Sakurai 1.17

  • #1
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Homework Statement



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

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

Yet they both commute with the Hamiltonian:

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

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

Are there exceptions?

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

The Attempt at a Solution



If the Hamiltonian operates on the ket, we get:

[tex]H|n\rangle = E_n|n\rangle[/tex]

If the [tex]A_n[/tex] operates operate on the ket:

[tex]A_n|n\rangle = a_n|n\rangle[/tex]

If the Hamiltonian operates on that:

[tex]H\left(A_n|n\rangle\right) = E_n\left(a_n|n\rangle\right)[/tex]

So:

[tex]H\left(A_1|n\rangle\right) = E_n\left(a_1|n\rangle\right)[/tex]
[tex]H\left(A_2|n\rangle\right) = E_n\left(a_2|n\rangle\right)[/tex]

These are non-degenerate, because [tex]a_1\ne a_2[/tex]

Given this:

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

That means [tex]A_1A_2-A_2A_1\ne 0[/tex]

Which means [tex]A_1A_2 \ne A_2A_1[/tex]

But

[tex]A_1A_2|n\rangle = a_1a_2|n\rangle[/tex]

And

[tex]A_2A_1|n\rangle = a_2a_1|n\rangle[/tex]

However: [tex]a_1a_2=a_2a_1[/tex], so

[tex]a_1a_2|n\rangle = a_2a_1|n\rangle[/tex]

But

[tex]A_1A_2|n\rangle \ne A_2A_1|n\rangle[/tex]

Now

[tex]H\left(A_1A_2|n\rangle\right) = E_n\left(a_1a_2|n\rangle\right)[/tex]

And

[tex]H\left(A_2A_1|n\rangle\right) = E_n\left(a_2a_1|n\rangle\right)[/tex]

So

[tex]E_n\left(a_1a_2|n\rangle\right) = E_n\left(a_2a_1|n\rangle\right)[/tex]

And therefore they are degenerate.
 
Last edited:

Answers and Replies

  • #2
338
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But I'm not really sure how to find an exception.
 

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