The Relativistic Interaction Hamiltonian

The interaction Hamiltonian for the Dirac equation can be deduced in several ways. The simplest for now is to just use the same interaction term that we had for electromagnetism

\begin{displaymath}\bgroup\color{black} H_{int}=-{1\over c}j_\mu A_\mu \egroup\end{displaymath}

and identify the probability current multiplied by the charge (-e) as the current that couples to the EM field.

\begin{displaymath}\bgroup\color{black} j^{(EM)}_\mu=-eic\bar{\psi}\gamma_\mu\psi \egroup\end{displaymath}

Removing the \bgroup\color{black}$\psi^\dagger$\egroup from the left and \bgroup\color{black}$\psi$\egroup from the right and dotting into \bgroup\color{black}$A$\egroup, we have the interaction Hamiltonian.

\begin{displaymath}\bgroup\color{black} H_{int}=ie\gamma_4\gamma_\mu A_\mu \egroup\end{displaymath}

Note the difference between this interaction and the one we used in the non-relativistic case. The relativistic interaction has just one term, is linear in \bgroup\color{black}$A$\egroup, and is naturally proportional to the coupling \bgroup\color{black}$e$\egroup. There is no longer an $A^2$ term with a different power of \bgroup\color{black}$e$\egroup. This will make our perturbation series also a series in powers of \bgroup\color{black}$\alpha$\egroup.

We may still assume that \bgroup\color{black}$A$\egroup is transverse and that \bgroup\color{black}$A_0=0$\egroup by choice of gauge.

\bgroup\color{black}$\displaystyle H_{int}=ie\gamma_4\gamma_k A_k$\egroup

Jim Branson 2013-04-22