Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 343 (Quantum Entanglement in Many-Body Physics)

Let $\Phi :M_d(\mathbb{C})\to M_D(\mathbb{C})$ be a linear map. It is positive when $X\ge 0$ implies $\Phi (X)\ge 0$, and completely positive whenis positive for every ancillary dimension $r$. In finite dimensions it is enough to check $r=d$.

A finite family of Kraus operators $A^{\alpha }:{\mathbb{C}}^d\to {\mathbb{C}}^D$ defines the Kraus representationThis map is completely positive because, for every positive semidefinite matrix $Y$ on the enlarged space,

Conversely, use the unnormalized maximally entangled vector $|\Omega ⟩={\sum }_{j=1}^d|j⟩|j⟩$. Complete positivity makes the Choi matrixpositive semidefinite. By the spectral theorem for normal operators, $C_{\Phi }={\sum }_{\alpha =1}^R|v_{\alpha }⟩⟨v_{\alpha }|$, where $R=\mathrm{rank}{C}_{\Phi }\le dD$. Reshape each $v_{\alpha }\in {\mathbb{C}}^D\otimes {\mathbb{C}}^d$ into a matrix $A^{\alpha }$ by $|v_{\alpha }⟩={\sum }_{\mu j}{A}_{\mu j}^{\alpha }|\mu ⟩|j⟩$. The Choi matrix inversion formulathen gives $\Phi (X)={\sum }_{\alpha }{A}^{\alpha }X(A^{\alpha }{)}^{†}$. Thus finite Kraus representations characterize finite-dimensional completely positive maps.

The additional normalizationmakes $\Phi$ trace preserving and hence a quantum channel; ${\sum }_{\alpha }{A}^{\alpha }(A^{\alpha }{)}^{†}=I_D$ instead makes it unital.

An open-boundary matrix product state with physical dimension $q$ and bond dimension $\chi$ iswhere the $A^i$ are $\chi \times \chi$ matrices and $|\ell ⟩,|r⟩$ are boundary vectors. For periodic boundary conditions, replace the boundary contraction by the matrix trace $\mathrm{Tr}(A^{i_1}\cdots A^{i_N})$.

The same matrices define the matrix product state transfer mapWhen ${\sum }_i(A^i{)}^{†}{A}^i=I$, the Stinespring dilationis an isometry. Repeatedly applying $V$ stores each Kraus label in a fresh physical register:Contracting the remaining virtual system with $⟨\ell |$ gives an MPS, while tracing over all recorded labels gives repeated application of the completely positive map $\mathcal{E}$. The MPS is therefore a coherent unravelling of the channel. Equivalently, retaining the Kraus-label registers realizes a purification of its output. A non-normalized MPS tensor gives the same construction with a general completely positive map; an appropriate canonical gauge normalizes the transfer map on its support.

Normalize the stated Pauli matrices asThe factor $3^{-1/2}$ changes only the overall normalization at fixed $N$. In the spin-one Cartesian basis, the associated periodic uniform matrix product state isThis is the Pauli-matrix representation of the Affleck--Kennedy--Lieb--Tasaki state.

The Pauli matrix multiplication lawshows that products on two neighboring sites span all of $M_2(\mathbb{C})$, so the tensor is an injective matrix product state after blocking two sites. More geometrically, its two-site image is the scalar plus antisymmetric subspace of $3\otimes 3$, namely the total-spin $J=0$ and $J=1$ sectors. By the Two-site support of the Pauli-matrix Affleck--Kennedy--Lieb--Tasaki tensor, the missing subspace is the five-dimensional symmetric traceless $J=2$ sector.

Let $P_{n,n+1}^{(2)}$ be the orthogonal projection onto that $J=2$ sector. The parent Hamiltonian of a matrix product state isEach term annihilates $|{\Psi }_N⟩$, so this is a frustration-free quantum Hamiltonian and the MPS is a ground state. Writing $x={\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$ and using the eigenvalues $-2,-1,1$ in the three total-spin sectors gives the explicit projectorThis is the Affleck--Kennedy--Lieb--Tasaki parent Hamiltonian. Injectivity implies that its periodic ground state is unique for every sufficiently long chain.

Write the spin-one-half Heisenberg antiferromagnet ason a bipartite lattice of coordination number $z$. A one-spin density operator is $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, where $\mathbf{r}$ is its Bloch vector and $|\mathbf{r}|\le 1$. In a two-sublattice product state,The minimum is $-1/4$, attained by pure antiparallel Bloch vectors, so the minimizing product state is a Néel state. In the $z\to \infty$ limit this mean-field approximation becomes exact and the ground-state energy per bond is therefore

The phrase “energy density” requires a coupling convention. For the unscaled Hamiltonian above, every site belongs to $z/2$ bonds andwhich diverges as $z\to \infty$. With the standard Kac normalizationthe finite energy density isIf the convention divides by the spatial dimension $d=z/2$ instead, the answer is $-J/4$ per site. A Hamiltonian written with ${\mathbf{\sigma }}_i\cdot {\mathbf{\sigma }}_j$ rather than ${\mathbf{S}}_i\cdot {\mathbf{S}}_j$ multiplies all these energies by four.

