I New experimental support for pilot wave theory?

[naima]

There are many things that i do not want to study. I do not want to waste time with the fractal QM of Nottale or to learn C++.
But when i wonder if there are recursive call in this language, i appreciate a yes/no answer.

My question was about the use of operators and eigenvectors in BM. Bhobba said that it is just an interpretation an thar it uses the Hilbert space machinary.
It seems that it is different. There a wave which obeys the Schrodinger equation but my question was mainly about the particles.
If there is no vector state in the Hilbert space associated to a position in the configuation space write it. Is there even a supperposition principle for particle states?

 

[bhobba]

My question was about the use of operators and eigenvectors in BM. Bhobba said that it is just an interpretation an thar it uses the Hilbert space machinary.

Your error in logic is assuming that it doesn't have more than operators and Hilbert space machinery.

In BM the usual QM formalism emerges from its assumptions rather than simply being assumed. It also has a different interpretation eg its probabilistic nature is due to lack of knowledge of initial conditions rather than being inherent and you don't have improper mixed states, they are all proper.

That's what interpretations do - they add more stuff to the formalism, but the formalism doesn't change.

Thanks
Bill

 

[naima]

Does it use superposition principle for particles? yes? no?
When you read a text it is not easy to find the words which are avoided

 

[bhobba]

Does it use superposition principle for particles? yes? no?
When you read a text it is not easy to find the words which are avoided

The superposition principle is simply that states form a vector space. Since that is part of the standard QM machinery it must.

Thanks
Bill

 

[Demystifier]

If there is no vector state in the Hilbert space associated to a position in the configuation space write it. Is there even a supperposition principle for particle states?

In Bohmian mechanics there is an eigenstate associated with the position operator, but there is no eigenstate associated with the actual position. Likewise, superposition principle is valid for wave functions, but not for actual positions. In Bohmian mechanics there are two kinds of "position" and your confusion seems to emerge from a failure to understand the difference between them.

 

[Demystifier]

Does it use superposition principle for particles? yes? no?

No. Only for wave functions.

 

[naima]

Likewise, superposition principle is valid for wave functions, but not for actual positions.

Thanks
So there is only superposition for the pilot wave.

 

[Demystifier]

Yes, exactly.

 

[stevendaryl]

John Bell wrote a short article about the relationship between Bohmian mechanics and Many-Worlds that was kind of interesting. But here's my take on it:

Many-Worlds starts with the not-too-radical assumption that quantum mechanics (or rather, the Schrodinger equation) applies to arbitrarily large systems, such as the entire universe. So there is no "wave function collapse" if you include the entire universe as your "system". That means that there is no obvious way to interpret the wave function as giving probabilities (via the Born rule), because for the universe as a whole, there are no measurements performed.

Bohmian mechanics, in a way of looking at it, starts in the same place, with a wave function for the universe that evolves unitarily. But it doesn't consider this universal wave function to be "the world". Instead, the world is basically a point in 3N-dimensional configuration space (configuration space being the set of simultaneous positions of each of the N particles in the universe --- this description might only make sense nonrelativistically, where we can think of the number of particles in the universe as a constant). The role of the wave function, then, is to give a probability distribution for the location of the world in phase space.

As an aside---something that's aesthetically unpleasing to me about Bohmian mechanics is the fact that it lacks the symmetry of standard quantum mechanics. The way that standard quantum mechanics is formulated using Dirac's abstract bra-ket notation, there is a kind of "coordinate independence"; the formalism has the same form in any basis. So you don't have to consider configuration space to be primary, you can work in momentum space, or you can work in a basis of harmonic oscillator eigenstates, or you can assume the existence of abstract internal variables such as spin that have no translation into configuration space. In contrast, Bohmian mechanics insists that configuration space is what's "real". The Born probability rules applied to observables besides position are not considered fundamental; if the observable does not relate to position, then it has no direct physical meaning.

Here's where Bell departs from the usual Bohmian mechanics. For the usual formulation of Bohmian mechanics, the relationship between the wave function and a probability distribution in configuration space is dynamic: You assume that that is true initially, and you propose equations of motion for configuration space that preserves this relationship. What Bell pointed out is that trajectories in configuration space are not (directly) observable. The only thing you know about the past history of the universe is whatever is recorded in persistent records. But those persistent records are part of the present state of the universe. So there is a sense in which trajectories are redundant. In light of this, Bell proposed an interpretation of quantum mechanics that was a kind of unification of the Bohmian and Many-Worlds interpretations. In his unified interpretation, the only dynamics is the unitary evolution of the universal wave function. There is no secondary equations of motion for configuration space. Instead, he proposed that at each moment, the world had a probability of being at any point in configuration space, according to the Born rule. So there are no trajectories, the universe just hops from point to point in configuration space randomly.

 

[N88]

… … I gave you the answer. I will repeat it. The spins are correlated that means you will get the same answer when multiplied regardless of outcome. It simple when you think about it. If you want me to go deeper you will need to explain precisely what you don't understand about it. The LHS is a statement about expectations. The RHS is what you must get regardless because the singlet state is 100% correlated.

But I again urge you to study the modern link I gave before delving into Bells paper.
Thanks, Bill

A. Delving into Bell's paper is neither relevant nor my intention here. 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). I expected your math would allow us to discuss where "non-locality" (allegedly) arises in such math.

B. As for the paper that you favour http://arxiv.org/pdf/1212.5214v2.pdf: it makes my point explicitly (p.2, with my emphasis):

"In other words, in a counterfactual-definite theory it is meaningful to assign a property to a system (e.g. the position of an electron) independently of whether the measurement of such property is carried out. [Sometime this counterfactual definiteness property is also called “realism”, but it is best to avoid such philosophically laden term to avoid misconceptions.]
Bell’s theorem can be phrased as “quantum mechanics cannot be both local and counterfactual-definite”. A logically equivalent way of stating it is “quantum mechanics is either non-local or non counterfactual-definite.”​

My interest is in learning about local and non counterfactual-definite QM.

 

[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?

 

[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.