Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 343 / 1

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.