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I have a question on the quantization of the gauge theory with minimal coupling term. What I understand is that if one is given an action $$ S=-\int d^4 x \frac{1}{4}F^2 \tag1 $$ Since this action has vanishing canonical momentum $\Pi_0^a \equiv \frac{\delta \mathcal{L}}{\delta \partial_0 A_0^a}=0$, one can use Faddeev-Popov method to find physically equivalent action $$ \int d^4 x -\frac{1}{4}F^2 -\partial_\mu \overline{c}\partial^\mu c + \frac{1}{2}(\partial_\mu A^\mu)^2 \tag2 $$ Then you can proceed with usual quantization because this action has non-vanishing canonical momentum. My question is: If instead we are given an action of the form $$ S=\int d^4 x -\frac{1}{4}F^2 +|D\phi|^2 -V(|\phi|^2)\tag3 $$ where $\phi$ is a scalar field and $D_\mu\phi = \partial_\mu \phi + i A^a_\mu \tau^a \phi$ where $\tau^a$ are generators of gauge group. Then do we need Faddeev-Popov method to rewrite the action $(1)$ as action $(2)$? Because action $(3)$ has non vanishing canonical momentum $\Pi_0^a \equiv \frac{\delta \mathcal{L}}{\delta \partial_0 A_0^a}$ coming from minimal coupling term anyway. $$ \int |D\phi|^2 = \int |\partial \phi|^2 + i (\phi^\dagger A\partial \phi-\partial\phi^\dagger A \phi)+\phi^\dagger A^2 \phi = \int |\partial \phi|^2 + i (-2\partial\phi^\dagger A \phi - \phi^\dagger \partial_\mu A^\mu \phi)+\phi^\dagger A^2 \phi $$ So canonical momentum is $\Pi^{a}_0 = \phi^\dagger \tau^a\phi $?

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  • $\begingroup$ Note that the Fadeev-Popov lagrangian is not gauge invariant. $\endgroup$ – my2cts Oct 21 '19 at 17:57
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Adding minimally coupled scalar matter $\phi$ does not remove the gauge symmetry. In particular, the Legendre transformation is still singular. The Faddeev-Popov method (or one of its equivalent formulations) should still be used.

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  • $\begingroup$ Thank you Qmechanic. I didn't check the Legendre transformation. I will do that. thx $\endgroup$ – user239970 Oct 21 '19 at 17:44

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