Aharonov-Bohm effect

Third update of aggregate notes for phy1520, Graduate Quantum Mechanics.

November 9, 2015 phy1520 1 comment , , , , , , , , , , , , , , , ,

I’ve posted a third update of my aggregate notes for PHY1520H Graduate Quantum Mechanics, taught by Prof. Arun Paramekanti. In addition to what was noted previously, this contains lecture notes up to lecture 13, my solutions for the third problem set, and some additional worked practice problems.

Most of the content was posted individually in the following locations, but those original documents will not be maintained individually any further.

Second update of aggregate notes for phy1520, Graduate Quantum Mechanics

October 20, 2015 phy1520 No comments , , , , , , , , , , , ,

I’ve posted a second update of my aggregate notes for PHY1520H Graduate Quantum Mechanics, taught by Prof. Arun Paramekanti. In addition to what was noted previously, this contains lecture notes up to lecture 9, my ungraded solutions for the second problem set, and some additional worked practise problems.

Most of the content was posted individually in the following locations, but those original documents will not be maintained individually any further.

PHY1520H Graduate Quantum Mechanics. Lecture 7: Aharonov-Bohm effect and Landau levels. Taught by Prof. Arun Paramekanti

October 16, 2015 phy1520 No comments , , ,

[Click here for a PDF of this post with nicer formatting]

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

fig. 1. Two slit interference with magnetic whisker

Aharonov-Bohm effect (cont.)

fig. 2. Energy vs flux

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

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

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

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

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

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.

PHY1520H Graduate Quantum Mechanics. Lecture 6: Electromagnetic gauge transformation and Aharonov-Bohm effect. Taught by Prof. Arun Paramekanti

October 6, 2015 phy1520 No comments , , ,

[Click here for a PDF of this post with nicer formatting]

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.

Particle with \( \BE, \BB \) fields

We express our fields with vector and scalar potentials

\begin{equation}\label{eqn:qmLecture6:20}
\BE, \BB \rightarrow \BA, \phi
\end{equation}

and apply a gauge transformed Hamiltonian

\begin{equation}\label{eqn:qmLecture6:40}
H = \inv{2m} \lr{ \Bp – q \BA }^2 + q \phi.
\end{equation}

Recall that in classical mechanics we have

\begin{equation}\label{eqn:qmLecture6:60}
\Bp – q \BA = m \Bv
\end{equation}

where \( \Bp \) is not gauge invariant, but the classical momentum \( m \Bv \) is.

If given a point in phase space we must also specify the gauge that we are working with.

For the quantum case, temporarily considering a Hamiltonian without any scalar potential, but introducing a gauge transformation

\begin{equation}\label{eqn:qmLecture6:80}
\BA \rightarrow \BA + \spacegrad \chi,
\end{equation}

which takes the Hamiltonian from

\begin{equation}\label{eqn:qmLecture6:100}
H = \inv{2m} \lr{ \Bp – q \BA }^2,
\end{equation}

to
\begin{equation}\label{eqn:qmLecture6:120}
H = \inv{2m} \lr{ \Bp – q \BA -q \spacegrad \chi }^2.
\end{equation}

We care that the position and momentum operators obey

\begin{equation}\label{eqn:qmLecture6:140}
\antisymmetric{\hat{r}_i}{\hat{p}_j} = i \Hbar \delta_{i j}.
\end{equation}

We can apply a transformation that keeps \( \Br \) the same, but changes the momentum

\begin{equation}\label{eqn:qmLecture6:160}
\begin{aligned}
\hat{\Br}’ &= \hat{\Br} \\
\hat{\Bp}’ &= \hat{\Bp} – q \spacegrad \chi(\Br)
\end{aligned}
\end{equation}

This maps the Hamiltonian to

\begin{equation}\label{eqn:qmLecture6:101}
H = \inv{2m} \lr{ \Bp’ – q \BA -q \spacegrad \chi }^2,
\end{equation}

We want to check if the commutator relationships have the desired structure, that is

\begin{equation}\label{eqn:qmLecture6:180}
\begin{aligned}
\antisymmetric{r_i’}{r_j’} &= 0 \\
\antisymmetric{p_i’}{p_j’} &= 0
\end{aligned}
\end{equation}

This is confirmed in \ref{problem:qmLecture6:1}.

Another thing of interest is how are the wave functions altered by this change of variables? The wave functions must change in response to this transformation if the energies of the Hamiltonian are to remain the same.

