I've read the paper 'Generalizations of Kochen and Specker’s theorem and the effectiveness of Gleason’s theorem', where it says that non-contextual hidden-variable theories are ruled out by a theorem of Bell (1966), which is stated as 'There does not exist a bi-valued probability function on the rays (onedimensional subspaces) of a Hilbert space of dimension greater than 2.' There is given no explanation how exactly non-contextual hidden variable theories are ruled out by it. It is clear to me that it has to do with the fact that hidden-variable theories would lead to deterministic outcomes, i.e. bi-valued probability functions. Somewhere else (unfortunately I've forgotten the source) I've read that the reasoning is that the mapping $u \rightarrow (\rho u, u)$ is continuous on the unit sphere of the Hilbert space for any density operator $\rho$, so due to Bell's 1966 theorem it cannot be deterministic.

So my first question is how to interpret the mapping $u \rightarrow (\rho u, u)$? Is $u$ a state? How does the specific observable and a specific outcome come into play in the formula? (I guess $(\rho u, u)$ is the probability for a specific outcome of a specific observable if the system is in state $u$?)

My second question is why non-contextual hidden variable theories are ruled out, but not contextual hidden variable theories?

If anything is unclear about my question, please feel free to ask. I'm looking forward to any answer or input.

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    $\begingroup$ I think you're probably confusing Bell's Theorem with the Kochen-Specker Theorem. The latter is the one which rules out non-contextual hidden variable models if the Hilbert space dimension is 3 or greater. Bell's Theorem has a broader reach. These theorems and others are discussed at length in Leifer's excellent review paper, linked below, along with much more conceptual discussion than the paper you mentioned. Your question would be clearer if you could refer to specific content in Lefier's paper: arxiv.org/pdf/1409.1570.pdf $\endgroup$ Commented Sep 16, 2023 at 23:53
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    $\begingroup$ Related. $\endgroup$
    – Kurt G.
    Commented Sep 17, 2023 at 5:13
  • $\begingroup$ @KenWharton Thanks for your comment. I don't think I'm confusing Bell's theorem with the KS Theorem. In the paper I'm referring to, Bell's theorem is stated as 'There does not exist a bi-valued probability function on the rays (onedimensional subspaces) of a Hilbert space of dimension greater than 2.' I know that what one normally understands under Bell's theorem is something different, that's why I stated it. But thanks for your link to Lefier's paper anyways. $\endgroup$
    – Studentu
    Commented Sep 17, 2023 at 20:51
  • $\begingroup$ @KurtG. Thanks for the link. $\endgroup$
    – Studentu
    Commented Sep 17, 2023 at 20:52

1 Answer 1


The Bell-Kochen-Specker theorem is one of the various no-go theorems against the existence of any "classical" theory capable to explain better the phenomenology of quantum mechanics and restoring realism and/or determinism.

These alternative theories are based on the assumption that a deeper explanation than QM exists. The deeper description is constructed out of a set of quantum hidden variables, usually denoted by $\lambda$, whose nature is unknown.

The considered theorem assumes certain quite general hypotheses on such theories, without entering into the details, and it proves that this class of theories cannot exist.

Note. The so called Kochen Specker (without Bell!) theorem was established after the one discussed here. The one discussed here can be found in one of the first papers by Bell (1966). It was already known to at least one of the other two authors. The later KS theorem has more or less the same statement, but it has a proof of much more elementary level, based on a direct and lengthy computation on a specific system of elementary Yes/No propositions. It is apparently independent of the Gleason theorem. The theoretical significance is however identical in my view.

The hypotheses of Bell-Kochen-Specker theorem are that, the hidden variable $\lambda$ defines a map associating each observable $A$ (any bounded selfadjoit operator in the Hilbert space of the system) with its value $v_\lambda(A)$ (a real number). This is the realism hypothesis: the hidden variable fixes the true values of all obvservables, including the incompatible ones!

The other explicit hypothesis is that, restricting ourselves to any pair of compatible observables $A$ and $A'$, the above map satisfies some functional requirement. Usually additivity $$v_\lambda(A+A')= v_\lambda(A)+ v_\lambda(A')\quad \mbox{if $A$,$A'$ are compatible}$$ and multiplicativity. $$v_\lambda(AA')= v_\lambda(A) v_\lambda(A')\quad \mbox{if $A$,$A'$ are compatible}$$

The thesis is that, if the Hilbert space has finite dimension >2, this map does not exist (as a topological consequence of the Gleason theorem firstly discovered by Bell) or it is the trivial one associating everything to $0$.

I stress that $\lambda$ has noting to do, at least directly, with any quantum state that can be defined on the physical system. In principle, a quantum state should correspond to some ensemble of values $\lambda$, i.e., a more approximate description. Quantum ramdomness should be explained in terms of classical randomness similary to statistical mechanics. However all those details are irrelevant in the theorem, and here is its powerfulness as a no-go theorem.

Coming back to the thesis of the theorem, a way out is that the map $v_\lambda$ is not a function only of the observable $A$ one measures, but it also depend on the other (compatible) observables one measures simultaneously with $A$.

$$v_\lambda(A| A_1, A_2, ....)$$

$A_1,A_2,...$ is the context of $A$. Each observable $A$ simultaneously has different values depending on its context when the hidden status $\lambda$ is given.

