# Does entanglement entropy follow a volume or an area law for 2D cluster states?

Consider a 2D cluster state defined on a rectangular lattice, which is universal for one way quantum computers. For a description of the state, see for example question 2 in this problem set.

Now, consider a bipartition of this state and look at the entanglement entropy of each partition. Does the entropy follow a volume law or an area law?

Every partitioning $$V=A\cup B$$ of the set of vertices $$V$$ of a 2D lattice $$G=(V,E)$$ into disjoint sets $$A$$ and $$B$$ induces the partitioning $$E=E_A\cup E_B\cup E'$$ of the set of edges $$E$$ into the set $$E_A$$ of edges connecting two vertices in $$A$$, the set $$E_B$$ of edges connecting two vertices in $$B$$ and the set $$E'$$ of edges straddling the two partitions. The set $$V'$$ of vertices incident on at least one edge in $$E'$$ partitions into $$V'=A'\cup B'$$ where $$A'=A\cap V'$$ and $$B'=B\cap V'$$.

A two-dimensional cluster state $$|G\rangle$$ is the result of applying a series of controlled-$$Z$$ gates to the product state $$\bigotimes_{w\in V}|+\rangle_w$$ with one controlled-$$Z$$ for each edge $$(u,v)\in E$$

$$|G\rangle = \prod_{(u,v)\in E} CZ_{u,v}\bigotimes_{w\in V}|+\rangle_w.$$

The entropy $$S$$ of $$\rho_A=\mathrm{tr}_B(|G\rangle\langle G|)$$ is the same as the entropy of $$\rho_B=\mathrm{tr}_A(|G\rangle\langle G|)$$, so any quantum operation that affects only one partition has no effect$$^1$$ on $$S$$. Therefore, undoing all the controlled-$$Z$$ gates associated with edges in $$E_A\cup E_B$$ has no effect on $$S$$. Moreover, since this operation returns all vertices in $$V-V'$$ to the state $$|+\rangle$$, discarding those vertices has no effect$$^2$$ on $$S$$. Thus, $$S$$ is equal to the entropy of $$\rho_{A'}=\mathrm{tr}_{B'}(|G'\rangle\langle G'|)$$ where $$|G'\rangle$$ is the graph state of the boundary region $$G'=(V',E')$$.

Finally, since the entropy $$S$$ cannot exceed the logarithm of the dimension of the smaller Hilbert space, we have

$$S\le\min(|A'|,|B'|)\le|E'|.\tag1$$

We conclude that $$S$$ follows the area$$^3$$ law rather than the volume law of bipartite entanglement entropy. More precisely, it grows at most proportionally to the length of the boundary of the partition.

$$^1$$ This can be proven more rigorously by considering Schmidt decomposition.

$$^2$$ This follows from the identities $$S(\rho_1\otimes\rho_2)=S(\rho_1)+S(\rho_2)$$ and $$S(|\psi\rangle\langle\psi|)=0$$.

$$^3$$ In the context of entanglement scaling laws for a two-dimensional system, the terms "area" and "volume" are misnomers since they refer to scaling with the length of the boundary and the area of the interior, respectively. A possible dimension-neutral names would be boundary law and interior law, but I have not seen these used in the literature.