Entanglement Subvolume Law for 2D Frustration-Free Spin Systems

Abstract

Let H be a frustration-free Hamiltonian describing a 2D grid of qudits with local interactions, a unique ground state, and local spectral gap lower bounded by a positive constant. For any bipartition defined by a vertical cut of length L running from top to bottom of the grid, we prove that the corresponding entanglement entropy of the ground state of H is upper bounded by ${\tilde{O}}(L^{5/3})$. For the special case of a 1D chain, our result provides a new area law which improves upon prior work, in terms of the scaling with qudit dimension and spectral gap. In addition, for any bipartition of the grid into a rectangular region A and its complement, we show that the entanglement entropy is upper bounded as ${\tilde{O}}(|\partial A|^{5/3})$ where $\partial A$ is the boundary of A. This represents a subvolume bound on entanglement in frustration-free 2D systems. In contrast with previous work, our bounds depend on the local (rather than global) spectral gap of the Hamiltonian. We prove our results using a known method which bounds the entanglement entropy of the ground state in terms of certain properties of an approximate ground state projector (AGSP). To this end, we construct a new AGSP which is based on a robust polynomial approximation of the AND function and we show that it achieves an improved trade-off between approximation error and entanglement.

Appendices

Robust AND Polynomial

We now provide a proof of Theorem 2.2, following Ref. [51].

Proof of Theorem 2.2

For $t\in \mathbbm {R}{\setminus }{\{0\}}$ let $\mathrm {sign}(t)=t/|t|$ denote the sign of t. We may equivalently write

$$\begin{aligned} \mathrm {sign}(t)=\frac{t}{\sqrt{1+(t^2-1)}}, \end{aligned}$$

and we may then use the binomial series to expand the denominator (see, e.g., Eq. 3.2 of [51]). This gives the following series expansion which converges for $0<|t|< \sqrt{2}$

$$\begin{aligned} \mathrm {sign}(t)=t\sum _{i=0}^{\infty } \left( -\frac{1}{4}\right) ^i {2i\atopwithdelims ()i} \left( t^2-1 \right) ^i \qquad \quad 0<|t|< \sqrt{2}. \end{aligned}$$

(82)

Now consider the following robust function for the Boolean monomial:

$$\begin{aligned} \mathrm {int}(x) = \frac{1+\mathrm {sign}(2x-1)}{2} = {\left\{ \begin{array}{ll} 1 &{} \text { if } x>\frac{1}{2}\\ 0 &{} \text { if } x< \frac{1}{2} \end{array}\right. } \end{aligned}$$

Define

$$\begin{aligned} S=\left\{ x\in \mathbbm {R}: 0<|2x-1| < \sqrt{2}\right\} . \end{aligned}$$

For $x\in S$ we may use Eq. (82) and separate out the $i=0$ term to express $\mathrm {int}(x)$ as

$$\begin{aligned} \mathrm {int}(x) =x+\frac{2x-1}{2}\sum _{i=1}^{\infty } {2i\atopwithdelims ()i} \left( x(1-x) \right) ^i=\sum _{i=0}^{\infty } A_i(x) \end{aligned}$$

where we define polynomials

$$\begin{aligned} A_0(x){\mathop {=}\limits ^{\mathrm {def}}}x \quad \text {and} \quad A_{i}(x) {\mathop {=}\limits ^{\mathrm {def}}}\frac{2x-1}{2}{2i\atopwithdelims ()i} \left( x(1-x) \right) ^i \text { for } i\ge 1. \end{aligned}$$

Observe that $A_i$ has real coefficients and degree $2i+1$, for all $i\ge 0$. For $(x_1, x_2, \ldots x_m) \in S^m$,

$$\begin{aligned} \mathrm {int}(x_1)\cdot \mathrm {int}(x_2)\ldots \mathrm {int}(x_m)&= \sum _{i_1, i_2, \ldots i_m} A_{i_1}(x_1)A_{i_2}(x_2)\ldots A_{i_m}(x_m)\nonumber \\&=\sum _{n=0}^{\infty }\sum _{i_1, i_2, \ldots i_m: i_1+\cdots + i_m=n} A_{i_1}(x_1)A_{i_2}(x_2)\ldots A_{i_m}(x_m)\nonumber \\&{\mathop {=}\limits ^{\mathrm {def}}}\sum _{n=0}^{\infty } \xi _{n}(x_1, \ldots x_m). \end{aligned}$$

(83)

Below we shall establish the following claim:

