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If, on the one hand, I were to paraphrase Gleason's theorem, it would loosely state that

if one can assign a truth value $p_k$ to each basis vector $\vec{u}_k$ such that $\sum_k p_k = 1$, then that assignment can only be produced by Born's rule using a density matrix $\hat{\rho}$.

while, on the other hand, Kochen-Specker states that

on the (3D) sphere, it is impossible to assign a truth value to each point among triples of points arising from orthogonal bases in such a way that those truth values behave as probabilities (namely, add up to one).

Based on the above, am I therefore to conclude that the link between Gleason and Kochen-Specker is that Kochen-Specker can only be violated by the Born's rule, i.e., by quantum mechanical assignments of truth values?

Please note: I don't have enough literacy in advanced algebra, so please keep the discussion intuitive and accessible to, say, a senior undergraduate, preferably starting from my (hopefully correct?) understanding of the theorems as I've stated them above. (This answer, for example, is currently beyond my grasp.)

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  • A direct consequence of Gleason's theorem is that when the Hilbert space $H$ has dimension at least three then there does not exist a value (aka frame) function $v$ that assigns only the values $0$ or $1$ to projections: $P:H\to H\,.$ Loosely speaking this means that it is not possible to assign true or false values to all elements of the Hilbert space that represent yes/no events (projections) in such a way that is compatible with their Boolean properties: e.g. $v(P_1+P_2)=v(P_1)+v(P_2)$ for $P_1\bot P_2\,.$

  • The Kochen-Specker theorem is similar in spirit. For simplicity I reproduce only the case $\dim(H)=4$ here: There exist $18$ projections $P_1,...,P_{18}$ from which one can form $9$ bases of $H\,,$ \begin{align} \{P_1,P_2,P_3,P_4\}\,,\dots\,,\{P_{15},P_{16},P_{17},P_{18}\} \end{align} such that it is not possible to assign the values $\{0,1\}$ to the projections so that in each of those bases exactly one projection has value $1\,.$ (Each projection occurs in exactly two of the $9$ bases.) This theorem rules out a value function that is independent of the chosen measurement basis.

  • Both theorems rule out non contextual hidden variables. I am not sure if your understanding of the Kochen-Specker theorem is correct. It deals with $\{0,1\}$-valuation, not just with summing to one.

A generalization of the Kochen-Specker theorem in $\dim(H)=3$ that looks fairly accessible you can find in

Voráček, V., Navara, M. Generalised Kochen–Specker Theorem in Three Dimensions. Found Phys 51, 67 (2021).

https://doi.org/10.1007/s10701-021-00476-3

https://link.springer.com/article/10.1007/s10701-021-00476-3

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  • $\begingroup$ From the way you phrased it, it sounds that Gleason's theorem is a no-go theorem (cf. "[...] it is not possible to assign true or false values to all elements of the Hilbert space [...]"). I'm not sure how you got to this conclusion. Isn't Gleason's theorem just the way I stated it? I can see how Kochen-Specker is a no-go theorem, but not with Gleason's. $\endgroup$
    – Tfovid
    Commented Feb 15, 2023 at 18:35
  • $\begingroup$ Gleason's theorem is originally how you stated it. In the wikipedia link I gave they apply a clever trick to draw the no-go conclusion. Essentially they argue that $v$ coming from a density matrix $\rho$ must be continuous on the unit-sphere. Since $v$ cannot assign one to both $P_1$ and $P_2$ when they are orthogonal this $v$ cannot just take the values $0$ or $1$ only. It must somewhere cross through the values in $(0,1)\,.$ $\endgroup$
    – Kurt G.
    Commented Feb 15, 2023 at 18:39
  • $\begingroup$ Sorry---this hasn't quite sunk in yet... if we agree on the premises, is it fair to conclude that the link between Gleason and Kochen-Specker is that Kochen-Specker can only be violated by the Born's rule, i.e., by quantum mechanical assignments of truth values? $\endgroup$
    – Tfovid
    Commented Mar 29, 2023 at 6:57
  • $\begingroup$ @Tfovid You are not alone. Those formalims are easier to undersand than their meanings. To me the Born rule is quite obvious or, if you want, a very natural assumption. I don't understand how it can violate any theorem. $\endgroup$
    – Kurt G.
    Commented Mar 30, 2023 at 4:05

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