# Electric field associated with a stationary electron

In the framework of QFT, quantum fields are the fundamental objects instead of point-like notion of particles. Particles, at least fundamental ones like electron, are understood to arise as excitations of the quantum fields.

We know from experiments that a stationary free electron has some electric field associated with it. How does QFT explain that? Sure there is the background electromagnetic field and the electron field permeating all space-time but how does that explain electron itself generating electric field? In the QFT framework, it doesn't appear that electron can do that.

Can someone please help me with this? I have taken a QFT I course but this issue wasn't addressed there as well. Perhaps I am overlooking something very trivial.

• Can you clarify precisely which observable you are trying to calculate in QED? The electron field itself couples to photons, of course. Commented Jul 23, 2020 at 15:19
• @CosmasZachos Apologies for not making that clear. I have the answer now. Commented Jul 25, 2020 at 4:30

As in any quantum theory, observables in QFT are represented by operators on a Hilbert space. In QFT, observables are constructed from field operators. Field operators, in turn, satisfy nice equations of motion that look like classical equations of motion — except that the field operators don't commute with each other.

In the simplest version of QED, we have two fields: the electron/positron field $$\psi$$, and the EM field. These are non-commuting operators, but they still satisfy familiar-looking equations of motion (warning — I didn't check the signs carefully):

• The Dirac equation: $$(\gamma^\mu(i\partial_\mu-e A_\mu)+m)\psi=0$$

• Maxwell's equations: $$\partial_\mu F^{\mu\nu}= e\overline\psi\gamma^\nu\psi$$ and $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$$.

The $$\nu=0$$ component of the first Maxwell's equation is the Gauss-law constraint $$\nabla\cdot\mathbf{E}= \rho$$ where $$\rho \equiv e\overline\psi\gamma^0\psi$$ is the charge-density operator (an observable) and $$\mathbf{E}$$ is the electric field operator (also an observable). This equation says that the observables $$\rho$$ and $$\nabla\cdot\mathbf{E}$$ are equal to each other: they are the same operator. Thus any state with a non-zero charge automatically has the associated Coulomb field.

For example, writing $$\langle\cdots\rangle$$ for the expectation value in any state, we always have $$\langle\nabla\cdot\mathbf{E}\rangle=\langle \rho\rangle$$. This holds for all states, no matter how many or how few electrons/positrons they happen to contain. In particular, it holds for single-electron states.

Whenever we write down equations of motion for the field operators like the ones I wrote above, we're using the Heisenberg picture. In the Heisenberg picture, the field operators at all times are just combinations of the field operators at any single time (say $$t=0$$), even though the interpretations depend on which time we're considering. The Gauss-law constraint is similar, except that time is not involved: it says that some combinations of field operators can be written in terms of others, in particular that $$\nabla\cdot\mathbf{E}$$ can be written in terms of $$\rho$$, even though their interpretations may be different. This, of course, is the whole point of the equations of motion: to tell us how different observables are related to each other.

I've glossed over some tricky technicalities related to the fact that the field operators don't commute with each other. In particular, constructing a representation of these operators as operators on a Hilbert space is tricky because of gauge invariance. However, it can be done, and then the Gauss-law constraint $$\nabla\cdot\mathbf{E}= \rho$$ can be viewed as the condition that physical states must be gauge invariant (at least up to a physically irrelevant overall constant phase factor).

In any case, here's the bottom line: any state that has a charge (as defined by the observable $$\rho$$) automatically also has the associated Coulomb field field (as defined by the observable $$\mathbf{E}$$), because the operators representing those two different observables are related to each other by the Gauss-law constraint. This is true for any (gauge-invariant) state, including single-electron states.

• This justifies the existence of an $\mathbf{E}$ as soon as some charge exists but my question was slightly different, in spirit only. I have now got the answer though. THE electromagnetic field when interacted with any number of quanta of THE electron field will lead to an associated $\mathbf{E}$. In this respect, the quanta of the $\mathbf{E}$ (among others) are responsible for quantizing the electromagnetic field. But : i) Is it possible for the EM field to be quantized without presence of any q-field with $U(1)$ symmetry? ii) What causes the quantization of any q-field with $U(1)$ symmetry? Commented Jul 25, 2020 at 4:27
• @asymptoticallyboundedgluon i) Yes, we can construct a model of the quantum EM field by itself, as a free field, in a pretend world with no charged matter. ii) We could consider a model in which only the electron/positron field is quantized and the EM field is just a prescribed classical background field. Again, this model is less realistic than fully-quantized QED, but mathematically, nothing forces us to quantize the EM field. (My answer assumed fully-quantized QED.) Is that what you're asking? Commented Jul 25, 2020 at 15:03

This is a comment on this part of your question:

that a stationary free electron

Free particles cannot be modeled by the plane waves fields of QFT, one has to go to the wave packet solution.

The mathematics of getting the electric field from the wave packet is beyond me.

• But what about free scalar particles modelled by plane wave solutions of Klein-Gordon Equation? Commented Jul 25, 2020 at 4:29
• It is a mathematical model that cannot be fitted to real particles, there are no free scalar real particles after all. It may be useful for particular disciplines(condensed physics?) but it is a model, that may be good in describing some data. It is not mathematics that defines reality, but data. Commented Jul 25, 2020 at 5:03
• Agreed. Would you say that solutions of Dirac Equation are real enough? We can construct physical observables using those and that justifies that they are not mere mathematical constructs. Now, consider this : components of solutions of Dirac Equation also satisfy the Klein-Gordon Equation. What would you say to this? Commented Jul 25, 2020 at 7:18
• @asymptoticallyboundedgluon What I have already said: it is not the mathematics. Mathematics is always real. It is whether it fits and predicts new data. Mathematics can predict an infinite number if things that have no connection with physical reality. Commented Jul 25, 2020 at 11:09