Do Empty Waves in Bohm Theory Evolve Differently Than Non-Empty Ones?

camboy
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I'm following a course in Bohm's theory of QM, and I'm trying to understand what happens to 'empty waves'.

Consider artificial universe: contains 1 H atom only and an external potential which is zero everywhere except at a little barrier.

The H atom particles are traveling along in their guiding wave packet, following the streamlines of the probability flow. The packet is incident on the barrier. The barrier is so shaped that the result is two identical packets, one reflected back, one transmitted through. The particles deterministically end up in one of the packets (50% chance either way).

Thus there is an actual H atom (particles + Psi) going one way, and an 'empty H atom' (Psi only) going the other.

My question is this:

Given that the positions of the particles appear in the Hamiltonian of the time-dependent Schrodinger equation, is the change in shape of the packet with time different for the actual and empty H atoms?

Cheers
 
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camboy said:
Given that the positions of the particles appear in the Hamiltonian of the time-dependent Schrodinger equation, is the change in shape of the packet with time different for the actual and empty H atoms?
This is an excellent question!
The answer is - no.
Instead, you calculate the wave function in exactly the same way as without the Bohmian interpretation. Otherwise, you could not achieve the observational equivalence between Bohmian theory and standard quantum theory.
 
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