Friday 21 February 2020

quantum field theory - Noether current for a local gauge transformation for the Klein-Gordon Lagrangian


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




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