\begin {align*} H & =-\mu \cdot B\\ & =\frac {eB}{m_{e}}S_{z}\\ & =\frac {eB\hbar }{2m_{e}}\begin {pmatrix} 1 & 0\\ 0 & -1 \end {pmatrix} \end {align*}
The eigenvalues are \(E_{+}=\frac {eB\hbar }{2m_{e}}m,E_{-}=-\frac {eB\hbar }{2m_{e}}m\)\begin {align*} i\hbar \frac {d}{dt}|X\rangle & =H|X\rangle \\ & =\frac {eB\hbar }{2m_{e}}\begin {pmatrix} 1 & 0\\ 0 & -1 \end {pmatrix} |X\rangle \end {align*}
Hence\begin {align*} i\begin {bmatrix} \dot {x}_{1}\relax (t) \\ \dot {x}_{2}\relax (t) \end {bmatrix} & =\frac {eB}{2m_{e}}\begin {bmatrix} x_{1}\relax (t) \\ x_{2}\relax (t) \end {bmatrix} \\ \hbar \dot {x}_{1}\relax (t) & =\frac {eB}{2m_{e}}x_{1}\relax (t) \\ \hbar \dot {x}_{2}\relax (t) & =-\frac {eB}{2m_{e}}x_{2}\left ( t\right ) \end {align*}
The solution is\begin {align*} x_{1}\relax (t) & =\frac {1}{\sqrt {2}}e^{-i\gamma t}\\ x_{2}\relax (t) & =\frac {1}{\sqrt {2}}e^{i\gamma t} \end {align*}
Or\[ |X\rangle =\frac {1}{\sqrt {2}}\begin {bmatrix} e^{-i\gamma t}\\ e^{i\gamma t}\end {bmatrix} \] Where \(\gamma =\frac {eB}{2m_{e}}\)\begin {align*} |X\rangle & =c_{+}|S_{x}=\frac {\hbar }{2}\rangle +c_{-}|S_{x}=-\frac {\hbar }{2}\rangle \\ c_{+} & =\langle S_{x}=\frac {\hbar }{2}|X\rangle \\ & =\frac {1}{\sqrt {2}}\frac {1}{\sqrt {2}}\begin {bmatrix} 1 & 1 \end {bmatrix}\begin {bmatrix} e^{-i\gamma t}\\ e^{i\gamma t}\end {bmatrix} \\ & =\frac {1}{2}\left (e^{i\gamma t}+e^{-i\gamma t}\right ) \\ & =\cos \gamma t \end {align*}
Probability to measure \(S_{x}=\frac {\hbar }{2}\) at \(t>0\) is \(P\relax (t) =\left \vert c_{+}\right \vert ^{2}=\cos ^{2}\gamma t\). And\begin {align*} |X\rangle & =c_{+}|S_{x}=\frac {\hbar }{2}\rangle +c_{-}|S_{x}=-\frac {\hbar }{2}\rangle \\ c_{-} & =\langle S_{x}=-\frac {\hbar }{2}|X\rangle \\ & =\frac {1}{\sqrt {2}}\frac {1}{\sqrt {2}}\begin {bmatrix} -1 & 1 \end {bmatrix}\begin {bmatrix} e^{-i\gamma t}\\ e^{i\gamma t}\end {bmatrix} \\ & =\frac {1}{2}\left (e^{i\gamma t}-e^{i\gamma t}\right ) \\ & =i\sin \gamma t \end {align*}
Probability to measure \(S_{x}=-\frac {\hbar }{2}\) at \(t>0\) is \(P\left ( t\right ) =\left \vert c_{-}\right \vert ^{2}=\sin ^{2}\gamma t\)