Discussion
An electron with mass \(m\) and charge \(q\) is in the spin up state \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\) at time \(t = 0\). A constant magnetic field is applied along the \(y\)-axis, \(B = B_{0}\hat{j}\), where \(B_{0}\) is a constant. The Hamiltonian of the system is \(H = -\hbar\omega\sigma_{y}\), where \(\omega = \frac{qB_{0}}{2 m} \gt 0\) and \(\sigma_{y} = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\). The minimum time after which the electron will be in the spin down state along the \(x\)-axis, i.e., \(\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}\), is
Answer: B
The minimum time is \(\frac{\pi}{4\omega}\).
The evolution operator for \(H = -\hbar\omega\sigma_{y}\) is \(U = e^{i\omega t\sigma_{y}} = \cos\omega t + i\sigma_{y}\sin\omega t\), and \(i\sigma_{y} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\). Acting on spin up:
\[ U\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \cos\omega t \\ -\sin\omega t \end{pmatrix} \]
This equals \(\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}\) first when \(\cos\omega t = \sin\omega t\), that is \(\omega t = \frac{\pi}{4}\).
The spin precesses about \(y\) at angular rate \(2\omega\), so turning from \(+z\) to \(-x\) (90°) takes \(\frac{\pi/2}{2\omega} = \frac{\pi}{4\omega}\). \(\frac{\pi}{2\omega}\) would take it to spin down along \(z\), and \(\frac{\pi}{\omega}\) back to spin up.
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