## Non-interacting particles in a box

December 17, 2015 phy1520 No comments , ,

### Q: [1] pr 4.1

Calculate the three lowest energy levels and their degeneracies for equal mass distinguishable spin half particles in a box of length $$L$$. Consider

(a) Two particles.
(b) Three particles.
(c) Four particles.

### A: part (a)

The problem statement doesn’t include the dimensionality of the box. The simplest case is the one dimensional box, for which the wave function of one particle is

\label{eqn:noninteractingParticlesInABox:20}
\psi_1(x) = \sqrt{\frac{2}{L}} \sin\lr{ \frac{n \pi x}{L} },

and the energy of that particle is

\label{eqn:noninteractingParticlesInABox:40}
E = \inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 n^2.

If the box is two dimensional the energy is

\label{eqn:noninteractingParticlesInABox:60}
E = \inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ n_1^2 + n_2^2 },

and if it’s a 3D box, we have

\label{eqn:noninteractingParticlesInABox:80}
E = \inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ n_1^2 + n_2^2 + n_3^2}.

Suppose we are considering the 3D box. In statistical mechanics when we are considering particles Fermions, they are indistinguishable, and thus not allowed to share the same spin state at a given energy level. However, for distinguishable particles, that restriction doesn’t exist, and we can have two (or more) such particles in the lowest order energy state. The lowest such energy is

\label{eqn:noninteractingParticlesInABox:100}
\begin{aligned}
E_{1,1,1 ;1,1,1}
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 6 \times 1^2 } \\
&=
\frac{6}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2.
\end{aligned}

The particle spin states can be any of $$\ket{++}, \ket{+-}, \ket{-+}, \ket{–}$$, so there is a four way degeneracy in the ground state.

the next lowest energy level is

\label{eqn:noninteractingParticlesInABox:120}
\begin{aligned}
E_{1,1,2 ;1,1,1}
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 5 \times 1^2 + 2^2 } \\
&=
\frac{9}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2,
\end{aligned}

where there are $$\binom{6}{1} = 6$$ ways to pick such a state for each variation of spin, for a total $$6 \times 4 = 24$$ way degeneracy. Finally, since $$2^2 + 2^2 < 3^2 + 1^2$$, the next lowest energy level is \label{eqn:noninteractingParticlesInABox:140} \begin{aligned} E_{1,2,2 ;1,1,1} &= \inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 4 \times 1^2 + 2 \times 2^2 } \\ &= \frac{12}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2, \end{aligned} with a $$\binom{6}{2} \times 4 = 15 \times 4 = 60$$ way degeneracy for this energy level.

### A: part (b)

For three particles (the two particle case wasn’t actually in the problem statement, but seemed an easier starting place), the lowest energy state for a 3D box is

\label{eqn:noninteractingParticlesInABox:160}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 9 \times 1^2 } \\
&=
\frac{9}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2.
\end{aligned}

There are now $$2^3 = 8$$ variations of spin $$\ket{+++}, \ket{++-}, \cdots$$, so the ground state is 8-way degenerate. Next up is

\label{eqn:noninteractingParticlesInABox:180}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 8 \times 1^2 + 2^2 } \\
&=
\frac{12}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2,
\end{aligned}

where there is a $$\binom{9}{1} \times 8 = 9 \times 8 = 72$$ way degeneracy in this energy level. Finally, the next lowest energy level is

\label{eqn:noninteractingParticlesInABox:200}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 7 \times 1^2 + 2 \times 2^2 } \\
&=
\frac{15}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2,
\end{aligned}

with a $$\binom{9}{2} \times 8 = 36 \times 8 = 288$$ way degeneracy for this energy level.

### A: part (c)

For four particles the lowest energy state for a 3D box is

\label{eqn:noninteractingParticlesInABox:220}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 12 \times 1^2 } \\
&=
\frac{12}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2.
\end{aligned}

There are now $$2^4 = 16$$ variations of spin $$\ket{++++}, \ket{+++-}, \cdots$$, so the ground state is 16-way degenerate. For the second level

\label{eqn:noninteractingParticlesInABox:240}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 11 \times 1^2 + 2^2 } \\
&=
\frac{15}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2,
\end{aligned}

where there is a $$\binom{12}{1} \times 16 = 12 \times 16 = 192$$ way degeneracy in this energy level. Finally, the next lowest energy level is

\label{eqn:noninteractingParticlesInABox:260}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 7 \times 1^2 + 2 \times 2^2 } \\
&=
\frac{15}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2,
\end{aligned}

with a $$\binom{12}{2} \times 16 = 66 \times 16 = 1056$$ way degeneracy for this energy level.

# References

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

## Totally asymmetric potential

December 16, 2015 phy1520 2 comments , , ,

### Q: [1] pr 4.11

(a) Given a time reversal invariant Hamiltonian, show that for any energy eigenket

\label{eqn:totallyAsymmetricPotential:20}
\expectation{\BL} = 0.

(b) If the wave function of such a state is expanded as

\label{eqn:totallyAsymmetricPotential:40}
\sum_{l,m} F_{l m} Y_{l m}(\theta, \phi),

what are the phase restrictions on $$F_{lm}$$?

### A: part (a)

For a time reversal invariant Hamiltonian $$H$$ we have

\label{eqn:totallyAsymmetricPotential:60}
H \Theta = \Theta H.

