Gauge invariance and covariant derivative

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SUMMARY

The discussion centers on the covariant derivative in scalar Quantum Electrodynamics (QED), specifically defined as ##D_{\mu}=\partial_{\mu}+ieA_{\mu}##. It is established that the covariant derivative ##D_{\mu}\phi## is invariant under the phase rotation ##\phi \rightarrow e^{i\Lambda}\phi## and the gauge transformation ##A_{\mu}\rightarrow A_{\mu}+\frac{1}{e}(\partial_{\mu}\Lambda)##. However, it is clarified that ##D_\mu \phi## transforms like the field ##\phi## itself, resulting in ##D_\mu \phi \to e^{i\Lambda} D_\mu \phi##, which ensures the invariance of the product ##D_\mu \phi D^\mu \phi^*##. The choice of gauge is emphasized as a critical aspect of problem-solving in QED.

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  • Basic grasp of phase rotations in quantum fields
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spaghetti3451
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Consider the covariant derivative ##D_{\mu}=\partial_{\mu}+ieA_{\mu}## of scalar QED.

I understand that ##D_{\mu}\phi## is invariant under the simultaneous phase rotation ##\phi \rightarrow e^{i\Lambda}\phi## of the field ##\phi## and the gauge transformation ##A_{\mu}\rightarrow A_{\mu}+\frac{1}{e}(\partial_{\mu}\Lambda)## of the vector potential ##A_{\mu}##.I was wondering if the phase rotation and the gauge transformation are related in any way? Does the gauge transformation necessarily lead to the phase rotation of the field?
 
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The correct transformation for the gauge field is ## A_\mu \to A_\mu - \frac 1 e (\partial_\mu \Lambda) ##.

## D_\mu \phi ## is not invariant under ## \phi \to e^{i\Lambda}\phi ## and ## A_\mu \to A_\mu - \frac 1 e (\partial_\mu \Lambda) ##. It transforms like ## \phi ## itself, i.e. ## D_\mu \phi \to e^{i\Lambda} D_\mu \phi ##. The point is that this will make ## D_\mu \phi D^\mu \phi^* ## invariant.

The gauge you use is a choice you make in a particular problem, like a coordinate system. You can choose any gauge you want, but you can't use one choice for some quantities and another choice for others. So all gauge dependent quantities should transform when you change gauge.
 

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