An electron is described by the Hamiltonian
$ H=\frac{e}{mc}\bar{S}\cdot\bar{B} $
where $\bar{S} =(S_x,S_y,S_z)$ is the spin operator and $\bar{B}$ the magnetic field.
For $t>0$ $\bar{B}=B_0\hat{x}$ and for $t=0$ the electron is in the state $|\psi\rangle=\frac{1}{\sqrt3}|+\rangle+\frac{\sqrt2}{\sqrt3}|-\rangle$, with $|+\rangle$ and $|-\rangle$ eigenvectors of the operator $S_z$ (with eigenvalues $\pm \frac{\hbar}{2}$).
I want to determine the time evolution of the state.
Since $B$ is along x, the dot product gives me $S_xB_0$ and I know that the operator $S_x$ is rapresented by the matrix $ \frac{\hbar}{2}\bigl(\begin{smallmatrix} 0 & 1 \\1 & 0\end{smallmatrix}\bigr)$.
For convenience $\frac{e\hbar B_0}{2mc}=\epsilon$.
Now, avoiding all the calculations (hoping I did them right), what I did is to find the eigenvalues and eigenvectors of my operator by setting the equation:
$ H(\alpha|+\rangle+\beta|-\rangle)=E(\alpha|+\rangle+\beta|-\rangle) $
From which the eigenvalues are $\epsilon$ and $-\epsilon$, and the respective eigenvectors (determining the relation between $\alpha$ and $\beta$ and normalizing) are:
$|\psi_1\rangle=\frac{1}{\sqrt2}|+\rangle+\frac{1}{\sqrt2}|-\rangle$
$|\psi_2\rangle=\frac{1}{\sqrt2}|+\rangle-\frac{1}{\sqrt2}|-\rangle$
Now I need to express the state $|\psi\rangle$ as a combination of the two eigenvectors:
$ |\psi\rangle=\frac{\sqrt2+2}{2\sqrt3}|\psi_1\rangle+\frac{\sqrt2-2}{2\sqrt3}|\psi_2\rangle $
And the time evolution is:
$ |\psi(t)\rangle=\frac{\sqrt2+2}{2\sqrt3}|\psi_1\rangle \exp[{-i\frac{\epsilon}{\hbar}t}]+\frac{\sqrt2-2}{2\sqrt3}|\psi_2\rangle \exp[{i\frac{\epsilon}{\hbar}t}] $
This is how I solved the problem but since I'm pretty new to the quantum mechanics I'd like to have some opinions.
Is this a reasonable procedure? Did I made some terrible mistakes or non-sense considerations?