= Solution
The factor $e^{ik_zz}$ is an eigenstate of longitudinal momentum with $p_z=\hbar k_z$. The integer $n$ is the azimuthal <angular momentum> quantum number, since $L_z=-i\hbar\partial_\theta$ has eigenvalue $n\hbar$.
For
$$
\phi_n(r)=r^\alpha e^{-\beta r^2/2},
$$
direct differentiation gives
$$
\frac1r(r\phi_n')'
=\left[
\frac{\alpha^2}{r^2}
-2\beta(\alpha+1)+\beta^2r^2
\right]\phi_n.
$$
Regularity and cancellation of the $r^{-2}$ term require $\alpha=n$ for $n\geq0$. Cancellation of the $r^2$ terms requires
$$
\boxed{\beta=\frac{eB}{2\hbar}},
$$
assuming $eB>0$. The remaining radial energy is independent of $n$:
$$
\boxed{
\alpha=n,\qquad
E=\frac{\hbar^2k_z^2}{2m}+\frac{\hbar eB}{2m}}.
$$
This is the lowest <Landau level>. If $eB<0$, the same family uses nonpositive angular-momentum quantum numbers and $|eB|$ replaces $eB$.
Back to article page
