# Problem with Bell's lambda - again

1. Aug 14, 2013

### harrylin

I have been bugged for a long time by EPR-Bell's hidden function "lambda" (Bell1964: lambda is a hidden 'variable or a set, or even a set of functions'). Here's my latest problem with it. Triggered by some recent discussions which brought up papers, in which is argued (and claimed to be proved) that Bell's inequality is equally applicable to non-local as to local influences, I suddenly remembered the following remarks by Bell:

BERTLMANN'S SOCKS AND THE NATURE OF REALITY

"It is important to note that to the limited degree to which determinism plays a role in the EPR argument, it is not assumed but inferred. What is held sacred is the principle of "local causality" - or "no action at a distance."

[..]

Let us suppose then that the correlations between A and B in the EPR experiment are likewise "locally explicable". That is to say we suppose that there are variables λ, which, if only we knew them, would allow decoupling of the fluctuations:

P(A,B¦a,b,λ) = P1(A¦a,λ) P2(B¦b,λ) ...(11)

[..]

It is notable that in this argument nothing is said about the locality, or even localizability, of the variables λ. These variables could well include, for example, quantum mechanical state vectors, which have no particular localization in ordinary space time. It is assumed only that the outputs A and B, and the particular inputs a and b, are well localized."

Indeed, as Bell stated, λ itself is not necessarily local and could well include parameters of QM; and it may be similarly stochastic, as long as it predicts with certainty the outcomes for certain settings.

But then his theorem should be equally valid for NON-local "quantum" influences: His same argument can be given for a probabilistic "quantum function" λ that fully determines the QM predictions, such that P(A|aλ) is not different from P(A|Bbaλ).

If not, why not??

Last edited: Aug 14, 2013
2. Aug 15, 2013

### harrylin

[edit: B and b swapped] OK... now I came up with an answer on my own question. I was thinking about the function for B - the measurement result at one location. However the function for b - the local temperature in his illustration, but the detector setting in experiments - is not included in the prediction, and is supposed to be independent due to free will of the experimenter (which is a rather foggy issue). Thus such a factoring out of b for a non-local model, by means of including it in λ, would imply super-determinism, which is not assumed.

Last edited: Aug 15, 2013
3. Aug 15, 2013

### wle

Actually that isn't necessary. You don't need to assume the probability distributions $P_{1}(A \mid a, \lambda)$ and $P_{2}(B \mid b, \lambda)$ are deterministic in order to derive Bell inequalities. The same Bell inequalities hold whether you assume that or not.

When you say that $P(A \mid a, \lambda)$ is not different from $P(A \mid B, b, a, \lambda)$, you are saying that, given $\lambda$, Bob's local measurement setting $b$ and local measurement outcome $B$ are redundant for making predictions about Alice's local outcome $A$. Or in other words, $b$ and $B$ don't causally influence $A$. This is how Bell defined locality.

4. Aug 15, 2013

### harrylin

Right. I meant the outcome for certain setting pairs - such as the assumed perfect anti-correlation.

Thanks for the clarification! As some have stressed, the requirement is subtly stronger than that: b and B should not influence the probability PA, given the specification function λ. And of course, what I overlooked is that b is assumed not to be included in λ, even with a non-local λ for a non-local model. I suppose that you had not seen my further reflection in post #2.

Last edited: Aug 15, 2013
5. Aug 15, 2013

### wle

Huh? You never need to assume perfect anti-correlation to derive most Bell inequalities. It's a special case. And even where you might be in that special case, it's something you could confirm directly by looking at your record of measurement outcomes, so even there it's not an assumption.

Isn't that the same thing? If $b$ and $B$ can causally influence $A$, that means that they can influence the probability that you will get a particular outcome for $A$.

More generally, derivations of Bell inequalities assume that the variable $\lambda$ is drawn from a probability distribution $\rho(\lambda)$ and is uncorrelated with the choice of settings $a$ and $b$. This assumption could be unjustified in the case of so-called "superdeterminism" (some variable $\lambda'$ further back in time causally influences all of $a$, $b$, and $\lambda$ in such a way that they end up correlated) or retrocausality ($a$ and $b$ causally influence $\lambda$).

