The Pros and Cons of Bohmian Mechanics

[ttn]

You really did not understand the subtle point I was trying to make. I clearly stated that the statistics concerning measurements of *quantum* observables in both formalisms are the same what is basically your argument. What I said moreover is that BM intrinsically allows for MORE than that (for example you can state that the particle *IS* standing still in BM, you cannot claim that in the Copenhagen formalism): we can couple dynamically the particle to a *classical* EM field and consider measurement of that field by detectors around the atom. So we are not measuring an observable of the H atom, neither do we have any trouble with collapse of the state and entanglement of the photon states to the electron state; we are simply measuring a classical field generated by a classical stationary charge cloud (think about Hartree here) which shifts of course a bit during the measurement. Of course you cannot do that in the Copenhagen approach, but the Bohmian approach allows for such possibility. You are of course free to say that this is a bad idea and that one should restrict to direct measurements of quantum observables, but I see no logical reason to do so.

Well then I guess I still don't understand the point. Are you saying that it is *correct* to do the sort of thing you're talking about (couple the Bohmian H atom to a classical EM field, and then "classically measure" the field in order to learn something about the state of the H atom)? Because I think it's clear that it isn't correct to do that. Basically, you are just hiding what amounts to the act of measurement "off stage", by assuming a bunch of "classical" fields which couple to the H-atom in some specified way and which can be measured without disturbance because, by assumption, they are "classical." In other words, you are smuggling in a bunch of things that are "illegal" according to Bohm's theory. And by the way it's exactly the same with orthodox QM. I could say something like: according to OQM, the electron in the H atom is a spherically symmetric smear of charge when it isn't being measured. So instead of measuring it, let me put in a "classical EM field" which couples to that smeary charge distribution, and then I'll learn something about the charge distribution by "classically measuring" the EM fields that are produced by it. But that's illegal, right? If you do things correctly, you'll find that all of this business about letting the atom couple with an EM field, etc., constitutes a *measurement* so you'll never get to directly observe the spherical symmetry of the ground state wave function.

Assuming you're talking about the real world. Of course, you can imagine a different universe in which these "classical EM fields" really exist. And maybe you're right that in that universe, Bohm and OQM make different predictions. I don't really know or care about that.

On the other hand, maybe your point was just that, because Bohm provides a different ontology than OQM, you can sort of imagine things happening according to Bohm's theory, and what you imagine is different from what OQM says. That's no doubt true. I think it's one of the (several) virtues of Bohm's theory that it permits this clear, unambiguous visualization of (even) unobserved quantum systems. That is, it's really a quantum theory without observers. But it's a *fact* that these different-looking ontologies give rise to all the same empirical predictions. And, despite denying that you're denying that, I still can't help thinking that you're denying it. Your last sentence suggests that it's somehow arbitrary (or unnecessarily restrictive) to require that we only talk about measuring quantum observables. But, if the theory is right, this is what we can do, period. Bohmian Mechanics is not classical mechanics. It's not at all arbitrary to say: if this theory is right, here's what we can do and what we can't do. In other words, if Bohm's theory is right (or for that matter if orthodox QM is right) there is an extremely "logical reason" why we can't do the sort of thing you're suggesting -- namely, it's *impossible*.

 

[Careful]

Well then I guess I still don't understand the point. Are you saying that it is *correct* to do the sort of thing you're talking about (couple the Bohmian H atom to a classical EM field, and then "classically measure" the field in order to learn something about the state of the H atom)? Because I think it's clear that it isn't correct to do that. Basically, you are just hiding what amounts to the act of measurement "off stage", by assuming a bunch of "classical" fields which couple to the H-atom in some specified way and which can be measured without disturbance because, by assumption, they are "classical." In other words, you are smuggling in a bunch of things that are "illegal" according to Bohm's theory. And by the way it's exactly the same with orthodox QM. I could say something like: according to OQM, the electron in the H atom is a spherically symmetric smear of charge when it isn't being measured. So instead of measuring it, let me put in a "classical EM field" which couples to that smeary charge distribution, and then I'll learn something about the charge distribution by "classically measuring" the EM fields that are produced by it. But that's illegal, right? If you do things correctly, you'll find that all of this business about letting the atom couple with an EM field, etc., constitutes a *measurement* so you'll never get to directly observe the spherical symmetry of the ground state wave function.

