Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 32D / Solution

The Pauli matrices obey $(\mathbf{n}\cdot \mathbf{\sigma }{)}^2=I$. Expanding the exponential into even and odd powers gives the time-evolution operatorFor orthogonal states the identity contribution vanishes, so the transition probability is $|⟨{\chi }^{\prime }|\mathbf{n}\cdot \mathbf{\sigma }|\chi ⟩{|}^2{\sin }^2(\gamma Bt/2)$. For spin up to spin down in a static field $(B_x,0,B_z)$, only ${\sigma }_x$ has the required matrix element. Hence

For the weak rotating transverse field, take $H_0=-\hslash \gamma B_z{\sigma }_z/2$. The unperturbed energies give $E_{\downarrow }-E_{\uparrow }=\hslash \gamma B_z$. The perturbation matrix element is $⟨\downarrow |V(t)|\uparrow ⟩=-\hslash \gamma A{e}^{i\alpha t}/2$. Thus first-order perturbation theory givesSquaring its magnitude yieldsThe hypothesis is $|\gamma A|\ll |\Omega |$, excluding resonance. At $\alpha =0$ this is the leading weak-$A$ expansion of the exact static-field formula with $B_x=A$. For extremely long times, a small higher-order frequency correction can accumulate phase, so first-order perturbation is not a uniform approximation on arbitrarily long time scales.