# Calculating Noether Current for Electromagnetic Current interacting with a Dirac Fermion

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.

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

• 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.