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Hamilton operator with moments of inertia : time - independence

  1. Aug 16, 2011 #1
    1. The problem statement, all variables and given/known data
    The Hamilton-operator is given as [tex]\hat{H}[/tex] and describes the movement of a free rigid object that has the moments of inertia [tex]I_{i}[/tex]
    Under what circumstances is
    [tex]<\Psi|\hat{L_{1}}|\Psi> [/tex]

    time-independent?


    2. Relevant equations
    [tex] \hat{H}=\frac{\hat{L_{1}^{2}}}{2I_{1}}+\frac{\hat{L_{2}^{2}}}{2I_{2}}+\frac{\hat{L_{3}^{2}}}{2I_{3}} [/tex]

    [tex][\hat{L_{j}},\hat{L_{k}}]=\iota\hbar\epsilon_{jkm}\hat{L_{m}} [/tex]
    [tex]<\Psi|\hat{L_{1}}|\Psi> [/tex]

    3. The attempt at a solution
    If it wasn't in the brac-kets, I would just try [tex]\frac{dL_{1}}{dt}=0[/tex] Also, I thought maybe I could use another picture to have the time-indepence in it automatically, but I think Schrödinger must be the right one as there the operators are constant.
     
  2. jcsd
  3. Aug 16, 2011 #2

    G01

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    HINT: Remember that the time dependence of the expectation value of an operator, O, is related to: [itex]<[O,H]>[/itex].
     
  4. Aug 18, 2011 #3
    Thanks! I think a found a solution. What do you think about that?
    I just need to know whether I can change the indices of the Levi-Civita tensor in a way so that I get a minus in front (other order of the indices) also with operators. If yes, this would not answer the question (see below).

    If "otherwise": Is this the end result or can you transform that even more?
    [PLAIN]http://img545.imageshack.us/img545/6031/timeindependencemomento.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  5. Aug 18, 2011 #4

    G01

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    No, you cannot switch the indices on the operators and just change the sign to compensate. That would mean the the angular momentum operators anti-commute, which they don't.

    You end result is correct(when you don't switch operator indices), but you missed a sign, I think. Your result should be:

    [tex]\frac{[L_2,L_3]_+}{I_2}=\frac{[L_2,L_3]_+}{I_3}[/tex]

    ([itex][A,B]_+[/itex] means anti-commutator) or when you simplify:

    [tex]I_2=I_3[/tex]

    Physically, this means you have an "axially symmetric rotator."


    You really went about this in the "brute-force" method! There's a much simpler way to get to the same result. You know you are looking for the condition when [itex]L_1[/itex] and [itex]H[/itex] commute. Now, you also know that you can choose any one angular momentum operator to commute with [itex]L^2[/itex]. Thus, if you can write [itex]H[/itex] in terms of only [itex]L^2[/itex] and [itex]L_1[/itex].

    You'll see that this is only possible in two cases: When [itex]I_1=I_2=I_3[/itex] and when [itex]I_2=I_3[/itex]
     
  6. Aug 19, 2011 #5
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