## Unimodular transformation

October 31, 2015 phy1520 No comments ,

### Q: Show that

This is ([1] pr. 3.3)

Given the matrix

\label{eqn:unimodularAndRotation:20}
U =
\frac
{a_0 + i \sigma \cdot \Ba}
{a_0 – i \sigma \cdot \Ba},

where $$a_0, \Ba$$ are real valued constant and vector respectively.

• Show that this is a unimodular and unitary transformation.
• A unitary transformation can represent an arbitary rotation. Determine the rotation angle and direction in terms of $$a_0, \Ba$$.

### A: unimodular

Let’s call these factors $$A_{\pm}$$, which expand to

\label{eqn:unimodularAndRotation:40}
\begin{aligned}
A_{\pm}
&=
a_0 \pm i \sigma \cdot \Ba \\
&=
\begin{bmatrix}
a_0 \pm i a_z & \pm \lr{ a_y + i a_x} \\
\mp (a_y – i a_x) & a_0 \mp i a_z \\
\end{bmatrix},
\end{aligned}

or with $$z = a_0 + i a_z$$, and $$w = a_y + i a_x$$, these are

\label{eqn:unimodularAndRotation:120}
A_{+}
=
\begin{bmatrix}
z & w \\
-w^\conj & z^\conj
\end{bmatrix}

\label{eqn:unimodularAndRotation:180}
A_{-}
=
\begin{bmatrix}
z^\conj & -w \\
w^\conj & z
\end{bmatrix}.

These both have a determinant of
\label{eqn:unimodularAndRotation:60}
\begin{aligned}
\Abs{z}^2 + \Abs{w}^2
&=
\Abs{a_0 + i a_z}^2 + \Abs{a_y + i a_x}^2 \\
&= a_0^2 + \Ba^2.
\end{aligned}

The inverse of the latter is
\label{eqn:unimodularAndRotation:200}
A_{-}^{-1}
=
\inv{ a_0^2 + \Ba^2 }
\begin{bmatrix}
z & w \\
-w^\conj & z^\conj
\end{bmatrix}

Noting that the numerator and denominator commute the inverse can be applied in either order. Picking one, the transformation of interest, after writing $$A = a_0^2 + \Ba^2$$, is

\label{eqn:unimodularAndRotation:100}
\begin{aligned}
U
&=
\inv{A}
\begin{bmatrix}
z & w \\
-w^\conj & z^\conj
\end{bmatrix}
\begin{bmatrix}
z & w \\
-w^\conj & z^\conj
\end{bmatrix} \\
&=
\inv{A}
\begin{bmatrix}
z^2 – \Abs{w}^2 & w( z + z^\conj) \\
-w^\conj (z^\conj + z ) & (z^\conj)^2 – \Abs{w}^2
\end{bmatrix}.
\end{aligned}

Recall that a unimodular transformation is one that has the form

\label{eqn:unimodularAndRotation:140}
\begin{bmatrix}
z & w \\
-w^\conj & z^\conj
\end{bmatrix},

provided $$\Abs{z}^2 + \Abs{w}^2 = 1$$, so \ref{eqn:unimodularAndRotation:100} is unimodular if the following sum is unity, which is the case

\label{eqn:unimodularAndRotation:160}
\begin{aligned}
\frac{\Abs{z^2 – \Abs{w}^2}^2}{\lr{ \Abs{z}^2 + \Abs{w}^2}^2 } + \Abs{w}^2 \frac{\Abs{z + z^\conj}^2 }{\lr{ \Abs{z}^2 + \Abs{w}^2}^2 }
&=
\frac{
\lr{ z^2 – \Abs{w}^2 } \lr{ (z^\conj)^2 – \Abs{w}^2 }
+ \Abs{w}^2 \lr{ z + z^\conj }^2
}{
\lr{ \Abs{z}^2 + \Abs{w}^2}^2
} \\
&=
\frac{
\Abs{z}^4 + \Abs{w}^4 – \Abs{w}^2 \lr{ {z^2 + (z^\conj)^2} }
+ \Abs{w}^2 \lr{ {z^2 + (z^\conj)^2} + 2 \Abs{z}^2 }
}{
\lr{ \Abs{z}^2 + \Abs{w}^2}^2
} \\
&= 1.
\end{aligned}

### A: rotation

The most general rotation of a vector $$\Ba$$, described by Pauli matrices is

\label{eqn:unimodularAndRotation:220}
e^{i \Bsigma \cdot \ncap \theta/2}
\Bsigma \cdot \Ba
e^{-i \Bsigma \cdot \ncap \theta/2}
=
\Bsigma \cdot \ncap + \lr{ \Bsigma \cdot \Ba – (\Ba \cdot \ncap) \Bsigma \cdot \ncap } \cos \theta + \Bsigma \cdot (\Ba \cross \ncap) \sin\theta.

If the unimodular matrix above, applied as $$\Bsigma \cdot \Ba’ = U^\dagger \Bsigma \cdot \Ba U$$ is to also describe this rotation, we want the equivalence

\label{eqn:unimodularAndRotation:240}
U = e^{-i \Bsigma \cdot \ncap \theta/2},

or

\label{eqn:unimodularAndRotation:260}
\inv{a_0^2 + \Ba^2}
\begin{bmatrix}
a_0^2 – \Ba^2 + 2 i a_0 a_z & 2 a_0 ( a_y + i a_x ) \\
-2 a_0( a_y – i a_x ) & a_0^2 – \Ba^2 – 2 i a_0 a_z
\end{bmatrix}
=
\begin{bmatrix}
\cos(\theta/2) – i n_z \sin(\theta/2) & (-n_y -i n_x) \sin(\theta/2) \\
-( – n_y + i n_x ) \sin(\theta/2) & \cos(\theta/2) + i n_z \sin(\theta/2)
\end{bmatrix}.

