Dirac Lagrangian invariance under chiral transformation

ppedro
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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}<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}[/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}<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}[/itex]

Can you help me finish this?
 
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Hint: ##\gamma_5## anti-commutes with all gamma matrices, including ##\gamma^0##.
 
Orodruin said:
Hint: ##\gamma_5## anti-commutes with all gamma matrices, including ##\gamma^0##.

I don't see it... Do you mean I should use that in (i) while also applying BCH's formula?
 
ppedro said:
I don't see it... Do you mean I should use the in (i) while also aplying BCH's formula?
There is no point in using the BCH formula. Just use the anti-commutativity.
 
How can I anti-commute with something that's in the exponent?
 
ppedro said:
How can I anti-commute with something that's in the exponent?
Use the series expansion of the exponent.
 

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