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In Bohmian Mechanics, do all particles have its own pilot wave, or can you have multiple particles on the same wave?

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    $\begingroup$ You know the pilot wave: it's simply the Schrodinger wave function $\endgroup$ – Ruben Verresen Apr 2 '17 at 23:42
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In Bohmian mechanics there is one wave function that permeates the Universe, so yes, all particles are 'on the same wave', and this is why this interpretation has such explicit nonlocality. For reasonably isolated systems you may treat them as having their own wave functions though.

This is sometimes used as an argument against BM suggesting it allows for superluminal signalling, however so long as Born's rule holds (known in BM as the quantum equilibrium hypothesis - which is no longer technically a rule), everything's effectively the same as other interpretations.

As others have mentioned, it's worth noting that the Bohmian wave function propagates on 3N-space instead of 3-space (so add three dimensions for every particle, known as a beable in BM, in the system). For this reason, BM isn't too far removed from the consistent histories or many-worlds interpretations (in my opinion - I'm sure many proponents of each theory will disagree).

Sources:

https://dx.doi.org/10.1007/BF01049004 (https://arxiv.org/abs/quant-ph/0308039)

http://arxiv.org/abs/quant-ph/9512031

http://dx.doi.org/10.1093/bjps/axt019 (https://arxiv.org/abs/1406.1371)

http://dx.doi.org/10.1088/1742-6596/701/1/012003 (open access)

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  • $\begingroup$ I'm in no way qualified to say which answer is correct, but I like this answer :). $\endgroup$ – Thanacles May 20 '17 at 3:52
  • $\begingroup$ Thanks :). It's really a case of perspective; if you look at papers on relativistic Bohmian mechanics you'll often find reference to the universal wave function. This was first alluded to in the conclusion of Bell's 1964 paper. It's a 'hidden variable' theory as the wave function describes everything in the Universe including ourselves, and we can't possibly account for that, so can never known the initial values in an experiment. A lot of the practical applications of BM are found within quantum chemistry (or similar) though, where this can be ignored. $\endgroup$ – Toby Hawkins May 20 '17 at 11:08
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That depends on what level you look. For a single atom in a gas, there is one Bohmian wave for each atom. Are you looking at the atom as a composite of particles, like nucleons and electrons, the Bohmian wave is more complicated like the Schrodinger wave function is. Ist's built of a combination of all these constituent Bohmian waves (wave functions). It depends on the distance you look at the particles.

The electrons in a metal all obey the Fermi statistics and an asymmetric (multi-particle, say N) pilot wave (wave function) will develop. At the moment the metal becomes superconducting the electrons become bosons (by a variety of quasi-particles and the Bohmian wave becomes a symmetric Bohmian wave of 1/2N particles.

Essentially it's the wave function that becomes the function of hidden variables which are deterministic variables and make it look as if the quantum function is inherently probabilistic. Like the movement of a Brown-particle which looks at first sight purely random, but when you look deeper (which is more difficult for hidden variables) it's just caused by the, in the time when the movements of Brown particles were first discovered, unknown atoms which surround the Brown-particle.

So in the double slit experiment, the particle that is sent to the two slits is part of one pilot wave (the wave function tied to the particle), but this pilot wave is built out of hidden variables which give the particle at any moment a well-defined position and momentum. The particle does not go through both slits at once but to one or another.

Not all particles in the Universe are on the same wave, only those that have once interacted with each other, which are a lot (hence the non-locality), but certainly not all.

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  • $\begingroup$ Hey, sorry but I didn't have the reputation previously to respond to your answer directly. Whilst I'm not sure everyone would agree on the universal wave function point (although I've not personally seen any disagreements), you can certainly find references to Bohm, Bell, and Dürr (probably three of the biggest figures within BM) stating this explicitly. For an example, see my edited answer. $\endgroup$ – Toby Hawkins Jun 28 '17 at 22:19
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Interesting question. I think from a theoretical standpoint, many to 1. From a practical standpoint, 1:1. For instance, as we shoot photons through a double slit, even as our setup remains the same, the universe itself evolves over time so each photon fired uses its own unique wave function. Firing many photons in sequence quickly, we do not see this because the wave functions of each photon fired are quite similar to the next, practically appearing the same when mathematically, they are distinct.

Just my two cents though.

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