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Sakurai 1.17 - Operators and Complete Eigenkets

  1. Dec 16, 2008 #1
    I'm pretty sure this is correct, but could someone verify for rigor?

    Two observables [tex]A_1[/tex] and [tex]A_2[/tex], which do not involve time explicitly, are known not to commute, yet we also know that [tex]A_1[/tex] and [tex]A_2[/tex] both commute with the Hamiltonian. Prove that the energy eigenstates are, in general, degenerate. Are there exceptions?

    The attempt at a solution
    Since [tex][H,A_1]=0[/tex], we know that there are complete eigenkets that [tex]A_1[/tex] and [tex]H[/tex] share. The same is true for [tex]A_2[/tex]. Now, generally, two observables that do not commute do not share eigenkets. Thus, we know that there exist two distinct eigenstates such that [tex]H|a\ket = e|a\ket[/tex]; particularly, each [tex]|a\ket[/tex] can be brought from the eigenkets of [tex]A_1,A_2[/tex].
  2. jcsd
  3. Dec 16, 2008 #2
    come on 25 views and no post? :(
  4. Dec 16, 2008 #3


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    Impatience is not a virtue. Can't A1 and A2 commute on a subspace without commuting on the whole space?
  5. Dec 16, 2008 #4
    Heh, it was more of a bump since it was already on the second page :) Yes, they can, but the question asks "generally," and "generally" this doesn't happen, right?
  6. Dec 16, 2008 #5


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    Ok, considered as a teasing bump. But the question asks, are there exceptions? I.e. could there be energy eigenstates that aren't degenerate? At least that's how I read it.
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