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

1. Oct 4, 2007

### 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:

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

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

$$\frac{d S_{z}}{dt}=\frac{1}{\imath \hbar}[S_{z}, -\omega S_{z}] =-\frac{\omega}{\imath \hbar}[S_{z},S_{z}] =0$$

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?

Last edited: Oct 4, 2007
2. Oct 4, 2007

### Avodyne

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?

3. Oct 4, 2007

### 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?

4. Oct 4, 2007

### meopemuk

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.

5. Oct 4, 2007

### 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.

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