The Wasserstein Distance of Order 1 for Quantum Spin Systems on Infinite Lattices

Abstract

We propose a generalization of the Wasserstein distance of order 1 to quantum spin systems on the lattice $\mathbb {Z}^d$, which we call specific quantum $W_1$ distance. The proposal is based on the $W_1$ distance for qudits of De Palma et al. (IEEE Trans Inf Theory 67(10):6627–6643, 2021) and recovers Ornstein’s $\bar{d}$-distance for the quantum states whose marginal states on any finite number of spins are diagonal in the canonical basis. We also propose a generalization of the Lipschitz constant to quantum interactions on $\mathbb {Z}^d$ and prove that such quantum Lipschitz constant and the specific quantum $W_1$ distance are mutually dual. We prove a new continuity bound for the von Neumann entropy for a finite set of quantum spins in terms of the quantum $W_1$ distance, and we apply it to prove a continuity bound for the specific von Neumann entropy in terms of the specific quantum $W_1$ distance for quantum spin systems on $\mathbb {Z}^d$. Finally, we prove that local quantum commuting interactions above a critical temperature satisfy a transportation-cost inequality, which implies the uniqueness of their Gibbs states.

Appendices

Properties of the Quantum $W_1$ Distance

Proposition A.1

([17, Proposition 2]). For any finite set $\Lambda$ and any $\Delta \in \mathcal {O}_\Lambda ^T$, we have

$$\begin{aligned} \frac{1}{2}\left\| \Delta \right\| _1 \le \left\| \Delta \right\| _{W_1} \le \frac{\left| \Lambda \right| }{2}\left\| \Delta \right\| _1. \end{aligned}$$

(A.1)

Proposition A.2

([17, Proposition 5]). Let $\Lambda '\subseteq \Lambda$ be finite sets. Then, for any $\Delta \in \mathcal {O}_\Lambda ^T$ such that $\textrm{Tr}_{\Lambda '}\Delta =0$ we have

$$\begin{aligned} \left\| \Delta \right\| _{W_1} \le \frac{q^2-1}{q^2}\left| \Lambda '\right| \left\| \Delta \right\| _1. \end{aligned}$$

(A.2)

Proposition A.3

(Superadditivity [17, Proposition 4]). The quantum $W_1$ distance is superadditive in general and additive for product states, i.e., for any two disjoint finite sets $\Lambda ,\,\Lambda '$ and any $\rho ,\,\sigma \in \mathcal {S}_{\Lambda \Lambda '}$ we have

$$\begin{aligned} \left\| \rho - \sigma \right\| _{W_1} \ge \left\| \rho _\Lambda - \sigma _\Lambda \right\| _{W_1} + \left\| \rho _{\Lambda '} - \sigma _{\Lambda '}\right\| _{W_1}, \end{aligned}$$

(A.3)

and for any $\rho _\Lambda ,\,\sigma _\Lambda \in \mathcal {S}_\Lambda$ and any $\rho _{\Lambda '},\,\sigma _{\Lambda '}\in \mathcal {S}_{\Lambda '}$ we have

$$\begin{aligned} \left\| \rho _\Lambda \otimes \rho _{\Lambda '} - \sigma _\Lambda \otimes \sigma _{\Lambda '}\right\| _{W_1} = \left\| \rho _\Lambda - \sigma _\Lambda \right\| _{W_1} + \left\| \rho _{\Lambda '} - \sigma _{\Lambda '}\right\| _{W_1}. \end{aligned}$$

(A.4)

Auxiliary Proofs

1.1 Proof of Proposition 2.1

Proposition

(2.1) The trace distance on $\mathcal {S}_{\mathbb {Z}^d}$ is the supremum of the trace distances between the marginal states: For any $\rho ,\,\sigma \in \mathcal {S}_{\mathbb {Z}^d}$,

$$\begin{aligned} T(\rho ,\sigma ) = \frac{1}{2}\sup _{\Lambda \in \mathcal {F}_{\mathbb {Z}^d}}\left\| \rho _\Lambda - \sigma _\Lambda \right\| _1, \end{aligned}$$

(B.1)

where $\Vert \cdot \Vert _1$ denotes the trace norm on $\mathfrak {U}_\Lambda$ given by

$$\begin{aligned} \left\| A\right\| _1 = \textrm{Tr}_\Lambda \sqrt{A^\dag A},\qquad A\in \mathfrak {U}_\Lambda . \end{aligned}$$

