= Solution
The <Quantum de Finetti theorem> says that the fixed-$k$ <reduced density matrix> of an exchangeable $N$-particle state approaches
$$
\rho^{(k)}=\int \sigma^{\otimes k}\,d\mu(\sigma)
$$
as $N\to\infty$. Finite versions bound the <trace norm> error by a constant of order $d_{\rm loc}^2k/N$. On a bipartite lattice, the corresponding two-sublattice form is a mixture
$$
\rho_{AB}=\int \rho_A\otimes\rho_B\,d\mu(\rho_A,\rho_B)+o(1).
$$
This is the <mean-field ansatz from the quantum de Finetti theorem>.
The bond energy is a <linear functional> of $\rho_{AB}$. A <convex combination> cannot have energy below its lowest product component, so it remains only to minimize
$$
\operatorname{Tr}\!\left[
(\rho_A\otimes\rho_B)\,
J\mathbf S_A\mathbin\cdot\mathbf S_B
\right]
=\frac J4\mathbf r_A\mathbin\cdot\mathbf r_B.
$$
The <Cauchy-Schwarz inequality> gives $\mathbf r_A\mathbin\cdot\mathbf r_B\geq-|\mathbf r_A||\mathbf r_B|\geq-1$, with equality for pure antiparallel vectors. This reproduces the <Néel state> and $-J/4$ per bond found in part (a).
Yes, the limiting state saturates the de Finetti mean-field lower bound: the minimizing product state belongs to the allowed de Finetti mixture, and the finite-de-Finetti error tends to zero as $z\to\infty$. At finite $z$ the theorem gives only an approximation; entanglement and correlated fluctuations can lower the energy below the product-state value by corrections that vanish in the infinite-coordination limit.
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