6. Aug 16, 2013

### harrylin

Oops, indeed, he used the certainty of some outcomes for certain settings the first time, but not this time.
Ehm no, his derivation concerns the theory of QM and not measurements.
In general it's only true one way, as in [statement A] => [statement B]. A physical influence is not in general a necessary condition for influencing a probability; Bell even refers to that fact in his introduction. However, it looks to me that Bell implied that by means of the added λ, that is the only possibility that is left.
Jaynes jumped on Bell's argument, as it looks as if Bell mistakenly thought that those are the same thing, but maybe Jaynes misunderstood Bell's reasoning. There was a long discussion about that on this forum.
Yes. And there has been quite some debate on the required probability distribution: according to Bell it may be anything ("whatever"), but not everyone agrees with that.

Last edited: Aug 16, 2013
7. Aug 17, 2013

### wle

No, Bell's theorem is an operational (i.e. experimentally testable) definition of locality. QM is simply an example of a theory that makes predictions that can be shown to be incompatible with Bell locality. If you perform a Bell experiment and you measure a Bell inequality violation, then you have evidence of nonlocality whether or not you accept QM.

In other words, a physical influence is a necessary condition for influencing the probability $P_{1}(A \mid a, \lambda)$ given a complete specification of the initial conditions $\lambda$, which is the scenario we're talking about here.

Huh? Bell's theorem certainly holds whatever probability distribution $\rho(\lambda)$ you use. The only requirement is that $\lambda$ is uncorrelated with the settings $a$ and $b$, i.e. if you attribute a joint probability distribution $P$ to $\lambda$, $a$, and $b$, then $P(\lambda, a, b) = \rho(\lambda) p_{1}(a) p_{2}(b)$.

8. Aug 18, 2013

### harrylin

The way he formulates it in that paper does make it sound like a "definition"; that would be a smart way out (abandoning the pretended purpose of adhering to Einstein's definition), but it is based on an unproven assumption. More elaborate explanation in the thread that I mentioned before, maybe most useful from here:

I agree with Peterdonis. Although Bell formulates it in that paper as a definition, it is in fact an assumption about a property of locality.
Huh?? No, it just sounds very plausible. A claim or assumption about something is not a fact about that something - it needs to be formally proved in order to be a fact.
According to a number of papers that I have read, his derivation is not valid for all possible probability distributions; and hopefully someone will be so helpful to elaborate this matter to us in this thread. It happens to be just one of the things that I would like to understand. If I'm not mistaken, some of those papers were linked in the other thread in which you have been participating. [edit: not sure now in which papers; if needed I will try to find at least one of them back]

Last edited: Aug 18, 2013
9. Aug 18, 2013

### wle

I don't see why you are putting "definition" in quotes. It is a definition. One that you can in principle agree or disagree with. This is why I sometimes refer specifically to "Bell locality", to be clear I'm not talking about some other possible definition or concept of locality.

For example, my understanding is that researchers with more of a quantum field theory background use "locality" to refer to what researchers in the quantum information community would more commonly call "no signalling", or in reference to a particular theory's structure (e.g. a quantum field theory with only local coupling terms in its Lagrangian). These are not the same thing as Bell locality. Quantum physics is non signalling but it is not Bell local, for instance.

Hardly. Bell locality is a definition of locality with a motivating argument behind it. It was originally specifically intended as a formalisation of the sort of locality that was alluded to (but not formally defined) in the EPR article for instance. Most people seem to think Bell's definition captures that idea very well -- that's why Bell's theorem is so widely accepted.

I don't see a meaningful distinction. If the probability of getting an outcome A in one place changes depending on something happening a great distance away, then for all intents and purposes that's what I'd call an "influence".

Well it's clear to me that Bell's theorem holds just fine regardless of the $\rho(\lambda)$ you put in.

The only subtlety I can think of (that I have seen referred to once or twice while I've been on this forum) is that actual Bell experiments estimate the Bell correlator from the results of measurements made on a very large number of entangled particle pairs. This leaves open the possibility that, for instance, the variable $\lambda$ might vary from one round of the experiment to the next and might even depend explicitly on previous measurement settings and outcomes. But even that can be accounted for.

10. Aug 18, 2013

### harrylin

I did the same in the other thread, to emphasize that Bell's opinion about what locality implies is not necessarily agreed upon by everyone. In particular, not all specialists in statistics agreed with his motivation.

Once more, assuming that you mean physical influence: that is fundamentally erroneous - even according to Bell.
This was also discussed in the section of thread that I linked for you...

I'll try to find back articles on that then. One is, I think, the one of DeRaedt that has a thread on it (Boole-Bell); another is I think a paper by Accardi that was linked in the hope for locality thread.

11. Aug 18, 2013

### audioloop

a gross oversimplification... and a rough and imprecise statement....

Contextuality and Non-Locality

https://www.physicsforums.com/showpost.php?p=3993579&postcount=5

.

