Undergrad New experiments supporting Bohmian mechanics?

[LeandroMdO]

Since the electron is in a stationary state with constant energy, that would seem to be what Bohmian mechanics would have to predict, since mathematically it is the same as standard QM.

The arguments that it isn't haven't been convincingly rebutted in my opinion.

 

[PeterDonis]

The arguments that it isn't

What arguments?

 

[LeandroMdO]

What arguments?

The ones relating to expectation values of time correlation of position observables. They have been discussed many times in these very boards.

 

[Denis]

Non-relativistic theories cannot be high-energy theories, because then the expansion in powers of $1/c$ is not valid anymore.

A theory is, I would guess, relativistic if it has Lorentz covariance or is, in case of GR, background-independent, not?

If we use this as a definition, then relativistic theories may be low energy limits of non-relativistic theories. These non-relativistic theories would differ from the $c\to\infty$ limit of relativistic theories, but would be, according to the definition above, non-relativistic if there would be effects violating Lorentz covariance resp. the equivalence principle.

As a cheap example theory that this is possible take standard condensed matter theory. Then, some large distance low energy approximation will describe sound waves with a standard wave equation $\square u = 0$. The symmetry group of this wave equation is the Poincare group. But the full theory, on the atomic level, does no longer have this Poincare symmetry group.

How can BM be a solution of any supposed problem of QT, if there are no other testable predictions than from QT?

Many problems are not problems of the physical theory itself, but of one or another interpretation. The measurement problem is such a problem. It exists only in some interpretations. In others, like the BM, it simply does not exist.

 

[Demystifier]

How can BM be a solution of any supposed problem of QT, if there are no other testable predictions than from QT?

Have you ever solved some problem without making a new testable prediction? I bet you have. In fact, you probably do that every day.

Or let me give some well-known examples you will probably appreciate:
1) Lagrangian or Hamiltonian mechanics make the same measurable predictions as Newtonian mechanics. Yet, they solve some problems of Newtonian mechanics. (Otherwise, why would anyone introduce them?)
2) Maxwell equations in the manifestly covariant F-form make the same measurable predictions as Maxwell equations in the non-covariant E-B-form. Yet, the covariant F-form solves some problems of the non-covariant form.
3) In QFT, a self-energy loop diagram is divergent, which is a problem. This divergence can be absorbed into a redefinition of the particle mass, which solves the problem. But this solution does not make any new testable prediction.

 

[PeterDonis]

The ones relating to expectation values of time correlation of position observables. They have been discussed many times in these very boards.

Then you should have no problem providing some specific references. Please do so.

 

[LeandroMdO]

Then you should have no problem providing some specific references. Please do so.

Here's a recent thread where this was discussed:
https://www.physicsforums.com/threads/an-argument-against-bohmian-mechanics.898028/

You were even an active participant in it.

 

[PeterDonis]

Here's a recent thread where this was discussed

Which is 24 pages long, and it was 5 months ago. If there are specific posts that illustrate what you're referring to, please link to them. I'm not going to wade through the whole thread to try and guess.

 

[LeandroMdO]

Which is 24 pages long, and it was 5 months ago. If there are specific posts that illustrate what you're referring to, please link to them. I'm not going to wade through the whole thread to try and guess.

Try the original post. It's really not that hard.

 

[PeterDonis]

Try the original post.

In other words, you are expressing your opinion that that thread never answered A. Neumaier's argument to your satisfaction? Ok. I don't see the point of rehashing a 24-page thread again. You've expressed your opinion.

It's really not that hard.

It's really not that hard to say what specific post you were referring to, yes. Then why did I have to request that three times before you did it?

Continuing to display this attitude will get you a warning.

 

[Demystifier]

The arguments that it isn't haven't been convincingly rebutted in my opinion.

Have you ever studied the theory of quantum measurements in BM? Try it, it's not that hard. You cannot understand why BM makes the same measurable predictions as standard QM if you don't understand what happens during a measurement. That's probably the main lesson of BM, but for some reason people repeatedly ignore it.

 

[LeandroMdO]

In other words, you are expressing your opinion that that thread never answered A. Neumaier's argument to your satisfaction? Ok. I don't see the point of rehashing a 24-page thread again. You've expressed your opinion.

What I would consider to be "an answer" would be a post pointing out the error in his calculation and how the calculation should be done correctly. There are many words in that thread, but no such calculation.

It's really not that hard to say what specific post you were referring to, yes. Then why did I have to request that three times before you did it?

