 Solution

The Quantum de Finetti theorem says that the fixed-$k$ reduced density matrix of an exchangeable $N$-particle state approachesas $N\to \infty$. Finite versions bound the trace norm error by a constant of order $d_{\mathrm{loc}}^{2}k/N$. On a bipartite lattice, the corresponding two-sublattice form is a mixtureThis 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 minimizeThe Cauchy-Schwarz inequality gives ${\mathbf{r}}_A\cdot {\mathbf{r}}_B\ge -|{\mathbf{r}}_A||{\mathbf{r}}_B|\ge -1$, 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.