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Quantum physics course of the University of Oxford

Past exam of the quantum physics course of the University of Oxford

2016

In the ordered <basis> $|+\rangle,|-\rangle$, the transverse <magnetic field> produces
$$
V(t)=\mu_BB_1(\sigma_x\cos\omega t+\sigma_y\sin\omega t)
=\boxed{\hbar\gamma\begin{pmatrix}0&e^{-i\omega t}\\e^{i\omega t}&0\end{pmatrix}},\qquad\gamma=\frac{\mu_BB_1}{\hbar}.
$$
The off-diagonal complex exponentials follow directly from $\cos\omega t\mp i\sin\omega t$. This <matrix> is <Hermitian> at every time.

Solution

Past exam of the quantum physics course of the University of Oxford / 2016 / 10 / b / Solution

Past exam of the quantum physics course of the University of Oxford / 2016 / 10 / b / Solution

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Past exam of the quantum physics course of the University of Oxford / 2016 / 10 / b / Solution

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