Context
We have a particle of spin $1/2$ and of magnetic moment $\vec{M} = \gamma\vec{S}$.
At time $t=0$, the state of the system is $$ |\psi(t=0) \rangle = |+\rangle_z $$
We let the system evolve under the influence of a magnetic field $B_0$ parallel to $Oy$.
We are asked to find the state of the system at time $t$.
Attempt
The energy of interaction between the $B$ field and the spin is $$\hat{H} = E_m \equiv -\vec{\mu} \cdot \vec{B} = - \gamma \vec{S} \cdot B_0 \hat{y} = - \gamma B_0 y S_y = \omega_0 S_y$$
where $\gamma$ denotes the gyromagnetic ratio and $S_y$ is the spin operator in the $y$ direction.
We can write the time-dependance of $\psi$ by applying the time-evolution operator on $|\psi(t=0)\rangle$, $$ \begin{align} |\psi(t)\rangle &= e^{-iHt/\hbar}|\psi(t=0)\rangle \\ &= e^{-i \omega_0 S_y t/\hbar}|+\rangle_z \end{align} $$
Now I would like to express $|+\rangle_z$ in the $\{|+\rangle_y, |-\rangle_y\}$ basis.
We can express $S_y$ in term of the Pauli matrix $\sigma_y$, $$ S_y = \frac{\hbar}{2} \underbrace{\begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}}_{\sigma_y} $$
Finding the eigenvalues of $S_y$, $$ det(S_y - \lambda I) = 0 \implies \lambda_{1,2} = \pm \frac{\hbar}{2} $$
Finding the eigenvectors, $$ (S_y - \lambda_{1,2}I)(\chi) = \vec{0} $$
- $\lambda_1 = \hbar/2$ $$ \frac{\hbar}{2} \begin{pmatrix}-1 & -i \\ i & -1\end{pmatrix} \begin{pmatrix}\chi_{_+} \\ \chi_{_-}\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix} \implies \begin{cases} -\chi_{_+} - i\chi_{_-} = 0 \\ i\chi_{_+} - \chi_{_{-}} = 0 \end{cases} $$
We find the normalised eigenvector $$ \chi_{_+} = \frac{1}{\sqrt{2}} \begin{pmatrix}1 \\ i\end{pmatrix} $$
We can write $$ |+\rangle_y = \frac{1}{\sqrt{2}}\left(|+\rangle_z + i|-\rangle_z \right) $$
- $\lambda_1 = -\hbar/2$ $$ \frac{\hbar}{2} \begin{pmatrix}1 & -i \\ i & 1\end{pmatrix} \begin{pmatrix}\chi_{_+} \\ \chi_{_-}\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix} \implies \begin{cases} \chi_{_+} - i\chi_{_-} = 0 \\ i\chi_{_+} + \chi_{_{-}} = 0 \end{cases} $$
We find the normalised eigenvector $$ \chi_{_-} = \frac{1}{\sqrt{2}} \begin{pmatrix}1 \\ -i\end{pmatrix} $$
We can write \begin{equation} |-\rangle_y = \frac{1}{\sqrt{2}}\left(|+\rangle_z - i|-\rangle_z \right) \end{equation}
Inverting the $|+\rangle_y$ and $|-\rangle_y$ relations we get $$ \begin{cases} |+\rangle_z = \frac{1}{\sqrt{2}}\left(|+\rangle_y + i|-\rangle_y \right) \\ |-\rangle_z = \frac{1}{\sqrt{2}}\left(|+\rangle_y - i|-\rangle_y \right) \end{cases} $$
Therefore, \begin{align} |\psi(t)\rangle &= e^{-i \omega_0 S_y t/\hbar}|+\rangle_z \\ &= e^{-i \omega_0 S_y t/\hbar} \frac{1}{\sqrt{2}}\left(|+\rangle_y + i|-\rangle_y \right) \end{align}
Now my question is how can i get rid of $S_y$ knowing that $S_y =\frac{\hbar}{2}\begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}$, I mean in class we saw that $$ |\psi(t)\rangle = e^{-i \frac{\omega_0}{2} t} \frac{1}{\sqrt{2}}\left(|+\rangle_y + i|-\rangle_y \right) $$ But I don't see how do we get that.