# Lagrangian is invariant under the transformation

1. Dec 7, 2014

### Chopin

I should mention that I'm self-studying this material, not taking it as part of a course, but since this is still a homework-style problem I figured it'd be best to post here.

1. The problem statement, all variables and given/known data
In Peskin and Schroeder problem #11.2, they ask us to consider the Lagrangian:
$$\mathcal{L} = \frac{1}{2}(\partial_\mu \phi^i)^2 + \frac{1}{2}\mu^2(\phi^i)^2 - \frac{\lambda}{4}((\phi^i)^2)^2 + \bar{\psi}(i\not{\partial})\psi - g\bar{\psi}(\phi^1 + i\gamma^5\phi^2)\psi$$
where $i=1,2$

They then ask us to show that this Lagrangian is invariant under the transformation:

$$\phi^1 \rightarrow \cos \alpha \phi^1 - \sin \alpha \phi^2\\ \phi^2 \rightarrow \sin \alpha \phi^1 + \cos \alpha \phi^2\\ \psi \rightarrow e^{-i\alpha \gamma^5/2}\psi$$

3. The attempt at a solution
It's easy to show that $(\partial_\mu \phi^i)^2 \rightarrow (\partial_\mu \phi^i)^2$ and $(\phi^i)^2 \rightarrow (\phi^i)^2$, so that handles all of the $\phi$ terms. I'm confused about the two $\psi$ terms, though.

1. For the $\bar{\psi}(i\not{\partial})\psi$ term, I have to commute the $e^{-i\alpha\gamma^5/2}$ past the $\gamma^\mu$ hidden in the $\not{\partial}$, so that I can cancel it against the $e^{i\alpha\gamma^5/2}$. My assumption is that this creates a non-trivial answer, though. Since $\{\gamma^5, \gamma^\mu\} = 0$, if we imagine Taylor-expanding the exponential, the terms with even powers of $\gamma^5$ will be unchanged, but the terms with odd powers will pick up an extra minus sign. I'm not sure how it's possible to make everything cancel out. Am I missing something stupid here?

2. For the $g\bar{\psi}(\phi^1 + i\gamma^5\phi^2)\psi$, I plugged in the transformations and got the following:

$$g\bar{\psi}(\phi^1 + i\gamma^5\phi^2)\psi \rightarrow g\bar{\psi}e^{i\alpha \gamma^5/2}((\cos \alpha \phi^1 - \sin \alpha \phi^2) + i\gamma^5(\sin \alpha \phi^1 + \cos \alpha \phi^2))e^{-i\alpha \gamma^5/2}\psi\\ = g\bar{\psi}e^{i\alpha \gamma^5/2}(\cos \alpha + i\gamma^5 \sin \alpha)(\phi^1 + i\gamma^5\phi^2)e^{-i\alpha \gamma^5/2}\psi$$

That gets me close, since at least the form of the $\phi$ terms is right. I'm pretty sure I can now commute the $e^{-i\alpha\gamma^5/2}$ term on the right hand end all the way over to the left to cancel the $e^{i\alpha\gamma^5/2}$ term, since the only thing in the middle is more $\gamma^5$ matrices. But that still leaves me with the $(\cos \alpha + i\gamma^5 \sin \alpha)$ term, and I don't see any way to get rid of that. Can anybody tell me what I'm doing wrong?

2. Dec 7, 2014

### king vitamin

Remember that you actually have $\bar{\psi} = \psi^{\dagger} \gamma^0$, so the transformation is

$$\bar{\psi} \rightarrow \psi^{\dagger}e^{i\alpha \gamma^5/2}\gamma^0 = \bar{\psi}e^{-i\alpha \gamma^5/2}.$$

This should take care of your first problem (and it's essential for the second part too of course). For your second problem, notice that the term $\cos\alpha + i\gamma^5 \sin\alpha$ resembles a complex exponential. See if you can write it as one to combine it with the other unwanted factors.

3. Dec 7, 2014

### Chopin

Doh, I always forget about the $\gamma^0$. Now I see how it all works out. Thanks very much!