Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 33A / Solution

In the interaction picture,At infinity this is $-(i\lambda /\hslash ){\sum }_k{\alpha }_k{W}_k{F}_k$, whereso $b_1=1,b_2=1,b_3=-1$ and ${\omega }_{f{i}_k}=(E_f-E_{i_k})/\hslash$. The probability is $({\lambda }^2/{\hslash }^2)|{\sum }_k{\alpha }_k{W}_k{F}_k{|}^2$, displaying interference between initial components. For $N$ equal response factors, Cauchy-Schwarz gives $|{\sum }_k{\alpha }_k{|}^2\le N$, attained by equal phases and $|{\alpha }_k|=N^{-1/2}$; the largest probability is $N{\lambda }^2|WF{|}^2/{\hslash }^2$ within perturbation theory.

Solved by gpt-5.6-sol high.