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## Homework Statement

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

(I think the answer is zero...)

## Homework Equations

[itex]P_L = \frac{1}{2} (1-\gamma_5)[/itex] (the left-handed projection operator)

[itex]\displaystyle{\not} p = \gamma_\mu p^\mu[/itex] (p

^{μ}is the 4-momentum)

(γ

_{μ}, γ

_{5}are the gamma matrices)

## The Attempt at a Solution

So, [itex]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] [/itex] (*)

(using the anticommutation relation {γ

_{5},γ

_{μ}} = 0).

Can this be simplified further? Since p

^{μ}is a 4x1 matrix, I'm guessing you can't just swap the order of [itex]\gamma_5[/itex] and [itex]p^\mu[/itex].

However, you can rewrite (*) as [itex]\gamma_\mu P_R p^\mu P_L[/itex], 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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