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My thought on this are somewhat scattered so I apologise in advance.

Maxwell's equations are gauge invariant. The physical Electric and Magnetic fields don't depend on whether we use $A_\mu$ or $A_\mu+\partial_\mu\Lambda$. We have the free EM Lagrangian $\mathcal{L}=\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ from which we can derive Maxwell's equations. It thus makes sense that the action stemming from this Lagrangian also be gauge invariant.

My issue arises when we couple to matter.

Let's take the Lagrangian for EM fields coupled to matter;

$$\mathcal{L}=\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+j^\mu A_\mu +\bar{\Psi}(i\gamma^\mu\partial_\mu-m)\Psi$$

Under $A_\mu\rightarrow A_\mu+\partial_\mu\Lambda$ the interaction term transforms as

$$\delta(j^\mu A_\mu)=j^\mu \partial_\mu\Lambda=\partial_\mu(\Lambda j^\mu)-\partial_\mu j^\mu\Lambda$$

Ignoring the total derivative term the invariance of the action then requires $\partial_\mu j^\mu=0$.

This is generally the point where textbooks say that we can find a $j^\mu$ made up of the matter fields using Noether's theorem; for spinor fields under global $U(1)$ transformations we would have $j^\mu=\bar{\Psi}\gamma^\mu\Psi$.

However, this conserved current is only obtained when the matter fields are on-shell, but we (presumably) require gauge invariance of the action even when the fields are off-shell. (Please correct me if this is incorrect). If the fields are off-shell we can cancel out the extra term we obtain by requiring that the matter fields transform in an appropriate way under local $U(1)$ transformation. We can then update our definition of gauge invariance to a local $U(1)$ transformation of the matter fields and the usual transformation of the EM fields.

My question; Why should we want to do this anyway? Even if we have the extra term $j^\mu\partial_\mu\Lambda$ Maxwell's equations are unaffected which was the reason we wanted gauge invarance in the first place. From this perspective, the extra term is not an issue.

Now I must admit my classical EM knowledge is a bit rusty so bear with me on this next point. Obviously the spinor equations of motion will not be invariant under the transformation of $A_\mu$ but is there any particular reason why the should be? Just because the 4-potential is not an observable in the pure EM sector is it necessarily like that in the spinor sector?

A thought has also occurred to me when writing this post; if we allow the term $j^\mu\partial_\mu\Lambda$ to appear in the Lagrangian then $\partial_\mu\Lambda$ itself becomes a dynamical field. I can imagine this could produce issues with regard to renormalizability (guessing) but a more immediate consequence I can see of this is the impossibility of interactions; the field equation for $\partial_\mu\Lambda$ is $j^\mu=0$, so we have no interaction. We could though claim that the field is $\Lambda$ instead of $\partial_\mu\Lambda$, in which case the field equation is $\partial_\mu j^\mu=0$.

I admit this post is a bit all over the place. My questions essentially boil down to;

  • Since the current is only conserved on-shell, why do we use Neother's theorem to create the current which we couple to the EM potential?

  • Even if we don't banish the extra term, how screwed are we? Does the extra term produce issues with regards to renormalizability or the resulting spinor equations have properties which disagree with experiments? We could just go from global to local transformations of the spinors and everything works out nicely, but why would we want to do this, apart from local transformations seeming like a more general alternative to global ones?

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However, this conserved current is only obtained when the matter fields are on-shell, but we (presumably) require gauge invariance of the action even when the fields are off-shell. (Please correct me if this is incorrect).

We only require gauge invariance on-shell. In an Abelian theory, something special happens and the equations $L_\mu = \partial^\nu F_{\nu\mu} - j_\mu = 0$ is gauge invariant, off-shell, i.e. $\delta L_\mu = 0$. However, in non-Abelian gauge theories, this is not true and there you have $\delta L_\mu = - i [ L_\mu , \Lambda ]$ which is only invariant on-shell.

Why should we want to do this anyway? Even if we have the extra term $j^\mu\partial_\mu\Lambda$ Maxwell's equations are unaffected which was the reason we wanted gauge invarance in the first place.

Why are the Maxwell equations unchanged by this term? For instance, this is not true in scalar QED, where the "conserved" current depends on the scalar field $\phi$ as well as the gauge field $A_\mu$. Also, the presence of this term will screw with the matter equations of motion.

A thought has also occurred to me when writing this post; if we allow the term $j^\mu\partial_\mu\Lambda$ to appear in the Lagrangian then $\partial_\mu\Lambda$ itself becomes a dynamical field.

Yes, you can do this. This is closely related to the Stueckelberg action.

Another related question that might help you is - Why we need gauge invariance in the first place? For this read this.

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  • $\begingroup$ Is there any particular reason why we only require gauge invariance on-shell? It, naivily, seems reasonable that it always be a symmetry of the action. Also, when you say $\partial^\nu F_{\nu\mu}-j_\mu=0$ is gauge invariant off shell - you mean that the matter fields are off-shell, yes? (since that equation is the on-shell condition for the photon field.) Why do we demand that the matter equations of motion be invariant under $A_\mu\rightarrow A_\mu+\partial_\mu\Lambda$? $\endgroup$ – Okazaki Dec 27 '15 at 18:54
  • $\begingroup$ Is there a typo in your first paragraph? Do you mean "$\delta L_\mu=-i[L_\mu,\Lambda]$ which is only invariant on-shell?", as opposed to off-shell? $\endgroup$ – Okazaki Dec 27 '15 at 18:55
  • $\begingroup$ I am being strict about the definition of a quantum theory. An action is simply a tool that is used (extremely efficiently, no doubt) to concisely describe information about the quantum theory. In the strictest sense, a quantum theory is completely described in terms of correlation functions of operators. All correlation functions satisfy Schwinger-Dyson equations - which are the quantum analog of EOM. We only require invariance of correlation functions (and there are other things, but those are details) which are ..... $\endgroup$ – Prahar Dec 27 '15 at 19:00
  • $\begingroup$ ...... due to the Schwinger-Dyson equations, on-shell objects. Thus, we only really need on-shell gauge invariance. $\endgroup$ – Prahar Dec 27 '15 at 19:00
  • $\begingroup$ @ryanp16 - Well, maybe I phrased it confusingly, but I meant that define $L_\mu = \partial^\nu F_{\nu\mu} - j_\mu$. The EOM is $L_\mu = 0$ but in general $L_\mu \neq 0$. The point is that even if $L_\mu = V_\mu$ for some arbitrary constant, field independent $V_\mu$ then this fact is preserved upon gauge transformations. However, note that the same is not true for non-abelian gauge theories, since only the statement $L_\mu = 0$ is preserved under gauge transformations, not the more general $L_\mu = V_\mu$. $\endgroup$ – Prahar Dec 27 '15 at 19:02

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