The Refutation of Bohmian Mechanics

[A. Neumaier]

It does give some information about the probability of finding a particle somewhere in spacetime, provided that you restrict the probability to a finite region of spacetime.

No. My argument remains essentially the same if one also imposes a Delta on space. The resulting probability still depends on the choice of T and vanishes in the limit T to infty.

 

[Demystifier]

My point is that, at the end, a real practical physical question one poses is either a question that refers to a finite portion of spacetime (in which case the answer is finite, even if T-dependent), or a question (like in the last paragraph of Sec. 2) the answer to which does not depend on T at all.

Besides, even if a priori probability of something vanishes, it does not mean that it is impossible. For example, the a priori probability that a random real number between 0 and 1 is equal to 0.5 vanishes. Yet, it is not impossible that a random real number between 0 and 1 is equal to 0.5.

 

[A. Neumaier]

My point is that, at the end, a real practical physical question one poses is either a question that refers to a finite portion of spacetime (in which case the answer is finite, even if T-dependent), or a question (like in the last paragraph of Sec. 2) the answer to which does not depend on T at all.

Besides, even if a priori probability of something vanishes, it does not mean that it is impossible. For example, the a priori probability that a random real number between 0 and 1 is equal to 0.5 vanishes. Yet, it is not impossible that a random real number between 0 and 1 is equal to 0.5.

That may well be the case.

But you cannot uphold the probability interpretation you give to (6), since as we have seen, it is manifestly violated in every stationary state. It results in a zero probability for _every_ bounded box in space-time, which is impossible for a meaningful probability.

 

[Demystifier]

No, but to explain why, I would need to repeat what I already said. I don't see a point in doing this.

 

[Varon]

Just to clarify something. Is it true that the wave function in de Broglie/Bohm mechanics is aware of all configurations of all particles in the entire universe at once. For example. A particle changing a spin 200 billion light years away can be detected by the bohmian wave function here on earth? Doesn't this establish an absolute space and time?

 

[Demystifier]

Just to clarify something. Is it true that the wave function in de Broglie/Bohm mechanics is aware of all configurations of all particles in the entire universe at once. For example. A particle changing a spin 200 billion light years away can be detected by the bohmian wave function here on earth? Doesn't this establish an absolute space and time?

Not necessarily. To see how it might NOT establish an absolute space and time see
http://xxx.lanl.gov/abs/1002.3226 [Int. J. Quantum Inf. 9 (2011) 367-377]

 

[A. Neumaier]

No, but to explain why, I would need to repeat what I already said. I don't see a point in doing this.

You could instead point ot the respective posts.

But I don't see any that addresses the discrepancy between the claim that (6) is the probability density for being in a given space-time region, and the zero result that one gets when applying it to an arbitrarty stationary state.

If your foundations lead to _some_ false conclusions then they are faulty, even if you can produce also some valid conclusions from them.

 

[akhmeteli]

Dear A. Neumaier,
Thank you for your reply.

I do follow the rules, of which I quote here the relevant part:

No, you don’t, sorry. In the phrase from the rules that you quoted: “It is against our Posting Guidelines to discuss, in most of the PF forums or in blogs, new or non-mainstream theories or ideas that have not been published in professional peer-reviewed journals or are not part of current professional mainstream scientific discussion” there is “or” in the phrase “or are not part of current professional mainstream scientific discussion”, not “and”. So what you do is you discuss your new theory or idea that has not been published in professional peer-reviewed journal, and this is definitely against the rules.

_Everything_ I say is my personal opinion (though it often agrees with established scientific fact), and when appropriate I give references to what I believe is a valid source. It is neither against the rules to voice a personal opinion (most contributors do that regularly) nor to refer to unpublished articles if they are ''part of current professional mainstream scientific discussion''

See above.

(Streater's book shows that my remarks on wrong signs in time correlations in BM is part of that).

Streater has no responsibility for your claim, he is neither your editor, nor your referee. No citation can turn your unpublished paper into a published one.

The remainder of the discussion has shown in which sense my statement was a fact.

And I insist that your claim is misleading in the best case, for reasons outlined, say, in my post 8 in this thread (with all due respect, your reply did not look satisfactory).

That a fact has no convenient reference doesn't make it a personal theory in the sense that it would belong only to the IR section of PF. It just takes more space to provide the evidence, and the discussion with Demystifier has provided it.

