mixture of coherent states, which means the phonons behave like a classical wave. In this way, non-classical behavior is indicated by Q < 0.</span><br />
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<span style="font-family: Georgia, Times New Roman, serif;">A numerical simulation of the dynamics between the vibrations and the excitation in pigments in algae with parameters matched to biological values show that for some time as the transfer occurs, the Q value dips to the negative region, indicating that non-classical effects are taking place. </span><span style="font-family: Georgia, 'Times New Roman', serif;">We note that the data involves short time scales (picoseconds)</span><span style="font-family: Georgia, 'Times New Roman', serif;"> since contact with a room temperature thermal bath leads to classical behavior when the interaction takes longer.</span><br />
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<span style="font-family: Georgia, Times New Roman, serif;">The observations made here apply to a vibrational mode for a pair of pigment molecules found in a majority of light harvesting structures. To rigorously assess the efficiency achieved with non-classical effects will require studying the dynamics of energy transfer across the whole structure, which involves interaction with many vibrational modes.</span><br />
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<span style="font-family: Georgia, Times New Roman, serif;">E. J. O'Reilly, A. Olaya-Castro, "<a href="http://www.nature.com/ncomms/2014/140109/ncomms4012/full/ncomms4012.html">Non-classicality of the molecular vibrations assisting exciton energy transfer at room temperature</a>," Nature Communications 5:3012 (2014).</span>Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-83848502231443874812012-04-20T17:11:00.000-07:002014-03-03T12:24:03.981-08:00Information and entanglement in DNA<div style="font-family: Georgia,"Times New Roman",serif; text-align: justify;">
Deoxyribonucleic acid or DNA is the biological molecule that carries genetic information in all known organisms except for certain viruses. It is a nucleic acid consisting of two long chains, or strands, of basic units called nucleotides. The two strands are arranged in a twisted ladder structure known as a double helix, and is shown below.</div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Each nucleotide is composed of a nitrogen-containing base attached to a sugar phosphate molecule. (The sugar in DNA is deoxyribose, which is a sugar with five carbon molecules, hence the name.)</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">There are four types of DNA bases: adenine(A), guanine (G), cytosine (C), and thymine (T). It is the sequence of these four bases that carry genetic information along segments of DNA called genes. That is, the 'language' of genes uses an alphabet with these four letters.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">Genes encode information for creating proteins responsible for various biological traits, and are passed onto offspring. (Those genes with hereditary variants are more properly referred to as alleles.) The main purpose of DNA is long-term storage of this genetic information. </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">The bases found in the two strands of DNA come together to form base pairs. The pairings go this way: A forms a base pair with T, while C forms a base pair with G. </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">For example, if one strand carries the sequence of bases </span><span style="font-family: Georgia, 'Times New Roman', serif;"><b>ATCGGACAGCAGTC</b>,<b> </b></span><span style="font-family: Georgia, 'Times New Roman', serif;">then the other strand will carry the sequence </span><span style="font-family: Georgia, 'Times New Roman', serif;"><b>TAGCCTGTCGTCAC</b>, in accordance</span><span style="font-family: Georgia, 'Times New Roman', serif;"> to the base pairing rule.</span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjWRUOdgj230IO43zzqWqGCX8SzxB-3xLZ6AbYmCakRuP_gq9A8_jHPqanrTbl9tJWh3XL42WAyTLLvkkKTNy3w-TVAqJzGSZIGxEgBlC6WV1LNK6FZZHsWo4MSCL0l0K4K9OiyE3_PX5-V/s1600/entangleDna0.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjWRUOdgj230IO43zzqWqGCX8SzxB-3xLZ6AbYmCakRuP_gq9A8_jHPqanrTbl9tJWh3XL42WAyTLLvkkKTNy3w-TVAqJzGSZIGxEgBlC6WV1LNK6FZZHsWo4MSCL0l0K4K9OiyE3_PX5-V/s640/entangleDna0.jpg" height="480" width="640" /></a></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">The double helix structure of DNA is held together primarily by two kinds of attractive forces. One is the force that creates the base pairs called hydrogen bonding. The other one is the base-stacking interaction between bases along the backbone of the double helix called the van der Waals interaction. </span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">Recent studies on energies involved in base-stacking suggest the interaction between neighboring bases along a single strand of DNA can be partly explained by the presence of quantum entanglement between the electron clouds of these adjacent bases. </span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">As was discussed in <a href="http://quantumgazette.blogspot.ca/2012/04/is-quantum-mechanics-relevant-for.html">an earlier post</a>, quantum effects in a biological system are possible as long as the ratio of the important interaction energies J to thermal energies k due to noise from the environment is large enough. For DNA, a rough estimate of J/k is about 20.