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Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] chap. 2 content.

problem set note.

In the problem set we’ll look at interference patterns for two slit electron interference like that of fig. 1, where a magnetic whisker that introduces flux is added to the configuration.

fig. 1. Two slit interference with magnetic whisker

Aharonov-Bohm effect (cont.)

fig. 2. Energy vs flux

Why do we have the zeros at integral multiples of $h/q$? Consider a particle in a circular trajectory as sketched in fig. 3

fig. 3. Circular trajectory

FIXME: Prof mentioned:

$$\begin{equation}\label{eqn:qmLecture7:20} \phi_{\textrm{loop}} = q \frac{ h p/ q }{\Hbar} = 2 \pi p \end{equation}$$

… I’m not sure what that was about now.

In classical mechanics we have

$$\begin{equation}\label{eqn:qmLecture7:40} \oint p dq \end{equation}$$

The integral zero points are related to such a loop, but the $q \BA$ portion of the momentum $\Bp – q \BA$ needs to be considered.

Superconductors

After cooling some materials sufficiently, superconductivity, a complete lack of resistance to electrical flow can be observed. A resistivity vs temperature plot of such a material is sketched in fig. 4.

fig. 4. Superconductivity with comparison to superfluidity

Just like \ce{He^4} can undergo Bose condensation, superconductivity can be explained by a hybrid Bosonic state where electrons are paired into one state containing integral spin.

The Little-Parks experiment puts a superconducting ring around a magnetic whisker as sketched in fig. 6.

fig. 6. Little-Parks superconducting ring

This experiment shows that the effective charge of the circulating charge was $2 e$, validating the concept of Cooper-pairing, the Bosonic combination (integral spin) of electrons in superconduction.

Motion around magnetic field

$$\begin{equation}\label{eqn:qmLecture7:140} \omega_{\textrm{c}} = \frac{e B}{m} \end{equation}$$

We work with what is now called the Landau gauge

$$\begin{equation}\label{eqn:qmLecture7:60} \BA = \lr{ 0, B x, 0 } \end{equation}$$

This gives

$$\begin{equation}\label{eqn:qmLecture7:80} \begin{aligned} \BB &= \spacegrad \cross \BA \\ &= \lr{ \partial_x A_y – \partial_y A_x } \zcap \\ &= B \zcap. \end{aligned} \end{equation}$$

An alternate gauge choice, the symmetric gauge, is

$$\begin{equation}\label{eqn:qmLecture7:100} \BA = \lr{ -\frac{B y}{2}, \frac{B x}{2}, 0 }, \end{equation}$$

that also has the same magnetic field

$$\begin{equation}\label{eqn:qmLecture7:120} \begin{aligned} \BB &= \spacegrad \BA \\ &= \lr{ \partial_x A_y – \partial_y A_x } \zcap \\ &= \lr{ \frac{B}{2} – \lr{ – \frac{B}{2} } } \zcap \\ &= B \zcap. \end{aligned} \end{equation}$$

We expect the physics for each to have the same results, although the wave functions in one gauge may be more complicated than in the other.

Our Hamiltonian is

$$\begin{equation}\label{eqn:qmLecture7:160} \begin{aligned} H &= \inv{2 m} \lr{ \Bp – e \BA }^2 \\ &= \inv{2 m} \hat{p}_x^2 + \inv{2 m} \lr{ \hat{p}_y – e B \xhat }^2 \end{aligned} \end{equation}$$

We can solve after noting that

$$\begin{equation}\label{eqn:qmLecture7:180} \antisymmetric{\hat{p}_y}{H} = 0 \end{equation}$$

means that

$$\begin{equation}\label{eqn:qmLecture7:200} \Psi(x,y) = e^{i k_y y} \phi(x) \end{equation}$$

The eigensystem

$$\begin{equation}\label{eqn:qmLecture7:220} H \psi(x, y) = E \phi(x, y) , \end{equation}$$

becomes

$$\begin{equation}\label{eqn:qmLecture7:240} \lr{ \inv{2 m} \hat{p}_x^2 + \inv{2 m} \lr{ \Hbar k_y – e B \xhat}^2 } \phi(x) = E \phi(x). \end{equation}$$

This reduced Hamiltonian can be rewritten as

$$\begin{equation}\label{eqn:qmLecture7:320} H_x = \inv{2 m} p_x^2 + \inv{2 m} e^2 B^2 \lr{ \xhat – \frac{\Hbar k_y}{e B} }^2 \equiv \inv{2 m} p_x^2 + \inv{2} m \omega^2 \lr{ \xhat – x_0 }^2 \end{equation}$$

where

$$\begin{equation}\label{eqn:qmLecture7:260} \inv{2 m} e^2 B^2 = \inv{2} m \omega^2, \end{equation}$$

or
$$\begin{equation}\label{eqn:qmLecture7:280} \omega = \frac{ e B}{m} \equiv \omega_{\textrm{c}}. \end{equation}$$

and

$$\begin{equation}\label{eqn:qmLecture7:300} x_0 = \frac{\Hbar}{k_y}{e B}. \end{equation}$$

But what is this $x_0$? Because $k_y$ is not really specified in this problem, we can consider that we have a zero point energy for every $k_y$, but the oscillator position is shifted for every such value of $k_y$. For each set of energy levels fig. 8 we can consider that there is a different zero point energy for each possible $k_y$.

fig. 8. Energy levels, and Energy vs flux

This is an infinitely degenerate system with an infinite number of states for any given energy level.

This tells us that there is a problem, and have to reconsider the assumption that any $k_y$ is acceptable.

To resolve this we can introduce periodic boundary conditions, imagining that a square is rotated in space forming a cylinder as sketched in fig. 9.

fig. 9. Landau degeneracy region

Requiring quantized momentum

$$\begin{equation}\label{eqn:qmLecture7:340} k_y L_y = 2 \pi n, \end{equation}$$

or

$$\begin{equation}\label{eqn:qmLecture7:360} k_y = \frac{2 \pi n}{L_y}, \qquad n \in \mathbb{Z}, \end{equation}$$

gives

$$\begin{equation}\label{eqn:qmLecture7:380} x_0(n) = \frac{\Hbar}{e B} \frac{ 2 \pi n}{L_y}, \end{equation}$$

with $x_0 \le L_x$. The range is thus restricted to

$$\begin{equation}\label{eqn:qmLecture7:400} \frac{\Hbar}{e B} \frac{ 2 \pi n_{\textrm{max}}}{L_y} = L_x, \end{equation}$$

or

$$\begin{equation}\label{eqn:qmLecture7:420} n_{\textrm{max}} = \underbrace{L_x L_y}_{\text{area}} \frac{ e B }{2 \pi \Hbar } \end{equation}$$

That is

$$\begin{equation}\label{eqn:qmLecture7:440} \begin{aligned} n_{\textrm{max}} &= \frac{\Phi_{\textrm{total}}}{h/e} \\ &= \frac{\Phi_{\textrm{total}}}{\Phi_0}. \end{aligned} \end{equation}$$

Attempting to measure Hall-effect systems, it was found that the Hall conductivity was quantized like

$$\begin{equation}\label{eqn:qmLecture7:460} \sigma_{x y} = p \frac{e^2}{h}. \end{equation}$$

This quantization is explained by these Landau levels, and this experimental apparatus provides one of the more accurate ways to measure the fine structure constant.

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.