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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


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    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.
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