3
$\begingroup$

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.)

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

  • 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

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.