Defect Modes for Dislocated Periodic Media


We study defect modes in a one-dimensional periodic medium perturbed by an adiabatic dislocation of amplitude \(\delta \ll 1\). If the periodic background admits a Dirac point—a linear crossing of dispersion curves—then the dislocated operator acquires a gap in its essential spectrum. For this model (and its honeycomb analog) Fefferman et al. (Proc Natl Acad Sci USA 111(24):8759–8763, 2014, Mem Am Math Soc 247(1173):118, 2017, Ann PDE 2(2):80, 2016, 2D Mater 3:1, 2016) constructed (at leading order in \(\delta \)) defect modes with energies within the gap. These bifurcate from the eigenmodes of an effective Dirac operator. Here we address the following open problems:

  • Do all defect modes arise as bifurcations from the Dirac operator eigenmodes?

  • Do these modes admit expansions to all order in \(\delta \)?

We respond positively to both questions. Our approach relies on (a) resolvent estimates for the bulk operators; (b) scattering theory for highly oscillatory potentials [Dr18a, Dr18b, Dr18c]. It has led to an understanding of the topological stability of defect states in continuous dislocated systems—in connection with the bulk-edge correspondence [Dr18d].

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This research was supported in part by National Science Foundation Grants DMS-1800086 (AD), DMS-1265524 (CF), DMS-1412560, DMS-1620418 (MIW) and Simons Foundation Math + X Investigator Award #376319 (MIW and AD). The authors thank A. B. Watson for very stimulating discussions and S. Dyatlov for suggesting a more conceptual proof to Proposition 4.2.

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Appendix A. From Quasimodes to Eigenvalues in Gapped Selfadjoint Problems

Proof of Lemma 3.1

If E is an eigenvalue of T, the lemma is proved. Otherwise \(T-E : \mathcal {H}\rightarrow \mathcal {H}\) is invertible; from the assumption, E is neither in the pure point spectrum nor in the essential spectrum. From (3.3), \(|(T-E)^{-1}|_{\mathcal {H}}\ge \epsilon ^{-1}\). But because of the spectral theorem,

$$\begin{aligned} \left\| (T-E)^{-1}\right\| _{\mathcal {H}} = \dfrac{1}{{{\text {dist}}}(E, \Sigma (T))}. \end{aligned}$$

We deduce \({{\text {dist}}}(E,\Sigma (T)) \le \epsilon \), therefore E is \(\epsilon \)-close to the spectrum of T. Since \(0< \epsilon < \epsilon _0\), and T has no essential spectrum in \([E-\epsilon _0,E+\epsilon _0]\), T must have an eigenvalue \(\epsilon \) in \([E-\epsilon ,E+\epsilon ]\).

Assume now that T has only one eigenvalue—say \(\lambda \)—in \([E-C\epsilon ,E+C\epsilon ]\). Let u be a corresponding normalized eigenvector. Define \(T'\) formally as T, but acting on the space \(\mathcal {H}' = \text {span}\{u\}^\perp \). The spectral theorem implies that

$$\begin{aligned} \left\| (T'-E)^{-1}\right\| _{\mathcal {H}'}= \frac{1}{{{\text {dist}}}(E,\Sigma (T'))}\le \dfrac{1}{C\epsilon }. \end{aligned}$$

Therefore we have

$$\begin{aligned} w\in \mathcal {H}' \ \Rightarrow \ C\epsilon \cdot |w|_{\mathcal {H}'} \le |(T'-E)w|_{\mathcal {H}'}. \end{aligned}$$

If v satisfies (3.3), write \(v = \alpha u + w\) where \(w \in \mathcal {H}'\) and \(\alpha \in \mathbb {C}\). Then,

$$\begin{aligned} (T-E) v = \alpha (T-E)u + (T-E)w = \alpha (\lambda -E)u + (T'-E)w. \end{aligned}$$

This is because \(Tu = \lambda u\) and \(w\in \mathcal {H}'\), hence \(Tw = T' w\). The RHS is an orthogonal decomposition of \((T-E) v\). Thus \(|(T-E)v|_{\mathcal {H}} \ge |(T'-E)w|_{\mathcal {H}}=|(T'-E)w|_{\mathcal {H}'}\). From (3.3) the lower bound (A.1) and the relation \(w=v-\alpha u\), we deduce

$$\begin{aligned} \epsilon \ge |(T-E)v|_{\mathcal {H}} \ge |(T'-E)w|_{\mathcal {H}'} \ge C\epsilon |w|_{\mathcal {H}'}=C\epsilon |v-\alpha u|_{\mathcal {H}'}. \end{aligned}$$

Thus, \(|v-\alpha u|_{\mathcal {H}} \le C^{-1}\); since \(\alpha u\) is an eigenvector of T, the proof is complete. \(\quad \square \)

Appendix B. Spectral Estimates for Highly Oscillatory Operators

Lemma B.1

Let \(F \in C^\infty (\mathbb {R}\times \mathbb {R})\) be 1-periodic in the first variable and such that there exists \(\theta > 0\) with

$$\begin{aligned} \text {for all}\quad \alpha \ge 0, \ \ \sup _{(x,y) \in \mathbb {R}^2} e^{\theta |y|} \cdot \left| {\partial }_y^\alpha F(x,y) \right| < \infty . \end{aligned}$$