(B.2)

Proof

Since $\mathfrak {U}_{\mathbb {Z}^d}^{loc}$ is dense in $\mathfrak {U}_{\mathbb {Z}^d}$, we have

$$\begin{aligned} 2\,T(\rho ,\sigma )&= \sup _{A\in \mathfrak {U}_{\mathbb {Z}^d}^{loc}:\Vert A\Vert _\infty \le 1}\left| \rho (A) - \sigma (A)\right| = \sup _{\Lambda \in \mathcal {F}_{\mathbb {Z}^d}}\sup _{A\in \mathfrak {U}_\Lambda :\Vert A\Vert _\infty \le 1}\left| \rho (A) - \sigma (A)\right| \nonumber \\&= \sup _{\Lambda \in \mathcal {F}_{\mathbb {Z}^d}}\sup _{A\in \mathfrak {U}_\Lambda :\Vert A\Vert _\infty \le 1}\left| \textrm{Tr}_\Lambda \left[ \left( \rho _\Lambda - \sigma _\Lambda \right) A\right] \right| = \sup _{\Lambda \in \mathcal {F}_{\mathbb {Z}^d}}\left\| \rho _\Lambda - \sigma _\Lambda \right\| _1\,. \end{aligned}$$

(B.3)

The claim follows. $\square$

1.2 Proof of Proposition 5.1

Proposition

(5.1) For any $\Lambda \in \mathcal {F}_{\mathbb {Z}^d}$, any $H\in \mathcal {O}_\Lambda$ and any $x\in \Lambda$, (3.3) and (5.1) are equivalent.

Proof

Let

$$\begin{aligned} \partial _x H = 2\min _{A\in \mathcal {O}_{\Lambda \setminus x}}\left\| H - A\right\| _\infty ,\qquad \tilde{\partial }_x H = 2\inf _{A\in \mathcal {O}_{\mathbb {Z}^d\setminus x}}\left\| H - A\right\| _\infty . \end{aligned}$$

(B.4)

We clearly have $\tilde{\partial }_x H \le \partial _x H$. Let $\omega _{\mathbb {Z}^d{\setminus }\Lambda }\in \mathcal {S}_{\mathbb {Z}^d{\setminus }\Lambda }$ be the uniform distribution on $\mathbb {Z}^d\setminus \Lambda$, and let $\Psi _\Lambda :\mathfrak {U}_{\mathbb {Z}^d}\rightarrow \mathfrak {U}_\Lambda$ be the completely positive unital linear map such that for any $A\in \mathfrak {U}_{\mathbb {Z}^d}$ and any $\rho _\Lambda \in \mathcal {S}_\Lambda$

$$\begin{aligned} \textrm{Tr}_\Lambda \left[ \rho _\Lambda \,\Psi _\Lambda (A)\right] = (\omega _{\mathbb {Z}^d\setminus \Lambda }\otimes \rho _\Lambda )(A). \end{aligned}$$

(B.5)

Let $A\in \mathcal {O}_{\mathbb {Z}^d\setminus x}$. We have for any $\rho _\Lambda \in \mathcal {S}_\Lambda$ and any unitary operator $U_x\in \mathfrak {U}_x$

$$\begin{aligned} \textrm{Tr}_\Lambda \left[ \rho _\Lambda \,U_x^\dag \,\Psi _\Lambda (A)\,U_x\right]&= \textrm{Tr}_\Lambda \left[ U_x\,\rho _\Lambda \,U_x^\dag \,\Psi _\Lambda (A)\right] = \left( \omega _{\mathbb {Z}^d\setminus \Lambda }\otimes U_x\,\rho _\Lambda \,U_x^\dag \right) (A)\nonumber \\&= (\omega _{\mathbb {Z}^d\setminus \Lambda }\otimes \rho _\Lambda )\left( U_x^\dag \,A\,U_x\right) = (\omega _{\mathbb {Z}^d\setminus \Lambda }\otimes \rho _\Lambda )(A)\nonumber \\ {}&= \textrm{Tr}_\Lambda \left[ \rho _\Lambda \,\Psi _\Lambda (A)\right] \,, \end{aligned}$$