Equating components, that is
\label{eqn:unimodularAndRotation:280}
\begin{aligned}
\cos(\theta/2) &= \frac{a_0^2 – \Ba^2}{a_0^2 + \Ba^2} \\
-n_x \sin(\theta/2) &= \frac{2 a_0 a_x}{a_0^2 + \Ba^2} \\
-n_y \sin(\theta/2) &= \frac{2 a_0 a_y}{a_0^2 + \Ba^2} \\
-n_z \sin(\theta/2) &= \frac{2 a_0 a_y}{a_0^2 + \Ba^2} \\
\end{aligned}

Noting that

\label{eqn:unimodularAndRotation:300}
\begin{aligned}
\sin(\theta/2)
&=
\sqrt{
1 – \frac{(a_0^2 – \Ba^2)^2}{(a_0^2 + \Ba^2)^2}
} \\
&=
\frac{
\sqrt{ (a_0^2 + \Ba^2)^2 – (a_0^2 – \Ba^2)^2 }
}
{
a_0^2 + \Ba^2
} \\
&=
\frac{\sqrt{ 4 a_0^2 \Ba^2 }}{a_0^2 + \Ba^2} \\
&=
\frac{2 a_0 \Abs{\Ba} }{a_0^2 + \Ba^2}
\end{aligned}

The vector normal direction can be written

\label{eqn:unimodularAndRotation:320}
\Bn
= – \frac{2 a_0}{(a_0^2 + \Ba^2) \sin(\theta/2)} \Ba,

or

\label{eqn:unimodularAndRotation:340}
\boxed{
\Bn = – \frac{\Ba}{\Abs{\Ba}}.
}

The angle of rotation is

\label{eqn:unimodularAndRotation:380}
\boxed{
\theta = 2 \tan^{-1} \frac{2 a_0 \Abs{\Ba}}{ a_0^2 – \Ba^2}.
}

# References

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

## Some spin problems

October 30, 2015 phy1520 No comments , ,

Problems from angular momentum chapter of [1].

### Q: $$S_y$$ eigenvectors

Find the eigenvectors of $$\sigma_y$$, and then find the probability that a measurement of $$S_y$$ will be $$\Hbar/2$$ when the state is initially

\label{eqn:someSpinProblems:20}
\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}

### A:

The eigenvalues should be $$\pm 1$$, which is easily checked

\label{eqn:someSpinProblems:40}
\begin{aligned}
0
&=
\Abs{ \sigma_y – \lambda } \\
&=
\begin{vmatrix}
-\lambda & -i \\
i & -\lambda
\end{vmatrix} \\
&=
\lambda^2 – 1.
\end{aligned}

For $$\ket{+} = (a,b)^\T$$ we must have

\label{eqn:someSpinProblems:60}
-1 a – i b = 0,

so

\label{eqn:someSpinProblems:80}
\ket{+} \propto
\begin{bmatrix}
-i \\
1
\end{bmatrix},

or
\label{eqn:someSpinProblems:100}
\ket{+} =
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
i
\end{bmatrix}.

For $$\ket{-}$$ we must have

\label{eqn:someSpinProblems:120}
a – i b = 0,

so

\label{eqn:someSpinProblems:140}
\ket{+} \propto
\begin{bmatrix}
i \\
1
\end{bmatrix},

or
\label{eqn:someSpinProblems:160}
\ket{+} =
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
-i
\end{bmatrix}.

The normalized eigenvectors are

\label{eqn:someSpinProblems:180}
\boxed{
\ket{\pm} =
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
\pm i
\end{bmatrix}.
}

For the probability question we are interested in

\label{eqn:someSpinProblems:200}
\begin{aligned}
\Abs{\bra{S_y; +}
\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}
}^2
&=
\inv{2} \Abs{
\begin{bmatrix}
1 & -i
\end{bmatrix}
\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}
}^2 \\
&=
\inv{2} \lr{ \Abs{\alpha}^2 + \Abs{\beta}^2 } \\
&=
\inv{2}.
\end{aligned}

There is a 50 % chance of finding the particle in the $$\ket{S_x;+}$$ state, independent of the initial state.

### Q: Magnetic Hamiltonian eigenvectors

Using Pauli matrices, find the eigenvectors for the magnetic spin interaction Hamiltonian

\label{eqn:someSpinProblems:220}
H = – \inv{\Hbar} 2 \mu \BS \cdot \BB.

### A:

\label{eqn:someSpinProblems:240}
\begin{aligned}
H
&= – \mu \Bsigma \cdot \BB \\
&= – \mu \lr{ B_x \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} + B_y
\begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} + B_z \begin{bmatrix} 1 & 0
\\ 0 & -1 \\ \end{bmatrix} } \\
&= – \mu
\begin{bmatrix}
B_z & B_x – i B_y \\
B_x + i B_y & -B_z
\end{bmatrix}.
\end{aligned}

The characteristic equation is
\label{eqn:someSpinProblems:260}
\begin{aligned}
0
&=
\begin{vmatrix}
-\mu B_z -\lambda & -\mu(B_x – i B_y) \\
-\mu(B_x + i B_y) & \mu B_z – \lambda
\end{vmatrix} \\
&=
-\lr{ (\mu B_z)^2 – \lambda^2 }
– \mu^2\lr{ B_x^2 – (iB_y)^2 } \\
&=
\lambda^2 – \mu^2 \BB^2.
\end{aligned}

That is
\label{eqn:someSpinProblems:360}
\boxed{
\lambda = \pm \mu B.
}

Now for the eigenvectors. We are looking for $$\ket{\pm} = (a,b)^\T$$ such that

\label{eqn:someSpinProblems:300}
0
= (-\mu B_z \mp \mu B) a -\mu(B_x – i B_y) b

or

\label{eqn:someSpinProblems:320}
\ket{\pm} \propto
\begin{bmatrix}
B_x – i B_y \\
B_z \pm B
\end{bmatrix}.

This squares to

\label{eqn:someSpinProblems:340}
B_x^2 + B_y^2 + B_z^2 + B^2 \pm 2 B B_z
= 2 B( B \pm B_z ),

so the normalized eigenkets are
\label{eqn:someSpinProblems:380}
\boxed{
\ket{\pm}
=
\inv{\sqrt{2 B( B \pm B_z )}}
\begin{bmatrix}
B_x – i B_y \\
B_z \pm B
\end{bmatrix}.
}

# References

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

## PHY1520H Graduate Quantum Mechanics. Lecture 11: Symmetries in QM. Taught by Prof. Arun Paramekanti

October 29, 2015 phy1520 No comments , ,

### 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 \textchapref{{4}} [1] content.

### Symmetry in classical mechanics

In a classical context considering a Hamiltonian

\label{eqn:qmLecture11:20}
H(q_i, p_i),

a symmetry means that certain $$q_i$$ don’t appear. In that case the rate of change of one of the generalized momenta is zero

\label{eqn:qmLecture11:40}
\ddt{p_k} = – \PD{q_k}{H} = 0,

so $$p_k$$ is a constant of motion. This simplifies the problem by reducing the number of degrees of freedom. Another aspect of such a symmetry is that it \underline{relates trajectories}. For example, assuming a rotational symmetry as in fig. 1.

fig. 1. Trajectory under rotational symmetry

the trajectory of a particle after rotation is related by rotation to the trajectory of the unrotated particle.

