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?


1 Answer 1


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


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.


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