# Do gauge fields not transform like functions of the coordinates under translations?

By "transform like a function of the coordinates," I mean that under an infinitesimal translation $$x^\mu \to x^\mu + \epsilon^\mu$$, to first order in $$\epsilon^\mu$$ the function $$f(t,\mathbf x)$$ becomes

$$f(t,\mathbf x) \to f(t,\mathbf x) + \epsilon^\nu (\partial_\nu f)(t,\mathbf x)$$

Suppose we start in the temporal gauge with $$A_0 = 0$$ and then perform a gauge transformation so that $$A_0 = \partial_0 \lambda$$ for some scalar field $$\lambda(t,\mathbf x)$$. If $$\lambda$$ transforms as a function of the coordinates under translations, then this becomes

$$\partial_0 \lambda \to \partial_0 (\lambda + \epsilon^\mu \partial_\mu \lambda) = \partial_0 \lambda + \partial_0(\epsilon^\mu \partial_\mu \lambda)$$

If $$A_0 \neq 0$$ before the gauge transformation and also transforms like an ordinary function, then it's not consistent

$$A_0 + \partial_0 \lambda \to (A_0 + \partial_0 \lambda) + \epsilon^\nu \partial_\nu (A_0 + \partial_0 \lambda) + (\partial_0 \epsilon^\nu) (\partial_\nu \lambda)$$

If $$A_\mu$$ instead transforms under translation as

$$A_\mu \to A_\mu + \partial_\mu(\epsilon^\nu A_\nu)$$

then consistency is restored:

$$(A_\mu+\partial_\mu\lambda) \to (A_\mu+\partial_\mu\lambda) + \partial_\mu(\epsilon^\nu (A_\nu+\partial_\nu\lambda))$$

If so, this would explain why the canonical momentum to position for the Dirac field includes the vector potential, whereas otherwise I can't see how to get the time derivative of the displacement to multiply $$A_k$$ in the transformed Lagrangian. Then when $$x^\mu \to x^\mu + \epsilon^\mu$$ we get

$$\mathcal L = \bar\psi \gamma^\mu (i \partial_\mu - e A_\mu) \psi$$

$$\mathcal L \to (\bar\psi + \epsilon^\nu \partial_\nu \bar\psi) \gamma^\mu (i \partial_\mu - e A_\mu - e \partial_\mu(\epsilon^\nu A_\nu)) (\psi + \epsilon^\nu \partial_\nu \psi)$$

$$\mathcal L \to \dot \epsilon^k \bar\psi \gamma^0 (i \partial_k - e A_k) \psi + \cdots$$

$$p_k = \frac{\mathrm d L'}{\mathrm d \dot\epsilon^k} = \int \mathrm d^3\mathbf x \, \psi^\dagger (i \partial_k - e A_k) \psi$$