- #1
Chopin
- 368
- 13
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.
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:
$$\phi^1 \rightarrow \cos \alpha \phi^1 - \sin \alpha \phi^2\\
\phi^2 \rightarrow \sin \alpha \phi^1 + \cos \alpha \phi^2\\
\psi \rightarrow e^{-i\alpha \gamma^5/2}\psi$$
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?
Homework Statement
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:
$$\phi^1 \rightarrow \cos \alpha \phi^1 - \sin \alpha \phi^2\\
\phi^2 \rightarrow \sin \alpha \phi^1 + \cos \alpha \phi^2\\
\psi \rightarrow e^{-i\alpha \gamma^5/2}\psi$$
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?