8
$\begingroup$

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 $\phi \to e^{i\alpha(x)}\phi$?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user7757 Jul 4 '13 at 2:25
  • $\begingroup$ @ramanujan_dirac: No, $\delta {\cal L}$ is not a total space-time divergence off-shell. $\endgroup$ – Qmechanic Jul 4 '13 at 12:53
1
$\begingroup$

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 $\partial_{\mu}J^{\mu}=0$

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Qmechanic Aug 17 '13 at 19:29

Your Answer

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