# Time evolution of state in rotating $B$-field spin system

I'm given a state $|\psi(t)\rangle$ that solves the time dependent Schroedinger equation with the Hamilton operator $\hat{H}(t)$. $\hat{H}(t)$ describes a Spin-$\frac{1}{2}$ in a rotating magnetic field ($\vec{B}(t) = B_1[\cos(\omega t)\vec{e_1} + \sin(\omega t) \vec{e_2}] + B_3 \vec{e_3}$) rotating around the z-axis with frequency $\omega$. Furthermore I have a transformed state $|\widetilde{\psi}(t)\rangle = \hat{D}(t) |\psi(t)\rangle$ with $\hat{D}(t)$ being the rotation operator $\hat{D}(t) = e^{i\omega t \hat{\sigma_3} /2}$. I have calculated that therefore the Hamiltonian solving the SG in this case becomes: \begin{align} \hat{\widetilde{H}}(t) = \frac{\hbar \Omega_1}{2} \hat{\sigma_1} + \frac{\hbar(\Omega_3-\omega)}{2}\hat{\sigma_3}, \quad \Omega_1 = -2\frac{\mu_0 B_1}{\hbar}, \quad \Omega_3 = -2\frac{\mu_0 B_3}{\hbar} \end{align} The task is to calculate the two time evolution operators $\hat{\widetilde{U}}(t)$ and $\hat{{U}}(t)$ given by $|\widetilde{\psi}(t)\rangle = \hat{\widetilde{U}}(t) |\widetilde{\psi}(0)\rangle$ and $|\psi(t)\rangle = \hat{{U}}(t) |\psi(0)\rangle$. I plugged in the standard formula for time evolution and tried to get a result by simply puttung in the Hamiltonians, but I don't get any further. I seem to have a blackout how to deal with this problem

The D operator you have accounts for a rotation of your state "around" the $\sigma_3$ matrix: if you think of this just as with vectors, you can see that D therefore accounts for the $B_x$ and $B_y$ components of your field. Therefore, you can use the $B_z$ part to solve the evolution of your rotating state.