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

[logic smogic]

Problem
Consider the spin precession problem in the Heisenberg picture. Using the Hamiltonian

H=-\omega S_{z}

where
\omega=\frac{eB}{mc}

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

Formulae

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

A_{H}=U^{\dagger}A_{S}U

U=e^{\frac{-\imath H t}{\hbar}}

Attempt
Well, computing the Heisenberg equations is pretty straitforward:

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

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

\frac{d S_{z}}{dt}=\frac{1}{\imath \hbar}[S_{z}, -\omega S_{z}]<br /> =-\frac{\omega}{\imath \hbar}[S_{z},S_{z}]<br /> =0

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

\frac{d^{2}S_{x}}{dt^{2}}=-\omega^{2} S_{x}

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

What “initial/boundary conditions” do I use to determine the two constants above? Normalization of some sort?

 

[Avodyne]

Normalization isn't the issue. Since initial conditions aren't specified, you should take them to just be S_i(0) (i=x,y,z), so that C₁+C₂=S_x(0). Can you see how to get C₁-C₂ from what you have so far?

 

[logic smogic]

I'm not quite sure.

I know of S_{x} as,

S_{x} = \frac{\hbar}{2} \sigma_{x}

...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 S_{x}(t=0) = 0, then I suppose...
C_{1} =-C_{2}
And perhaps we could set C_1 equal to 1 or hbar/2?

 

[meopemuk]

Attempt
Well, computing the Heisenberg equations is pretty straitforward:

\frac{d S_{x}}{dt}=\frac{1}{\imath \hbar}[S_{x}, -\omega S_{z}]<br /> =-\frac{\omega}{\imath \hbar}[S_{x},S_{z}]<br /> =\omega S_{y}...(1)


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

\frac{d^{2}S_{x}}{dt^{2}}=-\omega^{2} S_{x}

S_{x} = C_{1} e^{\omega t}+C_{2} e^{-\omega t}...(2)

What “initial/boundary conditions” do I use to determine the two constants above? Normalization of some sort?

I think your eq. (2) should read

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

It is more convenient to rewrite it as

S_{x}(t) = A \cos(\omega t) + B \sin(\omega t)

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

A = S_{x}(0)

and from eq. (1)

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

So, the full solution is

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

Eugene.

 

[logic smogic]

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.