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I'm working a lot on Bohmian mechanics related theories and just realised something that had escaped me until now but actually seems to throw a wrench in the core idea of the theory. I tried to clarify my doubt by reading Bohm's original paper but this did not really help, as I don't find his explanations satisfying. So, here's the problem.

In order for Bohmian mechanics to truly describe the same dynamics as regular quantum mechanics, we need the interactions to involve the wave functions, not the particles. For example, between two electrons, we would have something like:

$$ V_{ee} = \int\psi^*_1\psi_1\frac{1}{|\mathbf{r}_2-\mathbf{r}_1|}\psi^*_2\psi_2 d\mathbf{r}_{1}\mathbf{r}_{2} $$

All the particles do is coast along, following the guiding equation. Now, however, if the interactions are controlled by the wave function, so are the interactions between particles and measuring apparatus. Which means that if we had a wavefunction in a state of superposition, that superposition should still carry over to the apparatus. Sure, the actual Bohmian particles of the apparatus will slide into a definite, deterministic position. But that's not what we see. We see the wavefunction, and the wavefunction interacts with the EM field; the wavefunction reflects light that we see with our own eyes, light that is captured by our own retina's wavefunction, and so on. In all of this process, at no point comes a moment where the Bohmian particles finally step in and their deterministic outcome overrides the unitary quantum evolution of the wavefunction, causing the collapse. Because the interaction is one-way: the wavefunction drives the Bohmian particles, but the Bohmian particles never feed anything back. They're just along for the ride. Am I missing something or is this an actual problem?

EDIT: since I've received a comment that made me think my explanation wasn't clear, I'll try to express this with an example.

Suppose we have an experimental setup with two sensors, a "left" and a "right" sensor. Suppose also we prepare an electron in a way that its wavefunction is in a superposition between going left and going right:

$$ \psi = \frac{\sqrt{2}}{2}\psi_L+\frac{\sqrt{2}}{2}\psi_R $$

Now, the Bohmian particle, depending on its initial position, will actually ride only one of these two crests of the wavefunction - let's say the left one; the other is empty. Some time passes and the wavefunction hits the detector. It interacts by a coupling Hamiltonian:

$$ H_{meas} = H_L+H_R $$

that represents, for example, the Coulomb interaction of the electron with the atoms inside the detector, which will be bumped by it and thus measure the hit. Each term represents the interaction with one specific detector. Let's say the apparatus has a light that becomes red if the electron hits the left side, and blue if it hits the right. The overall wavefunction (instrument + electron) thus evolves into:

$$ \psi_{tot} = \psi_{det}\left(\frac{\sqrt{2}}{2}\psi_L+\frac{\sqrt{2}}{2}\psi_R\right) \rightarrow \frac{\sqrt{2}}{2}\psi_{red}\psi_L+\frac{\sqrt{2}}{2}\psi_{blue}\psi_R $$

At this point, we still have our measurement problem. Bohmian mechanics should solve it because the particle really is on the left. But how should the detector know? Measure isn't an abstract thing; it is the result of a specific series of interactions governed by the same quantum mechanical laws as everything else. And all interactions pass through an Hamiltonian, and affect the wavefunctions - not the Bohmian particles. If it weren't like this, then for example electron-electron interactions in a Helium atom would look very different! Coulomb interactions aren't described well between two point-like particles, and if they were, we would have a way to tell what the real position of the Bohmian particles is. Clearly, they aren't, and we don't.

In his 1952 paper [the second part, Phys. Rev. 85, 2 (1952) pp.180] Bohm makes a similar reasoning but then concludes that, basically, once the measurement is performed, our ignorance about which packet contains the particle disappears and we can then continue our following calculations by just restricting the wavefunction to that packet alone. But I'm not sure by which mechanism should our ignorance disappear. The theory doesn't really seem to provide any.

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    $\begingroup$ I'm not sure I understand your objection. Bohmian particles have definite trajectories and ultimately, it is position (which is definite) that is measured. While the state may not be a state of definite position and thus an ensemble of identical systems will not yield identical position measurements, for a given system the measured particle has a definite position at all times. $\endgroup$ – Alfred Centauri Aug 7 '18 at 12:38
  • $\begingroup$ The problem is, how is the position measured? Suppose you have a sensor with a lot of pixels, and an electron in superposition hits them. What will perturb the sensor? The wavefunction, in whose Hamiltonian figures the Coulomb interaction, not the Bohmian particle, which has no interaction with anything. Thus, how is the effect of the superposition resolved? The wavefunction is still delocalised, and will still produce a double peak. How does that turn into a single measurement, since the particle can't actually hit anything? $\endgroup$ – Okarin Aug 7 '18 at 13:07
  • $\begingroup$ For a discussion of some related issues see arxiv.org/abs/quant-ph/0403094 $\endgroup$ – alanf Aug 7 '18 at 14:17
  • $\begingroup$ Thanks @alanf! That seems to be exactly the criticism I had in mind. I see I am not alone in thinking this. $\endgroup$ – Okarin Aug 7 '18 at 14:31
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Strictly speaking, Bohmian mechanics does not solve the measurement problem. This is just my opinion, and, of course, there is no reason to believe me on my word. However, do proponents of Bohmian mechanics state that it solves the problem? I am not sure. You may wish to look at the article written by S. Goldstein in Stanford Encyclopedia of Philosophy (https://plato.stanford.edu/entries/qm-bohm/#MeasProb):

"What would nowadays be called effects of decoherence, which interaction with the environment (air molecules, cosmic rays, internal microscopic degrees of freedom, etc.) produces, make difficult the development of significant overlap between the component of the after-measurement wave function corresponding to the actual result of the measurement and the other components of the after-measurement wave function. (This overlap refers to the configuration space of the very large system that includes all systems with which the original system and apparatus come into interaction.) But without such overlap that component all by itself generates to a high degree of accuracy the future evolution of the configuration of the system and apparatus. The replacement is thus justified as a practical matter"

Thus, Goldstein claims only approximate solution of the problem. So, strictly speaking, the problem is not solved. And this may be good, because, strictly speaking, collapse

  1. contradicts unitary evolution (unitary evolution cannot produce irreversibility or turn a pure wave function into a mixture), and

  2. has no experimental confirmation (see a quote from Schlosshauer's article in my answer to Is the time of collapse of the wave function empirical? ).

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  • $\begingroup$ I've seen articles claim that it solves the measurement problem - in a way, isn't that its raison d'étre to begin with? To remove randomness and bring back good old determinism. Bohm himself seems to set out to that aim in the paper I mentioned. That said, I'm not convinced either, or I wouldn't have posted this question to begin with. The problem is, if you see it the way I wrote, Bohmian mechanics changes literally nothing. The particles existing or not is totally inconsequential. $\endgroup$ – Okarin Aug 7 '18 at 13:32
  • $\begingroup$ @Okarin : Maybe you should quote such articles, so that we could discuss them. I suspect that such articles either make some caveats (like those in the Goldstein's text) or they are wrong. I am not a fan of Bohmian mechanics, but it seems to me that, strictly speaking, it does offer a deterministic option. As I said, it may be good that Bohmian mechanics does not solve the measurement problem. $\endgroup$ – akhmeteli Aug 7 '18 at 13:45
  • $\begingroup$ Well, this one's one of the first hits, but sadly, also behind a paywall for me: jstor.org/stable/188499?seq=1#page_scan_tab_contents $\endgroup$ – Okarin Aug 7 '18 at 13:47
  • $\begingroup$ Anyway my problem starts at the origin. Check out this one, the original: cqi.inf.usi.ch/qic/bohm2.pdf and the way it addresses measure. It says clearly that after the electron makes contact with the detector, we can simply throw away the remaining branches. But that's the thing, I don't see how we could do that. We don't have any basis to do it. $\endgroup$ – Okarin Aug 7 '18 at 13:49
  • $\begingroup$ @Okarin : The Maudlin's article cites a Goldstein's article, so there is probably nothing there that Goldstein is not aware of. As for Bohm's article, I don't have time now to look into it, but again, I suspect that either he makes some caveats or he is, strictly speaking, wrong. $\endgroup$ – akhmeteli Aug 7 '18 at 14:05
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Bohmian mechanics cannot have a measurement problem: The word measurement does not figure in its formulation at all. It could be wrong, it could predict the wrong things, but it's actually quite correct, as one can show that for subsystems (!), Bohmian mechanics gives the same predictions as the textbook quantum mechanics. (Short comment to the other answer by akhmeteli: The "measurement problem" is not the same as "the collapse must be explained". The first one is the missing clarity and well-definedness of a theory that basically says "when we measure, something happens" without a clear equation for that. The second must be done in an approximate way since it's an effect that happens in many-particle-systems.)

The answer to your question about wavefunction collapse is basically given in the works of Dürr, Goldstein, and Zanghi, and is connected with the concept called effective wave function, an object different from the global wave function that always obeys the Schrödinger equation and never collapses. This is explained in quite some detail in the book Bohmian mechanics by Dürr and Teufel, or in this talk: youtube.com/watch?v=yr5yGxJg2X0.

Here is the basic point: It is actually sufficient to be sure that the Bohmian trajectory of a macroscopic object is (close to the) classical trajectory. If we look at a measurement apparatus, let's say for simplicity a pointer pointing left or right (this is a bit easier to explain that a red/blue light, but similar in spirit), we really see the Bohmian positions of the particles in it. You basically objected to this point since you said that all interactions take place on the wave function level. Sure, they do. That's why we need a measurement device: It is a system so large that we know that if I see the pointer on the left, we can be quite sure that the Bohmian particles consistuting the pointer are actually on the left. (We cannot see the particles themselves. But well, evidence is usually indirect.) So we are allowed to say that the pointer, in the case we see it on the left, actually points to the left (in contrast to usual quantum mechanics, where there are really no variables to say what actually happens). And with this knowledge, we can be sure that the -- maybe present -- wave function parts of a pointer on the right will never be relevant for the dynamics of the universe any more (this is explained in the talk and book, basically using decoherence). So, effectively, to find out what the actual particles do, we can throw away the right part of the wave function, and perform an effective collapse.

The explanations in the cited book and talk may be much better than what I wrote here, so look at them :-).

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  • $\begingroup$ I'll check them out, but on the face of it, this doesn't convince me. Unless we assume that for some reason our consciousness responds to the Bohmian particles, then I don't see why we should see those. And the way I understand it, a system being macroscopic doesn't protect it from being in a superposition state. It only means it will be decohered - quite different. The pointer will be in a left-right superposition, no matter how big it is. $\endgroup$ – Okarin Aug 8 '18 at 13:09
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    $\begingroup$ @Okarin, you seem to assume that there is no actual position of the pointer, only psi function and that implies superposition. This is explicitly something that is rejected in Bohmian mechanics. The pointer (and the particles that form it) does have specific position in Bohmian mechanics. The psi function is only an auxiliary mathematical function that gives probabilities in our calculations. $\endgroup$ – Ján Lalinský Aug 8 '18 at 13:36
  • $\begingroup$ But then, what is the mechanism through which this 'true' position of the pointer reaches us? How can we learn about it? If I look at the pointer, I see light reflected off it. And the EM field interacts with the wavefunction, not the Bohmian particles. So shouldn't I see the wavefunction rather than the true particles? $\endgroup$ – Okarin Aug 8 '18 at 13:40
  • $\begingroup$ @Okarin: Why do you keep claiming that it is actually possible to see a wavefunction? $\endgroup$ – D. Halsey Aug 8 '18 at 14:17
  • $\begingroup$ "Seeing" is just a shortcut for "receiving electromagnetic waves that are scattered by". What scatters the EM field, the wavefunction or Bohmian particles? I say the former; if it was the latter, then that would change completely the form of the electromagnetic potential in the Hamiltonian. $\endgroup$ – Okarin Aug 8 '18 at 14:31

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