Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.10.

The four solutions of the Dirac equation in relativistic quantum mechanics are

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (1)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (2)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (3)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (4)$

We’ve seen that these solutions are mutually orthogonal by taking the inner product of each solution with the complex conjugate transpose of another solution. You might think that we could derive a sort of conjugate transpose version of the Dirac equation by using the ‘bra’ versions of the solutions, but in fact it seems that it is more usual to define an adjoint of each solution by taking the conjugate transpose and then post-multiplying it by the matrix ${\gamma^{0}}$. That is, we define the adjoint solutions ${\left\langle \bar{\psi}^{\left(n\right)}\right|}$ by

$\displaystyle \left\langle \bar{\psi}^{\left(n\right)}\right|\equiv\left\langle \psi^{\left(n\right)}\right|\gamma^{0} \ \ \ \ \ (5)$

with

$\displaystyle \gamma^{0}=\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right] \ \ \ \ \ (6)$

We can write out the four adjoints explicitly by taking the conjugate transpose and doing the matrix multiplication. We get

 $\displaystyle \left\langle \bar{\psi}^{\left(1\right)}\right|$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{cccc} 1 & 0 & \frac{p^{3}}{E+m} & \frac{p^{1}-ip^{2}}{E+m}\end{array}\right]e^{ipx}\gamma^{0}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{cccc} 1 & 0 & -\frac{p^{3}}{E+m} & -\frac{p^{1}-ip^{2}}{E+m}\end{array}\right]e^{ipx}\ \ \ \ \ (8)$ $\displaystyle \left\langle \bar{\psi}^{\left(2\right)}\right|$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{cccc} 0 & 1 & -\frac{p^{1}+ip^{2}}{E+m} & \frac{p^{3}}{E+m}\end{array}\right]e^{ipx}\ \ \ \ \ (9)$ $\displaystyle \left\langle \bar{\psi}^{\left(3\right)}\right|$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{cccc} \frac{p^{3}}{E+m} & \frac{p^{1}-ip^{2}}{E+m} & -1 & 0\end{array}\right]e^{-ipx}\ \ \ \ \ (10)$ $\displaystyle \left\langle \bar{\psi}^{\left(4\right)}\right|$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{cccc} \frac{p^{1}+ip^{2}}{E+m} & -\frac{p^{3}}{E+m} & 0 & -1\end{array}\right]e^{-ipx} \ \ \ \ \ (11)$

Note that post-multiplying by ${\gamma^{0}}$ just changes the sign of the last two elements in ${\left\langle \psi^{\left(n\right)}\right|}$.