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Homework Help: Sakurai 1.17

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data

    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:


    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]

    3. 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]


    [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]


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


    [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]


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


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


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


    [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: Nov 1, 2009
  2. jcsd
  3. Nov 1, 2009 #2
    But I'm not really sure how to find an exception.
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