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!

Homework Help: Heisenberg Equations of Motion, Solving for S(t) in Spin Precession problem

  1. Oct 4, 2007 #1
    Consider the spin precession problem in the Heisenberg picture. Using the Hamiltonian

    [tex]H=-\omega S_{z}[/tex]


    write the Heisenberg equations of motion for the time dependent operators [tex]S_{x}(t)[/tex], [tex]S_{y}(t)[/tex], and [tex]S_{z}(t)[/tex]. Solve them to obtain [tex]\vec{S}[/tex] as a function of t.


    [tex]\frac{d A_{H}}{dt}=\frac{1}{\imath \hbar}[A_{H}, H][/tex]


    [tex]U=e^{\frac{-\imath H t}{\hbar}}[/tex]

    Well, computing the Heisenberg equations is pretty straitforward:

    [tex]\frac{d S_{x}}{dt}=\frac{1}{\imath \hbar}[S_{x}, -\omega S_{z}]
    =-\frac{\omega}{\imath \hbar}[S_{x},S_{z}]
    =\omega S_{y}[/tex]

    [tex]\frac{d S_{y}}{dt}=\frac{1}{\imath \hbar}[S_{y}, -\omega S_{z}]
    =-\frac{\omega}{\imath \hbar}[S_{y},S_{z}]
    =-\omega S_{x}[/tex]

    [tex]\frac{d S_{z}}{dt}=\frac{1}{\imath \hbar}[S_{z}, -\omega S_{z}]
    =-\frac{\omega}{\imath \hbar}[S_{z},S_{z}]

    But when it comes to solving for a function of t, I’m stuck with extra constants. My method here is to differentiate [tex]S_{x}[/tex] twice, and then solve the resulting differential equation.

    [tex]\frac{d^{2}S_{x}}{dt^{2}}=-\omega^{2} S_{x}[/tex]

    [tex]S_{x} = C_{1} e^{\imath \omega t}+C_{2} e^{-\imath \omega t}[/tex]

    What “initial/boundary conditions” do I use to determine the two constants above? Normalization of some sort?
    Last edited: Oct 4, 2007
  2. jcsd
  3. Oct 4, 2007 #2


    User Avatar
    Science Advisor

    Normalization isn't the issue. Since initial conditions aren't specified, you should take them to just be Si(0) (i=x,y,z), so that C1+C2=Sx(0). Can you see how to get C1-C2 from what you have so far?
  4. Oct 4, 2007 #3
    I'm not quite sure.

    I know of [tex]S_{x}[/tex] as,

    [tex]S_{x} = \frac{\hbar}{2} \sigma_{x}[/tex]

    ...but I'm not sure how that helps, considering both C's are complex coefficients, not operators, matrices, or vectors. At t=0, would S_x have any value at all, wouldn't it be zero considering the Hamiltonian? I was never really comfortable with spin...

    If [tex]S_{x}(t=0) = 0[/tex], then I suppose...
    [tex]C_{1} =-C_{2}[/tex]
    And perhaps we could set C_1 equal to 1 or hbar/2?
  5. Oct 4, 2007 #4
    I think your eq. (2) should read

    [tex]S_{x}(t) = C_{1} e^{i\omega t}+C_{2} e^{-i\omega t}[/tex]

    It is more convenient to rewrite it as

    [tex]S_{x}(t) = A \cos(\omega t) + B \sin(\omega t) [/tex]

    Supposedly you know the operator of spin at t=0 [itex] S_{x}(0), S_{y}(0), S_{z}(0)[/itex]. Then you obtain

    [tex]A = S_{x}(0) [/tex]

    and from eq. (1)

    [tex]B = \omega^{-1} d/dt [S_{x}(0)] = S_y(0)[/tex]

    So, the full solution is

    [tex]S_{x}(t) = S_x(0) \cos(\omega t) + S_y(0) \sin(\omega t) [/tex]

  6. Oct 4, 2007 #5
    Oh yes, forgot the i's. All fixed now.

    Using Euler's Formula and Eq. 1 to find B is very instructive. Thanks.
    Clearly, the vector precesses around the z-axis, as there's no change in S_z, and does so in an elliptical manner based on S_x(0) and S_y(0).

    We are not given S_i(0) for i=x,y,z. That is, I have given you all the information. I presume we cannot find them from the nature of the particle (and the subject of spin precession in a B-field, in general)? If so, I'll just leave them as is.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook