# Functional relations for Kochen-Specker proofs

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Many proofs of the Kochen-Specker theorem use some form of the following argument (from Mermin's "Simple Unified Form for the major No-Hidden-Variables Theorems" )

[I]f some functional relation $$f(A,B,C,\ldots)=0$$ holds as an operator identity among the observables of a mutually commuting set, then since the results of the simultaneous measurements of $A,B,C,\ldots$ will be one of the sets $a,b,c,\ldots$ of simultaneous eigenvalues of $A,B,C,\ldots$, the results of those measurements must also satisfy $$f(a,b,c,\ldots)=0$$

Parity-type contradictions (e.g., $1=-1$ or $0=1$) are then seen to arise when $a,b,c\ldots$ are assigned values independently of the context in which they are measured. The only explicit forms of $f$ that I have seen are either (i) $A+B+C+\ldots$ or (ii) $(A)(B)(C)\ldots$ (see e.g., "Generalized Kochen-Specker Theorem" by Asher Peres, where both forms are used).

My question, then, is: are there examples of parity-type proofs where $f$ is, necessarily, not of the above forms (i) or (ii)? For example one could consider $A+(B)(C)\ldots$ etc. Ideally, I'm looking for explicit examples where $f$ is spelled out, but I would also be interested in arguments where a different kind of $f$ is implicitly used.

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retagged Mar 18, 2014
Why are you interested in the form of $f$? Why would it be relevant? Do you want to know whether it is possible a no-contextuality proof where $f$ is neither additive nor multiplicative? I don't know of any over the top of my head, but I'd beat that it is easy to construct such an example.

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Like you, I believe it's quite possible that other forms of $f$ suffice to give proofs of quantum contextuality. Most interesting would be an example wherein $f$ of form (i) or (ii) does not exhibit contextuality, but a different $f$ does. As to why that's relevant? It may help provide greater understanding of quantum contextuality, and particularly Kochen-Specker sets (which are known to have various applications).

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First, a trivial example that might anger you:

Let $A_i$ be the observables of the Mermin-Peres square, and $a_i$ their non-contextual values. Then $\prod_i A_i = -\mathbb{1}$, but $\prod_i a_i = 1$, contradiction. In this case $f$ is multiplicative. But the same contradiction can be obtained considering $\prod_i A_i+\prod_i A_i = -2\mathbb{1}$ and $\prod_i a_i+\prod_i a_i = 2$, where $f$ is neither multiplicative nor additive.

Now, a more interesting example, that I've found in a paper by Adán Cabello about inequalities for testing state-independent contextuality:

Let $$A = \begin{pmatrix} Z \otimes \mathbb{1} & \mathbb{1} \otimes Z & Z \otimes Z \\ \mathbb{1} \otimes X & X \otimes \mathbb{1} & X \otimes X \\ Z \otimes X & X \otimes Z & Y \otimes Y \end{pmatrix}$$

be the Mermim-Peres square. If one ascribes non-contextual values $a_{ij} = \pm 1$ to the observables $A_{ij}$, one can then prove that $$a_{11} a_{12} a_{13} + a_{21} a_{22} a_{23} + a_{31} a_{32} a_{33} \\+ a_{11} a_{21} a_{31} + a_{12} a_{22} a_{32} - a_{13} a_{23} a_{33} \le 4,$$ whereas in quantum mechanics $$\langle A_{11} A_{12} A_{13}\rangle + \langle A_{21} A_{22} A_{23}\rangle + \langle A_{31} A_{32} A_{33}\rangle \\+ \langle A_{11} A_{21} A_{31}\rangle + \langle A_{12} A_{22} A_{32}\rangle - \langle A_{13} A_{23} A_{33}\rangle = 6.$$ The proof of the inequality may be done simply by enumerating the $2^9$ possibilities, if you're lazy, or by playing around with the triangle inequality. In either case, we have an $f$ that's not additive nor multiplicative. Of course, in this case the contradiction takes the form of an inequality, instead of a definite value for non-contextual values.

I guess then that they used always a multiplicative or additive $f$ because it's easier to construct these kind of contradictions, based on parity arguments. But I don't think there's anything fundamental to it.

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answered Mar 28, 2012 by (270 points)
Thanks Mateus. While these examples are not purely additive or multiplicative, they are trivially related to the multiplicative proof that is well known for the Peres-Mermin square. Reading your answer, I see a deficiency in how I phrased my question. Implicitly, I was looking for a contradiction of the $a \neq b$ kind (a generalized form of a parity proof, I suppose) rather than of the $a \not < b$ kind. Allowing for violation of inequalities, it seems fairly easy to cook up examples if one has an observable $O=-\mathbb{I}$ satisfying $v(O)=1$ as we do in the PM square.

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Both? Come on, the proof of the inequality is very different from the multiplicative proof of the Mermin-Peres square. But I think that the first trivial example is a proof of principle that one can mix addition or multiplication. Of course, an interesting question is whether every "mixed" proof can be reduced to a "pure" proof. I'd bet that the answer is yes and, furthermore, that you can always map them to an additive proof.

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