Energy Eigenstates and Degeneracy: Proving Non-Commuting Observables

[Bill Foster]

Homework Statement

Two observables A_1 and A_2, which do not involve time explicitly, are known not to commute,

\left[A_1,A_2\right]\ne 0

Yet they both commute with the Hamiltonian:

\left[A_1,H\right]=0
\left[A_2,H\right]=0

Prove that they energy eigenstates are, in general, degenerate.

Are there exceptions?

As an example, you may think of the central-force problem H=\frac{\textbf{p}^2}{2m}+V\left(r\right) with A_1 \rightarrow L_z, A_2 \rightarrow L_x

The Attempt at a Solution

If the Hamiltonian operates on the ket, we get:

H|n\rangle = E_n|n\rangle

If the A_n operates operate on the ket:

A_n|n\rangle = a_n|n\rangle

If the Hamiltonian operates on that:

H\left(A_n|n\rangle\right) = E_n\left(a_n|n\rangle\right)

So:

H\left(A_1|n\rangle\right) = E_n\left(a_1|n\rangle\right)
H\left(A_2|n\rangle\right) = E_n\left(a_2|n\rangle\right)

These are non-degenerate, because a_1\ne a_2

Given this:

\left[A_1,A_2\right]\ne 0

That means A_1A_2-A_2A_1\ne 0

Which means A_1A_2 \ne A_2A_1

But

A_1A_2|n\rangle = a_1a_2|n\rangle

And

A_2A_1|n\rangle = a_2a_1|n\rangle

However: a_1a_2=a_2a_1, so

a_1a_2|n\rangle = a_2a_1|n\rangle

But

A_1A_2|n\rangle \ne A_2A_1|n\rangle

Now

H\left(A_1A_2|n\rangle\right) = E_n\left(a_1a_2|n\rangle\right)

And

H\left(A_2A_1|n\rangle\right) = E_n\left(a_2a_1|n\rangle\right)

So

E_n\left(a_1a_2|n\rangle\right) = E_n\left(a_2a_1|n\rangle\right)

And therefore they are degenerate.

 

[Bill Foster]

But I'm not really sure how to find an exception.