[Click here for a PDF of an older version of post with nicer formatting]. Updates will be made in my old grad quantum notes.

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

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

but

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

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

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

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

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

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

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

The other eigenvectors for $A$ are

$$\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:420} \begin{aligned} \ket{a_1} &= \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \\ \ket{a_2} &= \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \end{aligned} \end{equation}$$

and for $B$

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

This clearly has the degeneracy sought.

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

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

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

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

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

$$\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:160} \begin{aligned} H \ket{1} &= \epsilon_1 \ket{1} \\ H \ket{2} &= \epsilon_2 \ket{2} \\ A \ket{1} &= a_1 \ket{1} \\ A \ket{2} &= a_2 \ket{2}, \end{aligned} \end{equation}$$
where $\ket{1}$, and $\ket{2}$ are not eigenkets of $B$. Because $B$ also commutes with $H$, we must have

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

and

$$\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:200} B H \ket{1} = B \epsilon_1 \ket{1} = \epsilon_1 \sum_n \ket{n}\bra{n} B \ket{1} = \epsilon_1 \sum_n \ket{n} B_{n 1}. \end{equation}$$

We can now compute the action of the commutators on $\ket{1}, \ket{2}$,

$$\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:220} \antisymmetric{B}{H} \ket{1} = \sum_n \lr{ \epsilon_1 – \epsilon_n } \ket{n} B_{n 1}. \end{equation}$$

Similarly

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

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

$$\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:280} \antisymmetric{B}{H} \ket{m} = B \epsilon_m \ket{m} – H b_m \ket{m} = b_m \epsilon_m \ket{m} – \epsilon_m b_m \ket{m} = 0, \end{equation}$$

The sums in $\ref{eqn:angularMomentumAndCentralForceCommutators:220}$ and $\ref{eqn:angularMomentumAndCentralForceCommutators:240}$ reduce to

$$\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:500} \antisymmetric{B}{H} \ket{1} = \sum_{n=1}^2 \lr{ \epsilon_1 – \epsilon_n } \ket{n} B_{n 1} = \lr{ \epsilon_1 – \epsilon_2 } \ket{2} B_{2 1}, \end{equation}$$
and
$$\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:520} \antisymmetric{B}{H} \ket{2} = \sum_{n=1}^2 \lr{ \epsilon_2 – \epsilon_n } \ket{n} B_{n 2} = \lr{ \epsilon_2 – \epsilon_1 } \ket{1} B_{1 2}. \end{equation}$$
Since the commutator is zero, the matrix elements of the commutator must all be zero, in particular
$$\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:260} \begin{aligned} \bra{1} \antisymmetric{B}{H} \ket{1} &= \lr{ \epsilon_1 – \epsilon_2 } B_{2 1} \braket{1}{2} = 0 \\ \bra{2} \antisymmetric{B}{H} \ket{1} &= \lr{ \epsilon_1 – \epsilon_2 } B_{2 1} \braket{1}{1} \\ \bra{1} \antisymmetric{B}{H} \ket{2} &= \lr{ \epsilon_2 – \epsilon_1 } B_{1 2} \braket{1}{2} = 0 \\ \bra{2} \antisymmetric{B}{H} \ket{2} &= \lr{ \epsilon_2 – \epsilon_1 } B_{1 2} \braket{2}{2}. \end{aligned} \end{equation}$$
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

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

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.