Locality of the windowed local density of states

Abstract

We consider a generalization of local density of states which is “windowed” with respect to position and energy, called the windowed local density of states (wLDOS). This definition generalizes the usual LDOS in the sense that the usual LDOS is recovered in the limit where the position window captures individual sites and the energy window is a delta distribution. We prove that the wLDOS is local in the sense that it can be computed up to arbitrarily small error using spatial truncations of the system Hamiltonian. Using this result we prove that the wLDOS is well-defined and computable for infinite systems satisfying some natural assumptions. We finally present numerical computations of the wLDOS at the edge and in the bulk of a “Fibonacci SSH model”, a one-dimensional non-periodic model with topological edge states.

Proofs of Lemma 3.1 and Proposition 3.2

In this section we give the proofs of Lemma 3.1 and Proposition 3.2.

1.1 Proof of Lemma 3.1

Before we can give the proof of Lemma 3.1 we require two preliminary Lemmas. Note that we take $x = E = 0$ for simplicity.

Lemma A.1

Suppose that A and B are bounded linear operators, with A Hermitian. Then

$$\begin{aligned} \Vert [ e^{i t A}, B ] \Vert \le |t| \Vert [ A, B ] \Vert . \end{aligned}$$

(A.1)

Proof

First note that

$$\begin{aligned} \frac{\text {d}}{\text {d}t} \Bigl ( e^{i t A} B e^{- i t A}\Bigr ) = i e^{i t A} [ A, B ] e^{- i t A}. \end{aligned}$$

(A.2)

Integrating from 0 to t, we get

$$\begin{aligned} e^{i t A} B e^{- i t A} - B = i \int _{0}^{t} \! e^{i s A} [ A, B ] e^{- i s A} \, \text {d}s . \end{aligned}$$

(A.3)

We now move to proving the estimate starting with

$$\begin{aligned} \Vert [ e^{i t A}, B ] \Vert = \Vert e^{i t A} B - B e^{i t A} \Vert = \Vert e^{i t A} B e^{- i t A} - B \Vert . \end{aligned}$$

(A.4)

Using (A.3) we now have

$$\begin{aligned} \Vert [ e^{i t A}, B ] \Vert = \left\| \int _{0}^{t} \! e^{i s A} [ A, B ] e^{- i s A} \, \text {d}s \right\| \le |t| \Vert [A,B] \Vert . \end{aligned}$$

(A.5)

$\square$

Lemma A.2

Let H and X be self-adjoint operators with H bounded. Let g and k be functions on ${\mathbb {R}}$ with $0 \le g \le 1$, $0 \le k \le 1$ and $k g = g$. Then

$$\begin{aligned} \Vert g^{\frac{1}{2}}(X) e^{i t H} g^{\frac{1}{2}}(X) {-} g^{\frac{1}{2}}(X) e^{i t k(X) H k(X)} g^{\frac{1}{2}}(X) \Vert {\le } |t| ( 1 + |t| \Vert H \Vert ) \Vert [k(X),H] \Vert .\nonumber \\ \end{aligned}$$

(A.6)

Proof

First note that

$$\begin{aligned}{} & {} \left\| g^{\frac{1}{2}}(X) e^{i t H} g^{\frac{1}{2}}(X) - g^{\frac{1}{2}}(X) e^{i t k(X) H k(X)} g^{\frac{1}{2}}(X) \right\| \nonumber \\{} & {} \quad \le \left\| g^{\frac{1}{2}}(X) - e^{- i t H} e^{i t k(X) H k(X)} g^{\frac{1}{2}}(X) \right\| \nonumber \\{} & {} \quad \le \int _{0}^{t} \! \left\| \frac{\text {d}}{\text {d}s} \left( g^{\frac{1}{2}}(X) - e^{- i s H} e^{i s k(X) H k(X)} g^{\frac{1}{2}}(X) \right) \right\| \, \text {d}s \nonumber \\{} & {} \quad = \int _{0}^{t} \! \left\| \frac{\text {d}}{\text {d}s} \left( e^{- i s H} e^{i s k(X) H k(X)} g^{\frac{1}{2}}(X) \right) \right\| \, \text {d}s .\nonumber \\ \end{aligned}$$

(A.7)

Clearly,

$$\begin{aligned} \frac{\text {d}}{\text {d}t} \left( e^{- i t H} e^{i t k(X) H k(X)} g^{\frac{1}{2}}(X) \right) {=} i e^{- i t H} \left( k(X) H k(X) {-} H \right) e^{i t k(X) H k(X)} g^{\frac{1}{2}}(X).\nonumber \\ \end{aligned}$$

(A.8)

