[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.

Marcos Basso2 years agoFirst, you forgot the summation over ‘n’ in Eqs. 11 and 12, otherwise we can put \ketbra{n} out of nowhere. But this does not interfere in the calculation.

However, your conclusion that, if B_21 = B_12 = 0 implies that |1>, |2> are eigenvectores of B is wrong, and this is due to the fact that you forgot the summation on n in Eq. 17.

If the claim ‘B_21 = B_12 = 0 implies that |1>, |2> are eigenvectores of B’ were true, then the hole exercise would be a theorem. But this is not true since there’s counter-examples. For instance, just consider the hydrogen atom, with observables L_x e L_z.

Peeter Joot2 years agoSummation was implied in 11, 12. Omitting those sums did lead to errors, which have been corrected. I believe the new end result is now correct.