# Quantisation of gauge theory with minimal coupling

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 $$\Pi^{a}_0 = \phi^\dagger \tau^a\phi$$?

• Note that the Fadeev-Popov lagrangian is not gauge invariant. – my2cts Oct 21 '19 at 17:57

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.