### Symmetry in quantum mechanics

Suppose that we have a symmetry operation that takes states from

\label{eqn:qmLecture11:60}
\ket{\psi} \rightarrow \ket{U \psi}

\label{eqn:qmLecture11:80}
\ket{\phi} \rightarrow \ket{U \phi},

we expect that

\label{eqn:qmLecture11:100}
\Abs{\braket{ \psi}{\phi} }^2 = \Abs{\braket{ U\psi}{ U\phi} }^2.

This won’t hold true for a general operator. Two cases where this does hold true is when

• $$\braket{\psi}{\phi} = \braket{ U\psi}{ U\phi}$$. Here $$U$$ is unitary, and the equivalence follows from

\label{eqn:qmLecture11:120}
\braket{ U\psi}{ U\phi} = \bra{ \psi} U^\dagger U { \phi} = \bra{ \psi} 1 { \phi} = \braket{\psi}{\phi}.

• $$\braket{\psi}{\phi} = \braket{ U\psi}{ U\phi}^\conj$$. Here $$U$$ is anti-unitary.

### Unitary case

If an “observable” is not changed by a unitary operation representing a symmetry we must have

\label{eqn:qmLecture11:140}
\bra{\psi} \hat{A} \ket{\psi}
\rightarrow
\bra{U \psi} \hat{A} \ket{U \psi}
=
\bra{\psi} U^\dagger \hat{A} U \ket{\psi},

so
\label{eqn:qmLecture11:160}
U^\dagger \hat{A} U = \hat{A},

or
\label{eqn:qmLecture11:180}
\boxed{
\hat{A} U = U \hat{A}.
}

An observable that is unchanged by a unitary symmetry commutes $$\antisymmetric{\hat{A}}{U}$$ with the operator $$U$$ for that transformation.

### Symmetries of the Hamiltonian

Given
\label{eqn:qmLecture11:200}
\antisymmetric{H}{U} = 0,

$$H$$ is invariant.

Given

\label{eqn:qmLecture11:220}
H \ket{\phi_n} = \epsilon_n \ket{\phi_n} .

\label{eqn:qmLecture11:240}
\begin{aligned}
U H \ket{\phi_n}
&= H U \ket{\phi_n} \\
&= \epsilon_n U \ket{\phi_n}
\end{aligned}

Such a state

\label{eqn:qmLecture11:260}
\ket{\psi_n} = U \ket{\phi_n}

is also an eigenstate with the \underline{same} energy.

Suppose this process is repeated, finding other states

\label{eqn:qmLecture11:280}
U \ket{\psi_n} = \ket{\chi_n}

\label{eqn:qmLecture11:300}
U \ket{\chi_n} = \ket{\alpha_n}

Because such a transformation only generates states with the initial energy, this process cannot continue forever. At some point this process will enumerate a fixed size set of states. These states can be orthonormalized.

We can say that symmetry operations are generators of a \underlineAndIndex{group}. For a set of symmetry operations we can

• Form products that lie in a closed set

\label{eqn:qmLecture11:320}
U_1 U_2 = U_3

• can define an inverse
\label{eqn:qmLecture11:340}
U \leftrightarrow U^{-1}.

• obeys associative rules for multiplication
\label{eqn:qmLecture11:360}
U_1 ( U_2 U_3 ) = (U_1 U_2) U_3.

• has an identity operation.

When $$H$$ has a symmetry, then degenerate eigenstates form \underlineAndIndex{irreducible} representations (which cannot be further block diagonalized).

## Example: Inversion.

{example:qmLecture11:1}

Given a state and a parity operation $$\hat{\Pi}$$, with the transformation

\label{eqn:qmLecture11:380}
\ket{\psi} \rightarrow \hat{\Pi} \ket{\psi}

In one dimension, the parity operation is just inversion. In two dimensions, this is a set of flipping operations on two axes fig. 2.

fig. 2. 2D parity operation

The operational effects of this operator are

\label{eqn:qmLecture11:400}
\begin{aligned}
\hat{x} &\rightarrow – \hat{x} \\
\hat{p} &\rightarrow – \hat{p}.
\end{aligned}

Acting again with the parity operator produces the original value, so it is its own inverse, and $$\hat{\Pi}^\dagger = \hat{\Pi} = \hat{\Pi}^{-1}$$. In an expectation value

\label{eqn:qmLecture11:420}
\bra{ \hat{\Pi} \psi } \hat{x} \ket{ \hat{\Pi} \psi } = – \bra{\psi} \hat{x} \ket{\psi}.

This means that

\label{eqn:qmLecture11:440}
\hat{\Pi}^\dagger \hat{x} \hat{\Pi} = – \hat{x},

or
\label{eqn:qmLecture11:460}
\hat{x} \hat{\Pi} = – \hat{\Pi} \hat{x},

\label{eqn:qmLecture11:480}
\begin{aligned}
\hat{x} \hat{\Pi} \ket{x_0}
&= – \hat{\Pi} \hat{x} \ket{x_0} \\
&= – \hat{\Pi} x_0 \ket{x_0} \\
&= – x_0 \hat{\Pi} \ket{x_0}
\end{aligned}

so

\label{eqn:qmLecture11:500}
\hat{\Pi} \ket{x_0} = \ket{-x_0}.

Acting on a wave function

\label{eqn:qmLecture11:520}
\begin{aligned}
\bra{x} \hat{\Pi} \ket{\psi}
&=
\braket{-x}{\psi} \\
&= \psi(-x).
\end{aligned}

What does this mean for eigenfunctions. Eigenfunctions are supposed to form irreducible representations of the group. The group has just two elements

\label{eqn:qmLecture11:540}
\setlr{ 1, \hat{\Pi} },

where $$\hat{\Pi}^2 = 1$$.

