I'm trying to confirm that the conserved current of the Lagrangian $$ {L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} -j^{\mu}A_{\mu}+\bar\psi(i\gamma^{\mu}\partial_{\mu}-m)\psi $$ associated with the transformation $$\psi(x) \rightarrow e^{-ieQa}\psi(x)$$ is $$j_{\mu} = eQ\bar\psi\gamma^{\mu}\psi.$$

Some working out showing the method for calculating this would be great.

I also wanted to know if $Q$ in this case is an operator (i.e. charge operator). If so it presumably has to be a scalar as for $\mu =0$, $\psi\gamma^{0}\psi$ would also be a scalar and an operator being applied to a scalar seems counter intuitive. However in QFT I was under the impression that after the first quantisation all physical quantities were represented by operators. Any resources on this topic would also also be appreciated.

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    $\begingroup$ this is covered in almost every QFT book in the obligatory "Noether's Theorem" section $\endgroup$ Feb 14, 2019 at 7:22

1 Answer 1


Some remarks:

  • Noether's theorem is a classical result, so you first apply Noether, and then quantize. So for the purposes of Noether's theorem, charge is a scalar (c-number).

  • Noether's theorem in it's ordinary form is used only for the "matter action", without the gauge field present. In that case, $j^\mu=\mathcal P^\mu_A\delta\psi^A$, where $A$ is a field index, and $\mathcal P^\mu_A=\partial \mathcal L/\partial\partial_\mu\psi^A$ is the lagrangian momentum. The matter action is $$ \mathcal L=\bar\psi(i\gamma^\mu\partial_\mu-m)\psi, $$ the variation is $\psi_\epsilon=e^{-i\epsilon Qa}\psi$, so $\delta\psi=-iQa\psi$, and $\bar\psi_\epsilon=e^{i\epsilon Qa}\bar\psi$, so $\delta\bar\psi=iQa\bar\psi$, the momenta are $\mathcal P^\mu=\bar\psi i\gamma^\mu$ and $0$, when varied with respect to the adjoint, so the current is $$j^\mu=Qa\bar\psi\gamma^\mu\psi,$$ which is what you have in your question.

  • If the coupled gauge-field/matter action is present, then a different, Noether-like procedure is used to deduce the conserved current. Sometimes this is referred to as Noether's second theorem. Basically, if $S_m[\psi,A]$ is the matter action extended by the gauge field, then we define $$ \mathcal J^\mu=\frac{\delta S_m}{\delta A^\mu} $$ as the current. Since the extended matter action is gauge-invariant, performing an infinitesimal gauge transformation leaves it invariant, so we have $$ \delta S_m=\int d^4x\ \left( \frac{\delta S_m}{\delta \psi}\delta\psi+\frac{\delta S_m}{\delta A_\mu}\delta A_\mu\right)\approx\int d^4x\ \frac{\delta S_m}{\delta A_\mu}\delta A_\mu=\int d^4x\ \mathcal J^\mu\delta A_\mu=0. $$ However under an infinitesimal gauge transformation, the gauge field $A$ transforms as $iQA^\prime_\mu=iQA_\mu+e^{-i\epsilon Qa(x)}\partial_\mu e^{i\epsilon Q a(x)}=iQA_\mu +\partial_\mu(i\epsilon Qa)=iQA_\mu+i\epsilon Q\partial_\mu a,$ and so $$ \delta A_\mu=\partial_\mu a, $$ hence $$ \delta S_m=\int d^4x\ \mathcal J^\mu\partial_\mu a=-\int d^4x\ \partial_\mu J^\mu a=0, $$ and since the gauge function $a$ is arbitrary, we obtain $\partial_\mu\mathcal J^\mu=0$.

  • Now all that remains to do is to calculate $\delta S_m/\delta A_\mu$. The gauge-extended matter action is $$ S_m[\psi,A]=\int d^4x\ \bar\psi(i\gamma^\mu D_\mu-m)\psi, $$ where $D_\mu=\partial_\mu+iQA_\mu$, from which one gets $$ S_m[\psi,A]=-\int d^4x\ Q\bar\psi\gamma^\mu A_\mu\psi+S_m[\psi,0], $$ and so $$ \frac{\delta S_m}{\delta A_\mu}=-Q\bar\psi\gamma^\mu\psi=\mathcal J^\mu. $$

  • As it is visible, the current $\mathcal J^\mu$ differs from $j^\mu$ by a sign, and the presence of $a$, however the sign difference can be explained by the fact that I wasn't careful enough with the conventions, and the factor of $a$ can be explained by the fact that currents can be rescaled, and I could have baked the Lie algebra coordinate $a$ into the variational parameter $\epsilon$ from the start.


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