# Simplifying gamma matrices

## Homework Statement

I'm trying to find $P_L \displaystyle{\not}p P_L$ for a left-handed particle.

(I think the answer is zero...)

## Homework Equations

$P_L = \frac{1}{2} (1-\gamma_5)$ (the left-handed projection operator)
$\displaystyle{\not} p = \gamma_\mu p^\mu$ (pμ is the 4-momentum)

μ, γ5 are the gamma matrices)

## The Attempt at a Solution

So, $P_L \displaystyle{\not} p P_L = \frac{1}{4} (1-\gamma_5) \gamma_\mu p^\mu (1-\gamma_5) = \frac{1}{4} [ \gamma_\mu p^\mu - \gamma_5 \gamma_\mu p^\mu - \gamma_\mu p^\mu \gamma_5 + \gamma_5 \gamma_\mu p^\mu \gamma_5 ] = \frac{1}{4} \gamma_\mu [ p^\mu + \gamma_5 p^\mu - p^\mu \gamma_5 - \gamma_5 p^\mu \gamma_5]$ (*)
(using the anticommutation relation {γ5μ} = 0).

Can this be simplified further? Since pμ is a 4x1 matrix, I'm guessing you can't just swap the order of $\gamma_5$ and $p^\mu$.

However, you can rewrite (*) as $\gamma_\mu P_R p^\mu P_L$, and then because it's a left-handed particle, P_R acting on the momentum gives zero?

Any help would be appreciated,

Thanks.

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