I respectfully reject this reasoning, as otherwise any independent research “just takes more space to provide the evidence”.

Our understanding of the rules is different, and your arguments did not convince me that your interpretation is better than mine. Only superior arguments than my own are suitable to convince me of something different from what I am already convinced of.

It is your sacred right to disagree with me on rules or on anything else. Anyway, I am no mentor. However, if I am to believe your phrase “I fully respect the rules as I understand them”, I have to believe that your understanding of the rules flip-flips on a weekly basis. Indeed, you referred to your paper in posts 4 and 7. That means that you believed then that it was OK. In posts 24 and 78 you wrote “But I cannot discuss my claim further because of the PF rules” or something to that effect. So perhaps you believed at that time that it was not quite OK. Then in post 114 you explained us that it was perfectly OK…

Look, A. Neumaier, you just made a mistake when you cited your unpublished paper. As I said, that happened to me too. Just don’t try to defend this mistake, you’re only making it worse. Move on.

 

[A. Neumaier]

And I insist that your claim is misleading in the best case, for reasons outlined, say, in my post 8 in this thread (with all due respect, your reply did not look satisfactory).

So let me try again:

Have you replied to the critique by Marchildon? (http://arxiv.org/PS_cache/quant-ph/pdf/0007/0007068v1.pdf )(It may well be that I just missed your reply).

I posted a link to the arxiv paper http://lanl.arxiv.org/pdf/quant-ph/0001011 claiming that Bohmian mechanics contradicts quantum mechanics since it gets the wrong time-correlations.

You posted a link to another arxiv paper (quoted above) rebutting my claim, not by pointing out a mistake in my exact calculation but by saying it doesn't matter, since things come out all right if one adds an approximate argument involving an observer.

Thus readers have all the information to decide for themselves and make up their mind about the truth in this matter. All has been said about it, so why should I add anything to what I had already said clearly?

Those who can judge for themselves will need no further information.

Those who need some authority to decide for themselves have multiple options, depending on what they trust:
Those who believe that truth is determined by peer review will disregard both papers since they are only arxiv preprints.
Those believing in the shining virtues of my science advisor medal or the title of a professor may perhaps give my arguments more weight.
Those believing that the last word is the final word may perhaps give preference to Marchildon's view.
Or they might read another arxiv paper: http://arxiv.org/abs/quant-ph/0008007 and conclude from it that Marchildon's credibility is perhaps not so high.

I just said that I don't care since I lost interest in interpreting things a la Bohm.

It also seems to me that unitary evolution of quantum mechanics is fully adopted by the Bohm interpretation. If you disagree, please advise. If, however, you agree, then one would expect that discrepancies between standard quantum mechanics and the Bohm interpretation (if any) can only arise from the theory of measurement

Unitary evolution of the state only caters for agreement of single-time expectations.
It says nothing about time-correlations, which cannot even be expressed in the Schroedinger picture underlying Bohmian mechanics. That's why I had looked at time correlations, and I wrote it up since no one had it done before.

If this is so, then it may be good for the Bohm interpretation if it fails to faithfully reproduce the theory of measurement of standard quantum mechanics? Furthermore, you yourself describe the measurement mechanism as ill-defined.

The measurement mechanism is ill-defined both in standard QM and in Bohmian QM.

 

[Demystifier]

Unitary evolution of the state only caters for agreement of single-time expectations.
It says nothing about time-correlations, which cannot even be expressed in the Schroedinger picture underlying Bohmian mechanics.

Let us ignore Bohmian mechanics for a moment!

Are you saying that Schroedinger picture of QM is not physically equivalent to some other (Heisenberg?) picture? That some measurable predictions of QM cannot be made by using Schroedinger picture?

By the way, I have an objection against your
http://lanl.arxiv.org/pdf/quant-ph/0001011
The crucial question is whether the time correlation you discuss is measurable or not. You admit that it is not easy to measure it, but still you argue that it is measurable at least in principle. More specifically, you argue that one can measure f in Eq. (23) because one can measure both Re f and I am f. However, my objection is that one cannot measure both Re f and I am f SIMULTANEOUSLY, because these two operators do not commute. Therefore, one cannot measure f either. More precisely, one cannot measure the product q(s)q(t).

What one can do, is to measure q(s) and q(t) separately, and then multiply the results. But for THAT measurement, standard QM in Heisenberg picture, standard QM in Schrodinger picture, and Bohmian QM all have the same measurable predictions.

