4
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.