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Matrix product state representation for the "infinitely repulsive hardcore boson" state

Physics Asked on July 21, 2021

Consider a one-dimensional spin-1/2 chain with $N$ spins, and let $|psirangle$ be the equal weight superposition of all states with no adjacent spin-ups, e.g.
for $N=3$ with open-boundary, $|psi_{N=3}rangle=|000rangle+|001rangle+|010rangle+|100rangle+|101rangle,$ up to a normalization factor. How to construct a MPS representation for $|psirangle$?

One Answer

Try to construct the MPS tensor such that it mediates the structure of your state through the virtual indices: If the physical state is $vert1rangle$, you want to choose the virtual state to the left and right such as to signal that, to avoid having another $vert1rangle$ state next to it.

One option would be to force the virtual state on the left to $0$ and on the right to $1$ if the physical state is $vert1rangle$: Then, it is impossible to have two $vert1rangle$'s adjacent to each other. On the other hand, a physical $vert0rangle$ should be compatible with both options, and thus treat a virtual $0$ or $1$ the same way.

This leads to a MPS tensor with $$ A^0=frac{1}{2}begin{pmatrix}1 &11 & 1end{pmatrix}=vert+ranglelangle+| , quad A^1=begin{pmatrix}0&2&0end{pmatrix}=(sqrt{2}vert0rangle)(langle1vertsqrt{2}) . $$ Clearly, $A^1A^1=0$, and thus, no two $1$'s can be adjacent to each other. On the other hand, any other configuration is admissible. More importantly - and this is how I chose the normalization (which I did in retrospect) - all configurations have the same weight (for periodic boundary conditions), since $$ langle+vert+rangle = sqrt{2}langle1vert+rangle = sqrt{2}langle+vert0rangle . $$ The same works for an open boundary condition MPS, if you choose left and right boundary conditions $vert+rangle$.

Correct answer by Norbert Schuch on July 21, 2021

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