\item $I = (\textrm{charge}) \cdot (\textrm{loops per time})
\stackrel{1/T=\omega_L/2\pi}{=}
(Ze)(\frac{1}{2\pi}\frac{-e}{2m}B)$\\
- $\mu=IA=I2\pi<\rho^2>=-\frac{Ze^2B}{4m}<\rho^2>$\\
+ $\mu=IA=I\pi<\rho^2>=-\frac{Ze^2B}{4m}<\rho^2>$\\
$<x^2>=<y^2>=<z^2> \Rightarrow <r^2>=3<x^2>=3<y^2>$\\
$<\rho^2>=<x^2>+<y^2>=\frac{2}{3}<r^2>$\\
$\mu=-\frac{Ze^2B}{6m}$
\left(0,0,B\right)$\\
Note: $\nabla_r{\bf A}=0$
\item $
- H_{kin}'=\frac{i\hbar e}{2m}\left(
+ H_{kin}'=\frac{i\hbar e}{2m}\frac{B}{2}\left(
x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}
\right)+\frac{e^2B^2}{8m}(x^2+y^2)
$\\
$
L_z=x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}
- \Rightarrow H_{kin}'=\frac{i\hbar e}{2m}L_z+
+ \Rightarrow H_{kin}'=\frac{i\hbar e}{2m}\frac{B}{2}L_z+
\frac{e^2B^2}{8m}(x^2+y^2)
$
\item $\chi=-\frac{1}{V}\mu_0\frac{\partial^2 E}{\partial B^2}