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">We demonstrate that entanglement exists by picturing a strand of DNA as a chain of oscillating dipoles. A dipole is just a separation of negative and positive charges. Looking at the diagram below, imagine the base to be made up of a positively charged molecule surrounded by a negatively charged electron cloud (oval). </span><br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiHMD3rQ1tul_rGyH6AhR6lxL-vcVjonF3PcVRfsK1e_tshy7EuxcqjAk55xXMLt9SYlcFqoAUP01pjpepMP60XQOHSxcN-zv0JOsSxq_DtfhMaPv0qquJJ2Wm_vLIgTnkzHG5BtLjS5v7G/s1600/entangleDna1.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiHMD3rQ1tul_rGyH6AhR6lxL-vcVjonF3PcVRfsK1e_tshy7EuxcqjAk55xXMLt9SYlcFqoAUP01pjpepMP60XQOHSxcN-zv0JOsSxq_DtfhMaPv0qquJJ2Wm_vLIgTnkzHG5BtLjS5v7G/s640/entangleDna1.jpg" height="480" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">The dipole on the left is labeled an <b>'up'</b> dipole. The one on the right is a <b>'down'</b> dipole.</td></tr>
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<span style="font-family: Georgia, 'Times New Roman', serif;">On the left, most of the electron cloud is distributed below the molecule; this would be a dipole with the positive charge on top and the negative charge in the bottom (an 'up' dipole). On the right, most of the electron cloud is distributed above the molecule; this would be a dipole with the charges reversed (a 'down' dipole). An oscillating dipole means that our base switches back and forth from being one kind of dipole to the other.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">(In physics terms, the non-permanent dipoles are modeled as harmonic oscillators and neighboring dipoles are made to interact via an dipole-dipole interaction called London dispersion force.)</span><br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdE0pm3nXOdl3QQLhiRyKDOBfzTJIJLJi3t6guP6qOmoiBXI-rhKuukhrDLa2P0HucIM2acQG0Uw00B6RoHdttHK5bYkc54QhNXzuxm5DiXsBdx7KnS1fdlRQz_yc6ec2yGAAfM3DtNRn0/s1600/entangleDna2.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdE0pm3nXOdl3QQLhiRyKDOBfzTJIJLJi3t6guP6qOmoiBXI-rhKuukhrDLa2P0HucIM2acQG0Uw00B6RoHdttHK5bYkc54QhNXzuxm5DiXsBdx7KnS1fdlRQz_yc6ec2yGAAfM3DtNRn0/s640/entangleDna2.jpg" height="480" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Aside from the interaction between dipoles within a single strand, we also have dipole-dipole interactions between dipoles in a base pair, since these are also neighbors.</td></tr>
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<span style="font-family: Georgia, 'Times New Roman', serif;">We can calculate the entanglement between electron clouds by checking correlations between adjacent dipoles </span><span style="font-family: Georgia, 'Times New Roman', serif;">in the oscillating dipole model</span><span style="font-family: Georgia, 'Times New Roman', serif;"> </span><span style="font-family: Georgia, 'Times New Roman', serif;">. </span><span style="font-family: Georgia, 'Times New Roman', serif;">The numerical results indicate that the amount of entanglement can be directly related to the binding energies in DNA.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">Binding energy refers to energy needed to break apart a composite system into its basic parts. For example, atomic binding energy is required to separate an atom into its constituent nucleus and electrons. The nucleus itself can be split into protons and neutrons, requiring nuclear binding energy.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">The two binding energies in DNA that indicate entanglement are the binding energy for adjacent dipoles in each strand, and the binding energy for dipoles found in base pairs. </span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">Furthermore, we can also identify the nature of the entanglement. For dipoles in a single strand, they tend to be correlated--that is, 'up' matches with 'up', 'down' matches with 'down'. For dipoles in a base pair, they are more likely to be anti-correlated; in this case, 'up' pairs with 'down'.</span><br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiPTKp6h3dGEfbt62qd7kGmQAFrUNmFoInuLXdZzv1-BFBmWZgymEGSGyndpB20RgyxkMAbQOf24QrqLsGJyIxoDXhWPsMGMDLPtVN4EJQbT_5HdNanXIC-ljl77-4IrLRdrel6wb6kYknK/s1600/entangleDna3.JPG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiPTKp6h3dGEfbt62qd7kGmQAFrUNmFoInuLXdZzv1-BFBmWZgymEGSGyndpB20RgyxkMAbQOf24QrqLsGJyIxoDXhWPsMGMDLPtVN4EJQbT_5HdNanXIC-ljl77-4IrLRdrel6wb6kYknK/s640/entangleDna3.JPG" height="393" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Neg is for negativity, a quantity that measures entanglement. Entanglement exists if Neg > 0. The distance between neighboring bases is r. The important lines are the top three ones (green, blue, red). S<span style="font-size: xx-small;">1</span> and S<span style="font-size: xx-small;">2</span> refer to correlation and anti-correlation. The colors refer to different trapping frequencies, which can be compared to experimental data.</td></tr>