Re-arranging the right-hand side of (A.8) and using $k g = k$ gives

$$\begin{aligned}{} & {} \frac{\text {d}}{\text {d}t} \left( e^{- i t H} e^{i t k(X) H k(X)} g^{\frac{1}{2}}(X) \right) \nonumber \\{} & {} \quad = i e^{- i t H} \left( [k(X),H] e^{i t k(X) H k(X)} \right) g^{\frac{1}{2}}(X) \nonumber \\{} & {} \quad \quad + i e^{- i t H} \left( k(X) H [ k(X), e^{i t k(X) H k(X)} ] + H [ k(X), e^{i t k(X) H k(X)} ] \right) g^{\frac{1}{2}}(X).\nonumber \\ \end{aligned}$$

(A.9)

The first term on the right-hand side of (A.9) can be bounded by $\Vert [k(X),H] \Vert$. Using Lemma A.1 we have

$$\begin{aligned} \Vert [ k(X), e^{i t k(X) H k(X)} ] \Vert \le |t| \Vert [ k(X), k(X) H k(X) ] \Vert = |t| \Vert [ k(X), H ] \Vert \end{aligned}$$

(A.10)

and hence the second two terms on the right-hand side of (A.9) are bounded by $2 |t| \Vert H \Vert \Vert [ k(X), H ] \Vert$. Substituting these estimates into (A.7) we have

$$\begin{aligned}{} & {} \left\| g^{\frac{1}{2}}(X) e^{i t H} g^{\frac{1}{2}}(X) - g^{\frac{1}{2}}(X) e^{i t k(X) H k(X)} g^{\frac{1}{2}}(X) \right\| \nonumber \\{} & {} \quad \le \int _{0}^{t} \! \Vert [k(X),H] \Vert + 2 |s| \Vert H \Vert \Vert [k(X),H] \Vert \, \text {d}s \nonumber \\{} & {} \quad \le |t| ( 1 + |t| \Vert H \Vert ) \Vert [k(X),H] \Vert \nonumber \\ \end{aligned}$$

(A.11)

as required. $\square$

We can now give the proof of Lemma 3.1.

Proof of Lemma 3.1

Let $\ell$ be the inverse Fourier transform of $f'$, so

$$\begin{aligned} f'(\xi )=\int _{-\infty }^{\infty }\ell (t)e^{it\xi }\,dt. \end{aligned}$$

(A.12)

For bounded Hermitian operators K, we have by functional calculus that

$$\begin{aligned} f(K)=f(0)I+\int _{-\infty }^{\infty }\frac{\ell (t)}{it}(e^{itK}-I)\,dt. \end{aligned}$$

(A.13)

It then follows that

$$\begin{aligned} g^{\frac{1}{2}}(X)f(K)g^{\frac{1}{2}}(X)=f(0)g(X)+\int _{-\infty }^{\infty }\frac{\ell (t)}{it}\left( g^{\frac{1}{2}}(X)e^{itK}g^{\frac{1}{2}}(X)-g(X)\right) \,dt.\nonumber \\ \end{aligned}$$

(A.14)

Comparing this identity with $K=H$ and $K=k(X)Hk(X)$ we find

$$\begin{aligned}&g^{\frac{1}{2}}(X)f(H)g^{\frac{1}{2}}(X)-g^{\frac{1}{2}}(X)f(k(X) H k(X))g^{\frac{1}{2}}(X)\\&\qquad =\int _{-\infty }^{\infty }\frac{\ell (t)}{it}\left( g^{\frac{1}{2}}(X)e^{itH}g^{\frac{1}{2}}(X)-g^{\frac{1}{2}}(X)e^{itk(X)Hk(X)}g^{\frac{1}{2}}(X)\right) \,dt. \end{aligned}$$

Therefore

$$\begin{aligned}&\left\| g^{\frac{1}{2}}(X)f(H)g^{\frac{1}{2}}(X)-g^{\frac{1}{2}}(X)f(k(X) H k(X) ) g^{\frac{1}{2}}(X)\right\| \\&\qquad \le \int _{-\infty }^{\infty }\frac{\left| \ell (t)\right| }{\left| t\right| }\left( |t| ( 1 + |t| \Vert H \Vert ) \Vert [k(X),H] \Vert \right) \,dt\\&\qquad = \left\| \left[ k(X),H\right] \right\| \int _{-\infty }^{\infty } ( 1 + | t | \Vert H \Vert ) \left| \ell (t)\right| \,dt. \end{aligned}$$

Since $\ell (t) = i t \widehat{f}(t)$ the statement follows.

1.2 Proof of Proposition 3.2

We now prove Proposition 3.2. We again start with a preliminary Lemma.

