In PWT, the wave function depends on the positions of all the particles. Therefore the velocity of a given particle depends of the positions of the other, it's why the theory is non local. In the model why there is a single wave for all the particle and not a wave for each particle? So that the velocity of a given particle does not depend on the other.
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$\begingroup$ Because as you said, if you want to keep this interpretation of quantum mechanics, there are then nonlocal effects and you need global hidden variables. This is due to Bell's theorem. $\endgroup$– MauricioCommented Aug 15, 2023 at 10:03
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1$\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$– Community BotCommented Aug 15, 2023 at 10:30
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$\begingroup$ In the article of Quantum equilibrium and the origin of absolute uncertainty by Durr, Golstein and Zanghi (arxiv.org/abs/quant-ph/0308039) at section 3-Bohmian mechanics (page 10) it says that the velocity of a given particle k depends on the gradient of the wave function which depends on the positions of all the other particles. But why the wave function depends on all the other positions and not only of a given particle. $\endgroup$– vincent woilineCommented Aug 15, 2023 at 12:26
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$\begingroup$ So in the double slit experience there is only one guiding wave which guide all the electron. Can a wave disappear as the electron touch the screen? And then when a new electron is fired a new wave guides him? $\endgroup$– vincent woilineCommented Aug 15, 2023 at 12:38
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1 Answer
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Short answer: There's a single pilot wave because there's a single wavefunction.
Before I get to the longer answer, it's important to clarify that your first sentence is slightly incorrect. Conventional Bohmian pilot wave theory has each particle represented by a single path in spacetime, $\vec{x}_i(t)$ and we don't know those exact paths. However, the wavefunction doesn't "depend on" those exact particle paths; the particles respond to the wavefunction, but the wavefunction doesn't respond to the particles. Learning which precise path a particle takes doesn't change the full wavefunction.
The other key piece to understand about pilot wave theory is that these particle paths are hypothesized as something new, in addition to the conventional quantum wavefunction. They don't exist in conventional quantum theory.
And this gets to the answer to your question. The map from conventional theory to pilot wave theory is that the conventional N-particle wavefunction $\psi(\vec{x}_i,\vec{x}_2,\vec{x}_3...\vec{x}_N,t)$ maps directly to the pilot wave, and the particle paths are then introduced as extra elements to the theory. So if you're asking why there's not a separate pilot wave for each "particle", first realize that those individual "particles" don't even exist in conventional quantum theory. Once you have that, the answer to your question is explained by the fact that you CANNOT (in general) decompose the N-particle wavefunction in 3N-dimensional configuration space into N single-particle wavefunctions, each in 3-dimensional space. Any entangled wavefunction cannot be decomposed in this manner. So there's no mathematical way to take the single configuration-space-pilot-wave and turn it into N real space pilot waves.
(There is a clever argument that you could turn the wavefunction into an infinite number of real-space pilot waves, thanks to Travis Norsen, but not in the way you have in mind.)
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$\begingroup$ Thank you Mr. Wharton for your answer. I have now a better understanding in pilot wave theory. Best regards Vincent $\endgroup$ Commented Aug 21, 2023 at 12:50