Lorentz transformation - Bjorken & Drell

[Ganesh Ujwal]

I'm trying to derive (14.25) in B&J QFT. This is

 

[latex]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}[/latex], where [latex]\lambda(x',\epsilon)[/latex] is an operator gauge function.

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

[latex]\epsilon[/latex] is an infinitesimal parameter of a Lorentz transformation [latex]\Lambda[/latex].

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

The unitary operator [latex]U[/latex] which generates the infinitesimal Lorentz transformation

[latex]x^{\mu} \rightarrow x'^{\mu} = x^{\mu} + \epsilon^{\mu}_{\nu}x^{\nu}[/latex] is

[latex]U(\epsilon)=1 - \frac{i}{2}\epsilon_{\mu\nu}M^{\mu\nu}[/latex]

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

[latex]U^{-1}(\epsilon)=1 + \frac{i}{2}\epsilon_{\mu\nu}M^{\mu\nu}[/latex]

Now I tried writing out [latex]U(\epsilon)A^\mu(x)U^{-1}(\epsilon)[/latex] explicitly but it didn't really get me anywhere. The answer is supposed to have [latex]x'[/latex] as the argument of [latex]A^\mu[/latex] on the RHS but I only get [latex]x[/latex]. 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 [latex]UA_0U^{-1}=0[/latex] since [latex]A_0=0[/latex]. 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 [latex]A_0[/latex].

I also want to find what [latex]\lambda[/latex] explicitly is.