= Solution
In the <single-excitation subspace>, part (a) turns each coupling into a hopping coefficient between adjacent sites. With $|\widetilde n\rangle$ denoting the excitation at site $N+n$, the two central links give
$$
\boxed{H_e|\widetilde0\rangle=\frac{J_0}{\sqrt2}(|\widetilde{-1}\rangle+|\widetilde1\rangle).}
$$
For $N\geq2$ the links from these neighbors to the next sites have strength $J_1$. Applying the <Hamiltonian operator> a second time yields
$$
\boxed{H_e^2|\widetilde0\rangle=J_0^2|\widetilde0\rangle+\frac{J_0J_1}{\sqrt2}(|\widetilde{-2}\rangle+|\widetilde2\rangle).}
$$
The central coefficient is $2(J_0/\sqrt2)^2=J_0^2$. On the comparison path, direct multiplication gives
$$
H_T|0\rangle=J_0|1\rangle,\qquad H_T^2|0\rangle=J_0^2|0\rangle+J_0J_1|2\rangle.
$$
Thus the two calculations match under $|0\rangle\mapsto|\widetilde0\rangle$ and $|j\rangle\mapsto(|\widetilde{-j}\rangle+|\widetilde j\rangle)/\sqrt2$ for $j=1,2$. If $N=1$, the terms involving $J_1$ and distance two are absent on both sides.
Back to article page