If $$\ket{\psi}$$ is an energy eigenstate with eigenvalue $$E$$, we have

\label{eqn:totallyAsymmetricPotential:80}
\begin{aligned}
H \Theta \ket{\psi}
&= \Theta H \ket{\psi} \\
&= \lambda \Theta \ket{\psi},
\end{aligned}

so $$\Theta \ket{\psi}$$ is also an eigenvalue of $$H$$, so can only differ from $$\ket{\psi}$$ by a phase factor. That is

\label{eqn:totallyAsymmetricPotential:100}
\begin{aligned}
\ket{\psi’}
&=
\Theta \ket{\psi} \\
&= e^{i\delta} \ket{\psi}.
\end{aligned}

Now consider the expectation of $$\BL$$ with respect to a time reversed state

\label{eqn:totallyAsymmetricPotential:120}
\begin{aligned}
\bra{ \psi’} \BL \ket{\psi’}
&=
\bra{ \psi} \Theta^{-1} \BL \Theta \ket{\psi} \\
&=
\bra{ \psi} (-\BL) \ket{\psi},
\end{aligned}

however, we also have

\label{eqn:totallyAsymmetricPotential:140}
\begin{aligned}
\bra{ \psi’} \BL \ket{\psi’}
&=
\lr{ \bra{ \psi} e^{-i\delta} } \BL \lr{ e^{i\delta} \ket{\psi} } \\
&=
\bra{\psi} \BL \ket{\psi},
\end{aligned}

so we have $$\bra{\psi} \BL \ket{\psi} = -\bra{\psi} \BL \ket{\psi}$$ which is only possible if $$\expectation{\BL} = \bra{\psi} \BL \ket{\psi} = 0$$.

### A: part (b)

Consider the expansion of the wave function of a time reversed energy eigenstate

\label{eqn:totallyAsymmetricPotential:160}
\begin{aligned}
\bra{\Bx} \Theta \ket{\psi}
&=
\bra{\Bx} e^{i\delta} \ket{\psi} \\
&=
e^{i\delta} \braket{\Bx}{\psi},
\end{aligned}

and then consider the same state expanded in the position basis

\label{eqn:totallyAsymmetricPotential:180}
\begin{aligned}
\bra{\Bx} \Theta \ket{\psi}
&=
\bra{\Bx} \Theta \int d^3 \Bx’ \lr{ \ket{\Bx’}\bra{\Bx’} } \ket{\psi} \\
&=
\bra{\Bx} \Theta \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} } \ket{\Bx’} \\
&=
\bra{\Bx} \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \Theta \ket{\Bx’} \\
&=
\bra{\Bx} \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \ket{\Bx’} \\
&=
\int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \braket{\Bx}{\Bx’} \\
&=
\int d^3 \Bx’ \braket{\psi}{\Bx’} \delta(\Bx- \Bx’) \\
&=
\braket{\psi}{\Bx}.
\end{aligned}

This demonstrates a relationship between the wave function and its complex conjugate

\label{eqn:totallyAsymmetricPotential:200}
\braket{\Bx}{\psi} = e^{-i\delta} \braket{\psi}{\Bx}.

Now expand the wave function in the spherical harmonic basis

\label{eqn:totallyAsymmetricPotential:220}
\begin{aligned}
\braket{\Bx}{\psi}
&=
\int d\Omega \braket{\Bx}{\ncap}\braket{\ncap}{\psi} \\
&=
\sum_{lm} F_{lm}(r) Y_{lm}(\theta, \phi) \\
&=
e^{-i\delta}
\lr{
\sum_{lm} F_{lm}(r) Y_{lm}(\theta, \phi) }^\conj \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{lm}(r)}^\conj Y_{lm}^\conj(\theta, \phi) \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{lm}(r)}^\conj (-1)^m Y_{l,-m}(\theta, \phi) \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{l,-m}(r)}^\conj (-1)^m Y_{l,m}(\theta, \phi),
\end{aligned}

so the $$F_{lm}$$ functions are constrained by

\label{eqn:totallyAsymmetricPotential:240}
F_{lm}(r) = e^{-i\delta} \lr{ F_{l,-m}(r)}^\conj (-1)^m.

# References

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

## Commutators for some symmetry operators

December 16, 2015 phy1520 No comments , , ,

### Q: [1] pr 4.2

If $$\mathcal{T}_\Bd$$, $$\mathcal{D}(\ncap, \phi)$$, and $$\pi$$ denote the translation, rotation, and parity operators respectively. Which of the following commute and why

• (a) $$\mathcal{T}_\Bd$$ and $$\mathcal{T}_{\Bd’}$$, translations in different directions.
• (b) $$\mathcal{D}(\ncap, \phi)$$ and $$\mathcal{D}(\ncap’, \phi’)$$, rotations in different directions.
• (c) $$\mathcal{T}_\Bd$$ and $$\pi$$.
• (d) $$\mathcal{D}(\ncap,\phi)$$ and $$\pi$$.

### A: (a)

Consider
\label{eqn:symmetryOperatorCommutators:20}
\begin{aligned}
\mathcal{T}_\Bd \mathcal{T}_{\Bd’} \ket{\Bx}
&=
\mathcal{T}_\Bd \ket{\Bx + \Bd’} \\
&=
\ket{\Bx + \Bd’ + \Bd},
\end{aligned}

and the reverse application of the translation operators
\label{eqn:symmetryOperatorCommutators:40}
\begin{aligned}
\mathcal{T}_{\Bd’} \mathcal{T}_{\Bd} \ket{\Bx}
&=
\mathcal{T}_{\Bd’} \ket{\Bx + \Bd} \\
&=
\ket{\Bx + \Bd + \Bd’} \\
&=
\ket{\Bx + \Bd’ + \Bd}.
\end{aligned}

so we see that

\label{eqn:symmetryOperatorCommutators:60}
\antisymmetric{\mathcal{T}_\Bd}{\mathcal{T}_{\Bd’}} \ket{\Bx} = 0,

for any position state $$\ket{\Bx}$$, and therefore in general they commute.

