Srednicki 58: EM current conservation & Gauge Symmetry

  1. Hi I am re-reading Srednicki's QFT.

    In chapter 58,
    he points out that the Noether current $$ j^\mu=e\bar{\Psi}\gamma^\mu\Psi$$ is only conserved when the fields are stationary, which is obvious from the derivation of the conservation law.
    Meanwhile he assumes that $$\partial _\mu j^\mu=0$$ always holds in deriving the free photon propagator in ch. 56-57

    However, he then suggests that "This issue can be resolved" by defining the U(1) gauge transformation.

    But I don't see how this solves the issue.
    I tried to use the gauge transformation in
    $$\partial _\mu j^\mu (x)=\delta \mathcal{L}(x)-(\delta S/\delta\psi_a(x)) \delta\psi_a(x)$$
    but I only got trivial result 0 = 0. (well the above equation is an identity so I don't expect otherwise.)

    Can anyone tell me what precisely what Srednicki means here?

    Thanks a lot
  2. jcsd
  3. You originally derived the conservation of current ∂μjμ= 0 from the equations of motion ∂μFμσ=jσ. The form that this current takes in terms of the Dirac fields ψ(x) was found by noting that the global U(1) transformations ψ(x)→e-ieΓψ(x), where Γ is a constant, leave the Lagrangian invariant.

    The point he is making, is that when you do a path integral over Aμ, you are assuming this invariance is always satisfied. This is false, however, since the whole idea of what a path integral does is vary over all the possible paths of Aμ---the equations of motion are only satisfied for that path that minimizes the action (the classical path). The way you make up for this is by imposing local U(1) invariance ψ(x)→e-ieΓ(x)ψ(x), in other words promoting U(1) to a gauge symmetry. Then the Lagrangian remains invariant, regardless of whether or not the fields are configured according to the classical path.

    I'm not sure I follow what you are doing here, but maybe I can try to resolve it if you provide some more detail. It looks like you are trying to confirm the invariance of the Lagrangian under the local U(1) transformations, but Srednicki does this explicitly from equation (91) to equation (99).
  4. Thanks for your reply.

    I understand what Srednicki meant. And my understanding is that gauging the U(1) makes the global EM U(1) current always conserved.

    What I wanted to show explicitly is $$\partial _\mu j^\mu=0$$ for the EM current (global U(1))

    So I will need to make use of $$\partial _\mu j^\mu (x)=\delta \mathcal{L}(x)-(\delta S/\delta\psi_a(x)) \delta\psi_a(x)$$
    If I set the field variation to be the infinitesimal global U(1) transformation, then it is what Srednicki said: "the equations of motion are only satisfied for that path that minimizes the action"

    If I set the field variation to be the infinitesimal local U(1) transformation, then the above equation (yes j_mu becomes the gauge current not the EM current of course) gives trivial result "0 = 0" even if I make use of $$\delta \mathcal{L}(x) = 0$$.

    To recapitulate, the statement is "imposing the U(1) gauge symmetry to the lagrangian--> the global U(1) current always conserved whether the fields obey the classical eom".

    But Srenicki only shows that the lagrangian is invariant.

    I don't understand explicitly how this leads to $$\partial _\mu j^\mu_{global/EM}=0$$
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?

Draft saved Draft deleted