Considering a plane wave specified by

\begin{equation}\label{eqn:qmLecture6:200}
e^{i \Bk \cdot \Br},
\end{equation}

where we alter the momentum by

\begin{equation}\label{eqn:qmLecture6:220}
\Bk \rightarrow \Bk – e \spacegrad \chi.
\end{equation}

This takes the plane wave to

\begin{equation}\label{eqn:qmLecture6:240}
e^{i \lr{ \Bk – q \spacegrad \chi } \cdot \Br}.
\end{equation}

We want to try to find a wave function for the new Hamiltonian

\begin{equation}\label{eqn:qmLecture6:260}
H’ = \inv{2m} \lr{ \Bp’ – q \BA -q \spacegrad \chi }^2,
\end{equation}

of the form

\begin{equation}\label{eqn:qmLecture6:280}
\psi'(\Br)
\stackrel{?}{=}
e^{i \theta(\Br)} \psi(\Br),
\end{equation}

where the new wave function differs from a wave function for the original Hamiltonian by only a position dependent phase factor.

Let’s look at the action of the Hamiltonian on the new wave function

\begin{equation}\label{eqn:qmLecture6:300}
H’ \psi'(\Br) .
\end{equation}

Looking at just the first action

\begin{equation}\label{eqn:qmLecture6:320}
\begin{aligned}
\lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi } e^{i \theta(\Br)} \psi(\Br)
&=
e^{i\theta}
\lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi }
\psi(\Br)
+
\lr{
-i \Hbar i \spacegrad \theta
}
e^{i\theta}
\psi(\Br) \\
&=
e^{i\theta}
\lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi
+ \Hbar \spacegrad \theta
}
\psi(\Br).
\end{aligned}
\end{equation}

If we choose

\begin{equation}\label{eqn:qmLecture6:340}
\theta = \frac{q \chi}{\Hbar},
\end{equation}

then we are left with

\begin{equation}\label{eqn:qmLecture6:360}
\lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi } e^{i \theta(\Br)} \psi(\Br)
=
e^{i\theta}
\lr{ -i \Hbar \spacegrad – q \BA }
\psi(\Br).
\end{equation}

Let \( \BM = -i \Hbar \spacegrad – q \BA \), and act again with \( \lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi } \)

\begin{equation}\label{eqn:qmLecture6:700}
\begin{aligned}
\lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi } e^{i \theta} \BM \psi
&=
e^{i\theta}
\lr{ -i \Hbar i \spacegrad \theta – q \BA – q \spacegrad \chi } e^{i \theta} \BM \psi
+
e^{i\theta}
\lr{ -i \Hbar \spacegrad } \BM \psi \\
&=
e^{i\theta}
\lr{ -i \Hbar \spacegrad -q \BA + \spacegrad \lr{ \Hbar \theta – q \chi} } \BM \psi \\
&=
e^{i\theta} \BM^2 \psi.
\end{aligned}
\end{equation}

Restoring factors of \( m \), we’ve shown that for a choice of \( \Hbar \theta – q \chi \), we have

\begin{equation}\label{eqn:qmLecture6:400}
\inv{2m} \lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi }^2 e^{i \theta} \psi = e^{i\theta}
\inv{2m} \lr{ -i \Hbar \spacegrad – q \BA }^2 \psi.
\end{equation}

When \( \psi \) is an energy eigenfunction, this means

\begin{equation}\label{eqn:qmLecture6:420}
H’ e^{i\theta} \psi = e^{i \theta} H \psi = e^{i\theta} E\psi = E (e^{i\theta} \psi).
\end{equation}

We’ve found a transformation of the wave function that has the same energy eigenvalues as the corresponding wave functions for the original untransformed Hamiltonian.

In summary
\begin{equation}\label{eqn:qmLecture6:440}
\boxed{
\begin{aligned}
H’ &= \inv{2m} \lr{ \Bp – q \BA – q \spacegrad \chi}^2 \\
\psi'(\Br) &= e^{i \theta(\Br)} \psi(\Br), \qquad \text{where}\, \theta(\Br) = q \chi(\Br)/\Hbar
\end{aligned}
}
\end{equation}

Aharonov-Bohm effect

Consider a periodic motion in a fixed ring as sketched in fig. 1.

fig. 1. particle confined to a ring

fig. 1. particle confined to a ring

Here the displacement around the perimeter is \( s = R \phi \) and the Hamiltonian

\begin{equation}\label{eqn:qmLecture6:460}
H = – \frac{\Hbar^2}{2 m} \PDSq{s}{} = – \frac{\Hbar^2}{2 m R^2} \PDSq{\phi}{}.
\end{equation}

