Time evolution with rotation Hamiltonian 
At $t=0$, the wave function of a particle with Hamiltonian
  $$\mathcal{H}=\mu B L_y \equiv \omega L_y$$ is given by $$\left
 \langle \mathbf{r}|\alpha \right \rangle \equiv \psi\left ( \mathbf{r}
 \right )= f\left ( r \right )\left ( r+z \right ).$$ Write the time
  evolution of the state.

I tried to write everything in Dirac notation; so, since
$$r+z=r\left ( 1+\frac{z}{r} \right )=r\sqrt{4\pi}\left ( Y_0^0 + \frac{1}{\sqrt{3}}Y_1^0\right ),$$
putting $R(r)\equiv rf(r)$, I wrote the normalized ket in the $\left \{ \left|L,L_z\right\rangle \right \}$ basis as
$$\left|\psi \right \rangle = \frac{\sqrt{3}}{2}R\left ( r \right )\left [ \left|00\right \rangle + \frac{1}{\sqrt{3}}\left|10\right \rangle \right ].$$
I've got trouble in writing $\left|\psi (t) \right\rangle$.
 A: This is but a classical rotation around the y-axis masquerading as a quantum problem. It amounts to rotation of z to x and back.
The evolution operator is $$\exp (-itH/\hbar)= \exp (-it\omega L_y/\hbar)=\exp(\theta K)\equiv R(\theta),$$ an orthogonal 3d rotation matrix, where $\theta= t\omega/\hbar$ and the Hermitean  spin (-one) matrix generator $L_y$ is proportional to the Cartesian antisymmetric rotation generator $K_y$,
$$
L_y= i     
\left[\begin{array}{ccc}
0 & 0 & 1 \\
0 & 0 & 0\\
-1 & 0 & 0
\end{array}\right]  =iK_y. 
$$
By the Rodrigues rotation formula, by inspection,
$$
R(\theta)=  I + (\sin\theta)  K_y  + (1-\cos\theta) K_y ^2= \left[\begin{array}{ccc}
\cos \theta & 0 & \sin \theta \\
0 & 1 & 0\\
-\sin \theta & 0 & \cos \theta
\end{array}\right].
$$
This is just a busy way of promoting (x,z) to $(x\cos \theta +z\sin\theta, z\cos\theta -x\sin\theta )$, and of course leaving the scalar r alone. Thus, the future wave function is $f(r)(r+z \cos\theta -x\sin \theta)$, with θ as defined above.


*

*To check unitarity, indeed, orthogonality, of R, simply note $K_y^3=-K_y$, and $K_y^4=-K_y^2$, so $R R^T=\mathbb{1}$.

*NB. The reason I use this classical-friendly Cartesian basis for the Ls and not the strictly equivalent more conventional QM spherical tensor one, T , is because the eigenvectors of the Ls are far more evident, instantly specified by the components of (x,y,z), just as they are in sophomore physics. 

*If you however insist on sticking to the inferior spherical basis, your evolution matrix turns into a monster,
$$
\left[\begin{array}{ccc}
(1+\cos\theta)/2 & -\sin \theta /\sqrt{2} & (1-\cos\theta)/2 \\
\sin \theta /\sqrt{2}  & \cos\theta & -\sin \theta /\sqrt{2} \\
(1-\cos\theta)/2 & \sin \theta /\sqrt{2}  & (1+\cos\theta)/2
\end{array}\right], 
$$
orthogonal and collapsing to the identity at t=0, naturally!
