Adjoint of Weyl Spinor

Question

Given a (Dirac), spinor in the Weyl basis, $\psi = \begin{pmatrix} \psi_{L}\\ \psi_{R} \end{pmatrix} $ , where $\psi_{L}$ and $\psi_{R}$ are Weyl spinors we define the adjoint of the Dirac spinor as;

$\bar{\psi}=\psi^{\dagger}\gamma^{0}=(\psi_{R}^{\dagger},\psi_{L}^{\dagger})$

I understand this, but recently I've ran into expressions like $\bar{\psi}_{L}$, and $\bar{\psi}_{R}$. I can't seem to find a definition for the adjoint of a Weyl spinor so this is confusing me. Is it as simple as; $$\bar{\psi}_{L}=P_{L}\bar{\psi}=P_{L}\psi^{\dagger}\gamma^{0}=\psi_{R}^{\dagger}$$ and likewise for the right handed Weyl spinor? This is the only definition that seems to make sense to me but I'd like to check to make sure.

Edit: In trying to make sense of this I've gone and confused myself further. Another suitable defintion seems to be;

$\bar{\psi}_{L}=\bar{P_{L}\psi}=(P_{L}\psi)^{\dagger}\gamma^{0}=(\psi_{L}^{\dagger},0)\gamma^{0}=(0,\psi_{L}^{\dagger})$ Then taking $\bar{\psi}_{L}=\psi_{L}^{\dagger}$. Which is contrary to my previous idea.

I would really appreciate some clarification on this, thanks.

Answer 1

Let's start with something we know must be true. Namely the result that we would get from acting with projection operators alone.

$$\bar{\psi}\psi = \bar{\psi}_L\psi_R + \bar{\psi}_R \psi_L $$

Take this as defining convention. This can only be true if the definition is

$$ \psi := {\psi_L \choose \psi_R}$$

and

$$\bar{\psi}:=(\bar{\psi}_R, \bar{\psi}_L)$$.

so the barred components get "reversed" in chirality. Were this not the case, then we would get the wrong expression. As you mentioned, since the $\gamma^0$ matrix is identity we must have that $\bar{\psi}_{R/L} = \psi_{R/L}^\dagger$.

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You can remember it with this heuristic: For barred spinors, the left and right components are "opposite" of the unbarred.