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Using operators and finding expectation value

  1. Aug 28, 2013 #1
    1. The problem statement, all variables and given/known data

    The expectation value of the time derivative of an arbitrary quantum operator [itex]\hat{O}[/itex] is given by the expression:

    d[itex]\langle[/itex][itex]\hat{O}[/itex][itex]\rangle[/itex]/dt[itex]\equiv[/itex][itex]\langle[/itex]d[itex]\hat{O}[/itex]/dt[itex]\rangle[/itex]=[itex]\langle[/itex]∂[itex]\hat{O}[/itex]/∂t[itex]\rangle[/itex]+i/hbar[itex]\langle[/itex][itex][[/itex][itex]\hat{H}[/itex],[itex]\hat{O}[/itex][itex]][/itex][itex]\rangle[/itex]​

    Obtain an expression for [itex]\langle[/itex]d[itex]\hat{L}[/itex]x/dt+d[itex]\hat{L}[/itex]y/dt[itex]\rangle[/itex] where [itex]\hat{H}[/itex]=[itex]\hat{H}[/itex]00Bz[itex]\hat{L}[/itex]z/hbar


    2. Relevant equations

    [itex][[/itex][itex]\hat{L}[/itex]x,[itex]\hat{L}[/itex]y[itex]][/itex]=i*hbar[itex]\hat{L}[/itex]z
    [itex][[/itex][itex]\hat{L}[/itex]y,[itex]\hat{L}[/itex]z[itex]][/itex]=i*hbar[itex]\hat{L}[/itex]x
    [itex][[/itex][itex]\hat{L}[/itex]z,[itex]\hat{L}[/itex]x[itex]][/itex]=i*hbar[itex]\hat{L}[/itex]y

    [itex][[/itex]A,B[itex]][/itex]=AB-BA


    3. The attempt at a solution

    [itex]\langle[/itex]d[itex]\hat{L}[/itex]x/dt+d[itex]\hat{L}[/itex]y/dt[itex]\rangle[/itex]=d[itex]\langle[/itex][itex]\hat{L}[/itex]x+[itex]\hat{L}[/itex]y[itex]\rangle[/itex]
    =[itex]\frac{1}{ih}[/itex] d[itex]\langle[/itex][itex][[/itex][itex]\hat{L}[/itex]y,[itex]\hat{L}[/itex]z[itex]][/itex]+[itex][[/itex][itex]\hat{L}[/itex]z,[itex]\hat{L}[/itex]x[itex]][/itex][itex]\rangle[/itex]/dt
    =[itex]\frac{1}{ih}[/itex][itex]\langle[/itex] ∂[itex][[/itex][itex]\hat{L}[/itex]y,[itex]\hat{L}[/itex]z[itex]][/itex]+[itex][[/itex][itex]\hat{L}[/itex]z,[itex]\hat{L}[/itex]x[itex]][/itex]/∂t[itex]\rangle[/itex]+[itex]\frac{i}{hbar}[/itex][itex]\langle[/itex][itex][[/itex][itex]\hat{H}[/itex],[itex][[/itex][itex]\hat{L}[/itex]y,[itex]\hat{L}[/itex]z[itex]][/itex]+[itex][[/itex][itex]\hat{L}[/itex]z,[itex]\hat{L}[/itex]x[itex]][/itex][itex]][/itex][itex]\rangle[/itex]

    I'm not sure how to continue on from this
     
  2. jcsd
  3. Aug 29, 2013 #2
    Just calculate the commutators [itex] [L_x, H] [/itex] and [itex] [L_y, H] [/itex]? You know that H0 should commute with all L's (why?), and the commutator with the extra term is easy to calculate.
     
  4. Aug 30, 2013 #3
    I calculated the commutators [itex] [L_x, H] [/itex] and [itex] [L_y, H] [/itex] but I don't see how this helps me answer the question. I also can't figured out why H0 should commute with all L's.

    Also [itex]\hat{H}[/itex]0=-hbar2/2mr2 [itex]\{[/itex][itex]\frac{∂}{∂r}[/itex](r2[itex]\frac{∂}{∂r}[/itex])+[itex]\frac{1}{sinθ}[/itex][itex]\frac{∂}{∂θ}[/itex](sinθ[itex]\frac{∂}{∂θ}[/itex])+[itex]\frac{1}{sin squared θ}[/itex][itex]\frac{∂ squared}{∂\phi squared}[/itex][itex]\}[/itex]+V(r)

    where the angular-dependent part of the Hamiltonian corresponds to the total angular momentum operator [itex]\hat{L}[/itex]2
     
    Last edited: Aug 30, 2013
  5. Aug 31, 2013 #4

    vela

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    You were given that
    $$\bigg\langle \frac{d\hat{O}}{dt} \bigg\rangle = \bigg\langle \frac{\partial \hat{O}}{\partial t} \bigg\rangle + \frac{i}{\hbar} \langle [\hat{H},\hat{O}] \rangle.$$ What do you get if you let ##\hat{O} = \hat{L}_x##?

    You should be able to prove that ##\hat{L}_i## commutes with ##\hat{L}^2## using the property [AB,C]=A[B,C]+[A,C]B and the commutation relations you listed above.
     
  6. Aug 31, 2013 #5
    But shouldn't I substitute [itex]\hat{O}[/itex]=[itex]\hat{L}[/itex]x+[itex]\hat{L}[/itex]y instead?

    Oh I understand now. Thanks
     
  7. Aug 31, 2013 #6

    vela

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    You asked why you want to calculate ##[\hat{H},\hat{L}_i]##. Do you see why?
     
  8. Aug 31, 2013 #7
    Yh I do. My final answer is

    [itex]\langle[/itex]d[itex]\hat{L}[/itex]x/dt+d[itex]\hat{L}[/itex]y/dt[itex]\rangle[/itex]=[itex]\langle[/itex]∂([itex]\hat{L}[/itex]x+[itex]\hat{L}[/itex]y)/∂t[itex]\rangle[/itex]+[itex]\frac{i}{hbar}[/itex][itex]\langle[/itex]iμ0Bz([itex]\hat{L}[/itex]y - [itex]\hat{L}[/itex]x)[itex]\rangle[/itex]

    Could I further simplify this?
     
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