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I have been in discussion with folks about the last 10 mins of the MIT Lecture 2 - Quantum Mechanics course.

Theorem. Given three propositional statements (about objects), A, B, C, and given a finite set of objects (that either have these properties or don't have these properties), then the following inequality is always true

$$n(\text{A}\ \&\ \text{B}) \le n(\text{A}\ \&\ \text{C}) + n(\text{B}\ \&\ \text{not C})$$

where $n$ means to count of the number of objects that satisfy the propositional statement.

The proof of this is incredible simple and follows the basics on Propositional Axioms, and Counting Axioms (like Peanos). It can be dervied directly through modus ponens, or it can be derived intuitively with a Venn Diagram.

Is MIT correct when they say, "This theorem is false in Quantum Mechanics."?

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    $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$ Commented Nov 21 at 21:19

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A “Non-Local Hidden Variable Theory” (NLHVT) formally means a hidden variable theory that violates one of Bell’s “Local Causality” assumptions. The idea is that if a model makes all of the assumptions in Bell’s “Local Causality”, you can then prove that such a model must respect Bell inequalities, and therefore the model can’t explain the experiments. But if you explicitly violate one of the core assumptions, you could get some hidden variable account of the experiments after all.

Unfortunately, NLHVT is often interpreted as meaning that the variables themselves are non-localized (say, not even existing at some point in spacetime, but only in some abstract Hilbert space). This doesn’t solve the problem for two reasons. One, standard QM falls right into this category, so it’s not what Bell was talking about in the first place. (He was following the chain of thought started by Einstein in EPR, asking if there was a realistic account, based in space and time, which might explain the experiments.). Another problem is that once a variable isn’t localized in space and time, usual notions of “faster than light influence” cease to have any meaning. You can’t talk about how fast an influence goes between two variables if those variables aren't even associated with spacetime locations.

So, back to the options for NLHVTs, in terms of spacetime-localized variables, such as the properties you describe in your question. The problem is not having those properties “exist”, the problem is the “Local Causality” assumptions which tell us that those properties should respect certain independence conditions. For example, one of the assumptions says that past hidden variables should never be correlated with future measurement settings. So, one class of NLHVTs are retrocausal models where different future settings correspond to different spaces of possible particle properties. Another class has physical faster-than-light influences. Another class has spacetime-jumping action-at-a-distance. In general, they come in four broad categories, as defined here: https://arxiv.org/abs/1906.04313 . (Class IA, IB, IIA, and IIB.)

Finally, you raise the question of how to analyze this in a standard QM framework -- although I think you should reformulate your "is BI false in QM?" question. Recall that Bell Inequalities (BI) can only be proven for a model which respects all of the assumptions in Bell's "Local Causality". QM does not respect all those assumptions, and so -- unsurprisingly -- it makes predictions which violate BI. The issue is not whether BI is "true" or "false" -- it can be derived from some sets of assumptions but not others. Why can't it be derived from QM? Because QM breaks Local Causality in an extremely dramatic manner. For an entangled state, between measurement and preparation, QM states don't seem to even have properties localized in space and time in the first place. NLHVTs try to fix this dramatic problem, but even so, thanks to Bell, we know they have to violate some other aspect of Local Causality (such as perhaps being retrocausal).

