# Specifics of Bell's lambda

• A
Science Advisor
In this paper, the locality assumption is "the vital assumption" after Eq. (1).
A side question. Can one be a bit more specific about what ##\lambda## is? He says that it doesn't matter and it could be many things, and he treats it as a single continuous variable, but then he integrates over it. So whatever the set of lambdas is, it has to be at least a measure space, no?

[Mentors' note: This thread was forked off from https://www.physicsforums.com/threads/can-locality-be-retrocausal.950401/. "This paper" refers to Bell's 1965 paper]

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## Answers and Replies

Demystifier
Science Advisor
Gold Member
So whatever the set of lambdas is, it has to be at least a measure space, no?
Yes. Moreover, Bell assumes that there is a probability measure over lambda.

martinbn
strangerep
Science Advisor
So whatever the set of lambdas is, it has to be at least a measure space, no?
IMHO, it's a loophole in the standard proof. If ##\lambda## denotes an infinite set of hidden variables, then he runs foul of the "no translation-invariant Lebesgue measure on an inf-dim space" gotcha.

And if one retreats to a Gaussian measure, one must explain why some hidden variables are more important (higher weighting) than others.

(It wouldn't be the first time a physicist blithely assumed that a limit as ##n\to\infty## could be taken with no problems for their integrals.)

martinbn
stevendaryl
Staff Emeritus
Science Advisor
IMHO, it's a loophole in the standard proof. If ##\lambda## denotes an infinite set of hidden variables, then he runs foul of the "no translation-invariant Lebesgue measure on an inf-dim space" gotcha.
Why would translation-invariance be important?

strangerep
Science Advisor
Why would translation-invariance be important?
Because, (as I said), otherwise...
strangerep said:
... one must explain why some hidden variables are more important than others.
I.e., why some (subsets) have different measure than others.

Science Advisor
That was my problem with his statement. He says that the set could be a set of functions.

Demystifier
Science Advisor
Gold Member
And if one retreats to a Gaussian measure, one must explain why some hidden variables are more important (higher weighting) than others.
But he doesn't need to assume a Gaussian measure. It is sufficient to assume that there exist some probability measure, without specifying what exactly this measure is. This is a very mild* assumption.

It is also illuminating to stress that Bell was inspired by Bohmian mechanics, in which it was known what lambda and the corresponding probability measure was. Since Bohmian mechanics was non-local, Bell wanted to see whether this is a specific property of Bohmian mechanics or whether "any" hidden-variable theory (satisfying some mild assumptions) must be non-local. So when reading his proof, it is useful to always have Bohmian mechanics in mind as a concrete example of what his abstract proof may represent.

*Mild assumptions are usually called "weak" assumptions, but I do not like to call them "weak" because it is not a weakness when an assumption is mild.

Boing3000
atyy
Science Advisor
IMHO, it's a loophole in the standard proof. If ##\lambda## denotes an infinite set of hidden variables, then he runs foul of the "no translation-invariant Lebesgue measure on an inf-dim space" gotcha.

And if one retreats to a Gaussian measure, one must explain why some hidden variables are more important (higher weighting) than others.

(It wouldn't be the first time a physicist blithely assumed that a limit as ##n\to\infty## could be taken with no problems for their integrals.)
No, it is not, because the "uniform distribution" is not invariant under arbitrary coordinate changes.

Demystifier
Demystifier
Science Advisor
Gold Member
No, it is not, because the "uniform distribution" is not invariant under arbitrary coordinate changes.
In other words, strictly speaking, an assumption of uniform distribution also requires an explanation.

strangerep
Science Advisor
But he doesn't need to assume a Gaussian measure. It is sufficient to assume that there exist some probability measure, without specifying what exactly this measure is. This is a very mild* assumption.
I picked Gaussian measure because it can be proved to give well-defined integrals in the inf-dim case. It's not sufficient to assume "there exists some probability measure" -- it must be a measure which is well-defined in the inf-dim case. (I wouldn't call this "weak", though perhaps that's a matter of taste.)

