# The commutations relations for left/right handed fermions

• A
decerto
I have a problem where I have to know the commutation relations for left handed fermions. I know
##\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.

Gold Member
it looks fine up to here...

decerto
it looks fine up to here...

I need to write the left handed commutator in terms of the unprojected commutator I don't see a nice way to do this? How can I commute the P_L operators past the psi's?

Gold Member
I want to say its the same, and will just be ##P_L \delta(x-y)##, but I'm not sure.

Maybe what you can do is write out the spinor indices explicitly. Normally you have :
$$\left\{\psi^a , \psi^{\dagger b} \right\} = \psi^a \psi^{\dagger b}+\psi^{\dagger b} \psi^a = \delta(x-y) \delta^{ab}$$
now we want to understand
$$\left\{\psi_L^a , \psi_L^{\dagger b} \right\} = P_L^{ai} \psi^i \psi^{\dagger j}P_L^{jb} + \psi^{\dagger j} P_L^{jb}P_L^{ai} \psi^i$$

we can combine and use the full commutator:

$$\left\{\psi_L^a , \psi_L^{\dagger b} \right\} = P_L^{ai}P_L^{jb} (\psi^i \psi^{\dagger j} + \psi^{\dagger j} \psi^i)\\ \left\{\psi_L^a , \psi_L^{\dagger b} \right\} = P_L^{ai}P_L^{jb} \delta(x-y) \delta^{ij}$$
you can change the indices with the delta function and use ##P_L^{ai}P_L^{ib} = P_L^{ab}## to combine the projections and get
$$\left\{\psi_L^a , \psi_L^{\dagger b} \right\} = P_L^{ab} \delta(x-y)$$
If ##a=b## then you get(in D then 4 dimensions)
$$P_L^{aa} = \frac{D}{2} = 2$$
This is because by definition the trace of gamma_5 is zero, and the trace of the identity matrix is D(4).

I think this is right....

decerto
Thanks, that makes sense, the whole dropping spinor indexes for simplicity makes things less clear sometimes.