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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.

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.