# I Chirality projection operator

#### Aleolomorfo

Hello everybody!
I have a doubt in using the chiral projection operators. In principle, it should be $P_L \psi = \psi_L$.
$$P_L = \frac{1-\gamma^5}{2} = \frac{1}{2} \begin{pmatrix} \mathbb{I} & -\mathbb{I} \\ -\mathbb{I} & \mathbb{I} \end{pmatrix}$$
If I consider $\psi = \begin{pmatrix} \phi \\ \chi \end{pmatrix}$, with $\phi$ and $\chi$ Weyl spinors.
$$P_L \psi = \frac{1}{2} \begin{pmatrix} \mathbb{I} & -\mathbb{I} \\ -\mathbb{I} & \mathbb{I} \end{pmatrix} \begin{pmatrix} \phi \\ \chi \end{pmatrix} = \frac{1}{2} \begin{pmatrix} \phi-\chi \\ \chi - \phi \end{pmatrix}$$
I get a spinor which still has the RH and the LH component. Shouldn't I get something with only the LH component and the RH's equal to zero?

Related Quantum Physics News on Phys.org

#### PeterDonis

Mentor
In principle, it should be $P_L \psi = \psi_L$.
Yes, but you have to be careful to use the same basis for the spinors and the matrices.

$$P_L = \frac{1-\gamma^5}{2} = \frac{1}{2} \begin{pmatrix} \mathbb{I} & -\mathbb{I} \\ -\mathbb{I} & \mathbb{I} \end{pmatrix}$$
This is written in the Dirac basis, not the Weyl basis, so the spinors in this basis do not have the $L$ part in the upper component and the $R$ part in the lower component. (It's a good exercise to figure out what pure left-handed and pure right-handed spinors do look like in this basis.)

In the Weyl basis, $\gamma^5$ is diagonal, so $P_L$ should be a matrix with only one nonzero component, in the "upper left" corner with the left-right convention you are using.

"Chirality projection operator"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving