Is the quantum probability density $\psi^*(x)\psi(x)$ a charge density? [closed]

Question

Although it is tempting to assume that $\psi^*(x)\psi(x)$ must manifest some substantive property beyond that of a mere calculating device--as advocated by the Copenhagen Interpretation, there is a simple way of rebutting its most obvious reification.

Assuming that, $\psi^*(x)\psi(x)$, is an actual density consisting of a "spread out particle", then for an electron in a hydrogen ground state, one might reasonably expect that the quantity $-e\psi^*(x)\psi(x)$ would play the role of a physically meaningful charge density. That is, such a charge density should be the source of an electric field. Moreover that field would be associated with a total energy equal to the repulsive self-interaction potential energy arising from the integral over all pairs of infinitesimal sub-volumes.

In other words, it would possess a capacitor like self-energy about equal to 1/2 e squared divided by a Bohr radius which plays the role of the effective capacitance. Since this energy would nearly cancel the negative binding energy, it would significantly decrease the Hydrogen ionization energy to the point of making the H atom (and the rest of the chemical elements) unstable --in contradiction of experiment. This is the reason why the Hartree approximation only includes inter-actions between different electron charge densities.

The main question, then, is how might it be possible to retain an effective charge density interpretation of $-e\psi^*(x)\psi(x)$ in light of the problem above.

Answer 1

It is wrong to think of $\psi^* \psi$ as a classical charge density; one needs to consider the quantum state more carefully. Suppose we had a superposition of positions of the form $$|x_1 \rangle + |x_2 \rangle.$$ It is not true that the electric field is simply that of charges $q/2$ at positions $x_1$ and $x_2$. In fact, there is no electric field value whatsoever. If you want to look at the electric field generated by a quantum particle, you must make the electric field quantum for consistency, so the true state of the system is $$|x_1, E \,\text{field sourced from } x_1 \rangle + |x_2, E \, \text{field sourced from } x_2 \rangle.$$ That is, the state of the quantum electromagnetic field and the quantum particle are entangled, so there is no notion of "the" electric field value at all.

This has definite physical consequences. Suppose you passed an electron through a double slit, and a proton sat behind and in between the slits. Under your picture, there would be equal charge density going through each slit, so the proton wouldn't move. But in reality, it ends up entangled with the electron, moving up if the electron goes through the top slit. If you don't make the electric field quantum, you can't get the right interaction between quantum charged particles.

Regarding electromagnetic field energy, this makes the problem much worse than you said; instead of computing the energy of a spread out particle, you need to compute the energy of the electromagnetic fields of a point charge. This is formally divergent, and it's taken care of by renormalization. The effects are already absorbed into the value of the electron mass by $E = mc^2$.

Answer 2

Schrödinger tried to interpret the square of the modulus of the wavefunction as the charge density and ran into problems. However, according to Barut, these problems can be overcome.

The following article may be relevant: A.O. Barut, Foundations of Physics Letters, VoL 1, No. 1, 1988, p. 47, The revival of Schrödinger's Interpretation of Quantum Mechanics

Apparently Schrödinger inserted in the nonlinear term in (2) only one term, the ground state wave function of the H atom, instead of the full Fourier expansion, and obtained a too large value of the energy shift. It turns out that this term is only part of the renormalization and not the observable energy shift.

Answer 3

The quantity $e|\psi|^2$ is the Noether charge probability distribution. Is is not a charge density but the probability of finding the entire charge $e$ at a specific position. This shows that there is an underlying assumption that it is a point charge.

Answer 4

The probability distributions of electrons in the orbitals around the nuclei is interpreted as a probable charge density, example the hydrogen atom:

Hydrogen 1s Radial Probability

The shape of the orbitals ( probability locuses of electrons bound into atoms) allow for the existence of transient electric and magnetic fields which lead to bonding into molecules and lattices.

The main question, then, is how might it be possible to retain an effective charge density interpretation of $−eψ∗(x)ψ(x)$ in light of the problem above.

By giving up classical modeling at the level of quantum mechanics, and accepting probability distributions. They make up the world as we see it macroscopically.