Suppose we have a Hamiltonian

\label{eqn:qmLecture11:560}
H = \frac{\hat{p}^2}{2m} + V(\hat{x}),

where $$V(\hat{x})$$ is even ( $$\antisymmetric{V(\hat{x})}{\hat{\Pi} } = 0$$ ). The squared momentum commutes with the parity operator

\label{eqn:qmLecture11:580}
\begin{aligned}
\antisymmetric{\hat{p}^2}{\hat{\Pi}}
&=
\hat{p}^2 \hat{\Pi}
– \hat{\Pi} \hat{p}^2 \\
&=
\hat{p}^2 \hat{\Pi}
– (\hat{\Pi} \hat{p}) \hat{p} \\
&=
\hat{p}^2 \hat{\Pi}
-(- \hat{p} \hat{\Pi}) \hat{p} \\
&=
\hat{p}^2 \hat{\Pi}
+ \hat{p} (-\hat{p} \hat{\Pi}) \\
&=
0.
\end{aligned}

Only two functions are possible in the symmetry set $$\setlr{ \Psi(x), \hat{\Pi} \Psi(x) }$$, since

\label{eqn:qmLecture11:600}
\begin{aligned}
\hat{\Pi}^2 \Psi(x)
&= \hat{\Pi} \Psi(-x) \\
&= \Psi(x).
\end{aligned}

This symmetry severely restricts the possible solutions, making it so there can be only one dimensional forms of this problem with solutions that are either even or odd respectively

\label{eqn:qmLecture11:620}
\begin{aligned}
\phi_e(x) &= \psi(x ) + \psi(-x) \\
\phi_o(x) &= \psi(x ) – \psi(-x).
\end{aligned}

# References

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

## Unionville public school. Acting even more like a jail.

October 26, 2015 Incoherent ramblings No comments , ,

I dropped off Karl’s lunch on the way to work (we didn’t have anything he would eat, so I made him something when we got back).

The school is more and more like a jail. In addition to the asinine security system, the secretary today wouldn’t even let me into the school office to drop off Karl’s lunch. She came to the door to get it, instead of letting me in for a few seconds.

I view the security system at the school as pandering to idiotic media fear porn. Implemented board wide, I’m sure that some security company is making bucket loads of cash at our expense.

Perhaps they wouldn’t let me into the school because I didn’t play their Oh Canada conformity training game, and have been open stating that their multiplication teaching methods are stupid. I’m definitely a bad influence. The new-math “four ways of multiplying” are great for making Karl (and other kids) confused, but are excellent ways of ensuring that we’ll have another generation of kids that have to use a calculator to do basic math, and will shortly live in a third world country with respect to the sciences.

## Degeneracy in non-commuting observables that both commute with the Hamiltonian.

October 22, 2015 phy1520 No comments ,

In problem 1.17 of [2] we are to show that non-commuting operators that both commute with the Hamiltonian, have, in general, degenerate energy eigenvalues. That is

\label{eqn:angularMomentumAndCentralForceCommutators:320}
[A,H] = [B,H] = 0,

but

\label{eqn:angularMomentumAndCentralForceCommutators:340}
[A,B] \ne 0.

### Matrix example of non-commuting commutators

I thought perhaps the problem at hand would be easier if I were to construct some example matrices representing operators that did not commute, but did commuted with a Hamiltonian. I came up with

\label{eqn:angularMomentumAndCentralForceCommutators:360}
\begin{aligned}
A &=
\begin{bmatrix}
\sigma_z & 0 \\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix} \\
B &=
\begin{bmatrix}
\sigma_x & 0 \\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
\end{bmatrix} \\
H &=
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
\end{aligned}

This system has $$\antisymmetric{A}{H} = \antisymmetric{B}{H} = 0$$, and

\label{eqn:angularMomentumAndCentralForceCommutators:380}
\antisymmetric{A}{B}
=
\begin{bmatrix}
0 & 2 & 0 \\
-2 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}

There is one shared eigenvector between all of $$A, B, H$$

\label{eqn:angularMomentumAndCentralForceCommutators:400}
\ket{3} =
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}.

The other eigenvectors for $$A$$ are
\label{eqn:angularMomentumAndCentralForceCommutators:420}
\begin{aligned}
\ket{a_1} &=
\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix} \\
\ket{a_2} &=
\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix},
\end{aligned}

and for $$B$$
\label{eqn:angularMomentumAndCentralForceCommutators:440}
\begin{aligned}
\ket{b_1} &=
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix} \\
\ket{b_2} &=
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix},
\end{aligned}

This clearly has the degeneracy sought.

Looking to [1], it appears that it is possible to construct an even simpler example. Let

\label{eqn:angularMomentumAndCentralForceCommutators:460}
\begin{aligned}
A &=
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix} \\
B &=
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix} \\
H &=
\begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}.
\end{aligned}

Here $$\antisymmetric{A}{B} = -A$$, and $$\antisymmetric{A}{H} = \antisymmetric{B}{H} = 0$$, but the Hamiltonian isn’t interesting at all physically.

A less boring example builds on this. Let

\label{eqn:angularMomentumAndCentralForceCommutators:480}
\begin{aligned}
A &=
\begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1
\end{bmatrix} \\
B &=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1
\end{bmatrix} \\
H &=
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}.
\end{aligned}

Here $$\antisymmetric{A}{B} \ne 0$$, and $$\antisymmetric{A}{H} = \antisymmetric{B}{H} = 0$$. I don’t see a way for any exception to be constructed.

### The problem

The concrete examples above give some intuition for solving the more abstract problem. Suppose that we are working in a basis that simulaneously diagonalizes operator $$A$$ and the Hamiltonian $$H$$. To make life easy consider the simplest case where this basis is also an eigenbasis for the second operator $$B$$ for all but two of that operators eigenvectors. For such a system let’s write

\label{eqn:angularMomentumAndCentralForceCommutators:160}
\begin{aligned}
H \ket{1} &= \epsilon_1 \ket{1} \\
H \ket{2} &= \epsilon_2 \ket{2} \\
A \ket{1} &= a_1 \ket{1} \\
A \ket{2} &= a_2 \ket{2},
\end{aligned}

where $$\ket{1}$$, and $$\ket{2}$$ are not eigenkets of $$B$$. Because $$B$$ also commutes with $$H$$, we must have

\label{eqn:angularMomentumAndCentralForceCommutators:180}
\begin{aligned}
H B \ket{1}
&= H \ket{n}\bra{n} B \ket{1} \\
&= \epsilon_n \ket{n} B_{n 1},
\end{aligned}

and
\label{eqn:angularMomentumAndCentralForceCommutators:200}
\begin{aligned}
B H \ket{1}
&= B \epsilon_1 \ket{1} \\
&= \epsilon_1 \ket{n}\bra{n} B \ket{1} \\
&= \epsilon_1 \ket{n} B_{n 1}.
\end{aligned}

The commutator is
\label{eqn:angularMomentumAndCentralForceCommutators:220}
\antisymmetric{B}{H} \ket{1}
=
\lr{ \epsilon_1 – \epsilon_n } \ket{n} B_{n 1}.

Similarily
\label{eqn:angularMomentumAndCentralForceCommutators:240}
\antisymmetric{B}{H} \ket{2}
=
\lr{ \epsilon_2 – \epsilon_n } \ket{n} B_{n 2}.

