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?
$$\bar{\psi}_L\gamma^{\mu}\psi_L=\bar{\psi}_L\gamma^{\mu}\psi_L$$