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<span style="font-family: Georgia, 'Times New Roman', serif;">What exactly is the implication of this entanglement? First, it must be said that the entanglement does not change the information encoded in the DNA bases. The bases are either A, C, T, or G.There are no superpositions like A+T or C+G so there is nothing quantum about how genetic information is stored. </span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">The quantum part comes from how the information is processed. </span><span style="font-family: Georgia, 'Times New Roman', serif;">Since the bases have entangled components (namely their electron clouds), it is strictly not correct to imagine them to be separate objects. </span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">In fact, the entanglement tells us that any single base contains some quantum information about its neighbors. </span><span style="font-family: Georgia, 'Times New Roman', serif;">This means that understanding the specific biochemistry of DNA transcription and replication might involve a picture where information is transmitted through quantum channels. </span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">For instance, in replication, as one base is being copied, it already possesses some information about the next base. This means that the DNA polymerase (the DNA-copying enzyme) can partially anticipate what comes next and it may be possible for it to adjust according to this advanced information.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">Of course, it must be said that our simple model above makes no quantitative claims. DNA is simply too complicated for our oscillating dipoles to capture all the necessary details. What is established is a qualitative picture of how the energy level structure DNA can be connected to some amount of quantum entanglement it possesses. At the moment, it is considered highly speculative, but nonetheless, it is an interesting sort of speculation.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;"><b>References:</b></span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">E. Rieper, J. Anders, V. Vedral, "<a href="http://arxiv.org/abs/1006.4053">Quantum entanglement between the electron clouds of nucleic acids in DNA</a>", Preprint, arXiv:1006.4053v2 [quant-ph] (2010).</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">E. Rieper, </span><span style="font-family: Georgia, 'Times New Roman', serif;">"<a href="http://www.youtube.com/watch?v=2nqHOnVTxJE">Classical and Quantum Information in DNA</a>", Google Workshop on Quantum Biology (2010).</span><br />
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Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-71075290093328528922012-04-18T16:33:00.001-07:002018-02-16T09:57:19.645-08:00A quantum compass for birds<div style="text-align: justify;">
<span style="font-family: "georgia" , "times new roman" , serif;">During migration, birds like the European robin use the earth's natural magnetic field to know which direction to go. Like with a compass, such migratory birds can sense if the earth's magnetic field changes direction as they fly. But beyond that, they can also detect subtle differences in the strength of magnetic field. This ability to perceive magnetic fields and use them for navigation is called <i>magnetoreception</i>. It is an ability shared by other organisms such as sea turtles, fruit flies, honey bees, and even certain bacteria. Because different organisms detect magnetic fields in different ways, we focus on the case of birds, where a simple hypothesis is known.</span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">The most widely accepted explanation for magnetoreception in birds is called the <i>radical pair (RP) mechanism</i>. If it's proven correct, it demonstrates one of the ways in which a living system exploits quantum effects to perform certain biological tasks.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">So what's the basic idea behind the RP mechanism? The RP refers to two molecules that are bound together, one is called a donor (D) while the other is an acceptor (A). The mechanism says that there is a pair of electrons-- one electron with the donor molecule, the other one being the electron shared with the acceptor molecule--where the combined quantum state of the electron spins will either be a singlet or a triplet. </span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">(Singlet and triplet states are distinguished by the total spin angular momentum they carry: a singlet state has total spin zero while the three triplet states have total spin one. )</span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">If there are many of these pairs of molecules, then a certain fraction of them has the electron pair in a singlet state while the rest are in a triplet state. The idea then is that birds can tell the fraction of singlet states and this tells them how strong the magnetic field is at each location. </span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">The quantum part has to do with the fact that the singlet state is an entangled quantum state, that is, the state describes both electrons together but not each electron individually. </span><span style="font-family: "georgia" , "times new roman" , serif;">This means that in order for the birds' internal compasses to work, the quantum entanglement between the electrons must persist for a time that's extraordinarily long for a noisy biochemical environment.