Set \(f_\delta (y) = F(y/\delta ,y)\). Then there exists \(C > 0\) such that for \(\delta \in (0,1)\),

$$\begin{aligned} |f_\delta - g|_{H^{-1/4}} \le C\delta ^{1/4}, \ \ \ \ g(y) = \int _0^1 F(x,y) dx. \end{aligned}$$


This is mostly contained in the proof of [Dr18b, Theorem 1]. Write a Fourier decomposition of F:

$$\begin{aligned} F(x,y) = \sum _{m \in 2\pi \mathbb {Z}} a_m(y) e^{imx}, \ \ \ \ a_m(y) = \int _0^1 F(x,y) e^{-imx} dx. \end{aligned}$$


$$\begin{aligned} \left| f_\delta - g\right| _{H^{-{1/4}}}= & {} \left| \left\langle D \right\rangle ^{-{1/4}}(f_\delta -g)\right| _{L^2} \le \sum _{m \in 2\pi \mathbb {Z}\setminus \{0\} } \left| \left\langle \ \cdot \ \right\rangle ^{-{1/4}} \widehat{a_m}(\ \cdot -m/\delta )\right| _{L^2} \\= & {} \sum _{m \in 2\pi \mathbb {Z}\setminus \{0\}} \left| \left\langle \ \cdot \ \right\rangle ^{-{1/4}} \left\langle \ \cdot - m/\delta \right\rangle ^{-{1/4}} \left\langle \ \cdot -m/\delta \right\rangle ^{{1/4}} \widehat{a_m}(\ \cdot -m/\delta )\right| _{L^2} \\\le & {} \sum _{m \in 2\pi \mathbb {Z}\setminus \{0\}} \left( \sup _{\zeta \in \mathbb {R}} \left\langle \zeta \right\rangle ^{-{1/4}} \left\langle \zeta -m/\delta \right\rangle ^{-{1/4}} \right) \cdot \left| \left\langle \ \cdot -m/\delta \right\rangle ^{{1/4}} \widehat{a_m}(\ \cdot -m/\delta )\right| _{L^2}. \end{aligned}$$

We now apply Peetre’s inequality to the first factor and deduce that

$$\begin{aligned} \left| f_\delta - g\right| _{H^{-{1/4}}} \le \sum _{m \in 2\pi \mathbb {Z}\setminus \{0\}} \left\langle m/\delta \right\rangle ^{-{1/4}} \cdot \left| \left\langle \ \cdot \ \right\rangle ^{{1/4}} \widehat{a_m}\right| _{L^2} \le C\delta ^{1/4} \sum _{m \in 2\pi \mathbb {Z}\setminus \{0\}} \dfrac{|a_m|_{H^{1/4}}}{m^{1/4}}. \end{aligned}$$

Because \(F \in C^\infty (\mathbb {R}\times \mathbb {R})\) decays sufficiently fast, the sum on the RHS is absolutely convergent. This completes the proof. \(\quad \square \)

Proof of Lemma 6.1

  1. 1.

    We first use the cyclicity of the determinant to write

    $$\begin{aligned} {{\text {Det}}}({{\text {Id}}}+ A \chi B) = {{\text {Det}}}\left( {{\text {Id}}}+ \left\langle D \right\rangle ^{-s} A \left\langle D \right\rangle ^{-s} \cdot \left\langle D \right\rangle ^s \chi B \left\langle D \right\rangle ^{s}\right) . \end{aligned}$$

    The trace-class norm controls the determinant difference—see e.g. [DZ18, Proposition B.26]. Hence,

    $$\begin{aligned}&\big | {{\text {Det}}}({{\text {Id}}}+A \chi B) - {{\text {Det}}}({{\text {Id}}}+A'\chi B) \big | \le C \left\| \left\langle D \right\rangle ^{-s} (A-A') \left\langle D \right\rangle ^{-s} \cdot \left\langle D \right\rangle ^s \chi B \left\langle D \right\rangle ^{s} \right\| _{{\mathcal {L}}(L^2)}\\&\quad \le C \left\| \left\langle D \right\rangle ^{-s} (A-A') \left\langle D \right\rangle ^{-s} \right\| _{L^2} \cdot \left\| \left\langle D \right\rangle ^s \chi B \left\langle D \right\rangle ^{s} \right\| _{{\mathcal {L}}(L^2)}. \end{aligned}$$

    The space of bounded operators from \(L^2\) to \(H^{2-2s}\) with range in function with fixed compact support embeds continuously in the space of trace-class operators on \(L^2\), because \(2-2s > 1\)—see e.g. [DZ18, Equation (B.3.9)]. Therefore,

    $$\begin{aligned} \big | {{\text {Det}}}({{\text {Id}}}+A \chi B) - {{\text {Det}}}({{\text {Id}}}+A'\chi B) \big | \le C \left\| \left\langle D \right\rangle ^{-s} (A-A') \left\langle D \right\rangle ^{-s} \right\| _{L^2} \cdot \left\| \chi B \right\| _{H^{-s} \rightarrow H^{2-s}}. \end{aligned}$$

This completes the proof of Lemma 6.1. \(\quad \square \)

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Drouot, A., Fefferman, C.L. & Weinstein, M.I. Defect Modes for Dislocated Periodic Media. Commun. Math. Phys. 377, 1637–1680 (2020).

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