New experimental support for pilot wave theory?

[martinbn]

...This means that we cannot say what is the state of the first particle at the position ${\bf x}_1$ or what is the state of the second particle at the position ${\bf x}_2$. In other words, the description of the system by $\psi({\bf x}_1, {\bf x}_2)$ is not local.

Is that what non-local means? To me it seems that it would be more accurate if it is called non-real, after all, the conclusion is that you cannot assign states to each particle.

 

[Demystifier]

Is that what non-local means? To me it seems that it would be more accurate if it is called non-real, after all, the conclusion is that you cannot assign states to each particle.

Perhaps this terminology would be more clear on the following purely mathematical example. Let $x$ be coordinate on the 1-dimensional manifold. Any point with a coordinate $x$ is a local object. Also any function $f(x)$ is a local object, in the sense that $f$ is assigned to any local point $x$. On the other hand, a pair of points with coordinates $(x_1,x_2)$ is not a local object. Or a functional such as $\int_{-\infty}^{\infty}dx\,f(x)$ is not a local object, because it is not assigned to a single point $x$.

Note that the pair $(x_1,x_2)$ could be reinterpreted as coordinates of a single point on a 2-dimensional manifold. With such reinterpretation, $(x_1,x_2)$ is local on the 2-dimensional manifold. But $(x_1,x_2)$ is not local on the initial 1-dimensional manifold. Similarly, the functional above could be reinterpreted as a local object on some infinite-dimensional space in functional analysis, but it is not local on the initial 1-dimensional space.

 

[bhobba]

Is that what non-local means?

I suppose its somewhat just semantics, but locality is the idea that only something infinitesimally close to an object can be affected by it in an infinitesimal time. Since the wave-function depends on x1 and x2 which indeed can be any distance apart its non-local. Its the same reason Newtons Law of gravitation, for example, is non-local.

Thanks
Bill

 

[Demystifier]

I suppose its somewhat just semantics, but locality is the idea that only something infinitesimally close to an object can be affected by it in an infinitesimal time.

When a mathematician explains something to a mathematician in terms of "infinitesimal" quantities, it cannot end good.

 

[naima]

There is obviously a false asumption in Bell's theorem.
is it in an equality where $\lambda$ appears or in the way to make it disappear with an integral?

 

[bhobba]

When a mathematician explains something to a mathematician in terms of "infinitesimal" quantities, it cannot end good.

Non standard analysis maybe.

Although I am formally trained in math I wouldn't necessarily describe myself as a mathematician these days. I was really into rigor at one time and my teachers said I most definitely had pure math tendencies even though my degree was in applied math. I would ask all these questions like how can you reverse integration there, you haven't proven what you did there - in the Heavisde function what is its value at the discontinuity (my lecturer said I knew you would ask that, just knew it - forget about it) - you get the picture. In exasperation one lecturer said he could show me books that took care of all that but you wouldn't read them. As time went by it has become clearer and clearer he was right and my attitude to rigor is now rather blase.

There is a notoriously difficult theorem to prove called the Feller-Erdös-Pollard theroem:
http://galton.uchicago.edu/~lalley/Courses/Summer/Renewal2.pdf

I came up with a really neat proof - only trouble was it relied on exchange of limits - and the caveat - 'and is not supported by any proper additive subgroup of the integers' wasn't required. It was wrong - so sometime rigor is required - trouble is knowing exactly when.

Thanks
Bill

 

[bhobba]

There is obviously a false asumption in Bell's theorem.

That would be news to the countless number of people that have gone through it and didn't notice it.

Thanks
Bill

 

[Demystifier]

There is obviously a false asumption in Bell's theorem.
is it in an equality where $\lambda$ appears or in the way to make it disappear with an integral?

Can you be more specific?

 

[naima]

In one of the Bell's proof we can find
P(a,b,$\lambda$) = P(a,$\lambda$)P(b,$\lambda$)
and
$P(a,b) = \int d\lambda f(\lambda) P(a,b,\lambda)$

 

[Demystifier]

As time went by it has become clearer and clearer he was right and my attitude to rigor is now rather blase.

So when you saw
https://www.physicsforums.com/threa...luding-dirac-delta.873711/page-2#post-5487662
I guess your instincts told you immediately that it can easily be made rigorous as sketched in
https://www.physicsforums.com/threa...luding-dirac-delta.873711/page-2#post-5488516
Unfortunately, as this thread has shown, some (otherwise smart) people never develop such instincts.

 

[Demystifier]

In one of the Bell's proof we can find
P(a,b,$\lambda$) = P(a,$\lambda$)P(b,$\lambda$)
and
$P(a,b) = \int d\lambda f(\lambda) P(a,b,\lambda)$

It is the first equation which is "false", in the sense that it is the assumption of statistical independence (which corresponds to the assumption of locality) contradicted by experiments and predictions of QM. The second equation cannot be false, because it is one of the basic general laws in the theory of probability.

 

[naima]

When several possibilities contribute to give a pattern Bell write that according to the theory of probabilities you have to sum them!
The Young device is a counterexample.
So i think that there are at less 2 wrong or "false" assumptions.

 

[N88]

Then you note wrong - I did. Why you don't get it beats me.

With respect Bill, I "noted" correctly. I asked for YOUR mathematical moves from LHS to RHS and you gave me no math!

For the last time the LHS is the expectation of the multiple of the the outcomes. The outcomes are 100% correlated so regardless you get the RHS ie its independent of probabilities, this follows immediately from what an expectation is.

? The LHS and the RHS both equal the expectation. I suspect you mean that the correlation ranges from 100% correlated to 100% anti-correlated? That explains very little about that eqn (3), but (see below) this issue can be put to rest now.

Dymystifyer told you, you can slog through the math if you like and do a tedious calculation. Do that if you don't get what I said. If you find that difficult then this is not the paper you should be studying - study the paper I suggested. If you want someone do actually do the calculation for you then start a separate thread - but don't be surprised if no one answers - most are like me and don't like doing and posting tedious calculations especially for things that are reasonably obvious. They will ask, at a minimum, for you to at least post your attempt at it and where you are stuck.
Thanks, Bill

The math can be written in one line. But I was keen to see your "tedious mathematical slog" to learn if you thought non-locality (NL) was anywhere involved. From your other recent comments here, I take it that you (like me) are not in Demystifier's camp when it comes to NL being involved in Bell's (1964) equation (3)? I'm OK with that.

 

[Demystifier]

When several possibilities contribute to give a pattern Bell write that according to the theory of probabilities you have to sum them!
The Young device is a counterexample.

First, you are doing a category mistake. You cannot use a physical experiment to prove or disprove a mathematical theorem.

Second, the relevant mathematical theorem in this case is the claim that probabilities should be summed within the same probability space. But two different experiments (one experiment with one open slit and the other experiment with the other open slit) correspond to two different probability spaces, so in the Young-device case the theory of probability does not imply (and Bell does not assume) that probabilities should be summed.

In QM, the same probability space means the same wave function. So with a single wave function $\psi(x)$, the probability density is $p(x)=|\psi(x)|^2$. Even in the two slit experiment these probabilities can be added as in standard probability theory. For instance,
$$\int_{-\infty}^{\infty}dx \, p(x)=1$$

 

[Demystifier]

The math can be written in one line.

I would like to see that line.

 

[bhobba]

I guess your instincts told you immediately that it can easily be made rigorous

Actually since we were discussing how to square the Dirac Delta function and its part of this enlarged space of generalized functions I simply took on board that since such a space exists what you wrote was valid.

Actually my suspicion is your sketched 'rigorous' proof may have subtle issues. This whole generalized function thing is full of deep and sometimes surprising stuff like nuclear spaces that some of the greatest mathematicians of the 20th century such as Grothendieck was involved in.

I still think unless such worries you just think of them as being FAPP the same as a test function in which case everything you did is fine.

I also have to say at one time it worried me and I did a long sojourn into Rigged Hilbert Spaces etc. I can say from that experience best to ignore it until you understand the physics reasonably well - it obscures what's important as far as the physics goes.

Thanks
Bill

 

[Demystifier]

I can say from that experience best to ignore it until you understand the physics reasonably well - it obscures what's important as far as the physics goes.

My experience is exactly the same.

 

[bhobba]

With respect Bill, I "noted" correctly. I asked for YOUR mathematical moves from LHS to RHS and you gave me no math!

Again, with respect, you were told its a tedious calculation. Its similar to the link I gave where I proved an entangled system acts like a mixed state. It's not hard but its tedious. What you generally do is what I did - see why its true rather than actually do the slog.

I haven't done that slog, but if you want to post your attempt at doing it we can go through it.

Its proof has got nothing to do with locality in the Bell sense, just in the sense Dymystifyer mentioned - it simply an application of quantum formalism.