I was very specific in what objection I was referring to: time correlations of position observables (which at first I assumed you would be familiar with since, as I said, you participated in threads that discussed it). The first post in the thread I linked to has a link to the paper with the argument. The issue of time correlations is also mentioned a few times in the first page of that thread. It takes only a trivial amount of effort to recognize this sort of thing. If you're unwilling to spend this minimal effort, that is fine: you don't even have to participate in any discussion if you don't want to.

Continuing to display this attitude will get you a warning.

Right, look, you were sounding a little exasperated there. If your reaction to the suggestion that you might be acting unreasonably is to threaten the use of mod powers, you might wish to ponder whether you should be wielding them at all. Little conflict of interest there.

 

[Demystifier]

What I would consider to be "an answer" would be a post pointing out the error in his calculation and how the calculation should be done correctly. There are many words in that thread, but no such calculation.

Here is a deal. I will make explicit correct calculation in Bohmian mechanics, provided that you first present a correct calculation of measurable time correlations in standard QM. In that way I will know at which level you understand measurable time correlations in standard QM, so that I can adjust my calculation to your level of understanding. And be careful, because the naive correlation function $$\langle\psi| x(t_1)x(t_2) |\psi\rangle$$ is not measurable.

 

[vanhees71]

Have you ever solved some problem without making a new testable prediction? I bet you have. In fact, you probably do that every day.

Or let me give some well-known examples you will probably appreciate:
1) Lagrangian or Hamiltonian mechanics make the same measurable predictions as Newtonian mechanics. Yet, they solve some problems of Newtonian mechanics. (Otherwise, why would anyone introduce them?)
2) Maxwell equations in the manifestly covariant F-form make the same measurable predictions as Maxwell equations in the non-covariant E-B-form. Yet, the covariant F-form solves some problems of the non-covariant form.
3) In QFT, a self-energy loop diagram is divergent, which is a problem. This divergence can be absorbed into a redefinition of the particle mass, which solves the problem. But this solution does not make any new testable prediction.

There is a difference between having different mathematical techniques to formulate and solve problems in theoretical physics. Concerning your examples it's very clear that it makes sense to study all three topics as a theoretical physicist, because

(1) Lagrangian and Hamiltonian mechanics (or in other words the least-action principle) offers analytical methods to solve Newtonian mechanics (but I don't see which problems it solves for Newtonian mechanics since it's in itself a closed consistent mathemtical theory) like Noether's theorem etc. From a practical point of view it also facilitates the derivation of equations of motion in arbitrary coordinates. For the progress of science it was very important in directing away the view from the murky concept of forces to more general dynamical laws in terms of local interactions mediated by fields, and it's applicable for a much wider range of theories than Newtonian mechanics (field theory, relativistic physics) and leads to a heuristic way to formulate QT (canonical quantization, group-theoretical approaches, $C^*$ algebras of observables etc. etc.)

(2) Whether you formulate Maxwell theory in terms of the manifestly covariant or the (1+3) formalism in which form it has been discovered doesn't make any difference. It's the very same mathematical theory.

(3) is a completely different case. It's part of the very definition of perturbative QFT as a physical theory. Without it you can't make contact to experiment beyond the tree-level results of perturbation theory. I'd say it's integral and undispensible part of the theory, without which you can't make any sense of the mathematically murky concept of ill-defined products of distribution valued operators.

In contradistinction to these 3 examples of different equivalent formulations of the same theory or the (admittedly only FAPP) solution of a mathematically ill-defined model Bohmian mechanics doesn't add anything to QT in terms of a physical theory. Everything you can observe according to QT (and in accordance with all hitherto made observations and stringent experimental tests of QT) you can derive from QT without adding any fictitious and unobservable trajectories. It's an empty academic exercise to calculate these trajectories without any physical consequences.

 

[Demystifier]

(2) Whether you formulate Maxwell theory in terms of the manifestly covariant or the (1+3) formalism in which form it has been discovered doesn't make any difference. It's the very same mathematical theory.

So what's the point of introducing a manifestly covariant formalism once you already know the traditional non-covariant one? It doesn't make new predictions, so what's the point of it? (I am just trying to force you to give a reason which, with a minor modification, can also be applied to BM. )

 

[vanhees71]

Well, the point of it is to have a much better understanding of the structure of the theory and it helps to solve the puzzle of "Electrodynamics for moving bodies" (as Eisntein's famous paper about SRT was titled in 1905). BTW you can write classical E&M in manifest covariant form also in terms of the Riemann-Silberstein vector which is very elegant, but as I said, all this is still the very same theory as has been written down by Maxwell after a cumbersome detour through very complicated mechanical models ;-).

