I A skeptic's view on Bohmian Mechanics

[Demystifier]

As part of my efforts there I look at what each interpretations contributes to the understanding of the questions that I find good foundations should settle.

It seems to me that you don't catch the idea of Anderson's "More is Different".

 

[A. Neumaier]

Bohmian mechanics needs to model the complete system, including the apparatus and the environment, in order to (perhaps) agree with the QM predictions.

Yes. In Bohmian mechnaics, one always need to study the whole universe, in order to make a statement about even the tiniest system. It is a waste of virtual complexity.

Fortunately, nobody needs to actually do it, as the statistical predictions are (under the standard assumptions) identical to ordinary quantum mechanics. Thus the whole stuff can be cut away with Ockham's razor, without losing anything of value.

There is no such fine tuning of initial conditions in BM.

Of course there is. You must assume the Bohmian particles are in some sort of equilibrium that gives the presupposed statistical state used to draw conclusions. Since there is no theory showing that this equilibrium is approached in finite time significantly less than the age of the universe, it is a fine-tuning assumption.

 

[rubi]

I'm glad that you asked it. Indeed, there is no free choice in BM because it is a fully deterministic theory. However, it is not superdeterministic theory. In a superdeterministic theory the initial conditions are fine tuned in order to simulate a law which does not really exist as a law. (For instance, 't Hooft studies superdeterministic local hidden variables for QM, where the appearance of non-locality is simulated by fine tuned initial conditions for local hidden variables.) There is no such fine tuning of initial conditions in BM. Similar to classical mechanics, BM is fully deterministic but not superdeterministic.

So why then can't I come up with a non-superdeterministic local hidden variable model by allowing $P(\lambda,\vec a,\vec b)$ to depend on the polarizer angles?

 

[zonde]

I'm glad that you asked it. Indeed, there is no free choice in BM because it is a fully deterministic theory. However, it is not superdeterministic theory. In a superdeterministic theory the initial conditions are fine tuned in order to simulate a law which does not really exist as a law. (For instance, 't Hooft studies superdeterministic local hidden variables for QM, where the appearance of non-locality is simulated by fine tuned initial conditions for local hidden variables.) There is no such fine tuning of initial conditions in BM. Similar to classical mechanics, BM is fully deterministic but not superdeterministic.

Does it mean that in BM $P(\lambda,\vec a,\vec b) = P(\lambda)$ is correct FAPP as far as entangled particles before measurements are considered?

 

[Demystifier]

Since there is no theory showing that this equilibrium is approached in finite time significantly less than the age of the universe

Yes there is, see the papers by Valentini.

 

[Demystifier]

So why then can't I come up with a non-superdeterministic local hidden variable model by allowing $P(\lambda,\vec a,\vec b)$ to depend on the polarizer angles?

You can't, due to the Bell theorem.

 

[Demystifier]

Does it mean that in BM $P(\lambda,\vec a,\vec b) = P(\lambda)$ is correct FAPP as far as entangled particles before measurements are considered?

No.

 

[rubi]

You can't, due to the Bell theorem.

Bell's theorem requires the $P(\lambda,\vec a,\vec b)=P(\lambda)$ assumption. The derivation is blocked, if you allow the correlations to be of this form:
$$\left<A(\vec a)B(\vec b)\right>=\int A(\vec a,\lambda) B(\vec b,\lambda) P(\lambda,\vec a,\vec b)\mathrm d\lambda$$
So if you are right that $P(\lambda,\vec a,\vec b)=P(\lambda)$ only implies determinism, but not superdeterminism, one should be able to come up with a local hidden variable theory as well.

 

[zonde]

No.

Why? Can't I describe source of entangled particles (including all the positions of these particles) without reference to whole universe?

 

[A. Neumaier]

Yes there is, see the papers by Valentini.

I had hoped for a substantial theorem, but again in vain. Since you didn't give a particular reference I checked some of Valentini's papers on the arXiv and looked at two recent papers:
https://arxiv.org/pdf/1310.1899.pdf
https://arxiv.org/abs/1609.04485
I found only numerical evidence for very simple systems (two oscillator degrees of freedom), and not even unequivocally positive one. In the second paper, his most recent on the subject, Valentini writes towards the bottom of p.2,

[...] an incomplete relaxation. This was shown to occur when a significant fraction of the trajectories remain confined to
sub-regions and do not explore the full support of |ψ| ^2.

The 2013 paper has a similar statement even in its abstract.

We initiate the study of relaxation to quantum equilibrium [...] can decay to a large nonequilibrium residue exceeding 10% of its initial value or it can become indistinguishable from zero (the equilibrium value).

Since this seems to be just the beginning of research of the topic ''We initiate'', and Valentini himself does not substantiate your claim, I consider your answer to my query unfounded.

