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

• sebomba
In summary: So in summary, the solution to the given problem is to use the anti-commutation relation to simplify the expression and then use the properties of the projectors to get the desired form of the Lagrangian.
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: $$\Psi= \Psi_L +\Psi_R$$
The left- and right handed fields can be expressed through the projectors ##P_L## and ##P_R## so that:
$$\Psi_L=P_L \Psi \quad \Psi_R=P_R \Psi$$
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?

berkeman
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.

## 1. What is the QED Lagrangian in terms of left- and right-handed spinors?

The QED Lagrangian, also known as the quantum electrodynamics Lagrangian, is a mathematical expression that describes the dynamics of the electromagnetic field and the interactions between charged particles. It is written in terms of left- and right-handed spinors, which are mathematical objects that describe the spin and angular momentum of particles.

## 2. How are left- and right-handed spinors related in the QED Lagrangian?

In the QED Lagrangian, left- and right-handed spinors are related through the chiral projection operator, which separates the particles with different chirality (left- or right-handedness). This is necessary because the QED Lagrangian only describes the interactions between particles with the same chirality.

## 3. What is the role of left- and right-handed spinors in the QED Lagrangian?

The left- and right-handed spinors in the QED Lagrangian represent the different types of particles that interact with the electromagnetic field. Left-handed particles are affected by the weak nuclear force, while right-handed particles are not. The QED Lagrangian describes the interactions between these particles and the electromagnetic field.

## 4. How does the QED Lagrangian account for the mass of particles?

The QED Lagrangian includes a term called the mass term, which accounts for the mass of particles. This term is responsible for the interactions between particles and the Higgs field, which gives particles their mass. Without this term, particles would be massless and would not interact with the Higgs field.

## 5. What other fundamental forces does the QED Lagrangian describe?

The QED Lagrangian only describes the interactions between charged particles and the electromagnetic field. It does not account for the other fundamental forces, such as the strong and weak nuclear forces or gravity. However, it can be combined with other Lagrangians to create a more comprehensive theory that includes all of the fundamental forces.

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