Solving the Mystery of ##\bar{\psi}_L \psi_L = \bar{\psi} \psi_L##

[Monaliza Smile]

Hi,

It's known that $\bar{\psi}_L \psi_L = \bar{\psi} \psi_L$ I tried to work this out but i do not reach that
Here what I do : since $\bar{\psi} = \psi^\dagger \gamma^0$, and $\gamma_5 \gamma_0 = - \gamma_0 \gamma_5$

then

$\bar{\psi}_L \psi_L = \frac{1}{4} (1-\gamma_5 ) \psi^\dagger \gamma^0 (1-\gamma_5 ) \psi = \frac{1}{4} \psi^\dagger \gamma^0 (1+\gamma_5 ) (1-\gamma_5 ) \to 0$

I get this equals zero ! since (1+\gamma_5 ) (1-\gamma_5 ) = 0

so what's wrong I made ?

Best ..

 

[Monaliza Smile]

I knew the answer $\bar{\psi}_L \gamma^\mu \psi_L$ which equals $\bar{\psi} \gamma^\mu \psi_L$

 

[saybrook1]

Not sure I understand what you're trying to do. You want to prove this?
$$\bar{\psi}_L\gamma^{\mu}\psi_L=\bar{\psi}_L\gamma^{\mu}\psi_L$$