Note also that the equilibrium assumption must be made for the system consisting of all particles in the universe, since these determine the fate of every subsystem. To remove the assumption one must show that it automatically holds generically for many-particle systems.

There appears to be no general argument covering the typical situation for many particles. But maybe I missed the correct paper that would demonstrate this. You can perhaps fill in this gap!

 

[Demystifier]

So if you are right that $P(\lambda,\vec a,\vec b)=P(\lambda)$ only implies determinism, but not superdeterminism,

I didn't say that $P(\lambda,\vec a,\vec b)=P(\lambda)$ implies determinism.

 

[Demystifier]

Can't I describe source of entangled particles (including all the positions of these particles) without reference to whole universe?

You can, but I don't see how this contradicts what I said.

 

[rubi]

I didn't say that $P(\lambda,\vec a,\vec b)=P(\lambda)$ implies determinism.

Ok, I didn't express myself properly, but that doesn't answer my question. The derivation of Bell's inequality is blocked if $P(\lambda)$ can depend on $\vec a$ and $\vec b$, so why can't I exploit this to come up with a local hidden variable theory? The usual answer is that such a theory would have to be superdeterministic, but apparently you disagree with this. If BM is not superdeterministic despite its violation of this property, why would a local hidden variable theory that exploits this property be superdeterministic?

 

[Demystifier]

I had hoped for a substantial theorem

You know my opinion about theorems for many-particle phenomena.

 

[Demystifier]

The derivation of Bell's inequality is blocked if $P(\lambda)$ can depend on $\vec a$ and $\vec b$, so why can't I exploit this to come up with a local hidden variable theory? The usual answer is that such a theory would have to be superdeterministic,

Can you pinpoint to the paper (preferably by Bell) where it is claimed that such a theory would be superdeterministic?

 

[A. Neumaier]

You know my opinion about theorems for many-particle phenomena.

The opinion I gather you have is that it is pointless to give a derivation of something so fundamental as showing that the major assumption put into the whole Bohmian edifice, namely that the Bohmian particles (including the measurement devices which are needed to guarantee consistence with quantum mechanics) are in quantum equilibrium.

You argue (by citing Valentini in this context) that the appropriateness of this assumption is a trivial consequence of having done some simulations with two position degrees of freedom that show that this assumption sometimes follows dynamically, and in other cases ostensibly fails to do so.

What everyone else would conclude from the same evidence is that the Bohmian edifice rests on very shaky foundations, unless God has initiated the universe in a specially fine-tuned configuration that satisfies this quantum equilibrium.

 

[rubi]

Can you pinpoint to the paper (preferably by Bell) where it is claimed that such a theory would be superdeterministic?

Bell, "La nouvelle cuisine", the paragraph after equation (14).

 

[zonde]

You can, but I don't see how this contradicts what I said.

Maybe it does not contradict.
Let me ask further questions. Let's say we specify the source with all the positions. For particular specification of source there is some freedom about possible specifications of the rest of the universe (there are many configurations of the rest of the universe that are consistent with particular specification of the source).
Is it fine so far?
So within this freedom we can specify different configurations of the rest so that entangled particles will (deterministically) end up being measured with different measurement settings, right?

Hmm, maybe I should ask my questions later as it seems that you have come under the fire of questions.

 

[Demystifier]

Maybe it does not contradict.
Let me ask further questions. Let's say we specify the source with all the positions. For particular specification of source there is some freedom about possible specifications of the rest of the universe (there are many configurations of the rest of the universe that are consistent with particular specification of the source).
Is it fine so far?
So within this freedom we can specify different configurations of the rest so that entangled particles will (deterministically) end up being measured with different measurement settings, right?

Hmm, maybe I should ask my questions later as it seems that you have come under the fire of questions.

Yes.

 

[Demystifier]

Ok, I didn't express myself properly, but that doesn't answer my question. The derivation of Bell's inequality is blocked if $P(\lambda)$ can depend on $\vec a$ and $\vec b$, so why can't I exploit this to come up with a local hidden variable theory? The usual answer is that such a theory would have to be superdeterministic, but apparently you disagree with this. If BM is not superdeterministic despite its violation of this property, why would a local hidden variable theory that exploits this property be superdeterministic?

The assumption that $P(\lambda)$ does not depend on $\vec a$ and $\vec b$ leads to Bell inequalities, which contradict experiments. Therefore, in an acceptable theory of hidden variables $P$ must depend on $\vec a$ and $\vec b$. The point is that such a dependence on $\vec a$ and $\vec b$ can be achieved in at least 3 different ways, namely
1) non-locality (which Bell himself preferred),
2) superdeterminism, or
3) signaling backwards in time.
The Bohmian theory exploits possibility 1). 't Hooft exploits possibility 2). Transactional interpretation exploits possibility 3).