 

[A. Neumaier]

Let us ignore Bohmian mechanics for a moment. Are you saying that Schroedinger picture of QM is not physically equivalent to some other (Heisenberg?) picture? That some measurable predictions of QM cannot be made by using Schroedinger picture?

For single-time dynamics, the Schroedinger picture of QM is equivalent to the Heisenberg picture. Time-correlations can be formulated _only_ in the Heisenberg picture, hence the question of equivalence doesn't arise.

But one can conclude that the Heisenberg picture is more general than the Schroedinger picture. The Schroedinger picture only extends classical deterministic mechanics to the quantum case, while the Heisenberg picture extends classical stochastic processes.

 

[Demystifier]

Thanks! In the meantime, I have edited (extended) my post. Can you comment the added parts as well?

 

[akhmeteli]

Dear A. Neumaier,
Thank you for your reply.

So let me try again:

I posted a link to the arxiv paper http://lanl.arxiv.org/pdf/quant-ph/0001011 claiming that Bohmian mechanics contradicts quantum mechanics since it gets the wrong time-correlations.

You posted a link to another arxiv paper (quoted above) rebutting my claim, not by pointing out a mistake in my exact calculation but by saying it doesn't matter, since things come out all right if one adds an approximate argument involving an observer.

Thus readers have all the information to decide for themselves and make up their mind about the truth in this matter. All has been said about it, so why should I add anything to what I had already said clearly?

Those who can judge for themselves will need no further information.

Those who need some authority to decide for themselves have multiple options, depending on what they trust:
Those who believe that truth is determined by peer review will disregard both papers since they are only arxiv preprints.
Those believing in the shining virtues of my science advisor medal or the title of a professor may perhaps give my arguments more weight.
Those believing that the last word is the final word may perhaps give preference to Marchildon's view.
Or they might read another arxiv paper: http://arxiv.org/abs/quant-ph/0008007 and conclude from it that Marchildon's credibility is perhaps not so high.

I just said that I don't care since I lost interest in interpreting things a la Bohm.

I mentioned Marchildon’s paper just because I was trying to determine the status of your claim. You did give a direct (negative) reply to my question (“did you reply to the critique?”), but that made the status of your claim even more dubious.

Unitary evolution of the state only caters for agreement of single-time expectations.
It says nothing about time-correlations, which cannot even be expressed in the Schroedinger picture underlying Bohmian mechanics.

Why do you say that? There is a well-known relation between wavefunctions and operators in the Heisenberg picture and wavefunctions and operators in the Schroedinger picture. If you believe that time correlations can be expressed in the Heisenberg picture, you can express the wavefunctions and operators in the relevant expression via wavefunctions and operators in the Schroedinger picture. Furthermore, the expressions will be the same in standard quantum mechanics and in the Bohm interpretation, as the unitary evolution is the same and the wavefunction is the same. Am I missing something simple?

The measurement mechanism is ill-defined both in standard QM and in Bohmian QM.

Suppose I agree with that, for the sake of discussion (though I prefer the wording of my post 8: “unitary evolution and the theory of measurement of quantum mechanics, strictly speaking, contradict each other”). Then we seem to agree that 1. Unitary evolution is the same in standard quantum mechanics and the Bohm interpretation, and 2. The measurement mechanism is ill-defined both in standard QM and in Bohmian QM.
But there is nothing in standard QM but unitary evolution and the measurement mechanism. Therefore, if there is indeed any inconsistency between the Bohm interpretation and standard QM, it cannot arise from anywhere but the measurement mechanism. And the latter is, by the way, ill-defined. So can you reasonably blame the Bohm interpretation for inconsistency (if any) with the ill-defined part of standard QM? That is why I insist that your claim (Bohmian mechanics is inconsistent with standard QM once time correlations are taken into account), while looking damning for the Bohm interpretation, is misleading in the best case.

 

[vanhees71]

I haven't followed this thread in detail, but I think there's a common misconception showing up in your arguments concerning the choice of the "picture" of quantum-theoretical time evolution.

In the abstract Hilbert space the observables are represented by self-adjoint operators and the states by Statistical Operators (or state operators) which are self-adjoint positive semi-definite operators with trace 1. The time dependence of the observable operators and state operators is determined only up to a (time-dependent) unitary transformation.

