Does one electron in superposition repel itself?

Question

Consider Quantum Electrodynamics, and consider the electron field to be in a state which is a superposition of two wavepackets, each located in a different spatial position. Explicitly:

$$|\psi\rangle = \sum_{s=1}^2\int d^3 p ( f^{1}_s(\vec{p})+ f^{2}_s(\vec{p}))\hat{b}_{s,\vec{p}}^\dagger |0\rangle,$$

where the functions $f^{1}_s(\vec{p}) = e^{i\vec{k}\cdot \vec{x}_0}f^{2}_s(\vec{p})$, so that these are shifted in space by $\vec{x}_0$. $\hat{b}_{s,\vec{p}}$ is the electron annihilation operator, and $|0\rangle$ is the vacuum state of the electron field.

The question is whether each part of the electron wavefunction will interact with the electromagnetic field sourced by the other. In other words, does each branch of the superposition experience the electric field sourced by the other branch?

Answer 1

There are two issues in your question.

First: do two parts of a superposition interact with each other? The answer is no. I think, implicit in your question, is an idea of the quantum state of an electron as a wavefunction in 3-dimensional space. If you visualize the quantum state that way, then you can be led to wrong ideas, like two bumps in the wavefunction being analogous to two particles that interact.

Really, the quantum state is defined over the entire configuration space of the system. If you have one particle, then the configuration space is just 3-dimensional space, but you get into trouble if you don't remember that the dimension of the configuration space grows if you introduce more degrees of freedom. In this case, you want to consider the electron, plus the electromagnetic field. So the state is a superposition of: "electron near position A, generating an electrostatic field centered at A", and "electron near position B, generating an electrostatic field centered at B." There is no basis state in the superposition corresponding to an electron near position A, responding to a field centered on position B.

Second, there is a sense in which the electron does interact with itself, although the origin of this effect is not superposition. This happens even classically. Classically, the problem is to consider the effect that the electric field an electron generates, on itself. This creates a paradox, classically, because if the electron is a point particle, then the field is infinite at the location of the particle, and so there's no way to calculate this effect. In quantum field theory, this self-energy can be understood in Feynman diagram language as a loop diagram, where the electron emits a photon and then absorbs that same photon some time later (at least in the simplest 1-loop case). This interaction renormalizes (shifts) the mass of the electron.

Answer 2

The electron has the same propagator as a free uncharged fermion. The interaction only gives the loop correction to the propagator and this correction renormalize only mass. Actually the dispersion law does not changed. So in that sense electron would not repel itself in the superposition state.

Answer 3

Second quantized case
There are different levels on which this question could be answered. A two-particle interaction has form: $$ \hat{U}=\int d1 \int d2 \hat{\psi}^\dagger(1)\hat{\psi}^\dagger(2)v(1,2)\hat{\psi}(2)\hat{\psi}(1) $$ This is a two-particle operator gives zero when acting on a one-particle state.

Basic QM
If we revert to non-second-quantized quantum mechanics, the Coulomb interaction is included in a two-particle Schrödinger equation as $$ \left[-\frac{\hbar^2\nabla_1^2}{2m}-\frac{\hbar^2\nabla_2^2}{2m} + v(1,2)\right]\varphi(1,2)=E\varphi(1,2), $$ where ansatz $\varphi(1,2)=\phi(1)\chi(2)-\phi(2)\chi(1)$ gives the Hartree-Fock equations, including conventional Coulomb interaction (Hartree term) and the exchange interaction (Fock term). One still needs two particles here, as particles are considered the elementary building blocks for describing interactions

"Naïve"/"intuitive" reasoning
If we reason in (admittedly classical) way of a particle as a charge distribution, which can be split into small elements, and which thereby interact between themselves, we could indeed write the energy of a particle interacting with itself $$ U_{self}=\int d1\int d2 \phi^*(1)\phi^*(2)v(1,2)\phi(2)\phi(1), $$ which would mean adding to the one-particle Hamiltonian a potential term like $$ \int d2 \phi^*(2)v(1,2)\phi(2)\phi(1)=u(1)\phi(1), $$ Note that this term is divergent, since $v(1,1)$ is undefined for Coulomb interaction, whereas the wave function cannot be identically zero everywhere. We could then argue that adding such a term amounts to shifting the energies of all the particles by the same amount, and thus does not affect the physics. In more QFT language - we dismiss the divergent energy of self-interaction.

Answer 4

I think there is a much simpler answer. If two branches of superpositions interact, it would destroy the linearity of quantum mechanics. As far as we now know, both QFT and QM are linear theories in the sense that all operators (evolution, hamiltonian, etc) are linear. Thus, it is impossible to have interaction between two branches. However, there are modifications of QM where this kind of interactions are present. As an example, see this.