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Bill Foster
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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.
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