# Where is the orthogonality center of the AKLT ground state matrix product state?

It is a known fact that the ground state of the 1D Affleck-Kennedy-Lieb-Tasaki model (AKLT) can be represented as a matrix product state (MPS) of the following form: $$\left|{\psi}\right\rangle = \sum_{\sigma_1\cdots \sigma_N}{\rm Tr}\left(A^{\sigma_1} \cdots A^{\sigma_N}\right) \left|{\sigma}\right\rangle$$

Where the tensors $$A^\sigma$$ are: $$\begin{array}{lcr} A^{+1}=\begin{pmatrix}0 & \sqrt{2/3}\\ 0 & 0 \end{pmatrix} & A^{0}=\begin{pmatrix}-\sqrt{1/3} & 0\\ 0 & \sqrt{1/3} \end{pmatrix} & A^{-1}=\begin{pmatrix}0 & 0\\ -\sqrt{2/3} & 0 \end{pmatrix}\end{array}$$

It is a result of treating each site as composed of two spin-1/2 virtual particles, where each pair of spins of different sites are anti-symmetric, and the whole virtual state is projected onto the whole spin-1 space.

Assuming these exact tensors, where is the orthogonality center of the MPS?

First, if I am correct, the matrix elements $$\sqrt{3/2}$$ should be instead $$\sqrt{2/3}$$. Then you can check that both left and right normalization conditions are satisfied on every site, i.e., $$\sum_{\sigma} {A^{\sigma}}^{\dagger} A^{\sigma} = \sum_{\sigma} A^{\sigma} {A^{\sigma}}^{\dagger} = \mathbb{I}$$. This means that all the sites are the orthognality center of this translationally invariant MPS.
• Oh, you're right, its indeed 2/3. Probably a typo, so I fixed it. Anyways, what does it mean that the orthogonality center is all of the sites? For example, assuming I want to contract the MPS around a local operator (e.g. $S_z$ on site $i$), does that mean I don't have to move the orthogonality center to the site? Commented May 7 at 5:41