Does the quantum mechanical wave function work as a charge distribution? The quantum mechanical wave function is traditionally interpreted as a probability distribution of the particles position. It is possible to interact with an electron in a conductor, without collapsing the wave function, which can be distributed over the entire conductor. Can the wave function be understood as a charge distribution in this case? And if so, wouldn’t that mean that the particle actually takes the shape of the wave function?
 A: The charge distribution is related to the probability of the electron being in a given location. That is, the amplitude-squared of the wave fucnction, $\bar{\psi}  \psi $.
Under some circumstances this can work as a charge distribution. For example, in an atom the charge distribution is given that way, including the nucleus.
The interaction between some other particle and an electron in a conductor is going to be with the wave function, not the charge distribution. So, if the other thing is a photon, there will be some sort of $A \psi$ type interaction. That is, it will sample the wave function, not specifically the charge distribution.
A: Schrödinger initially believed that the wave function is related to actual charge distribution density, rather than to probability density. Some objections were raised to such interpretation. One of the objections is based on wave packet dispersion. I offered some modification of the interpretation (Entropy 2022, 24(2), 261), where it is assumed that the wave function is related to coarse-grained charge density, and one-particle wave functions are modeled as plasma-like collections of a large number of particles and antiparticles. This modification seems to be immune to the problem of wave packet dispersion.
