= Solution
For physical <spin ladder operators>, $S_j^\pm=S_j^x\pm iS_j^y$. Operators at distinct sites commute. Expanding the two products cancels their cross terms and gives
$$
\frac12(S_j^+S_{j+1}^-+S_j^-S_{j+1}^+)=S_j^xS_{j+1}^x+S_j^yS_{j+1}^y.
$$
Adding the remaining component proves
$$
\boxed{\mathbf S_j\cdot\mathbf S_{j+1}=\frac12(S_j^+S_{j+1}^-+S_j^-S_{j+1}^+)+S_j^zS_{j+1}^z.}
$$
For <spin one-half>, $S^+|\downarrow\rangle=\hbar|\uparrow\rangle$, $S^-|\uparrow\rangle=\hbar|\downarrow\rangle$, and raising an up <spin> or lowering a down <spin> gives zero. These formulas restore the physical $\hbar$ factors implicit in the dimensionless ladder hint.
Let $|n\rangle$ abbreviate the state with the only down <spin> at site $n$, keeping $|\psi_0\rangle$ for the fully up state. If $n$ lies on neither end of bond $(j,j+1)$, that bond contains two up <spins>. The ladder terms vanish, while $S_j^zS_{j+1}^z$ gives $\hbar^2/4$. Thus
$$
\mathbf S_j\cdot\mathbf S_{j+1}|n\rangle=\frac{\hbar^2}{4}|n\rangle\qquad(n\ne j,j+1).
$$
For a down <spin> at $j$, the $z$ product is instead $-\hbar^2/4$, and $S_j^+S_{j+1}^-$ moves it to $j+1$ with <amplitude> $\hbar^2$. For a down <spin> at $j+1$, the other ladder product moves it back. Therefore
$$
\boxed{\mathbf S_j\cdot\mathbf S_{j+1}|j\rangle=\frac{\hbar^2}{4}(2|j+1\rangle-|j\rangle),\qquad \mathbf S_j\cdot\mathbf S_{j+1}|j+1\rangle=\frac{\hbar^2}{4}(2|j\rangle-|j+1\rangle).}
$$
Every bond in $|\psi_0\rangle$ is aligned, so summing its $N$ bond energies gives
$$
\boxed{H|\psi_0\rangle=E_0|\psi_0\rangle,\qquad E_0=-\frac{NJ}{4}.}
$$
This is indeed a <ground-state energy>: a pair of spin-one-half particles has $\mathbf S_j\cdot\mathbf S_{j+1}$ <eigenvalues> $\hbar^2/4$ and $-3\hbar^2/4$, so each ferromagnetic bond is bounded below by $-J/4$, simultaneously attained by the fully aligned state. The isotropic <Heisenberg model> has a whole maximum-total-spin ground multiplet, not a unique all-up <ground state>.
In the <single-excitation subspace>, only the two bonds adjacent to the down <spin> differ from the aligned background. Their diagonal <energy> increase is $J$, and each neighboring hop has <amplitude> $-J/2$. Hence, with periodic site indices,
$$
\boxed{H|n\rangle=(E_0+J)|n\rangle-\frac J2(|n-1\rangle+|n+1\rangle).}
$$
This is an exact finite-dimensional hopping problem. Choose the active one-site translation convention $T|n\rangle=|n+1\rangle$. A periodic plane-wave coefficient pattern requires $e^{iq(n+N)a}=e^{iqna}$, so
$$
\boxed{q=\frac{2\pi\ell}{Na},\qquad \ell\in\mathbb Z,\qquad q\ \text{identified modulo }\frac{2\pi}{a}.}
$$
There are $N$ independent such <wavevectors>, for example one set of $N$ consecutive integers $\ell$ in a <Brillouin zone>. The normalized <one-magnon Bloch states on a ring> are $|q\rangle=N^{-1/2}\sum_{n=1}^Ne^{iqna}|n\rangle$. Reindexing with the periodic coefficient condition gives $T|q\rangle=e^{-iqa}|q\rangle$ and hence $T^N|q\rangle=|q\rangle$.
There is a logical qualification to the condition stated in the paper. On an $N$-site ring $T^N=I$ already holds for every state. In particular, the finite sum written with an arbitrary real $q$ also returns to itself after $N$ translations; that condition alone does not quantize $q$. The displayed allowed values are the ones for which the ansatz is a single one-site translation <eigenstate>, equivalently a consistently periodic <plane wave>. This is the additional condition needed for the subsequent energy-eigenstate argument. Other $q$ values generally describe superpositions of the allowed <momentum> modes.
Apply the hopping <Hamiltonian operator> and shift the summation index using <periodic boundary condition>. The coefficients from $|n-1\rangle$ and $|n+1\rangle$ are respectively $e^{iqa}$ and $e^{-iqa}$ times the original coefficient. Thus
$$
H|q\rangle=\left(E_0+J-\frac J2e^{iqa}-\frac J2e^{-iqa}\right)|q\rangle,
$$
and the <nearest-neighbour ferromagnetic magnon dispersion> is
$$
\boxed{E(q)=E_0+J[1-\cos(qa)]=E_0+2J\sin^2\frac{qa}{2}.}
$$
The unnormalized state in the question has the same <energy>. Distinct allowed <momenta> are <orthogonal>, since $N^{-1}\sum_ne^{i(q-q')na}=\delta_{qq'}$ over one set of representatives.
\Image[/past-exam-of-the-advanced-quantum-physics-course-of-the-university-of-cambridge/2018-magnon-dispersion.png]
{title=Exact one-magnon dispersion and its discrete ring momenta over two reciprocal periods}
For large $N$ the sampled <wavevectors> become dense, and the requested sketch is the periodic curve $[E(q)-E_0]/J=1-\cos(qa)$ over $-2\pi<qa<2\pi$. It has a zero at $q=0$ and tends to the equivalent zero at either open endpoint; maxima $2J$ occur at $q=\pm\pi/a$. Near any equivalent zero,
$$
\boxed{E(q)-E_0=\frac{Ja^2q^2}{2}+O(q^4),}
$$
using $q$ measured from that zero. The low-momentum <ferromagnetic magnon> is quadratic, with <group velocity> $\hbar^{-1}dE/dq=(Ja/\hbar)\sin(qa)$. The $q=0$ state is proportional to the total <spin lowering operator> acting on $|\psi_0\rangle$, so it belongs to the same ground multiplet and has exactly zero excitation <energy>. Extending the plot beyond one reciprocal period repeats states rather than adding new independent modes. The lattice spacing $a$ is the nearest-neighbor chord length specified in the paper; $Na$ is the translation length in this lattice description, not necessarily the continuous circumference of a circular arrangement.
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