Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 67 / 1 / b / ii / Solution

Fix an integer radius $m\ge 0$ and put $r=\lfloor m/2\rfloor$. Starting with $A_x$, scan the $2r$ layers of $K^r$ from right to left. In each layer retain precisely those bond orthogonal projections whose supports meet the current support; multiply them onto the current operator and enlarge that support to include their bonds. Discard every other projection: it commutes through the current operator and acts as the identity on $|{\psi }_0⟩$, by frustration freeness.

If $L_r$ denotes the ordered product of the retained projections, defineEach layer expands the support by at most one lattice spacing. Therefore the local projector cone in a frustration-free chain lies within radius $2r\le m$ of $x$, andThe last equality holds because every retained projection also fixes the ground-state bra.

For $r\ge 1$, the stronger estimate from (i) givesFor $r=0$, choose $A_{x(m)}=A_x$; the error is at most $\Vert A_x\Vert$, so the same bound holds. HenceHere $\alpha$ is exactly the value in (i), and the prefactor is independent of $m$ and chain length. Counting $K^r$ as $r$ rather than $2r$ layers would give an incorrect support claim; the powers of a product of two orthogonal projections are what avoid a lost factor of two in the decay rate.