How do I expand the Hermitian conjugate in the Lagrangian?

Question

I am trying to expand the following equation:

$$\mathcal{L} = \partial_\mu \phi^\dagger_1\partial^\mu \phi_1 - m^2_1 \phi_1^\dagger \phi_1 - \lambda (\phi^\dagger _1 \phi_1)^2\tag{1}$$

using the following:

$$\phi_1 \to \phi'_1 = e^{i\theta_1}\phi_1\tag{2}$$

but I am not sure how to simplify it.

My take on it is that the Hermitian conjugate transforms equation (2) into:

$$\phi_1^\dagger \to (\phi'_1)^{\dagger} = (e^{i\theta_1}\phi_1)^\dagger = \phi_1^\dagger e^{-i\theta} \tag{3} $$

and when trying to apply this to the first part of equation (1), $\partial_\mu \phi^\dagger_1\partial^\mu \phi_1 $, I think this leads to

$$\partial_\mu \phi^\dagger_1\partial^\mu \phi_1 = e^{i\theta} \partial_\mu \phi_1 \partial^\mu \phi_1^\dagger e^{-i\theta} \tag{4}$$

I think that this could be further simplified but I am not sure how, or even if what I have done is actually correct...any help would be appreciated.

Answer 1

It should be $\partial_\mu(\phi_1^\dagger e^{-i\theta})\partial^\mu(e^{i\theta}\phi_1)$, which by the product rule is a sum of four terms which for spacetime-constant $\theta$ simplifies to just one of them, $\partial_\mu\phi_1^\dagger\partial^\mu\phi_1$. For $\theta$ varying across spacetime, the result is more complicated, motivating gauge covariant derivatives.

Edit: as requested, I'll show how the local symmetry works. Define $D_\mu\phi_1:=\partial_\mu\phi_1+iqA_\mu\phi_1$. We don't just replace of $\phi_1$ with $\phi_1^\prime:=e^{i\theta}\phi_1$; we also replace $A_\mu$ with whatever value of $A_\mu^\prime$ obtains$$D_\mu^\prime\phi_1^\prime=e^{i\theta}D_\mu\phi_1\implies (D_\mu^\prime\phi_1^\prime)^\dagger D^{\mu\prime}\phi_1^\prime=(D_\mu\phi_1)^\dagger e^{-i\theta}e^{i\theta}D^\mu\phi_1=(D_\mu\phi_1)^\dagger D^\mu\phi_1.$$So$$\begin{align}0&=(\partial_\mu+iqA_\mu^\prime)(e^{i\theta}\phi_1)-e^{i\theta}(\partial_\mu+iqA_\mu)\phi_1\\&=\partial_\mu(e^{i\theta}\phi_1)+iq(A_\mu^\prime-A_\mu)e^{i\theta}\phi_1-e^{i\theta}\partial_\mu\phi_1\\&=iq(A_\mu^\prime-A_\mu+\tfrac1q\partial_\mu\theta)e^{i\theta}\phi_1,\\A_\mu^\prime&=A_\mu-\tfrac1q\partial_\mu\theta.\end{align}$$