Is the form of Schrödinger equation without fields/potentials neglecting self-interaction?

Question

I understand that a charged particle (like electron or proton) in some potential $V(x,y,z)$ would be described by the following form of the Schrödinger equation:

\begin{equation} i \hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^2}{2m}\nabla^2 \psi + V(x,y,z) \psi. \end{equation}

I also know that a particle in an electromagnetic field is described by the following form of the Schrödinger equation:

\begin{equation} i \hbar \frac{\partial \psi}{\partial t}= \frac{1}{2m} \left(\frac{\hbar}{i} \nabla -q \vec A \right)\cdot \left(\frac{\hbar}{i} \nabla -q \vec A \right)\psi+q\phi\psi + V(x,y,z) \psi. \end{equation} This describes a semi-classical model where the field is treated classically and the particle quantum-mechanically. $\vec A$ and $\phi$ are associated with the electromagnetic field:

\begin{equation} \vec E = -\frac{\partial A}{\partial t} - \nabla \phi \end{equation}

\begin{equation} \vec B = \nabla \times \vec A \end{equation}

So, as I understand, the electromagnetic field can come from both:

1. External sources
2. The particle itself, since it definitely has a charge and can also have current

For the second point, see for example

C. J. Ryu, A. Y. Liu, W. E. I. Sha and W. C. Chew, "Finite-Difference Time-Domain Simulation of the Maxwell–Schrödinger System," in IEEE Journal on Multiscale and Multiphysics Computational Techniques, vol. 1, pp. 40-47, 2016, doi: 10.1109/JMMCT.2016.2605378

where charge density associated with the particle is $\rho = q |\psi|^2$ and $\vec J$ is an expression obeying $\nabla \cdot \vec J = -\frac{\partial \rho}{\partial t}$)

But for some reason, unless someone talks about external applied electromagnetic field, nobody mentions the form of the Schrödinger equation involving $\vec A$ and $\phi$. So my question is this:

1. Is it true that the form of the Schrödinger equation without $\vec A$ or $\phi$ is only approximate and neglects the self-interaction through electromagnetic field?
2. If it is approximate, do you have any insight on why (and when?) it is Okay to neglect the self-interaction? Intuitively, the charge density given by $\rho = q |\psi|^2$ is the "closest" charge density to the particle itself. So how come we can neglect it and not the field/potentials due to the other charged particles?

Answer 1

1. Is it true that the form of the Schrödinger equation without $\vec A$ or $\phi$ is only approximate and neglects the self-interaction through electromagnetic field?

Even with $\vec A$ and/or $\vec \phi$ the Schrodinger equation is only approximate.

The equation you wrote down is a single-particle equation. There is no dynamic interaction of particles, there is only an "interaction" with the fixed single-particle potential.

2. If it is approximate, do you have any insight on why (and when?) it is Okay to neglect the self-interaction?

Pretty much always.

If you want another perspective on this, you could check our Schwinger's book on Electrodynamics "Classical Electrodynamics" at Section 1.1 (the very first section). Already in classical electrodynamics you have to be sensible about the self interaction.

As a preview of that section of Schwinger's book, consider his equation 1.6: $$ E = \frac{1}{2}\sum_a q_a\phi(\vec r_a) - \sum_a E_a\;, $$ about which he says "The last part of (1.6) is not to be understood numerically, but rather as an injunction to remove those terms in the first sum that refer to a single particle. In other words, we remove the 'self-action,' leaving the mutual interactions between particles." (Emphasis added.)

Quantum mechanically, we describe electrodynamics with QED, which is not the same as a single-particle Schrodinger equation description.

So how come we can neglect it and not the field/potentials due to the other charged particles?

Because you can't pull yourself up by your own bootstraps.