Can a pilot wave be instantaneous?

[Terry M]

I don't understand why it cannot be. If a particle is "riding" a pilot wave and is guided by it, then the pilot wave must precede the particle in some way. But at what speed? Isn't the simplest explanation that the pilot wave is instantaneous, with the particle simply being a phase phenomenon of this wave? Specifically, the group velocity of the pilot wave? This group velocity is limited in some way to c, so there are no causality issues. And it seems to solve the non-locality issues of QM very nicely. I've looked at the literature a bit, including Bell's "Speakable and unspeakable in quantum mechanics," and can't seem to find any discussion of this.

I'm also slugging my way through very basic quantum field theory, and see Fourier transforms being used to go back and forth between first and second quantization. Within a box, and using the boundary conditions given by this box, the Fourier summation doesn't pose a problem. But without a box, it seems that the boundary conditions could require the entire universe. Again, an instantaneous pilot wave seems to address this.

 

[Demystifier]

The pilot wave is described by the Schrodinger equation or its relativistic cousin, so no, it is not instantaneous.

 

[Ilja]

Hm, the pilot wave is a function on the configuration space, and time. [itex]\Psi=\Psi(q, t)[/itex]. The notion of time here is absolute time, not some relativistic proper time or so, and not a measurable quantity (each clock has even a nonzero probability to go backward in time).

It guides the configuration, which is some global object - the configuration of the whole universe. And one part of the configuration in one part of the universe defines which part of the wave function is relevant to all the other parts - and this happens instantaneously. So, even if the evolution of the wave function may be defined by local Hamilton operators, it is a very global object and the whole theory is inherently nonlocal.

 

[Demystifier]

Yes, but the speed of wave propagation is finite. In that sense (which seems to me to be closer to what Terry M asked) it is not instantaneous.

 

[Terry M]

Thanks very much. I think I see my problem now. It is about under what conditions one may use the time-independent Schrodinger equation vs. the time-dependent one.
More for me to study!