Hi tnt,

Calm down ... I want to argue that there is something fishy about the reality aspect of BM. Of course I am aware that if you apply it ``correctly´´ there is no contradiction with observation, but I want to argue that this ``correctness´´ conflicts with the very meaning of reality. I shall explain in full detail what I mean. In Bohmian mechanics the wave function gives rise to an extra force (the quantal guidance mechanism); of course this force is not classical in the sense that it depends on the possible motion of the particle would it have been somewhere else. Still, the particle exists and follows a well defined trajectory determined by a solution of the Hamilton Jacobi equation with some energy. So, consider the stationary ground state of the H-atom, we know the particle stands still (which is impossible to tell in OQM). We can ask ourselves now, what is the external EM field produced by this configuration? We can take two points of view: (a) the H-atom is quantum mechanical and whatever influence it has is considered to be classical (b) the EM field itself is quantum mechanical. Now, there is *nothing* in QM which tells us that a classical massless field needs to be quantized (stricly speaking the Schroedinger equation only applies for massive particles.). First, consider (a), the reality criterion of BM tells us that the electron is standing still and therefore the EM field is *not* rotationally invariant (wrt. to rotations with the nucleus as centre), we do not know the axis of the dipole, but we do know that there is a dipole and a such a measurement shall reveal this. Standard quantum mechanics does *not* give us such information, and the best we can do is to compute the Coulomb potential corresponding to the nucleus and the charge density of the electron determined by the wave function which *is* for all practical considerations zero at a distance scales of 10^{-9} meters from the nucleus (this is the Hartree approach which is correct in this case) and the experimental outcome is *no* electric dipole. In case you do second quantization, both theories agree (perfect symmetry). I do not see why (a) is illegal; the classical EM field produces a kind of measurement in BM but is incapable of doing so in OQM.

You may not like this, but it is a logical reasoning. But again, you can just exclude this reasoning by hand, but for me it is a sign that there is something wrong with the reality criterion on BM.

Cheers,

Careful.

 

[ttn]

In Bohmian mechanics the wave function gives rise to an extra force (the quantal guidance mechanism); of course this force is not classical in the sense that it depends on the possible motion of the particle would it have been somewhere else.

I don't think that really captures what's non-classical about the force, but I don't think that's a main point.

Still, the particle exists and follows a well defined trajectory determined by a solution of the Hamilton Jacobi equation with some energy.

Yes, fine.

So, consider the stationary ground state of the H-atom, we know the particle stands still (which is impossible to tell in OQM). We can ask ourselves now, what is the external EM field produced by this configuration? We can take two points of view: (a) the H-atom is quantum mechanical and whatever influence it has is considered to be classical (b) the EM field itself is quantum mechanical. Now, there is *nothing* in QM which tells us that a classical massless field needs to be quantized (stricly speaking the Schroedinger equation only applies for massive particles.). First, consider (a), the reality criterion of BM tells us that the electron is standing still and therefore the EM field is *not* rotationally invariant (wrt. to rotations with the nucleus as centre), we do not know the axis of the monopole, but we do know that there is a monopole and a such a measurement shall reveal this. Standard quantum mechanics does *not* give us such information, and the best we can do is to compute the Coulomb potential corresponding to the nucleus and the charge density of the electron determined by the wave function which *is* for all practical considerations zero at a distance scales of 10^{-9} meters from the nucleus (this is the Hartree approach which is correct in this case) and the experimental outcome is *no* electric monopole.

OK, so then isn't it orthodox QM which is refuted by experiment? Because when you actually look for the edm (i.e., measure the EM field configuration near the atom) you *find* an edm.

Of course, OQM isn't really refuted, because what you calculated was just the expectation value for the edm operator. But my point is, if there's a problem here for anything, it's OQM -- which has to invoke a kind of "magic" (the collapse of the wf) to get from what it says is happening before we look (the charge is distributed symmetrically around the nucleus) to what we see when we do look (the electron is at some specific point, giving rise to a nonzero edm).

I just don't see what you think is the problem for Bohm here.

In case you do second quantization, both theories agree (perfect symmetry). I do not see why (a) is illegal; the classical EM field produces a kind of measurement in BM but is incapable of doing so in OQM.

I agree, I don't see any a priori reason why (a) is illegal. (Of course, it might turn out that EM fields aren't really classical, and that to get the right answer you need to quantize that too, presumably in a Bohmian kind of way if we're talking about extending the Bohmian picture of the H atom. But that's a different point, right?)

So the real question is: does an H atom produce an EM field in its vicinity that is the kind of field produced by an edm? If not, we need to go deeper into the question and maybe think about quantizing the field. But if so, Bohm has a ready explanation. OQM also has a ready explanation, though it is a bit fishier as it involves the collapse postulate.

