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Simultaneous eigenstate of angular momentum and hamiltonian

  1. May 6, 2017 #1
    1. The problem statement, all variables and given/known data
    b.jpg
    The red box only
    2. Relevant equations


    3. The attempt at a solution
    I suppose we have to show
    L_3 (Π_1) | E,m> = λ (Π_1) | E,m>
    and
    H (Π_1) | E,m> = μ (Π_1) | E,m>
    And I guess there is something to do with the formula given? But they are in x_1 direction so what did they have to do with | E,m> ?
    Any hints given will be much appreciated.
    Thanks:smile:
     
  2. jcsd
  3. May 6, 2017 #2

    blue_leaf77

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    Figure out the behavior of ##L_3## under the transformation induced by ##\Pi_1##.
    First show that the Hamiltonian commutes with ##\Pi_1##.
     
  4. May 6, 2017 #3
    1.I don't know what Π_1 is representing , I mean,like position and momentum operator you have Xψ = xψ , Pψ = -ih d/dx ψ ,but here you have got something like ΠXΠ^-1 = -X which is not the usual form of Kψ = kψ. Then what can I do to compute [ H,Π } ?

    2.Since L_3 is acting in the z-direction, Π L_3 ∏^-1 = L_3 ? so Π L_3 = L_3 Π and
    L_3 Π | E,m> = Π L_3 | E,m> = m Π | E,m> ?

    However ,the question said Π X ∏^-1 is valid for position and momentum operator, but here we are dealing with angular momentum operator?
     
  5. May 6, 2017 #4

    blue_leaf77

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    ##\Pi_1## is the parity operator in ##x## direction. It is defined by its action in position space on a wavefunction ##\psi(x_1,x_2,x_3)## by ##\Pi_1 \psi(x_1,x_2,x_3) = \psi(-x_1,x_2,x_3)##. But this is not necessary in the present problem since you are already given by its transformation properties.
    By expanding ##L_3 = x_1 p_2 - x_2 p_1## and using the properties of ##\Pi_1## as given in the question, calculate ##L_3 \Pi_1##.
    Don't compare them, one is the product between operators and the other one is the application of an operator on a state.
    No, it doesn't. It acts on the azimuthal coordinate instead, please review again your QM notes.
     
    Last edited: May 6, 2017
  6. May 6, 2017 #5
    Thanks I got the first part of red box:smile:. For the remaining part, I need to show that the Hamiltonian is degenerate. By definition that means E corresponds to more than 1 eigenstate. I am not sure , but from the form of eigenstate | E,m > and the equation H | E,m > = E | E,m > , we see that m does not appear in the eigenvalue of H so I guess m can play a role here? Like E corresponds to | E,1> , | E,0> and | E,-1> ; though I don't know the correct way of saying this.
     
  7. May 6, 2017 #6

    blue_leaf77

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    Yes in an indirect way.
    To do this part, you can show that ##|E,m\rangle## and ##\Pi_1|E,m\rangle## correspond to the same energy but different ##m## values.
     
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