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Dirac Lagrangian invariance under chiral transformation

  1. Jun 22, 2017 #1
    Consider the Dirac Lagrangian,

    [itex] L =\overline{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi, [/itex]

    where [itex] \overline{\psi}=\psi^{\dagger}\gamma^{0} [/itex], and show that, for [itex] \alpha\in\mathbb{R} [/itex] and in the limit [itex] m\rightarrow0 [/itex], it is invariant under the chiral transformation

    [itex] \psi\rightarrow\psi'=e^{i\alpha\gamma_{5}}\psi [/itex]

    [itex] \psi^{\dagger}\rightarrow\left(\psi^{\dagger}\right)'=\psi^{\dagger}e^{-i\alpha\gamma_{5}} [/itex]

    Attempt at a solution


    [itex] \begin{array}{ll}
    L' & =\overline{\psi}'\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi'\\
    & =\left(\psi^{\dagger}\right)'\gamma^{0}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi'\\
    & =\psi^{\dagger}e^{-i\alpha\gamma_{5}}\gamma^{0}\left(i\gamma^{\mu}\partial_{\mu}-m\right)e^{i\alpha\gamma_{5}}\psi\\
    & =\underset{(i)}{\underbrace{i\psi^{\dagger}e^{-i\alpha\gamma_{5}}\gamma^{0}\gamma^{\mu}\partial_{\mu}e^{i\alpha\gamma_{5}}\psi}}-\underset{(ii)}{\underbrace{m\psi^{\dagger}e^{-i\alpha\gamma_{5}}\gamma^{0}e^{i\alpha\gamma_{5}}\psi}}\\
    & =
    \end{array} [/itex]

    For (ii) I tried using [itex] \exp\left(s\hat{X}\right)\hat{Y}\exp\left(-s\hat{X}\right)=\hat{Y}+s\left[\hat{X},\hat{Y}\right] [/itex] to get

    [itex] \begin{array}{ll}
    (ii) & =m\psi^{\dagger}e^{-i\alpha\gamma_{5}}\gamma^{0}e^{i\alpha\gamma_{5}}\psi\\
    & =m\psi^{\dagger}\left(\gamma^{0}-i\alpha\left[\gamma_{5},\gamma^{0}\right]\right)\psi\\
    & =
    \end{array} [/itex]

    Can you help me finish this?
     
  2. jcsd
  3. Jun 22, 2017 #2

    Orodruin

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    Hint: ##\gamma_5## anti-commutes with all gamma matrices, including ##\gamma^0##.
     
  4. Jun 22, 2017 #3
    I don't see it... Do you mean I should use that in (i) while also applying BCH's formula?
     
  5. Jun 22, 2017 #4

    Orodruin

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    There is no point in using the BCH formula. Just use the anti-commutativity.
     
  6. Jun 22, 2017 #5
    How can I anti-commute with something that's in the exponent?
     
  7. Jun 22, 2017 #6

    Orodruin

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    Use the series expansion of the exponent.
     
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