# Dirac Lagrangian invariance under chiral transformation

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1. Jun 22, 2017

### ppedro

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

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

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

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

Can you help me finish this?

2. Jun 22, 2017

### Orodruin

Staff Emeritus
Hint: $\gamma_5$ anti-commutes with all gamma matrices, including $\gamma^0$.

3. Jun 22, 2017

### ppedro

I don't see it... Do you mean I should use that in (i) while also applying BCH's formula?

4. Jun 22, 2017

### Orodruin

Staff Emeritus
There is no point in using the BCH formula. Just use the anti-commutativity.

5. Jun 22, 2017

### ppedro

How can I anti-commute with something that's in the exponent?

6. Jun 22, 2017

### Orodruin

Staff Emeritus
Use the series expansion of the exponent.