### A: (b)

That rotations do not commute when they are in different directions (like any two orthogonal directions) need not be belaboured.

### A: (c)

We have
\label{eqn:symmetryOperatorCommutators:80}
\begin{aligned}
\mathcal{T}_\Bd \pi \ket{\Bx}
&=
\mathcal{T}_\Bd \ket{-\Bx} \\
&=
\ket{-\Bx + \Bd},
\end{aligned}

yet
\label{eqn:symmetryOperatorCommutators:100}
\begin{aligned}
\pi \mathcal{T}_\Bd \ket{\Bx}
&=
\pi \ket{\Bx + \Bd} \\
&=
\ket{-\Bx – \Bd} \\
&\ne
\ket{-\Bx + \Bd}.
\end{aligned}

so, in general $$\antisymmetric{\mathcal{T}_\Bd}{\pi} \ne 0$$.

### A: (d)

We have

\label{eqn:symmetryOperatorCommutators:120}
\begin{aligned}
\pi \mathcal{D}(\ncap, \phi) \ket{\Bx}
&=
\pi \mathcal{D}(\ncap, \phi) \pi^\dagger \pi \ket{\Bx} \\
&=
\pi \mathcal{D}(\ncap, \phi) \pi^\dagger \pi \ket{\Bx} \\
&=
\pi \lr{ \sum_{k=0}^\infty \frac{(-i \BJ \cdot \ncap)^k}{k!} } \pi^\dagger \pi \ket{\Bx} \\
&=
\sum_{k=0}^\infty \frac{(-i (\pi \BJ \pi^\dagger) \cdot (\pi \ncap \pi^\dagger) )^k}{k!} \pi \ket{\Bx} \\
&=
\sum_{k=0}^\infty \frac{(-i \BJ \cdot \ncap)^k}{k!} \pi \ket{\Bx} \\
&=
\mathcal{D}(\ncap, \phi) \pi \ket{\Bx},
\end{aligned}

so $$\antisymmetric{\mathcal{D}(\ncap, \phi)}{\pi} \ket{\Bx} = 0$$, for any position state $$\ket{\Bx}$$, and therefore these operators commute in general.

# References

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

## Plane wave and spinor under time reversal

### Q: [1] pr 4.7

1. (a)
Find the time reversed form of a spinless plane wave state in three dimensions.

2. (b)
For the eigenspinor of $$\Bsigma \cdot \ncap$$ expressed in terms of polar and azimuthal angles $$\beta$$ and $$\gamma$$, show that $$-i \sigma_y \chi^\conj(\ncap)$$ has the reversed spin direction.

### A: part (a)

The Hamiltonian for a plane wave is

\label{eqn:timeReversalPlaneWaveAndSpinor:20}
H = \frac{\Bp^2}{2m} = i \PD{t}.

Under time reversal the momentum side transforms as

\label{eqn:timeReversalPlaneWaveAndSpinor:40}
\begin{aligned}
\Theta \frac{\Bp^2}{2m} \Theta^{-1}
&=
\frac{\lr{ \Theta \Bp \Theta^{-1}} \cdot \lr{ \Theta \Bp \Theta^{-1}} }{2m} \\
&=
\frac{(-\Bp) \cdot (-\Bp)}{2m} \\
&=
\frac{\Bp^2}{2m}.
\end{aligned}

The time derivative side of the equation is also time reversal invariant
\label{eqn:timeReversalPlaneWaveAndSpinor:60}
\begin{aligned}
\Theta i \PD{t}{} \Theta^{-1}
&=
\Theta i \Theta^{-1} \Theta \PD{t}{} \Theta^{-1} \\
&=
-i \PD{(-t)}{} \\
&=
i \PD{t}{}.
\end{aligned}

Solutions to this equation are linear combinations of

\label{eqn:timeReversalPlaneWaveAndSpinor:80}
\psi(\Bx, t) = e^{i \Bk \cdot \Bx – i E t/\Hbar},

where $$\Hbar^2 \Bk^2/2m = E$$, the energy of the particle. Under time reversal we have

\label{eqn:timeReversalPlaneWaveAndSpinor:100}
\begin{aligned}
\psi(\Bx, t)
\rightarrow e^{-i \Bk \cdot \Bx + i E (-t)/\Hbar}
&= \lr{ e^{i \Bk \cdot \Bx – i E (-t)/\Hbar} }^\conj \\
&=
\psi^\conj(\Bx, -t)
\end{aligned}

### A: part (b)

The text uses a requirement for time reversal of spin states to show that the Pauli matrix form of the time reversal operator is

\label{eqn:timeReversalPlaneWaveAndSpinor:120}
\Theta = -i \sigma_y K,

where $$K$$ is a complex conjugating operator. The form of the spin up state used in that demonstration was