Again, I don't see any merit of the additions of BM to QT compared to the standard formulation. To the contrary it makes the standard formulation way more complicated and adds (unnecessary!) problems in the case of relativistic QFT, which is problematic enough without additional useless problems by itself.

 

[Demystifier]

Well, the point of it is to have a much better understanding of the structure of the theory and it helps to solve the puzzle of "Electrodynamics for moving bodies" (as Eisntein's famous paper about SRT was titled in 1905). BTW you can write classical E&M in manifest covariant form also in terms of the Riemann-Silberstein vector which is very elegant, but as I said, all this is still the very same theory as has been written down by Maxwell after a cumbersome detour through very complicated mechanical models ;-).

So, you agree that a new theoretical insight does not always need to give new measurable predictions? Good, keep it in mind when thinking about BM!

Again, I don't see any merit of the additions of BM to QT compared to the standard formulation. To the contrary it makes the standard formulation way more complicated and adds (unnecessary!) problems in the case of relativistic QFT, which is problematic enough without additional useless problems by itself.

I'm sure there are physicists (perhaps working in condensed matter) who do not see any merit in introducing covariant F-formalism in electrodynamics, but see it as making electrodynamics look more complicated. If somebody does not see a merit, it doesn't mean that there isn't any. To see the merit, one must first look at things from a different perspective. Perhaps covariant electrodynamics does not have a merit from condensed matter perspective, but it does from high-energy perspective. Perhaps BM does not have a merit from a perspective of making predictions, but it does from what-is-going-on-when-we-do-not-observe perspective. The latter perspective is perhaps more philosophical than scientific, but it is a legitimate perspective used by some people, even if you are not one of them.

 

[vanhees71]

Well, I don't know, what sense it makes to discuss about things you can never check. You cannot check what happens with something when you are not looking without looking. As you see, already the experiment description is contradictory in itself. This has nothing to do specifically with quantum theory but for observations within any model of nature.

 

[Denis]

Well, I don't know, what sense it makes to discuss about things you can never check.

Different interpretations are different starting points for future theory development. Because different interpretations have different weak points, and attempts to solve these problems often lead to proposals to modify the theory.

Say, one solution of the measurement problem is to introduce a physical collapse into the theory. Which is a modification of quantum theory and can be tested.

Similarly, BM also has weak points where one can look for modifications of QT. One place is quantum equilibrium. There may be some non-equilibrium effects somewhere. Another one is that the velocity becomes infinite near the zeros of the wave function. This asks for a regularization. And such a regularization will probably get rid of the zeros.

 

[vanhees71]

I'll not argue about the interpretational issues. The final statement is strange to me. What do you define as "velocity" here? I'm not aware with anything that's in anyway problematic with velocity in QM, which is defined as $p/m$, I'd guess.

 

[Denis]

I mean the velocity of the Bohmian particle, as defined by the guiding equation. So, no, it has nothing to do with p/m measurements.

 

[vanhees71]

Is it an observable quantity? If not, why is it a problem? Is it the same issue as in the WKB approximation at the "classical turning point"? If so than it's no issue at all, because the classical approximation breaks down there.

 

[Denis]

Is it an observable quantity? If not, why is it a problem?

For positivists like you it is not a problem. For those who think that a physical theory has to describe reality, it is a problem if something supposed to describe something real becomes, somewhere, infinite. Even if the infinity itself is only at a harmless place where the probability to appear is zero.

And my point was that the solution of this problem - even if only a problem for those interested in reality instead of observables only - will probably lead to a different sub-quantum theory with equations and solutions which differ from QT, even in its observable effects, so that, after this, you can come back and find out if that modified theory is better than QT.

With your self-restriction not to care about non-observable problems you restrict yourself from participation in the creation of such sub-quantum theories, which solve problems of different interpretations. Ignorance of problems means, first of all, refusal to participate in their solution. Your choice.

 

[berkeman]

If your reaction to the suggestion that you might be acting unreasonably is to threaten the use of mod powers, you might wish to ponder whether you should be wielding them at all. Little conflict of interest there.

Please rest assured that all the Mentors are watching your posts in this thread. It may not be Peter that gives that warning...