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  • $\begingroup$ Thanks for the detailed response. Classically, Bell's inequality is theorem in logic and counting. I understand that we may wish to "change" what we mean by property, or having a property, or things, etc. Doing so elevates the discussion to a new theory, with time, measurement, etc. Things are now like functions, and properties something else. We talk about their commutability, etc. However, none of this undoes the experiamental violation of Bell's classical inequality, right? Like we all believe (I assume) the proof and theorem of Bell's inequality at a macro level, like with students, right? $\endgroup$ Commented Nov 20 at 22:55
  • $\begingroup$ If one accepts that "BI is false in QM" (again, quoting the MIT video), then I agree that one is going to want to extend what we mean by "thing", "measurement", etc. While doing so could create a well formulated axiomatic system for which there is a Bell like inequality that "goes the other way" and which is sound and consistent, this does nothing to address the classical Bell's inequality, right? It is still violated. I am still unable to count things that have properties, a position that is incredibly unreasonable to believe and hold at a macro level, right? $\endgroup$ Commented Nov 20 at 23:00
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    $\begingroup$ All theorems have assumptions. You can't prove anything without any assumptions, BI included. There's no such thing as a "theorem" that stands on its own with no package of axioms or assumptions behind it. The name of the package in the case of BI is "Local Causality", sometimes called "local realism" or just "locality". (And yes, BIs are experimentally violated; any model of the experiments therefore cannot maintain all the Local Causality assumptions.). But if there is any "belief" involved, it should be belief about the assumptions, not the proof. $\endgroup$ Commented Nov 20 at 23:07
  • $\begingroup$ Interesting....so in my head we have propositional logic with its axioms, complete and consistent. We then need some counting, so something like peano's axioms. We don't have to make the system incomplete, so we could use a softer version of peano's axioms, like all numbers less than 10^1000 exist. A simpler version of Bell's inequality would be, "for any 3 properties, any number of things less than 10^1000, we have n(A&B) <= n(A&C) + n(B&not C)". Provable in this system. I am confused where "locality" is assumed? Or is it that what we define "locality" as....that collection of these axioms? $\endgroup$ Commented Nov 20 at 23:19
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    $\begingroup$ @BobbyOcean The first paragraph of the Bell paper (referenced in different comment) mentions the nonlocal side of things explicitly. QM does not assign a value to an observable (your A B C) outside the context of a measurement. That’s the primary source of conflict. Ken has provided an excellent answer +1 but you keep returning to a video. Because you backed into the subject and skip EPR (“elements of reality”) and Bell (which as a theorem is always true), your questions (seemingly reasonable) are poorly grounded. Trying to parse out a video makes no sense, look to the fundamental principles. $\endgroup$ Commented Nov 21 at 13:42
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The question is ambiguously stated because the physical/practical/operational interpretation of the conjunction $\&$ is not provided.

In QT there are binary propositions (aka elementary observables), say $A$ and $B$ on a quantum system $S$ which are incompatible. This means no physical instrument is capable of simultaneously giving them a truth value when measuring properties of $S$. If you try to measure first $A$, next $B$ (after an arbitrarily small interval of time) and next $A$, the two values of $A$ are different in general, so that there is no way to give truth values to $A$ and $B$ simultaneously in order to use the truth table. If at least one of the couples $\{A,B\}$, $\{A,C\}$, $\{B,C\}$ is made of quantistically incompatible propositions, the obstruction to achieve your inequality is obvious: at least one of the conjunctions you consider has no truth value. (Notice that, if $A$ and $C$ are incompatible, then also $A$ and $\neg C$ are).

I stress that if $A,B,C$ are mutually compatible then they can be simultaneously measured on $S$ and your inequality holds and the proof is the one you presented.

When dealing with incompatible propositions you have at least two possibilities to try to tackle the issue.

  1. Using an ensemble interpretation.
  2. Using the structure of the orthomodular (bounded, atomic, separable) lattice of quantum propositions instead of the Boolean lattice to give a meaning to $\&$ and $\neg$.

In the first case you have to consider an ensemble of systems $S$ identically prepared and you measure $A$ and $B$, separately, on different elements of the ensemble.

$A \& B$ is true if you always obtain YES in both separated measurements on two elements of the ensemble.

(I stress that, in a generic quantum state of the ensemble, in general $A$ can be true for some elements and false for others and the same fact is true for $B$: so in a generic state $A \& B$ is false.)

At this juncture, your inequality is true, according to your proof.

It is worth stressing that, with this interpretation, properties are attributed to the state of the ensemble and not of the single chosen element of the ensemble.

Here a crucial difference with the classical world shows up. If the system is classical, it is possible to prepare the ensemble in a state where each of the properties $A,B,C$ is true (or only one or two of them are true etc.) In the quantum case, it is generally impossible to prepare the ensemble in a state where all elementary properties (not only $A,B,C$) are true or false with certainty: every element of the ensemble gives rise to YES or NO when measuring say $A$, but in general not all give the same outcome. This is consequence of Gleason's and Bell-Kochen-Specker's theorems.

