The Noether current corresponding to the transformation $\phi \to e^{i\alpha} \phi$ for the Klein-Gordon Lagrangian density

$$\mathcal{L}~=~|\partial_{\mu}\phi|^2 -m^2 |\phi|^2$$

by finding $\delta S$, and setting it to zero. The general formula for a global transformation is

$$j^{\mu}=\frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}\Delta \phi-\mathcal{J}^{\mu},$$

where $ \partial_{\mu} \mathcal{J}^{\mu}$ is the change in the Lagrangian density due to the transformation. (See Peskin section 2.2).

How do I find the Noether's current corresponding to a local transformation $\phi \to e^{i\alpha(x)}\phi$?

For a local gauge transformation $\delta \phi = i \alpha(x)\phi$, and $\delta \phi^*=-i\alpha(x)\phi$.

These equations imply $\delta(\partial_{\mu}\phi)=i(\partial_{\mu} \alpha)\phi+i \alpha (\partial_{\mu}\phi)$ and $\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi^*-i \alpha (\partial_{\mu}\phi^*)$

Therefore, $\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*)-m^2 \delta(\phi* \phi)=\partial_{\mu}\alpha J^{\mu}$, where $J^{\mu}=i(\phi \partial^{\mu}\phi^*-\phi^* \partial^{\mu}\phi)$

Setting $\delta S=\int \delta \mathcal{L}=0$, one gets $\partial_{\mu}J^{\mu}=0$

  • 2
    Comment to the answer (v3): The current $d_{\mu}J^{\mu}\approx 0$ is only conserved on-shell (reflecting the fact that the corresponding global transformation $\delta$ is a symmetry via Noether's first theorem). The local transformation $\delta$ is not a quasi-symmetry, since $\delta {\cal L}$ is not a total space-time divergence off-shell. – Qmechanic Aug 17 '13 at 19:29

Comment to the question (v4): OP is talking about a local complex phase transformation for a complex massive scalar (KG) theory. But a generic local complex phase transformation is not a quasi-symmetry$^1$ of the KG action, and hence Noether's (2nd) theorem does not apply.

--

$^1$ A infinitesimal transformation $\delta$ is a quasi-symmetry if the Lagrangian density ${\cal L}$ is preserved $\delta {\cal L}= d_{\mu} f^{\mu}$ modulo a total space-time divergence off-shell, cf. this Phys.SE answer. If the total space-time divergence $d_{\mu} f^{\mu}$ is zero off-shell, we speak of a symmetry.

  • Though the Lagrangian is not invariant, it differs by a derivative of $\alpha$, and hence $\delta S$, can be set to 0. Please look at the answer I have posted, and tell me if anything is wrong. – user7757 Jul 4 '13 at 2:25
  • @ramanujan_dirac: No, $\delta {\cal L}$ is not a total space-time divergence off-shell. – Qmechanic Jul 4 '13 at 12:53

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.