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In a discussion in another question, a user named @Claudiu showed me his own version of the derivation of CHSH inequality that does not need hidden variables (reproduced below). I can just wonder if this proof is valid and if so why aren't we introduced to this version of the derivation that rules out locality and avoids the whole need to discuss for hidden variables.

I leave here the derivation, but you may skip it and go directly to the next section (it is just to show "previous work").

Derivation

This derivation is based in J.S. Bell's Speakeable and Unspekeable in Quantum Mechanics, first edition, pp. 156 Appendix 2. However, the idea in this question is to carry out the same derivation as Bell did in the book, but removing all integrals related to hidden variables; steps that I reproduce here:

Let us assume (1) statistical independence (no superdeterminism) (2) factorabizability and the usual (3) locality so that we can write the correlations as

$$E(\theta,\phi)=[P_1(\uparrow|\theta)-P_1(\downarrow|\theta)][P_2(\uparrow|\phi)-P_2(\downarrow|\phi) ]=A(\theta)B(\phi)$$ where $A$ and $B$ stand for the first and second expressions in square brackets, respectively. Here $\theta$ and $\phi$ are the the angles of measurement of two detectors 1 and 2 in a Bell test with two entangled spin-1/2 particles. The results of the experiment can be $\updownarrow=\uparrow,\downarrow$ and $P_i(\updownarrow|\varphi)$ is the conditional probability of measuring $\updownarrow$ with in detector $i=1,2$ measuring with an angle $\varphi$.

Note that some of you will be familiar with this expression, but it would normally look like $\int \rho(\lambda)A(\theta,\lambda)B(\phi,\lambda)d\lambda$ but I just got rid of the hidden variable $\lambda$.

Without assuming determined values of $0$ and $1$ for $P_i$ (determinism), we can at least admit $0\leq P_1\leq 1$ and $0\leq P_2\leq 1$, and thus $|A(\theta)|\leq 1$ and $|B(\phi)|\leq 1$.

We can use that to write $$|E(\theta,\phi)\pm E(\theta,\phi')|=|A(\theta)B(\phi)\pm A(\theta) B(\phi')|\leq |B(\phi)\pm B(\phi')|$$ and analogously we have $$|E(\theta',\phi)\mp E(\theta',\phi')|=|A(\theta')B(\phi)\mp A(\theta) B(\phi')|\leq |B(\phi)\mp B(\phi')|$$ where I took $\theta\to \theta'$ and $\pm\to\mp$.

Summing the two last equations, we have $$|E(\theta,\phi)\pm E(\theta,\phi')|+|E(\theta',\phi)\pm E(\theta',\phi')|\leq |B(\phi)\pm B(\phi')|+|B(\phi)\mp B(\phi')|\leq 2$$

This last result includes the famous CHSH inequality $$|E(\theta,\phi)+ E(\theta,\phi')+E(\theta',\phi)- E(\theta',\phi')|\leq 2$$

Nowhere in this derivation, I needed to use hidden variables. For the version with the integrals over the hidden variable distribution, check the reference above.

The only "classical" physical assumption above was locality there rest seems to come from probability theory. Of course this derivation does not address how you argue for assumption (2) [which is very natural], but that is sometimes defined through "determinism" (fixing some probabilities to be strictly 0 or 1).

[Assumption (1) is out of the scope of this question].

Comment and question

A derivation of Bell's inequalities without all the integrals associated to hidden variables is to me a stronger argument for those that advocate violations of locality as it avoids the unnecessary discussion about the nature of hidden variables. Maybe somebody with more background could provide a perspective on this issue. It seems that such a version should be taught more in introductory courses as it is a much simpler demonstration "quantum nonlocality" if you wish.

Note that this boils down to Occam's razor, why have a more complicated version of the theorem if it adds nothing to the result.

I will try not ask if the derivation above is right (as that would probably get the question closed). However I will ask, is it possible to derive CHSH without ever making appeal to hidden variables as I did here?