In other words, superdeterminism implies $P(\lambda,\vec a,\vec b)$, but $P(\lambda,\vec a,\vec b)$ does not imply superdeterminism.

Does $P(\lambda,\vec a,\vec b)$ imply non-locality? Strictly speaking it does not. But it does if you exclude other alternatives such as 2) and 3).

If there are at least 3 options, why did Bell prefer option 1)? Probably because Bohmian mechanics looked like a very natural theory to him.

 

[Demystifier]

Bell, "La nouvelle cuisine", the paragraph after equation (14).

Thanks, see the reply above!

 

[rubi]

The assumption that $P(\lambda)$ does not depend on $\vec a$ and $\vec b$ leads to Bell inequalities, which contradict experiments. Therefore, in an acceptable theory of hidden variables $P$ must depend on $\vec a$ and $\vec b$.

No, it doesn't lead to Bell's inequalities. The joint assumption of both
(1) $A(\lambda,\vec a,\vec b) = A(\lambda,\vec a)$, $B(\lambda,\vec a,\vec b) = B(\lambda,\vec b)$
(2) $P(\lambda,\vec a,\vec b) = P(\lambda)$
leads to Bell's inequality. In a hidden variable theory, we can deny (1), which leads to non-locality or we can deny (2), which (according to Bell) leads to superdeterminism, or we can of course deny both. Apparently, Bohmian mechanics does not only violate (1), but it also violates (2), which would mean that it is not just non-local, but also superdeterministic (according to Bell).

Does $P(\lambda,\vec a,\vec b)$ imply non-locality? Strictly speaking it does not. But it does if you exclude other alternatives such as 2) and 3).

By using an expression for the correlations, given by $\left<A(\vec a) B(\vec b)\right> = \int A(\lambda,\vec a) B(\lambda,\vec b) P(\lambda, \vec a, \vec b)\mathrm d\lambda$, we can in principle write down a local hidden variable theory. The usual argument is that because it violates (2), it must be superdeterministic. I don't understand, why this argument does not equally apply to Bohmian mechanics, if BM violates (2), which it does.

 

[Demystifier]

unless God has initiated the universe in a specially fine-tuned configuration that satisfies this quantum equilibrium.

Equilibrium does not need fine tuning. It is non-equilibrium that does. One of the greatest mysteries in classical statistical mechanics is why the universe is so far from the thermodynamic equilibrium.

The positive side of thermodynamic non-equilibrium is that life is possible, so there are us who can be puzzled about it. If we lived in a quantum non-equilibrium we would have one more puzzle to solve, but the positive side would be that then we could use non-locality to communicate faster than light.

 

[A. Neumaier]

Equilibrium does not need fine tuning. It is non-equilibrium that does. One of the greatest mysteries in classical statistical mechanics is why the universe is so far from the thermodynamic equilibrium.

The positive side of thermodynamic non-equilibrium is that life is possible, so there are us who can be puzzled about it. If we lived in a quantum non-equilibrium we would have one more puzzle to solve, but the positive side would be that then we could use non-locality to communicate faster than light.

The Bohmian quantum equilibrium has nothing to do with the thermal equilibrium in quantum mechanics to which your remarks refer. Unlike the latter, which is a special state of some pieces of matter, the former is a universal assumption required to make Bohmian mechanics agree on average with quantum mechanics.

The overwhelming majority of ensembles violate the assumption of Bohmian quantum equilibrium, hence for them no theory guarantees that the resulting statistics is the same as in quantum mechanics. Thus the results from quantum statistical mechanics are also not guaranteed to hold, as they assume the laws of quantum mechanics.

To exclude all these from consideration and impose the requirement that the true collection of particles satisfies Bohmian quantum equilibrium is therefore a highly fine-tuned situation. Worse, the Bohmian universe is claimed to be deterministic, hence not represented by an ensemble but by a single collection of Bohmian particles. This is a pure deterministic state grossly violating the Bohmian quantum equilibrium hypothesis. Thus one needs to show that the predictions from this single deterministic state are nevertheless the same as that obtained from the assumption of Bohmian quantum equilibrium. This requires an ergodic theorem to be valid. But already the 2D simulations show that ergodicity cannot be expected.

Thus the Bohmian foundations are self-contradictory.

 

[Demystifier]

Worse, the Bohmian universe is claimed to be deterministic, hence not represented by an ensemble but by a single collection of Bohmian particles. This is a pure deterministic state grossly violating the Bohmian quantum equilibrium hypothesis.

As usual, you use double standards. Classical mechanics also claims to be deterministic, hence not represented by an ensemble but by a single collection of classical particles. Nevertheless, you do not say that themodynamics is incompatible with classical mechanics.

 

[rubi]

As usual, you use double standards. Classical mechanics also claims to be deterministic, hence not represented by an ensemble but by a single collection of classical particles. Nevertheless, you do not say that themodynamics is incompatible with classical mechanics.

