Recall our usual example of a qubit represented by the spin of an electron. Typically, write the state

*|E)*of such an electron as a superposition of the spin pointing up, which we write as the state

*|u),*and spin pointing down, which we write as the state

*|d)*:

*|E) = a |u) + b |d)*,

where a and b are numbers such that

*|a|^2*is the probability of measuring the spin as up and

*|b|^2*is the probability of measuring the spin as down.

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

*|u)*and |

*d)*, respectively.

If we choose to always write states in terms of

*|u)*and |

*d),*then we may write the state

*|E)*in the shorthand form

*|E) := (a, b)*

where the symbol := means that it is understood that the first number refers to spin up and the second number to spin down.

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.

For example, a qutrit is a 3-dimensional quantum system, with distinguishable states commonly labeled as

*|0), |1),*and

*|2).*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.

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

*a |0) + b|1) + c|2) := (a, b, c).*

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

*Q*and

*R*, with states

*|Q) := (a, b, c),*

*|R) := (x, y, z).*

The inner product between states

*|Q)*and

*|R)*, is computed by multiplying corresponding numbers to each other and adding the products:

*(Q|R) = ax + by + cz.*

[If we have complex numbers, we have to take the complex conjugate of

*|Q)*before multiplying to

*|R)*. In this case

*(R|Q)*is the complex conjugate of

*(Q|R)*. If we only have real numbers, then

*(Q|R)*and

*(R|Q)*are equal.]

The significance of the inner product

*(Q|R)*is that it can be used to find the probability of measuring the state

*(Q|*when the state is initially

*|R)*. More accurately, the probability is given by

*| (Q|R) |^2*.

The inner product is actually something we have implicitly used with our discussion with qubits before. Recall the state

*|+)*, which is an equal superposition state of

*|u)*and

*|d)*. In terms of

*|u)*and

*|d)*, we would write it in shorthand form as

*|+) = (1/sqrt(2) , 1/sqrt(2) ).*

Now the probability of measuring an electron with spin in state

*|u) := (1, 0),*given that it was prepared in state

*|+),*is given by the absolute-value square of the inner product

*(u|+) = [1 x 1/sqrt(2)] + [0 x 1/sqrt(2)] = 1/sqrt(2),*

so that

*| (u|+)|^2 = 1/2*.

The main reason we introduced the inner product here is to be able to make the following statement: If

*|Q)*and

*|R)*represent two quantum states that are perfectly distinguishable, then

*|Q)*and

*|R)*are orthogonal, that is,

*(Q|R) = 0*.

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

*|u)*or

*|d)*, then you can perform a "

*|u)*or

*|d)*" 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.

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:

(1, 0, 0) |
(0, 1, 1) |
(0, 1, -1) |

(0, 1, 0) |
(1, 0, 1) |
(1, 0, -1) |

(0, 0, 1) |
(1, 1, 0) |
(1, -1, 0) |

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.

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

*|u)*is ON and

*|d)*is OFF since

*|u)*and

*|d)*are orthogonal.

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

*|0)*, then it is definitely not in state

*|1)*or state

*|2)*because these are states that are orthogonal to each other.

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

(1, 0, 0) |
(0, 1, 1) |
(0, 1, -1) |

(0, 1, 0) |
(1, 0, 1) |
(1, 0, -1) |

(0, 0, 1) |
(1, 1, 0) |
(1, -1, 0) |

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.

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).

Another valid color scheme in our game would be

(1, 0, 0) |
(0, 1, 1) |
(0, 1, -1) |

(0, 1, 0) |
(1, 0, 1) |
(1, 0, -1) |

(0, 0, 1) |
(1, 1, 0) |
(1, -1, 0) |

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".

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.

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.

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.

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.

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).

A simple example for proving the theorem involves the following 4-dimensional quantum states:

|A)
:= (0, 0, 0, 1) |
|G)
:= (0, 1, 0, 0) |
|M)
:= (1, 0, 0, 1) |

|B)
:= (1, -1, 1, -1) |
|H)
:= (1, 1, 1, 1) |
|N)
:= (1, 0, 0, -1) |

|C)
:= (0, 0, 1, 0) |
|I)
:= (-1, 1, 1, 1) |
|O)
:= (1, -1, 0 ,0) |

|D)
:= (1, -1, -1, 1) |
|J)
:= (1, 1, 0, 0) |
|P)
:= (0, 0, 1, 1) |

|E)
:= (1, 1, -1, 1) |
|K)
:= (1, 0, 1, 0) |
|Q)
:= (0, 1, 0, -1) |

|F)
:= (1, 1, 1, -1) |
|L)
:= (1, 0, -1, 0) |
|R)
:= (0, 1, -1, 0) |

which can be grouped into columns of orthogonal states like this:

|A) |
|A) |
|B) |
|B) |
|C) |
|D) |
|E) |
|E) |
|F) |

|C) |
|G) |
|D) |
|H) |
|G) |
|H) |
|F) |
|I) |
|I) |

|J) |
|K) |
|J) |
|L) |
|M) |
|N) |
|O) |
|K) |
|M) |

|O) |
|L) |
|P) |
|Q) |
|N) |
|R) |
|P) |
|Q) |
|R) |

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.

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

*|A), |B),*and

*|E)*green:

|A) |
|A) |
|B) |
|B) |
|C) |
|D) |
|E) |
|E) |
|F) |

|C) |
|G) |
|D) |
|H) |
|G) |
|H) |
|F) |
|I) |
|I) |

|J) |
|K) |
|J) |
|L) |
|M) |
|N) |
|O) |
|K) |
|M) |

|O) |
|L) |
|P) |
|Q) |
|N) |
|R) |
|P) |
|Q) |
|R) |

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. A few more attempts at coloring should convince you it can not be done.

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.

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:

Kochen and Speckers 117 states for proving their theorem with qutrits |

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

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.

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.

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. |

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.

Asher Peres was fond of saying "unperformed experiments have no results." Echoing those words, we could say "unmeasured (quantum) properties have no (definite) values."

**References:**

A. Cabello, J. M. Estebaranz, and G. G. Alcaine, "Bell-Kochen-Specker theorem: A proof with 18 vectors"

*Physics Letters A*

**212**(1996) 183.

A. Peres, "Two simple proofs of the Kochen-Specker theorem"

*Journal of Physics A: Math. and Gen*.

**24**(1991) L175.

The Kochen-Specker Theorem. Stanford Encyclopedia of Philosophy.

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