- #1
ppedro
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Consider the Dirac Lagrangian,
L =\overline{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi,
where \overline{\psi}=\psi^{\dagger}\gamma^{0}, and show that, for \alpha\in\mathbb{R} and in the limit m\rightarrow0, it is invariant under the chiral transformation
\psi\rightarrow\psi'=e^{i\alpha\gamma_{5}}\psi
\psi^{\dagger}\rightarrow\left(\psi^{\dagger}\right)'=\psi^{\dagger}e^{-i\alpha\gamma_{5}}
Attempt at a solution
\begin{array}{ll}<br /> L' & =\overline{\psi}'\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi'\\<br /> & =\left(\psi^{\dagger}\right)'\gamma^{0}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi'\\<br /> & =\psi^{\dagger}e^{-i\alpha\gamma_{5}}\gamma^{0}\left(i\gamma^{\mu}\partial_{\mu}-m\right)e^{i\alpha\gamma_{5}}\psi\\<br /> & =\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}}\\<br /> & =<br /> \end{array}
For (ii) I tried using \exp\left(s\hat{X}\right)\hat{Y}\exp\left(-s\hat{X}\right)=\hat{Y}+s\left[\hat{X},\hat{Y}\right] to get
\begin{array}{ll}<br /> (ii) & =m\psi^{\dagger}e^{-i\alpha\gamma_{5}}\gamma^{0}e^{i\alpha\gamma_{5}}\psi\\<br /> & =m\psi^{\dagger}\left(\gamma^{0}-i\alpha\left[\gamma_{5},\gamma^{0}\right]\right)\psi\\<br /> & =<br /> \end{array}
Can you help me finish this?
L =\overline{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi,
where \overline{\psi}=\psi^{\dagger}\gamma^{0}, and show that, for \alpha\in\mathbb{R} and in the limit m\rightarrow0, it is invariant under the chiral transformation
\psi\rightarrow\psi'=e^{i\alpha\gamma_{5}}\psi
\psi^{\dagger}\rightarrow\left(\psi^{\dagger}\right)'=\psi^{\dagger}e^{-i\alpha\gamma_{5}}
Attempt at a solution
\begin{array}{ll}<br /> L' & =\overline{\psi}'\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi'\\<br /> & =\left(\psi^{\dagger}\right)'\gamma^{0}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi'\\<br /> & =\psi^{\dagger}e^{-i\alpha\gamma_{5}}\gamma^{0}\left(i\gamma^{\mu}\partial_{\mu}-m\right)e^{i\alpha\gamma_{5}}\psi\\<br /> & =\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}}\\<br /> & =<br /> \end{array}
For (ii) I tried using \exp\left(s\hat{X}\right)\hat{Y}\exp\left(-s\hat{X}\right)=\hat{Y}+s\left[\hat{X},\hat{Y}\right] to get
\begin{array}{ll}<br /> (ii) & =m\psi^{\dagger}e^{-i\alpha\gamma_{5}}\gamma^{0}e^{i\alpha\gamma_{5}}\psi\\<br /> & =m\psi^{\dagger}\left(\gamma^{0}-i\alpha\left[\gamma_{5},\gamma^{0}\right]\right)\psi\\<br /> & =<br /> \end{array}
Can you help me finish this?
