= Solution
Normalize the stated <Pauli matrix>[Pauli matrices] as
$$
A^x=\frac{\sigma_x}{\sqrt3},
\qquad
A^y=\frac{\sigma_y}{\sqrt3},
\qquad
A^z=\frac{\sigma_z}{\sqrt3}.
$$
The 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> is
$$
|\Psi_N\rangle
=\sum_{i_1,\ldots,i_N\in\{x,y,z\}}
\operatorname{Tr}(A^{i_1}\cdots A^{i_N})
|i_1\cdots i_N\rangle.
$$
This is the <Pauli-matrix representation of the Affleck--Kennedy--Lieb--Tasaki state>.
The <Pauli matrix multiplication law>
$$
\sigma_i\sigma_j=\delta_{ij}I+i\varepsilon_{ijk}\sigma_k
$$
shows that products on two neighboring sites span all of $M_2(\mathbb C)$, so the tensor is an <injective matrix product state> after <blocking a matrix product state>[blocking] two sites. More geometrically, its two-site image is the scalar plus antisymmetric subspace of $3\otimes3$, 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 $P^{(2)}_{n,n+1}$ be the <orthogonal projection> onto that $J=2$ sector. The <parent Hamiltonian of a matrix product state> is
$$
H=\sum_nP^{(2)}_{n,n+1}.
$$
Each term annihilates $|\Psi_N\rangle$, so this is a <frustration-free quantum Hamiltonian> and the MPS is a ground state. Writing $x=\mathbf S_n\mathbin\cdot\mathbf S_{n+1}$ and using the <spin-dot-product eigenvalue>[eigenvalues] $-2,-1,1$ in the three <total-spin sector>[total-spin sectors] gives the explicit projector
$$
P^{(2)}_{n,n+1}
=\frac{(x+2)(x+1)}6.
$$
This is the <Affleck--Kennedy--Lieb--Tasaki parent Hamiltonian>. Injectivity implies that its periodic ground state is unique for every sufficiently long chain.
Back to article page
