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Lagrangian is invariant under the transformation

  1. Dec 7, 2014 #1
    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. jcsd
  3. Dec 7, 2014 #2

    king vitamin

    User Avatar
    Gold Member

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

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

    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 [itex]\cos\alpha + i\gamma^5 \sin\alpha[/itex] resembles a complex exponential. See if you can write it as one to combine it with the other unwanted factors.
     
  4. Dec 7, 2014 #3
    Doh, I always forget about the ##\gamma^0##. Now I see how it all works out. Thanks very much!
     
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