Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.11.

The Dirac equation in condensed form is

$\displaystyle i\gamma^{\mu}\partial_{\mu}\left|\psi\right\rangle =m\left|\psi\right\rangle \ \ \ \ \ (1)$

where the gamma matrices have been defined earlier. The Hermitian conjugate of the gamma matrix ${\gamma^{\mu}}$ is given by the Hermiticity condition

$\displaystyle \gamma^{\mu\dagger}=\gamma^{0}\gamma^{\mu}\gamma^{0} \ \ \ \ \ (2)$

To get the adjoint form of the Dirac equation, we use the adjoint solutions

$\displaystyle \left\langle \bar{\psi}^{\left(n\right)}\right|\equiv\left\langle \psi^{\left(n\right)}\right|\gamma^{0} \ \ \ \ \ (3)$

The Hermitian conjugate of 1 is

$\displaystyle -i\partial_{\mu}\left\langle \psi\right|\gamma^{\mu\dagger}=m\left\langle \psi\right| \ \ \ \ \ (4)$

Remember that when we take the Hermitian conjugate of a matrix equation we must take the Hermitian conjugate of each matrix and also reverse the order of matrix multiplication, which is why the ${\gamma^{\mu\dagger}}$ term appears at the end on the LHS. The ${\partial_{\mu}}$ is a differential operator, not a matrix, so it retains its position in front of the ${\left\langle \psi\right|}$.

Using 2, we can write this as

$\displaystyle -i\partial_{\mu}\left\langle \psi\right|\gamma^{0}\gamma^{\mu}\gamma^{0}=m\left\langle \psi\right| \ \ \ \ \ (5)$

Then, by post-multiplying by ${\gamma^{0}}$ and using ${\left(\gamma^{0}\right)^{2}=I}$, the identity matrix, we get, using the definition 3

 $\displaystyle -i\partial_{\mu}\left\langle \psi\right|\gamma^{0}\gamma^{\mu}\left(\gamma^{0}\right)^{2}$ $\displaystyle =$ $\displaystyle m\left\langle \psi\right|\gamma^{0}\ \ \ \ \ (6)$ $\displaystyle -i\partial_{\mu}\left\langle \bar{\psi}\right|\gamma^{\mu}$ $\displaystyle =$ $\displaystyle m\left\langle \bar{\psi}\right|\ \ \ \ \ (7)$ $\displaystyle i\partial_{\mu}\left\langle \bar{\psi}\right|\gamma^{\mu}+m\left\langle \bar{\psi}\right|$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (8)$

Note that although the terms ${\left\langle \bar{\psi}\right|}$ in this adjoint equation are adjoint solutions, the gamma matrices ${\gamma^{\mu}}$ are the original (that is, not the Hermitian conjugate) gamma matrices.