The time evolution of observable quantities, however, is not affected by chosing a certain picture of time dependence, and the whole quantum dynamics can be formulated in an invariant way. E.g., let's look at a particle without spin. Then the funcamental observable operators are the postion and momentum operators of the particle, and any other observable is given as a function of these fundamental observables and perhaps the time. This possible time dependence is called explicit time dependence. The Hamilton operator is usually given by

\hat{H}(\hat{x},\hat{p})=\frac{\hat{p}^2}{2m} +V(\hat{x}).

Then, if \hat{A}(t,\hat{x},\hat{p}) is an observable operator, the operator associated with the time derivative of the corresponding observable is given by the "covariant time derivative",

\mathrm{D}_t \hat{A}=\frac{1}{\mathrm{i}} [\hat{A},\hat{H}] + \frac{\partial \hat{A}}{\partial t}.

The Statistical operator fulfills the von Neumann equation,

\mathrm{D}_t \hat{\rho}=0.

The actual mathematical time dependence can be shuffled nearly arbitrarily between the (fundamental) observable operators and the Statistical Operator. The extreme cases are the Heisenberg picture (full time dependence at the observable operators) and the Schrödinger pictures (full time dependence at the Stat. Op.). Everything in between is known as the one or other Dirac picture, among which the interaction picture is the mostly used one for time-dependent perturbation theory.

Observable are probabilities for the outcome of measurements of observables at any time. In the case of a pure state, where the Statististical Operator is given by a projection operator \hat{\rho}=|\psi(t) \rangle \langle \psi(t)|, the wave function is the probability amplitude to find a particle at a position x. If \ket{x,t} are the generalized common eigenvectors of the position operators \hat{x}(t) at time t,

\hat{x}(t)|x,t \rangle=x |x,t \rangle,

the wave function is given by

\psi(t,x)=\langle x,t|\psi(t) \rangle.

This wave function is determined by the time evolution of the observable operators and that of the state ket up to a (time-dependent) phase. The probability distribution for position at time, t is thus unique, i.e., independent of the choice of the picture of time evolution:

P(t,x)=|\psi(t,x)|^2.

For a general Statistical Operator this probability distribution is given by

P(t,x)=\langle x,t|\hat{\rho}|x,t \rangle,

whose time dependence is also unique, i.e., independent of the choice of picture.

No matter which "interpretation of quantum theory" one follows, it cannot depend on a particular choice of the picture of time evolution!

 

[Demystifier]

No matter which "interpretation of quantum theory" one follows, it cannot depend on a particular choice of the picture of time evolution!

Neumaier seems to think that the quantity <Psi|q(s)q(t)|Psi>, here expressed in the Heisenberg picture, cannot be calculated in the Schrodinger picture. But if he does, he is obviously wrong.

 

[A. Neumaier]

Thanks! In the meantime, I have edited (extended) my post. Can you comment the added parts as well?

Sure!

By the way, I have an objection against your
http://lanl.arxiv.org/pdf/quant-ph/0001011
The crucial question is whether the time correlation you discuss is measurable or not. You admit that it is not easy to measure it, but still you argue that it is measurable at least in principle. More specifically, you argue that one can measure f in Eq. (23) because one can measure both Re f and I am f. However, my objection is that one cannot measure both Re f and I am f SIMULTANEOUSLY, because these two operators do not commute. Therefore, one cannot measure f either. More precisely, one cannot measure the product q(s)q(t).

As I mentioned in the paper: What one actually measures when probing time correlations are responses to a small external force, which, by standard statistical mechanics, give the time-correlations.

One can also compute the expectations of q(s)q(t)q(s), which is Hermitian, so that your objection does not apply. I'd be very surprised if the answer obtained from Bohmian mechanics were the same as that from standard QM.

 

[A. Neumaier]

No matter which "interpretation of quantum theory" one follows, it cannot depend on a particular choice of the picture of time evolution!

One cannot define time correlations in the Schroedinger picture since they involve operators at different times, which is meaningless in the Schroedinger picture.

Of course, one can rewrite any expression (and therefore also <q(s)q(t)> in the Heisenberg picture as an equivalent expression in the Schroedinger picture, but one loses the interpretation. Moreover, one loses uniqueness of the representation since it is not clear which time one should use for the Schroedinger state.

Thus as soon as one considers multiple times, the Heisenberg picture is essential for the interpretation of the formulas.

