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;


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.


1 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

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


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

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 $\bar{\psi}_{R/L} = \psi_{R/L}^\dagger$.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.