I Chirality projection operator

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?
 
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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.
 

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