I New experimental support for pilot wave theory?

[N88]

So you do accept that there is some extra entity, which here we call C, while Bell calls it $\lambda$. And like Bell, you assume that this extra quantity is local. But unlike Bell, you don't see a contradiction with QM. Am I right?So are you questioning this particular mathematical step in the Bell's derivation? Fine, now we know where exactly do you disagree with Bell. But Eq. (3) is a consequence of standard QM. People, like Zeilinger, who question the Bell's conclusions, do not question Eq. (3). So are you sure that Eq. (3) is the crucial issue for you? In other words, if you could prove (3), would you then accept nonlocality?

Anyway, the proof of (3) is straightforward but slightly tedious and boring. So let me just give you a few hints. You should use the singlet state defined in
https://en.wikipedia.org/wiki/Singlet_state
and properties of $\sigma$ matrices presented in
https://en.wikipedia.org/wiki/Pauli_matrices

I am NOT in any way questioning Bell (1964), equation (3). I was questioning the way that YOU personally fill in the missing details. So, to be clear: Are you saying that non-locality is implicit in the way Bell's equation (3) is worked out? If so, could you show me your calculation and where the non-locality arises? Thanks.

 

[bhobba]

But I again note that, despite my providing the context, you do not give me the way that you mathematically link the LHS to the RHS of Bell (1964), eqn (3)

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

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.

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

 

[bhobba]

I am NOT in any way questioning Bell (1964), equation (3). I was questioning the way that YOU personally fill in the missing details. So, to be clear: Are you saying that non-locality is implicit in the way Bell's equation (3) is worked out? If so, could you show me your calculation and where the non-locality arises? Thanks.

It has nothing to do with locality. As has been explained to you, its simply the result of a tedious calculation from the formalism of QM, although the result is fairly obvious as I have indicated.

Its now rather obvious you do not have the background to understand the paper otherwise you would simply do the calculation and move on.

Thanks
Bill

 

[Demystifier]

I am NOT in any way questioning Bell (1964), equation (3). I was questioning the way that YOU personally fill in the missing details. So, to be clear: Are you saying that non-locality is implicit in the way Bell's equation (3) is worked out? If so, could you show me your calculation and where the non-locality arises? Thanks.

Yes, the non-locality is implicit in equation (3). To make it explicit, one needs to make a few steps. Some of those steps may look obvious to you, but let me present them all just to be as explicit as possible.
- First, the notation <something> really means $<\psi|something|\psi>$.
- Second, it is said that it is for the singlet state, which means
$$|\psi>=|\uparrow>|\downarrow>-|\downarrow>|\uparrow>$$
- Third, this $|\psi>$ cannot define the full wave function, because the full wave function depends also on positions. So the above is just a short-hand notation for something like
$$\psi({\bf x}_1, {\bf x}_2)=\psi_A({\bf x}_1)|\uparrow>\otimes\psi_B({\bf x}_2)|\downarrow>-
\psi_A({\bf x}_1)|\downarrow>\otimes\psi_B({\bf x}_2)|\uparrow>$$
- Forth, we see that it cannot be written in the form
$$\psi_A({\bf x}_1)|something_1>\otimes\psi_B({\bf x}_2)|something_2>$$
so we cannot say that the first particle is in the state $\psi_A({\bf x}_1)|something_1>$ and the second particle in the state $\psi_B({\bf x}_2)|something _2>$. 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.

 

[bhobba]

$$\psi({\bf x}_1, {\bf x}_2)=\psi_A({\bf x}_1)|\uparrow>\otimes\psi_B({\bf x}_2)|\downarrow>-

Hmmmm. I thought there should be a 1/root 2 there.

Also I don't think there is any locality or non locality involved in equation 3. I know exactly what you are saying eg (see post 22):
https://www.physicsforums.com/threads/is-the-cat-alive-dead-both-or-unknown.819497/page-2

Its just basic QM.

To the OP the above math is rather close to the math in deriving equation 3 instead of just seeing it must be like that. If you still can't do it post your attempt.

Thanks
Bill

 

[Demystifier]

Hmmmm. I thought there should be a 1/root 2 there.

It should (my bad), but it's not important for understanding the origin of nonlocality.

Also I don't think there is any locality or non locality involved in equation 3.

Then why did you like my post #38?

 

[bhobba]

Actually, they are. All mathematical approaches use a wave function (or something equivalent), and wave function for entangled particles is a nonlocal object. For a 2-particle entangled wave function $\Psi({\bf x}_1,{\bf x}_2)$, you cannot say what is the value of wave function at the position ${\bf x}_1$.

It should (my bad), but it's not important for understanding the origin of nonlocality.

Absolutely - its simply tedium.

Then why did you like my post #38?

For reference here is post 38

Actually, they are. All mathematical approaches use a wave function (or something equivalent), and wave function for entangled particles is a nonlocal object. For a 2-particle entangled wave function $\Psi({\bf x}_1,{\bf x}_2)$, you cannot say what is the value of wave function at the position ${\bf x}_1$.

My bad. I was responding to it being an entangled 'single' object and should have mentioned such doesn't really imply locality or non locality.

Thanks
Bill

 

[Demystifier]

My bad. I was responding to it being an entangled 'single' object and should have mentioned such doesn't really imply locality or non locality.

So do you or do you not agree that entangled wave function by itself is a non-local object?

Of course, if you say that it is a non-local object, it does not mean that this implies that nature itself is non-local. For instance, the statistics of classical Bertlmann socks can be described by a non-local object, and yet the nature of Bertlmann socks is local. If wave function by itself is not a fundamental object, then the question of non-locality of nature is a question of non-locality of the fundamental objects, whatever they are. In the case of Bertlmann socks, the fundamental objects are the socks themselves, not their statistical description. The power of the Bell theorem is precisely in the fact that he does not assume that wave function is a fundamental object. Instead, fundamental objects in his theorem are some very general objects called $\lambda$. His theorem refers to those general objects. If he assumed that wave function was a fundamental object, then the proof of non-locality of nature would be trivial. If wave function were fundamental, then my post #38 would already be a proof that nature is not local.

 

[bhobba]

If wave function by itself is not a fundamental object, then the question of non-locality of nature is a question of non-locality of the fundamental objects, whatever they are.

Ahhhh. Now I see your point. Yes its non local in your sense (ie the wave-function depends on both positions and in general can not be factored). But since the ontological status of a state the formalism is silent on its of no moment.

Thanks
Bill

 

[Demystifier]

Ahhhh. Now I see your point. Yes its non local in your sense (ie the wave-function depends on both positions and in general can not be factored). But since the ontological status of a state the formalism is silent on its of no moment.

Exactly!

 

[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?