The Conserved Probability Current

We now return to the nagging problem of the probability density and current which prompted Dirac to find an equation that is first order in the time derivative. We derived the equation showing conservation of probability for 1D Schrödinger theory by using the Schrödinger equation and its complex conjugate to get an equation of the form

\begin{displaymath}\bgroup\color{black} {\partial P(x,t)\over\partial t}+{\partial j(x,t)\over\partial x}=0. \egroup\end{displaymath}

We also extended it to three dimensions in the same way.

Our problem to find a similar probability and flux for Dirac theory is similar but a little more difficult. Start with the Dirac equation.

\begin{displaymath}\bgroup\color{black} \left(\gamma_\mu{\partial\over\partial x_\mu}+{mc\over\hbar}\right)\psi=0 \egroup\end{displaymath}

Since the wave function is a 4 component spinor, we will use the Hermitian conjugate of the Dirac equation instead of the complex conjugate. The \bgroup\color{black}$\gamma$\egroup matrices are Hermitian.

\gamma_\mu{\partial\psi\over\partial x_\mu}+{mc\over\hbar}\psi...\partial (x_\mu)^*}\gamma_\mu+{mc\over\hbar}\psi^\dagger=0 \\

The complex conjugate does nothing to the spatial component of \bgroup\color{black}$x_\mu$\egroup but does change the sign of the fourth component. To turn this back into a 4-vector expression, we can change the sign back by multiplying the equation by \bgroup\color{black}$\gamma_4$\egroup (from the right).

{\partial\psi^\dagger\over\partial x_k}\gamma_k+{\partial\psi^...
...r\partial x_4}\gamma_4
+{mc\over\hbar}\psi^\dagger\gamma_4=0 \\

Defining \bgroup\color{black}$\bar{\psi}=\psi^\dagger\gamma_4$\egroup, the adjoint spinor, we can rewrite the Hermitian conjugate equation.

-{\partial\bar{\psi}\over\partial x_k}\gamma_k-{\partial\bar{\...}\over\partial x_\mu}\gamma_\mu+{mc\over\hbar}\bar{\psi}=0 \\

This is the adjoint equation. We now multiply the Dirac equation by \bgroup\color{black}$\bar{\psi}$\egroup from the left and multiply the adjoint equation by \bgroup\color{black}$\psi$\egroup from the right, and subtract.

\bar{\psi}\gamma_\mu{\partial\psi\over\partial x_\mu}+{mc\over...
...i}\gamma_\mu\psi\right)=0 \\
j_\mu=\bar{\psi}\gamma_\mu\psi \\

\bgroup\color{black}$\displaystyle {\partial\over\partial x_\mu}j_\mu=0$\egroup
We have found a conserved current. Some interpretation will be required as we learn more about the solutions to the Dirac equation and ultimately quantize it. We may choose an overall constant to set the normalization. The fourth component of the current should be \bgroup\color{black}$ic$\egroup times the probability density so that the derivative with respect to \bgroup\color{black}$x_4$\egroup turns into \bgroup\color{black}${\partial P\over\partial t}$\egroup. Therefore let us set the properly normalized conserved 4-vector to be
\bgroup\color{black}$\displaystyle j_\mu=ic\bar{\psi}\gamma_\mu\psi .$\egroup

Jim Branson 2013-04-22