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##U(\epsilon)A^\mu(x)U^{-1}(\epsilon) = A^\mu(x') - \epsilon^{\mu\nu}A_\nu(x') + \frac{\partial \lambda(x',\epsilon)}{\partial x'_\mu}##, where ##\lambda(x',\epsilon)## is an operator gauge function.

This is all being done in the radiation gauge, i.e. ##A_0 = 0## and ##\partial_i A^i=0##, with ##i \in {1,2,3}##.

##\epsilon## is an infinitesimal parameter of a Lorentz transformation ##\Lambda##.

Under this transformation, ##A^\mu(x) \rightarrow A'^\mu(x')=U(\epsilon)A^\mu(x)U^{-1}(\epsilon)##.

The unitary operator ##U## which generates the infinitesimal Lorentz transformation

##x^{\mu} \rightarrow x'^{\mu} = x^{\mu} + \epsilon^{\mu}_{\nu}x^{\nu}## is

##U(\epsilon)=1 - \frac{i}{2}\epsilon_{\mu\nu}M^{\mu\nu}##

where ##M## are the generators of Lorentz transformations. (I guess really I should have ##M^{\mu\nu}=(M^{\rho\sigma})^{\mu\nu}##. M is a hermitian operator, so

##U^{-1}(\epsilon)=1 + \frac{i}{2}\epsilon_{\mu\nu}M^{\mu\nu}##

Now I tried writing out ##U(\epsilon)A^\mu(x)U^{-1}(\epsilon)## explicitly but it didn't really get me anywhere. The answer is supposed to have ##x'## as the argument of ##A^\mu## on the RHS but I only get ##x##. I'm not sure how to Lorentz transform the function and the argument at the same time.

Underneath the formula in B&J it says the gauge term is necessary because ##UA_0U^{-1}=0## since ##A_0=0##. I don't see why this warrants the need of a gauge term I'm guessing it's needed because otherwise there will be no conjugate momenta for the ##A_0##. Okay I get that, but still don't understand where the initial formula comes from.