D.74 Orbital motion in a magnetic field

This note derives the energy of a charged particle in an external magnetic field. The field is assumed constant.

According to chapter 13.1, the Hamiltonian is

\begin{displaymath}
H = \frac{1}{2m}\left({\skew 4\widehat{\skew{-.5}\vec p}}- q\skew3\vec A\right)^2 + V
\end{displaymath}

where $m$ and $q$ are the mass and charge of the particle and the vector potential $\skew3\vec A$ is related to the magnetic field $\skew2\vec{\cal B}$ by $\skew2\vec{\cal B}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\nabla$ $\times$ $\skew3\vec A$. The potential energy $V$ is of no particular interest in this note. The first term is, and it can be multiplied out as:

\begin{displaymath}
H = \frac{1}{2m}{\skew 4\widehat{\skew{-.5}\vec p}}^{\,2}
...
...,\right)
+ \frac{q^2}{2m} \left(\skew3\vec A\,\right)^2 + V
\end{displaymath}

The middle two terms in the right hand side are the changes in the Hamiltonian due to the magnetic field; they will be denoted as:

\begin{displaymath}
H_{{\cal B}L} \equiv - \frac{q}{2m}\left({\skew 4\widehat{...
...\cal B}D} \equiv \frac{q^2}{2m} \left(\skew3\vec A\,\right)^2
\end{displaymath}

Now to simplify the analysis, align the $z$-​axis with $\skew2\vec{\cal B}$ so that $\skew2\vec{\cal B}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${{\hat k}}{\cal B}_z$. Then an appropriate vector potential $\skew3\vec A$ is

\begin{displaymath}
\skew3\vec A= -{\hat\imath}{\textstyle\frac{1}{2}} y {\cal B}_z + {\hat\jmath}{\textstyle\frac{1}{2}} x {\cal B}_z.
\end{displaymath}

The vector potential is not unique, but a check shows that indeed $\nabla$ $\times$ $\skew3\vec A$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${{\hat k}}{\cal B}_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\skew2\vec{\cal B}$ for the one above. Also, the canonical momentum is

\begin{displaymath}
{\skew 4\widehat{\skew{-.5}\vec p}}= \frac{\hbar}{{\rm i}}...
...+
{\hat k}\frac{\hbar}{{\rm i}}\frac{\partial}{\partial z}
\end{displaymath}

Therefore, in the term $H_{{{\cal B}}L}$ above,

\begin{displaymath}
H_{{\cal B}L} = - \frac{q}{2m} ({\skew 4\widehat{\skew{-.5...
...al}{\partial x}
\right)
= - \frac{q}{2m} {\cal B}_z \L _z
\end{displaymath}

the latter equality being true because of the definition of angular momentum as ${\skew0\vec r}$ $\times$ ${\skew 4\widehat{\skew{-.5}\vec p}}$. Because the $z$-​axis was aligned with $\skew2\vec{\cal B}$, ${\cal B}_z\L _z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\skew2\vec{\cal B}\cdot{\skew 4\widehat{\vec L}}$, so, finally,

\begin{displaymath}
H_{{\cal B}L} = - \frac{q}{2m} \skew2\vec{\cal B}\cdot{\skew 4\widehat{\vec L}}.
\end{displaymath}

Similarly, in the part $H_{{{\cal B}}D}$ of the Hamiltonian, substitution of the expression for $\skew3\vec A$ produces

\begin{displaymath}
\frac{q^2}{2m} \left(\skew3\vec A\right)^2 =
\frac{q^2}{8m} {\cal B}_z^2 \left(x^2+y^2\right),
\end{displaymath}

or writing it so that it is independent of how the $z$-​axis is aligned,

\begin{displaymath}
H_{{\cal B}D} = \frac{q^2}{8m} \left(\skew2\vec{\cal B}\times{\skew0\vec r}\right)^2
\end{displaymath}