# Sakurai 1.17

1. Oct 31, 2009

### Bill Foster

1. The problem statement, all variables and given/known data

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 $$A_1 \rightarrow L_z$$, $$A_2 \rightarrow L_x$$

3. 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.

Last edited: Nov 1, 2009
2. Nov 1, 2009

### Bill Foster

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