For those kets $$\ket{m} \in \setlr{ \ket{3}, \ket{4}, \cdots }$$ that are eigenkets of $$B$$, with $$B \ket{m} = b_m \ket{m}$$, we have

\label{eqn:angularMomentumAndCentralForceCommutators:280}
\begin{aligned}
\antisymmetric{B}{H} \ket{m}
&=
B \epsilon_m \ket{m} – H b_m \ket{m} \\
&=
b_m \epsilon_m \ket{m} – \epsilon_m b_m \ket{m} \\
&=
0.
\end{aligned}

If the commutator is zero, then we require all its matrix elements
\label{eqn:angularMomentumAndCentralForceCommutators:260}
\begin{aligned}
\bra{1} \antisymmetric{B}{H} \ket{1} &= \lr{ \epsilon_1 – \epsilon_1 } B_{1 1} \\
\bra{2} \antisymmetric{B}{H} \ket{1} &= \lr{ \epsilon_1 – \epsilon_2 } B_{2 1} \\
\bra{1} \antisymmetric{B}{H} \ket{2} &= \lr{ \epsilon_2 – \epsilon_1 } B_{1 2} \\
\bra{2} \antisymmetric{B}{H} \ket{2} &= \lr{ \epsilon_2 – \epsilon_2 } B_{2 2},
\end{aligned}

to be zero. Because of (15) only the matrix elements with respect to states $$\ket{1}, \ket{2}$$ need be considered. Two of the matrix elements above are clearly zero, regardless of the values of $$B_{1 1}$$, and $$B_{2 2}$$, and for the other two to be zero, we must either have

• $$B_{2 1} = B_{1 2} = 0$$, or
• $$\epsilon_1 = \epsilon_2$$.

If the first condition were true we would have

\label{eqn:angularMomentumAndCentralForceCommutators:300}
\begin{aligned}
B \ket{1}
&=
\ket{n}\bra{n} B \ket{1} \\
&=
\ket{n} B_{n 1} \\
&=
\ket{1} B_{1 1},
\end{aligned}

and $$B \ket{2} = B_{2 2} \ket{2}$$. This contradicts the requirement that $$\ket{1}, \ket{2}$$ not be eigenkets of $$B$$, leaving only the second option. That second option means there must be a degeneracy in the system.

# References

[1] Ronald M. Aarts. Commuting Matrices, 2015. URL http://mathworld.wolfram.com/CommutingMatrices.html. [Online; accessed 22-Oct-2015].

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

## Greetings to new Markham-Unionville conservative rep Bob Saroya.

The Honourable Bob Saroya,

Congratulations for your success winning your position in parliament for our district. Unfortunately for you, this means that you are also obligated to represent me. I did not vote for you. I effectively voted none-of-the-above, by voting for the Green Party representative Elvin Kao, knowing full well that he was too inexperienced to be successful. I’m not sad that he was not successful, because his emails expressed a belligerence in foreign policy matters that turned my stomach. He got my vote in the end because he answered my mailed questions, unlike yourself and unlike your like your Liberal running mate Mrs. Jiang. Your Liberal predecessor in Markham Unionville, Mr John McCallum, as a representative, has a 2/5 score for answering correspondence. However, that small non-zero portion of his positive score can really be attributed to his administrative assistant, Mr. Nicholson.

I have a very poor opinion of politics, and politicians, and as my representative, you have an infinitesimal chance of changing that position. Given that you did not answer correspondence sent while running, I do not expect there is much chance of that occurring. What I do expect is to see more voting for increased government when you have the chance. Take bill-C51, the police state bill, tabled by members of Minister Blainey’s hierarchy that effectively gave themselves paychecks and power. Despite knowing that there was a massive objection to this bill, so much that it probably cost the Conservatives their majority this election, it was still forced through. It seems to me that this bill owes its success to the putrid non-democratic policy of the party whip. Because the Liberal leadership was also deluded into thinking there was rational for a new Canadian police state, the average joe like me will slowly start to see how big government in Canada will exploit this to spy on all it’s could-be-terrorist citizens. Welcome to the new Canada, where fear mongering trumps rationality.

I am not surprised that fear mongering is so successful here. We are conditioned to be conformist and patriotic. Our schools are like jails, locked down, with active shooter drills, and require police checks of parent volunteers that discourage involvement of non-school personal in raising the little obedient soldiers who stand for their dose of Oh Canada every morning.

The desired product seems to be unthinking patriots that will be willing to go off to war to bomb the current bad-guy brown people de-jour. Such people are guilty of having been selected as targets by the USA and their North bordered military lackey, but this always occurs in a predictable way. First they are sold or given weapons, then later declared to be enemies. It’s a beautiful scheme to keep the armaments industry going.

What do we not see taught in schools? We don’t see anybody taught to think for themselves. We don’t see basics taught. We are raising a generation of kids that have to use a calculator because they haven’t learned their timetables, and have had the new math shoved down their throats. They won’t be able to multiply the way our generation, our parents, and grandparents learned, but by the time they have been subjected to the matrix method of multiplication, the line crossing method of multiplication, and the array method of multiplication … they will give up. Math will be viewed as too confusing, and we will have a generation of math illiteracy. Our generation has front row seats to watching our first world status get flushed down the toilet.

Ranting aside, I have a couple questions.

1) Should Mr Trudeau follow through with his unlikely promise to repeal parts of bill-C51, imagine that the party whip was given the day off, and you were given the chance to vote in a democratic fashion instead of having to obediently follow the party line like a good Oh Canada trained compliant patriot, how would you vote?

Yes, I know that it is unlikely that the party whip would be given the day off. A more likely scenario is that he/she gets correctly identified as a source of anti-free-speech and anti-democracy, and gets shot by terrorists inspired by Canadian bombing throughout the world.

2) The Quirks and Quarks radio show hosted by Bob McDonald hosted a political debate on science topics. This was a pretty putrid affair, like all politician debates, and the point of the debaters seemed to be to win points for most spin, and least fact.

One point debated seemed to be fact checkable. Opposition members brought up the destruction of one or more Canadian research libraries by the conservative party. The conservative party rep claimed that they were lying, and said that the libraries were not destroyed but digitized to be made available for all. The opposition then predictably claimed that the conservative was also lying.

How many research libraries were destroyed? Where are the digitized copies of all the books available? Was that a 100% digitization, or a partial digitization. If partial, where is the policy used to decide what was destroyed, and are there records showing what was destroyed?

Sincerely,

Peeter Joot

## PHY1520H Graduate Quantum Mechanics. Lecture 10: 1D Dirac scattering off potential step. Taught by Prof. Arun Paramekanti

October 20, 2015 phy1520 No comments , ,

### 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.

### Dirac scattering off a potential step

For the non-relativistic case we have

\label{eqn:qmLecture10:20}
\begin{aligned}
E < V_0 &\Rightarrow T = 0, R = 1 \\ E > V_0 &\Rightarrow T > 0, R < 1.
\end{aligned}

What happens for a relativistic 1D particle?

