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Spin expectation values in x and y direction

  1. May 30, 2013 #1
    I have found what I think is the correct answer I just want to check an assumption. The magnetic field points in the +ve z-direction. We are given the initial state vector

    [itex]\left| A \right\rangle_{initial}=\frac{1}{5}\left[ \begin{array}{c}3\\4\end{array} \right][/itex]

    Am I right in thinking that this is in the z-direction?
    So that

    [itex]\left| A \right\rangle_{initial}= a_1 \left| \uparrow_z \right\rangle + a_2 \left| \downarrow_z \right\rangle = \frac{3}{5}\left[ \begin{array}{c}1\\0\end{array} \right] + \frac{4}{5}\left[ \begin{array}{c}0\\1\end{array} \right][/itex]
  2. jcsd
  3. May 30, 2013 #2


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    hi bobred! :wink:
    if you mean, do the notations (1 0) and (0 1) mean the +ve and -ve z directions, then yes :smile:

    (that is the general convention, and you can assume it unless stated otherwise)
  4. May 30, 2013 #3
    Yes that's what I meant. We are then asked to find the expectation values at a later time which I have.

    We are then asked using the generalized Ehrenfest theorem to find the rate of change of [itex]\left\langle S_x \right\rangle[/itex] and [itex]\left\langle S_y \right\rangle[/itex] in terms of [itex]\left\langle S_x \right\rangle[/itex] and [itex]\left\langle S_y \right\rangle[/itex] and ω. Using the generalized Ehrenfest theorem I find them in terms of [itex]\hat{S_x}[/itex] and [itex]\hat{S_y}[/itex] and ω. Any ideas?
  5. May 30, 2013 #4


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    How far have you gotten? Are you able to get an expression for [Sx, H] in terms of Sx and/or Sy?
    Last edited: May 30, 2013
  6. May 30, 2013 #5
    Hi, I get (we are told [itex]\omega = -\gamma_s B[/itex])

    [itex]\left[ \hat{S_x},\hat{H} \right]=-\frac{\gamma_s B \hbar^2}{2}\left[ \begin{array}{cc}0&-1\\1& 0 \end{array} \right]=\frac{\omega \hbar^2}{2}\left[ \begin{array}{cc}0&-1\\1& 0 \end{array} \right][/itex]


    [itex]\displaystyle\dfrac{d\left\langle S_x \right\rangle}{dt}=-\frac{\omega \hbar}{2}\left[ \begin{array}{cc}0&-i\\i& 0 \end{array} \right]=-\omega \hat{S_y}[/itex]
  7. May 30, 2013 #6


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    Looks pretty good. But note that Ehrenfest's theorem states

    [itex]\displaystyle\dfrac{d\left\langle S_x \right\rangle}{dt}= \frac{1}{i\hbar}\langle\left[ \hat{S_x},\hat{H} \right]\rangle [/itex]

    The right hand side contains an expectation value of ##\left[ \hat{S_x},\hat{H} \right]##.
  8. May 31, 2013 #7
    From the commutator I get the expectation value of the spin y operator... can't seem to see where to go from there.
  9. May 31, 2013 #8


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    Doesn't that provide the answer to the question? That is, you now have the rate of change of ##\langle S_x \rangle## expressed in terms of ##\omega## and ##\langle S_y \rangle##. Or have I misunderstood what you are trying to find?
    Last edited: May 31, 2013
  10. Jun 4, 2013 #9
    Hi, my fault I had confused myself. Thanks
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