12. Aug 19, 2013

### stevendaryl

Staff Emeritus
I followed that link to the article written by Jaynes, here:
http://bayes.wustl.edu/etj/articles/cmystery.pdf

I feel that Jaynes is misunderstanding Bell's reasoning. Bell certainly was not saying that the failure of the joint probability distribution for two events to factor means that there is a causal influence between the two events. He certainly was not saying that conditional probability implies causal influence. So Jaynes counterargument (the urns example) is attacking a straw man. He's not addressing Bell, at all.

The way that I understand Bell is that he is talking about the propagation of information. If Bob at event $B$ gets information about the results of an experiment performed by Alice at event $A$, then that information must have propagated from the common causal past of events $A$ and $B$. Jaynes' urn example is not a counter-example.

I'm not sure if it is possible to prove that Bell's notion captures the intuitive notion of "locality". But it seems to me that that's beside the point. The real issue is whether the predictions of quantum mechanics are compatible with a particular classical notion of locally realistic theories. Roughly speaking, that the universe consists of fields and particles that are only influenced by local conditions.

13. Aug 19, 2013

### harrylin

Your understanding of what Bell meant is somewhat different from mine; but we do agree that Jaynes seems to misunderstand Bell, who even gave an example somewhat similar to Jaynes' urn example in the introduction of the Bertlmanns paper.

That's right; however the usefulness of Bell's theorem depends on that starting assumption. If it's not proven that it perfectly catches the problem that he wanted to describe, then Bell's theorem falls flat; and we're back to the start (with however improved understanding of the issues).

14. Aug 19, 2013

### stevendaryl

Staff Emeritus
Einstein never explicitly stated what sort of realistic theories he would have accepted, but assuming that they are like the ones that physicists looked for prior to the discovery of quantum mechanics, they have these features:

1. At each point $P$ in spacetime, there are a number of variables $V_j(P)$ characterizing local conditions at that point. These variables might include: the values of fields, the presence or absence of various particles, physical properties of the particles (such as charge, momentum, angular momentum, etc.)
2. These variables evolve in a local way (e.g., by local differential equations, where the evolution of $V_j(P)$ depends only on $V_j(P')$ for nearby values of $P'$)
3. A measurement is just a macroscopic, coarse-grained, record of the variables $V_j(P)$.

Bell's factorizability condition would hold for any theory of this sort.

15. Aug 19, 2013

### harrylin

It appears to me that point 3 may be in direct conflict with QM... I doubt that Einstein would have insisted on such a pre-QM requirement. The discovery of QM effects required non-classical features, somewhat different from before QM.

16. Aug 19, 2013

### stevendaryl

Staff Emeritus
You doubt it, but to me that's exactly what Einstein believed. That's what he's talking about when he talks about hidden variables---they are revealing pre-existing information about a hidden state of the particle.

17. Aug 19, 2013

### stevendaryl

Staff Emeritus
I don't think point 3 is in conflict with QM. When you measure that an electron has spin-up, the state of the measuring device + recording device + observer + particle + whatever is DIFFERENT than if you had measured spin-down. Measurements change the state of the measuring device/observer. Knowing the microscopic state of the measuring device/observer is sufficient (in principle, anyway) to know what the measurement result was.

18. Aug 19, 2013

### harrylin

That doesn't exclude a very different state upon interaction with a detector than without detection...
That was exactly my point: the state upon measurement doesn't have to be pre-existing according to common "local reality" concepts.

19. Aug 19, 2013

### stevendaryl

Staff Emeritus
I'm not sure that there's a real disagreement here. After one has made a measurement, that measurement reflects local information about the microscopic state. It's not clear what it shows about the local microscopic state before the measurement was made (although if you have a complete enough theory of how local microscopic states evolve, then you should be able to get at least a probability distribution on what the microscopic state could have been prior to the measurement.

I understand the distinction you're talking about, but I don't think it makes any difference to the main point, which is whether all such "local" theories obey Bell's factorizability assumption. I think they do.

20. Aug 20, 2013

### wle

I don't see how point 3 is necessary. The "outcomes" in a Bell experiment are simply macroscopic events and can be anything: whether a particle went up or down in a Stern-Gerlach apparatus, whether photon detector 1 or photon detector 2 clicked, or whether a red light or a green light turned on, for instance. You neither need to know nor care that these events are being generated by "measurements" on "photons" or "spin states" or whatever. If you have a local theory about photons, for instance, then it is the responsibility of that theory to explain how and why a photon being in a particular initial state ends up influencing the probability of a particular detector clicking some time later.