\label{eqn:timeReversalPlaneWaveAndSpinor:140}
\begin{aligned}
\ket{\ncap ; +}
&= e^{-i S_z \beta/\Hbar} e^{-i S_y \gamma/\Hbar} \ket{+} \\
&= e^{-i \sigma_z \beta/2} e^{-i \sigma_y \gamma/2} \ket{+} \\
&= \lr{ \cos(\beta/2) – i \sigma_z \sin(\beta/2) }
\lr{ \cos(\gamma/2) – i \sigma_y \sin(\gamma/2) } \ket{+} \\
&= \lr{ \cos(\beta/2) – i \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \sin(\beta/2) }
\lr{ \cos(\gamma/2) – i \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \sin(\gamma/2) } \ket{+} \\
&=
\begin{bmatrix}
e^{-i\beta/2} & 0 \\
0 & e^{i \beta/2}
\end{bmatrix}
\begin{bmatrix}
\cos(\gamma/2) & -\sin(\gamma/2) \\
\sin(\gamma/2) & \cos(\gamma/2)
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
&=
\begin{bmatrix}
e^{-i\beta/2} & 0 \\
0 & e^{i \beta/2}
\end{bmatrix}
\begin{bmatrix}
\cos(\gamma/2) \\
\sin(\gamma/2) \\
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\gamma/2)
e^{-i\beta/2}
\\
\sin(\gamma/2)
e^{i \beta/2}
\end{bmatrix}.
\end{aligned}

The state orthogonal to this one is claimed to be

\label{eqn:timeReversalPlaneWaveAndSpinor:180}
\begin{aligned}
\ket{\ncap ; -}
&= e^{-i S_z \beta/\Hbar} e^{-i S_y (\gamma + \pi)/\Hbar} \ket{+} \\
&= e^{-i \sigma_z \beta/2} e^{-i \sigma_y (\gamma + \pi)/2} \ket{+}.
\end{aligned}

We have

\label{eqn:timeReversalPlaneWaveAndSpinor:200}
\begin{aligned}
\cos((\gamma + \pi)/2)
&=
\textrm{Re} e^{i(\gamma + \pi)/2} \\
&=
\textrm{Re} i e^{i\gamma/2} \\
&=
-\sin(\gamma/2),
\end{aligned}

and
\label{eqn:timeReversalPlaneWaveAndSpinor:220}
\begin{aligned}
\sin((\gamma + \pi)/2)
&=
\textrm{Im} e^{i(\gamma + \pi)/2} \\
&=
\textrm{Im} i e^{i\gamma/2} \\
&=
\cos(\gamma/2),
\end{aligned}

so we should have

\label{eqn:timeReversalPlaneWaveAndSpinor:240}
\ket{\ncap ; -}
=
\begin{bmatrix}
-\sin(\gamma/2)
e^{-i\beta/2}
\\
\cos(\gamma/2)
e^{i \beta/2}
\end{bmatrix}.

This looks right, but we can sanity check orthogonality

\label{eqn:timeReversalPlaneWaveAndSpinor:260}
\begin{aligned}
\braket{\ncap ; -}{\ncap ; +}
&=
\begin{bmatrix}
-\sin(\gamma/2)
e^{i\beta/2}
&
\cos(\gamma/2)
e^{-i \beta/2}
\end{bmatrix}
\begin{bmatrix}
\cos(\gamma/2)
e^{-i\beta/2}
\\
\sin(\gamma/2)
e^{i \beta/2}
\end{bmatrix} \\
&=
0,
\end{aligned}

as expected.

The task at hand appears to be the operation on the column representation of $$\ket{\ncap; +}$$ using the Pauli representation of the time reversal operator. That is

\label{eqn:timeReversalPlaneWaveAndSpinor:160}
\begin{aligned}
\Theta \ket{\ncap ; +}
&=
-i \sigma_y K
\begin{bmatrix}
e^{-i\beta/2} \cos(\gamma/2) \\
e^{i \beta/2} \sin(\gamma/2)
\end{bmatrix} \\
&=
-i \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
\begin{bmatrix}
e^{i\beta/2} \cos(\gamma/2) \\
e^{-i \beta/2} \sin(\gamma/2)
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
e^{i\beta/2} \cos(\gamma/2) \\
e^{-i \beta/2} \sin(\gamma/2)
\end{bmatrix} \\
&=
\begin{bmatrix}
-e^{-i \beta/2} \sin(\gamma/2) \\
e^{i\beta/2} \cos(\gamma/2) \\
\end{bmatrix} \\
&= \ket{\ncap ; -},
\end{aligned}

which is the result to be demononstrated.

# References

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

## Another aggregation of notes for phy1520, Graduate Quantum Mechanics.

December 15, 2015 phy1520, Uncategorized No comments

I’ve posted a fourth (pre-exam) update of my aggregate notes for PHY1520H Graduate Quantum Mechanics, taught by Prof. Arun Paramekanti. In addition to what was noted previously, this contains the remainder of my lecture notes, more problem set solutions (not posted separately), and 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.

## Time reversal behavior of solutions to crystal spin Hamiltonian

### Q: [1] pr 4.12

Solve the spin 1 Hamiltonian
\label{eqn:crystalSpinHamiltonianTimeReversal:20}
H = A S_z^2 + B(S_x^2 – S_y^2).

Is this Hamiltonian invariant under time reversal?

How do the eigenkets change under time reversal?

In spinMatrices.nb the matrix representation of the Hamiltonian is found to be
\label{eqn:crystalSpinHamiltonianTimeReversal:40}
H =
\Hbar^2
\begin{bmatrix}
A & 0 & B \\
0 & 0 & 0 \\
B & 0 & A
\end{bmatrix}.

The eigenvalues are
\label{eqn:crystalSpinHamiltonianTimeReversal:60}
\setlr{ 0, A – B, A + B},

and the respective eigenvalues (unnormalized) are

\label{eqn:crystalSpinHamiltonianTimeReversal:80}
\setlr{
\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix},
\begin{bmatrix}
1 \\
0 \\
1 \\
\end{bmatrix}
}.

