Quantization of Conductance in Quasi-periodic Quantum Wires

Abstract

We study charge transport in the Peierls–Harper model with a quasi-periodic cosine potential. We compute the Landauer-type conductance of the wire. Our numerical results show the following: (i) When the Fermi energy lies in the absolutely continuous spectrum that is realized in the regime of the weak coupling, the conductance is quantized to the universal conductance. (ii) For the regime of localization that is realized for the strong coupling, the conductance is always vanishing irrespective of the value of the Fermi energy. Unfortunately, we cannot make a definite conclusion about the case with the critical coupling. We also compute the conductance of the Thue–Morse model. Although the potential of the model is not quasi-periodic, the energy spectrum is known to be a Cantor set with zero Lebesgue measure. Our numerical results for the Thue–Morse model also show the quantization of the conductance at many locations of the Fermi energy, except for the trivial localization regime. Besides, for the rest of the values of the Fermi energy, the conductance shows a similar behavior to that of the Peierls–Harper model with the critical coupling.

Appendices

Appendix A: Derivation of of Linear Response Formula (2.4)

In this appendix, we recall the derivation of the expansion (2.4) for wavefunctions by following Kato [21].

Let $\Psi (s)$ be a solution of the Schrödinger equation (2.3). Then, one has

$$\begin{aligned} \frac{d}{ds}U_0(t,s)\Psi (s)= & {} \left[ \frac{d}{ds}U_0(t,s)\right] \Psi (s)+U_0(t,s)\frac{d}{ds}\Psi (s)\\= & {} iU_0(t,s)H_0\Psi (s)-iU_0(t,s)H(s)\Psi (s)\\= & {} -i\mu U_0(t,s)W(s)\Psi (s). \end{aligned}$$

Integrating over s from $-T$ to t, one obtains

$$\begin{aligned} \Psi (t)-U_0(t,-T)\Psi (-T)=-i\mu \int _{-T}^t ds\; U_0(t,s)W(s)\Psi (s). \end{aligned}$$

Therefore, we have

$$\begin{aligned} \Psi (t)=U_0(t,-T)\Psi (-T)+{\mathcal {O}}(\mu )=U_0(t,0)\Phi +{\mathcal {O}}(\mu ), \end{aligned}$$

where we have used the initial condition, $\Psi (-T)=U_0(-T,0)\Phi$. Substituting this into the above integrand, we obtain the desired result (2.4).

Appendix B: Proof of Proposition 2.1

In this appendix, we prove that the conductance $g_j$ of (2.8) is equal to the standard form, which is expressed in terms of the current–current correlation function as in (2.12) with (2.9), under the assumption (2.11).

Consider the summand in the right-hand side of (2.7). By integration by parts, one has

$$\begin{aligned}&\int _{-T}^0 ds\; e^{\eta s}\langle \Phi _k,[J_j,\chi _{[j,j+\ell ]}(s)]\Phi _k\rangle \nonumber \\&\quad =\left[ s e^{\eta s}\langle \Phi _k,[J_j,\chi _{[j,j+\ell ]}(s)]\Phi _k\rangle \right] _{-T}^0 -\int _{-T}^0 ds\; s\frac{d}{ds} e^{\eta s}\langle \Phi _k,[J_j,\chi _{[j,j+\ell ]}(s)]\Phi _k\rangle \nonumber \\&\quad =T e^{-\eta T}\langle \Phi _k,[J_j,\chi _{[j,j+\ell ]}(-T)]\Phi _k\rangle -\int _{-T}^0 ds\; \eta s e^{\eta s}\langle \Phi _k,[J_j,\chi _{[j,j+\ell ]}(s)]\Phi _k\rangle \nonumber \\&\qquad -\int _{-T}^0 ds\; s e^{\eta s}\langle \Phi _k,[J_j,\frac{d}{ds}\chi _{[j,j+\ell ]}(s)]\Phi _k\rangle . \end{aligned}$$

(B.1)

In the following three subsections, we will estimate the three terms in the right-hand side.