Claim A.1

For $(x_1, x_2,\ldots x_m)\in \left( \left[ -\frac{1}{20}, \frac{1}{20} \right] \cup [1-1/20,1+1/20] \right) ^m$,

$$\begin{aligned} |\xi _n(x_1, x_2, \ldots x_m)| \le 3^m \left( \frac{3}{5} \right) ^n. \end{aligned}$$

Let us define the robust polynomial $p_{AND}$ by truncating the sum in Eq. (83) to $n\le 5m$:

$$\begin{aligned} p_{AND}(x_1, \ldots x_m) {\mathop {=}\limits ^{\mathrm {def}}}\sum _{n=0}^{5m} \xi _{n}(x_1, \ldots x_m). \end{aligned}$$

(84)

Since each $A_i$ is a univariate polynomial with real coefficients and degree $2i+1$, $p_{AND}$ has real coefficients and degree

$$\begin{aligned} \mathrm {max}_{i_1+i_2+\cdots i_m \le 5m} \left( (2i_1+1)+(2i_2+1)+\cdots (2i_m+1) \right) = 2(5m) + m = 11m. \end{aligned}$$

(85)

In addition,

$$\begin{aligned} p_{AND}(1,1,\ldots 1) = \sum _{n=0}^{5m} \xi _{n}(1,1 \ldots 1) = \xi _0(1,1,\ldots 1) =1, \end{aligned}$$

(86)

where we used the identity $A_i(1)=0$ for $i\ge 1$ and $A_0(1)=1$. Finally, suppose $x=(x_1,x_2,\ldots , x_m)=y+\epsilon$ where $y\in \{0,1\}^m$ and $\epsilon \in [-1/20, 1/20]^m$. Then for each $1\le i\le m$ we have

$$\begin{aligned} x_i\in [-1/20,1/20]\cup [1-1/20, 1+1/20] \subset S \end{aligned}$$

and

$$\begin{aligned} |p_{AND}(y+\epsilon ) - y_1y_2\ldots y_m|=|p_{AND}(x_1,x_2,\ldots , x_m)-\mathrm {int}(x_1)\mathrm {int}(x_2)\ldots \mathrm {int}(x_m)|. \end{aligned}$$

Using Eqs. (83, 84) and the triangle inequality to bound the right-hand side gives

$$\begin{aligned} |p_{AND}(y+\epsilon ) - y_1y_2\ldots y_m|&\le \sum _{n=5m+1}^{\infty } |\xi _{n}(x_1, \ldots x_m)|\\&\le 3^m\sum _{n=5m+1}^{\infty } \left( \frac{3}{5} \right) ^n\\&=3^m \left( \frac{3}{5}\right) ^{5m} \cdot \frac{3}{2}\\&\le \left( 3\cdot (3/5)^5\cdot (3/2)\right) ^m. \end{aligned}$$

Noting that $3\cdot (3/5)^5\cdot (3/2)\le e^{-1}$ we arrive at Eq. (7) and complete the proof. $\square$

Proof of Claim A.1

Define $J=[-1/20,1/20]\cup [1-1/20, 1+1/20]$ and note that for all $i\ge 1$ we have

$$\begin{aligned} \mathrm {max}_{x\in J}|A_i(x)| = {2i\atopwithdelims ()i}\mathrm {max}_{x\in J}\left| \frac{2x-1}{2} \left( x(1-x) \right) ^i\right| \le 4^i\cdot \left( \frac{1}{20}\right) ^i\left( \frac{21}{20}\right) ^i \le \left( \frac{21}{100} \right) ^i, \end{aligned}$$

(87)

where we used the fact that ${2i\atopwithdelims ()i}\le 4^i$, $\mathrm {max}_{x\in J} |\frac{2x-1}{2}|\le 1$, and $\mathrm {max}_{x\in J}|x(1-x)|\le (1/20)(21/20)$. Furthermore,

$$\begin{aligned} \mathrm {max}_{x\in J}|A_0(x)| = \mathrm {max}_{x\in J}|x| \le \frac{21}{20}. \end{aligned}$$

(88)

Combining Eqs. (87, 88) we see that for all $i\ge 0$,

$$\begin{aligned} \mathrm {max}_{x\in J}|A_i(x)| \le \left( \frac{21}{20}\right) \left( \frac{21}{100} \right) ^i. \end{aligned}$$

(89)