 

[Demystifier]

One cannot define time correlations in the Schroedinger picture since they involve operators at different times, which is meaningless in the Schroedinger picture.

This is simply wrong. The quantity
<Psi| q(s) q(t) |Psi>
written in the Heisenberg picture is equal to
<Psi(s)| q U(s) U^*(t) q |Psi(t)>
written in the Schrodinger picture (at DIFFERENT times), where
U(t)=exp{-i \hbar H t}
and U^* is the hermitian conjugate of U.

 

[A. Neumaier]

This is simply wrong. The quantity
<Psi| q(s) q(t) |Psi>
written in the Heisenberg picture is equal to
<Psi(s)| q U(s) U^*(t) q |Psi(t)>
written in the Schrodinger picture, where
U(t)=exp{-i \hbar H t}
and U^* is the hermitian conjugate of U.

You quoted me out of context. I had added

Of course, one can rewrite any expression (and therefore also <q(s)q(t)> in the Heisenberg picture as an equivalent expression in the Schroedinger picture, but one loses the interpretation. Moreover, one loses uniqueness of the representation since it is not clear which time one should use for the Schroedinger state.

You just rewrote the expression, as I said one always could. But I also said that one loses the meaning. Multi-time correlations have a natural meaning in the Heisenberg picture, but the associated operators in the Schroedinger picture are just formal expressions without any meaning.

For example, if you rewrite <q(t)q(s)q(t)> along the lined indicated by you, one gets psi(t)^*A(t,s) psi(t), with a Hermitian operator
A(t,s):=q U(t-s) q U(s-t) q.

This immediately raises several issues:

(i) This is not the Schroedinger picture since A(t,s) still depends on $t$, whereas in the
Schroedinger picture, observables are supposed to be independent of t.

(ii) The Hermitian operator A(s,t) has no discernible meaning, except that inherited from the Heisenberg picture, which must therefore be regarded as the fundamental picture.

(iii) How would you measure A(s,t)? There is no associated measurement theory.

 

[Demystifier]

(i) This is not the Schroedinger picture since A(t,s) still depends on $t$, whereas in the
Schroedinger picture, observables are supposed to be independent of t.

I see your point, but I would interpret it differently. The formal manipulations that lead to this expression show that it is not really true that observables (assuming that any hermitian operator is an observable) in the Schrodinger picture are independent of time. In other words, I think it is more appropriate to generalize the Schrodinger picture itself, rather than to complain that the Schrodinger picture is not general enough.

(ii) The Hermitian operator A(s,t) has no discernible meaning, except that inherited from the Heisenberg picture, which must therefore be regarded as the fundamental picture.

You may say so, but I would prefer to say that this quantity has the same meaning in both pictures, without insisting that one picture is more fundamental than the other.

(iii) How would you measure A(s,t)? There is no associated measurement theory.

Is there an associated measurement theory in the Heisenberg picture? If yes, then I can use the same theory (suitably rewritten) in the Schrodinger picture. If no (which probably is the right answer) then there is a physical reason for this, which has nothing to do with a choice of picture.

The paper
http://xxx.lanl.gov/abs/quant-ph/0509019 [Found. Phys. 36, 1601 (2006)]
may also be relevant here. (It cites your paper.)

 

[A. Neumaier]

I see your point, but I would interpret it differently. The formal manipulations that lead to this expression show that it is not really true that observables (assuming that any hermitian operator is an observable) in the Schrodinger picture are independent of time. In other words, I think it is more appropriate to generalize the Schrodinger picture itself, rather than to complain that the Schrodinger picture is not general enough.

Then you need to define the Schroedinger meaning of a measurement of a time-dependent observable A(t) in a time-dependent state psi(t) as t varies over time (and of A(t,s), which is also dependent on a former or later time).

This is not a trivial generalization of the standard textbook meaning of the Schroedinger picture, since one cannot appeal to ensembles of identically prepared systems anymore. So what is your proposal for defining this meaning?

The paper
http://xxx.lanl.gov/abs/quant-ph/0509019 [Found. Phys. 36, 1601 (2006)]
may also be relevant here. (It cites your paper.)

This is a thick paper and I need more time reading it.

 

[A. Neumaier]

Is there an associated measurement theory in the Heisenberg picture?

In the paper, I referred to linear response theory, about which you can read in any book on nonequilibrium statistical mechanics. (I am using Reichl.)

