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34 {\LARGE {\bf Materials Physics II}\\}
39 {\Large\bf Tutorial 1 - proposed solutions}
42 \section{Diamagnetism}
44 \item Magnetic field ${\bf B}$
45 \item Magnetization ${\bf M}$
46 \item Suscebtibility $\chi=\frac{\mu_0 {\bf M}}{{\bf B}}$
50 \item {\bf Classical approach:}
52 \item Maxwell: $\oint_{\partial A} E \, ds
53 = -\frac{d}{dt}(\int_A B \, dA)
54 \stackrel{B(r)=B}{=}-\frac{d}{dt}(BA)$\\
55 $-\frac{d(BA)}{dt}=-\pi r^2 \dot{B}=U_{ind}$\\
56 $U_{ind}=\oint_{\partial A} E \, ds
57 \stackrel{E(s)=E}{\Rightarrow}
58 E=-\frac{\pi r^2}{2\pi r}\dot{B}$\\
59 $\dot{v}=a=\frac{e}{m}E=-\frac{e}{2m}r\dot{B}
60 \Rightarrow v=-\frac{e}{2m}rB$\\
61 $\omega_L=\frac{v}{r}=-\frac{e}{2m}B$
62 \item $I = (\textrm{charge}) \cdot (\textrm{loops per time})
63 \stackrel{1/T=\omega_L/2\pi}{=}
64 (Ze)(\frac{1}{2\pi}\frac{-e}{2m}B)$\\
65 $\mu=IA=I\pi<\rho^2>=-\frac{Ze^2B}{4m}<\rho^2>$\\
66 $<x^2>=<y^2>=<z^2> \Rightarrow <r^2>=3<x^2>=3<y^2>$\\
67 $<\rho^2>=<x^2>+<y^2>=\frac{2}{3}<r^2>$\\
68 $\mu=-\frac{Ze^2B}{6m}$
69 \item $\chi=\frac{\mu_0N\mu}{B}=-\frac{\mu_0NZe^2}{6m}<r^2>$
71 \item {\bf Quantum mechanical theory:}
73 \item vector potential ${\bf A}$
74 \item ${\bf B}=\nabla\times{\bf A}$
76 H_{kin}=\frac{1}{2m}(-i\hbar\nabla_{r}-e{\bf A})^2
81 \item \begin{eqnarray}
82 H_{kin}&=&\frac{1}{2m}(-\hbar^2\nabla_{r}^2+e^2{\bf A}^2
83 +i\hbar \nabla_{r}e{\bf A}
84 +e{\bf A}i\hbar \nabla_{r})\nonumber\\
85 H_{kin}^0&=&\frac{-\hbar^2}{2m}\nabla_r^2\nonumber\\
86 H_{kin}'&=&\frac{i\hbar e}{2m}(\nabla_r{\bf A}+{\bf A}\nabla_r)+
87 \frac{e^2{\bf A}^2}{2m}\nonumber
89 Terms in $H_{kin}'$ can be treated as small perturbation.
90 \item ${\bf A}=\left(-\frac{1}{2}yB,\frac{1}{2}xB,0\right)$, since:
91 $\nabla_r\times{\bf A}=\left(0,0,\frac{1}{2}B+\frac{1}{2}B\right)=
93 Note: $\nabla_r{\bf A}=0$
95 H_{kin}'=\frac{i\hbar e}{2m}\frac{B}{2}\left(
96 x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}
97 \right)+\frac{e^2B^2}{8m}(x^2+y^2)
100 L_z=x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}
101 \Rightarrow H_{kin}'=\frac{i\hbar e}{2m}\frac{B}{2}L_z+
102 \frac{e^2B^2}{8m}(x^2+y^2)
104 \item $\chi=-\frac{1}{V}\mu_0\frac{\partial^2 E}{\partial B^2}
106 only second term contributes to $\chi$!\\
107 $\chi=-\frac{1}{V}\mu_0\frac{e^2}{4m}<\phi|(x^2+y^2)|\phi>$
108 \item $<\phi|x^2|\phi>=<\phi|y^2|\phi>=\frac{1}{3}<\phi|r^2|\phi>$\\
109 $\Rightarrow \chi=-\frac{1}{V}\mu_0\frac{e^2}{6m}
111 Consider all $Z$ electrons and all atoms per volume:\\
112 $\chi=-\frac{\mu_0NZe^2}{6m}<\phi|r^2|\phi>$