Under time reversal, the Hamiltonian is

\label{eqn:crystalSpinHamiltonianTimeReversal:100}
H \rightarrow A (-S_z)^2 + B ( (-S_x)^2 – (-S_y)^2 ) = H,

so we expect the eigenkets for this Hamiltonian to vary by at most a phase factor. To check this, first recall that the time reversal action on a spin one state is

\label{eqn:crystalSpinHamiltonianTimeReversal:120}
\Theta \ket{1, m} = (-1)^m \ket{1, -m},

or

\label{eqn:crystalSpinHamiltonianTimeReversal:140}
\begin{aligned}
\Theta \ket{1,1} &= -\ket{1,-1} \\
\Theta \ket{1,0} &= \ket{1,0} \\
\Theta \ket{1,-1} &= -\ket{1,1}.
\end{aligned}

Let’s write the eigenkets respectively as

\label{eqn:crystalSpinHamiltonianTimeReversal:160}
\begin{aligned}
\ket{0} &= \ket{1,0} \\
\ket{A-B} &= -\ket{1,-1} + \ket{1,1} \\
\ket{A+B} &= \ket{1,-1} + \ket{1,1}.
\end{aligned}

Under the reversal operation, we should have

\label{eqn:crystalSpinHamiltonianTimeReversal:180}
\begin{aligned}
\Theta \ket{0} &\rightarrow \ket{1,0} \\
\Theta \ket{A-B} &= +\ket{1,-1} – \ket{1,1} \\
\Theta \ket{A+B} &= -\ket{1,-1} – \ket{1,1}.
\end{aligned}

Up to a sign, the time reversed states match the unreversed states, which makes sense given the Hamiltonian invariance.

# References

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

## Spin three halves spin interaction

### Q: [1] pr 3.33

A spin $$3/2$$ nucleus subjected to an external electric field has an interaction Hamiltonian of the form

\label{eqn:spinThreeHalvesNucleus:20}
H = \frac{e Q}{2 s(s-1) \Hbar^2} \lr{
\lr{\PDSq{x}{\phi}}_0 S_x^2
+\lr{\PDSq{y}{\phi}}_0 S_y^2
+\lr{\PDSq{z}{\phi}}_0 S_z^2
}.

Show that the interaction energy can be written as

\label{eqn:spinThreeHalvesNucleus:40}
A(3 S_z^2 – \BS^2) + B(S_{+}^2 + S_{-}^2).

Find the energy eigenvalues for such a Hamiltonian.

### A:

Reordering
\label{eqn:spinThreeHalvesNucleus:60}
\begin{aligned}
S_{+} &= S_x + i S_y \\
S_{-} &= S_x – i S_y,
\end{aligned}

gives
\label{eqn:spinThreeHalvesNucleus:80}
\begin{aligned}
S_x &= \inv{2} \lr{ S_{+} + S_{-} } \\
S_y &= \inv{2i} \lr{ S_{+} – S_{-} }.
\end{aligned}

The squared spin operators are
\label{eqn:spinThreeHalvesNucleus:100}
\begin{aligned}
S_x^2
&=
\inv{4} \lr{ S_{+}^2 + S_{-}^2 + S_{+} S_{-} + S_{-} S_{+} } \\
&=
\inv{4} \lr{ S_{+}^2 + S_{-}^2 + 2( S_x^2 + S_y^2 ) } \\
&=
\inv{4} \lr{ S_{+}^2 + S_{-}^2 + 2( \BS^2 – S_z^2 ) },
\end{aligned}

\label{eqn:spinThreeHalvesNucleus:120}
\begin{aligned}
S_y^2
&=
-\inv{4} \lr{ S_{+}^2 + S_{-}^2 – S_{+} S_{-} – S_{-} S_{+} } \\
&=
-\inv{4} \lr{ S_{+}^2 + S_{-}^2 – 2( S_x^2 + S_y^2 ) } \\
&=
-\inv{4} \lr{ S_{+}^2 + S_{-}^2 – 2( \BS^2 – S_z^2 ) }.
\end{aligned}

This gives
\label{eqn:spinThreeHalvesNucleus:140}
\begin{aligned}
H &= \frac{e Q}{2 s(s-1) \Hbar^2} \biglr{ \inv{4} \lr{\PDSq{x}{\phi}}_0 \lr{ S_{+}^2 + S_{-}^2 + 2( \BS^2 – S_z^2 ) }
-\lr{\PDSq{y}{\phi}}_0 \lr{ S_{+}^2 + S_{-}^2 – 2( \BS^2 – S_z^2 ) }
+\lr{\PDSq{z}{\phi}}_0 S_z^2 } \\
&= \frac{e Q}{2 s(s-1) \Hbar^2} \biglr{ \inv{4} \lr{ \lr{\PDSq{x}{\phi}}_0 -\lr{\PDSq{y}{\phi}}_0 } \lr{ S_{+}^2 + S_{-}^2 }
+ \inv{2} \lr{ \lr{\PDSq{x}{\phi}}_0 + \lr{\PDSq{y}{\phi}}_0 } \BS^2
+ \lr{ \lr{\PDSq{z}{\phi}}_0 – \inv{2} \lr{\PDSq{x}{\phi}}_0 – \inv{2} \lr{\PDSq{y}{\phi}}_0 } S_z^2
}.
\end{aligned}

For a static electric field we have

\label{eqn:spinThreeHalvesNucleus:160}

but are evaluating it at a point away from the generating charge distribution, so $$\spacegrad^2 \phi = 0$$ at that point. This gives