</span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">Now for some specifics. Here we consider only the situation where we have a pair of electrons and the nucleus of one of the molecules, say the donor molecule. We also suppose that the nucleus (N<span style="font-size: x-small;">1</span>) of the donor molecule interacts only with one electron (e<span style="font-size: x-small;">1</span>). This simplified picture captures all the essential features of the bird's quantum compass.</span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">The magnetic field is sensed through the bird's eye, that is, the orientation and strength of the magnetic field is received as light particles or photons in the bird's eye. The photon hits the radical pair found in a so-far-unknown protein (believed to be a chryptochrome sensitive to blue light) in the bird's retina, the tissue in the back wall of the eye where light triggers nerve impulses needed for vision.</span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">When the donor molecule is hit by a photon, it gets excited (D*) and gives away one of its electrons to the acceptor molecule, in such a way that there is a pair of electrons shared by the donor and acceptor whose spins forms either a singlet or triplet state. Which state it chooses depends on the interaction of the electron spins (e<span style="font-size: x-small;">1</span> and e<span style="font-size: x-small;">2</span>) with the surrounding magnetic field (B) in what's called the <i>Zeeman effect</i>. To make things easy, we may assume that the electron spins always start in a singlet state.</span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">Recall that we also have the interaction between the nucleus and an electron, which is called the <i>hyperfine interaction</i>. This hyperfine interaction causes the singlet state to change to any of the triplet states and back again. But the average effect of the interaction is to orient the nucleus so that it aligns with the direction of the magnetic field. So while the magnetic field strength is obtained from the fraction of singlet states, its direction is obtained from the orientation of the nuclear spin. </span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">(More precisely, birds only know the inclination of the magnetic field, that is, the orientation of the line where the north and south poles are. However, it does not perceive which pole is north or south. In practice, a bird has to check if it's upright or upside-down relative to the ground to know where the magnetic north pole is.)</span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">How do we know the RP model makes sense? It helps to look at the numerical results shown above. </span><span style="font-family: "georgia" , "times new roman" , serif;">(</span><span style="font-family: "georgia" , "times new roman" , serif;">The picture included in the graph shows a European robin.)</span><span style="font-family: "georgia" , "times new roman" , serif;"> </span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">The number k in just refers to a decay rate that's related to the effect of having other sources of magnetic fields. The blue line is the one that uses the known value (B</span><span style="font-family: "georgia" , "times new roman" , serif; font-size: x-small;">0</span><span style="font-family: "georgia" , "times new roman" , serif;"> = 0.5 Gauss) for the strength of earth's magnetic field. The red lines introduce an extra oscillatory magnetic field on top of the earth's field. The graph illustrates that for high enough decay rate, even if there are other sources of magnetic field, the fraction of singlet states (singlet yield) can be used to know the relative field strength. </span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">It should be mentioned that the explanation derived from this model agrees well to the fair amount of experimental observations made on these birds.</span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;"><b>References:</b></span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">E. Gauger, E. Rieper, J. J. L. Morton, S. C. Benjamin, V. Vedral, </span><span style="font-family: "georgia" , "times new roman" , serif;">"<a href="http://arxiv.org/abs/0906.3725v5">Sustained Quantum Coherence and Entanglement in the Avian Compass</a>", Physical Review Letters 106, (2011) 040503.</span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;">K. Schulten, "<a href="http://pitpas1.phas.ubc.ca/varchive/StJohnsCollegeTalks/pitp_2011_04_schulten.pdf">Quantum Biology of Animal Navigation</a>", Lecture at University of British Columbia (April 2011).</span></div>
