The Dirac adjoint for Dirac spinors is defined as,
$$
\bar{u} = u^{\dagger} \gamma^{0} \, .
$$
However I have come across this,
$$
\overline{\gamma^{\mu}} = \gamma^{\mu} \, , \tag{1}
$$
(where $\gamma^{\mu}$ are the $4\times4$ gamma matrices). Naively applying the same rules as for the Dirac spinor clearly does not get us anywhere,
$$
\overline{\gamma^{\mu}} = \gamma^{\mu \dagger} \gamma^{0} = \gamma^{0} \gamma^{\mu} \gamma^{0} \gamma^{0} = \gamma^{0} \gamma^{\mu} \neq \gamma^{\mu} \, .
$$
So it seems that the Dirac adjoint for a matrix is defined differently, so in trying to figure this out I make the following reasoning, let $A$ be a $4 \times 4$ matrix and $u$ a Dirac spinor so that $Au$ is again a Dirac spinor. Taking the Dirac conjugate (which is defined) gives,
$$
\overline{A u} = (A u)^{\dagger} \gamma^{0} = u^{\dagger} A^{\dagger} \gamma^{0} = u^{\dagger} \gamma^{0} \gamma^{0} A^{\dagger} \gamma^{0} = \bar{u} \;\underbrace{\gamma^{0} A^{\dagger} \gamma^{0}}_{ = \bar{A} ? } \, .
$$
So my guess is that $\bar{A} = \gamma^{0} A^{\dagger} \gamma^{0}$. If this is the case it is straightforward to show that $\overline{\gamma^{\mu}} = \gamma^{\mu} $.
My question is the following, is the above statement correct? Is it so that the Dirac adjoint is actually only defined for Dirac spinors but it can be sort of extended to $4 \times 4$ matrices as above (allowing one to write $\overline{A u} = \bar{u} \bar{A}$)?
Link where I found eq. (1) (page 93, eq. 3.249)
Link where I found eq. (1) and the claim $ \overline{X} = \gamma^{0} X \gamma^{0} $ which appears to be missing a "$^{\dagger}$"? (page 9, eq. 5.54)
Best Answer
In your first reference, page $58$, equation $(3.55)$, there is a personal definition by the author of what it calls "spinor adjoint of a matrix":
$\overline M \stackrel {def}{\equiv} \gamma_0 M^\dagger \gamma_0$
With this definition, as you noticed, you have obviously $\overline {\gamma^\mu}= \gamma^\mu$
The above definition of "spinor adjoint of a matrix" is compatible with the definition of the adjoint $(3.54)$:
$\overline \psi = \psi ^\dagger \gamma_0$
in the following way :
$\overline {M\psi} = (M\psi)^\dagger \gamma_0 = \psi^\dagger M^\dagger \gamma_0 = (\psi^\dagger \gamma_0 ) (\gamma_0 M^\dagger \gamma_0 ) = \overline {\psi}~ \overline {M}$