\label{eqn:spinThreeHalvesNucleus:180}
H
=
\frac{e Q}{4 s(s-1) \Hbar^2}
\biglr{
\inv{2} \lr{ \lr{\PDSq{x}{\phi}}_0 -\lr{\PDSq{y}{\phi}}_0
} \lr{ S_{+}^2 + S_{-}^2 }
+
\lr{
\lr{\PDSq{x}{\phi}}_0 + \lr{\PDSq{y}{\phi}}_0
} (\BS^2 – 3 S_z^2)
},

so
\label{eqn:spinThreeHalvesNucleus:200}
A =
-\frac{e Q}{4 s(s-1) \Hbar^2} \lr{
\lr{\PDSq{x}{\phi}}_0 + \lr{\PDSq{y}{\phi}}_0
}

\label{eqn:spinThreeHalvesNucleus:220}
B =
\frac{e Q}{8 s(s-1) \Hbar^2}
\lr{ \lr{\PDSq{x}{\phi}}_0 – \lr{\PDSq{y}{\phi}}_0 }.

### A: energy eigenvalues

Using sakuraiProblem3.33.nb, matrix representations for the spin three halves operators and the Hamiltonian were constructed with respect to the basis $$\setlr{ \ket{3/2}, \ket{1/2}, \ket{-1/2}, \ket{-3/2} }$$

\label{eqn:spinThreeHalvesNucleus:240}
\begin{aligned}
S_{+} &=
\Hbar
\begin{bmatrix}
0 & \sqrt{3} & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & \sqrt{3} \\
0 & 0 & 0 & 0 \\
\end{bmatrix} \\
S_{-} &=
\Hbar
\begin{bmatrix}
0 & 0 & 0 & 0 \\
\sqrt{3} & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & \sqrt{3} & 0 \\
\end{bmatrix} \\
S_x &=
\Hbar
\begin{bmatrix}
0 & \sqrt{3}/2 & 0 & 0 \\
\sqrt{3}/2 & 0 & 1 & 0 \\
0 & 1 & 0 & \sqrt{3}/2 \\
0 & 0 & \sqrt{3}/2 & 0 \\
\end{bmatrix} \\
S_y &=
i \Hbar
\begin{bmatrix}
0 & -\ifrac{\sqrt{3}}{2} & 0 & 0 \\
\ifrac{\sqrt{3}}{2} & 0 & -1 & 0 \\
0 & 1 & 0 & -\ifrac{\sqrt{3}}{2} \\
0 & 0 & \ifrac{\sqrt{3}}{2} & 0 \\
\end{bmatrix} \\
S_z &=
\frac{\Hbar}{2}
\begin{bmatrix}
3 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -3 \\
\end{bmatrix} \\
H &=
\begin{bmatrix}
3 A & 0 & 2 \sqrt{3} B & 0 \\
0 & -3 A & 0 & 2 \sqrt{3} B \\
2 \sqrt{3} B & 0 & -3 A & 0 \\
0 & 2 \sqrt{3} B & 0 & 3 A \\
\end{bmatrix}.
\end{aligned}

The energy eigenvalues are found to be

\label{eqn:spinThreeHalvesNucleus:260}
E = \pm \Hbar^2 \sqrt{9 A^2 + 12 B^2 },

with two fold degeneracies for each eigenvalue.

# References

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

## Angular momentum expectation values

December 14, 2015 phy1520 No comments ,

### Q: [1] pr 3.18

Compute the expectation values for the first and second powers of the angular momentum operators with respect to states $$\ket{lm}$$.

### A:

We can write the expectation values for the $$L_z$$ powers immediately

\label{eqn:angularMomentumExpectation:20}
\expectation{L_z}
= m \Hbar,

and

\label{eqn:angularMomentumExpectation:40}
\expectation{L_z^2} = (m \Hbar)^2.

For the x and y components first express the operators in terms of the ladder operators.

\label{eqn:angularMomentumExpectation:60}
\begin{aligned}
L_{+} &= L_x + i L_y \\
L_{-} &= L_x – i L_y.
\end{aligned}

Rearranging gives

\label{eqn:angularMomentumExpectation:80}
\begin{aligned}
L_x &= \inv{2} \lr{ L_{+} + L_{-} } \\
L_y &= \inv{2i} \lr{ L_{+} – L_{-} }.
\end{aligned}

The first order expectations $$\expectation{L_x}, \expectation{L_y}$$ are both zero since $$\expectation{L_{+}} = \expectation{L_{-}}$$. For the second order expectation values we have

\label{eqn:angularMomentumExpectation:100}
\begin{aligned}
L_x^2
&= \inv{4} \lr{ L_{+} + L_{-} } \lr{ L_{+} + L_{-} } \\
&= \inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} + L_{+} L_{-} + L_{-} L_{+} } \\
&= \inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} + 2 (L_x^2 + L_y^2) } \\
&= \inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} + 2 (\BL^2 – L_z^2) },
\end{aligned}

and
\label{eqn:angularMomentumExpectation:120}
\begin{aligned}
L_y^2
&= -\inv{4} \lr{ L_{+} – L_{-} } \lr{ L_{+} – L_{-} } \\
&= -\inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} – L_{+} L_{-} – L_{-} L_{+} } \\
&= -\inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} – 2 (L_x^2 + L_y^2) } \\
&= -\inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} – 2 (\BL^2 – L_z^2) }.
\end{aligned}

Any expectation value $$\bra{lm} L_{+} L_{+} \ket{lm}$$ or $$\bra{lm} L_{-} L_{-} \ket{lm}$$ will be zero, leaving

\label{eqn:angularMomentumExpectation:140}
\begin{aligned}
\expectation{L_x^2}
&=
\expectation{L_y^2} \\
&=
\inv{4} \expectation{2 (\BL^2 – L_z^2) } \\
&=
\inv{2} \lr{ \Hbar^2 l(l+1) – (\Hbar m)^2 }.
\end{aligned}

Observe that we have
\label{eqn:angularMomentumExpectation:160}
\expectation{L_x^2}
+
\expectation{L_y^2}
+
\expectation{L_z^2}
=
\Hbar^2 l(l+1)
=
\expectation{\BL^2},

which is the quantum mechanical analogue of the classical scalar equation $$\BL^2 = L_x^2 + L_y^2 + L_z^2$$.