Consequently, for $(x_1, \ldots x_m)\in J^m$, using the definition of $\xi _n$ and the triangle inequality, we get

$$\begin{aligned} |\xi _n(x_1, \ldots x_m)|&\le \sum _{i_1, i_2, \ldots i_m: i_1+\cdots i_m=n} \left| A_{i_1}(x_1)A_{i_2}(x_2)\ldots A_{i_m}(x_m)\right| \nonumber \\&\le \left( \frac{21}{20}\right) ^m \sum _{i_1, i_2, \ldots i_m: i_1+\cdots i_m=n} \left( \frac{21}{100} \right) ^{i_1+i_2+\cdots i_m} \end{aligned}$$

(90)

$$\begin{aligned}&= \left( \frac{21}{20}\right) ^m \left( \frac{21}{100}\right) ^n {m+n-1 \atopwithdelims ()n-1}. \end{aligned}$$

(91)

where we used Eq. (89) and the fact that the number of tuples $(i_1,i_2,\ldots , i_m)$ of nonnegative integers satisfying $i_1+i_2+\cdots +i_m=n$ is given by ${m+n-1 \atopwithdelims ()n-1}$. Finally, we substitute the bound ${m+n-1 \atopwithdelims ()n-1}\le 2^{m+n}$ into Eq. (91) to arrive at

$$\begin{aligned} |\xi _n(x_1, \ldots x_m)|\le \left( \frac{42}{20}\right) ^m \left( \frac{42}{100}\right) ^n\le 3^m (3/5)^n. \end{aligned}$$

$\square$

Proof of Lemma 3.1

Fig. 4

The local projectors can be divided into 4 groups, where the projectors in each group commute with each other.

The proof is similar to that given in [7], which uses a Chebyshev polynomial function of the detectability operator, as suggested in [25]. The projectors $\{P_{ij}\}$ can be divided into 4 groups as follows (see Fig. 4), with the property that the projectors in each group commute with each other:

$$\begin{aligned}&{\mathcal {G}}_1{\mathop {=}\limits ^{\mathrm {def}}}\{P_{ij}: i=\mathrm {odd}, j=\mathrm {odd}\}, \quad {\mathcal {G}}_2{\mathop {=}\limits ^{\mathrm {def}}}\{P_{ij}: i=\mathrm {even}, j=\mathrm {odd}\},\\&{\mathcal {G}}_3{\mathop {=}\limits ^{\mathrm {def}}}\{P_{ij}: i=\mathrm {odd}, j=\mathrm {even}\}, \quad {\mathcal {G}}_4{\mathop {=}\limits ^{\mathrm {def}}}\{P_{ij}: i=\mathrm {even}, j=\mathrm {even}\}. \end{aligned}$$

We also define

$$\begin{aligned} DL_k&{\mathop {=}\limits ^{\mathrm {def}}}&\prod _{P_{ij}\in {\mathcal {G}}_k} \left( \mathbbm {1}-P_{ij} \right) , \qquad \quad 1\le k\le 4, \end{aligned}$$

(92)

and define $DL {\mathop {=}\limits ^{\mathrm {def}}}DL_4\cdot DL_3\cdot DL_2\cdot DL_1$. From [7, Corollary 3], it holds that for any $\psi$ satisfying $\langle \psi |\Omega \rangle =0$, we have

$$\begin{aligned} \Vert DL \left| \psi \right\rangle \Vert ^2 \le \frac{1}{1+\frac{\gamma }{8^2}} = \frac{1}{1+\frac{\gamma }{64}}. \end{aligned}$$

(93)

Here we used the fact that, for every projector $P_{ij}$, at most 8 projectors do not commute with it. Now, we have the following claim, which is proved towards the end. It uses the ‘light cone’ argument from [3].

Claim B.1

Let F be any univariate polynomial of degree at most t/6 satisfying $F(0)=1$. Then

$$\begin{aligned} DL(t)= \left( Q'_{2t}\cdot Q'_{8t}\cdot Q'_{14t}\cdot \cdots \right) \cdot F\left( \mathbbm {1}- DL^{\dagger }DL\right) \cdot \left( Q'_{5t}\cdot Q'_{11t}\cdot Q'_{17t}\cdot \cdots \right) . \end{aligned}$$

(94)

