Eigenstates of a Hamiltonian For a particle with a spin of 1/2, which was exposed to both magnetic fields 
$B_{0}=B_{z}e_z$
 and 
$B_1=B_xe_x$
I already found the eigenvalues of its Hamiltonian which is given by 
\begin{equation}
H = -\gamma (B_0+B_1)\cdot S=-\dfrac{\gamma \hbar}{2} (B_x\sigma_x+B_z\sigma_z)
\end{equation}
and they were 
$E_{\pm}=\mp \dfrac{\gamma \hbar}{2} \sqrt{B_z^2+B_x^2}$
can anyone guide me to find its eigenstates?
 A: Hint: Consider a matrix $\mathbf{A}$ with eigenvalue $\lambda$ and eigenvector $\mathbf{u}$.
\begin{align}
\mathbf{A} \mathbf{u} \; = \; \lambda \mathbf{u}
\end{align}
Since your question is for $2 \times 2$ matrix, we explicitly write it as
\begin{align}
\underbrace{\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}}_{\mathbf{A}}
\;
\underbrace{
\begin{pmatrix}
u_{1}\\
u_{2}
\end{pmatrix}
}_{\mathbf{u}}
\; = \;
\lambda
\;
\underbrace{
\begin{pmatrix}
u_{1}\\
u_{2}
\end{pmatrix}
}_{\mathbf{u}}
\end{align}
For each eigenvalue $\lambda$, we are ready to solve $u_{1}$ and $u_{2}$
\begin{align}
au_{1} + bu_{2} = \lambda u_{1} \tag{1}\\
cu_{1} + du_{2} = \lambda u_{1} \tag{2}
\end{align}
Notice that Eq. (1) and (2) are linearly dependent in this case. So you actually need one equation only (say Eq. (1)). Thus, we have
\begin{align}
b u_{2} \; = \; (\lambda - a)u_{1}
\end{align}
Now, it seems that we don't have unique solution. Yes, remember that eigenvector is in general not unique since the vector norm is not fixed.
We pick $u_{1} = 1$, then we have $u_{2} = (\lambda - a)/b$. Since we are talking a quantum state, we have to normalize the eigenvector by $\sqrt{u_{1}^{2} + u_{2}^{2}}$. 
