Dirac spinor in the chiral basis

Question

In the chiral basis, the gamma matrices take the form $$ \gamma^0=\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}, \quad \gamma^j=\begin{bmatrix}0 & -\sigma^j \\ \sigma^j & 0\end{bmatrix} $$ and therefore one can calculate what the left and right projectors look like: $$ P_R=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}, \quad P_L=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}. $$ Given a Dirac spinor with components $\psi=(\psi_1,\psi_2,\psi_3,\psi_4)^T$, it is pretty clear that the Weyl spinors should become $$ \psi_R:=P_R\psi=\begin{pmatrix}\psi_1 \\ \psi_2 \\ 0 \\ 0\end{pmatrix}, \quad \psi_L:=P_L\psi=\begin{pmatrix}0 \\ 0 \\ \psi_3 \\ \psi_4\end{pmatrix} $$ and one can reconstruct the spinor by summing over both of them, as $\psi=\psi_R+\psi_L$. I've been told however that in this basis we can decompose the Dirac spinor in terms of the Weyl spinors as $$ \psi=\begin{bmatrix}\psi_R \\ \psi_L \end{bmatrix}. $$ This can't be possinle, if $\psi_R$ and $\psi_L$ are the objects with four components defined above. So it is probably a notational issue; who are these $\psi_R,\psi_R$ and what is their relation with $P_R\psi, P_L\psi$?

Answer 1

The projectors $P_R, P_L$ project $\psi \in \mathcal{H}\cong \mathbb{R}^4$ onto the right- and left-handed sectors of the representation of the Lorentz algebra, which are each a two-dimensional vector space, hence (locally) isomorphic to $\mathbb{R}^2$.

What "sum" is meant here is the direct sum $\oplus$: $$\begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \oplus \begin{pmatrix} \psi_3 \\ \psi_4 \end{pmatrix} = \begin{pmatrix} \psi_1 \\ \psi_2 \\ \psi_3 \\ \psi_4 \end{pmatrix}.$$ You correctly observed that the projector operators drop the two entries not in the projected space, so $P_L \mathcal{H} \cong \mathbb{R}^2$, likewise for $P_R \mathcal{H}$. So the statement "$\psi = \psi_1 + \psi_2$" really means $$\mathcal{H} \cong P_L \mathcal{H} \oplus P_R \mathcal{H}.$$ To answer your last question, then, $\psi_R$ and $\psi_L$ are the members of $\mathbb{R}^2$ which can be identified (via isomorphism) to the images of the projections of $\mathcal{H}$.