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QM - angular momentum

  1. Dec 6, 2005 #1

    quasar987

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    The question:

    I am given a system (a cylinder made of particles), and its hamiltonian:

    [tex]\hat{H} = \frac{L_z^2}{2MR^2} , \ \ \ \ \ \ L_z = -ih\frac{\partial}{\partial \phi}[/tex]

    a) What are the possible values for the angular momentum?

    b) What are the eigenvalues of the energy?

    c) What is the difference btw the energy of the ground state of angular momentum 0 and the first rotational state?


    My dilema: The answer to a) is found by solving the eigenvalue equation for the operator L_z and applying the boundary condition that a change of 2pi in the argument of the wave function must not change its value. I find that the possible values are [itex]L_z = n\hbar[/itex], where n is an integer (I saw no reason not to allow negative n).

    The answer to b) is found in the same way: first solve the eigenvalue equation, then apply the same boundary condition. I find

    [tex]E_k = \frac{\hbar^2k^2}{2MR^2}[/tex]

    where k is an integer but not 0.

    But question c) seems contradictory with the principles of QM. Here's why: If I say that the angular momentum is 0. then it means that the wave function is collapsed to an eigenfunction corresponding to the eigenvalue L = 0. And if, after that, I want the energy, then for each eigen energy, I can only assert that there is a cerain probability that my measurement will yield each of these energies. This probability is the square of the norm of the coefficient a_n associated with the n-th energy eigenfunction in the expansion of the wave function (in this case, the angular momentum eigenfunction of eigenvalue L = 0) in terms of the energy eigenfunctions.

    So there is no energy associated with a given angular momentum eigenfunction, so I cannot compute a difference.

    Where's the problem? Thx!
     
    Last edited: Dec 6, 2005
  2. jcsd
  3. Dec 6, 2005 #2

    quasar987

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    Unless L_z and H commute! In this case they have the same eigenfunction I think...... but I'd have made a mistake in solving the eigenvalue equations then.... gotta find that commutation relation, brb.
     
  4. Dec 6, 2005 #3

    Galileo

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    Have you actually checked the eigenstates of the angular momentum are also energy eigenstates?

    In general, you're right, but in this case H is basically the square of L_z, so an eigenfunction of L_z will also be an eigenfunction of H. The energy associated with it is ofcourse simply the square of the angular momentum divided by 2MR^2 (moment of inertia).

    Another way to see this is that H and L_z trivially commute, so they have a common basis of eigenstates. Since the spectrum of L_z is nondegenerate, the set of eigenstates you found is the eigenbasis common to H and L_z.

    EDIT: AH, you had the lightbulb flashing moment while I was posting :tongue2:
     
  5. Dec 6, 2005 #4

    quasar987

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    yes, they do trivially commute.. :P
     
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