Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 57 / 3 / b / Solution

In the single-excitation subspace, part (a) turns each coupling into a hopping coefficient between adjacent sites. With $|\tilde{n}⟩$ denoting the excitation at site $N+n$, the two central links giveFor $N\ge 2$ the links from these neighbors to the next sites have strength $J_1$. Applying the Hamiltonian operator a second time yieldsThe central coefficient is $2(J_0/\sqrt{2}{)}^2=J_0^2$. On the comparison path, direct multiplication givesThus the two calculations match under $|0⟩\mapsto |\tilde{0}⟩$ and $|j⟩\mapsto (|\widetilde{-j}⟩+|\tilde{j}⟩)/\sqrt{2}$ for $j=1,2$. If $N=1$, the terms involving $J_1$ and distance two are absent on both sides.