Coupling of a massless vector to a conserved current

Question

In order to describe one-particle states of spin-1 in a Lagrangian description, we need to use a field $A_\mu$. This is a 4-vector up to gauge transformations, which means that under Lorentz transformations $\Lambda^\mu_\nu$ the field $A_\mu$ transforms as

$$ A_\mu \rightarrow \Lambda^\nu_{\,\,\mu}\, A_\nu + \partial_\mu\Omega(x) $$

where $\Omega(x)$ is a function of creation and annihilation operators. Equivalently, the polarizations shift analogously

$$ \epsilon^\mu(p,\sigma) \rightarrow \Lambda^\nu_{\,\,\mu}\,\epsilon_\nu(p,\sigma) + p^\mu(...) $$

where the detailed expression of $(...)$ is unimportant here.

The net result of this discussion is that, in order to have manifestly Lorentz invariant S-matrix elements (and also Lorentz invariant theory), we need to write the theory in a gauge invariant way.

The most economic way to do so is to couple the photon field $A_\mu$ to a conserved current

$$ \mathcal{L}_A = -\frac{1}{4}F_{\mu\nu}^2 + A_\mu J^\mu $$ and the invariance under gauge transformation is recovered on-shell (i.e. on the equation of motion)

$$ \delta \mathcal{L}_A = -\Omega(x)\partial_\mu J^\mu\Big|_\text{on-shell}=0 $$

The most famous example is the electronic current $J^\mu =e\,\bar{\psi}\gamma^\mu \psi$, because it is always quoted that $\bar{\psi}\gamma^\mu\partial_\mu \psi=0$.

Here is my confusion. When we gauge the $U(1)$ global symmetry $\psi\rightarrow e^{ie\alpha}\psi$, the equation of motions get modified by the coupling of $\psi$ with $A_\mu$

$$ (\gamma^\mu \partial_\mu + e A_\mu) \psi= 0 $$

and this seems to tell that when the fermion and gauge sectors communicate, it is any more not true that the current is conserved. Indeed,

$$ \partial_\mu J^\mu \Big|_\text{on-shell} = \partial_\mu(\bar{\psi})\gamma^\mu\psi + \bar{\psi}\gamma^\mu\partial_\mu\psi = -A_\mu \gamma^\mu \psi - \bar{\psi}\gamma^\mu A_\mu $$

Stupid question

Is this last quantity zero? In other words, is the global conserved current still conserved once we gauge the $U(1)$ symmetry?

Answer 1

Your equation of motion does not make sense. Try instead

$$ \gamma^{\mu} \left( \partial_{\mu} + i A_{\mu} \right) \psi = 0 $$

Then follows

$$ \partial_{\mu} J^{\mu} = \partial_{\mu} \bar{\psi} \gamma^{\mu} \psi + \bar{\psi} \gamma^{\mu} \partial_{\mu} \psi = + i A_{\mu} \bar{\psi} \gamma^{\mu} \psi - i\bar{\psi}\gamma^{\mu} A_{\mu} \psi =0 $$

Answer 2

The current $j^\mu\equiv\bar{\psi}\gamma^\mu\psi$ is conserved because the Lagrangian is invariant under the global U(1) symmetry $\psi\rightarrow e^{i\alpha}\psi$.

The QED Lagrangian is invariant under the local U(1) symmetry $\psi\rightarrow e^{i\alpha(x)}\psi$.

Since the global U(1) symmetry is a subset of the local U(1) symmetry--i.e. the QED Lagrangian is invariant under the global U(1) symmetry--the current $j^\mu$ must still be conserved for the QED Lagrangian by Noether's Theorem.