GATE 2025 PH – Question 45
An electron of mass $m$ and charge $q$ is initially spin up, $\begin{pmatrix}1\\0\end{pmatrix}$. A constant field $\mathbf B=B_0\hat j$ is applied. The Hamiltonian is $H=-\hbar\omega\sigma_y$, where $\omega=qB_0/(2m)>0$ and $$\sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix}.$$ The minimum time to reach spin down along x, $\frac1{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix}$, is
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Correct answer: (B) $\pi/(4\omega)$
Explanation
**Time evolution.** For $H=-\hbar\omega\sigma_y$ the evolution operator is
$$U(t)=e^{-iHt/\hbar}=e^{i\omega t\sigma_y}=\cos(\omega t)\,I+i\sin(\omega t)\,\sigma_y .$$
With $\sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix}$, $i\sigma_y=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. Acting on the initial state $(1,0)^T$:
$$\psi(t)=\begin{pmatrix}\cos\omega t\\-\sin\omega t\end{pmatrix}.$$
**Target state** (spin down along x): $\dfrac1{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix}$. This is reached when $\cos\omega t=\sin\omega t=\dfrac1{\sqrt2}$, i.e. $\omega t=\dfrac\pi4$.
The minimum time is $t=\dfrac{\pi}{4\omega}$ (option B).