# How do I expand the Hermitian conjugate in the Lagrangian?

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.

• Is the $\theta$ a constant or a function that depends on space and time? Also, is the $\theta$ a scalar? A c-number? A real number? Or is it an operator/matrix?
– hft
Commented Nov 15, 2021 at 18:18

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}

• Do you possess any link that explains how to get from that product to the $\partial_\mu\phi_1^\dagger\partial_\mu\phi_1$ answer? I haven't done this in many years and seem to have forgotten most of the rules. I am looking for a good website or book that explains these sort of calculations. Commented Dec 2, 2021 at 16:20
• @user7077252 See e.g. Sec. 4.4.2 here.
– J.G.
Commented Dec 2, 2021 at 16:34
• Once expanded the terms end up looking like: $\phi^\dagger e^{i\theta} (\partial _\mu e^{-i \theta})(\partial^\mu \phi) + \phi^\dagger \phi (\partial_\mu e^{-i\theta})(\partial^\mu e^{i\theta}) + (\partial_\mu \phi^\dagger)(\partial^\mu \phi) + e^{-i\theta}\phi (\partial_\mu \phi^\dagger)(\partial^\mu e^{i\theta})$ So I do obtain the term that you have mentioned above, but there doesn't seem to be away to cancel the other factors. In the source you provided above, in equation (4.129) an equation is used to solve this, but I didn't quite understand where that equality comes from. Commented Dec 3, 2021 at 10:54
• @user7077252 I'll edit in more detail later when I have free time.
– J.G.
Commented Dec 3, 2021 at 11:02
• Thank you very much, I have been trying this for a while, and I know the solution is probably right before my eyes, but it isn't making sense. Commented Dec 3, 2021 at 11:06