It is also illuminating to stress that Bell was inspired by Bohmian mechanics, [...]
I've always been given to understand that the above can be regarded as an explanation of why Bohmian mechanics seems to be an exception to Bell's theorem.

... in which it was known what lambda and the corresponding probability measure was.
I'm not an expert on Bohmian mechanics, and it's been a long time since I looked at it. Could you please remind me what its lambda and corresponding probability measure are?

strangerep
Science Advisor
No, it is not, because the "uniform distribution" is not invariant under arbitrary coordinate changes.
... in which case any non-uniformities are an artifact of the coordinate change, and not an intrinsic feature of the physics.

strangerep
Science Advisor
In other words, strictly speaking, an assumption of uniform distribution also requires an explanation.
I don't think your statement is equivalent to atyy's. You're talking about an intrinsic feature of the physics, whereas atyy was talking about arbitrary coordinate changes.

Anyway, I would have thought that having zero knowledge about any of the hidden variables would naturally correspond to a uniform distribution, (but perhaps a Bayesian will correct me on that point).

Demystifier
Science Advisor
Gold Member
I've always been given to understand that the above can be regarded as an explanation of why Bohmian mechanics seems to be an exception to Bell's theorem.
No, Bohmian mechanics is not an exception to Bell's theorem. The Bell's theorem, in his own view, is best interpreted as - it proves that hidden variables cannot be local, so, since non-existence of any hidden variables at all would look absurd, it shows that nature is non-local, as suggested also by Bohmian mechanics.

I'm not an expert on Bohmian mechanics, and it's been a long time since I looked at it. Could you please remind me what its lambda and corresponding probability measure are?
In Bohmian mechanics of ##n## particles, ##\lambda## is the set of the actual particle positions ##{\bf x}_a(t)## (##a=1,\ldots ,n##) at any time ##t##. The probability measure is
$$\rho({\bf x}_1,\ldots, {\bf x}_n;t)=|\psi({\bf x}_1,\ldots, {\bf x}_n,t)|^2$$

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Demystifier
Science Advisor
Gold Member
Anyway, I would have thought that having zero knowledge about any of the hidden variables would naturally correspond to a uniform distribution, (but perhaps a Bayesian will correct me on that point).
I think a Bayesian would say that there is no such thing as zero knowledge. After all, a knowledge that a distribution is uniform is a knowledge too.

Or another example, if I tell you that I just imagined a real number, and challenge you to guess which one is that, you would probably try with something like 1, ##\sqrt{2}## or ##\pi##, and not with something like 56324078941123,7654339908. You know that with an attempt of the first kind you have much better chances, even if you don't know how to formalize this knowledge.

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Science Advisor
The Bell's theorem, in his own view, is best interpreted as - it proves that hidden variables cannot be local, so, since non-existence of any hidden variables at all would look absurd, it shows that nature is non-local, as suggested also by Bohmian mechanics.
Well "looks absurd, therefore..." is not a very convincing argument. Also your statement needs to be more precise. When you say "hidden variables cannot be local" you mean hidden variables for which there is a probability measure and so on are non-local. Also, ignoring all that, the conclusion cannot be nature is non-local as suggested by BM, it should be nature is non-local (where local means so and so), which might be as BM suggests.

By the way why is it so absurd to suggest that there are no hidden variables? Do you consider all no-go theorems the same way?

Demystifier
Science Advisor
Gold Member
By the way why is it so absurd to suggest that there are no hidden variables?
In my opinion, that is the central question. Hidden variables, by definition, are quantities existing (in the ontological sense) even when they are not observed. If nothing exists until observed, then I don't see how to avoid solipsism. Yet, there are no many physicists or philosophers who think that solipsism is a viable possibility.

When confronted with this type of argument, many physicists try to escape by objecting that solipsism is philosophy that cannot be verified by experiment. But just because it's philosophy doesn't make it less valid. If philosophy is based on sound logic, then it's conclusions are even more reliable then conclusions based on experiments.

Science Advisor
But I don't think it implies that nothing exists. Of course, things as particles and fields exist. What it means is that certain dynamical quantities of the mathematical description do not have values at all times. It isn't that strange given that these quantities don't have values in the set of real numbers but operators.

