## Unimodular transformation

October 31, 2015 phy1520 ,

### 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 , ,

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 , ,

### 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 , ,

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 ,

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 simultaneously 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}
H B \ket{1}
= H \sum_n \ket{n}\bra{n} B \ket{1}
= \sum_n \epsilon_n \ket{n} B_{n 1},

and
\label{eqn:angularMomentumAndCentralForceCommutators:200}
B H \ket{1}
= B \epsilon_1 \ket{1}
= \epsilon_1 \sum_n \ket{n}\bra{n} B \ket{1}
= \epsilon_1 \sum_n \ket{n} B_{n 1}.

We can now compute the action of the commutators on $$\ket{1}, \ket{2}$$,
\label{eqn:angularMomentumAndCentralForceCommutators:220}
\antisymmetric{B}{H} \ket{1}
=
\sum_n \lr{ \epsilon_1 – \epsilon_n } \ket{n} B_{n 1}.

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

However, 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}
\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,

The sums in \ref{eqn:angularMomentumAndCentralForceCommutators:220} and \ref{eqn:angularMomentumAndCentralForceCommutators:240} reduce to
\label{eqn:angularMomentumAndCentralForceCommutators:500}
\antisymmetric{B}{H} \ket{1}
=
\sum_{n=1}^2 \lr{ \epsilon_1 – \epsilon_n } \ket{n} B_{n 1}
=
\lr{ \epsilon_1 – \epsilon_2 } \ket{2} B_{2 1},

and
\label{eqn:angularMomentumAndCentralForceCommutators:520}
\antisymmetric{B}{H} \ket{2}
=
\sum_{n=1}^2 \lr{ \epsilon_2 – \epsilon_n } \ket{n} B_{n 2}
=
\lr{ \epsilon_2 – \epsilon_1 } \ket{1} B_{1 2}.

Since the commutator is zero, the matrix elements of the commutator must all be zero, in particular
\label{eqn:angularMomentumAndCentralForceCommutators:260}
\begin{aligned}
\bra{1} \antisymmetric{B}{H} \ket{1} &= \lr{ \epsilon_1 – \epsilon_2 } B_{2 1} \braket{1}{2} = 0 \\
\bra{2} \antisymmetric{B}{H} \ket{1} &= \lr{ \epsilon_1 – \epsilon_2 } B_{2 1} \braket{1}{1} \\
\bra{1} \antisymmetric{B}{H} \ket{2} &= \lr{ \epsilon_2 – \epsilon_1 } B_{1 2} \braket{1}{2} = 0 \\
\bra{2} \antisymmetric{B}{H} \ket{2} &= \lr{ \epsilon_2 – \epsilon_1 } B_{1 2} \braket{2}{2}.
\end{aligned}

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}
B \ket{1}
=
\ket{n}\bra{n} B \ket{1}
=
\ket{n} B_{n 1}
=
\ket{1} B_{1 1},

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.