How Do Heisenberg Equations Describe Spin Dynamics in a Magnetic Field?

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Homework Statement


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}##

Homework Equations

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.
 
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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.
 
DrDu said:
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.

I'm still having trouble solving this. I can't see how to find ##\hat S_y (0)##
 
At t=0, the operators in the Schroedinger and Heisenberg picture coincide.
 
DrDu said:
At t=0, the operators in the Schroedinger and Heisenberg picture coincide.

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?
 
I think so, what do you get?
 
DrDu said:
I think so, what do you get?

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)##
 
BOAS said:
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)##
Looks good!
 
DrDu said:
Looks good!

Thank you for your help
 
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