Referring to fig. 1.

fig. 1. Potential step

the region I Hamiltonian is

\label{eqn:qmLecture10:40}
H =
\begin{bmatrix}
\hat{p} c & m c^2 \\
m c^2 & – \hat{p} c
\end{bmatrix},

for which the solution is

\label{eqn:qmLecture10:60}
\Phi = e^{i k_1 x }
\begin{bmatrix}
\cos \theta_1 \\
\sin \theta_1
\end{bmatrix},

where
\label{eqn:qmLecture10:80}
\begin{aligned}
\cos 2 \theta_1 &= \frac{ \Hbar c k_1 }{E_{k_1}} \\
\sin 2 \theta_1 &= \frac{ m c^2 }{E_{k_1}} \\
\end{aligned}

To consider the $$k_1 < 0$$ case, note that

\label{eqn:qmLecture10:100}
\begin{aligned}
\cos^2 \theta_1 – \sin^2 \theta_1 &= \cos 2 \theta_1 \\
2 \sin\theta_1 \cos\theta_1 &= \sin 2 \theta_1
\end{aligned}

so after flipping the signs on all the $$k_1$$ terms we find for the reflected wave

\label{eqn:qmLecture10:120}
\Phi = e^{-i k_1 x}
\begin{bmatrix}
\sin\theta_1 \\
\cos\theta_1
\end{bmatrix}.

FIXME: this reasoning doesn’t entirely make sense to me. Make sense of this by trying this solution as was done for the form of the incident wave solution.

The region I wave has the form

\label{eqn:qmLecture10:140}
\Phi_I
=
A e^{i k_1 x}
\begin{bmatrix}
\cos\theta_1 \\
\sin\theta_1 \\
\end{bmatrix}
+
B e^{-i k_1 x}
\begin{bmatrix}
\sin\theta_1 \\
\cos\theta_1 \\
\end{bmatrix}.

By the time we are done we want to have computed the reflection coefficient

\label{eqn:qmLecture10:160}
R =
\frac{\Abs{B}^2}{\Abs{A}^2}.

The region I energy is

\label{eqn:qmLecture10:180}
E = \sqrt{ \lr{ m c^2}^2 + \lr{ \Hbar c k_1 }^2 }.

We must have
\label{eqn:qmLecture10:200}
E
=
\sqrt{ \lr{ m c^2}^2 + \lr{ \Hbar c k_2 }^2 } + V_0
=
\sqrt{ \lr{ m c^2}^2 + \lr{ \Hbar c k_1 }^2 },

so

\label{eqn:qmLecture10:220}
\begin{aligned}
\lr{ \Hbar c k_2 }^2
&=
\lr{ E – V_0 }^2 – \lr{ m c^2}^2 \\
&=
\underbrace{\lr{ E – V_0 + m c }}_{r_1}\underbrace{\lr{ E – V_0 – m c }}_{r_2}.
\end{aligned}

The $$r_1$$ and $$r_2$$ branches are sketched in fig. 2.

fig. 2. Energy signs

For low energies, we have a set of potentials for which we will have propagation, despite having a potential barrier. For still higher values of the potential barrier the product $$r_1 r_2$$ will be negative, so the solutions will be decaying. Finally, for even higher energies, there will again be propagation.

The non-relativistic case is sketched in fig. 3.

fig. 3. Effects of increasing potential for non-relativistic case

For the relativistic case we must consider three different cases, sketched in fig 4, fig 5, and fig 6 respectively. For the low potential energy, a particle with positive group velocity (what we’ve called right moving) can be matched to an equal energy portion of the potential shifted parabola in region II. This is a case where we have transmission, but no antiparticle creation. There will be an energy region where the region II wave function has only a dissipative term, since there is no region of either of the region II parabolic branches available at the incident energy. When the potential is shifted still higher so that $$V_0 > E + m c^2$$, a positive group velocity in region I with a given energy can be matched to an antiparticle branch in the region II parabolic energy curve.

Fig 4. Low potential energy

fig. 5. High enough potential energy for no propagation

fig 6. High potential energy

### Boundary value conditions

We want to ensure that the current across the barrier is conserved (no particles are lost), as sketched in fig. 7.

fig. 7. Transmitted, reflected and incident components.

Recall that given a wave function

\label{eqn:qmLecture10:240}
\Psi =
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix},

the density and currents are respectively

\label{eqn:qmLecture10:260}
\begin{aligned}
\rho &= \psi_1^\conj \psi_1 + \psi_2^\conj \psi_2 \\
j &= \psi_1^\conj \psi_1 – \psi_2^\conj \psi_2
\end{aligned}

Matching boundary value conditions requires

1. For both the relativistic and non-relativistic cases we must have\label{eqn:qmLecture10:280}
\Psi_{\textrm{L}} = \Psi_{\textrm{R}}, \qquad \mbox{at $$x = 0$$.}
2. For the non-relativistic case we want
\label{eqn:qmLecture10:300}
\int_{-\epsilon}^\epsilon -\frac{\Hbar^2}{2m} \PDSq{x}{\Psi} =
{\int_{-\epsilon}^\epsilon \lr{ E – V(x) } \Psi(x)}.
The RHS integral is zero, so

\label{eqn:qmLecture10:320}
-\frac{\Hbar^2}{2m} \lr{ \evalbar{\PD{x}{\Psi}}{{\textrm{R}}} – \evalbar{\PD{x}{\Psi}}{{\textrm{L}}} } = 0.

We have to match

For the relativistic case

\label{eqn:qmLecture10:460}
-i \Hbar \sigma_z \int_{-\epsilon}^\epsilon \PD{x}{\Psi} +
{m c^2 \sigma_x \int_{-\epsilon}^\epsilon \psi}
=
{\int_{-\epsilon}^\epsilon \lr{ E – V_0 } \psi},

the second two integrals are wiped out, so

\label{eqn:qmLecture10:340}
-i \Hbar c \sigma_z \lr{ \psi(\epsilon) – \psi(-\epsilon) }
=
-i \Hbar c \sigma_z \lr{ \psi_{\textrm{R}} – \psi_{\textrm{L}} }.

so we must match

\label{eqn:qmLecture10:360}
\sigma_z \psi_{\textrm{R}} = \sigma_z \psi_{\textrm{L}} .

It appears that things are simpler, because we only have to match the wave function values at the boundary, and don’t have to match the derivatives too. However, we have a two component wave function, so there are still two tasks.

