Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 343 (Quantum Entanglement in Many-Body Physics)

The generator of every finite-dimensional, time-homogeneous Markovian quantum channel has Lindblad equation formHere $H=H^{†}$. The Lindblad operators $L_{\alpha }$ may be arbitrary bounded operators; positivity of their rates is already incorporated by rescaling them. They need not be Hermitian, traceless, normalized, or mutually orthogonal. Their description is nonunique: unitary mixing of the $L_{\alpha }$, and shifts by scalar multiples of the identity accompanied by a compensating change of $H$, leave the same Lindbladian.

The exact spectral condition is that $0$ be a simple eigenvalue of $\mathcal{L}$ and that every other eigenvalue satisfy $\mathrm{Re}\lambda <0$. Then $e^{t\mathcal{L}}\rho$ converges for every initial density operator to the unique stationary state ${\rho }_{\ast }$ satisfying $\mathcal{L}({\rho }_{\ast })=0$. This is the unique stationary state of a Lindbladian condition; the smallest nonzero value of $-\mathrm{Re}\lambda$ is the Lindbladian gap.

A standard operator criterion is irreducibility: the only subspaces invariant under $H$, all $L_{\alpha }$, and all $L_{\alpha }^{†}$ are the zero and full spaces, equivalently their common commutant consists only of scalar multiples of the identity. With no nonzero imaginary-axis eigenvalues, this makes the semigroup relaxing. The spectral statement is the safest general answer because uniqueness of a fixed point alone does not exclude persistent oscillatory modes.

Let $|s⟩=|{\psi }^{-}⟩$ be the spin-one-half singlet state, complete it to an orthonormal basis $\{|s⟩,|e_1⟩,|e_2⟩,|e_3⟩\}$, and chooseEach excited population decays into the singlet. On the matrix-unit basis, $|e_j⟩⟨e_k|$ has decay eigenvalue $-\gamma$ after separating off its contribution to the stationary population, while $|s⟩⟨e_j|$ and its adjoint have eigenvalue $-\gamma /2$. The sole zero mode is $|s⟩⟨s|$, so every initial state converges toThe nonzero eigenvalue closest to the origin is therefore

In a quantum-trajectory unravelling, a pure state evolves between jumps underand a jump of type $\alpha$ sends $|\varphi ⟩$ to $L_{\alpha }|\varphi ⟩/\Vert L_{\alpha }|\varphi ⟩\Vert$. Averaging $|{\varphi }_t⟩⟨{\varphi }_t|$ over the stochastic jump records recovers $\rho (t)$.

Equivalently, over a short interval $dt$ use the operators in a Kraus representationand the Stinespring dilation $V={\sum }_{\mu }{K}_{\mu }\otimes |\mu {⟩}_E$. Iterating with fresh environment systems produces a pure system-environment history whose partial trace is the Lindblad evolution.

If the fixed point is pure, stationarity requires $L_{\alpha }|{\psi }_{\ast }⟩={\ell }_{\alpha }|{\psi }_{\ast }⟩$ for every $\alpha$, together with preservation of its ray by the adjusted effective Hamiltonian. After shifting the jumps one may take it to be a dark state of a Lindblad equation. Once a trajectory reaches that ray, neither no-jump evolution nor a jump takes it away; in the example of part (c), a jump transfers any excited component directly into the singlet.

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.

For a unique gapped ground state, one-dimensional bosonic phases protected by an on-site symmetry $G$ are classified by the cohomology classof a projective virtual symmetry of a matrix product state. After blocking finitely many sites, an injective matrix product state tensor satisfiesChanging the virtual gauge or rephasing $V(g)$ changes $\omega$ by a coboundary and leaves $[\omega ]$ fixed.

If symmetry breaking and degenerate ground states are allowed, the additional data are the unbroken subgroup $H\subseteq G$, the way $G$ permutes the broken-symmetry sectors, and a class in $H^2(H,U(1))$ within one sector. Acting with ${⨂}_i{U}_i(g)$ preserves a symmetric ground state up to phase in the unbroken case; in a broken phase it generally permutes the different ground states.

A uniform matrix product state is specified by matrices $A^i\in M_D$ and, on a periodic chain, has amplitudesIt is injective when, after some blocking length $\ell$, the products $A^{i_1}\cdots A^{i_{\ell }}$ span $M_D$.

The fundamental theorem of matrix product states says that two injective tensors generating the same states for all sufficiently large $N$ satisfyfor one invertible matrix $X$; conversely this relation plainly gives the same periodic states up to the overall phase $e^{iN\theta }$.

For the proof, block enough sites that both tensors are injective. Injectivity gives left inverses from physical blocks to arbitrary virtual matrices. Equality of the states then implies that replacing one blocked tensor inside any sufficiently long network defines an invertible linear map on its two virtual boundary indices. Applying the replacement at two adjacent blocks in either order shows that this boundary map preserves multiplication: $\Phi (MN)=\Phi (M)\Phi (N)$. Every automorphism of the full matrix algebra is inner, so $\Phi (M)=XM{X}^{-1}$. Undoing the blocking gives $A^i=e^{i\theta }X{B}^i{X}^{-1}$, with only an $N$th-root phase left by periodic closure. Equality for consecutive sufficiently large lengths makes that phase independent of $N$ and completes the result.

Apply the theorem to a symmetric injective MPS. The physically transformed tensor $A_g^i={\sum }_{j}U(g{)}_{ij}{A}^j$ generates the same state, so the theorem forcesApplying $g$ and then $h$ shows that $V(g)V(h)$ and $V(gh)$ implement the same gauge transformation. Injectivity makes that gauge unique up to a scalar, hence $V(g)V(h)=\omega (g,h)V(gh)$. Associativity gives the two-cocycle equation for $\omega$.

