I presented a way to derive Coulomb force via the canonical mechanism.(adsbygoogle = window.adsbygoogle || []).push({});

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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# QED in Coulomb gauge

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