Understanding Simple Chirality Equations with Dirac Spinors

[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.

 

[strangerep]

Can anyone explain to me the math behind this simple step:
$$ P_L \psi = \psi P_R $$

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

[...] And why is $\bar{\psi }P_R \gamma^{\mu } \psi = \bar{\psi } \gamma^{\mu } P_L \psi$

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

 

[StephvsEinst]

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

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.

Hint: what is the anticommutation relation between # \gamma_5 # and # \gamma_{\mu } # ?

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.

 

[ChrisVer]

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

 

[ChrisVer]

I see now that [ γμ,γ5]=0 \gamma_{\mu }, \gamma_5 ] = 0 so it's true that γ5γμ=γμγ5 \gamma_5 \gamma_{\mu } = \gamma_{\mu } \gamma_5 .

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

 

[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.