2
$\begingroup$

The momentum of particles in pilot wave theory is dependent on the phase of the spatial wave function, which is 0 for stationary states. This means that electrons in all excited stationary states of a hydrogen atom would have a momentum of 0. However actual measurements reveal a nonzero momentum. I have read that pilot wave theory can explain this by using the “time-of-flight” approach to measurement. By turning the potential “off” and letting the particle freely fly away (due to Gaussian wave function) and then measuring its position again, we can calculate the momentum. This would be nonzero since it involved changing the potential.

My question is, how is this a measurement at all? When we measure things in the real world, we don’t magically turn off the potential. And when a particle is free, it’s wave function need not be a Gaussian, it can be any normalized function that’s a superposition of momenta eigenstates. So how can pilot wave theory explain nonzero momentum measurements?

$\endgroup$
2
  • 4
    $\begingroup$ "However actual measurements reveal a nonzero momentum" can you give a supportive link for this statement?. I know no experiments that can measure momentum of bound electrons. $\endgroup$
    – anna v
    Commented Feb 9 at 11:58
  • $\begingroup$ Either way, the momentum distribution should be nonzero, according to the statistical approach. $\endgroup$
    – user310742
    Commented Feb 9 at 17:16

0