2
$\begingroup$

It appears to me that the Kochen-Specker theorem, if not Gleason’s theorem already, seals the fate of realism / value definiteness (with possibly the additional assumption of non-contextuality, depending on how one precisely defines the above terms). Unlike Bell's theorem (though like some of what is commonly considered within Bell-type inequality offshoots such as CHSH Inequality) the Kochen-Specker theorem appears to clarify the line between classical and quantum, when one finally has to accept that quantum cannot be imagined apart on the boundary (Hilbert space / measurement).

Bell's theorem in itself, on the other hand, does not appear the resolve clearly (I am speaking for myself) which of the components of the local realism is wrong if the inequality is violated (locality or realism or both). Yet, for some reason, it appears that the Bell's Theorem has garnered far more fascination among the physics community. (Q1) Is it because of the appearance of the faster-than-light communication, or this is just a historical accident, helped by the fame of the EPR paradox?

On the basis of (numerous) experimental evidence, we know that we need to deny its premises, namely, either (or both) of the locality or realism (that many physicists will argue had known always to be true, even without any experimental evidence). Logically, in itself, this may convey the idea that it is possible to have non-locality (with realism in place).

(Q2) However, since we already know (e.g., by the Kochen-Specker, etc) that the realism assumption is incorrect, then the Bell's theorem per se does not appear to resolve the validity of locality vs non-locality assumption?

In fact, it appears to me that the experimental evidence (that shows the violation of the inequality) may be safely interpreted only to confirm the invalidity of the assumption of realism. Further confusion may be introduced when one understands (or, so it seems) that the locality in the Bell's theorem or Bell-type family of statements -- at least those that I have seen -- is defined and intrinsically linked with the realism assumption so much so that it may at least be profoundly misleading to state that one can dispose of realism whilst retaining locality; it is not clear (to me) what kind of notion of locality it would be. To the extent that it is the idea of locality intrinsically linked with realism (value definiteness), then the Bell's theorem does not appear to imply anything about it (fortunately, relativity makes it clear); and to the extent that it is some other notion of locality, then it appears it could be anything, including some type of quantum locality that, when represented in our standard spacetime, are actually quite non-local (but, of course, due to absence of realism or value definiteness, it cannot be used to deterministically send signals faster than light).

Thanks.

$\endgroup$
3
  • $\begingroup$ The only component of realism that is wrong, is the assumption that a particle cannot oscillate. Consider the oscillating frequency of many coherent photons and you can classically derive any quantum phenomena. You can mathematically and physically explain what’s going on and do not need to rely on the in completeness of QM. $\endgroup$ Jan 22 at 0:19
  • $\begingroup$ I am not sure if I understand these statements correctly, however, I will try to relate them to something I gave a thought to a while ago. Concretely within the premises of the Kochen-Specker theorem, if you are referring to the possibility that, by virtue of bestowing some kind of time-varying value definiteness to spin directions (to avoid the simultaneity of the assignment because no two observables can be measured at the truly same, logical time), one can perhaps dodge the conclusion of the theorem. I do not think so, I remember you still end up with a contradiction in the assignment. $\endgroup$ Jan 22 at 1:55
  • $\begingroup$ yes an oscillation does add a time variant or more importantly (linear dependent). This is important because every particle passes through the slit with different proximities to the edges, especially when the slit is rotated. This can decide if they pass or not. This also matches cos2theta $\endgroup$ Jan 22 at 2:52

1 Answer 1

2
$\begingroup$

I am not sure to understand well the issues you raised. I am sure that some of the following comments are already included in your questions.

I stress that the KS theorem and the Bell one have a very different nature.

The KS theorem does assume part of the quantum phenomenology and theory of quantum observables.

Taking advantage of Gleason's theorem -- which assumes the orthomodular structure of the lattice of quantum observables -- it proves that, for a quantum system whose Hilbert space has finite dimension and larger than $2$, its phenomenology cannot be described in terms of a realistic non-contextual hidden variable theory. It happens provided the outcomes of compatible observables satisfy some natural functional relations.

The choice between realism or non-contextuality is made by the standard interpretation of quantum theory, if one assumes it in toto. Indeed, the outcomes of measurements of an observable in QT do not depend on the choice of any other compatible observable simultaneously measured: QT is non-contextual. On the other hand, realism is not valid in standard QT, since the values of observables are not predetermined. In principle however there could exist a realistic contextual hidden variable theory compatible with the quantum phenomenology, different from the standard interpretation of QT. The Bohm interpretation is considered such from some perspective.

Differently from the KS theorem, Bell's theorem does not use any quantum description or assumption. It proves that the measurements of some properties of a system made of two causally-separated parts must satisfy a certain inequality. It happens provided some realism and locality assumptions are satisfied.

There is no possibility to decide which assumption does not hold in case these inequalities are violated in the framework of Bell's proof.

Quantum systems violate those inequalities and, as before, the most common interpretation of quantum theory is that realism is violated and locality is safe. However, in principle there could be another hidden variable theory compatible with the observed violation which is realistic an non local or even non realistic and non local.

Personally, like most physicists I guess, I lean towards the validity of locality in spite of realism, but I don't think the debate is really over.

$\endgroup$
4
  • $\begingroup$ I understand that Bell's inequality, at least in its original form, is completely classical in nature. Therefore, as you may be implying, one may argue that it is not fair to assume the Kochen-Specker (KS) result to be complementary in any sense to it. $\endgroup$ Jan 23 at 16:21
  • $\begingroup$ Nevertheless, by appealing to only basic, "kinematic structure" of the quantum observable(s) which, for all that we can see, is (so far) utterly consistent with, and sufficient for, what we see in nature (and I think most physicists would agree with it, though some may add that it is not, strictly logically, necessary), it seems to me that the KS does provide the needed prod that allows one to consider the question of realism without getting ensnared by the notion of locality. $\endgroup$ Jan 23 at 16:22
  • $\begingroup$ Hence, once one accepts its conclusion as the additional backdrop within which to examine the Bell-type inequality, then purely by virtue of the intrinsic connection between the locality and the realism as defined within the premises of the Bell's inequality (again, at least in its original form), it appears to me that (once the idea of realism is denied) the question of locality cannot be settled by the Bell's inequality. $\endgroup$ Jan 23 at 16:24
  • $\begingroup$ Perhaps that explains why the experiments that are interpreted thru language of the Bell-type inequalities continue to produce exegeses on both sides of the "local vs non-local" debate. $\endgroup$ Jan 23 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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