QED in Coulomb gauge

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  • #1
tom.stoer
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I presented a way to derive Coulomb force via the canonical mechanism.

One uses Coulomb gauge

[tex]\partial_i A^i = 0[/tex]

derives

[tex]\Delta A_0 = -4\pi\,\rho[/tex]

which can be inverted formally

[tex]A_0 = -4\pi\,\Delta^{-1}\,\rho[/tex]

and calculates the interaction term in the Hamiltonian density

[tex]\mathcal{H}_\text{int} = - 4 \pi \, \rho \,\Delta^{-1}\,\rho + \ldots[/tex]

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).

[tex]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[/tex]

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

[tex]\Delta \, a = 0[/tex]

[tex]A_0 = -4\pi\,\Delta^{-1}\,\rho + a[/tex]

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

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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Answers and Replies

  • #3
tom.stoer
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Gauge transformation, I think.
OK, let's try that.

A(x) respects Coulomb gauge

[tex]\partial_i A^i = 0[/tex]

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

[tex]A_\mu \to A_\mu^\prime = A_\mu - e\partial_\mu \chi[/tex]

which respect Coulomb gauge

[tex]\partial_i (A^i)^\prime = 0\;\to\;\partial_i (\partial^i \chi) = \Delta\chi = 0[/tex]

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

[tex]-e\partial_0\chi = a;\;\Delta a = 0[/tex]

which can be solved

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

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

[tex]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)[/tex]

Fine, thanks for the idea.
 
  • #4
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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"?
 
  • #5
tom.stoer
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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).
 
  • #6
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a(x) is indeed the background, not necessarily removable by gauge transformations. In other words, it's just propagating EM waves. With the additional [itex]a(x)[/itex] term [itex]A_0[/itex] is non-vanishing even when [itex]\rho=0[/itex], which means it's just the vacuum solution of [itex]\Delta A_0=4\pi\rho=0[/itex], i.e. free EM waves.

P.S. OK, as Tom points out, what I said was wrong.
 
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  • #7
tom.stoer
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a(x) is indeed the background, not necessarily removable by gauge transformations. In other words, it's just propagating EM waves. With the additional [itex]a(x)[/itex] term [itex]A_0[/itex] is non-vanishing even when [itex]\rho=0[/itex], which means it's just the vacuum solution of [itex]\Delta A_0=4\pi\rho=0[/itex], 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

[tex]\partial_\mu A^\mu = 0[/tex]

with the wave equation

[tex]\Box A^\text{Lorentz}_0 = \rho[/tex]

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

[tex]\Delta A^\text{Coulomb}_0 = \rho[/tex]

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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  • #8
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Thanks Tom! Your explanation is very clear.
 

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