\[ H=\frac {e\hslash B}{2m}\begin {pmatrix} 0 & -i\\ i & 0 \end {pmatrix} \] The eigenvalues can be found to be \(E_{+}=\frac {e\hslash B}{2m}=\hslash \lambda ,E_{-}=-\frac {e\hslash B}{2m}=-\hslash \lambda \) where \(\lambda =\frac {eB}{2m}\). The associated normalized eigenvectors are \begin {align*} v_{1} & =\frac {1}{\sqrt {2}}\begin {pmatrix} 1\\ i \end {pmatrix} \\ v_{2} & =\frac {1}{\sqrt {2}}\begin {pmatrix} i\\ 1 \end {pmatrix} \end {align*}
To find the spin state for \(t>0\), we need to solve the Schrodinger equation \[ i\hslash \frac {\partial }{\partial t}|\psi \rangle =H|\psi \rangle \] In the basis of \(H\) the above becomes\begin {align*} i\hslash \begin {pmatrix} x_{1}^{\prime }\relax (t) \\ x_{2}^{\prime }\relax (t) \end {pmatrix} & =\begin {pmatrix} E_{+} & 0\\ 0 & E_{-}\end {pmatrix}\begin {pmatrix} x_{1}\relax (t) \\ x_{2}\relax (t) \end {pmatrix} \\ & =\begin {pmatrix} \hslash \lambda & 0\\ 0 & -\hslash \lambda \end {pmatrix}\begin {pmatrix} x_{1}\relax (t) \\ x_{2}\relax (t) \end {pmatrix} \end {align*}
Hence\begin {align*} i\hslash x_{1}^{\prime }\relax (t) & =\hslash \lambda x_{1}\left ( t\right ) \\ i\hslash x_{2}^{\prime }\relax (t) & =-\hslash \lambda x_{2}\left ( t\right ) \end {align*}
Or
\begin {align*} x_{1}^{\prime }\relax (t) & =-i\lambda x_{1}\relax (t) \\ x_{2}^{\prime }\relax (t) & =i\lambda x_{2}\relax (t) \end {align*}
The solution is\begin {align*} x_{1}\relax (t) & =x_{1}\relax (0) e^{-i\lambda t}\\ x_{2}\relax (t) & =x_{2}\relax (0) e^{i\lambda t} \end {align*}
In the original basis, this becomes\begin {align} \begin {pmatrix} X_{1}\relax (t) \\ X_{2}\relax (t) \end {pmatrix} & =x_{1}\relax (t) \vec {v}_{1}+x_{2}\relax (t) \vec {v}_{2}\nonumber \\ & =\frac {1}{\sqrt {2}}x_{1}\relax (0) e^{-i\lambda t}\begin {pmatrix} 1\\ i \end {pmatrix} +\frac {1}{\sqrt {2}}x_{2}\relax (0) e^{i\lambda t}\begin {pmatrix} i\\ 1 \end {pmatrix} \tag {1} \end {align}
What is left is to determine \(x_{1}\relax (0) ,x_{2}\relax (0) \). We are told that at \(t=0\), \(\begin {pmatrix} X_{1}\relax (0) \\ X_{2}\relax (0) \end {pmatrix} =\begin {pmatrix} 1\\ 0 \end {pmatrix} \) since \(\begin {pmatrix} 1\\ 0 \end {pmatrix} \) is the eigenstate associated with \(\frac {\hslash }{2}\) of \(S_{z}\). Therefore the above becomes (at \(t=0\)) \begin {align*} \begin {pmatrix} 1\\ 0 \end {pmatrix} & =\frac {1}{\sqrt {2}}x_{1}\relax (0) \begin {pmatrix} 1\\ i \end {pmatrix} +\frac {1}{\sqrt {2}}x_{2}\relax (0) \begin {pmatrix} i\\ 1 \end {pmatrix} \\ & =\frac {1}{\sqrt {2}}\begin {pmatrix} 1 & i\\ i & 1 \end {pmatrix}\begin {pmatrix} x_{1}\relax (0) \\ x_{2}\relax (0) \end {pmatrix} \end {align*}
Hence\begin {align*} \begin {pmatrix} x_{1}\relax (0) \\ x_{2}\relax (0) \end {pmatrix} & =\sqrt {2}\begin {pmatrix} 1 & i\\ i & 1 \end {pmatrix} ^{-1}\begin {pmatrix} 1\\ 0 \end {pmatrix} \\ & =\sqrt {2}\frac {\begin {pmatrix} 1 & -i\\ -i & 1 \end {pmatrix}\begin {pmatrix} 1\\ 0 \end {pmatrix} }{1-\left (i^{2}\right ) }\\ & =\frac {\sqrt {2}}{2}\begin {pmatrix} 1 & -i\\ -i & 1 \end {pmatrix}\begin {pmatrix} 1\\ 0 \end {pmatrix} \\ & =\frac {\sqrt {2}}{2}\begin {pmatrix} 1\\ -i \end {pmatrix} \\ & =\frac {1}{\sqrt {2}}\begin {pmatrix} 1\\ -i \end {pmatrix} \end {align*}
Therefore\begin {align*} x_{1}\relax (0) & =\frac {1}{\sqrt {2}}\\ x_{2}\relax (0) & =\frac {-i}{\sqrt {2}} \end {align*}
Hence the solution (1) becomes\begin {align*} \begin {pmatrix} X_{1}\relax (t) \\ X_{2}\relax (t) \end {pmatrix} & =\frac {1}{\sqrt {2}}\frac {1}{\sqrt {2}}e^{-i\lambda t}\begin {pmatrix} 1\\ i \end {pmatrix} +\frac {1}{\sqrt {2}}\frac {-i}{\sqrt {2}}e^{i\lambda t}\begin {pmatrix} i\\ 1 \end {pmatrix} \\ & =\frac {1}{2}e^{-i\lambda t}\begin {pmatrix} 1\\ i \end {pmatrix} -\frac {i}{2}e^{i\lambda t}\begin {pmatrix} i\\ 1 \end {pmatrix} \end {align*}
Therefore\begin {align*} X_{1}\relax (t) & =\frac {1}{2}e^{-i\lambda t}-\frac {i^{2}}{2}e^{i\lambda t}\\ X_{2}\relax (t) & =i\frac {1}{2}e^{-i\lambda t}-\frac {i}{2}e^{i\lambda t} \end {align*}
Or\begin {align*} X_{1}\relax (t) & =\frac {1}{2}e^{-i\lambda t}+\frac {1}{2}e^{i\lambda t}\\ X_{2}\relax (t) & =-\frac {1}{2i}e^{-i\lambda t}+\frac {1}{2i}e^{i\lambda t} \end {align*}
Or\begin {align*} X_{1}\relax (t) & =\cos \left (\lambda t\right ) \\ X_{2}\relax (t) & =\sin \left (\lambda t\right ) \end {align*}
Or\begin {align*} X_{1}\relax (t) & =\cos \left (\frac {eB}{2m}t\right ) \\ X_{2}\relax (t) & =\sin \left (\frac {eB}{2m}t\right ) \end {align*}
Hence\[ |\psi \rangle =\begin {pmatrix} \cos \left (\frac {eB}{2m}t\right ) \\ \sin \left (\frac {eB}{2m}t\right ) \end {pmatrix} \]