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.

Letwhere the two signs describe the Heisenberg antiferromagnet and Heisenberg ferromagnet. A standard Lieb-Robinson bound is obtained by iterating the Heisenberg picture equation and bounding nested commutators by operator norms. Its constants depend on the interaction only through quantities such asChanging $J$ to $-J$ changes neither the supports nor the norms $\Vert \Phi (X)\Vert$. Every term in the nested-commutator estimate acquires at most an irrelevant sign before its absolute value is taken. Therefore both chains obey exactly the same estimatewith the same $C,\mu$, and Lieb-Robinson velocity $v_{\mathrm{LR}}$. No unitary equivalence of the two Hamiltonians is required.

Consider a depth-$D$ circuit whose gates have range at most $r$. Under backward Heisenberg evolution, a one-site observable has a backward light cone of a local quantum circuit of radius at most $rD$. Choose sites $i,j$ separated by $L>2rD$. Their two backward light cones are disjoint. Since the input is a product state, expectations factorize, and hence every connected correlation function between the two output observables vanishes.

For the GHZ statehowever,at every separation. The light cones must therefore overlap, which forcesFor opposite ends of a one-dimensional chain, $L=\Theta (N)$, so $D=\Omega (N)$ and no constant-depth local circuit can prepare the GHZ state. The continuous-time version follows directly from the Lieb-Robinson bound, with preparation time at least $L/(2v_{\mathrm{LR}})$ up to exponentially small tails.

This is the GHZ-state circuit-depth lower bound. Finite-depth local circuits define equivalence within a gapped phase, so a state with this long-range order cannot lie in the same circuit phase as a product state. The GHZ state is the finite-size cat state associated with spontaneous symmetry breaking; it is not a unique short-range-entangled ground state. The persistent distant correlation is precisely the obstruction.

For the injective translationally invariant case, the fundamental theorem of matrix product states states that two tensors $A=\{A^i\}$ and $B=\{B^i\}$ of the same minimal bond dimension generate the same periodic MPS for every sufficiently large length if and only iffor every physical index $i$, with one invertible matrix $X$. Equality as normalized rays permits the phase $e^{i\theta }$; equality as vectors for all lengths restricts the resulting factor $e^{iN\theta }$ accordingly. The converse is immediate from cyclicity of the matrix trace:

For the nontrivial direction, block enough sites that both tensors are injective and definewith ${\Gamma }_B^L$ defined similarly. Injectivity means that ${\Gamma }_A^L$ and ${\Gamma }_B^L$ have trivial kernels. Equality of all sufficiently long periodic states implies equality of the local support spaces $\mathrm{im}{\Gamma }_A^L=\mathrm{im}{\Gamma }_B^L$, so there is an invertible linear map $F$ on the virtual matrix algebra satisfying

Compare two adjacent blocks and contract arbitrary environments on their left and right. Because both block maps are injective, equality of the physical contractions forcesThus $F$ is a unital algebra automorphism of $M_{\chi }(\mathbb{C})$. By every automorphism of a full matrix algebra is inner, $F(X)=X_{0}X{X}_0^{-1}$ for an invertible $X_0$. Applying this relation to a block with one physical site exposed givesThis proves the theorem and identifies the freedom as the gauge equivalence of injective matrix product state tensors. For noninjective tensors, their canonical forms first split into injective blocks; equality then permits a permutation of equivalent blocks together with a similarity transformation and phase on each block.

The Affleck--Kennedy--Lieb--Tasaki state places two virtual spin-one-half degrees of freedom at every site, puts neighboring virtual spins into singlets, and projects the two virtual spins at each site onto their symmetric spin-one triplet. Its local tensor is equivalently proportional to ${\sigma }_x,{\sigma }_y,{\sigma }_z$ in the spin-one Cartesian basis, and its Affleck--Kennedy--Lieb--Tasaki parent Hamiltonian is

The one-dimensional cluster state is the simultaneous $+1$ eigenstate of the commuting stabilizersOn a periodic chain it is obtained by applying a controlled-$Z$ gate on every neighboring pair of the product state $|+{⟩}^{\otimes N}$.

After suitable blocking, the two states have the same nontrivial projective virtual symmetry of a matrix product state for the protecting group ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$. The two virtual symmetry generators can be represented by anticommuting Pauli matrices, so they realize the nontrivial projective class in group cohomology. Consequently the AKLT and cluster states can be connected by a symmetry-preserving gapped path, or equivalently by a symmetry-preserving finite-depth local circuit: this is the Symmetry-protected equivalence of the Affleck--Kennedy--Lieb--Tasaki state and cluster state.

Both states therefore exhibit one-dimensional symmetry-protected topological order. On an open chain their nontrivial virtual representation produces protected edge degrees of freedom and a characteristic degeneracy in the entanglement spectrum of a matrix product state. They do not have intrinsic topological order: if the protecting symmetry is discarded, either state can be connected to a product state by a finite-depth local circuit.