Graph state of an ordered finite simple graph

Let $G=(V,E)$ be a finite simple graph with a fixed vertex order. Let ${\mathcal{H}}_V={⨂}_{v\in V}{\mathbb{C}}^2$ use that order and its ordered computational basis, as in Finite registers and their ordered computational bases, Big-endian encoding of ordered bit strings, and Ordered tensor registers, their inner product, and wire permutations. For each edge $\{u,v\}$ let $C{Z}_{uv}$ be the unique full-register local embedding from A local gate has a unique global wire embedding of the controlled phase gate with angle $\pi$ from Controlled Phase Rotation. Order the endpoints so that $u$ precedes $v$ in the fixed vertex order; the embedded two-qubit matrix is $\mathrm{diag}(1,1,1,-1)$ in the ordered pair basis $\mathrm{\mid }00⟩,\mathrm{\mid }01⟩,\mathrm{\mid }10⟩,\mathrm{\mid }11⟩$, and every other tensor factor acts as the identity. Thus the two local factors have the same order as the edge endpoints, within the full register ordered by $V$. Order the finite edge set lexicographically by its ordered endpoint indices and define

$\mathrm{\mid }G⟩:=\left({\prod }_{\{u,v\}\in E}^{\mathrm{l}\mathrm{e}\mathrm{x}}C{Z}_{uv}\right)\mathrm{\mid }+{⟩}^{\otimes V},\mathrm{\mid }+⟩=2^{-1\mathrm{/}2}(\mathrm{\mid }0⟩+\mathrm{\mid }1⟩)\mathrm{.}$

The lexicographic order makes the product a specified finite composition. If $V=\mathrm{\varnothing }$, the tensor product and empty product act on the one-dimensional trivial register and $\mathrm{\mid }G⟩$ is its unit vector.

The separate result Controlled-Z edge gates commute proves that this vector is independent of edge order.