# References

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

## Expectation of spherically symmetric 3D potential derivative

### Q: [1] pr 5.16

For a particle in a spherically symmetric potential $$V(r)$$ show that

\label{eqn:symmetricPotentialDerivativeExpectation:20}
\Abs{\psi(0)}^2 = \frac{m}{2 \pi \Hbar^2} \expectation{ \frac{dV}{dr} },

for all s-states, ground or excited.

Then show this is the case for the 3D SHO and hydrogen wave functions.

### A:

The text works a problem that looks similar to this by considering the commutator of an operator $$A$$, later set to $$A = p_r = -i \Hbar \PDi{r}{}$$ the radial momentum operator. First it is noted that

\label{eqn:symmetricPotentialDerivativeExpectation:40}
0 = \bra{nlm} \antisymmetric{H}{A} \ket{nlm},

since $$H$$ operating to either the right or the left is the energy eigenvalue $$E_n$$. Next it appears the author uses an angular momentum factoring of the squared momentum operator. Looking earlier in the text that factoring is found to be

\label{eqn:symmetricPotentialDerivativeExpectation:60}
\frac{\Bp^2}{2m}
= \inv{2 m r^2} \BL^2 – \frac{\Hbar^2}{2m} \lr{ \PDSq{r}{} + \frac{2}{r} \PD{r}{} }.

With
\label{eqn:symmetricPotentialDerivativeExpectation:80}
R = – \frac{\Hbar^2}{2m} \lr{ \PDSq{r}{} + \frac{2}{r} \PD{r}{} }.

we have

\label{eqn:symmetricPotentialDerivativeExpectation:100}
\begin{aligned}
0
&= \bra{nlm} \antisymmetric{H}{p_r} \ket{nlm} \\
&= \bra{nlm} \antisymmetric{\frac{\Bp^2}{2m} + V(r)}{p_r} \ket{nlm} \\
&= \bra{nlm} \antisymmetric{\inv{2 m r^2} \BL^2 + R + V(r)}{p_r} \ket{nlm} \\
&= \bra{nlm} \antisymmetric{\frac{-\Hbar^2 l (l+1)}{2 m r^2} + R + V(r)}{p_r} \ket{nlm}.
\end{aligned}

Let’s consider the commutator of each term separately. First

\label{eqn:symmetricPotentialDerivativeExpectation:120}
\begin{aligned}
\antisymmetric{V}{p_r} \psi
&=
V p_r \psi

p_r V \psi \\
&=
V p_r \psi

(p_r V) \psi

V p_r \psi \\
&=

(p_r V) \psi \\
&=
i \Hbar \PD{r}{V} \psi.
\end{aligned}

Setting $$V(r) = 1/r^2$$, we also have

\label{eqn:symmetricPotentialDerivativeExpectation:160}
\antisymmetric{\inv{r^2}}{p_r} \psi
=
-\frac{2 i \Hbar}{r^3} \psi.

Finally
\label{eqn:symmetricPotentialDerivativeExpectation:180}
\begin{aligned}
\antisymmetric{\PDSq{r}{} + \frac{2}{r} \PD{r}{} }{ \PD{r}{}}
&=
\lr{ \partial_{rr} + \frac{2}{r} \partial_r } \partial_r

\partial_r \lr{ \partial_{rr} + \frac{2}{r} \partial_r } \\
&=
\partial_{rrr} + \frac{2}{r} \partial_{rr}

\lr{
\partial_{rrr} -\frac{2}{r^2} \partial_r + \frac{2}{r} \partial_{rr}
} \\
&=
-\frac{2}{r^2} \partial_r,
\end{aligned}

so
\label{eqn:symmetricPotentialDerivativeExpectation:200}
\antisymmetric{R}{p_r}
=-\frac{2}{r^2} \frac{-\Hbar^2}{2m} p_r
=\frac{\Hbar^2}{m r^2} p_r.

Putting all the pieces back together, we’ve got
\label{eqn:symmetricPotentialDerivativeExpectation:220}
\begin{aligned}
0
&= \bra{nlm} \antisymmetric{\frac{-\Hbar^2 l (l+1)}{2 m r^2} + R + V(r)}{p_r} \ket{nlm} \\
&=
i \Hbar
\bra{nlm} \lr{
\frac{\Hbar^2 l (l+1)}{m r^3} – \frac{i\Hbar}{m r^2} p_r +
\PD{r}{V}
}
\ket{nlm}.
\end{aligned}

Since s-states are those for which $$l = 0$$, this means

\label{eqn:symmetricPotentialDerivativeExpectation:240}
\begin{aligned}
\expectation{\PD{r}{V}}
&= \frac{i\Hbar}{m } \expectation{ \inv{r^2} p_r } \\
&= \frac{\Hbar^2}{m } \expectation{ \inv{r^2} \PD{r}{} } \\
&= \frac{\Hbar^2}{m } \int_0^\infty dr \int_0^\pi d\theta \int_0^{2 \pi} d\phi r^2 \sin\theta \psi^\conj(r,\theta, \phi) \inv{r^2} \PD{r}{\psi(r,\theta,\phi)}.
\end{aligned}

Since s-states are spherically symmetric, this is
\label{eqn:symmetricPotentialDerivativeExpectation:260}
\expectation{\PD{r}{V}}
= \frac{4 \pi \Hbar^2}{m } \int_0^\infty dr \psi^\conj \PD{r}{\psi}.