(B.6)

therefore $U_x^\dag \,\Psi _\Lambda (A)\,U_x = \Psi _\Lambda (A)$, hence $\Psi _\Lambda (A)\in \mathcal {O}_{\Lambda \setminus x}$. We then have

$$\begin{aligned} \partial _x H \le 2\left\| H - \Psi _\Lambda (A)\right\| _\infty = 2\left\| \Psi _\Lambda (H-A)\right\| _\infty \le 2\left\| H-A\right\| _\infty , \end{aligned}$$

(B.7)

where the last inequality follows since $\Psi _\Lambda$ is completely positive and unital. We then have $\partial _x H \le \tilde{\partial }_x H$. The claim follows. $\square$

Auxiliary Lemmas

Lemma C.1

(Multidimensional Fekete’s lemma [100]). Let $f:\mathbb {N}_+^d\rightarrow \mathbb {R}$ be superadditive with respect to each variable, i.e.,

$$\begin{aligned} f(x_1,\,\ldots ,\,x_i+t,\,\ldots ,\,x_d) \ge f(x_1,\,\ldots ,\,x_i,\,\ldots ,\,x_d) + f(x_1,\,\ldots ,\,t,\,\ldots ,\,x_d)\nonumber \\ \end{aligned}$$

(C.1)

for any $x_1,\,\ldots ,\,x_d,\,t\in \mathbb {N}$ and any $i=1,\,\ldots ,\,d$. Then,

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{f(x)}{x_1\ldots x_d} = \sup _{x\in \mathbb {N}_+^d}\frac{f(x)}{x_1\ldots x_d}. \end{aligned}$$

(C.2)

Lemma C.2

Let $H\in \mathcal {O}_\Lambda$ be positive semi-definite. Then, for any $x\in \Lambda$,

$$\begin{aligned} \partial _x H \le \left\| H\right\| _\infty . \end{aligned}$$

(C.3)

Proof

We have

$$\begin{aligned} -\frac{\left\| H\right\| _\infty }{2}\,\mathbb {I} \le H - \frac{\left\| H\right\| _\infty }{2}\,\mathbb {I} \le \frac{\left\| H\right\| _\infty }{2}\,\mathbb {I}, \end{aligned}$$

(C.4)

therefore

$$\begin{aligned} \partial _x H \le 2\left\| H - \frac{\left\| H\right\| _\infty }{2}\,\mathbb {I}\right\| _\infty \le \left\| H\right\| _\infty . \end{aligned}$$

(C.5)

The claim follows. $\square$

Proposition C.1

Let $\Lambda _1,\,\ldots ,\,\Lambda _k$ be k copies of the finite set $\Lambda$. Then, for any $\rho \in \mathcal {S}_{\Lambda _1\ldots \Lambda _k}$ and any $\sigma \in \mathcal {S}_\Lambda$ we have

$$\begin{aligned} \left\| \rho - \sigma ^{\otimes k}\right\| _{W_1}^2 \le 2k\left| \Lambda \right| ^2\,S\left( \rho \left\| \sigma ^{\otimes k}\right. \right) . \end{aligned}$$

(C.6)

Proof

The proof follows the same lines as the proof of [17, Theorem 2]. We have

$$\begin{aligned} \left\| \rho - \sigma ^{\otimes k}\right\| _{W_1}&\le \sum _{i=1}^k\left\| \sigma ^{\otimes \left( i-1\right) }\otimes \rho _{\Lambda _i\ldots \Lambda _k} - \sigma ^{\otimes i}\otimes \rho _{\Lambda _{i+1}\ldots \Lambda _k}\right\| _{W_1} \nonumber \\&\overset{\mathrm {(a)}}{\le } \left| \Lambda \right| \sum _{i=1}^k\left\| \rho _{\Lambda _i\ldots \Lambda _k} - \sigma \otimes \rho _{\Lambda _{i+1}\ldots \Lambda _k}\right\| _1 \nonumber \\&\overset{\mathrm {(b)}}{\le } \left| \Lambda \right| \sum _{i=1}^k\sqrt{2\,S\left( \rho _{\Lambda _i\ldots \Lambda _k}\left\| \sigma \otimes \rho _{\Lambda _{i+1}\ldots \Lambda _k}\right. \right) } \nonumber \\&= \left| \Lambda \right| \sum _{i=1}^k\sqrt{2\left( S(\rho _{\Lambda _i}) + S(\rho _{\Lambda _{i+1}\ldots \Lambda _k}) - S(\rho _{\Lambda _i\ldots \Lambda _k}) + S(\rho _{\Lambda _i}\Vert \sigma )\right) } \nonumber \\&\overset{\mathrm {(c)}}{\le } \left| \Lambda \right| \sqrt{2k\sum _{i=1}^k\left( S(\rho _{\Lambda _i}) + S(\rho _{\Lambda _{i+1}\ldots \Lambda _k}) - S(\rho _{\Lambda _i\ldots \Lambda _k}) + S(\rho _{\Lambda _i}\Vert \sigma )\right) } \nonumber \\&=\left| \Lambda \right| \sqrt{2k}\sqrt{\sum _{i=1}^k\left( S(\rho _{\Lambda _i}) + S(\rho _{\Lambda _i}\Vert \sigma )\right) - S(\rho )} =\left| \Lambda \right| \sqrt{2k\,S\left( \rho \left\| \sigma ^{\otimes k}\right. \right) }\,. \end{aligned}$$

