问题 1. Consider a two-dimensional strip of material of length $L$ and width $W$, placed in a magnetic field perpendicular to the strip and with an electric field established in the direction of the length $L$.
(a) Suppose that the resistivity matrix is given by the classical result
$$
\rho=\left(\begin{array}{cc}
\rho_0 & -\rho_H \
\rho_H & \rho_0
\end{array}\right)
$$
where $\rho_H=B /$ nec is the Hall resistivity and $\rho_0$ is the usual Ohmic resistivity. Find the conductivity matrix, $\sigma=\rho^{-1}$. Write it in the form:
$$
\sigma=\left(\begin{array}{cc}
\sigma_0 & \sigma_H \
-\sigma_H & \sigma_0
\end{array}\right) .
$$
What are $\sigma_0$ and $\sigma_H$ ?
(b) Suppose $B=0$, so the Hall resistivity is zero. Notice that the Ohmic conductivity, $\sigma_0$, is just $1 / \rho_0$. In particular, note that $\sigma_0 \rightarrow \infty$ as $\rho_0 \rightarrow 0$. Now suppose $\rho_H \neq 0$. Show that $\sigma_0 \rightarrow 0$ as $\rho_0 \rightarrow 0$, so it is possible to have both $\sigma_0$ and $\rho_0$ equal to zero

问题 2.

This problem asks you to give a complete presentation of a calculation that is almost the same as one you saw in lecture.

Consider a constant electric field, $\vec{E}=\left(0, E_0, 0\right)$ and a constant magnetic field, $\vec{B}=\left(0,0, B_0\right)$.
(a) Choose an electrostatic potential $\phi$ and a vector potential $\vec{A}$ which describe the $\vec{E}$ and $\vec{B}$ fields, and write the Hamiltonian for a charged particle of mass $m$ and charge $q$ in these fields. Assume that the particle is restricted to move in the $x y$-plane.
(b) What are the allowed energies as a function of $B_0$ and $E_0$ ? Draw a figure to show how the Landau levels (energy levels when $E_0=0$ ) change as $E_0$ increases.

问题 3.

You will see the “standard presentation” of the Aharonov-Bohm effect in lecture, on the day that this problem set is due. The standard presentation has its advantages, and in particular is more general than the presentation you will work through in this problem. However, students often come away from the standard presentation of the Aharonov-Bohm effect thinking that the only way to detect this effect is to do an interference experiment. This is not true, and the purpose of this problem is to disabuse you of this misimpression before you form it.

As Griffiths explains on pages 385-387 (344-345 in 1st Ed.), the Aharonov-Bohm effect modifies the energy eigenvalues of suitably chosen quantum mechanical systems. In this problem, I ask you to work through the same physical example that Griffiths uses, but in a different fashion which makes more use of gauge invariance.

Imagine a particle constrained to move on a circle of radius $b$ (a bead on a wire ring, if you like.) Along the axis of the circle runs a solenoid of radius $a<b$, carrying a magnetic field $\vec{B}=\left(0,0, B_0\right)$. The field inside the solenoid is uniform

and the field outside the solenoid is zero. The setup is depicted in Griffiths’ Fig. 10.10. (10.12 in 1st Ed.)
(a) Construct a vector potential $\vec{A}$ which describes the magnetic field (both inside and outside the solenoid) and which has the form $A_r=A_z=0$ and $A_\phi=\alpha(r)$ for some function $\alpha(r)$. I am using cylindrical coordinates $z, r$, $\phi$.
(b) Since $\vec{\nabla} \times \vec{A}=0$ for $r>a$, it must be possible to write $\vec{A}=\vec{\nabla} f$ in any simply connected region in $r>a$. [This is a theorem in vector calculus.] Show that if we find such an $f$ in the region
$$
r>a \text { and }-\pi+\epsilon<\phi<\pi-\epsilon,
$$
then
$$
f(r, \pi-\epsilon)-f(r,-\pi+\epsilon) \rightarrow \Phi \text { as } \epsilon \rightarrow 0 .
$$
Here, the total magnetic flux is $\Phi=\pi a^2 B_0$. Now find an explicit form for $f$, which is a function only of $\phi$.
(c) Now consider the motion of a “bead on a ring”: write the Schrödinger equation for the particle constrained to move on the circle $r=b$, using the $\vec{A}$ you found in (a). Hint: the answer is given in Griffiths.
(d) Use the $f(\phi)$ found in (b) to gauge transform the Schrödinger equation for $\psi(\phi)$ within the angular region $-\pi+\epsilon<\phi<\pi-\epsilon$ to a Schrödinger equation for a free particle within this angular region. Call the original wave function $\psi(\phi)$ and the gauge-transformed wave function $\psi^{\prime}(\phi)$.
(e) The original wave function $\psi$ must be single-valued for all $\phi$, in particular at $\phi=\pi$. That is, $\psi(\pi-\epsilon)-\psi(-\pi+\epsilon) \rightarrow 0$ and $\frac{\partial \psi}{\partial \phi}(\pi-\epsilon)-\frac{\partial \psi}{\partial \phi}(-\pi+\epsilon) \rightarrow 0$ as $\epsilon \rightarrow 0$. What does this say about the gauge-transformed wave function? I.e., how must $\psi^{\prime}(\pi-\epsilon)$ and $\psi^{\prime}(-\pi+\epsilon)$ be related as $\epsilon \rightarrow 0$ ?
[Hint: because the $f(\phi)$ is not single valued at $\phi=\pi$, the gauge transformed wave function $\psi^{\prime}(\phi)$ is not single valued there either.]
(f) Solve the Schrödinger equation for $\psi^{\prime}$, and find energy eigenstates which satisfy the boundary conditions you derived in (e). What are the allowed energy eigenvalues?
(g) Undo the gauge transformation, and find the energy eigenstates $\psi(\phi)$ in the original gauge. Do the energy eigenvalues in the two gauges differ?
(h) Plot the energy eigenvalues as a function of the enclosed flux, $\Phi$. Show that the energy eigenvalues are periodic functions of $\Phi$ with period $\Phi_0$, where you must determine $\Phi_0$. For what values of $\Phi$ does the enclosed magnetic field have no effect on the spectrum of a particle on a ring? Show that the

Aharonov-Bohm effect can only be used to determine the fractional part of $\Phi / \Phi_0$.
[Note: you have shown that even though the bead on a ring is everywhere in a region in which $\vec{B}=0$, the presence of a nonzero $\vec{A}$ affects the energy eigenvalue spectrum. However, the effect on the energy eigenvalues is determined by $\Phi$, and is therefore gauge invariant. To confirm the gauge invariance of your result, you can compare your answer for the energy eigenvalues to Griffiths’ result, obtained using a different gauge.]