 

[Demystifier]

Then you need to define the Schroedinger meaning of a measurement of a time-dependent observable A(t) in a time-dependent state psi(t) as t varies over time (and of A(t,s), which is also dependent on a former or later time).

This is not a trivial generalization of the standard textbook meaning of the Schroedinger picture, since one cannot appeal to ensembles of identically prepared systems anymore. So what is your proposal for defining this meaning?

There are probably many possibilities, but there is one possibility which seems reasonable to me:
Not every hermitian operator is an observable. In particular, q(s)q(t)q(s) in the Heisenberg picture is not an observable, and A(s,t) in the Schrodinger picture is not an observable. Observables are only those hermitian operators that in the Heisenberg picture depend on one time only. Equivalently, observables are only those hermitian operators that in the Schrodinger picture do not depend on time (except when there is an explicit time-dependence involved).

Of course, one can make separate measurements at different times and then multiply the results, but such measurements can be reduced to a measurement of observables defined as above.

In both pictures there are hermitian operators that depend on many times, but these mathematical objects are not observables.

The confusing part is the fact that the quantity <Psi|q(s)q(t)|Psi> is called a correlation function, suggesting that this quantity represents a measurable correlation between q(s) and q(t). Yet, despite the suggestion, <Psi|q(s)q(t)|Psi> is not measurable. If you measure q(s) and q(t), multiply the results for each pair of measurements, and then repeat the procedure many times to average over the statistical ensemble of equally prepared measurements, the result you will get in general will NOT be equal to <Psi|q(s)q(t)|Psi>.

 

[Demystifier]

In the paper, I referred to linear response theory, about which you can read in any book on nonequilibrium statistical mechanics. (I am using Reichl.)

I believe that all equations of linear response theory can be formally rewritten in the Schrodinger picture. Perhaps it will make the theory look mathematically more ugly, but it does not mean that one cannot apply the same physical interpretation of the final equations that describe physically measurable quantities.

 

[vanhees71]

Sure, everything can be formulated also in the Schrödinger picture. E.g., the QFT book by Hatfield does this for relativistic vacuum QFT. What you have to define are the Green's or correlation functions (in the most general case as expectation values of Schwinger-Keldysh-contour ordered field-operator products).

 

[A. Neumaier]

If you measure q(s) and q(t), multiply the results for each pair of measurements, and then repeat the procedure many times to average over the statistical ensemble of equally prepared measurements, the result you will get in general will NOT be equal to <Psi|q(s)q(t)|Psi>.

It is not supposed to be that; nobody ever claimed that. Time correlations are probed by linear response theory, not by your simplistic classical multi-time recipe.

 

[A. Neumaier]

I believe that all equations of linear response theory can be formally rewritten in the Schrodinger picture. Perhaps it will make the theory look mathematically more ugly, but it does not mean that one cannot apply the same physical interpretation of the final equations that describe physically measurable quantities.

A formalism in which things look ugly is inferior to one in which things look nice.

The power of physical explanations often lies in finding the formulation that gives insight, and with time, the formulation giving the most insight is viewed as the ''real'' way to look at things.

Otherwise we could still today adhere to the view that the sun revolves around the Earth - it is fully equivalent to the conventional view, just a change in coordinates.

 

[A. Neumaier]

Sure, everything can be formulated also in the Schrödinger picture. E.g., the QFT book by Hatfield does this for relativistic vacuum QFT. What you have to define are the Green's or correlation functions (in the most general case as expectation values of Schwinger-Keldysh-contour ordered field-operator products).

But the field-operator products are multi-time correlations in the Heisenberg picture,
not in the Schroedinger picture.

 

[A. Neumaier]

Why do you say that?

I explained it in the discussion with Demystifier.

I prefer the wording of my post 8: “unitary evolution and the theory of measurement of quantum mechanics, strictly speaking, contradict each other”).

This contradiction is harmless. Unitary evolution holds only for an isolated system,
while an observed system is never isolated since it must interact with any instrument
that measures it.

But there is nothing in standard QM but unitary evolution and the measurement mechanism.

You silently equate QM with the Schroedinger picture. But QM is more. In the Heisenberg picture, the operators transform unitariily, and operators at different times exist side by side in the Heisenberg picture and can be composed. This iis _not_ matched by the Schroedinger picture, and hence not by Bohmian mechanics. That makes a world of differences.