Thanks
Bill

 

[N88]

There is obviously a false asumption in Bell's theorem.
is it in an equality where $\lambda$ appears or in the way to make it disappear with an integral?

My view goes something like this: In so far as our quantum world is concerned, there is an unrealistic assumption in Bell's (1964) theorem; i.e., the attribution of classicality (via $\lambda$) to quantum objects. So Bell Inequalities are consistent with classical objects and Bell's unrealistic assumption. But, via many experiments (fully consistent with quantum theory), Bell Inequalities are not consistent with quantum theory or quantum objects or our quantum world.

 

[stevendaryl]

My view goes something like this: In so far as our quantum world is concerned, there is an unrealistic assumption in Bell's (1964) theorem; i.e., the attribution of classicality (via $\lambda$) to quantum objects. So Bell Inequalities are consistent with classical objects and Bell's unrealistic assumption. But, via many experiments (fully consistent with quantum theory), Bell Inequalities are not consistent with quantum theory or quantum objects or our quantum world.

Isn't that what Bell was trying to prove? That QM is inconsistent with certain classical assumptions?

 

[Demystifier]

But I was keen to see your "tedious mathematical slog" to learn if you thought non-locality (NL) was anywhere involved. From your other recent comments here, I take it that you (like me) are not in Demystifier's camp when it comes to NL being involved in Bell's (1964) equation (3)? I'm OK with that.

So you don't agree with my post #38? May I know why?

 

[Demystifier]

In so far as our quantum world is concerned, there is an unrealistic assumption in Bell's (1964) theorem; i.e., the attribution of classicality (via λ) to quantum objects.

But Bell's $\lambda$ is equivalent to my C in post #28. Any yet, you said that my C is OK for you. So you are not being consistent.

 

[bhobba]

Isn't that what Bell was trying to prove? That QM is inconsistent with certain classical assumptions?

Indeed.

What I don't understand is why he is looking at Bells original paper. Dr Chinese's write up is much simpler:
http://drchinese.com/David/Bell_Theorem_Easy_Math.htm

Once that is understood then you can look at more advanced treatments.

Thanks
Bill

 

[Demystifier]

What I don't understand is why he is looking at Bells original paper. Dr Chinese's write up is much simpler:

He wants to prove that mainstream understanding is wrong. For that purpose it is much more cool to prove that Bell was wrong than to prove that Dr Chinese is wrong.

Similarly, people who want to prove that theory of relativity is wrong often look at Einstein's original papers. Physicists who accept theory of relativity rarely look at Einstein's original papers.

 

[Demystifier]

Isn't that what Bell was trying to prove? That QM is inconsistent with certain classical assumptions?

It looks as if some people don't understand the concept of reductio ad absurdum, i.e. making correct conclusion by taking a false assumption.

 

[naima]

I read the elegant paper of Dr Chinese.
Have hidden variables to give outputs to not measured things?
I think that it would be enough if they could predict them for all measurements actually done.

 

[Ilja]

About $P(a,b)=\int d\lambda f(\lambda)P(a,b,\lambda)$:

. The second equation cannot be false, because it is one of the basic general laws in the theory of probability.

Not exactly, it contains the assumption that there is no superdeterminisms. Else, this could be $P(a,b)=\int d\lambda f(a,b,\lambda)P(a,b,\lambda)$

 

[Ilja]

Have hidden variables to give outputs to not measured things?
I think that it would be enough if they could predict them for all measurements actually done.

It is, indeed, enough. And in particular dBB theory does not define outputs to not measured things. Except for positions. But for everything else, the "measurement result", even if it is defined in a deterministic way, depends also on the unknown position of the "measurement device". So, without measurement being done there is also no hidden state of the "measurement device", and, therefore, no predicted output.

This property is known as contextuality.

 

[Demystifier]

About $P(a,b)=\int d\lambda f(\lambda)P(a,b,\lambda)$:

Not exactly, it contains the assumption that there is no superdeterminisms. Else, this could be $P(a,b)=\int d\lambda f(a,b,\lambda)P(a,b,\lambda)$

Interesting! Is there a reference for that, or is it your own conclusion?

 

[Ilja]

No, this is my own remark. But it seems quite trivial. That superdeterminism means that the preparation is allowed to know in advance what will be decided by the experimenters is clear. Their decisions what to measure are a and b. Superdeterminism would allow the probability distribution of the hidden variables to depend on a and b. And with this additional possibility you would be unable to proof the theorem.