This is why something like the ergodic hypothesis is needed. A classical mechanical system is definitely only in one pure state and all observables are determined by this state. In order to predict the correct thermodynamics, it must be the case that the particles of this one state are distributed in a chaotic way, such that averages turn out to be the same as if we had computed the ensemble average.

 

[A. Neumaier]

As usual, you use double standards. Classical mechanics also claims to be deterministic, hence not represented by an ensemble but by a single collection of classical particles. Nevertheless, you do not say that themodynamics is incompatible with classical mechanics.

This is because statistical mechanics does not make the assumption of equilibrium to deduce results.

The equilibrium assumption is made in statistical mechanics only for special systems known to be in equilibrium, such as crystals, water at rest, or a gas at uniform pressure and temperature. For all other cases (the majority of real world situations), there is nonequilibrium statistical mechanics which derives hydrodynamics and elasticity theory under much weaker assumptions, or kinetic theory for very dilute gases. The latter give a fairly realistic description of the universe at last, which is not in equilibrium.

On the other hand, Bohmian mechanics cannot even start without assuming Bohmian quantum equilibrium of the whole universe! This is quite a different sort of assumption. And it is assumed without any justification except that it is needed to reproduce Born's rule.

Thus the same standards result in a very different appraisal of the two theories.

 

[Demystifier]

No, it doesn't lead to Bell's inequalities. The joint assumption of both
(1) $A(\lambda,\vec a,\vec b) = A(\lambda,\vec a)$, $B(\lambda,\vec a,\vec b) = B(\lambda,\vec b)$
(2) $P(\lambda,\vec a,\vec b) = P(\lambda)$
leads to Bell's inequality. In a hidden variable theory, we can deny (1), which leads to non-locality or we can deny (2), which (according to Bell) leads to superdeterminism, or we can of course deny both. Apparently, Bohmian mechanics does not only violate (1), but it also violates (2), which would mean that it is not just non-local, but also superdeterministic (according to Bell).By using an expression for the correlations, given by $\left<A(\vec a) B(\vec b)\right> = \int A(\lambda,\vec a) B(\lambda,\vec b) P(\lambda, \vec a, \vec b)\mathrm d\lambda$, we can in principle write down a local hidden variable theory. The usual argument is that because it violates (2), it must be superdeterministic. I don't understand, why this argument does not equally apply to Bohmian mechanics, if BM violates (2), which it does.

Ah, now I see the source of confusion. When I was referring to a quantity $P(\lambda)$ for the first time, it was not the same quantity as you defined it above. What you call $P(\lambda)$, Bell calls $\rho(\lambda)$.

So let me explain it all over again, now using your conventions. In BM we have
$$A(\lambda,\vec a,\vec b) \neq A(\lambda,\vec a)$$
$$B(\lambda,\vec a,\vec b) \neq B(\lambda,\vec b)$$
So when the measurement setup changes, then $A(\lambda,\vec a,\vec b)$ and $B(\lambda,\vec a,\vec b)$ change. Due to this, there is no need to change $P(\lambda)$ and introduce superdeterminism in BM.

 

[Demystifier]

This is because statistical mechanics does not make the assumption of equilibrium to deduce results.

The equilibrium assumption is made in statistical mechanics only for special systems known to be in equilibrium, such as crystals, water at rest, or a gas at uniform pressure and temperature. For all other cases (the majority of real world situations), there is nonequilibrium statistical mechanics which derives hydrodynamics and elasticity theory under much weaker assumptions, or kinetic theory for very dilute gases. The latter give a fairly realistic description of the universe at last, which is not in equilibrium.

On the other hand, Bohmian mechanics cannot even start without assuming Bohmian quantum equilibrium of the whole universe! This is quite a different sort of assumption. And it is assumed without any justification except that it is needed to reproduce Born's rule.

Thus the same standards result in a very different appraisal of the two theories.

As I already said, there is no point in arguing about BM as we (you and me) cannot even agree of the foundations of classical statistical mechanics, or even on the concept of probability. I wish you a good luck with your thermal interpretation inspired by your own interpretation of thermodynamics and statistical mechanics.

 

[rubi]

Ah, now I see the source of confusion. When I was referring to a quantity $P(\lambda)$ for the first time, it was not the same quantity as you defined it above.

What other quantity $P(\lambda)$ did you refer to then?

Due to this, there is no need to change $P(\lambda)$ and introduce superdeterminism in BM.

It seems like $P(\lambda)$ depends on $\vec a$, $\vec b$ through $\lambda$, which determines the settings of the measurement apparata. In order to compute expectation values like $\left<A(\vec a)B(\vec b)\right>$, you need to tune $\lambda$ such that it reflects the particular settings.