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