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<span style="font-family: "georgia" , "times new roman" , serif;">C. T. Rodgers and P. J. Hore, "<a href="http://www.pnas.org/content/106/2/353.full.pdf+html">Chemical magnetoreception in birds: the radical pair mechanism</a>", PNAS <b>106</b>, 2 (2009) 353-360.</span></div>
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Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-25029116936944591292012-04-18T08:02:00.000-07:002014-03-03T12:23:11.111-08:00Photosynthesis is partly a quantum process<br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">Photosynthesis is the process by which plants use energy from sunlight to convert carbon dioxide into organic compounds such as sugars. It is also the first biochemical process where the presence of quantum effects has been experimentally verified. </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Photosynthesis involves two kinds of reactions called light and dark reactions. </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Light reactions are the chemical processes where sunlight is needed. In particular, sunlight is absorbed by light-harvesting pigments such as chlorophyll, and substances within those pigments react with light and water to produce the standard energy-storage molecule ATP, oxygen and the co-enzyme NADPH. </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Dark reactions are processes that do not depend on light and part of what is known as the Calvin cycle, a series of biochemical processes which use NADPH and ATP from light reactions to turn carbon dioxide into sugar-phosphate molecules. </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">The truly quantum effects, as far as we know, only appear in light reactions.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">One of the most studied organisms for studying photosynthesis are green sulphur bacteria. In these bacteria, the basic structure of their light-harvesting apparatus is shown in the diagram below. </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Light is absorbed by an antenna region of densely packed pigments that converts light into chemical energy. This chemical energy is in the form of an excitation, which for our purposes can be thought of as an electron freed from one of the antenna molecules using energy from sunlight. (Technically speaking, the excitation is called a Frenkel exciton, which is a bound state of an electron and hole.) </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">The excitation (or electron) gets transferred to any one of three proteins in what is called the Fenna-Matthews-Olson (FMO) complex. The FMO complex is a structure composed three monomers (the 3 blue ovals) that function independently. Each FMO monomer consists of 7 major sites where chlorophyll molecules are found. These sites are arranged roughly in a circle (blue hexagons) with one in the middle, all of them surrounded by proteins (yellow bands) that act like a scaffold. </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">The excitation moves from the antenna to the FMO protein into either site 1 or 6, jumps around the other sites, and then goes into the reaction center through site 3 or 4. The reaction center is where the dark reactions occur. </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">So now we are ready to understand where quantum coherence plays a role. The thing to think about is how exactly does the excitation travel within the FMO complex? It is easy to imagine that maybe it jumps from one site to the next, for example, it might go from 1-7-2-3 or maybe 6-5-4. Which path it chooses might be random or it might depend on how the different sites are oriented and spaced at any particular time.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">However, research studies using spectroscopy have determined that the transfer rates are incredibly fast, in the order of picoseconds, which is 0.000000000001 seconds. If it jumps from site to site in random fashion, that would take too much time. What it observed is that the excitation travels in a wave-like manner. It is not as though you can pinpoint specifically as being on one site or the other; rather, it scans the paths in simultaneous fashion. </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">This is possible because the excitation is not in the state of being, for example, in site 1 or site 6 but it can be in a state that is both partly in site 1 and site 6. In quantum mechanics, this is called a superposition and it indicates quantum coherent behaviour. In fact, what the excitation does as it journeys through the FMO complex is to solve the problem of finding a most efficient path using a quantum walk. </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">To gain some idea of what a quantum walk is, we can ask the following problem: suppose we have a “drunk” electron and as the electron walks, it is equally likely to go left or right from a position we will call x = 0. Suppose a left step counts as -1 and a right step counts as +1. After a total of 16 steps taken, where do you expect the electron to be on average? </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">It might sound tough but if we are thinking of a random chance of going left and right, then on average you take the same number of steps to the left and steps to the right. That means we expect the electron to be at x = 0 after 16 steps. And that’s what a classical random walk would tell you.