### Solving the system

Let’s look for a solution for the $$E + m c^2 > V_0$$ case on the right branch, as sketched in fig. 8.

fig. 8. High potential region. Anti-particle transmission.

While the right branch in this case is left going, this might work out since that is an antiparticle. We could try both.

Try

\label{eqn:qmLecture10:480}
\Psi_{II} = D e^{i k_2 x}
\begin{bmatrix}
-\sin\theta_2 \\
\cos\theta_2
\end{bmatrix}.

This is justified by

\label{eqn:qmLecture10:500}
+E \rightarrow
\begin{bmatrix}
\cos\theta \\
\sin\theta
\end{bmatrix},

so

\label{eqn:qmLecture10:520}
-E \rightarrow
\begin{bmatrix}
-\sin\theta \\
\cos\theta \\
\end{bmatrix}

At $$x = 0$$ the exponentials vanish, so equating the waves at that point means

\label{eqn:qmLecture10:380}
\begin{bmatrix}
\cos\theta_1 \\
\sin\theta_1 \\
\end{bmatrix}
+
\frac{B}{A}
\begin{bmatrix}
\sin\theta_1 \\
\cos\theta_1 \\
\end{bmatrix}
=
\frac{D}{A}
\begin{bmatrix}
-\sin\theta_2 \\
\cos\theta_2
\end{bmatrix}.

Solving this yields

\label{eqn:qmLecture10:400}
\frac{B}{A} = – \frac{\cos(\theta_1 – \theta_2)}{\sin(\theta_1 + \theta_2)}.

This yields

\label{eqn:qmLecture10:420}
\boxed{
R = \frac{1 + \cos( 2 \theta_1 – 2 \theta_2) }{1 – \cos( 2 \theta_1 – 2 \theta_2)}.
}

As $$V_0 \rightarrow \infty$$ this simplifies to

\label{eqn:qmLecture10:440}
R = \frac{ E – \sqrt{ E^2 – \lr{ m c^2 }^2 } }{ E + \sqrt{ E^2 – \lr{ m c^2 }^2 } }.

Filling in the details for these results part of problem set 4.

## Second update of aggregate notes for phy1520, Graduate Quantum Mechanics

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.

## Plane wave ground state expectation for SHO

Problem [1] 2.18 is, for a 1D SHO, show that

\label{eqn:exponentialExpectationGroundState:20}
\bra{0} e^{i k x} \ket{0} = \exp\lr{ -k^2 \bra{0} x^2 \ket{0}/2 }.

Despite the simple appearance of this problem, I found this quite involved to show. To do so, start with a series expansion of the expectation

\label{eqn:exponentialExpectationGroundState:40}
\bra{0} e^{i k x} \ket{0}
=
\sum_{m=0}^\infty \frac{(i k)^m}{m!} \bra{0} x^m \ket{0}.

Let

\label{eqn:exponentialExpectationGroundState:60}
X = \lr{ a + a^\dagger },

so that

\label{eqn:exponentialExpectationGroundState:80}
x
= \sqrt{\frac{\Hbar}{2 \omega m}} X
= \frac{x_0}{\sqrt{2}} X.

Consider the first few values of $$\bra{0} X^n \ket{0}$$

\label{eqn:exponentialExpectationGroundState:100}
\begin{aligned}
\bra{0} X \ket{0}
&=
\bra{0} \lr{ a + a^\dagger } \ket{0} \\
&=
\braket{0}{1} \\
&=
0,
\end{aligned}

\label{eqn:exponentialExpectationGroundState:120}
\begin{aligned}
\bra{0} X^2 \ket{0}
&=
\bra{0} \lr{ a + a^\dagger }^2 \ket{0} \\
&=
\braket{1}{1} \\
&=
1,
\end{aligned}

\label{eqn:exponentialExpectationGroundState:140}
\begin{aligned}
\bra{0} X^3 \ket{0}
&=
\bra{0} \lr{ a + a^\dagger }^3 \ket{0} \\
&=
\bra{1} \lr{ \sqrt{2} \ket{2} + \ket{0} } \\
&=
0.
\end{aligned}

Whenever the power $$n$$ in $$X^n$$ is even, the braket can be split into a bra that has only contributions from odd eigenstates and a ket with even eigenstates. We conclude that $$\bra{0} X^n \ket{0} = 0$$ when $$n$$ is odd.

Noting that $$\bra{0} x^2 \ket{0} = \ifrac{x_0^2}{2}$$, this leaves

\label{eqn:exponentialExpectationGroundState:160}
\begin{aligned}
\bra{0} e^{i k x} \ket{0}
&=
\sum_{m=0}^\infty \frac{(i k)^{2 m}}{(2 m)!} \bra{0} x^{2m} \ket{0} \\
&=
\sum_{m=0}^\infty \frac{(i k)^{2 m}}{(2 m)!} \lr{ \frac{x_0^2}{2} }^m \bra{0} X^{2m} \ket{0} \\
&=
\sum_{m=0}^\infty \frac{1}{(2 m)!} \lr{ -k^2 \bra{0} x^2 \ket{0} }^m \bra{0} X^{2m} \ket{0}.
\end{aligned}

This problem is now reduced to showing that

\label{eqn:exponentialExpectationGroundState:180}
\frac{1}{(2 m)!} \bra{0} X^{2m} \ket{0} = \inv{m! 2^m},

or

\label{eqn:exponentialExpectationGroundState:200}
\begin{aligned}
\bra{0} X^{2m} \ket{0}
&= \frac{(2m)!}{m! 2^m} \\
&= \frac{ (2m)(2m-1)(2m-2) \cdots (2)(1) }{2^m m!} \\
&= \frac{ 2^m (m)(2m-1)(m-1)(2m-3)(m-2) \cdots (2)(3)(1)(1) }{2^m m!} \\
&= (2m-1)!!,
\end{aligned}

where $$n!! = n(n-2)(n-4)\cdots$$.

