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Heisenberg equations for spin

  1. Apr 10, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider a spin ##\frac{1}{2}## particle at rest in a B-field ##\vec B = B_0\vec e_z##.

    (a) Find the Heisenberg equations for ##\hat S_x## and ##\hat S_y##.

    (b) Obtain from the Heisenberg equations two decoupled second-order differential equations for ##\langle \hat S_x \rangle_{\psi}## and ##\langle \hat S_y \rangle_{\psi}## for a general state ##| \psi \rangle## of the particle.

    (c) Solve the equations for a particle that initially (at t = 0) is in a state of ##S_x = \frac{\hbar}{2}##

    2. Relevant equations


    3. The attempt at a solution

    For part (a) I have done the following;

    ##\langle \psi (t) | \hat B_S | \psi (t) \rangle = \langle e^{-\frac{i \hat H t}{\hbar}} \psi (0) | \hat B_S | e^{-\frac{i \hat H t}{\hbar}} \psi (0) \rangle = \langle \psi (0) | e^{\frac{i \hat H t}{\hbar}} \hat B_S e^{-\frac{i \hat H t}{\hbar}} | \psi (0) \rangle##

    ##B_H = e^{\frac{i \hat H t}{\hbar}} \hat B_S e^{-\frac{i \hat H t}{\hbar}}##

    where the H and S subscripts correspond to the Heisenberg and Schrodinger pictures respectively.

    The hamiltonian that I have is ##\hat H = - \gamma B_0 \hat S_z##

    which leads me to the conclusion that my spin operators are unchanged in the Heisenberg picture.

    for part (b) I use the Heisenberg equation of motion on both the ##\hat S_x## and ##\hat S_y## operators

    ##\frac{d}{dt} \hat S_x = \frac{i}{\hbar} [\hat H, \hat S_x] = \gamma B_0 \hat S_y##

    ##\frac{d}{dt} \hat S_y = \frac{i}{\hbar} [\hat H, \hat S_y] = - \gamma B_0 \hat S_x##

    I then differentiate both again, and substitute the original equations into the result to find

    ##\frac{d^2}{dt^2} \hat S_y = -(\gamma B_0)^2 \hat S_y##

    ##\frac{d^2}{dt^2} \hat S_x = -(\gamma B_0)^2 \hat S_x##

    These two equations are SHM equations with solutions of the form ##y = A \cos \omega t + B \sin \omega t ##

    solving these I find that ##\hat S_y (t) = \hat S_y (0) \cos (\gamma B_0 t) - \hat S_x (0) \sin (\gamma B_0 t)## and ##\hat S_x (t) = \hat S_x (0) \cos (\gamma B_0 t) + \hat S_y (0) \sin (\gamma B_0 t)##

    I don't really understand what the question is trying to get me to do here. Any guidance would be much appreciated.
     
  2. jcsd
  3. Apr 13, 2017 #2

    DrDu

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    Looks all ok, I think the only thing you are assumed still to do is to form the expectations values with the psi(0) specified.
     
  4. Apr 28, 2017 #3
    I'm still having trouble solving this. I can't see how to find ##\hat S_y (0)##
     
  5. Apr 28, 2017 #4

    DrDu

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    At t=0, the operators in the Schroedinger and Heisenberg picture coincide.
     
  6. Apr 28, 2017 #5
    Thanks, I see.

    Ok, so I know that ##\psi (s_x = \hbar / 2) = \frac{1}{\sqrt{2}} \begin{pmatrix} 1\\ 1 \end{pmatrix}##. Does "solving" these equations mean taking the expectation value for my state?
     
  7. Apr 28, 2017 #6

    DrDu

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    I think so, what do you get?
     
  8. Apr 28, 2017 #7
    I find that ##\langle \hat S_y \rangle_{\psi} = - \frac{\hbar}{2} \sin (\gamma B_0 t)## and ##\langle \hat S_x \rangle_{\psi} = \frac{\hbar}{2} \cos (\gamma B_0 t)##
     
  9. Apr 29, 2017 #8

    DrDu

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    Looks good!
     
  10. Apr 29, 2017 #9
    Thank you for your help
     
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