Before proving this claim, we show how it can be used to establish Lemma 3.1. We apply the Claim with $F=\mathrm {Step}_{\frac{t}{6}, \frac{\gamma }{64+\gamma }}$ where the right-hand side is the polynomial from Fact 2.1. From this we see that for any $\psi \in G_{\perp }$

$$\begin{aligned} \Vert DL(t) \left| \psi \right\rangle \Vert ^2 \le \Vert \mathrm {Step}_{\frac{t}{6}, \frac{\gamma }{64+\gamma }} \left( \mathbbm {1}- DL^{\dagger }DL \right) \left| \psi ' \right\rangle \Vert ^2 \end{aligned}$$

where $\psi '\in G_{\perp }$ is the state

$$\begin{aligned} \left| \psi ' \right\rangle = \left( Q'_{5t}\cdot Q'_{11t}\cdot Q'_{17t}\cdot \cdots \right) \left| \psi \right\rangle /\Vert \left( Q'_{5t}\cdot Q'_{11t}\cdot Q'_{17t}\cdot \cdots \right) \left| \psi \right\rangle \Vert . \end{aligned}$$

But Eq. (93) ensures that the eigenvalues of $DL^{\dagger }DL$ in $G_\perp$ are at most $\frac{1}{1+ \frac{\gamma }{64}} = 1- \frac{\gamma }{64+\gamma }$. Using Fact 2.1 and the fact that $\gamma \le 1$, we get

$$\begin{aligned} \Vert DL(t) \left| \psi \right\rangle \Vert \le 2e^{-\frac{t}{3}\sqrt{\frac{\gamma }{64+\gamma }}}\le 2e^{-\frac{t}{3}\sqrt{\frac{\gamma }{65}}}\le 2e^{-\frac{t\sqrt{\gamma }}{25}} \end{aligned}$$

Proof of Claim B.1

For every $i\in [n-1]$ and $k\in [4]$, let

$$\begin{aligned} \Pi _{i,k}{\mathop {=}\limits ^{\mathrm {def}}}\prod _{\mathbbm {1}-P_{ij}\in {\mathcal {G}}_k: Supp(P_{ij})\in \{i,i+1\}}(\mathbbm {1}-P_{ij}) \end{aligned}$$

be the product of projectors from ${\mathcal {G}}_k$ that are supported only on columns $\{i,i+1\}$. Since all projectors in ${\mathcal {G}}_k$ commute, $\Pi _{i,k}$ is also a projector and we can write $DL_k = \prod _{i\in [n-1]}\Pi _{i,k}$. For any $S\subset [n]$, define

$$\begin{aligned} DL_k^S{\mathop {=}\limits ^{\mathrm {def}}}\prod _{i:Supp(\Pi _{i,k})\cap S \ne \phi } \Pi _{i,k} \end{aligned}$$

as the product of projectors $\Pi _{i,k}$ that have their support overlapping with S.

Fig. 5

Graphical description of Eq. (97). a The operators $DL_1, DL_2, DL_3, DL_4$ correspond to the dark yellow, light yellow, dark blue and light blue layers, respectively. Within each layer, all the projectors (small rectangles representing $\Pi _{i,k}$) mutually commute, although they need not have disjoint support. Two projectors from different layers may not commute if they have overlapping support. b Some projectors in $DL_1$ are ‘absorbed’ by the red coarse-grained layer. Resulting operator is $DL_1^{S_0}$ from Eq. (95). c The same process occurs for 4 steps, with projectors from $DL_2, DL_3, DL_4$ absorbed in the red coarse-grained layer. The resulting operator is $DL_1^{S_0}DL_2^{S_1}DL_3^{S_2}DL_4^{S_3}$ and the support of ‘unabsorbed’ projectors increases its boundary by one at each step. All the remaining projectors can be absorbed in the green coarse-grained layer, as they are contained in its support

The argument below has been illustrated in Fig. 5. Let $S_0$ be the complement of the support of $\left( Q'_{2t}\cdot Q'_{8t}\cdot Q'_{14t}\cdot \cdots \right)$. Observe, using frustration-freeness, that for any $\Pi _{i,k}$ whose support is contained in the support of $\left( Q'_{2t}\cdot Q'_{8t}\cdot Q'_{14t}\cdot \cdots \right)$, we have

$$\begin{aligned} \left( Q'_{2t}\cdot Q'_{8t}\cdot Q'_{14t}\cdot \cdots \right) \Pi _{i,k} = \left( Q'_{2t}\cdot Q'_{8t}\cdot Q'_{14t}\cdot \cdots \right) . \end{aligned}$$

This implies the following identity (c.f. Fig. 5b):

$$\begin{aligned} \left( Q'_{2t}\cdot Q'_{8t}\cdot Q'_{14t}\cdot \cdots \right) DL_1 = \left( Q'_{2t}\cdot Q'_{8t}\cdot Q'_{14t}\cdot \cdots \right) DL_1^{S_0}. \end{aligned}$$

(95)

