Local gauge transformations and Noether Current

Question

I am trying to derive the conserved Noether current, corresponding to a local gauge transformation in the theory of charged matter, coupled to the electromagnetic field: $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+A_{\mu}j^{\mu},\tag{1}$$ where $j^{\mu}(x)$ is the matter current and $A_{\mu}(x)$ is the gauge field. I know that there are several relevant posts, but to the best of my knowledge, none of them answers my question(s).

So, I will try to follow the standard (according to my understanding) procedure in deriving Noether currents: first, I will claim that the Lagrangian can vary up to the divergence of some vector quantity and then I will derive the variation of the Lagrangian wrt the fields. Finally, I will derive the conserved vector quantity as the difference between the two, as the divergence of the latter difference vanishes.

1. I start with the transformation $\delta A_{\mu}(x)=-\partial_{\mu}\alpha(x)$.
2. The variation in the Lagrangian is $\delta\mathcal{L}=\partial_{\mu}\mathcal{J}^{\mu}$, where $\mathcal{J}^{\mu}$ can be specified from the fact that when $\alpha(x)$ drops its spacetime dependence (i.e. becomes constant), $\delta\mathcal{L}=0$. That can be achieved if $\mathcal{J}^{\mu}(x)=-\alpha(x)j^{\mu}(x)$!
3. Varying wrt the gauge field (assuming that the EoMs hold) yields $$\delta\mathcal{L}=\partial_{\nu} \bigg(\frac{\delta\mathcal{L}}{\delta (\partial_{\nu}A_{\mu})}\delta A_{\mu}\bigg)= -\partial_{\nu}[F^{\nu\mu}(x)\partial_{\mu}\alpha(x)]$$
4. Equating the two expressions for the variation yields $$-\partial_{\nu}[F^{\nu\mu}(x)\partial_{\mu}\alpha]=-\partial_{\nu}[\alpha(x)j^{\nu}(x)] \Rightarrow -\partial_{\nu}[F^{\nu\mu}(x)\partial_{\mu}\alpha]=\partial_{\nu}[\alpha(x)\partial_{\mu}F^{\mu\nu}(x)] \Rightarrow \partial_{\nu}\partial_{\mu}[\alpha(x)F^{\mu\nu}(x)]=0$$ From the latter we conclude that the conserved Noether current, associated to local gauge transformations is $J_{\alpha}^{\mu}=\partial_{\nu}[\alpha(x)F^{\mu\nu}]$.

So, my questions are the following:

• Is this the correct Noether current?
• Is this referred to the literature as Noether's first theorem or the second one? (Some references about first and second Noether's theorem would be nice-preferably for physics background students!). For example when do we use the former and when do we use the latter?
• Do we assume that $j^{\mu}$ contains just the QED interaction? What happens to our derivation if we add the Dirac kinetic term in the Lagrangian? (We know that the kinetic term of the Dirac Lagrangian contains a derivative and we know that the latter derivative, when acted upon the gauge transformed matter field produces one additional term proportional to the derivative of the gauge parameter)
• Is there a reason why I should include a minus sign in $\mathcal{J}^{\mu}=-\alpha(x)j^{\mu}(x)$? (I have read this from Weinberg's book on quantum theory of fields, Vol.1 p.342). Moreover, could someone provide futher elaboration on why this is the only allowed form of the Lagrangian variation wrt spacetime points? (I mean, I can see that this choice works, but why is it the only one that works is not clear)
• What happens on the boundary? We know that $\int d^4x \partial_{\mu}J_{\alpha}^{\mu}(x)=0$ equals to the flux through the boundary of the spacetime. Is this telling us anything about the behavior of fields/gauge parameter on that boundary?

Answer 1

1. A global/$n$-parameter quasisymmetry is the realm of Noether's 1st theorem.

   Example: Global gauge symmetry in E&M with matter sectors implies electric charge conservation, cf. e.g. this Phys.SE post.

   Technically, to view OP's gauge transformation within$^1$ the framework of Noether's 1st theorem, one writes it (up to sign conventions) as$^2$ $$A^{\prime}_{\mu}(x)-A_{\mu}(x)~=~\epsilon d_{\mu}\Lambda(x,A,\partial A),\tag{A}$$ where $\epsilon$ is a free 1-parameter. Note that the function $\Lambda$ in this application is fixed but arbitrary. The 1st Noether current is$^3$ $$ J^{\mu}~=~-F^{\mu\nu}d_{\nu}\Lambda - j^{\mu}_{\rm m}\Lambda~\approx~d_{\nu}(F^{\nu\mu}\Lambda). \tag{B}$$ The corresponding on-shell continuity equation $$ d_{\mu}J^{\mu}~\approx~d_{\mu}d_{\nu}(F^{\nu\mu}\Lambda)~=~0\tag{C} $$ is trivial. See also e.g. this & this related Phys.SE posts.