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Now since we have an electron, then instead of choosing to go left and right, it can also move in a superposition, where in some sense it partly moves in both directions. We can then ask, on average, where will the electron be after performing this quantum walk for 16 steps. It turns out the average will be roughly the square root of the total number of steps, so the electron will most likely be found either at x = -4 if it takes more left steps or x = +4 if it takes more right steps. </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Evidence for the fact that the excitation undergoes a quantum walk can also be obtained from analyzing the amount of quantum entanglement in the 7 chlorophyll sites in the FMO complex. In this case, the coherent transport of the electron through the FMO complex and the entanglement between chlorophyll states represent the same quantum phenomenon.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Quantum entanglement describes how parts of a quantum system can have properties that are strongly correlated, more so that what you would normally expect. A bit more precisely, entanglement has to do with the “state” you use to describe a system. It is a statement about how independent various parts of a system are from each other. If two systems are entangled then there is a state that provides good description of both systems together but a poor description of each system separately. </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Entanglement should strike you as quite weird: how can you know two systems well but not know them well individually? But we know such systems exist. There is even a standard procedure for creating entangled photons in the lab.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">This suggests that the chlorophyll sites do not function independent of each other. It means that the FMO complex is more accurately described by an aggregate state rather than by 7 individual states. It has been shown in numerical models that entanglement between certain pairs of sites persists as long as the excitation has not reached the reaction center. The graph below shows the result for the amount of entanglement (concurrence) between some pairs of sites. For instance, the blue line corresponds to entanglement between sites 1 and 2. </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Why is presence of quantum coherence important in the study of photosynthetic systems? Since quantum coherence has been observed in a photosynthetic role, this might lead us to a further understanding of why pigments and proteins in the light-harvesting units of plants and bacteria are organized in the way they are. As is always the question with proteins, it might elucidate the relationship between structure and function in these pigment-protein complexes. In terms of physics, it opens avenues for research in quantum biology and perhaps organic systems can tell us more about how to use quantum effects in ways we have not yet imagined.</span></div>
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<b><span style="font-family: Georgia, 'Times New Roman', serif;">References:</span></b><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">G. S. Engel, T. R. Calhoun, E. L. Read,T.-K. Ahn, T. Mančal, Y.-C. Cheng, R. E. Blankenship, and G. R. Fleming, "<a href="http://www.physics.gla.ac.uk/~dtngo/Article/Nature_446_782_2007.pdf">Evidence for wavelike energy transfer through quantum coherence in photosynthetic systems</a>", Nature <b>446</b>, (2007) 782-786.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">M. Sarovar, A. Ishizaki, G. R. Fleming, and K. B. Whaley, "<a href="http://arxiv.org/abs/0905.3787v2">Quantum entanglement in photosynthetic light harvesting complexes</a>", Nature Physics <b>6</b> (2010) 462.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">M. Mohseni, P. Rebentrost, S. Lloyd, A. Aspuru-Guzik, "<a href="http://arxiv.org/abs/0805.2741">Environment-Assisted Quantum Walks in Photosynthetic Energy Transfer</a>", Journal of Chemical Physics <b>129</b> (2008) 174106.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">M. Sarovar, "<a href="http://www.youtube.com/watch?v=CnHM-PyN0gg">Quantum Mechanics of Photosynthetic Light Harvesting Machinery</a>", Google Workshop on Quantum Biology (2010).</span><br />
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Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0tag:blogger.com,1999:blog-1593607942505794705.post-62549016830783234062012-04-18T08:00:00.000-07:002014-03-03T12:22:06.854-08:00The quantum physics of smell<div style="text-align: justify;">
