The Schrödinger-Pauli Hamiltonian

In the homework on electrons in an electromagnetic field, we showed that the Schrödinger-Pauli Hamiltonian gives the same result as the non-relativistic Hamiltonian we have been using and automatically includes the interaction of the electron's spin with the magnetic field.
\bgroup\color{black}$\displaystyle H={1\over 2m}\left(\vec{\sigma}\cdot[\vec{p}+{e\over c}\vec{A}(\vec{r},t)]\right)^2-e\phi(\vec{r},t)$\egroup
The derivation is repeated here. Recall that \bgroup\color{black}$\{\sigma_i,\sigma_j\}=2\delta_{ij}$\egroup, \bgroup\color{black}$[\sigma_i,\sigma_j]=2\epsilon_{ijk}\sigma_k$\egroup, and that the momentum operator differentiates both \bgroup\color{black}$\vec{A}$\egroup and the wavefunction.

\left(\vec{\sigma}\cdot[\vec{p}+{e\over c}\vec{A}(\vec{r},t)]\...
...}\vec{A}\cdot\vec{p}+{e\hbar\over c}\vec{\sigma}\cdot\vec{B} \\

H&=&{p^2\over 2m}+{e\over mc}\vec{A}\cdot\vec{p}+{e^2\over 2mc...
...c{r},t)+{e\hbar\over 2mc}\vec{\sigma}\cdot\vec{B}(\vec{r},t) \\

We assume the Lorentz condition applies. This is a step in the right direction. The wavefunction now has two components (a spinor) and the effect of spin is included. Note that this form of the NR Hamiltonian yields the coupling of the electron spin to a magnetic field with the correct $g$ factor of 2. The spin-orbit interaction can be correctly derived from this.

Jim Branson 2013-04-22