So I come from a chemistry background, where the electronic structure of atoms and molecules is central. For practical purposes, we usually work with a charge density operator

$$ \hat{\rho}(r) = q \delta (r-r') \; ,$$

with $r'$ being the "particle coordinate" so that the expectation value is

$$ \langle \Psi|\hat{\rho}(r)|\Psi\rangle = \int \left| \Psi (r')\right|^2 q \delta (r-r') \;dr' = q \left| \Psi (r)\right|^2 \; ,$$

in accordance with the idea of the probability being proportional to the wavefunction squared.

Then, in some electrodynamics exercises (eg. in Jackson) we take this charge density, then solve Gauss's equation for the electric field, and conclude that the electric field of the quantum mechanical system is given by our result.

I'm wondering: is there some operator $\hat{E}(r)$ that would give us back this semiclassical electric field as $\langle \Psi|\hat{E}(r)|\Psi\rangle $?

Maybe I am using semiclassical wrong in this context, in my understanding if I get my charge density from quantum mechanics but solve the classical Maxwell equations, that can be called semiclassical. I am also aware that the electric field operator is defined in quantum electrodynamics as acting on the modes of the field, but I don't think I am looking for that operator (as I am not interested in field-matter interactions), nor can I understand QED derivations with a chemistry background sadly.

I've found this paper which introduced a polarization operator $\hat{P}(r)$ so that the polarization field will be the solution of Gauss's equation for the charge density (Eq. 3 of the paper), which seems to be similar to what I am looking for. I am wondering: is this polarization field the same as the electric field in this context? My understanding is that we can only get a well defined longitudinal field component by solving Gauss's equation, so it might very well be that the longitudinal component of this polarization field is the same as of the electric field that I am looking for.

So is there such an operator that would have the electric field of the charge density corresponding to our wavefunction as the expectation operator? And if yes, how is that different from the polarization field operator?

EDIT for clarity, let's consider the example of a hydrogen atom.

The electronic wavefunction of the ground state of the hydrogen atom is (in the simplest units)

$$\Psi(r) = \frac{1}{\sqrt{\pi}}\exp{(-r)} \; .$$

I want to find the electric field of this electron. In order to do this, I'll first evaluate the classical charge density, then solve Gauss's law for the field.

The charge density - and here we employ the semiclassical idea - will be given as the expectation value of the charge density operator

$$ \rho(r) = \int \left| \Psi (r')\right|^2 q \delta (r-r') \;dr' = \frac{q}{\pi} \exp{(-2r)}\; .$$

To get the field from this classical charge density I just solve Gauss's law for the field

$$\nabla \cdot \vec{E} = \rho \; \rightarrow \; \vec{E}(r) = -\frac{q e^{-2 r} \left(2 r^2+2 r+1\right)}{4 \pi r^2 } \hat{e}_r \; .$$

This idea is clear to me.

My first question is:

  1. Is there an operator directly for the electric field, so that instead of evaluating first the expectation value of $\hat{\rho}$ and solving Gauss's law I could directly evaluate $\hat{E}$?

While researching this question I've came across the concept of the polarization density. I am not familiar with polarization calculations at all, but to me it seems like there's an intimate connection between the electric field and the polarization field in this context. To be more exact, in this paper Eq. 29 is exactly what I find for the electric field for the hydrogen atom, however in the paper it is claimed that it is the polarization field of the atom.

My second question is:

  1. Is this electric field operator that I am looking for exactly the same as the polarization field operator? If not, what's the difference?
  • $\begingroup$ "Semi-classical" usually means fields are wave-like; i.e., no field operators. $\endgroup$
    – Yejus
    Nov 26, 2021 at 13:23
  • 1
    $\begingroup$ Formally we may write $E:=-\nabla_x \phi$, where $\phi(x,x'')=\int d^3x' \frac{q \delta(x''-x')}{|x-x'|}$ which, when averaged over a given state, looks like $\left<\psi|E(x)|\psi\right>=\int d^3x''\ \psi^*(x'')\left[-\nabla_x \phi \right] \psi(x'')$. No new information has been added with this formalism, this still amounts to finding the field due to the current density defined in your post. $\endgroup$
    – Sal
    Nov 26, 2021 at 18:23

1 Answer 1


As long as we talk about quantum mechanics, the electromagnetic field is usually taken to be non-quantized (unlike the electrons and nuclei of the atoms). That is, there is no need for introducing field operators and wave functions. The situation changes in quantum field theory, where one quantized the particles and the elecytromagnetic field on equal footing, and one can introduce the operators in the standard way, see Quantization of the electromagnetic field (and also this answer). However, this is largely beyond the scope of what seems to be required in the case of the OP.

Update: The particular ways of solving the Poisson equation for the electric potential, $$\nabla\cdot\mathbf{E}(\mathbf{r})=\nabla^2\phi(\mathbf{r})=\rho(\mathbf{r})$$ is by using Green's function, that is the solution of the equation for a point charge: $$\nabla^2G(\mathbf{r}, \mathbf{r}')=\delta(\mathbf{r}-\mathbf{r}'),$$ which gives us $$ G(\mathbf{r}, \mathbf{r}')=\frac{1}{4\pi}\frac{1}{|\mathbf{r}- \mathbf{r}'|}. $$ The potential from charge distribution $\rho(\mathbf{r})$ can now be written as $$ \phi(\mathbf{r})=\frac{1}{4\pi}\int d^3\mathbf{r}'\frac{\rho(\mathbf{r}')}{|\mathbf{r}- \mathbf{r}'|}.$$ One could also obtain a similar solution directly for the field, by defining a function $$\mathbf{F}(\mathbf{r},(\mathbf{r})')=\nabla_\mathbf{r}\cdot G(\mathbf{r}, \mathbf{r}')$$ (Working directly with the field is however less practical, since one needs to take care of all its components.)

See also this answer.

  • $\begingroup$ Thank you for the clarification; it seems like my understanding of the semiclassical idea was in the right ballpark. My question - that is, asking for an operator for the nonquantized electric field - still makes sense though, right? $\endgroup$
    – user312272
    Nov 26, 2021 at 13:40
  • $\begingroup$ In classical electrodynamics one usually doesn't deal wuth operators... In Schrödinger equation we simply include $E(x,t)$, as if it is a function - liek potential. Tht is, it is an operator in respect to electrons, as any function dependent on $x$. Btw, isn't "croissant" masculine - "le beau croissant"? $\endgroup$
    – Roger V.
    Nov 26, 2021 at 13:46
  • $\begingroup$ I think including E(x,t) into the Schrodinger equation is usually done when my quantum system is in an external field. However, if I am interested in the field generated by the system, I think that should be able to be expressed with an operator, just as I can express the charge density of my system with an operator. $\endgroup$
    – user312272
    Nov 26, 2021 at 13:53
  • $\begingroup$ You need not make the external field into the operator - however now your external field is not a given function, but obeys its own equations, which are the classical Maxwell equations. $\endgroup$
    – Roger V.
    Nov 26, 2021 at 14:08
  • $\begingroup$ Sadly I don't see how that helps me to find the electric field of a quantum system without any external field present. $\endgroup$
    – user312272
    Nov 26, 2021 at 14:12

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