1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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
    Hi
    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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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]

    and

    [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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Spin expectation values in x and y direction
Loading...