The Hamiltonian in terms of B

Start with the Hamiltonian

\begin{displaymath}\bgroup\color{black}H={1\over 2\mu}\left(\vec{p} + {e\over c}\vec{A}\right)^2 - e\phi\egroup\end{displaymath}

Now write the Schrödinger equation.

{1\over 2\mu}\left({\hbar\over i} \vec{\nabla} + {e\over c}\ve...
+ {e^2\over{2mc^2}}A^2\psi
& = & \left(E + e \phi\right)\psi

The second term vanishes in the Coulomb gauge i.e., \bgroup\color{black}$\vec{\nabla}\cdot\vec{A}=0$\egroup, so

- {ie\...
= \left(E + e \phi\right)\psi \egroup\end{displaymath}

Now for constant \bgroup\color{black}$B_z$\egroup, we choose the vector potential

\begin{displaymath}\bgroup\color{black}\vec{A}=-{1\over 2}\vec{r}\times\vec{B}\egroup\end{displaymath}


& = &{\partial\over{\... 2}B_k \left(\sum_i \sum_j\varepsilon^2_{ijk}\right) = B_k \\

it gives the right field and satisfies the Coulomb gauge condition.

Substituting back, we obtain

+ {ie\...
= \left(E + e \phi\right)\psi \egroup\end{displaymath}

Now let's work on the vector arithmetic.

= -{i\over\hbar}\vec{B}\cdot\vec{L}\psi \egroup\end{displaymath}

& = & r_iB_j\varepsilon_{i...
...ght) \\
& = & r^2 B^2 - \left(\vec{r}\cdot\vec{B}\right)^2 - 0

So, plugging these two equations in, we get

+ {e\o...
= \left(E + e \phi\right)\psi. \egroup\end{displaymath}

We see that there are two new terms due to the magnetic field. The first one is the magnetic moment term we have already used and the second will be negligible in atoms.

Jim Branson 2013-04-22