Continuum limit of the lattice quantum graph Hamiltonian

Abstract

We consider the quantum graph Hamiltonian on the square lattice in Euclidean space, and we show that the spectrum of the Hamiltonian converges to the corresponding Schrödinger operator on the Euclidean space in the continuum limit, and that the corresponding eigenfunctions and eigenprojections also converge in some sense. We employ the discrete Schrödinger operator as the intermediate operator, and we use a recent result by the second and third authors on the continuum limit of the discrete Schrödinger operator.

Appendix A. Proof of Corollary 2.4

Proof of Corollary 2.4

Since we already have

$$\begin{aligned} d_\mathrm {H}(\sigma ((H+M)^{-1}), \sigma ((H_2+M)^{-1}))\rightarrow 0, \quad \text {as }\ell \rightarrow 0, \end{aligned}$$

(see [7]), it suffices to show

$$\begin{aligned} d_\mathrm {H}(\sigma ((H_2+M)^{-1}), \sigma ((\nu H_1+M)^{-1}))\rightarrow 0, \quad \text {as }\ell \rightarrow 0. \end{aligned}$$

(A.1)

We note that Corollary 4.3 implies

$$\begin{aligned} \bigl | \Vert I^*\varphi \Vert _{{\mathcal {H}}_2}-\Vert \varphi \Vert _{{\mathcal {H}}_1} \bigr | \le C {\ell } \Vert \varphi \Vert _{H^1(\Gamma )}. \end{aligned}$$

(A.2)

In fact, by Corollary 4.3, we immediately have

$$\begin{aligned} \bigl | \Vert I^*\varphi \Vert ^2_{{\mathcal {H}}_2}-\Vert \varphi \Vert ^2_{{\mathcal {H}}_1} \bigr |&=|\langle \varphi ,(I I^*-1)\varphi \rangle _{{\mathcal {H}}_1}|\\&\le \Vert I^*I-1 \Vert _{{\mathcal {B}}(H^1(\Gamma ),{\mathcal {H}}_1)} \Vert \varphi \Vert _{H^1(\Gamma )} \Vert \varphi \Vert _{{\mathcal {H}}_1}\\&\le C\ell \Vert \varphi \Vert _{H^1(\Gamma )}\Vert \varphi \Vert _{{\mathcal {H}}_1}. \end{aligned}$$

Hence, we have

$$\begin{aligned} \bigl | \Vert I^*\varphi \Vert _{{\mathcal {H}}_2}-\Vert \varphi \Vert _{{\mathcal {H}}_1} \bigr |&= \frac{\bigl | \Vert I^*\varphi \Vert ^2_{{\mathcal {H}}_2}-\Vert \varphi \Vert ^2_{{\mathcal {H}}_1} \bigr | }{{\Vert I^*\varphi \Vert _{{\mathcal {H}}_2}+\Vert \varphi \Vert _{{\mathcal {H}}_1}} } \le \frac{\bigl | \Vert I^*\varphi \Vert ^2_{{\mathcal {H}}_2}-\Vert \varphi \Vert ^2_{{\mathcal {H}}_1} \bigr | }{\Vert \varphi \Vert _{{\mathcal {H}}_1}}\\&\le C\ell \Vert \varphi \Vert _{H^1(\Gamma )}. \end{aligned}$$

Let $z\in {\mathbb {C}}\setminus {\mathbb {R}}$. We note that $(\nu H_1-z)^{-1}$ is bounded from ${\mathcal {H}}_1$ to $H^1(\Gamma )$, uniformly in $\ell$. By (A.2), we have

$$\begin{aligned} \bigl | \Vert I^*(\nu H_1-z)^{-1} I\varphi \Vert _{{\mathcal {H}}_2} - \Vert (\nu H_1-z)^{-1}I\varphi \Vert _{{\mathcal {H}}_1} \bigr |\\ \qquad \le C{\ell }\Vert (\nu H_1-z)^{-1}I\varphi \Vert _{H^1(\Gamma )} \le C'{\ell } \Vert \varphi \Vert _{{\mathcal {H}}_2}. \end{aligned}$$

This implies

$$\begin{aligned} \bigl | \Vert I^*(\nu H_1-z)^{-1} I \Vert _{{\mathcal {B}}({\mathcal {H}}_2)} - \Vert (\nu H_1-z)^{-1}I \Vert _{{\mathcal {B}}({\mathcal {H}}_2,{\mathcal {H}}_1)} \bigr | \le C{\ell }. \end{aligned}$$

Then, we consider

$$\begin{aligned} \bigl | \Vert (\nu H_1-z)^{-1} I \Vert _{{\mathcal {B}}({\mathcal {H}}_2,{\mathcal {H}}_1)} - \Vert (\nu H_1-z)^{-1} \Vert _{{\mathcal {B}}({\mathcal {H}}_1)} \bigr |\\ =\bigl | \Vert I^*(\nu H_1-{\bar{z}})^{-1} \Vert _{{\mathcal {B}}({\mathcal {H}}_1,{\mathcal {H}}_2)} - \Vert (\nu H_1-{\bar{z}})^{-1} \Vert _{{\mathcal {B}}({\mathcal {H}}_1)} \bigr |, \end{aligned}$$

but the right-hand side is bounded by $C{\ell }$ as well as the above argument, simply by replacing z by ${\bar{z}}$, and $I\varphi$ by $\varphi$. Thus, we have

$$\begin{aligned} \bigl | \Vert I^*(\nu H_1-z)^{-1} I \Vert _{{\mathcal {B}}({\mathcal {H}}_2)} - \Vert (\nu H_1-z)^{-1} \Vert _{{\mathcal {B}}({\mathcal {H}}_1)} \bigr | \le C{\ell } \end{aligned}$$

Combining this with Theorem 4.6, we have

$$\begin{aligned} \bigl | \Vert (H_2-z)^{-1} \Vert _{{\mathcal {B}}({\mathcal {H}}_2)} - \Vert (\nu H_1-z)^{-1} \Vert _{{\mathcal {B}}({\mathcal {H}}_1)} \bigr | \le C{\ell } \end{aligned}$$

(A.3)

where $0<\ell \le 1$.

