Property of hidden variable theories which produce the same predictions regarding outcomes of measurements as standard quantum mechanics, but which are supposed to be realistic differently than quantum mechanics. The outcomes of an observable depend on the choice of the other observables simultaneously measured.
Questions tagged [contextuality]
7 questions
16
votes
3 answers
What is contextuality?
I have been trying to get and idea of contextuality in quantum mechanics for a while but I still do not get it.
Reading Wikipedia lead on quantum contextuality,
Quantum contextuality is a feature of the phenomenology of quantum mechanics whereby…
Mauricio
- 6,886
8
votes
3 answers
Is Pilot Wave Theory contextual? How?
The Kochen Specker theorem says that hidden variable theories must be contextual. I'm not seeing anything in the definition of Bohmian mechanics that makes the hidden variable variable assignments dependent on the measurements. Bohmian mechanics…
Ryder Rude
- 6,915
3
votes
1 answer
How are Gleason's and Kochen-Specker's theorems related?
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…
Tfovid
- 1,425
1
vote
1 answer
Doubt regarding the trivial construction in the original Kochen-Specker article
I'm trying to understand a construction in the first section of the original Kochen-Specker article (S. Kochen; E. P. Specker (1967). "The problem of hidden variables in quantum mechanics". Journal of Mathematics and Mechanics. 17 (1): 59–87.…
ErrorPropagator
- 111
1
vote
1 answer
Bell-Kochen-Specker theorem impact on realistic hidden-variable theories
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…
Studentu
- 140
1
vote
1 answer
Is degeneracy of eigenvalues required for the Kochen-Specker theorem?
I'm wondering why the operators for the Kochen-Specker theorem are 3-dimensional while they only produce two eigenvalues $\{0,1\}$. Is this degeneracy always needed regardless of the dimensionality of the Hilbert space or is it an artifact in the…
Tfovid
- 1,425
1
vote
1 answer
What does it mean for two compatible observables to be a "coarse-graining" of a third?
In reading about quantum contextuality, I've encountered the statement that
if [A,B] = 0, then there exists another observable C such that the
spectral projections of A and B are a coarse-graining of those of C
and, thus, measuring C allows one to…
Tfovid
- 1,425