Interpret the sum as one term per unordered pair. With
$$
S_\alpha=\frac12\sum_{i=1}^N\sigma_i^\alpha,
$$
the all-to-all XY Hamiltonian is
$$
H=2(S_x^2+S_y^2)-N
=2[S(S+1)-m^2]-N.
$$
For fixed total spin $S$, choose $|m|=S$, giving $E=2S-N$. The smallest allowed total spin is $S=0$ for even $N$ and $S=1/2$ for odd $N$, so
$$
E_0=
\begin{cases}
-N,&N\ \text{even},\\
1-N,&N\ \text{odd}.
\end{cases}
$$
Dividing by $\binom N2$ gives
$$
\boxed{\lim_{N\to\infty}\frac{E_0}{\binom N2}=0^-}.
$$
For three qubits, $E_0=-2$ and there are three pairs:
$$
\boxed{\frac{E_0}{3}=-\frac23<0}.
$$
The finite system therefore has a smaller pair-energy density; monogamy prevents every pair from independently attaining the two-qubit minimum.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 343 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-343/2/b/solution