The Gauss-Faraday law, in the covariant form, reads

$ \epsilon^{\alpha\beta\gamma\delta} F_{\gamma\delta,\alpha} = 0, $

while the vacuum field equation is

$ \partial_\mu F^{\mu\nu} = 0. $

When it comes to quantize the electromagnetic field $A^\mu$, only the field equation is considered (as far as I know). So my question is: does the Gauss-Faraday law play any role in QFT?


The Gauss-Faraday law can be viewed as the integrability condition for the existence of the electro-magnetic $4$-potential $A_{\mu}$ via Poincare Lemma.

Or conversely, assuming a globally defined electro-magnetic $4$-potential $A_{\mu}$, the Gauss-Faraday law is trivially satisfied. See also this related Phys.SE post.

So yes, Gauss-Faraday law plays a role already before quantization.

  • $\begingroup$ Correct me if I'm wrong, but mathematically this law is just $$\partial_\mu \star F^{\mu\nu} = 0$$ isn't it? Where $\star F^{\mu\nu}$ is the Hodge dual. Here's an explicit proof that this condition is always satisfied when the EM tensor exists, in case it helps the OP. $\endgroup$ – Philip Aug 1 '17 at 12:06
  • $\begingroup$ $\uparrow$ You are right. $\endgroup$ – Qmechanic Aug 1 '17 at 12:18

The Gauss-Faraday law in the QFT plays essential role, as well as $\partial_{\mu}F^{\mu\nu} = 0$. In fact, this is related to the fact that in the QFT quantum fields represent irreducible representations of the Poincare group, i.e., the particles. The photon is massless particle with helicities 1, -1, therefore it has only two degrees of freedom, and it can be shown that the corresponding field is $F_{\mu\nu}$ satisfyng above two equations.


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