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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-60/3/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 60 / 3 / b / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Use

t_{0} = 0

as in the displayed carrier-phase convention. The drift is diagonal, and conjugation gives

H_{1} (t) = (0 e^{- iω t} e^{iω t} 0) .

Multiplying by

A (t) cos (ω t + ϕ)

and expanding the cosine into exponentials yields

f (t) H_{1} (t) = \frac{A ( t )}{2} (0 e^{i ϕ} + e^{- i (2 ω t + ϕ)} e^{- i ϕ} + e^{i (2 ω t + ϕ)} 0) .

The rotating-wave approximation averages away the components oscillating at

2 ω

, retaining

H_{I}^{RWA} (t) = \frac{A ( t )}{2} (0 e^{i ϕ} e^{- i ϕ} 0) = \frac{A ( t )}{2} (cos ϕ σ_{x} + sin ϕ σ_{y}) .

The envelope is slowly varying and the effective coupling is weak compared with the carrier frequency. The pulse phase selects the transverse rotation axis on the Bloch sphere; its pulse area determines the angle. Counter-rotating corrections are neglected under this approximation, not identically zero.

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