Question on Peskin & Schroeder's QFT: Noether's Theorem; finding conserved currents

Question

So, I am trying to learn how to find conserved currents using what I see in P&S's "Intro to QFT"; sec.2.2 pg.17-18. Particularly, I am working through the example in the text for the complex Klein-Gordon field subject to the transformation:$$\phi \rightarrow e^{i\alpha }\phi. \tag{p.18}$$ It seems that the Noether currents can be found utilizing equation 2.12 for $j^\mu (x)$:$$j^\mu (x)={\partial \mathcal {L} \over \partial (\partial_\mu \phi )}\Delta \phi - \mathcal{J}^\mu.\tag{2.12}$$ So far as I can tell $\partial_\mu\mathcal {J}^\mu$ can be found from calculating:$\Delta \mathcal{L}$, since these two are set equal on page 17. Thus, for the given Lagrangian density we have:$$\partial_\mu\mathcal{J}^\mu=\Delta \mathcal {L}=\partial_\mu ({\partial \mathcal {L} \over \partial (\partial_\mu \phi )} \Delta \phi )+\partial_\mu({\partial \mathcal {L} \over \partial (\partial_\mu \phi^* )}\Delta\phi^*)$$ $$=i\alpha\partial_\mu[(\partial^\mu\phi^*)\phi-(\partial^\mu\phi)\phi^*].$$ However, if I insert this into the above expression for $j^\mu(x)$ I will get zero. What's going on?

Answer 1

1. In the notation of P&S ${\cal J}^{\mu}$ denotes the improvement term in the Noether current $j^{\mu}$. The improvement term ${\cal J}^{\mu}$ is used when a transformation is only a quasi-symmetry (and not a strict symmetry) of the Lagrangian density.

2. The Lagrangian density for the complex Klein-Gordon field has a strict $U(1)$ symmetry, so ${\cal J}^{\mu}\!=\!0$ is zero in this case.

Answer 2

What you have done to find the vector $J^\mu$ is almost exactly the same as how we derive the Noether's first theorem. That is why you arrived at a zero. Equating equation 2.12 to itself will not lead you to new physics.

The appropriate way to find the $\delta L$ is to check the behaviour of the Lagrangian when you "substitute" the transformation "explicitly" into it. In your case, the U(1) phase symmetry is an internal symmetry, and the Lagrangian is invariant ($\delta L=\partial_\mu (\phi^* e^{-i\alpha})\partial^\mu (\phi e^{i\alpha} )-V(\phi^* e^{-i\alpha})(\phi e^{i\alpha} )-L=0$). Then the Noether current $j^\mu=i(\partial^\mu \phi^{*})\phi-i\phi^{*}(\partial^\mu \phi)$. I do not have Peskin's book by hand at the moment, but I hope my guess of the exact form of the Lagrangian is not far from what he used.