I would like to understand the mathematical content of Kochen-Specker theorem. This theorem states the following:

If the dimension of a Hilbert space $\mathcal{H}$ is $>2$ then there is no valuation $\lambda:\mathcal{B}(\mathcal{H}) \to \mathbb{C}$.

Recall that a valuation is the function with the following properties:
(1) for each observable (meaning: self-adjoint operator) $A$ the value $\lambda(A)$ belongs to the spectrum of $A$
(2) if two observables $B,A$ are such that $B=h(A)$ for some Borel real valued function then $\lambda(B)=h(\lambda(A))$.

Q1: What is the significance of the condition (1) above?

Am I right saying that the reason for condition (1) is only due to the fact that if $B=h(A)$ then $h$ is defined on the spectrum of $A$ thus in order to write $h \circ \lambda(A)$ we assume that $\lambda(A) \in \sigma(A)$? This enables us to define valuation as a function which is an algebra homomorphism on each commutative subalgebra.

The second question concerns the proof: each proof which I have found assumes that $\mathcal{H}$ is finite dimensional (usually of dimension 4 or 8).

Q2: Is there a simple way to proof this theorem for the $n$-dimensional space $\mathcal{H}$ provided we have proved it for some dimension less than $n$? In particular, does this theorem follow for infinite dimensional $\mathcal{H}$ once we have proved it for dimension $4$?

I will be very grateful if someone could clarify those issues for me


1 Answer 1


The standard hypotheses for the valuation function $v: B(H)_{sa} \to \mathbb{R}$, where $B(H)_{sa}$ denotes the real linear subspace of the selfadjoint operators of $B(H)$ and $2< \dim(H) < +\infty$, are that

(i) $v(A+B) = v(A) +v(B)$ for all $A,B\in B(H)_{sa}$ with $AB=BA$;

(ii) $v(AB) = v(A)v(B)$ for all $A,B\in B(H)_{sa}$ with $AB=BA$.

The thesis of this version of the KS theorem asserts that,

Bell-Kochen-Specker Theorem

Assuming (i) and (ii) with $2< \dim(H) < +\infty$, there is no non-vanishing function $v$.

The proof for a generic $+\infty >\dim(H) = n >2$ is the following one. From (i) and (ii) one sees that $v$ restricted to the lattice of the orthogonal projectors in $H$, on the one hand assumes only values in $\{0,1\}$, on the other hand it satisfies the additivity property

(iii) $v(P_1+\cdots+P_k) = \sum_{j=1}^kv(P_k)$ if $P_aP_b=0$ if $a \neq b$.

This is in contraddiction with a topological corollary of Gleason's theorem (discovered by Bell) that asserts that the above function must be the zero function.

The said proof is not valid in the infinite dimensional case since, to take advantage of Gleason's theorem, the additivity requirment (iii) should be assumed valid also for an infinitely countable set of mutually orthogonal projectors (and the sum of them is intepreted in the strong operator topology). All that is instead automatic in the finite dimensional case.

A direct attempt to produce a contraddiction with the finite dimensional case, passing from $B(H)_{sa}$ to $B(H_0)_{sa}$ for a finite dimensional subspace $H_0\subset H$ faces a continuity problem: an valuation function $v: B(H)_{sa} \to \mathbb{R}$ necessarily restricts to a valuation function on every $B(H_0)_{sa}$, so that the finite-dimensional proof implies that this restriction vanishes. However, without a suitable continuity property, we cannot conclude that this means that $v$ vanishes everywhere on $B(H)_{sa}$.

Frankly speaking, I do not know if, in the absence of a continuity hypothesis on $v$ in some operator topology, the simplest statement of KS theorem is also valid when $\dim(H) = +\infty$.

REMARK It is not necessary that (a) $v(A) \in \sigma(A)$ nor (b) $f(v(A))= v(f(A))$. Requirment (a) is however physically meaningful, since the theorem is used to rule out some hidden variable theories (called realistic and non-contextual) that try to describe the quantum phenomenology in classical stochastic terms. There the valuation function $v=v_\lambda$, where $\lambda$ is the hidden variable, assigns values attained by the quantum observables. Since these values are the same as those of the standard formulation, it must hold $v_\lambda(A) \in \sigma(A)$.

  • 1
    $\begingroup$ Thank you very much for a great answer! Could you please give me some reference to Gleason theorem which was invoked in your argument (or for the corolllary of it)? $\endgroup$
    – truebaran
    Commented Sep 4, 2020 at 11:08
  • 1
    $\begingroup$ Chapter 5 of this book of mine springer.com/it/book/9783030183455#aboutBook discusses KS Thm as I wrote in my answer above. $\endgroup$ Commented Sep 4, 2020 at 11:10
  • $\begingroup$ One again, many thanks! $\endgroup$
    – truebaran
    Commented Sep 4, 2020 at 11:23

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