Bohmian Mechanics: When Does Pilot Wave Action Cease?

[DrChinese]

Under the Bohmian Interpretation (BM or BMI):
Suppose we have a traditional pair of polarization entangled photons A and B going to Alice and Bob, who will test their polarizations in the same reference frame. To be specific, let's say A and B are entangled |HH> + |VV>. In the traditional view of this interpretation, a measurement of A by Alice occurring first (in all reference frames) instantaneously leads to an update of B due to action of a pilot wave. This is action at a distance as a way to explain entangled systems.

Bob is certain to observe the same result as Alice when the angle settings are the same, say on the H/V basis. This result is not in question, and all viable Interpretations predict the same. I also note that while BM does not feature spin as an intrinsic property of a photon, it is an observable. So I am assuming - perhaps incorrectly - that polarization can be discussed here in the same terms as with orthodox QM and other interpretations.

Q1) After Alice measures A, and A no longer exists: has the pilot wave completed its action on B? Does the pilot wave have any residual effect on B afterwards? Presumably B is now of the same polarization as A.

Q2) A photon can be polarized in at least 3 mutually unbiased manners: H/V, 1/0, L/R. Does a measurement on the H/V basis of A fix the photon's unmeasured value on the 1/0 or L/R bases? Ditto for B photon? This is confusing to me, because presumably under BM: particles have definite values for all observables simultaneously - which is different than most other interpretations. So... if A and B now have specific static values for their H/V polarization, does that also imply their other polarization observables are fixed and static? Or are they free variables/observables available to be measured?

Q3) Is there still a pilot wave connecting A and B? My understanding is that the pilot wave exists in a configuration space. Once Alice knows the outcome of her measurement on A, nothing else occurring in configuration space in the entire rest of the universe can change the certain outcome of Bob's future measurement on B. B must be static as to H/V polarization once A has been measured. No other particles can affect B, certainly at least not on the H/V basis.

Q4) If B is now fixed as to its H/V basis observable (depending on the answer to Q3): That almost implies that an exact (but unknown) position measurement has been performed on A. Wouldn't that be necessary to cause B to take on a static and known value for its future H/V measurement outcome?

My apologies for so many questions at once. Any enlightenment from BM advocates and/or other knowledgeable members is welcome.

-DrC

Addendum:

I am supplying this reference more as a point of explanation of why I am asking the above questions.

Revisiting Entanglement within the Bohmian Approach to Quantum Mechanics (2018)

"In spite of the intense research work that has been devoted to Bohmian dynamics and its applications, relatively little attention has been paid to the quantitative analysis of entanglement within the Bohmian approach."

The usual approach to BM vis a vis entanglement is: The math makes the same predictions as orthodox QM at certain fundamental levels, therefore BM must be equivalent in all respects. Yet there are obvious differences: Orthodox QM has no pilot wave, for example! Orthodox QM features spin! Orthodox QM can be upgraded to a relativistic version. So this is what leads me to confusion about the Bohmian perspective on entanglement as to a variety of specifics. How does basic entanglement work in Bohmian Mechanics? The usual spots I've researched over the past 10-20 years really don't ask/answer deep questions.

 

[PeterDonis]

As I understand it, the simple answer to the title question, when does the pilot wave action cease? is never. The pilot wave is always part of the equation of motion. Remember that in BM there is no collapse; there is just particle motion according to the equation of motion that includes the pilot wave (quantum potential) in the Hamiltonian. Measurement results are determined by the motion of the particles.

 

[PeterDonis]

Orthodox QM has no pilot wave

Yes it does; "pilot wave" is just another name for the wave function. What BM has that orthodox QM does not have is the unobservable particles that move according to the equation of motion that includes the quantum potential in the Hamiltonian.

Orthodox QM features spin!

So does BM. But BM's account of spin measurements is very, very different from the orthodox QM account, yes. On BM's account, spin is not a separate, independent degree of freedom; what we call "spin measurements" are measurements of particle positions--for example, which arm of a polarizing beam splitter a photon comes out of. On BM's account, the photon has a definite position at all times, and moves according to the equation of motion that includes the quantum potential in the Hamiltonian; but we can't observe that position except at isolated times when we detect the photon, say, in a particular output arm of a polarizing beam splitter.

Orthodox QM can be upgraded to a relativistic version.

@Demystifier would be able to say more about how BM relates to relativity.

 

[DrChinese]

As I understand it, the simple answer to the title question, when does the pilot wave action cease? is never. The pilot wave is always part of the equation of motion. Remember that in BM there is no collapse; there is just particle motion according to the equation of motion that includes the pilot wave (quantum potential) in the Hamiltonian. Measurement results are determined by the motion of the particles.

That description would be incompatible with A having a measured polarization, which causes B to take on a static value. The pilot wave action must have ceased for B to have a static polarization value.

 

[PeterDonis]

The pilot wave action must have ceased for B to have a static polarization value.

No, it doesn't. It just has to act on the B particle in such a way as to ensure that it comes out of the proper arm of whatever polarizing beam splitter is measuring it.

I know there is literature that goes into much more detail on how this actually works out mathematically, but unfortunately I'm not familiar enough with it to suggest a particular reference. Perhaps @Demystifier can suggest one. But I do think that BM is not an interpretation that one can just apply one's intuition to, even if one's intuition has been well trained on orthodox QM. One has to take the time to look at the actual math, as proponents of BM formulate it.

 

[DrChinese]

No, it doesn't. It just has to act on the B particle in such a way as to ensure that it comes out of the proper arm of whatever polarizing beam splitter is measuring it.

That value cannot change once A is measured. So the nonlocal pilot wave/guiding wave/quantum potential wave is no longer dynamic.

"In Bohmian mechanics, one can understand that the quantum potential is responsible for the instantaneous nonlocal changes on the trajectories of quantum particles."

From: Overview of Bohmian Mechanics (2019), page 20 of 87.

My whole point here is that BM proponents say "it's the same" as orthodox QM except for "X". It can't be both a separate theory and the same theory. Either there is a nonlocal action at a distance (not present in orthodox theory) or there isn't, and proponents say there is. We can call it whatever it takes to differentiate it from orthodox QM's "Schrödinger wave equation plus collapse", but I am asking for details on the whatever it is we want to call it. If "Pilot wave" doesn't fit the bill, let's pick "quantum potential" or whatever.

But regardless: Once A is measured as H>, B will certainly be measured as H>. And nothing in the universe changes that. So as far as I can see, the Pilot wave has no further impact on B and can be ignored. Or?

 

[PeterDonis]

That value cannot change once A is measured.

You're missing the point. In BM, the "value" of spin is not something separate from the particle's position. When we measure "spin" in BM, we are actually measuring particle position and interpreting it as a measurement of "spin", according to, for example, which output arm of a polarizing beam splitter the particle moves into. So when you say B's "spin value" is determined once A's is measured, that's not the way BM would describe it. BM would say that your measurements are simply revealing a pre-existing correlation between the positions of particles A and B, mediated by the pilot wave.

Remember that BM is a nonlocal realistic interpretation of QM: particle positions always have well-defined values (we just can't observe what they are most of the time), and all other observables are determined by particle positions, so all observables always have well-defined values. That's the "realistic" part. The "nonlocal" part just means that correlations between positions (and therefore between all observables) are mediated by the pilot wave, and the pilot wave is nonlocal: it instantaneously updates through the entire universe. (Note that this is a non-relativistic description; someone like @Demystifier would be much better able than I to explain how all this is affected by relativity.)

 

[Matterwave]

Either there is a nonlocal action at a distance (not present in orthodox theory) or there isn't, and proponents say there is.

Can you elaborate on "not present in orthodox theory" and what you think orthodox theory has to say (or doesn't say) about "nonlocal action at a distance"?

If you are talking about the no-communication theorem then I would think even BM would have to obey that.

 

[Demystifier]

For details about spin in Bohmian mechanics I recommend https://arxiv.org/abs/1305.1280

For a relativistic version of Bohmian mechanics see my https://arxiv.org/abs/2205.05986

 

[Demystifier]

Q1) After Alice measures A, and A no longer exists: has the pilot wave completed its action on B?

To describe the situation in which A no longer exists you need particle destruction, and for that you need QFT. Thus the usual Bohmian interpretation of nonrelativistic QM is not an appropriate theory to describe it. Anyway, without going into details about Bohmian interpretation of relativistic QFT, I can say that, in this case, the pilot wave of A has completed its action.

 

[Demystifier]

presumably under BM: particles have definite values for all observables simultaneously

No they don't. Once you get rid of this misconception, perhaps your other confusions will go away too.

Just because the results of measurements of all observables can in principle be predicted, doesn't mean that all observables have values when they are not measured. The values of observables in BM are contextual, meaning that they are not defined without a measuring apparatus. If there is no measurement, than there is no value. And yet, the value can in principle be predicted because in principle it can be predicted which observable will be measured. That's because the measuring apparatus, and the whole universe for that matter, is a quantum system described by certain deterministic equations of motion.

 

[Demystifier]

Q3) Is there still a pilot wave connecting A and B? My understanding is that the pilot wave exists in a configuration space.

Once the A has been destroyed, for which you need QFT, there is no longer pilot wave connecting A and B. But note that "configuration space" in QFT is a very different thing from the "configuration space" in nonrelativistic QM. So what you imagine in this context when you say "configuration space" may be completely wrong.