It looks like $$\bra{0} X^{2m} \ket{0}$$ can be expanded by inserting an identity operator and proceeding recursively, like

\label{eqn:exponentialExpectationGroundState:220}
\begin{aligned}
\bra{0} X^{2m} \ket{0}
&=
\bra{0} X^2 \lr{ \sum_{n=0}^\infty \ket{n}\bra{n} } X^{2m-2} \ket{0} \\
&=
\bra{0} X^2 \lr{ \ket{0}\bra{0} + \ket{2}\bra{2} } X^{2m-2} \ket{0} \\
&=
\bra{0} X^{2m-2} \ket{0} + \bra{0} X^2 \ket{2} \bra{2} X^{2m-2} \ket{0}.
\end{aligned}

This has made use of the observation that $$\bra{0} X^2 \ket{n} = 0$$ for all $$n \ne 0,2$$. The remaining term includes the factor

\label{eqn:exponentialExpectationGroundState:240}
\begin{aligned}
\bra{0} X^2 \ket{2}
&=
\bra{0} \lr{a + a^\dagger}^2 \ket{2} \\
&=
\lr{ \bra{0} + \sqrt{2} \bra{2} } \ket{2} \\
&=
\sqrt{2},
\end{aligned}

Since $$\sqrt{2} \ket{2} = \lr{a^\dagger}^2 \ket{0}$$, the expectation of interest can be written

\label{eqn:exponentialExpectationGroundState:260}
\bra{0} X^{2m} \ket{0}
=
\bra{0} X^{2m-2} \ket{0} + \bra{0} a^2 X^{2m-2} \ket{0}.

How do we expand the second term. Let’s look at how $$a$$ and $$X$$ commute

\label{eqn:exponentialExpectationGroundState:280}
\begin{aligned}
a X
&=
\antisymmetric{a}{X} + X a \\
&=
\antisymmetric{a}{a + a^\dagger} + X a \\
&=
\antisymmetric{a}{a^\dagger} + X a \\
&=
1 + X a,
\end{aligned}

\label{eqn:exponentialExpectationGroundState:300}
\begin{aligned}
a^2 X
&=
a \lr{ a X } \\
&=
a \lr{ 1 + X a } \\
&=
a + a X a \\
&=
a + \lr{ 1 + X a } a \\
&=
2 a + X a^2.
\end{aligned}

Proceeding to expand $$a^2 X^n$$ we find
\label{eqn:exponentialExpectationGroundState:320}
\begin{aligned}
a^2 X^3 &= 6 X + 6 X^2 a + X^3 a^2 \\
a^2 X^4 &= 12 X^2 + 8 X^3 a + X^4 a^2 \\
a^2 X^5 &= 20 X^3 + 10 X^4 a + X^5 a^2 \\
a^2 X^6 &= 30 X^4 + 12 X^5 a + X^6 a^2.
\end{aligned}

It appears that we have
\label{eqn:exponentialExpectationGroundState:340}
\antisymmetric{a^2 X^n}{X^n a^2} = \beta_n X^{n-2} + 2 n X^{n-1} a,

where

\label{eqn:exponentialExpectationGroundState:360}
\beta_n = \beta_{n-1} + 2 (n-1),

and $$\beta_2 = 2$$. Some goofing around shows that $$\beta_n = n(n-1)$$, so the induction hypothesis is

\label{eqn:exponentialExpectationGroundState:380}
\antisymmetric{a^2 X^n}{X^n a^2} = n(n-1) X^{n-2} + 2 n X^{n-1} a.

Let’s check the induction
\label{eqn:exponentialExpectationGroundState:400}
\begin{aligned}
a^2 X^{n+1}
&=
a^2 X^{n} X \\
&=
\lr{ n(n-1) X^{n-2} + 2 n X^{n-1} a + X^n a^2 } X \\
&=
n(n-1) X^{n-1} + 2 n X^{n-1} a X + X^n a^2 X \\
&=
n(n-1) X^{n-1} + 2 n X^{n-1} \lr{ 1 + X a } + X^n \lr{ 2 a + X a^2 } \\
&=
n(n-1) X^{n-1} + 2 n X^{n-1} + 2 n X^{n} a
+ 2 X^n a
+ X^{n+1} a^2 \\
&=
X^{n+1} a^2 + (2 + 2 n) X^{n} a + \lr{ 2 n + n(n-1) } X^{n-1} \\
&=
X^{n+1} a^2 + 2(n + 1) X^{n} a + (n+1) n X^{n-1},
\end{aligned}

which concludes the induction, giving

\label{eqn:exponentialExpectationGroundState:420}
\bra{ 0 } a^2 X^{n} \ket{0 } = n(n-1) \bra{0} X^{n-2} \ket{0},

and

\label{eqn:exponentialExpectationGroundState:440}
\bra{0} X^{2m} \ket{0}
=
\bra{0} X^{2m-2} \ket{0} + (2m-2)(2m-3) \bra{0} X^{2m-4} \ket{0}.

Let

\label{eqn:exponentialExpectationGroundState:460}
\sigma_{n} = \bra{0} X^n \ket{0},

so that the recurrence relation, for $$2n \ge 4$$ is

\label{eqn:exponentialExpectationGroundState:480}
\sigma_{2n} = \sigma_{2n -2} + (2n-2)(2n-3) \sigma_{2n -4}

We want to show that this simplifies to

\label{eqn:exponentialExpectationGroundState:500}
\sigma_{2n} = (2n-1)!!

The first values are

\label{eqn:exponentialExpectationGroundState:540}
\sigma_0 = \bra{0} X^0 \ket{0} = 1

\label{eqn:exponentialExpectationGroundState:560}
\sigma_2 = \bra{0} X^2 \ket{0} = 1

which gives us the right result for the first term in the induction

\label{eqn:exponentialExpectationGroundState:580}
\begin{aligned}
\sigma_4
&= \sigma_2 + 2 \times 1 \times \sigma_0 \\
&= 1 + 2 \\
&= 3!!
\end{aligned}

For the general induction term, consider

\label{eqn:exponentialExpectationGroundState:600}
\begin{aligned}
\sigma_{2n + 2}
&= \sigma_{2n} + 2 n (2n – 1) \sigma_{2n -2} \\
&= (2n-1)!! + 2n ( 2n – 1) (2n -3)!! \\
&= (2n + 1) (2n -1)!! \\
&= (2n + 1)!!,
\end{aligned}

which completes the final induction. That was also the last thing required to complete the proof, so we are done!

# References

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

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

### 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:

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

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

In classical mechanics we have

\label{eqn:qmLecture7:40}
\oint p dq

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

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

We work with what is now called the Landau gauge

\label{eqn:qmLecture7:60}
\BA = \lr{ 0, B x, 0 }

This gives

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

An alternate gauge choice, the symmetric gauge, is

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

that also has the same magnetic field

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

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

\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}

We can solve after noting that

\label{eqn:qmLecture7:180}
\antisymmetric{\hat{p}_y}{H} = 0

means that

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

The eigensystem

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

becomes

\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).

This reduced Hamiltonian can be rewritten as

\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

where

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

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

and

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

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

\label{eqn:qmLecture7:340}
k_y L_y = 2 \pi n,

or

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

gives

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

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

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

or

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

That is

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

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

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

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.