# Simple chirality question

1. Jul 30, 2015

### StephvsEinst

Hi!
Can anyone explain to me the math behind this simple step:
$$P_L \psi = \psi P_R$$ where $$P_L = \frac{1}{2} ( 1 + \gamma_5 )$$ and $$P_R = \frac{1}{2} ( 1 - \gamma_5 )$$.

And why is $$\bar{\psi }P_R \gamma^{\mu } \psi = \bar{\psi } \gamma^{\mu } P_L \psi ,$$
where $$\gamma_5$$ and $$\gamma_\mu$$ are Dirac matrices.

Can anyone help??

Edit: The psi's $$\psi$$ represent Dirac spinors.

Last edited: Jul 30, 2015
2. Jul 30, 2015

### strangerep

Where did that come from? It doesn't look quite right.

Hint: what is the anticommutation relation between $\gamma_5$ and $\gamma_\mu$ ?

3. Jul 31, 2015

### StephvsEinst

It's true that $\bar{\psi }_L \gamma^{ \mu } \psi_L = \bar{\psi }P_R \gamma^{\mu } P_L \psi$ but I can't prove mathematically this step.

I see now that [ $\gamma_{\mu }, \gamma_5 ] = 0$ so it's true that $\gamma_5 \gamma_{\mu } = \gamma_{\mu } \gamma_5$ .

I am still not understing the first step, though.

4. Jul 31, 2015

### ChrisVer

what does the bar notation stand for? If you apply its meaning on the expression below you will have it.
$\bar{\psi_L} = \bar{(P_L \psi)} = ...$

5. Jul 31, 2015

### ChrisVer

That is wrong. The anticommutation is zero... {A,B}= AB+BA

6. Jul 31, 2015

### Hepth

First one should have a bar over the second Psi. You get it by taking the complex conjugate of the left hand side, then inserting 1=GAMMA0.GAMMA0, where needed, then commuting them through the P.