Mauricio
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  • If I recall correctly, this paper might be of interest to you – Níckolas Alves Oct 31 '22 at 20:49
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    More importantly, I don't understand the point of this question: Bell's theorem, in a more general fashion, excludes local realist theories - if you look at the "toy" version stated in the Wiki article, you should notice it doesn't use any hidden variables, either. It's not surprising at all to me that the inequality still holds when you drop the hidden variable. after all, if we could've explained quantum phenomena with a local realist picture without hidden variables, why would we have ever tried to introduce hidden variables? – ACuriousMind Oct 31 '22 at 20:58
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    @ACuriousMind I guess my point is that saying that it excludes "local realism" is just adding "realism" as if it was possible to violate realism and not locality. It is not, it violates locality. On the contrary I guess this version is almost supporting the case for realistic theories (Bohmian or superdeterminism) as the only nonlocal nonrealistic one I could think of seems to be objective-collapse. – Mauricio Oct 31 '22 at 21:38
  • @Mauricio You seem to be under the impression that "realism" and "hidden variables" are synonymous. They are not. Classical mechanics does not have hidden variables, but it is realist and local. In contrast, the many-worlds interpretation is usually understood to be non-realist, but local, and Bohmian mechanics is realist, but non-local (although outside the context of Bell's theorem we often split "realism" into finer distinctions). I don't know where your idea that we necessarily need to violate locality comes from, it is not evident from your question. – ACuriousMind Oct 31 '22 at 21:52
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    @ACuriousMind I agree with all the distinction you gave for Bohmian, MWI and classical mechanics. Then it's about realism. To me "realism" means that you have hidden variables somewhere that explain your theory, if you could somehow predict those variables you could know the result of the experiment before measuring it. In classical statistical mechanics we talk about probability due to the difficulties of measuring/following the thin trajectories of chaotic many body systems in phase space. Maybe could you provide what is the "realism" in the usual CHSH derivation? – Mauricio Oct 31 '22 at 22:03
  • @Mauricio "To me "realism" means that you have hidden variables somewhere that explain your theory" Why would a theory with $n$ (>0) hidden variables be more "realist" than an equally powerful theory with 0 hidden variables? – JiK Nov 01 '22 at 17:05
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    @JiK you are using realist there in a different whay what I would have used. A theory like Bohr's version of Copenhague is not "real" in the sense that there is no way to predict the exact result of measuring a superposition, only probabilities. Bohm on the other hand would be "real" in the sense that if you describe the pilot-wave then you could exactly predict the results of QM without probabilities. Note that I am not supporting any version of QM here. This Bell's theorem is hidden variables agnostic, however it seems that locality cannot be solved either way that's the point. – Mauricio Nov 01 '22 at 20:19
  • Physically one wouldn't one expect that the result depend on the incoming photon ? Hence a polarization angle of each photon $\varphi_n$. But the power of the theorem is that the incoming datas can be everything, so it is probably why it was generalized and not citing explicitly which physical quantity it refers to ? – QuantumPotatoïd Nov 03 '22 at 05:36
  • @Cretin2 here $\updownarrow$ can be the photon polarization, or the spin of an electron, or any property of 2-level quantum system. – Mauricio Nov 03 '22 at 08:13

2 Answers2

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There is no mystery here.

You've simply shown Bell's theorem for the special case of no meaningful hidden variable at all, i.e. $\rho(\lambda) = \delta(\lambda - \lambda_0)$ and $A(\phi,\lambda_0) = A(\phi)$ etc. where $\lambda_0$ is just the single value the "non-existent hidden variable" always has. You can turn any physical theory into a "hidden variable theory" by making everything formally dependent on some $\lambda$ that never varies from a constant $\lambda = \lambda_0$.

That, as a special case, Bell's theorem includes the case where the local realist theory it excludes doesn't actually involve any hidden variable should not be particularly surprising - after all, there are no hidden variables in classical mechanics and if classical mechanics didn't obey the inequality, what would be the point of it?

The point of Bell's theorem is to exclude the larger set of theories where we try to reject the weirdness of quantum mechanics by trying to reproduce it by assuming the world is just like classical mechanics just with additional variables we don't easily have access to, i.e. are "hidden". The "hidden" is an attribute that refers to these quantities being unknown to our naive experimentalists' eyes, not to some special property these variables have that makes Bell's theorem work - they are realist variables just like the "visible" variables in a local realist theory.

