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A norm-convergent observable with no finite localization region

Statement

Take $d = 1$ and $q = 2$ . For $n \geq 1$ , let $Z_{n}$ be the one-site Pauli matrix $Z = diag (1, - 1)$ embedded at site $n \in Z$ . The sequence

$H_{N} : = \sum_{n = 1}^{N} 2^{- n} Z_{n}$

converges in norm in the quasi-local algebra to an element $H$ . This element is self-adjoint but has no finite localization region. Thus norm completion adds observables that are not strictly local.

Facts & Assumptions

Given: $d = 1$ , $q = 2$ , the local algebras and factor inclusions, and the norm completion $A$ .

[F1]

A localization region for $d = 1$ is a finite subset $Λ \subseteq Z$ (Local spin matrix algebras and their factor inclusions). Integer addition and $1$ are defined by Arithmetic on the integers, the integer order is defined by Order on the integers, and it is a total order compatible with addition by The integers form a totally ordered ring.

[F2]

The integer $1 = [(1, 0)]$ is positive: by Order on the integers this is equivalent to $0 < 1$ in $N$ , which holds because $1 = σ (0) = {0}$ (Arithmetic on the integers, The integers as equivalence classes of pairs of naturals, The natural numbers $N$ (von Neumann), On $N$ the order is membership: $m < n ⟺ m \in n$ ).

[F3]

Each one-site Pauli matrix $Z$ is self-adjoint and has operator norm one; the factor inclusions are isometric (Local spin matrix algebras and their factor inclusions, Local spin factor inclusions preserve the operator norm).

[F4]

$A_{l o c}$ is dense in its norm completion $A$ , multiplication and adjoint extend continuously, and disjoint local factors commute (Norm-completed quasi-local quantum spin algebra, Quasi-local algebra completion and locality).

Proof

technique · direct

Given: The sequence $(H_{N})$ and facts [F1]–[F4].

1.1givenF1F2constructalgebra

Fix any finite region $Λ \subseteq Z$ . Enumerate the finite set $F : = Λ \cap {n \in Z : 0 < n}$ as $n_{1}, \dots, n_{r}$ , and put $S : = \sum_{j = 1}^{r} n_{j}$ , taking $S = 0$ when $r = 0$ . By induction on the number of terms, every finite sum of positive integers is nonnegative: the empty sum is $0$ , and if a partial sum $R \geq 0$ then adding a positive integer $n$ gives $R + n > R \geq 0$ , since $0 < n$ translates to $R + 0 \leq R + n$ by [F1], and equality would cancel $R$ in the additive group $Z$ to give $0 = n$ . If $k \in F$ , write $S = k + R$ , where $R$ is the sum of the remaining positive terms; the same induction gives $R \geq 0$ , and order-compatibility yields $S \geq k$ . Set $m : = S + 1$ . Since $1 > 0$ by [F2] and $S \geq 0$ , order-compatibility gives $m > S \geq 0$ , so $m > 0$ . If $m \in Λ$ , then $m \in F$ and the preceding bound gives $S \geq m$ , contradicting $m > S$ . Thus every finite region omits a positive integer site.

1.2givenF3F4algebra

If $M > N$ , the triangle inequality and isometric inclusions give $∥ H_{M} - H_{N} ∥ \leq \sum_{n = N + 1}^{M} 2^{- n} < 2^{- N}$ . Thus $(H_{N})$ is Cauchy in $A_{l o c}$ and converges to some $H \in A$ by completion. Since each $H_{N}$ is self-adjoint and the adjoint extends continuously, $H^{*} = H$ .

2.1F3F4step 1.2algebra

Fix $k \geq 1$ and define $X = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ on $C^{2}$ , and let $X_{k}$ be its one-site embedding at site $k$ . For $N \geq k$ , every $Z_{n}$ with $n \neq k$ commutes with $X_{k}$ by disjoint locality, while direct multiplication gives $[Z, X] = Z X - X Z = 2 i Y$ , where $Y = (\begin{array}{cc} 0 & - i \\ i & 0 \end{array})$ and $Y^{*} Y = I$ . Hence $[H_{N}, X_{k}] = 2^{- k} (2 i Y)_{k}$ for all $N \geq k$ . By continuity of multiplication and norm convergence, $[H, X_{k}] = 2^{1 - k} i Y_{k}$ , whose norm is $2^{1 - k} > 0$ .

3.1F1F4step 1.1step 2.1∎

If $H$ had a finite localization region $Λ$ , step 1.1 supplies $k \geq 1$ outside $Λ$ . Then [F4] would give $[H, X_{k}] = 0$ , contradicting step 2.1. Therefore $H$ has no finite localization region.

A norm-convergent observable with no finite localization region

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources