# How does pilot wave theory explain "identical particle" interference?

Pilot wave theory says that there exist waves in 3D space which carry particles. This explains, say, the double slit experiment.

But this does not explain the behavior of identical particles. According to standard QM, a system of two identical particles will have quantum interference. But this interference does not take place in the real world 3D space, but rather in an abstract space. The wavefunction of two identical particles looks like $$\psi (x_1,x_2)$$. But the points $$(x_1,x_2)$$ live in an abstract space, as in, $$(x_1,x_2)$$ is not to be identified with a location in the real world 3D space. Rather, it is to be identified with a configuration of the system.

So, since the waves in pilot wave theory live in the real world space, how can it explain the wave-like interference in abstract spaces?

• The waves in pilot wave theory are exactly the same as wavefunctions in ordinary quantum mechanics. When you have two identical bosons, for example, they are symmetric functions on 6d configuration space. When you have a field, they're wavefunctionals. Commented Aug 25, 2022 at 4:26
• By the way, this is also the reason non-physicists absolutely love pilot wave theory and physicists generally don't. Non-physicists get the impression that you can reduce all of quantum theory to waves in real space, so that it's no weirder than other classical field theories such as electromagnetism. But that's not actually how it works. Real pilot wave theory is just the exact same as ordinary QM plus a hidden variable to track the "real" positions, which adds complexity without making anything more intuitive. Commented Aug 25, 2022 at 4:28
• @knzhou wait, is this why this theory is non-local? If you have waves in abstract space, then, they can affect particles faster than light. Commented Aug 25, 2022 at 4:30

To quote Bohm's 1952 paper:

In the two-body problem, the system is described therefore by a six-dimensional Schroedinger wave and by a six-dimensional trajectory, specifying the actual location of each of the two particles. The velocity of this trajectory has components $$\nabla_1 S/m$$ and $$\nabla_2 S/m$$, respectively, in each of the three-dimensional surfaces associated with a given particle.

The "Schrodinger wave" is what we now call the pilot wave, which is written $$\psi= e^{iS} R$$ for $$\mathbb R$$-valued functions $$S$$ and $$R$$.

So the straightforward answer to your question is that it's founded on an incorrect premise - the pilot wave is defined on the configuration space of the system (in this case, $$\mathbb R^3\times \mathbb R^3$$) not simply on $$\mathbb R^3$$. If you read the paper, it is made quite plain that Bohm is proposing an alternative interpretation of the mathematics which underlies the standard formulation of quantum mechanics, not a new theory. The equations you need to solve in de Broglie-Bohm picture are exactly equivalent to the equations you need to solve in ordinary QM, you're just wrapping different words around them.

• But this theory is non-local, which makes it different. Is the reason this theory is non-local is that it employs waves in abstract space? I mean, two points can be close in abstract space, but far away in real space. Now that I think of it, why doesn't standard QM suffer from this same non-locality? Standard QM also has abstract space interference, which means that part of the wavefunction that are close in abstract space, can affect each other, even if they're far in real space Commented Aug 25, 2022 at 4:39
• @RyderRude It's not different in terms of measurable predictions. The sense in which the theory is nonlocal is that it postulates an additional "quantum potential" (see my answer here for a summary) which, in the case of multiple particles, constitutes an interaction between particles at different locations in space. In essence, the state of motion of each particle depends on the positions of all of the other particles. Commented Aug 25, 2022 at 4:52
• @RyderRude Standard quantum mechanics can also exhibit this feature, but in relativistic QM we typically impose the requirement of locality on the interactions - which in this case means that wavefunctions couple to each other pointwise ($H_{int} \sim \psi(x)\phi(y) \delta(x-y)$) or via an intermediate field. The same imposition cannot be made on de Broglie-Bohm because the gradient of $S$ at a point is understood to the velocity of a particle located there, but $S$ depends on the location of each particle. Commented Aug 25, 2022 at 5:02
• I can only agree with J. Murray. To support his answer I will add: The Bohmian theory has explicitly non-local interactions, but due to randomness in the theory experienced by the experimenter's lack of information, it is not possible to send information faster than light. Just as in QM, even though we instantaneously update the singlet state in measurements of systems like in Bell's Theorem, this instantaneous update cannot be used to send information faster than light. The difference is, in QM, the details a bit more unresolved so maybe it can be explained locally, e.g. w/ superdeterminism Commented Aug 25, 2022 at 13:05

The pilot wave theory is mostly a strategy for denying the implications of quantum theory. The pilot wave theory takes the wavefunction and adds particles on top of it. Pilot wave theorists then have two options. (1) They can deny that the wavefunction is real, in which case it makes no sense that the particles are influenced by it and the theory can't explain single particle interference let alone more complicated issues like EPR correlations. (2) They say the wavefunction is real in which case they still have to deal with the issue of how to understand the fact that it looks approximately like a collection of parallel universes under conditions of decoherence. So pilot wave theory doesn't solve the measurement problem and also adds structure to quantum theory so it has parallel universes and other stuff on top.

Also, contrary to one of the answers given above, pilot wave theory does make predictions different from those of quantum theory for the added particles.

• In the link you gave with the description "pilot wave theory doesn't solve the measurement problem", the first few words of the paper's abstract is "The quantum theory of de Broglie and Bohm solves the measurement problem". Your source therefore is in contradiction with your answer. The paper then just takes issue with BM through a claim that the resolution to the msmt problem doesn't have to do with bohmian particles. In my understanding of the resolution to the msmt problem in BM, this is not the case, and I could not understand the point of the paper, though I am open to be corrected. Commented Aug 25, 2022 at 12:54
• In addition, your answer doesn't seem to address the question, rather just pilot wave theory in general. Commented Aug 25, 2022 at 12:59
• If pilot wave theory doesn't solve the measurement problem for any quantum system, then it doesn't solve the measurement problem for systems of identical particles. Also, the abstract of the paper I linked states that the solution has nothing to do with the particles in the Bohm theory, which are the only thing that sets it apart from quantum theory. So its sole contribution is useless. Commented Aug 26, 2022 at 0:56
• The point of the paper is that if the wavefunction decoheres then the measurement problem is solved by quantum theory without the extra particles. If the wavefunction doesn't decohere then the presence of the particles won't matter because without decoherence there aren't stable branches of the wavefunction and the particles wouldn't form stable configurations. Commented Aug 26, 2022 at 1:01
• "The point of the paper is that if the wavefunction decoheres then the measurement problem is solved by quantum theory without the extra particles" - This is an uninformed take on the measurement problem. Even the biggest advocates and researchers of decoherence, e.g. Wojciech Zurek, are very clear in communicating that it does not resolve the measurement problem. The reason that it does not is that it does not take the final step of predicting that any specific outcome happens at all. "So its sole contribution is useless" - Again, very confident opinion that is missing a lot of understanding. Commented Aug 26, 2022 at 1:12