2. A local/gauge/$x$-dependent symmetry is the realm of Noether's 2nd theorem. There are no non-trivial codimension-1 (=volume) Noether charges. However there are codimension-2 (=surface) charges, cf. e.g. this Phys.SE.

   In OP's E&M example the 2nd Noether identity reads $$ 0~=~d_{\mu}\frac{\delta S}{\delta A_{\mu}}~=~d_{\mu}\left(j^{\mu}_{\rm m} +d_{\nu} F^{\nu\mu}\right). \tag{D} $$ The 2nd Noether current is $$ {\cal J}^{\mu}~=~J^{\mu}+\frac{\delta S}{\delta A_{\mu}}\Lambda ~=~d_{\nu}(F^{\nu\mu}\Lambda),\tag{E}$$ which satisfies a trivial off-shell continuity equation.

References:

1. N. Miller, arXiv:2112.05289; Appendix B.3.

2. A. Strominger, Lectures on the Infrared Structure of Gravity and Gauge Theory, arXiv:1703.05448.

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$^1$ Ref. 1 calls this Noether's 1.5th theorem.

$^2$ Since OP treats the matter current $j^{\mu}_{\rm m}$ as a background, it must for consistency satisfy the continuum equation $$d_{\mu}j^{\mu}_{\rm m}~=~0.$$ Else OP's action (1) will not have a gauge-quasisymmetry, for starters.

$^3$ We use $(-,+,+,+)$ sign convention.

Answer 2

Noether's theorem is about global symmetries. Gauge "symmetries" are not physical symmetries. They are local redundancies. In quantum field theory, a gauge transformation is a do-nothing transformation in the Hilbert space (after gauge fixing) because physical states should be gauge invariant. In other words, the gauge group action in the Hilbert space should be a trivial representation of the gauge group.

On the other hand, physical symmetries transform physical states into other physical states. In other words, physical symmetries should be (projective)-unitary representations in the Hilbert space.

Also, charges (aka Hermitian operators corresponding to classical Noether charges associated with global symmetries) should be gauge invariant. But these charges change accordingly under the adjoint action of global symmetries.

Of course nothing can stop one from plugging the variation of local gauge transformation into the Noether current formula. The result is indeed conserved, but physically meaningless. This is because the conserved charge is not gauge invariant. It explicitly depends on the choice of gauges. As a result, one obtains an infinite family of redundant charges related by gauge transformations, and so they serve no practical purposes.

Please check my answer here for more details.

Answer 3

Funnily enough this has only really been understood in recent years in the connection to large gauge transformations and soft theorems. See https://arxiv.org/abs/2112.05289 for a recent in depth discussion. To answer your questions in order,

• That is the correct Noether current
• This should be thought of as Noether's first theorem. The fact that the current reduces to a total divergence $j^\mu = \partial_\nu(\alpha F^{\mu \nu})$ is a basic feature of all Noether currents derived from gauge symmetries. In the linked article this is referred to as "Noether's 1.5th theorem," which is proven in appendix B.
• Yes, you can couple this to the Dirac lagrangian. The part of the current from the Dirac kinetic term will just be the usual Dirac charge current weighted by $\alpha$.
• The sign is just a matter of convention and different sources will disagree.
• The behavior at the boundary is very interesting. The reason why not much was done with the charges arising from this gauge symmetry in EM until recently is that you need to push the surfaces on which you are evaluating them to null infinity in order to get anything interesting. If you think about your charges as living on $t = const$ surfaces (which is legitimate) you will not be able to derive anything interesting. However, if you think about a closed surface which covers spatial infinity, null infinity, and timelike infinity, and consider the total flux passing through this surface, it turns out that the contribution from spatial infinity is 0 as long as $\alpha$ is "large" (i.e., it doesn't die off at infinity) and satisfies a certain antipodal matching condition. The portion of the surface from null infinity will give information on the radiation emitted by charges in the spacetime. Taken to its logical conclusion, the conservation law from your current will become the soft photon theorem.