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Rotational Analog to Ehrenfest's Theorem

  1. Apr 3, 2006 #1

    eep

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    Hi,

    I'm trying to prove that for a particle in a potential V(r), the rate of change of the expectation value of the orbital angular momentum L is equal to the expectation value of the torque:

    [tex]
    \frac{d}{dt}<L> = <N>
    [/tex]

    where

    [tex] N = r \times (-\bigtriangledown{V}) [/tex]

    Basically, I'm having problems calculating the commutor of the Hamiltonian and the angular momentum operator, as

    [tex]
    \frac{d}{dt}<L> = \frac{i}{\hbar}<[H,L]> + <\frac{\partial{L}}{\partial{t}}>
    [/tex]

    Any hints on how I can calculate this?
     
  2. jcsd
  3. Apr 3, 2006 #2

    Physics Monkey

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    As always, when in doubt about where to start, go back to the basics. Use the position representation where [tex] \vec{p} = - i \hbar \vec{\nabla} [/tex] and simply calculate the commutator explicitly.

    Some things to consider:
    Clearly the partial derivative term vanishes. You can argue one of the terms in the Hamiltonian commutator is zero based on rotational symmetry. You also need only calculate a single component. If these points are not obvious to you, think about them until you see clearly why they are true.
     
    Last edited: Apr 3, 2006
  4. Apr 4, 2006 #3

    eep

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    Thanks for the tips. I want to say that the momentum term in the Hamiltonian commuter is zero as the linear momentum should have constant magnitude. A change in one direction will be counteracted by changes in the other directions. Then I got into trouble trying to commute V(r) with L. Now you said that I only need to calculate one component, and the way this equation is written makes me think of it as a vector equation. That is I have 3 equations in the form

    [tex]
    \frac{d<L_x>}{dt} = \frac{i}{\hbar}<[H,L_x]> = <N_x>
    [/tex]

    But I don't think this can be right as as the Hamiltonian commutes with each component of angular momentum. What am I doing wrong here? And what's wrong with my tex code?
     
    Last edited: Apr 4, 2006
  5. Apr 4, 2006 #4

    Physics Monkey

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    Hi eep,

    1) The commutator with the kinetic energy is zero because this term is rotationally invariant.

    2) You do indeed have a vector equation there. However, the components of the angular momentum don't necessarily commute with the Hamiltonian. Otherwise, angular momentum would always be conserved for every system! Earlier I gave you some tips on proving that the angular momentum does commute with the Hamiltonian if the potential is central, but here you're supposed to be looking at a more general case (I think!).

    Oh yeah, don't worry about the latex, it seems the whole site is broken right now.
     
    Last edited: Apr 4, 2006
  6. Apr 4, 2006 #5

    eep

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    Well this is what I don't get. In the previous problem I proved that the Hamiltonian commutes with all three components of L, provided that V depends only on r. In this problem, the question states "Prove that for a particle in a potential V(r)...". I'm not seeing what the difference in the potentials are. I would go through the same steps to calculate the commuter of H with Lx which leads me to [H,L] = 0...
     
  7. Apr 4, 2006 #6

    Physics Monkey

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    I think the difference is that they want you to assume that V is general function of x, y, and z, not just a function of r. If your potential is only a function of r, then certainly the commutator vanishes. However, since you are being asked to prove a general theorem, I think that you are supposed to leave the potential completely general.
     
  8. Apr 4, 2006 #7

    eep

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    Well thank you. If you leave the potential completely general then everything works out correctly. I guess there's a big difference between the potential depends only on r (scalar) and the function V(r) where r is a vector. Thanks, you gave me a much better understanding of the situation. One more thing, though. By saying that the commutator of the kinetic energy with angular momentum is zero because the term is rotationally invariant, you're saying that kinectic energy doesn't change if you rotate your coordinates, right? Why does this guarantee that the commutator will be zero? I must need more sleep!
     
  9. Apr 4, 2006 #8

    Physics Monkey

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    You're welcome!

    Regarding the kinetic energy term, the brief answer to your question is that the commutator with L represents a kind of first order change caused by L. Now, what sort of changes does L cause? Well, L is the generator of rotations, so if the kinetic energy is rotationally invariant, L shouldn't be able to change it.
     
  10. May 24, 2009 #9
    I'm having trouble with the mathematical argument that [T,L_i] = 0. When I compute it, using T = -(h-bar)^2/(2m) * Laplacian I get 2*d/d(x_i) * d/d(x_j).
     
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