QED Lagrangian in terms of left- and right-handed spinors

[sebomba]

Homework Statement

I'm stuck at my particle physics exercise about 4-component chiral fields.

The following problem is given: "Derive the expression for the QED Lagrangian in terms of the four component right-handed and left-handed Dirac fields $\Psi_R(x)$ and $\Psi_L(x)$, respectively."

What I understand is that I need to get from the QED Lagrangian:

$$\mathscr{L}_{QED}=-\frac{1}{4} F_{\rho \sigma} F^{\rho \sigma}+\bar{\Psi} i \gamma^{\mu} \partial \Psi - m \bar{\Psi} \Psi - e A_\mu \bar{\Psi} \gamma^{\mu} \Psi$$

to here:

$$\mathscr{L}_{QED}=-\frac{1}{4} F_{\rho \sigma} F^{\rho \sigma}+\bar{\Psi}_L i \gamma^{\mu} \partial \Psi_L + \bar{\Psi}_R i \gamma^{\mu} \partial \Psi_R - m \bar{\Psi}_L \Psi_R - m \bar{\Psi}_R \Psi_L - e A_\mu [ \bar{\Psi}_L \gamma^{\mu} \Psi_L + \bar{\Psi}_R \gamma^{\mu} \Psi_R ]$$

Homework Equations

The operators $\Psi$ can be written in terms of the left- and right handed fields: \begin{equation}\Psi= \Psi_L +\Psi_R\end{equation}
The left- and right handed fields can be expressed through the projectors $P_L$ and $P_R$ so that:
\begin{equation}\Psi_L=P_L \Psi \quad \Psi_R=P_R \Psi\end{equation}
The projectors are:
$$P_L=\frac{1}{2}(1-\gamma^5) \quad P_R=\frac{1}{2}(1+\gamma^5)$$
With the properties:
$$P_{L,R}^2=P_{L,R} \quad P_L P_R=0 \quad P_L + P_R = 1$$
All properties of the $\gamma$-matrices are given too.

The Attempt at a Solution

My first step was to take the second term of the lagrangian and rewrite it with help of equation (1) and (2):
$$\bar{\Psi} (P_L + P_R) i \gamma^{\mu} \partial (P_L +P_R) \Psi$$
Then I tried it with resolving the statement between the $\Psi$'s and looking where it leads, but I ended up with long calculation and a nonsense statement in the end.
Then I realized that the parenthesis are scalars, so i could write:
$$\bar{\Psi} (P_L + P_R)(P_L +P_R) i \gamma^{\mu} \partial \Psi$$
With the properties of the projectors I then get:
$$\bar{\Psi} (P_L + P_R)^2 i \gamma^{\mu} \partial \Psi$$
$$\bar{\Psi} (P_L^2 + P_R^2) i \gamma^{\mu} \partial \Psi$$
$$\bar{\Psi} P_L^2 i \gamma^{\mu} \partial \Psi +\bar{\Psi} P_R^2i \gamma^{\mu} \partial \Psi$$
As the projectors do not commute with the $\gamma$-matrices (I checked that myself), I can't get the projectors to the other $\Psi$'s on the right side, which would lead to the wanted end result.

I think that I am missing something important or very trivial. Maybe I need another approach for the problem. Could someone help me there?
Thanks in advance.

 

[sebomba]

I just solved the puzzle by myself.

The projectors and the $\gamma$-matrices may not commute directly, but you can consider something that follows from the anti-commutation relation $$\{\gamma^5,\gamma^\mu\}=0$$. If we multiply $\gamma^\mu$ and $P_L=\frac{1}{2}(1-\gamma^5)$ we get of course:
$$\gamma^\mu P_L=\frac{1}{2}(\gamma^\mu-\gamma^\mu \gamma^5) $$
But from the anticommutator we know that $\gamma^\mu \gamma^5=-\gamma^5 \gamma^\mu$. We can plug this in and pull the $\gamma^\mu$ out of the parenthesis. And so we get the following:
$$\gamma^\mu P_L=\frac{1}{2}(\gamma^\mu-\gamma^\mu \gamma^5) = \frac{1}{2}(\gamma^\mu+\gamma^5 \gamma^\mu) =\frac{1}{2}(1+\gamma^5)\gamma^\mu=P_R \gamma^\mu$$
$$\rightarrow \gamma^\mu P_L=P_R \gamma^\mu$$
Of course this works the other way around too: $P_L \gamma^\mu=\gamma^\mu P_R$
Let's take a attempt simmilar to my second attempt from my post above and use it on $\bar{\Psi} \gamma^\mu\Psi$:
\begin{align*} \bar{\Psi} \gamma^\mu\Psi &= \Psi^{\dagger} \gamma^0 \gamma^\mu\Psi \\
&=\Psi^{\dagger}(P_L+P_R) \gamma^0 \gamma^\mu (P_L+P_R)\Psi \\
&=\Psi^{\dagger}(P_L^2+P_R^2)\gamma^0 \gamma^\mu \Psi \\
&=\Psi^{\dagger}P_L P_L \gamma^0 \gamma^\mu \Psi +\Psi^{\dagger}P_R P_R\gamma^0 \gamma^\mu \Psi \\
&=\Psi^{\dagger}P_L \gamma^0 P_R \gamma^\mu \Psi +\Psi^{\dagger}P_R \gamma^0 P_L \gamma^\mu \Psi \\
&=\Psi^{\dagger}P_L \gamma^0 \gamma^\mu P_L \Psi +\Psi^{\dagger}P_R \gamma^0 \gamma^\mu P_R\Psi \\
&=\Psi^{\dagger}_L \gamma^0 \gamma^\mu \Psi_L +\Psi^{\dagger}_R \gamma^0 \gamma^\mu \Psi_R \\
&=\bar{\Psi}_L \gamma^\mu \Psi_L +\bar{\Psi}_R \gamma^\mu \Psi_R
\end{align*}
This works for all the terms in the Lagrangian analogously.