# QED in Coulomb Gauge: Deriving the Coulomb Force and Questions

• tom.stoer
In summary, the conversation discussed a way to derive Coulomb force using the canonical mechanism and Coulomb gauge. The derivation involves using the Greens function of the Laplacian and calculating the interaction term in the Hamiltonian density. The conversation also touched upon the physical meaning of a background field a(x) and its relationship to gauge transformations and electromagnetic radiation. It was concluded that a(x) is not radiation, but a pure gauge and that the Lorentz gauge can be problematic in the canonical formulation of QED.

#### tom.stoer

I presented a way to derive Coulomb force via the canonical mechanism.

One uses Coulomb gauge

$$\partial_i A^i = 0$$

derives

$$\Delta A_0 = -4\pi\,\rho$$

which can be inverted formally

$$A_0 = -4\pi\,\Delta^{-1}\,\rho$$

and calculates the interaction term in the Hamiltonian density

$$\mathcal{H}_\text{int} = - 4 \pi \, \rho \,\Delta^{-1}\,\rho + \ldots$$

where ... represents the coupling of the 3-potential A to the 3-vector current density.

Using the Greens function of the Laplacian one immediately finds the well-known Coulomb potential term (from which in classical electrodynamics the Coulomb force for point charges can be derived).

$$H_\text{int} = \int_{\mathbb{R}^3 \times \mathbb{R}^3} d^3r\,d^3r^\prime \frac{\rho(\mathbf{r})\,\rho(\mathbf{r}^\prime)}{|\mathbf{r} - \mathbf{r}^\prime|} + \ldots$$

That means that the Coulomb potential can be derived w/o using the dynamical equations of the theory and w/o any restriction like electrostatics.

Problem:

In principle I can generalize this as follows: I add a 'background' field a(x), i.e. a harmonic function

$$\Delta \, a = 0$$

$$A_0 = -4\pi\,\Delta^{-1}\,\rho + a$$

$$\mathcal{H}_\text{int} = - 4 \pi \, \rho \,\Delta^{-1}\,\rho - 4 \pi \, \rho a + \ldots$$

Questions:
1) what would be the physical meaning of a(x) - which is source-free, i.e. neither generated by charges nor by dynamical el.-mag. fields?
2) are there mathematical reasons for a(x) = 0
3) what would it mean to introduce such fields in the canonical quantization of QED?

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Gauge transformation, I think.

Bill_K said:
Gauge transformation, I think.

OK, let's try that.

A(x) respects Coulomb gauge

$$\partial_i A^i = 0$$

Adding this new term a(x) shall not violate the Coulomb gauge condition which means we look for a residual gauge transformation

$$A_\mu \to A_\mu^\prime = A_\mu - e\partial_\mu \chi$$

which respect Coulomb gauge

$$\partial_i (A^i)^\prime = 0\;\to\;\partial_i (\partial^i \chi) = \Delta\chi = 0$$

but at the same time generates a(x). That means we need

$$-e\partial_0\chi = a;\;\Delta a = 0$$

which can be solved

$$\chi(x,t) = -\frac{1}{e}\int_{t_0}^t dt^\prime\,a(x,t^\prime)$$

For a an arbitrary harmonic function a(x,t) the gauge transformation is again a harmonic function which means it respects Coulomb gauge.

The last step is to calculate the new vector potential

$$A_i \to A_i^\prime = A_i - e\partial_i \chi = A_i + \int_{t_0}^t dt^\prime\,\partial_i a(x,t^\prime)$$

Fine, thanks for the idea.

Am I missing something important, or isn't the only difference between the addition of your field a(x) and a simple gauge transformation that you call it a "background field" instead of a "gauge transformation"?

My problem was the following. a(x) enters the interaction part of H b/c it modifies the coupling j°A°. That means it seems to have physical consequences. Of course one can add background fields not generated by charges, but they are difficult to study in the canonical formalism.

No I see that a(x) can be understood as a gauge transformation. That means once I accept a(x) in A°, I will have to transform the vector potential as well. That may change the form of H, but certainly not physics.

On the other hand a(x) van be understood as a background field. That means I change A° but NOT the vector potential. This change will certainly affect physics.

(I know how to avoid this problem: work in the A°=0 gauge, quantize QED, fix Coulomb gauge by a unitary transformation acting on A and E; E is constrained by div E in the Gauß law which obviously does NOT constrain the (dynamical) zero mode; doing it that way is slightly more involved but closer to Dirac's quantization procedure, which I believe is correct).

a(x) is indeed the background, not necessarily removable by gauge transformations. In other words, it's just propagating EM waves. With the additional $a(x)$ term $A_0$ is non-vanishing even when $\rho=0$, which means it's just the vacuum solution of $\Delta A_0=4\pi\rho=0$, i.e. free EM waves.

P.S. OK, as Tom points out, what I said was wrong.

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petergreat said:
a(x) is indeed the background, not necessarily removable by gauge transformations. In other words, it's just propagating EM waves. With the additional $a(x)$ term $A_0$ is non-vanishing even when $\rho=0$, which means it's just the vacuum solution of $\Delta A_0=4\pi\rho=0$, i.e. free EM waves.
As I have shown above in Coulomb gauge it is always possible to remove the harmonic function a(x) using a residual gauge transformation and therefore you can set this a(x) = 0.

Usually you treat el.-mag. radiation in the Lorentz gauge

$$\partial_\mu A^\mu = 0$$

with the wave equation

$$\Box A^\text{Lorentz}_0 = \rho$$

which seems as if A_0 is dynamical, but this a gauge-artefact (see below)! Lorentz gauge is unphysical and therefore problematic in the canonical formulation of QED.

In Coulomb gauge the field equations differ from Lorentz gauge. The 0-component reads

$$\Delta A^\text{Coulomb}_0 = \rho$$

which is not a wave equation; the wave is not represented by A_0 but by A_i. The A_0 equation is by no means a dynamical equation, it is a time-indep. constraint, and therefore neither A_0 nor a(x) are dynamical, propagating degrees of freedom (A_0 is a Lagrangian multiplier) due to the fact that there is no time-derivative for A_0 and therefore no canonical conjugate momentum in the Lagrangian.

A 'background field' is different from 'radiation'; radiation couples to charges in principle, whereas the time-depencency of a background field is arbitrary even in the presence of charges, as long as the constrained is satisified; that's what happens for a(x). Therefore a(x) is not radiation (but pure gauge).

Remark: Lorentz gauge and its treatment of A_0 is problematic, especially in the canonical formulation. You can see this by looking at the original Lagrangian: it does not contain a time-derivative for A_0, therefore the equations of motion cannot contain a time-derivative-squared. The fact that the d'Alembertian operator is present and is acting on A_0 with a time-derivative-squared has been introduced by the Lorentz gauge condition only; it's a gauge artefact, A_0 is still unphysical.

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Thanks Tom! Your explanation is very clear.