ACuriousMind
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  • It would helpful if you explain where "realism" enters the derivation, according to you. Also I disagree with Bell's theorem excluding a "larger set" or this being a "special case", it is not, the derivation above excludes the same number of theories. – Mauricio Nov 01 '22 at 23:38
  • @Mauricio 1. The precise meaning of "realism" is something a lot of ink has been spilled over. The only uncontroversial statement is that it is related to the factorizability condition. The Stanford Encyclopedia has a good discussion of where it gets murky after that. 2. I agree that your version in spirit applies to hidden variable theories, too, but to see that one has to recognize the meaning of your "without assuming determined values" and convince oneself that no other step is affected by the presence of hidden variables. – ACuriousMind Nov 02 '22 at 00:13
  • Oh! I have not look at Standford Encyclopedia in a while, they do not use hidden variables in their first derivation (hurray!) I am now more convinced (maybe you could add this link to the answer). However, I now get how the murky $\lambda$ is necessary, if you assume "outcome determinism" as given a specific $\lambda$ all the probabilities are fixed to 0 or 1. – Mauricio Nov 02 '22 at 08:13
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Your assumption that the expectation value is factorisable is not true in general. Consider the simple example of a Bell state $|00\rangle+|11\rangle$ measured in the computational basis, and you'll already see it fail. It will only hold for product states, where there's no correlation at all.

Unless that's your point? What I would consider a standard way to derive Bell-like inequalities is in fact to observe that for uncorrelated outcome distributions, because the expectation values factorise as in your example, we have $|S|\le2$. One can then conclude the general result from this by convexity: any "local realist theory" is a convex combination of such uncorrelated distributions (by definition), and being $S$ a linear function of the probabilities, the expression $|S|\le2$ remains true when computed on convex combinations of probabilities. This is the approach I use e.g. here or here (what I write as $S_\lambda$ there would be the $S$ factor computed on an uncorrelated distribution).

glS
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  • By factorizability, I do not necessarily mean separability, I just mean $\mathcal{P}(AB|xy)=\sum_\lambda \mathcal P(AB|xy\lambda)\mathcal{P}(\lambda|xy)$. Of course I am attacking this from a classical perspective, the quantum version of course violates Bell's theorem. Does your version need hidden variables? I will take a look . – Mauricio Nov 02 '22 at 12:38
  • well, if $\lambda$ depends on $x,y$, then no it's a different thing. But then you're just defining some "intermediate variable" $\lambda$ which explains the conditional prob distribution from the measurement choices. You can always write a decomposition like that using a single value of $\lambda$, with $P(\lambda|xy)=1$ and $P(AB|xy\lambda)\equiv P(AB|xy)$. And if you have more values of $\lambda$, you're saying measurement choices don't deterministically determine the outcome prob distribution. Which is the same as saying the outcome distro has to have some minimum amount of noise. – glS Nov 02 '22 at 12:42
  • but more generally I don't really understand the objection about hidden variables I think. The whole point of Bell's nonlocality discussions is to study if (when) you can (have to) rule out explanations of the form $p(ab|xy)=\sum_\lambda p(a|x\lambda)p(b|y\lambda)p_\lambda$ (and then you call $\lambda$ a "local hidden variable"). If you don't want to consider decompositions of this form, you need to completely redefine the problem, but then what's the connection with Bell nonlocality? – glS Nov 02 '22 at 12:43
  • Sorry, I mispoke, by factorizability I meant $(|)=(|xy)(|x)$. No need of hidden variable $\lambda$ to formulate Bell's theorem as seen above. – Mauricio Nov 02 '22 at 20:00
  • @Mauricio if you consider outcomes on each side as conditioned by both measurement choices, then you're not discussing locality at all. You might as well say that there is a single person making the choice of measurement, calling it say $z$ rather than "$xy$", so you're again completely changing the context. The point of nonlocality results is always that there's two parties that can make different (independent) measurement choices – glS Nov 02 '22 at 20:35
  • I thinking we are missing each other here, if I put all assumptions together here (statistical freedom, factorizability and locality) I have $p(ab|xy)=p(a|x)p(b|y)$ and with this Bell's theorem follows directly and QM violates it. This seems to imply that we should throw out locality (or accept superdetermism). No need of $\lambda$ – Mauricio Nov 02 '22 at 22:20
  • @Mauricio no, you don't have that. That's only true for outcome distributions that are uncorrelated for any measurement choice. That's not what QM predicts, but it's also not true for classical systems. A simple correlated classical state does not have give outcome factorisable distributions of that form – glS Nov 02 '22 at 23:12
  • I agree that QM does not apply here. Just to be clear you are saying that we indeed need $\lambda$ to argue for factorisability? – Mauricio Nov 03 '22 at 07:55
  • Let us continue this discussion in chat. – glS Nov 03 '22 at 08:26