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<span style="font-family: Georgia, 'Times New Roman', serif;">When I first heard about how quantum effects might be involved in the process of smelling, I was immediately fascinated by it. So much so that I searched for the original paper making such a proposal and studied what role quantum mechanics plays in the nose's ability to recognize odors. This entry is an attempt to explain what I've learned, in the simplest possible terms that I can make it.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">To start, it is worth mentioning that much is already known about the smelling process. The 2004 Nobel Prize in Physiology or Medicine was awarded to Richard Axel and Linda Buck for their discoveries on odor receptors and how the signals from these receptors are amplified and processed. The basic idea is that odor molecules, or odorants, attach onto receptors that match their size and shape, in what is called a lock-and-key mechanism. Just as a lock can only be opened with a key of the right shape, an odor receptor is activated only if the right odor molecules comes along and 'plugs' it.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">Despite the nice and simple picture it provides for how odors are recognized, the lock-and-key mechanism does have its limitations. One big problem is that it does not perfectly match with observations: some molecules of widely different structures carry similar odors (e.g., borane, thiols) while some molecules with just a few atoms changed or rearranged can have totally different smell characteristics (e.g., enantiomers).</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">In 1996, the biophysicist Luca Turin proposed that ultimately, the odor receptors respond not to the shape of molecules but to their molecular vibrations. He provided a detailed and plausible mechanism for how this happens via a quantum effect known as inelastic phonon-assisted electron tunneling. Before I get to that, I will try to explain quantum tunneling first.</span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi9ysvBy5C-iXhDxyC8sA6r3Gx7uNKKozQHxzoFISkoG8tHCKnUYxQtwW4XBAWElIFOsWUht3t-c78rVjzRVIMIXXh3F6CPyf5CBRMR_LKPHldR0YfGYvmIJdz1Jn0Mu_GTqsz3gePM-gdk/s1600/smell1.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi9ysvBy5C-iXhDxyC8sA6r3Gx7uNKKozQHxzoFISkoG8tHCKnUYxQtwW4XBAWElIFOsWUht3t-c78rVjzRVIMIXXh3F6CPyf5CBRMR_LKPHldR0YfGYvmIJdz1Jn0Mu_GTqsz3gePM-gdk/s400/smell1.JPG" height="292" width="400" /></a></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">One can think of the tunneling process using the picture shown above. Imagine the yellow ball is at point A and it wishes to go to point B. However, there is a barrier that blocks its path. The question is, how can it get to the other side of the wall?</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">If the yellow ball is something like a tennis ball, that is, a classical object, then we know from experience that it can never go straight through the wall--it will just bounce off it. The only way to get it beyond the barrier is to provide it with enough energy so that it can jump over the wall and land on the other side.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Now if we imagine the yellow ball to be something like an electron, that is, a quantum object. Then something truly bizarre happens. Certainly, we can still make the electron scale the wall by giving it enough energy to jump the barrier. But even without any extra energy, there is a very small but non-vanishing chance that it will shoot past the barrier and appear on the other side, at point B. This is because quantum mechanics tells us that the probability of finding an electron in some location is given by a wave function ψ. As indicated in the picture, this wave function has a much smaller amplitudes at point B than at point A (if the ball starts at point A) but it is nonetheless not zero. This ability of quantum objects to penetrate barriers this way is what's called the quantum tunneling effect. This effect seems to play a key role in explaining how the smell of odorants are related to their molecular vibrations.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">You may ask, why is the motion of quantum objects governed by this so-called wave function? Quite frankly, we don't exactly know why. We just know that our equations in quantum physics say this is so, and we know it works because it has been observed and verified in experiments many times. For example, the decay of radioactive material such as uranium is, in fact, a quantum tunneling process.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Finally, we show what it means to have inelastic quantum tunneling. In regular cases, when an electron shoots past the barrier and appears on the other side, it retains its energy. This is what's called elastic tunneling. Now it may be the case that the electron at point A loses some energy while crossing the barrier and ends up with a lower energy at point B. This is called inelastic tunneling. Because energy must be conserved, the lost energy must be emitted somewhere in the barrier and in many situations, it does so in the form of vibrations, which in particle form are called phonons. Such a process is called phonon-assisted tunneling, since it needs a phonon to take the 'garbage' energy out of the tunneling electron.