So can you reasonably blame the Bohm interpretation for inconsistency (if any) with the ill-defined part of standard QM?

Standard QM is _not_ ill-defined if one works in the shut-up-and-calculate interpretation, which is enough for all real life predictions. The ill-definedness comes in only through obscuring the foundations with the measurement process.

 

[akhmeteli]

I explained it in the discussion with Demystifier.

And I just cannot accept your “explanation”. Indeed, the Heisenberg picture and the Schroedinger picture are unitarily equivalent (for a finite number of degrees of freedom). Therefore, the expression for time correlation in the Heisenberg picture can and should be used in the Schroedinger picture (after proper transforms of wavefunctions and operators). Small talk about “loss of interpretation” is just that – small talk. I believe that unitary equivalence mandates this choice of time correlations. If you want to smuggle in something different and call it “time correlations”, it’s your choice, but don’t expect me to accept this arbitrary choice.

Multi-time correlations have a natural meaning in the Heisenberg picture, but the associated operators in the Schroedinger picture are just formal expressions without any meaning.

They have the same meaning as in the Heisenberg picture, as the two pictures are unitarily equivalent.

For example, if you rewrite <q(t)q(s)q(t)> along the lined indicated by you, one gets psi(t)^*A(t,s) psi(t), with a Hermitian operator
A(t,s):=q U(t-s) q U(s-t) q.

This immediately raises several issues:

(i) This is not the Schroedinger picture since A(t,s) still depends on $t$, whereas in the
Schroedinger picture, observables are supposed to be independent of t.

Do you mean that no operators explicitly dependent on time can exist in the Schroedinger picture?

(ii) The Hermitian operator A(s,t) has no discernible meaning, except that inherited from the Heisenberg picture, which must therefore be regarded as the fundamental picture.

And that inherited meaning is quite enough, as the pictures are equivalent.

(iii) How would you measure A(s,t)? There is no associated measurement theory.

I would suspect that if some procedure is used to measure time correlation, the results of the measurements can be used/described in either of the equivalent pictures. Why should I invent some extra measurement procedure?

This contradiction is harmless. Unitary evolution holds only for an isolated system, while an observed system is never isolated since it must interact with any instrument that measures it.

You can consider an isolated system including the instrument (and the observer, if you wish). Do you mean unitary evolution does not hold for such a system? And unitary evolution predicts something different from what the theory of measurements predicts, as unitary evolution cannot turn a superposition into a mixture or introduce irreversibility. I disagree that this contradiction is harmless; however, harmless or not, it’s still a contradiction.

You silently equate QM with the Schroedinger picture. But QM is more. In the Heisenberg picture, the operators transform unitariily, and operators at different times exist side by side in the Heisenberg picture and can be composed. This iis _not_ matched by the Schroedinger picture, and hence not by Bohmian mechanics. That makes a world of differences.

The two pictures are equivalent, so how can one of them have more content than the other? Again, your small talk about operators is just that – small talk. Everything you say about the Heisenberg picture can be translated into the language of the Schroedinger picture. The pictures are equivalent, remember? Some phrase can sound clumsier in the language of the other picture, so what?

Standard QM is _not_ ill-defined if one works in the shut-up-and-calculate interpretation, which is enough for all real life predictions. The ill-definedness comes in only through obscuring the foundations with the measurement process.

Do you mean the Bohm interpretation is ill-defined even if one works in the shut-up-and-calculate mode? I guess this is just a double standard: where standard QM stinks (i.e. contains mutually contradictory components), you are trying to explain to us that actually it does not stink, but its fragrance is just a bit unusual, whereas any problem with the Bohm interpretation stinks to heaven. Again, in general, the Bohm interpretation’s problems are not my problems, but I do have a problem with your claim. Having the same unitary evolution, the Bohm interpretation could possibly produce predictions different from those of standard QM only due to some difference in the theories of measurement. And can you blame the theory of measuremn of the Bohm interpretation for inconsistency with the theory of measurement of the standard QM, if the latter theory contradicts unitary evolution?

Another thing. Even if I believed your words about the Schroedinger picture being deficient compared to the unitarily equivalent Heisenberg picture, your claim would still be misleading, as it turns out that the deficiencies of the Bohm interpretation that you claimed in post 7 are actually also deficiencies of the Schroedinger picture, so it looks like the Bohm interpretation may be in good company?