(C.7)

(a) follows from Proposition A.2 observing that

$$\begin{aligned} \textrm{Tr}_{\Lambda _i}\left[ \sigma ^{\otimes \left( i-1\right) }\otimes \rho _{\Lambda _i\ldots \Lambda _k} - \sigma ^{\otimes i}\otimes \rho _{\Lambda _{i+1}\ldots \Lambda _k}\right] = 0; \end{aligned}$$

(C.8)

(b) follows from Pinsker’s inequality; (c) follows from the concavity of the square root. The claim follows. $\square$

Lemma C.3

We have

$$\begin{aligned} \dim \mathcal {W}_k \le D_k\dim \mathcal {V}. \end{aligned}$$

(C.9)

Proof

Let $A_0,\,\ldots ,\,A_{q^2-1}$ be a basis of $\mathbb {C}^{q\times q}$ with $A_0 = \mathbb {I}$. For any $x\in \left\{ 0,\,\ldots ,\,q^2-1\right\} ^{\Lambda }$, let

$$\begin{aligned} A_x = \bigotimes _{i\in \Lambda }A_{x_i}, \end{aligned}$$

(C.10)

where each $A_{x_i}$ acts on the site i. We have

$$\begin{aligned} \mathcal {W}_k \subseteq \textrm{span}\left\{ A_x|\psi \rangle :|\psi \rangle \in \mathcal {V},\,H(x)\le k\right\} . \end{aligned}$$

(C.11)

We also have

$$\begin{aligned} \mathcal {W}_{k-1} \subseteq \textrm{span}\left\{ A_x|\psi \rangle :|\psi \rangle \in \mathcal {V},\,H(x)\le k-1\right\} , \end{aligned}$$

(C.12)

and since $\mathcal {W}_k\perp \mathcal {W}_{k-1}$, we have

$$\begin{aligned} \mathcal {W}_k \subseteq \textrm{span}\left\{ A_x|\psi \rangle :|\psi \rangle \in \mathcal {V},\,H(x) = k\right\} . \end{aligned}$$

(C.13)

Therefore,

$$\begin{aligned} \dim \mathcal {W}_k \le \left| H^{-1}(k)\right| \dim \mathcal {V}. \end{aligned}$$

(C.14)

The claim follows. $\square$

Lemma C.4

Let $\Phi \in \mathcal {B}_{\mathbb {Z}^d}^r$. Then, for any $\Lambda \in \mathcal {F}_{\mathbb {Z}^d}$ and any $x\in \Lambda$ we have

$$\begin{aligned} \partial _x H^\Phi _\Lambda \le 2\left\| \Phi \right\| _r, \end{aligned}$$

(C.15)

and

$$\begin{aligned} \left\| H^\Phi _\Lambda \right\| _L \le 2\left\| \Phi \right\| _r. \end{aligned}$$

(C.16)

Proof

We have

$$\begin{aligned} \partial _x H^\Phi _\Lambda\le & {} 2\sum _{x\in X \subseteq \Lambda }\left\| \Phi (X)\right\| _\infty \overset{\mathrm {(a)}}{=} 2\sum _{0\in X \subseteq \Lambda -x}\left\| \Phi (X)\right\| _\infty \nonumber \\\le & {} 2\sum _{0\in X \in \mathcal {F}_{\mathbb {Z}^d}}\left\| \Phi (X)\right\| _\infty \le 2\left\| \Phi \right\| _r, \end{aligned}$$

(C.17)

where (a) follows from the translation invariance of $\Phi$. The claim follows. $\square$