The Lagrangian of QED is

\begin{equation} \mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\bar{\psi}\big(i\not{D}-m\big)\psi \end{equation}

where $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ and $\not{D}=\gamma^\mu\big(\partial_\mu+iqA_\mu)$. Here $A_\mu$ is the EM gauge field and $q$ is the charge of the electron. This lagrangian has a $U(1)$ gauge symmetry

\begin{equation} \begin{cases} \psi\rightarrow\psi'=e^{iq\alpha(x)}\psi \\ \bar{\psi}\rightarrow\bar{\psi}'=e^{-iq\alpha(x)}\bar{\psi} \\ A_\mu\rightarrow A_\mu'=A_\mu-\partial_\mu\alpha(x) \end{cases} \end{equation}

where $\alpha(x)$ is a smooth arbitrary function of spacetime coordinate $x^\mu$. Now, we apply the definition of the Noether Current

\begin{equation} J^\mu=\frac{\delta \mathcal{L}}{\delta (\partial_\mu\phi_a)}\delta \phi_a \end{equation} (where I'm not adding the extra term due to the lagrangian changing in a total derivative since $\delta\mathcal{L}=0$). So we get

\begin{equation} J^\mu=-q\alpha(x)\bar{\psi}\gamma^\mu\psi+F^{\mu\nu}\partial_\nu\alpha(x) \end{equation}

When $\alpha(x)=constant$ we recover the usual electric charge. Now, I want to address two questions regarding the complete current $J^\mu$ for arbitrary $\alpha(x)$.

First Question: It is said in many textbooks that global symmetries give conservation laws that are satisfied on-shell while local symmetries give conservation laws satisfied off-shell. It is clear for the global part that this is true: The current is just (setting $\alpha=1$)

\begin{equation} J^\mu|_{\alpha=1}=-q\bar{\psi}\gamma^\mu \psi \end{equation}

Since the equation of motion for the gauge fields is

\begin{equation} \partial_\nu F^{\mu\nu} =-q\bar{\psi}\gamma^\mu \psi \end{equation}

we inmediately get that $\partial_\mu J^\mu=\partial_\mu\partial_\nu F^{\mu\nu}=0$. So the current get's conserved when the EoM are satisfied. Now, the question is: How do we check that either $J^\mu$ is identically zero off-shell or that $\partial_\mu J^\mu$ is zero off-shell?

Second Question: If we keep working on the full current we can use Leibniz rule to get

\begin{equation} J^\mu=-q\alpha(x)\bar{\psi}\gamma^\mu\psi+\partial_\nu\big[F^{\mu\nu}\alpha(x)\big]-\partial_\nu F^{\mu \nu}\alpha(x) \\ \ \ \ \ \ \ \ \ \ \ =\alpha(x)\Big[-q\bar{\psi}\gamma^\mu\psi-\partial_\nu F^{\mu \nu}\Big]+\partial_\nu\big[F^{\mu\nu}\alpha(x)\big] \end{equation}

Now, if we apply the EoM, the first term vanishes. We are left with a total derivative. If we find the total charge in a given volume $V$ we get

\begin{equation} Q=\int_V J^0 d^3x = \int_V \partial_\nu\big[F^{0 \nu}\alpha(x)\big] d^3x= \int_S \vec{E}\cdot \hat{n} \ \alpha(x) dS \end{equation}

Now, if the volume we choose is the whole universe and if the electric field and the gauge condition $\alpha(x)$ decay nicely to infinity (see my other question where I ask why does $\alpha(x)$ need to decay at infinity [1]) then the total charge in the universe is zero. However, that doesn't mean that this symmetry is not real or not useful. I could also choose any other volume and get a perfectly functional conservation law for a new quantity whose charge is the electric field at the boundary of the desired volume weighted by an arbitrary function $\alpha(x)$. So the question is: Why is this conserved quantity not physical?

[1] Why do we require that the gauge condition $\alpha(x)$ falls off at infinity?


1 Answer 1

  1. OP's current $J^{\mu}$ only satisfies the continuity equation on-shell. In contrast, the current from Noether's 2nd theorem is instead ${\cal J}^{\mu}=\partial_{\nu}F^{\nu\mu}$, which satisfies the continuity equation off-shell, cf. e.g. my Phys.SE answer here.

  2. OP is correct: A gauge-symmetry [of the form $\alpha(x)=\varepsilon f(x)$, where $\varepsilon$ is an infinitesimal constant (=$x$-independent) parameter] leads [via Noether's first theorem] to infinitely many on-shell continuity equations by choosing different functions $f(x)$. At least in the case of EM, there are infinitely many manifestly gauge-invariant (and hence physical) continuity equations.

  • $\begingroup$ Hello! Thanks for the answer. Regarding 1., could you expand a bit mor on this on this other Noether current? The link you provided doesn’t give much information. Why do we have two different currents? Are they both conserved? Are they related? Regarding 2., the conserved charge I found depends on the gauge parameter alpha so I don’t see how’s gauge invariant. $\endgroup$ Dec 3, 2019 at 14:58
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Dec 3, 2019 at 16:21

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