A continuous symmetry-preserving gapped path changes the tensor and $V(g)$ continuously but cannot change the discrete cohomology class $[\omega ]$ without losing injectivity, breaking the symmetry, or closing the gap. Thus the fundamental theorem of matrix product states turns the virtual projective representation into the invariant classifying one-dimensional symmetry-protected topological phases. With broken symmetry, one first records the permuted ground-state sectors and then applies the same argument to their unbroken subgroup.

For the Affleck--Kennedy--Lieb--Tasaki state, take the dihedral subgroup of spin rotations generated by $\pi$ rotations about the $x$ and $z$ axes. On each physical spin-one site these generators commute and give an ordinary representation of ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$. On the virtual spin-one-half space they may be represented, up to phases, byThey anticommute,so the virtual action is the nontrivial projective representation of ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$.

The reduced density operator across a cut commutes with this irreducible projective action. By Schur lemma it is proportional to the identity within each virtual doublet. Consequently every level in the entanglement spectrum of a matrix product state is at least twofold degenerate, the characteristic protected edge-spin degeneracy of the AKLT phase.

The Transverse-field Ising modelcommutes with translations, reflections, and the global spin flipFor $\lambda \gg 1$, the unique product ground state has $Z_i=-1$ at every site; for $\lambda \ll -1$, it has $Z_i=+1$. Both are symmetric paramagnets. For $|\lambda |\ll 1$, the two thermodynamic ground states are ferromagnets $|+{⟩}_X^{\otimes L}$ and $|-{⟩}_X^{\otimes L}$, exchanged by $G$; on a finite ring their even and odd cat combinations have an exponentially small splitting.

The large-field phase has a unique symmetry-preserving ground state, whereas the small-field phase spontaneously breaks the ${\mathbb{Z}}_2$ symmetry. These distinct symmetry realizations cannot be connected while retaining both a nonzero thermodynamic gap and the symmetry, so a phase transition must intervene. Kramers--Wannier duality locates the self-dual transitions at $|\lambda |=1$.

Use the $X$ eigenbasis $|s_i{⟩}_X$, $s_i\in {\mathbb{Z}}_2$, and put a dual qubit on every edge with $t_i=s_i+s_{i+1}(\mathrm{mod}2)$. The Kramers--Wannier intertwiner isAn explicit matrix product operator tensor isContracting neighboring virtual indices ${\beta }_i={\alpha }_{i+1}$ around the ring is the graphical MPO: each tensor copies its input bit to the left virtual leg and outputs the XOR of its two virtual legs.

Directly from the domain-wall definition,Therefore $DH(\lambda )=\tilde{H}(\lambda )D$, whereAfter exchanging $X$ and $Z$ and rescaling, this is the same Ising family at reciprocal coupling, so corresponding symmetry sectors have the same spectrum and $H(\lambda )$ is dual to $|\lambda |H(1/\lambda )$ up to the elementary sign conventions. In particular the Ising spectrum is self-dual at $|\lambda |=1$.

The original symmetry flips every $X$-basis bit, $s_i\mapsto s_i+1$, without changing any domain wall. Hencewith the present normalization. On the dual side, periodic domain walls obeyso the original global symmetry becomes a constraint on the dual symmetry sector.

Periodic versus antiperiodic boundary conditions determine whether the product of dual domain walls is $+1$ or $-1$. Conversely, the original even and odd $G$ sectors correspond to choices of dual boundary twist. Keeping all sectors therefore requires summing over both symmetry charges and both boundary conditions; within each matched sector the intertwiner is invertible up to normalization.

If an operator $O$ is symmetric, $[O,G]=0$, it descends consistently to a dual operator satisfying $DO=\tilde{O}D$. If it is nonsymmetric, it changes the global charge and cannot be represented by a local operator within one fixed dual boundary sector; its dual either changes the twist, acquires a disorder string, or is annihilated by the projected intertwiner.

The two-dimensional Kramers--Wannier projected entangled pair operator puts an input bit $s_v$ at every vertex and an output bit $t_e$ at every edge. A COPY tensor at each vertex sends $s_v$ to every incident virtual leg, and an XOR tensor on $e=(uv)$ imposesContracting the vertex-edge virtual legs gives the requested PEPO. Algebraically it is

The output domain walls are not independent. Around every plaquette $p$, each vertex bit occurs twice, soProducts of these loop operators generate a ${\mathbb{Z}}_2$ one-form symmetry: its charged objects are line operators, and contractible closed loops act trivially on the PEPO image. This symmetry is unavoidable because edge variables obtained as discrete gradients have zero flux around every contractible loop.

On the square lattice the PEPO intertwines the local operators asIts image also obeys the automatic zero-flux constraints $B_p={\prod }_{e\in \partial p}{\tilde{Z}}_e=1$. Thus the transverse-field Ising Hamiltonian maps to a ${\mathbb{Z}}_2$ lattice-gauge Hamiltonian generated by edge fields and stars in the zero-flux sector.

At the commuting-projector fixed point, promoting the image constraints to energetic terms givesthe toric code Hamiltonian. Conversely, within a simply connected zero-flux sector one may solve ${\tilde{Z}}_{uv}=X_u{X}_v$ and recover vertex Ising variables. On a torus, the remaining noncontractible loop eigenvalues correspond to the different Ising boundary twists, which accounts for the toric code's topological ground-state sectors.