In principle, maps of this form are not forbidden by the BKS theorem.

From this perspective, the original maps that do not depend on the context are called non-contextual.

Therefore, the BKS theorem rules out hidden variable theories which are (a) realistic (b) non-contextual, and (3) they satisfy some natural functional relations only referring to pairs of compatible observables.

ADDENDUM Sketch of proof of BKS theorem. This is not the originary proof by Bell, but it similarly uses the continuity argument due to the Gleason theorem into a more straightforward way.

THEOREM Let ${\cal H}$ be a finite dimensional real or complex Hilbert space with dimension $>2$. Let $B_{sa}({\cal H})$ the real linear space of everywhere defined selfadjoint operators on ${\cal H}$. There are no maps $v: B_{sa}({\cal H}) \to \mathbb{R}$, different of the zero map, such that

(i) $v(A+B)= v(A)+v(B)$ if $AB=BA$,

(ii) $v(AB) = v(A)v(B)$ if $AB=BA$.


Notice that orthogonal projectors are selfadjoint operators $P:{\cal H}\to {\cal H}$ such that $PP=P$. We can restrict $v$ to the space (lattice) of orthogonal projectors).

From the hypotheses of additivity and multiplicativity, taking $PP=P$ into account, it is easy to conclude that (a) $v(P) \in \{0,1\}$ for every orthogonal projector $P$ and that (b) $v_\lambda(P_1+...+P_k)= v(P_1)+...+v(P_k)$ if $P_kP_h=0$ when $h\neq k$. Furthermore, $v(I)=1$ otherwise $v$ is the trivial map $v(P)=0$ for all orthogonal projectors. The spectral theorem would imply that $v(A)=0$ for every $A\in B_{sa}({\cal H})$ and this is not permitted.

$\dim{\cal H}>2$, (a), and (b) through the Gleason theorem (a part of the proof) imply that there exists a unique mixed state, $\rho$, such that $v(P) = tr (P\rho)$ for every orthogonal projector $P\in B_{sa}({\cal H})$.

Let us restrict this map to the set $S$ of the one-dimensional orthogonal projectors (that is the rays $p= |\psi\rangle \langle \psi|$).

$$S \ni |\psi\rangle \langle \psi| \mapsto \langle \psi, \rho_\lambda\psi \rangle \in \{0,1\}$$

This map -- viewed as a map to $\mathbb{R}$ -- is continuous trivially, $S$ is connected and thus the image must be connected as well. However $\{0,1\}$ is a disconnected subset of $\mathbb{R}$ whose connected components are $\{0\}$ and $\{1\}$. Hence either

$\langle \psi, \rho_\lambda\psi \rangle =0$ for all unit vectors $\psi \in {\cal H}$


$\langle \psi, \rho_\lambda\psi \rangle =1$ for all unit vectors $\psi \in {\cal H}$.

Notice that it must be $tr(\rho) =1$ since $\rho$ represents a mixed state. However, in the first case $tr(\rho) =0$, and in the second case (for $\dim({\cal H}) >2$) $tr(\rho) >1$. In both cases $tr(\rho) \neq 1$ as instead requested by the Gleason theorem. In summary, $\rho$ -- and thus $v$ -- does not exists. QED

For some further discussion see chapter 5 of this book of mine.

  • $\begingroup$ Thank you for your answer! How does the map $\lambda: A \mapsto v_{\lambda}(A)$ take the state which the system is in into account? What's the value $v_{\lambda}(A)$, does it represent the outcome that one measures with certainty? Is this map in any way related to Born's rule? And how exactly is this map ruled out by the BKS theorem ('There does not exist a bi-valued probability function on the rays (onedim subspaces) of a Hilbert space of dim > 2')? I mean, the map $\lambda$ you mentioned is not a map on the rays (i.e. states?), but a map on the observables (which are not rays?)? $\endgroup$
    – Studentu
    Commented Sep 17, 2023 at 21:00
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    $\begingroup$ @Studentu The map $v_\lambda$ has, in principle, nothing to do with the values of $A$ obtained by the quntum theory. In fact every observable always has its own value even if when it is not possible in the QM approach (because some observable is not defined in the given quantum state). There is no relation, in principle with the Born rule. We are only assuming that there is a deeper theory than QM where a hidden a deeper state $\lambda$ fixes the values of all observables. How this state is related with quantum states is irrelevant here. $\endgroup$ Commented Sep 18, 2023 at 9:28
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    $\begingroup$ In this view, a quantum state presumably is an expectation value with respect an ensemble of hidden variables $\lambda$. Only the quantum nature of the observables (they are selfadjoint operators in a finite dimensional Hilbert space) is assumed here. $\endgroup$ Commented Sep 18, 2023 at 9:30
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    $\begingroup$ I added a sketch of proof where the connection between the map $v_\lambda$ and the Gleason theorem shows up. $\endgroup$ Commented Sep 18, 2023 at 11:22
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    $\begingroup$ @Studentu I think you have uderstood well $\endgroup$ Commented Sep 18, 2023 at 11:45

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