Lemma A.3

Suppose that A and B are bounded matrices, with A Hermitian, and suppose $k \in C^1(\mathbb {R})$ is such that

$$\begin{aligned} k'(x) = \int _{-\infty }^{\infty } \! \ell (t) e^{i t x} \, \text {d}t \end{aligned}$$

(A.15)

for some $\ell (t) \in L^1(\mathbb {R})$. Then

$$\begin{aligned} \Vert [ k(A), B ] \Vert \le \Vert \ell \Vert _{L^1} \Vert [ A, B ] \Vert . \end{aligned}$$

(A.16)

Proof

Using (A.15), we have

$$\begin{aligned} k(x){} & {} = k(0) + \int _{0}^{x} \! k'(y) \, \text {d}y \nonumber \\{} & {} = k(0) + \int _{0}^{x} \! \int _{-\infty }^{\infty } \! \ell (t) e^{i t y} \, \text {d}t \, \text {d}y \nonumber \\{} & {} = k(0) + \int _{-\infty }^{\infty } \! \ell (t) \int _{0}^{x} \! e^{i t y} \, \text {d}y \, \text {d}t \nonumber \\{} & {} = k(0) + \int _{-\infty }^{\infty } \! \ell (t) \left( \frac{ e^{i t x} - 1 }{ i t } \right) \, \text {d}t . \nonumber \\ \end{aligned}$$

(A.17)

Hence

$$\begin{aligned} k(A) = k(0) I + \int _{-\infty }^{\infty } \! \frac{ \ell (t) }{ i t } \left( e^{i t A} - 1 \right) \, \text {d}t , \end{aligned}$$

(A.18)

where the integral is well-defined because

$$\begin{aligned} \Vert e^{i t A} - 1 \Vert \le | t | \Vert A \Vert \end{aligned}$$

(A.19)

(to see this differentiate the operator on the left-hand side). From (A.18), we have that

$$\begin{aligned}{}[ k(A), B ] = \int _{-\infty }^{\infty } \! \frac{ \ell (t) }{ i t } [ e^{i t A}, B ] \, \text {d}t . \end{aligned}$$

(A.20)

The result now follows by Lemma A.1. $\square$

We are now in a position to give an explicit construction of $k_\alpha (\xi )$ equalling 1 for all $\xi \in [-L,L]$ and satisfying (3.7).

Proof of Proposition 3.2

Fix $L > 0$, and define $M_L$ to be the smallest integer such that $2 M_L > L$. Define $k(\xi )$ as in (2.12), i.e.

$$\begin{aligned} k(\xi ) = {\left\{ \begin{array}{ll} 0 &{} \xi \le -2 \\ \frac{1}{2} (\xi + 2)^2 &{} -2< \xi \le -1 \\ 1 - \frac{1}{2} \xi ^2 &{} -1< \xi \le 1 \\ \frac{1}{2} (\xi - 2)^2 &{} 1 < \xi \le 2 \\ 0 &{} 2 \le \xi . \end{array}\right. } \end{aligned}$$

(A.21)

It is clear that

$$\begin{aligned} k(\xi ) + k(\xi -2) = {\left\{ \begin{array}{ll} 0 &{} \xi \le -2 \\ \frac{1}{2}(\xi +2)^2 &{} -2< \xi \le -1 \\ 1 - \frac{1}{2}\xi ^2 &{} -1< \xi \le 0 \\ 1 &{} 0< \xi \le 2 \\ 1 - \frac{1}{2} (\xi - 2)^2 &{} 2 \le \xi< 3 \\ \frac{1}{2}(\xi - 4)^2 &{} 3< \xi \le 4 \\ 0 &{} 4 < \xi , \end{array}\right. } \end{aligned}$$

(A.22)

and more generally,

$$\begin{aligned} k_1(\xi ):= \sum _{l = -M}^M k(\xi - 2 l) = {\left\{ \begin{array}{ll} 0 &{} \xi \le - 2 M - 2 \\ \frac{1}{2}(\xi + 2 M + 2)^2 &{} - 2 M - 2 \le \xi \le - 2 M - 1 \\ 1 - \frac{1}{2}(\xi + 2M)^2 &{} - 2 M - 1 \le \xi \le - 2 M \\ 1 &{} -2 M \le \xi \le 2 M \\ 1 - \frac{1}{2} (\xi - 2 M)^2 &{} 2 M \le \xi \le 2 M + 1 \\ \frac{1}{2}(\xi - 2 M - 2)^2 &{} 2 M + 1 \le \xi \le 2 M + 2 \\ 0 &{} \xi \ge 2 M + 2. \end{array}\right. } \end{aligned}$$

(A.23)

Using the fact that $2 M > L$ by assumption, we have that $k_1(\xi )$ acts by 1 over the whole interval $[-L,L]$. For positive $\alpha$, we now define

$$\begin{aligned} k_\alpha (\xi ):= k_1(\alpha \xi ). \end{aligned}$$

(A.24)