Now assume that there is a magnetic field squeezed into the point at the origin, by virtue of a flux at the origin

\begin{equation}\label{eqn:qmLecture6:480}
\BB = \Phi_0 \delta(\Br) \zcap.
\end{equation}

We know that

\begin{equation}\label{eqn:qmLecture6:500}
\oint \BA \cdot d\Bl = \Phi_0,
\end{equation}

so that

\begin{equation}\label{eqn:qmLecture6:520}
\BA = \frac{\Phi_0}{2 \pi r} \phicap.
\end{equation}

The Hamiltonian for the new configuration is

\begin{equation}\label{eqn:qmLecture6:540}
\begin{aligned}
H
&= – \lr{ -i \Hbar \spacegrad – q \frac{\Phi_0}{2 \pi r } \phicap }^2 \\
&= – \inv{2 m} \lr{ -i \Hbar \inv{R} \PD{\phi}{} – q \frac{\Phi_0}{2 \pi R } }^2.
\end{aligned}
\end{equation}

Here the replacement \( r \rightarrow R \) makes use of the fact that this problem as been posed with the particle forced to move around the ring at the fixed radius \( R \).

For this transformed Hamiltonian, what are the wave functions?

\begin{equation}\label{eqn:qmLecture6:560}
\psi(\phi)’
\stackrel{?}{=}
e^{i n \phi}.
\end{equation}

\begin{equation}\label{eqn:qmLecture6:580}
\begin{aligned}
H \psi
&= \inv{2 m}
\lr{ -i \Hbar \inv{R} (i n) – q \frac{\Phi_0}{2 \pi R } }^2 e^{i n \phi} \\
&=
\underbrace{\inv{2 m}
\lr{ \frac{\Hbar n}{R} – q \frac{\Phi_0}{2 \pi R } }^2}_{E_n} e^{i n \phi}.
\end{aligned}
\end{equation}

This is very unclassical, since the energy changes in a way that depends on the flux, because particles are seeing magnetic fields that are not present at the point of the particle.

This is sketched in fig. 2.

fig. 2. Energy variation with flux.

fig. 2. Energy variation with flux.

we see that there are multiple points that the energies hit the minimum levels

Question:

Show that after a transformation of position and momentum of the following form

\begin{equation}\label{eqn:qmLecture6:600}
\begin{aligned}
\hat{\Br}’ &= \hat{\Br} \\
\hat{\Bp}’ &= \hat{\Bp} – q \spacegrad \chi(\Br)
\end{aligned}
\end{equation}

all the commutators have the expected values.

Answer

The position commutators don’t need consideration. Of interest is the momentum-position commutators

\begin{equation}\label{eqn:qmLecture6:620}
\begin{aligned}
\antisymmetric{\hat{p}_k’}{\hat{x}_k’}
&=
\antisymmetric{\hat{p}_k – q \partial_k \chi}{\hat{x}_k} \\
&=
\antisymmetric{\hat{p}_k}{\hat{x}_k} – q \antisymmetric{\partial_k \chi}{\hat{x}_k} \\
&=
\antisymmetric{\hat{p}_k}{\hat{x}_k},
\end{aligned}
\end{equation}

and the momentum commutators

\begin{equation}\label{eqn:qmLecture6:640}
\begin{aligned}
\antisymmetric{\hat{p}_k’}{\hat{p}_j’}
&=
\antisymmetric{\hat{p}_k – q \partial_k \chi}{\hat{p}_j – q \partial_j \chi} \\
&=
\antisymmetric{\hat{p}_k}{\hat{p}_j}
– q \lr{ \antisymmetric{\partial_k \chi}{\hat{p}_j} + \antisymmetric{\hat{p}_k}{\partial_j \chi} }.
\end{aligned}
\end{equation}

That last sum of commutators is

\begin{equation}\label{eqn:qmLecture6:660}
\begin{aligned}
\antisymmetric{\partial_k \chi}{\hat{p}_j} + \antisymmetric{\hat{p}_k}{\partial_j \chi}
&=
– i \Hbar \lr{ \PD{k}{(\partial_j \chi)} – \PD{j}{(\partial_k \chi)} } \\
&= 0.
\end{aligned}
\end{equation}

We’ve shown that

\begin{equation}\label{eqn:qmLecture6:680}
\begin{aligned}
\antisymmetric{\hat{p}_k’}{\hat{x}_k’} &= \antisymmetric{\hat{p}_k}{\hat{x}_k} \\
\antisymmetric{\hat{p}_k’}{\hat{p}_j’} &= \antisymmetric{\hat{p}_k}{\hat{p}_j}.
\end{aligned}
\end{equation}

All the other commutators clearly have the desired transformation properties.

References

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