That integral is

\label{eqn:symmetricPotentialDerivativeExpectation:280}
\int_0^\infty dr \psi^\conj \PD{r}{\psi}
=
\evalrange{\Abs{\psi}^2}{0}{\infty} – \int_0^\infty dr \PD{r}{\psi^\conj} \psi.

With the hydrogen atom, our radial wave functions are real valued. It’s reasonable to assume that we can do the same for other real-valued spherical potentials. If that is the case, we have

\label{eqn:symmetricPotentialDerivativeExpectation:300}
2 \int_0^\infty dr \psi^\conj \PD{r}{\psi}
=
\Abs{\psi(0)}^2,

and

\label{eqn:symmetricPotentialDerivativeExpectation:320}
\boxed{
\expectation{\PD{r}{V}}
= \frac{2 \pi \Hbar^2}{m } \Abs{\psi(0)}^2,
}

which completes this part of the problem.

### A: show this is the case for the 3D SHO and hydrogen wave functions

For a hydrogen like atom, in atomic units, we have

\label{eqn:symmetricPotentialDerivativeExpectation:360}
\begin{aligned}
\expectation{
\PD{r}{V}
}
&=
\expectation{
\PD{r}{} \lr{ -\frac{Z e^2}{r} }
} \\
&=
Z e^2
\expectation
{
\inv{r^2}
} \\
&=
Z e^2 \frac{Z^2}{n^3 a_0^2 \lr{ l + 1/2 }} \\
&=
\frac{\Hbar^2}{m a_0} \frac{2 Z^3}{n^3 a_0^2} \\
&=
\frac{2 \Hbar^2 Z^3}{m n^3 a_0^3}.
\end{aligned}

On the other hand for $$n = 1$$, we have

\label{eqn:symmetricPotentialDerivativeExpectation:380}
\begin{aligned}
\frac{2 \pi \Hbar^2}{m} \Abs{R_{10}(0)}^2 \Abs{Y_{00}}^2
&=
\frac{2 \pi \Hbar^2}{m} \frac{Z^3}{a_0^3} 4 \inv{4 \pi} \\
&=
\frac{2 \Hbar^2 Z^3}{m a_0^3},
\end{aligned}

and for $$n = 2$$, we have

\label{eqn:symmetricPotentialDerivativeExpectation:400}
\begin{aligned}
\frac{2 \pi \Hbar^2}{m} \Abs{R_{20}(0)}^2 \Abs{Y_{00}}^2
&=
\frac{2 \pi \Hbar^2}{m} \frac{Z^3}{8 a_0^3} 4 \inv{4 \pi} \\
&=
\frac{\Hbar^2 Z^3}{4 m a_0^3}.
\end{aligned}

These both match the potential derivative expectation when evaluated for the s-orbital ($$l = 0$$).

For the 3D SHO I verified the ground state case in the Mathematica notebook sakuraiProblem5.16bSHO.nb

There it was found that

\label{eqn:symmetricPotentialDerivativeExpectation:420}
\expectation{\PD{r}{V}}
= \frac{2 \pi \Hbar^2}{m } \Abs{\psi(0)}^2
= 2 \sqrt{\frac{m \omega ^3 \Hbar}{ \pi }}

# References

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

### Q: [1] pr. 5.18

Work out the quadratic Zeeman effect for the ground state hydrogen atom due to the usually neglected $$e^2 \BA^2/2 m_e c^2$$ term in the Hamiltonian.

### A:

The first order energy shift is

For a z-oriented magnetic field we can use

\BA = \frac{B}{2} \setlr{ -y, x, 0 },

so the perturbation potential is

\begin{aligned}
V
&= \frac{e^2 \BA^2}{2 m_e c^2} \\
&= \frac{e^2 \BB^2 (x^2 + y^2)}{8 m_e c^2} \\
&= \frac{ e^2 \BB^2 r^2 \sin^2\theta }{8 m_e c^2}
\end{aligned}

The ground state wave function is

\begin{aligned}
\psi_0
&= \braket{\Bx}{0} \\
&= \inv{\sqrt{\pi a_0^3}} e^{-r/a_0},
\end{aligned}

so the energy shift is

\begin{aligned}
\Delta
&= \bra{0} V \ket{0} \\
&= \inv{ \pi a_0^3 } 2 \pi \frac{ e^2 \BB^2 }{8 m_e c^2} \int_0^\infty r^2 \sin\theta e^{-2r/a_0} r^2 \sin^2\theta dr d\theta \\
&=
\frac{ e^2 \BB^2 }{4 a_0^3 m_e c^2}
\int_0^\infty r^4 e^{-2r/a_0} dr \int_0^\pi \sin^3\theta d\theta \\
&= –
\frac{ e^2 \BB^2 }{4 a_0^3 m_e c^2}
\frac{4!}{(2/a_0)^{4+1} } \evalrange{\lr{u – \frac{u^3}{3}}}{1}{-1} \\
&=
\frac{ e^2 a_0^2 \BB^2 }{4 m_e c^2}.
\end{aligned}

If this energy shift is written in terms of a diamagnetic susceptibility $$\chi$$ defined by