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">The crucial point here is that a phonon with precise energy is essential in completing the tunneling process: if the barrier doesn't permit vibrations with the required energy then the electron can't emit its excess energy and won't be able to tunnel through. This would be the case if region of point B is such that it can't hold electrons that are too energetic.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">This is why this scenario has been referred to as the swipe card model--the electron can't access point B from point A without a valid swipe card, which in this case is a phonon with just the right amount of energy. </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">(The equation in the diagram just says that the required energy V of the phonon is equal to the difference between the energy of the electron at points A and B--no more, no less.) </span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">If you're wondering what the heck a particle of vibration is, don't worry about it. Quantum physics is just weird that way--anything that can carry energy can be treated as a particle sometimes. In fact, that's related to the original meaning of the word 'quantum': a smallest discrete unit. Phonons just indicate that even vibrational energies come in indivisible chunks.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Now we are ready to explain how the tunneling effect might be involved in smelling. A diagram of an odor receptor binding site is shown above. The barrier in this case is actually the empty space within the binding site, the place where odor molecules of the correct shape and size fit into. This shows that the lock-and-key model stills plays a role in the process. Like in regular tunneling scenarios, we have two opposite sides for this barrier: a donor and an acceptor region of the receptor protein. The exact biological nature of the donor and acceptor is presently unknown.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">A source of external electrons is needed (only free, excess electrons get teleported) and this is speculated to be some soluble electron donor in the cell fluid, perhaps a substance called NADPH (nicotinamide adenine dinucleotide phosphate). An excess electron is obtained from such a source and goes into the donor region, where it sits with energy E[D].</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">If the binding site has no odorant, the electron won't be able to tunnel through for the same reasons mentioned above: it has to be able to get rid of its excess energy by creating vibrations of the right frequency. The electron can do this if a suitable odorant is inserted in the binding site. (By suitable, we mean odorants with the right vibrational energy). If that is the case, the electron sweeps through the odor molecule, releases its excess energy V = E[D] - E[A] in the form of a phonon, and moves onto the acceptor with some lower energy E[A]. The presence of the electron in the acceptor then triggers the release of the G-protein, which initiates further signals that are eventually transmitted to the brain and registers a smell.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Again, just to reiterate: if the odorant has the wrong vibrational energies then the electron won't be able to produce a phonon with an energy that matches the energy difference between the donor and acceptor, in which case, the access to tunneling is denied.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">For some extra technical detail, the G-protein (guanine nucleotide-binding protein, which plays a major role in transmitting chemical signals beyond cells) is connected to acceptor by a disulphide bridge (R-S-S-G). Because of the incoming electron, the disulphide bridge is chemically reduced (R-S-S-G goes to R-S-H + H-S-G), freeing up the G-protein for further transduction activity.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;"><b>References: </b></span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">L. Turin, <a href="http://www.blogger.com/goog_1473237349">"</a></span><span style="font-family: Georgia, 'Times New Roman', serif;"><a href="http://chemse.oxfordjournals.org/content/21/6/773.full.pdf+html">A Spectroscopic Mechanism for Primary Olfactory Reception</a></span><span style="font-family: Georgia, 'Times New Roman', serif;">", Chemical Senses 21 (1996) 773-91. </span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">M. I. Francoa, L. Turina, A. Mershin, and E. M. C. Skoulakisa, "<a href="http://www.pnas.org/content/early/2011/02/08/1012293108.full.pdf+html">Molecular vibration-sensing component in Drosophila melanogaster olfaction</a>", PNAS 108 (2011) 3797–3802.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">J. C. Brookes, F. Hartoutsiou, A. P. Horsfield, and A. M. Stoneham, "<a href="http://arxiv.org/abs/physics/0611205">Could Humans Recognize Odor by Phonon Assisted Tunneling?"</a>, Physical Review Letters 98, (2007) 038101.</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">L. Turin, "<a href="http://www.youtube.com/watch?v=yzOcvINn8Iw">The science of scent</a>", TED Talks (2005). </span><br />
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Gelohttp://www.blogger.com/profile/08753321753026213723noreply@blogger.com0