Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 343 / 2

Entanglement monogamy says that maximal entanglement with one independent system excludes entanglement with another. For three qubits this is quantified by the Coffman--Kundu--Wootters inequalityFor example, if $AB$ is a maximally entangled state, then $C_{AB}=1$ and the global state factorizes as $|{\psi }^{-}{⟩}_{AB}\otimes |\varphi {⟩}_C$ up to local unitaries, so $C_{AC}=C_{BC}=0$. The three-qubit GHZ state instead has entanglement across every one-versus-two cut but no pairwise concurrence after the third qubit is traced out.

Taking the sum once over each unordered pair, write the All-to-all Heisenberg model asFor even $N$, the minimum total spin is $S=0$, henceFor odd $N$, $S=1/2$ and $E_0=3/2-3N/2$. If the paper's ${\sum }_{ij}$ counts ordered pairs or includes $i=j$, the corresponding harmless factors and additive constant change, but the minimizing total-spin sector is the same.

Now partition an even number of spins into disjoint pairs and put every pair in a spin-one-half singlet state. Each pair has total spin zero, so their tensor product also has $S=0$ and is itself a ground state. Its internal pair contributes $-3$, while correlations between different singlets vanish, giving $-3N/2$ in total. ThereforeThe large degeneracy is special to equal all-to-all coupling: the Hamiltonian sees only total spin and cannot distinguish different singlet coverings.

In a local gapped ground state, correlations and the entanglement that lower local interaction energies are concentrated within a finite correlation length. A degree of freedom deep inside a region uses most of its entanglement with nearby degrees of freedom that are also inside. Entanglement monogamy limits its simultaneous entanglement with the exterior, so only degrees of freedom within a correlation-length layer of the boundary can contribute extensively across the cut. Their number scales as the boundary area, suggesting the entanglement area lawThis is a physical argument rather than a proof from monogamy alone; locality and suitable ground-state assumptions are essential.

Cutting an open-boundary matrix product state across one virtual bond of dimension $\chi$ gives Schmidt rank at most $\chi$, so its entanglement entropy obeysA periodic interval cuts two bonds and obeys $S(A)\le 2\log \chi$. Since the boundary of a one-dimensional interval has a constant number of points, this is an area law.

For a projected entangled pair state, a bipartition crossing $n_{\partial }$ virtual bonds has Schmidt rank at most ${\chi }^{n_{\partial }}$. The tensor-network area-law bound is thereforeBecause $n_{\partial }$ is proportional to the lattice boundary area, every fixed-bond-dimension PEPS satisfies an area law.