1.1 Appendix B.1: Estimating the First Term in the Right-Hand Side of (B.1)

To begin with, we note that

$$\begin{aligned} \frac{d}{ds}\chi _{[j,j+\ell ]}(s)=e^{isH_0}(J_j-J_{j+\ell +1})e^{-isH_0}=J_j(s)-J_{j+\ell +1}(s) \end{aligned}$$

(B.2)

which is derived from the definition of the current operator $J_j$. Integrating over s, one has

$$\begin{aligned} \chi _{[j,j+\ell ]}(s)=\chi _{[j,j+\ell ]}(0)+\int _0^s d\tau \; [J_j(\tau )-J_{j+\ell +1}(\tau )]. \end{aligned}$$

(B.3)

Substituting this into the first term in the right-hand side of (B.1), we have

$$\begin{aligned}&T e^{-\eta T}\langle \Phi _k,[J_j,\chi _{[j,j+\ell ]}(-T)]\Phi _k\rangle \nonumber \\&\quad =T e^{-\eta T}\langle \Phi _k,[J_j,\chi _{[j,j+\ell ]}]\Phi _k\rangle +T e^{-\eta T}\int _0^{-T}d\tau \; \langle \Phi _k,[J_j,J_{j}(\tau )]\Phi _k\rangle \nonumber \\&\qquad -T e^{-\eta T}\int _0^{-T}d\tau \; \langle \Phi _k,[J_j,J_{j+\ell +1}(\tau )]\Phi _k\rangle . \end{aligned}$$

(B.4)

Since the commutator $[J_j,\chi _{[j,j+\ell ]}]$ is local, the contribution of the first term in the right-hand side is vanishing in the limit $T\rightarrow \infty$. By using Schwarz inequality, one can show that the second and third terms are also vanishing in the limit because the current operator $J_j$ is local. Thus, the contribution of the first term in the right-hand side of (B.1) is vanishing.

1.2 Appendix B.2: Estimating the Second Term in the Right-Hand Side of (B.1)

To begin with, we write

$$\begin{aligned} f_j(s):=\lim _{\ell \nearrow \infty }\lim _{L\nearrow \infty }i\sum _{k:E_k\le E_{\mathrm{F}}} \langle \Phi _k,[J_j,\chi _{[j,j+\ell ]}(s)\Phi _k\rangle . \end{aligned}$$

This limit exists thanks to (B.3). Then, one has

$$\begin{aligned} g_j(\eta ,T)=\int _{-T}^0 ds\; e^{\eta s}f_j(s) \end{aligned}$$

from (2.7) and (2.10).

Let $\Delta \alpha$ be a real number whose absolute value $|\Delta \alpha |$ is small. Note that

$$\begin{aligned} g_j((1+\Delta \alpha )\eta ,T)-g_j(\eta ,T) =\int _{-T}^0 ds\;(e^{\Delta \alpha \eta s}-1)e^{\eta s}f_j(s). \end{aligned}$$

Further, we introduce

$$\begin{aligned} {\tilde{g}}_j(\eta ,T):=\int _{-T}^0ds\; \eta se^{\eta s}f_j(s), \end{aligned}$$

in order to estimate the second term in the right-hand side of (B.1). This quantity is nothing but the contribution corresponding to the second term. From these two equations, one has

$$\begin{aligned} \frac{g_j((1+\Delta \alpha )\eta ,T)-g_j(\eta ,T)}{\Delta \alpha }-{\tilde{g}}_j(\eta ,T) =\int _{-T}^0 ds\; \sum _{n=2}^\infty \frac{1}{n!}(\Delta \alpha )^{n-1} (\eta s)^ne^{\eta s}f_j(s).\qquad \end{aligned}$$

(B.5)

In order to estimate this right-hand side, we write

$$\begin{aligned} F_j(s):=-\int _s^0 d\tau \; e^{\eta \tau /2}f_j(\tau ) \end{aligned}$$

for short. Clearly, $F_j(s)=-g_j(\eta /2,-s)$, and hence this is uniformly bounded with respect to $\eta$ and s from the assumption (2.11) of Proposition 2.1. By integration by parts, the right-hand side of (B.5) can be calculated as

$$\begin{aligned}&\sum _{n=2}^\infty \frac{1}{n!}\int _{-T}^0 ds\; (\Delta \alpha )^{n-1} (\eta s)^n e^{\eta s}f_j(s)\\&\quad =\sum _{n=2}^\infty \frac{1}{n!}\left[ (\Delta \alpha )^{n-1} (\eta s)^n e^{\eta s/2}F_j(s)\right] _{-T}^0\\&\qquad -\sum _{n=2}^\infty \frac{1}{n!}\int _{-T}^0 ds\; \left[ (\Delta \alpha )^{n-1}n\eta (\eta s)^{n-1}+(\Delta \alpha )^{n-1}(\eta s)^{n}\cdot \frac{\eta }{2}\right] e^{\eta s/2}F_j(s). \end{aligned}$$