In the second case, $A$ and $B$ are to be interpreted as orthogonal projectors in the Hilbert space of the quantum system. This set has a lattice structure which generalize the standard Boolean one and two connectives $\vee$ and $\wedge$, an orthocomplement $\perp$ etc. arise. When dealing with a maximal set of compatible propositions these operations have the same algebra as in classical logic: the maximal set is isomorphic to an abstract $\sigma$-algebra. As I said previously, your inequality can be proved true (independently from the state of the system and with the caveat I wrote above).

If some couple is made of incompatible propositions then the full orthomodular structure has to be exploited. It is possible to give a truth value to each conjunction $A \wedge B$, etc. as a function of the quantum state, but I do not know if the inequality is valid with this interpretation of connectives. (I should check).

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  • $\begingroup$ Sure, it sounds like we are in agreement. That the CBI theorem of PLC is false in QM. In this case you are changing "&" to be more complicated, thereby preventing the Theorem from holding in QM. That is, axiomatically "&" is defined for all propositions in PLC and it is hard to imagine otherwise. Again, we would consider anyone irrational if they said PLC didn't hold for boats, cars, houses, etc. Very unintuitive. I would be curious if your definitions for "&" and "not" prove the law of excluded middle, i.e., Thm "A or not A" is true for all objects and properties in PLC. $\endgroup$ Commented Nov 21 at 17:39
  • $\begingroup$ To reject the law of excluded middle would be incredibly unintuitive indeed. $\endgroup$ Commented Nov 21 at 17:51
  • $\begingroup$ Yes, “A or not A” is always true: it is the tautology, the top of the lattice, also in the quantum orthomodular case. Similarly “A and not A” is the bottom, always false. $\endgroup$ Commented Nov 21 at 17:52
  • $\begingroup$ Given N things, the following are theorems n(A&not A)=0, n(A|not A)=N, n(A|B)<=n(A)+n(B), we also have the distribution properties of & and |, A&(B|C)=(A&B)|(A&C) and A|(B&C)=(A|B)&(A|C). I assume you agree through modus ponens, CBI is a proven theorem of these principles. $\endgroup$ Commented Nov 21 at 19:37
  • $\begingroup$ After more thought, a little more push back on the unintuitive incompatible definition. The introduction of time does not seem to play a role, in that I could just as easily say A = "student with brown hair at time T". Properties at or about time still follow CBI. The contention is that in QM, properties are not simultaneously knowable, which is incredibly unintuitive, and would be irrational outside QM. In fact, I assume you are arguing that they are not even simultaneously knowable in principle; i.e., not even god could tell us what the simultaneous values are. $\endgroup$ Commented Nov 21 at 20:15
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Your post here is mostly a rant on NLHVT that is very much your own misunderstanding of what it is.

First of all, there are extremely few physicists in support of NLHVT, and their existence is struggling to survive; there is no hegemony of them dominating the conversation and needing any opposition.

Instead, your rant demonstrates that you did not understand that your binary table (colouring problem) representation of BI required locality to produce this representation at all. Without locality, there is no such binary table, and so there is neither experimental nor logical refutation of NLHVT.

In the realm of quantum interpretations, there are scarcely few contenders that are not explicitly shown false, either logically or experimentally, and so it is the norm to try and keep all the open possibilities going as much as is possible. Again, it is not like their existence is silencing other viewpoints, so there is no need to go out of one's way to attack them.

Most physicists consider non-locality as too much of a weirdness, too much of a price to pay for realism.

Also, you did not understand that NLHVT physicists also look at superdeterminists as if they are lunatics.

The least you could do is to not lump the different minorities together, and do some investigation to figure out what they are actually advocating for, and why the ones that survive the logical scrutiny and experimental testing are surviving as they are.

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  • $\begingroup$ Comments purged. Be kind, folks. Or use some other website. Those are the options. $\endgroup$ Commented Nov 21 at 21:27

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