It is easy to see that $k_\alpha (\xi )$ acts as 1 over the interval $\left[ -\frac{2 M}{\alpha },\frac{2 M}{\alpha }\right]$ and hence acts as 1 over the interval $\left[ -\frac{L}{\alpha },\frac{L}{\alpha }\right]$, which contains $[-L,L]$ for $0< \alpha < 1$. The support of $k_\alpha (\xi )$ is clearly confined to $\left[ - \frac{2\,M+2}{\alpha }, \frac{2\,M + 2}{\alpha } \right]$. Using the definition of M as the smallest integer such that $2 M > L$ we see that $L \le 2\,M \le L + 2 \le 2\,M + 2 \le L + 4$ and hence the support of $k_\alpha (\xi )$ is confined to $\left[ - \frac{ 2\,M + 4 }{ \alpha }, \frac{ 2\,M + 4 }{ \alpha } \right]$.

We will now prove that $k_\alpha (\xi )$ satisfies the bound (3.7) using Lemma A.3. Our strategy is to build up to a bound on the Fourier transform of $k_\alpha '(\xi )$ from a bound on the Fourier transform of $k'(\xi )$.

We start by noting that if $k(\xi )$ is defined by (A.21), then

$$\begin{aligned} k'(\xi ) = {\left\{ \begin{array}{ll} 0 &{} \xi \le -2 \\ \xi + 2 &{} -2< \xi \le -1 \\ - \xi &{} -1< \xi \le 1 \\ \xi - 2 &{} 1 < \xi \le 2 \\ 0 &{} 2 \le \xi . \end{array}\right. } \end{aligned}$$

(A.25)

Since $k'(\xi )$ is odd, its Fourier transform $\ell (t)$ equals

$$\begin{aligned} \ell (t){} & {} := - \frac{i}{\pi } \int _{0}^{\infty } \! k'(\xi ) \sin (t \xi ) \, \text {d}\xi \nonumber \\{} & {} = - \frac{i}{\pi } \left[ \int _{0}^{1} \! (- \xi ) \sin (t \xi ) \, \text {d}\xi + \int _{1}^{2} \! (\xi - 2) \sin (t \xi ) \, \text {d}\xi \right] . \end{aligned}$$

(A.26)

Integrating by parts in these integrals we have

$$\begin{aligned} \int _{0}^{1} \! (- \xi ) \sin (t \xi ) \, \text {d}\xi= & {} \frac{ \cos (t) }{ t } - \frac{ \sin (t) }{ t^2 } \end{aligned}$$

(A.27)

$$\begin{aligned} \int _{1}^{2} \! (\xi - 2) \sin (t \xi ) \, \text {d}\xi= & {} - \frac{ \cos (t) }{ t } + \frac{ \sin (2 t) }{ t^2 } - \frac{ \sin (t) }{ t^2 } \end{aligned}$$

(A.28)

and hence

$$\begin{aligned} \ell (t) = - \frac{i}{\pi } \frac{ \sin (2 t) - 2 \sin (t) }{ t^2 }, \end{aligned}$$

(A.29)

which is clearly in $L^1(\mathbb {R})$. In fact, numerical computation shows that $\Vert \ell \Vert _1 \approx 1.27$ (3sf). Now let $\ell _1(t)$ denote the Fourier transform of $k_1'(\xi )$. Using linearity and a change of variables we have

$$\begin{aligned} \ell _1(t) = \sum _{l = - M}^M e^{- 2 i l t} \ell (t). \end{aligned}$$

(A.30)

Using the triangle inequality we have

$$\begin{aligned} \Vert \ell _1 \Vert _{L^1} \le (2 M + 1) \Vert \ell \Vert _{L^1}. \end{aligned}$$

(A.31)

Finally, let $\ell _\alpha (t)$ denote the Fourier transform of $k_\alpha '(\xi )$. Since $k_\alpha '(\xi ) = \alpha k_1'(\alpha \xi )$, we see that

$$\begin{aligned} \ell _\alpha (t) = \frac{1}{2 \pi } \int _{-\infty }^{\infty } \! \alpha k_1'(\alpha \xi ) e^{- i t \xi } \, \text {d}\xi = \ell _1 \left( \frac{t}{\alpha }\right) , \end{aligned}$$

(A.32)

from which it follows immediately that

$$\begin{aligned} \Vert \ell _\alpha (t) \Vert _{L^1} = \alpha \Vert \ell _1(t) \Vert _{L^1} \le (2 M + 1) \alpha \Vert \ell (t) \Vert _{L^1}. \end{aligned}$$

(A.33)

Applying Lemma A.3 now proves Proposition 3.2.