Since the first sum in the right-hand side is vanishing in the limit $T\nearrow \infty$, one has

$$\begin{aligned}&\left| \int _{-\infty }^0 ds\; \sum _{n=2}^\infty \frac{1}{n!}(\Delta \alpha )^{n-1} (\eta s)^n e^{\eta s}f_j(s)\right| \\&\quad \le \int _{-\infty }^0 ds\cdot \eta \; \left[ \sum _{n=2}^\infty \frac{1}{(n-1)!}\big |\Delta \alpha \big |^{n-1}\big |\eta s\big |^{n-1}+\frac{1}{2}\sum _{n=2}^\infty \frac{1}{n!} \big |\Delta \alpha \big |^{n-1}\big |\eta s\big |^{n}\right] e^{\eta s/2}\big |F_j(s)\big |\\&\quad \le \big |\Delta \alpha \big | \int _{-\infty }^0 ds\cdot \eta \;\left[ \big |\eta s\big |e^{|\Delta \alpha \eta s|} +\frac{1}{2}|\eta s|^2 e^{|\Delta \alpha \eta s|}\right] e^{\eta s/2}|F_j(s)|\\&\quad \le {\mathcal {C}}_1|\Delta \alpha |\int _{-\infty }^0 ds'\;\left[ |s'|+\frac{1}{2}|s'|^2\right] e^{|\Delta \alpha ||s'|}e^{s'/2} \le \mathrm{Const.}|\Delta \alpha |, \end{aligned}$$

where we have used the assumption (2.11) with $F_j(s)=-g_j(\eta /2,-s)$. Combining this with (B.5), we obtain

$$\begin{aligned} \left| \lim _{T\nearrow \infty }\left[ \frac{g_j((1+\Delta \alpha )\eta ,T)-g_j(\eta ,T)}{\Delta \alpha }-{\tilde{g}}_j(\eta ,T)\right] \right| \le \mathrm{Const.}|\Delta \alpha |. \end{aligned}$$

We recall one of the two assumptions of Proposition 2.1 which requires the existence of $\lim _{\eta \searrow 0}\lim _{T\nearrow \infty }g_j(\eta ,T)$. From this assumption, one has

$$\begin{aligned} \lim _{\eta \searrow 0}\lim _{T\nearrow \infty }g_j((1+\Delta \alpha )\eta ,T)=\lim _{\eta \searrow 0}\lim _{T\nearrow \infty }g_j(\eta ,T). \end{aligned}$$

Combining this with the above inequality, we have

$$\begin{aligned} \left| \lim _{\eta \searrow 0}\lim _{T\nearrow \infty }{\tilde{g}}_j(\eta ,T)\right| \le \mathrm{Const.}|\Delta \alpha |. \end{aligned}$$

Since this holds for any small $|\Delta \alpha |$, we obtain the desired result,

$$\begin{aligned} \lim _{\eta \searrow 0}\lim _{T\nearrow \infty }{\tilde{g}}_j(\eta ,T)=0. \end{aligned}$$

1.3 Appendix B.3: Estimating the Third Term in the Right-Hand Side of (B.1)

Using the identity (B.2), the third term in the right-hand side of (B.1) is written as

$$\begin{aligned}&-\int _{-T}^0 ds\; s e^{\eta s}\langle \Phi _k,[J_j,\frac{d}{ds}\chi _{[j,j+\ell ]}(s)]\Phi _k\rangle \nonumber \\&\quad =\,-\int _{-T}^0 ds\; s e^{\eta s}\langle \Phi _k,[J_j,J_{j,}(s)]\Phi _k\rangle +\int _{-T}^0 ds\; s e^{\eta s}\langle \Phi _k,[J_j,J_{j+\ell +1}(s)]\Phi _k\rangle .\qquad \qquad \end{aligned}$$

(B.6)

The first term in the right-hand side is nothing but the desired contribution of the current–current correlation. Using Lieb–Robinson bound [47,48,49,50], one can show that the contribution of the second term is vanishing in the limit $\ell \rightarrow \infty$.

Appendix C: Proof of Theorem 2.2

In this appendix, we prove that the conductance is quantized to the universal conductance for the periodic potential v under the assumptions in Theorem 2.2.

