- #1

- 87

- 2

##\psi_L=\frac{1}{2}(1-\gamma^5)\psi##

##\psi^\dagger_L=\psi^\dagger_L\frac{1}{2}(1-\gamma^5)##

and

## \left\{ \psi(x) , \psi^\dagger(y)\right\} = \delta(x-y)##

So writing

## \left\{P_L\psi(x) , \psi^\dagger(y)P_L\right\} ##

##=P_L\psi(x) \psi^\dagger(y)P_L + \psi^\dagger(y)P_LP_L\psi(x)##

And I don't see how I can go any further.