 

[LeandroMdO]

Here is a deal. I will make explicit correct calculation in Bohmian mechanics, provided that you first present a correct calculation of measurable time correlations in standard QM. In that way I will know at which level you understand measurable time correlations in standard QM, so that I can adjust my calculation to your level of understanding. And be careful, because the naive correlation function $$\langle\psi| x(t_1)x(t_2) |\psi\rangle$$ is not measurable.

I have no doubt that you could show how to calculate a different correlation function and have the results agree between BM and QM. But I'm not interested in other correlation functions: I'm interested in $$\langle\psi| x(t_1)x(t_2) |\psi\rangle$$. It's not like it's some esoteric thing, after all, $$\langle 0| T x(0)x(t_1) |0\rangle$$ for instance is just the two-point function for a 0+1-dimensional field theory, a quantity that can certainly be probed.

You told me that the A. Neumaier's calculation for the expectation value of this observable in BM was incorrect, and that if the calculation is done correctly the discrepancy disappears. That's certainly a possibility I'm willing to entertain. But now you tell me that not all observables are "measurable". I don't know what that word means. When you say "$$\langle\psi| x(t_1)x(t_2) |\psi\rangle$$ is not measurable", do you mean that you can't think of a way to measure it, or do you have a demonstration that it cannot be measured? Can you show me how to classify observables between "measurable" and "not measurable" ones, and then show that all of those in the first group agree between BM and QM?

Saying "you're asking a bad question" might be a perfectly fine response, but only with some more substantiation. Right now what I have is an indication that there is a class of observables whose expectation values disagree between BM and QM. I don't much care what that observable is because once a single one is found, even if it is "not measurable", the door is open for others, some of which might well be straightforward to measure.

Please rest assured that all the Mentors are watching your posts in this thread. It may not be Peter that gives that warning...

Of course. And I am watching you. You are entitled to think this irrelevant, but I question the seriousness of a physics forum that censors based on personal feelings and in-group bias. My impression was that this is a serious forum, but please do correct me if that impression was mistaken.

 

[berkeman]

My impression was that this is a serious forum, but please do correct me if that impression was mistaken.

No correction needed.

 

[PeterDonis]

You are entitled to think this irrelevant, but I question the seriousness of a physics forum that censors based on personal feelings and in-group bias.

Just to be clear: my objection was that you did not provide a specific reference (link to a specific post) when asked. It had nothing to do with "personal feelings" or "in-group bias". A request for a specific reference is in accordance with the PF rules, and the proper response is to provide one, not to say "it really isn't so hard" while refusing to specify what exactly you were referring to.

 

[Demystifier]

When you say "$$\langle\psi| x(t_1)x(t_2) |\psi\rangle$$ is not measurable", do you mean that you can't think of a way to measure it, or do you have a demonstration that it cannot be measured?

You can measure $x(t_1)$, and after that you can measure $x(t_2)$, and finally you can multiply the two values obtained by those two measurements. But this procedure is not the same as measurement of $x(t_1)x(t_2)$. In fact, the operator $x(t_1)x(t_2)$ is not even a hermitian operator, so it is not an observable. Only hermitian operators are observables, and only observables can be measured.

 

[Demystifier]

Well, I don't know, what sense it makes to discuss about things you can never check.

How about discussing things which can be checked in principle, but not in practice? E.g. string theory at the Planck scale, the nature of black-hole interior, or a macro system in which the entropy decreases? Does that make sense to you?

 

[LeandroMdO]

You can measure $x(t_1)$, and after that you can measure $x(t_2)$, and finally you can multiply the two values obtained by those two measurements. But this procedure is not the same as measurement of $x(t_1)x(t_2)$.

Indeed, but nobody suggested that one measure x(t_1) and then x(t_2). That is obviously inequivalent. However, there doesn't appear to be any fundamental difficulty in measuring or probing the correlation between x(t_1) and x(t_2) while leaving the positions themselves unmeasured. It's a matter of being creative in designing the proper experiment, unless it can be demonstrated that this is impossible. I don't think it's at all obvious that it should be impossible to measure a correlation between x(t_1) and x(t_2) without measuring either individually.

In fact, the operator $x(t_1)x(t_2)$ is not even a hermitian operator, so it is not an observable. Only hermitian operators are observables, and only observables can be measured.

Right, you have to use an appropriate symmetrization. x(t_1)x(t_2) + x(t_2)x(t_1) is Hermitian, and its naive expectation value in BM differs from that of QM. So the lack of Hermiticity is not the source of the discrepancy.