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I wanted to verify two intuitions about the entanglement entropy of quantum states.

Consider an $n$ qubit quantum state, prepared by a depth $d$ circuit acting on $|0\rangle^{\otimes n}$ and a bipartition of the state into two systems, $A$ and $B$. Let the boundary between $A$ and $B$ be of size $M$.

The von Neumann entanglement entropy of the reduced density matrix, for each bipartition, should be at most $Md$. This is because there can be at most $Md$ entangling gates across the bipartition ($M$ gates for each time-step), and each gate would increase the entanglement entropy by at most $1$.

Hence, for all 1D circuits of depth $d$, where $M$ is a constant, the entanglement entropy is at most a constant times $d$.

Are these intuitions correct?


The source of the confusion is this paper, where on page 19, they exhibit a $\text{poly}\log$ depth 1D circuit with entropy ~$n$.

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It is not true that the increase in von Neumann entropy of entanglement between the two partitions due to a single cross-partition two-qubit gate is at most one. See below for an example where the increase is two. However, we can use log Schmidt rank to justify a different asymptotically linear upper bound.

Product states

A two-qubit gate cannot produce more than one ebit of entanglement when applied to a product state. This follows from Schmidt decomposition and the fact that the Hilbert space of one qubit is two-dimensional.

Counterexample

However, the argument fails when the qubits to which the gate is applied have preexisting entanglement with other qubits. For example, consider four qubits $A_1A_2B_1B_2$ in the state

$$ |\psi_0\rangle = \frac12(|0_{A_1}0_{A_2}\rangle+|1_{A_1}1_{A_2}\rangle)\otimes(|0_{B_1}0_{B_2}\rangle+|1_{B_1}1_{B_2}\rangle).\tag1 $$

Clearly, the initial entanglement entropy between the partition $A=A_1A_2$ and $B=B_1B_2$ is zero. However, following a single SWAP gate applied to qubits $A_2$ and $B_1$, we get

$$ |\psi_1\rangle = \frac12(|0_{A_1}0_{B_1}\rangle+|1_{A_1}1_{B_1}\rangle)\otimes(|0_{A_2}0_{B_2}\rangle+|1_{A_2}1_{B_2}\rangle)\tag2 $$

with the reduced state of partition $A$

$$ \rho=\mathrm{tr}_{B}(|\psi_1\rangle\langle\psi_1|)=\frac{I_2}{2}\otimes\frac{I_2}{2}=\frac{I_4}{4}\tag3 $$

and entanglement entropy $S(\rho)=2$.

Schmidt rank

By Schmidt decomposition, a state $|\phi\rangle$ of partitions $A$ and $B$ can be written as

$$ |\phi\rangle = \sum_{i=1}^m\lambda_i|i_A\rangle|i_B\rangle\tag4 $$

where $\lambda_i>0$ are Schmidt coefficients and $|i_X\rangle$ are orthonormal states of partition $X=A,B$. The number $m$, often denoted by $\mathrm{Sch}(|\phi\rangle)$ and called the Schmidt rank of the bipartite state $|\phi\rangle$, is another important measure of entanglement.

For any $m$ positive real numbers $s_1, \dots, s_m$ summing to $1$, the Shannon entropy $H(s_1,\dots,s_m)\le\log m$. Therefore, the logarithm of the Schmidt rank provides an upper bound on von Neumann entropy

$$ S(\mathrm{tr}_B(|\psi\rangle\langle\psi|)) \le \log\mathrm{Sch}(|\psi\rangle)\tag5 $$

for any pure bipartite state $|\psi\rangle$.

Increase in log Schmidt rank due to one gate

We can derive an upper bound on the increase in log Schmidt rank due to a single two-qubit cross-partition gate. Suppose the gate $U$ has operator Schmidt rank $r$, i.e. can be written as

$$ U=\sum_{j=1}^rR_j\otimes S_j\tag6 $$

for some single-qubit operators $R_j$ and $S_j$ acting on a qubit in partition $A$ and $B$, respectively and orthogonal with respect to Hilbert-Schmidt inner product (see e.g. $6.4.2$ in Nielsen's PhD thesis or this paper for more details on operator variant of Schmidt decomposition). Then $U|\phi\rangle$ may be written as

$$ \begin{align} |\phi'\rangle=U|\phi\rangle&=\sum_{i=1}^m\sum_{j=1}^r\lambda_iR_j|i\rangle S_j|i'\rangle \\ &= \sum_{k=1}^{mr}\mu_k|k''\rangle|k'''\rangle \end{align}\tag7 $$

so$^1$ $|\phi'\rangle$ has Schmidt rank at most $mr$.

Now, $R_j$ and $S_j$ belong to the complex vector space of operators on $\mathbb{C}^2$ which is four dimensional. Therefore, for a two-qubit gate $r\le 4$. We conclude that a two qubit gate can increase the log Schmidt rank by at most two.

Linear bound on von Neumann entropy

Therefore, by $(5)$ the final von Neumann entropy $S_f$ of entanglement between the two partitions after running a circuit with $Md$ cross-partition two-qubit gates can be bounded above as

$$ S_f\le 2Md.\tag8 $$

Even though $(8)$ is not as tight as $Md$, it is still asymptotically linear.


$^1$ Every representation of a bipartite state $|\psi\rangle$ as a sum of product states has at least $\mathrm{Sch}(|\psi\rangle)$ terms. See e.g. Problem $2.2$ on page $117$ in Nielsen & Chuang.
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    $\begingroup$ Thanks, as always, for a fantastic answer! :) $\endgroup$
    – BlackHat18
    Dec 20, 2021 at 10:07
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    $\begingroup$ You're welcome! Thank you for the kind words and an interesting question! Note that it may be possible to prove the tighter bound $Md$ since the input is a product state. I don't know how to do this rigorously right now, though. A special case where the above arguments imply $S_f\le Md$ occurs if the boundary qubits spend half of the time interacting with neighbors across the boundary and half the time interacting with neighbors in their home partition. $\endgroup$ Dec 20, 2021 at 22:07

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