Is the Dirac adjoint in the representation dual to Dirac spinor?

Question

As seen in this Wikipedia page, the Lorentz group is not compact and the Dirac spinor (spin $\frac{1}{2}$) representation is NOT unitary.

Therefore, the complex conjugate representation does NOT coincide with the dual representation in general, cf. this Wiki page.

Now, I am looking at the definition of the Dirac adjoint. Using the notations in this link, I think that $\psi^{\dagger}$ is the complex conjugate representation of the Dirac spinor $\psi$.

Then, the Dirac adjoint $\overline{\psi}:= \psi^\dagger \gamma^0$ transforms as follows: \begin{equation} \overline{\psi} \to \overline{\psi} \lambda^{-1} \end{equation} for any Lorentz transformation $\lambda$ in the spin $\frac{1}{2}$ representation.

This transformation law seems to be in the form of the dual representation to $\lambda$. That is,

The Dirac adjoint $\overline{\psi}:= \psi^\dagger \gamma^0$ is in the representation dual to the Dirac spinor $\psi$,

Is this correct? After all, $\overline{\psi} \psi$ is well-known to be a Lorentz scalar. So, I guess these all looks consistent. Still, I would like to check for sure.

Answer 1

In order for $\overline{\psi} \psi$ to be a Lorentz scalar, we should have $$\lambda^{\dagger}\gamma^0\lambda = \gamma^0$$ or equivalently (assuming $(\gamma^0)^2 = 1$) $$\lambda^{-1} = \gamma^0\lambda^{\dagger}\gamma^0$$ what else do you need to know?

The non-unitarity is caused by $\gamma^0$ in $\lambda^{\dagger}\gamma^0\lambda = \gamma^0$. Without $\gamma^0$, $\lambda$ as in $$\lambda^{\dagger}\lambda = 1$$ would be unitary.