Lorentz Transformation in Bjorken & Drell QFT

1. Nov 15, 2014

Maybe_Memorie

1. The problem statement, all variables and given/known data

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

$U(\epsilon)A^\mu(x)U^{-1}(\epsilon) = A^\mu(x') - \epsilon^{\mu\nu}A_\nu(x') + \frac{\partial \lambda(x',\epsilon)}{\partial x'_\mu}$, 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 $\Lambda$.

2. Relevant equations

3. The attempt at a solution

$\epsilon$ 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.

Edit: Oh wait, 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.

Last edited: Nov 15, 2014
2. Nov 20, 2014