Solving the <PERSON> equation of motion for the spin operators
This question is motivated by the first problem in the second chapter of Modern Quantum Mechanics (by <PERSON>).
I want to find the time evolution of the spin operators ($S_x,S_y,S_z$) in the <PERSON> picture of an electron ($e<0$) subject to the following Hamiltonian: $$ H = -\frac{eB}{mc}S_z=\omega S_z, \quad \omega \equiv -\frac{eB}{mc}S_z$$
I want to do it by solving the Heisenberg equation of motion.
On the notation used: $A^H$ is an operator in the <PERSON> formalism and $A$ in the <PERSON> formalism.
The <PERSON> equation of motion is:
$$ \frac{dA^H}{dt} = \frac{1}{i\hbar} [A^H,U^\dagger H U]$$
For any $S_i$: $$\frac{dA^H}{dt} = (1/i\hbar)[U^\dagger S_i U,U^\dagger H U]$$ $$ = (1/i\hbar) \omega [U^\dagger S_i U,U^\dagger S_z U] $$ $$ =(\omega/i\hbar) (U^\dagger S_i U U^\dagger S_z U - U^\dagger S_i U U^\dagger S_z U )$$ usung $UU^\dagger =1$ $$ = (\omega/i\hbar) U^\dagger (S_iS_z - S_zS_i) U = (\omega/i\hbar) U^\dagger [S_i,S_z] U$$ $$ = \omega U^\dagger \epsilon_{i,z,j}S_j U$$
For the case $S_z$, we clearly have $d_tS_z^H = 0$. Therefore, $ S_z^H (t) =S_z^H (0) =S_z^S $.
For the other cases, let's consider the time-evolution operator: $U=exp(-iHt/\hbar) = exp(-i\omega S_zt/\hbar)$. Then, we have:
$$ \frac{dS_i^H}{dt} = \omega \epsilon_{i,z,j} e^{-i \omega S_z t/\hbar} S_j e^{i \omega S_z t/\hbar}= \omega \epsilon_{i,z,j} S_j^H$$
which gives a system of two coupled equations:
$$ \frac{dS_x^H}{dt} = -\omega S_y^H $$ $$ \frac{dS_y^H}{dt} = \omega S_x^H $$
which leads to: $$ \frac{d^2S_x^H}{dt^2} = -\omega^2 S^H_x $$
and
$$ \frac{d^2S_y^H}{dt^2}= -\omega^2 S^H_y $$
I am not sure how to proceed, since $S^H_{x,y}$ are operators, hence a solution of the type $S^H_{x,y} = Ae^{i\omega t}+Be^{-i\omega t}$ doesn't make sense.
Is this approach correct?
If not (most likely), where is the flaw?
If it makes sense, what is the best way to go forward? 955