6
$\begingroup$

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?
$\endgroup$

3 Answers 3

4
$\begingroup$
  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.


$^1$ 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.

$\endgroup$
3
  • 1
    $\begingroup$ This is actually incorrect. If one considers the 1-parameter family of transformations $A_\mu \to A_\mu + \tau \partial_\mu \alpha$, for $\tau \in \mathbb{R}$ (a global symmetry) and computes the conserved current from Noether's 1st theorem, one gets OP's expression. Noether's 2nd theorem can also be applied but one reaches a different equation which does not depend on $\alpha$. If $\alpha$ is large, the charge is non-zero. The fact that the underlying symmetry is a gauge transformation implies that the current is a divergence of a 2-form , see appendix B.3 arxiv.org/abs/2112.05289 $\endgroup$ Jul 1 at 16:29
  • 1
    $\begingroup$ I plan an update. $\endgroup$
    – Qmechanic
    Jul 1 at 17:58
  • 2
    $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jul 2 at 14:15
1
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ Actually, while small gauge transformations are not physical, large gauge transformations, given by $\alpha$'s which do not die off at infinity, actually are physical transformations. Using the symplectic form of electromagnetism, small gauge transformations are degenerate while large gauge transformations are not. The corresponding conservation law is the soft photon theorem. In the context of general relativity, the corresponding "large" diffeomorphisms called supertranslations similarly act physically on the Hilbert space of asymptotically flat gravity. arxiv.org/abs/1703.05448 $\endgroup$ Jul 1 at 16:05
  • 1
    $\begingroup$ @user1379857 Thanks for the link. $\endgroup$ Jul 1 at 16:09
  • 1
    $\begingroup$ See also the article linked in my answer for a discussion involving conserved currents and Noether's theorem arxiv.org/abs/2112.05289. Strominger's lectures do not take the path of Noether currents, instead reaching their conclusions a different way. $\endgroup$ Jul 1 at 16:23
  • 1
    $\begingroup$ @user1379857 Okay now I find why I was wrong. The charge is actually a "surface charge". Am I understanding it correctly? Is this like the charge of asymptotic $AdS_{3}$ of Brown and Henneaux? $\endgroup$ Jul 1 at 16:36
  • 1
    $\begingroup$ Indeed it a surface charge. Noether charges arising from gauge symmetries are always surface charge because the Noether currents are always divergences of 2-forms, $j^\mu = \partial_\nu K^{\mu \nu}$ for some $K^{\mu \nu}$ which is locally constructed out of the fields and gauge parameter functions. $\endgroup$ Jul 1 at 16:38
1
$\begingroup$

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.
$\endgroup$
5
  • 1
    $\begingroup$ The conserved charge depends on the choice of gauge. What's the physical meaning of such a conserved current? $\endgroup$ Jul 1 at 16:00
  • 1
    $\begingroup$ Amazingly, the corresponding charge does not depend on the choice of gauge. Notice that the charge current $j_\mu = \partial_\mu(\alpha F^{\mu \nu} )$ is gauge invariant for the simple reason that $F^{\mu \nu}$ is gauge invariant. The physical consequence of this conservation, when using a "large" gauge transformation $\alpha$ which satisfies an antipodal matching condition (and is $u$ and $v$ independent at $\mathcal{I}^{\pm}$) is that it constrains the amount of radiation emitted at each angle given the initial and final momenta of charged particles in the bulk. $\endgroup$ Jul 1 at 16:08
  • 1
    $\begingroup$ It can also be understood as specifying the exact amount of soft radiation released from a scattering event, which in this context is a classical version of the soft photon theorem. $\endgroup$ Jul 1 at 16:09
  • 1
    $\begingroup$ So even different choices of $\alpha$ do not lead to different charges? $\endgroup$ Jul 1 at 16:13
  • 1
    $\begingroup$ We need to be careful. $\alpha$ is an external parameter, it doesn't depend on $A_\mu$. $A_\mu$ is the field which specifies the state of the EM field. Gauge transformations, of the form $A_\mu \to A_\mu + \partial_\mu \lambda$, do not change $F_{\mu \nu}$. $\alpha$ on the other hand is just an independent free function which can be thought of as a parameter in $j_\mu$. Because $j_\mu[A] = j_\mu[A + \partial \lambda]$, the current, and hence the charge, is gauge invariant. $\endgroup$ Jul 1 at 16:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.