It is useful to understand the effect of a parity inversion on a Dirac spinor. Again work with the Dirac equation and its parity inverted form in which \bgroup\color{black}$x_j\rightarrow -x_j$\egroup and \bgroup\color{black}$x_4$\egroup remains unchanged (the same for the vector potential).

\gamma_\mu{\partial\over\partial x_\mu}\psi(x)+{mc\over\hbar}\...
...rtial\over\partial x_4}\right)S_P\psi+{mc\over\hbar}\psi&=&0 \\

Since \bgroup\color{black}$\gamma_4$\egroup commutes with itself but anticommutes with the \bgroup\color{black}$\gamma_i$\egroup, it works fine.

\begin{displaymath}\bgroup\color{black} S_P=\gamma_4 \egroup\end{displaymath}

(We could multiply it by a phase factor if we want, but there is no point to it.)

Therefore, under a parity inversion operation

\bgroup\color{black}$\displaystyle \psi'=S_P\psi=\gamma_4\psi$\egroup
Since \bgroup\color{black}$\gamma_4=\pmatrix{1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & -1 & 0 \cr 0 & 0 & 0 & -1 \cr}$\egroup, the third and fourth components of the spinor change sign while the first two don't. Since we could have chosen \bgroup\color{black}$-\gamma_4$\egroup, all we know is that components 3 and 4 have the opposite parity of components 1 and 2.

Jim Branson 2013-04-22