So again, what's the point exactly? What's the problem with the "reality criterion on BM"?

 

[Careful]

**I don't think that really captures what's non-classical about the force, but I don't think that's a main point.**

I think it exactly captures it apart from some faster than speed of light motion due to a quasi local coupling between the phase and amplitude in the hyperbolic equations for the phase and amplitude.

The point is that, to my knowledge, the H atom in the ground state is measured *not* to have a permanent electric dipole moment (as I said the result from the Hartree approximation is correct here). To be entirely explicit; I you let this one H-atom stand for a long time between the detectors, then there will be a dipole detected *on average* (that is: just count the average field strengts measured by each detector and compare them) in scenario (a) in the Bohmian thought experiment, since a classical measurement on the EM *classical* field is not going to have any influence on the electron position in the H atom (remember: classical measurements on classical fields do not disturb).

 

[DrChinese]

If position does not commute with momentum in Bohmian Mechanics, how is that reconciled with the idea that position is fundamental?

 

[Careful]

If position does not commute with momentum in Bohmian Mechanics, how is that reconciled with the idea that position is fundamental?

Hi, this depends upon your notion of reality. Your position seems to be: the reality of a property of the particle IS the value of the corresponding observable obtained in a (future) experiment (the position Bohr takes). This is also the notion of reality used in the Kochen Specker theorems as far as I remember. My notion of reality is much more extended: for example a particle may have a well defined postion and momentum at the same time, but it may be that I cannot *measure* both at once since in a position measurement where I fire a ``photon´´ to the particle, the momentum is going to be disturbed. However, I can still build in a split between the quantum world and the classical one (such as happens in BM/they do not dispose of the Reduction). Of course, it would be preferred that the number which is produced (i.e. outcome of experiment) must result only from *physical* interactions and not through some magical collapse of the state. That is: I would like to precisely know how (and how much) the momentum is disturbed (which is not possible in BM) !

Cheers,

Careful

 

[ttn]

If position does not commute with momentum in Bohmian Mechanics, how is that reconciled with the idea that position is fundamental?

Following up on Careful's remarks... it really just isn't clear what you mean when you say that "position does not commute with momentum in Bohmian Mechanics." As I'm sure you know, BM isn't fundamentally framed in terms of operators. "Operators as observables" emerges from the basic structure of the theory as a theorem. So it is possible, I guess, to talk about position and momentum operators (which of course don't commute) but then we are just talking about one of the standard pieces of the quantum formalism which is in no way unique to (and is definitely not fundamental to) Bohm's theory.

But maybe the thrust of your question is simply this: how do we reconcile the uncertainty principle (which follows from the non-commutation of the x and p operators) with the fact that, in Bohm's theory, particles are postulated to always have definite positions -- and, for that matter, to be following definite trajectories and hence to have always some definite velocity (hence also mass times velocity!). Doesn't this violate the HUP? No, it doesn't, because in BM the HUP is a statement about the spread of *measured* (or possibly-measured) values, not a statement about the *actual* values. In the case of *position*, these aren't different, since measurement simply reveals the pre-existing position. But in the case of *momentum* they definitely are different (as Careful pointed out). If at some instant a Bohmian particle of mass m is moving with velocity v and then you make a "momentum measurement", you will (in general) *not* get the value mv as the outcome of the measurement.

The trapped particle-in-the-box that I mentioned before is the clearest example of this. If the wf is an energy eigenfunction, the particle will be stationary. But if you measure the momentum, you won't get zero (which one can visualize by imagining the measurement is made by removing the walls of the box and letting the particle fly free). So one gets, as the measurement outcome, (roughly) either plus or minus h-bar times k (where k is the wave number associated with the initial wf). So the "uncertainty in p" associated with the original state is just hbar*k. And this even though, in that initial state, the particle was *definitely* at rest!

Note by the way that the smallest possible value of k is pi/L, where L is the length of the box. And for that wf, the quantum equilibrium hypothesis yields that the "uncertainty in x" is (roughly) L/2. So the product of the "uncertainty in x" and the "uncertainty in p" is pi/2 -- just as the HUP says it should be.

Anyway, the point is: how this consistency with the HUP comes about in Bohm's theory is rather interesting and subtle, because the "uncertainties" in general refer not to any kind of objective indefiniteness in the initial state, but simply to our ignorance about the precise state -- in this case, our ignorance of the position of the particle in the wave and the *resulting* ignorance about what value will be obtained for momentum if a momentum measurement is performed.