Let $R>0$, fixed. We consider the resolvents at $z=\mu +i$, $\mu \in [-R,R]$. Here we show the estimate (A.3) holds uniformly in such z. By the resolvent equation, we know

$$\begin{aligned} \Vert (A-(\mu +i))^{-1} - (A-(\mu '+i))^{-1} \Vert \le |\mu -\mu '|, \quad \mu ,\mu '\in {\mathbb {R}}, \end{aligned}$$

in general, where A is a self-adjoint operator. Let $\varepsilon >0$, and let $\{\mu _j\}_{j=1}^N =\{n\varepsilon /3\in [-R,R]\mid n\in {\mathbb {Z}}\}$. We choose $\ell _0>0$ so small that

$$\begin{aligned} \bigl | \Vert (H_2-(\mu _j+i))^{-1} \Vert _{{\mathcal {B}}({\mathcal {H}}_2)} - \Vert (\nu H_1-(\mu _j+i))^{-1} \Vert _{{\mathcal {B}}({\mathcal {H}}_1)} \bigr | < \varepsilon /3 \end{aligned}$$

for $0<\ell \le \ell _0$ and $j=1,\dots , N$. Then, by the $\varepsilon /3$-argument, we learn

$$\begin{aligned} \bigl | \Vert (H_2-(\mu +i))^{-1} \Vert _{{\mathcal {B}}({\mathcal {H}}_2)} - \Vert (\nu H_1-(\mu +i))^{-1} \Vert _{{\mathcal {B}}({\mathcal {H}}_1)} \bigr | < \varepsilon \end{aligned}$$

for all $\mu \in [-R,R]$, $0<\ell \le \ell _0$, and this proves the uniform bound.

We now prove the local convergence of the spectrum with respect to the Hausdorff distance. We fix $R>0$, and consider $\sigma (\nu H_1)$ and $\sigma (H_2)$ in $[-R,R]$. Let $\varepsilon >0$ and we set

$$\begin{aligned} \rho _\varepsilon = [-R,R]\cap \{\mu \in {\mathbb {R}}\mid \mathrm {dist}(\mu ,\sigma (\nu H_1))\ge \varepsilon \}. \end{aligned}$$

We note, for each $\mu \in \rho _\varepsilon$,

$$\begin{aligned} \Vert (\nu H_1-(\mu +i))^{-1} \Vert _{{\mathcal {B}}({\mathcal {H}}_1)}\le 1/\sqrt{1+\varepsilon ^2}. \end{aligned}$$

We choose $\ell _0>0$ so small that

$$\begin{aligned} \bigl | \Vert (H_2-(\mu +i))^{-1} \Vert _{{\mathcal {B}}({\mathcal {H}}_2)} - \Vert (\nu H_1-(\mu +i))^{-1} \Vert _{{\mathcal {B}}({\mathcal {H}}_1)} \bigr | \le \frac{1}{\sqrt{1+\varepsilon ^2/4}} -\frac{1}{\sqrt{1+\varepsilon ^2}} \end{aligned}$$

for $\mu \in [-R,R]$ and $0<\ell \le \ell _0$. Then, this implies

$$\begin{aligned} \Vert (H_2-(\mu +i))^{-1} \Vert _{{\mathcal {B}}({\mathcal {H}}_2)} \le \frac{1}{\sqrt{1+\varepsilon ^2/4}} \end{aligned}$$

if $\mu \in \rho _\varepsilon$ and $0<\ell \le \ell _0$, and hence, $(\mu -\varepsilon /2,\mu +\varepsilon /2)\subset \rho (H_2)$. In particular, $\mu \in \rho (H_2)$ if $\mu \in \rho _\varepsilon$, i.e., $\rho _\varepsilon \subset \rho (H_2)$, provided $0<\ell \le \ell _0$. This, in turn, implies $\sigma (H_2)\cap [-R,R]$ is included in the $\varepsilon$-neighborhood of $\sigma (\nu H_1)$.

Replacing $\nu H_1$ and $H_2$, we also learn $\sigma (\nu H_1)\cap [-R,R]$ is included in the $\varepsilon$-neighborhood of $\sigma (H_2)$ if $\ell$ is sufficiently small. In other words, we now have

$$\begin{aligned} \sup _{\mu \in \sigma (\nu H_1)\cap [-R,R]} d(\mu ,\sigma (H_2))<\varepsilon , \quad \sup _{\mu \in \sigma (H_2)\cap [-R,R]} d(\mu ,\sigma (\nu H_1))<\varepsilon , \end{aligned}$$

if $\ell >0$ is sufficiently small. This local convergence implies (A.1) by the spectrum mapping theorem. This completes the proof.