From the assumptions, we can choose the length L of the chain to be $L=pM$ with a large positive integer M, where p is the period of the periodic potential, and we impose the periodic boundary condition for the unperturbed Hamiltonian $H_0$. Since the validity of the relation (2.12) in the present case is proved in “Appendix D”, it is enough to calculate the right-hand side of (2.12).

We write $\Phi _{m,k}$ for the eigenvectors of the unperturbed Hamiltonian $H_0$ with the energy eigenvalue $E_{m,k}$, where m and k are, respectively, the band index and the wavenumber. Then, one has

$$\begin{aligned} {\hat{g}}_j(\eta ,L)=\sum _{m,k:E_{m,k}\le E_\mathrm{F}}\; \sum _{m',k':E_{m',k'}>E_{\mathrm{F}}} \frac{4\eta (E_{m',k'}-E_{m,k})}{[(E_{m',k'}-E_{m,k})^2+\eta ^2]^2}|\langle \Phi _{m,k},J_j\Phi _{m',k'}\rangle |^2\qquad \end{aligned}$$

(C.1)

from the definition (2.9) of ${\hat{g}}_j(\eta ,L)$.

To begin with, we note that $|\langle \Phi _{m,k},J_j\Phi _{m',k'}\rangle |={\mathcal {O}}(M^{-1})$ because the current operator $J_j$ is local. Let $\delta$ be a small positive number. Then, the contributions of the double sums in (C.1) which satisfy $E_{m',k'}-E_{m,k}\ge \delta$ are vanishing in the double limit $\eta \searrow 0$ and $M\nearrow \infty$. Actually, one has

$$\begin{aligned} \frac{\eta |E_{m',k'}-E_{m,k}|}{[(E_{m',k'}-E_{m,k})^2+\eta ^2]^2}\le \frac{\eta |E_{m',k'}-E_{m,k}|}{|E_{m',k'}-E_{m,k}|^4} \le \frac{\eta }{\delta ^3}. \end{aligned}$$

Combining this with the above $|\langle \Phi _{m,k},J_j\Phi _{m',k'}\rangle |={\mathcal {O}}(M^{-1})$, one can easily show that the corresponding contributions are vanishing. Thus, it is enough to treat the energies near the Fermi energy $E_{\mathrm{F}}$. Without loss of generality, it is sufficient to deal with the following four cases:

• (i) $k=k_{\mathrm{F}}-\kappa , \quad k'=k_{\mathrm{F}}+\kappa '$

• (ii) $k=-k_{\mathrm{F}}+\kappa , \quad k'=-k_{\mathrm{F}}-\kappa '$

• (iii) $k=k_{\mathrm{F}}-\kappa , \quad k'=-k_{\mathrm{F}}-\kappa '$

• (iv) $k=-k_{\mathrm{F}}+\kappa , \quad k'=k_{\mathrm{F}}+\kappa '$,

where $k_{\mathrm{F}}$ is the Fermi wavenumber, and $\kappa$ and $\kappa '$ are a positive small variable.

Consider first Case (i). The energies near the Fermi energy are given by

$$\begin{aligned} E_{m,k}\sim E_{\mathrm{F}}+\frac{d}{dk}E_{m,k}(k_{\mathrm{F}})\cdot (-\kappa )=E_{\mathrm{F}}-v_{\mathrm{F}}\kappa \end{aligned}$$

and

$$\begin{aligned} E_{m,k'}\sim E_{\mathrm{F}}+\frac{d}{dk}E_{m,k}(k_{\mathrm{F}})\cdot \kappa '=E_{\mathrm{F}}+v_{\mathrm{F}}\kappa '. \end{aligned}$$

Therefore, the difference is

$$\begin{aligned} E_{m,k'}-E_{m,k}\sim v_{\mathrm{F}}(\kappa '+\kappa ). \end{aligned}$$

Because of the translational invariance, the eigenvectors can be written as

$$\begin{aligned} \Phi _{m,k}(n)=\frac{1}{\sqrt{M}}e^{ikr}u_{m,k}(q) \end{aligned}$$

in Bloch form, where we have written $n=pr+q$ with $r=0,1,2,\ldots ,M-1$ and $q=1,2,\ldots ,p$. The matrix elements in the right-hand side of (C.1) are estimated as