Demystifier
Science Advisor
Gold Member
What it means is that certain dynamical quantities of the mathematical description do not have values at all times.
Fine, but then one has to specify when do those quantities acquire values. In other words, one must answer the following question: What conditions must be fulfilled in order for a quantity to acquire a value? In particular, does it require a conscious observer, or perhaps only a human made apparatus, or perhaps a sufficient amount of decoherence is enough? In the latter case, precisely how much decoherence is sufficient? You may decide to remain agnostic about such questions, but as soon as you try to give an explicit answer, you either get a sort of solipsism or a sort of non-locality.

Science Advisor
Fine, but then one has to specify when do those quantities acquire values. In other words, one must answer the following question: What conditions must be fulfilled in order for a quantity to acquire a value? In particular, does it require a conscious observer, or perhaps only a human made apparatus, or perhaps a sufficient amount of decoherence is enough? In the latter case, precisely how much decoherence is sufficient? You may decide to remain agnostic about such questions, but as soon as you try to give an explicit answer, you either get a sort of solipsism or a sort of non-locality.
Or one can acknowledge that there are open problems, that need to be solved. Look Fermat took centuries to solve, no need to hurry with "solutions" as solipsism or non-locality.

Demystifier
Science Advisor
Gold Member
Or one can acknowledge that there are open problems, that need to be solved. Look Fermat took centuries to solve, no need to hurry with "solutions" as solipsism or non-locality.
Even before Wiles finally settled the Fermat problem, most mathematicians thought, with a good reason, that Fermat's conjecture is probably right. Similarly, I can say that at the moment we have good reasons to think that nature is probably non-local. But I certainly agree that it is an open problem. Even if we take non-locality for granted, we still don't know how exactly this non-locality is realized.

Boing3000
Gold Member
no need to hurry with "solutions" as solipsism or non-locality.
True, but solipsism (as an attitude) is presented as a solution, while in reality it's a cop-out.
And non-locality is logically sound and verified experimentally. There are phenomenon like entanglement that can only be described as a unique quantity that span a huge swath of space

DarMM
Science Advisor
Gold Member
If you want to get a grip on ##\lambda## and the common assumptions on it I recommend the following paper by Matt Leifer:

https://arxiv.org/abs/1409.1570

It goes through the common assumptions on ##\lambda##, those assumptions defining the so called "ontological models" framework, within which Bell's theorem, Kochen Specker etc are proved.

His slides here are a quicker introduction and present the assumptions more plainly (but the first paper is more complete):

https://foundations.ethz.ch/wp-content/uploads/2017/06/leifer1.pdf

strangerep
atyy
Science Advisor
I think a Bayesian would say that there is no such thing as zero knowledge. After all, a knowledge that a distribution is uniform is a knowledge too.
For continuous variables, there is no such "real" thing as a uniform distribution, since a uniform distribution will become non-uniform under a coordinate change. Thus, uniformity or non-uniformity are equally real.

Supporting your point, one can find many discussions along the lines of https://www.stat.auckland.ac.nz/~millar/Bayesian/Handouts/Ch4Priors.pdf.

Furthermore, one point of view from traditional kinetic theory, is that it is dynamics that determines the distributions https://arxiv.org/abs/0807.1268, so the need for a non-informative prior is irrelevant.

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Boing3000
Science Advisor
For continuous variables, there is no such "real" thing as a uniform distribution, since a uniform distribution will become non-uniform under a coordinate change. Thus, uniformity or non-uniformity are equally real.
What do you mean by that?

stevendaryl
Staff Emeritus
Science Advisor
What do you mean by that?
If you have a variable ##X## that takes on values from 0 to 1, you could either choose the flat distribution on ##X##. Or you could let ##Y = X^2##, and choose a flat distribution on ##Y##, or you could let ##Z = sin(\frac{\pi}{2} X)##, and choose a flat distribution on ##Z##, etc. If you know nothing about ##X##, then does that mean you know something about ##Y##?

Demystifier and strangerep