How should the electric field produced by an electron look like if electron is described by a wave function?

In quantum mechanics, the position of an object is not definitely known, but instead is described by a probability density function of where it would be located.

Then what should the electric field produced by an electron look like if its position is described a probability density? Is the electric field described by a probability density function as well?

• This requires using quantum field theory, because the electric field is a field. If we want the model to be self-consistent, then treating the charged particles as quantum requires treating the electric field as quantum, too -- hence quantum field theory. If a charged particle is not in an eigenstate of the position operator, then its associated electric field will not be in an eigenstate of the electric field operators either. The question asks what it "looks like," but I don't know what that means in this context, which is why this is only a comment and not an answer. – Chiral Anomaly Oct 18 '19 at 15:57
• @ChiralAnomaly I am looking for a mathematical description of quantum electric field. Like, wave function is a mathematical description of where the electron is and how it moves, a vector field is a mathematical description of classical electric field, there should be some mathematical object that describes quantum electric field. – edm Oct 18 '19 at 16:09

In nonrelativistic QM the charge density is $$\rho = e|\psi|^2$$. From this you find V and E using Coulomb's law.

In quantum mechanics, the field act as an operator acting on a state of electron.

The electrons state is described by some mathematical object (vector in some hilbert space) that encodes all the information of the electron we have (its state). There is also procedure how to get from this abstract mathematical object the physically meaningful information. For example, if you project the state of your electron on a vector that describes state of definite position you get amplitude that tells you how probable is finding the electron in that position.

Now, what should the field do to a state of some electron? Well, change it. So the field will be some operator that maps electron state to another.

In quantum mechanics the evolution of system is described by operator called Hamiltonian. This Hamiltonian operator can be vied in some sense as energy. You can look into Hamiltonian formulation of classical mechanics to gain deeper insight of connection between time evolution and energy, without the complications of QM. Anyway, the point is, that Hamiltonian is closely related to energy and it is operator that governs time evolution.

Now, the electric energy of electron in classical mechanics is given by the potential that is function of position of this electron. Specifically, the classical energy of system of two electrons is: $$E=\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}+2k\frac{e^2}{\left(r_1-r_2\right)^2}$$

It turns out that in QM the Hamiltonian of the same system will have exactly this form. But now, ps and rs are not numbers, but operators that are changing the state of electron. In particular, the ps act as derivatives on wave function (in position space) and rs act as numbers you multiply the wave function (in position space) by.

Why the ps and rs are "derivatives/multiplier" operators can be answered elsewhere (or found elsewhere), the main point of my answer is that the electric field in QM is operator that maps some state of the system to another and curiously enaugh, this operator has the same form as formula for energy in classical physics.