For all integers $\alpha \ge 1$, recursively define $S_{\alpha }$ as the set of all columns at distance at most 1 from $S_{\alpha -1}$. Clearly, we have the inclusion $S_0 \subset S_1 \subset S_2 \ldots$. Similar to Eq. (95), we can ‘absorb’ some of the projectors in $DL_1DL_2$ and obtain the identity:

$$\begin{aligned} \left( Q'_{2t}\cdot Q'_{8t}\cdot Q'_{14t}\cdot \cdots \right) DL_1DL_2= \left( Q'_{2t}\cdot Q'_{8t}\cdot Q'_{14t}\cdot \cdots \right) DL_1^{S_0}DL_2^{S_1}. \end{aligned}$$

(96)

Applying the same argument recursively, and using the fact that

$$\begin{aligned} \left( DL^{\dagger }DL \right) ^p = \left( DL_1\cdot DL_2\cdot DL_3\cdot DL_4\cdot DL_3\cdot DL_2 \right) ^p\cdot DL_1, \end{aligned}$$

we conclude (c.f. Fig. 5c)

$$\begin{aligned}&\left( Q'_{2t}\cdot Q'_{8t}\cdot Q'_{14t}\cdot \cdots \right) \left( DL^{\dagger }DL \right) ^{p}\nonumber \\&\quad = \left( Q'_{2t}\cdot Q'_{8t}\cdot Q'_{14t}\cdot \cdots \right) DL_1^{S_0}\cdot DL_2^{S_1}\cdot DL_3^{S_2}\ldots DL_1^{S_{6p}}. \end{aligned}$$

(97)

If $6p\le t$, the set $S_{6p}$ is contained in the support of $\left( Q'_{5t}\cdot Q'_{11t}\cdot Q'_{17t}\cdot \cdots \right)$. Furthermore, if $\Pi _{i,k}$ is in the support of $\left( Q'_{5t}\cdot Q'_{11t}\cdot Q'_{17t}\cdot \cdots \right)$, we have

$$\begin{aligned} \Pi _{i,k} \left( Q'_{5t}\cdot Q'_{11t}\cdot Q'_{17t}\cdot \cdots \right) = \left( Q'_{5t}\cdot Q'_{11t}\cdot Q'_{17t}\cdot \cdots \right) . \end{aligned}$$

Thus, all the projectors in $DL_1^{S_0}\cdot DL_2^{S_1}\cdot DL_3^{S_2}\ldots DL_1^{S_{6p}}$ can be ‘absorbed’ in $\left( Q'_{5t}\cdot Q'_{11t}\cdot Q'_{17t}\cdot \cdots \right)$, which can be formalized as:

$$\begin{aligned} DL_1^{S_0}\cdot DL_2^{S_1}\cdot DL_3^{S_2}\ldots DL_1^{S_{6p}} \left( Q'_{5t}\cdot Q'_{11t}\cdot Q'_{17t}\cdot \cdots \right) = \left( Q'_{5t}\cdot Q'_{11t}\cdot Q'_{17t}\cdot \cdots \right) . \end{aligned}$$

(98)

Combining Eqs. (97) and (98), we find that

$$\begin{aligned}&\left( Q'_{2t}\cdot Q'_{8t}\cdot Q'_{14t}\cdot \cdots \right) \left( DL^{\dagger }DL \right) ^{p} \left( Q'_{5t}\cdot Q'_{11t}\cdot Q'_{17t}\cdot \cdots \right) \nonumber \\&\quad = \left( Q'_{2t}\cdot Q'_{8t}\cdot Q'_{14t}\cdot \cdots \right) \left( Q'_{5t}\cdot Q'_{11t}\cdot Q'_{17t}\cdot \cdots \right) \end{aligned}$$

(99)

for any $p\le t/6$. Thus, any such power $(DL^{\dagger }DL)^p$ can be replaced by 1 whenever it is sandwiched between the products of projectors in Eq. (99). This implies that for a polynomial F of degree at most t/6, we have

$$\begin{aligned}&\left( Q'_{2t}\cdot Q'_{8t}\cdot Q'_{14t}\cdot \cdots \right) F(I-DL^{\dagger }DL) \left( Q'_{5t}\cdot Q'_{11t}\cdot Q'_{17t}\cdot \cdots \right) \nonumber \\&\quad = \left( Q'_{2t}\cdot Q'_{8t}\cdot Q'_{14t}\cdot \cdots \right) F(0) \left( Q'_{5t}\cdot Q'_{11t}\cdot Q'_{17t}\cdot \cdots \right) , \end{aligned}$$

(100)

and using the fact that $F(0)=1$ completes the proof. $\quad \square$