$$\begin{aligned} \langle \Phi _{m,k},J_j\Phi _{m,k'}\rangle \sim \langle \Phi _{m,k_\mathrm{F}},J_j\Phi _{m,k_{\mathrm{F}}}\rangle= & {} \frac{1}{M}\sum _{r=0}^{M-1}\langle \Phi _{m,k_{\mathrm{F}}},J_{j+pr}\Phi _{m,k_{\mathrm{F}}}\rangle \\\sim & {} \frac{1}{M}\langle u_{m,k_{\mathrm{F}}},{\tilde{J}}(k_{\mathrm{F}})u_{m,k_{\mathrm{F}}}\rangle \\= & {} \frac{1}{M}\langle u_{m,k_{\mathrm{F}}},\frac{\partial {\tilde{H}}_0(k_{\mathrm{F}})}{\partial k}u_{m,k_{\mathrm{F}}}\rangle \\= & {} \frac{1}{M}\frac{d E_{m,k}}{d k}(k_{\mathrm{F}})=\frac{1}{M}v_{\mathrm{F}}, \end{aligned}$$

where ${\tilde{J}}$ and ${\tilde{H}}_0$ are, respectively, the Fourier transform. Substituting these into the right-hand side of (C.1), the corresponding contribution is written

$$\begin{aligned} g_+(\eta ):=\frac{v_{\mathrm{F}}^2}{\pi ^2}\int _0^\delta d\kappa \int _0^\delta d\kappa '\; \frac{v_\mathrm{F}(\kappa +\kappa ')\eta }{[v_{\mathrm{F}}^2(\kappa +\kappa ')^2+\eta ^2]^2} \end{aligned}$$

in the limit $M\nearrow \infty$. Changing the variables in the integrals as ${\tilde{\kappa }}=v_{\mathrm{F}}\kappa /\eta$ and ${\tilde{\kappa }}'=v_{\mathrm{F}}\kappa '/\eta$, one has

$$\begin{aligned} g_+(0)=\frac{1}{\pi ^2}\int _0^\infty d{\tilde{\kappa }} \int _0^\infty d\tilde{\kappa '}\; \frac{{\tilde{\kappa }}+{\tilde{\kappa }}'}{[({\tilde{\kappa }}+{\tilde{\kappa }}')^2+1]^2} \end{aligned}$$

in the limit $\eta \searrow 0$. Further, by setting $w={\tilde{\kappa }}+{\tilde{\kappa }}'$, we obtain

$$\begin{aligned} g_+(0)=\frac{1}{\pi ^2}\int _0^\infty dw\; \frac{w^2}{(w^2+1)^2}=\frac{1}{4\pi }. \end{aligned}$$

Although this is only the leading term, all of the higher corrections can be proved to be vanishing in the limit $\eta \searrow 0$ by relying on the fact that the cutoff $\delta$ can be chosen to be an arbitrary small positive number.

Similarly, one can deal with Case (ii). The resulting contribution for the conductance is given by

$$\begin{aligned} g_-(0)=\frac{1}{4\pi }. \end{aligned}$$

In order to treat Cases (iii) and (iv), we note that

$$\begin{aligned} \langle \Phi _{m,k_{\mathrm{F}}},J_j\Phi _{m,-k_{\mathrm{F}}}\rangle = (-i)\left[ \Phi _{m,k_{\mathrm{F}}}^*(j-1)\Phi _{m,-k_\mathrm{F}}(j)-\Phi _{m,k_{\mathrm{F}}}^*(j)\Phi _{m,-k_{\mathrm{F}}}(j-1)\right] , \end{aligned}$$

which is obtained from the expression (2.5) of the current operator $J_j$. Since the Hamiltonian $H_0$ is a real symmetric matrix, one has $\Phi _{m,k_{\mathrm{F}}}^*(n)=\Phi _{m,-k_{\mathrm{F}}}(n)$ for all the sites n. Substituting this into the above right-hand side yields the vanishing of the matrix element. Therefore, the corresponding contributions for the conductance is vanishing, too, in the same way.

Adding all the contributions, we obtain the desired result,

$$\begin{aligned} \lim _{\eta \searrow 0}\lim _{M\nearrow \infty }{\hat{g}}_j(\eta ,N)=g_+(0)+g_-(0)=\frac{1}{2\pi }. \end{aligned}$$

Appendix D: Proof of (2.12) in the Case of the Periodic Potentials

In this appendix, we prove that the equality (2.12) is valid in the case of the periodic potentials without relying on the assumptions of Proposition 2.1. For this purpose, it is sufficient to show that the contribution of the second term in the right-hand side of (B.1) is vanishing because the contribution of the first term in the right-hand side of (B.1) can be proved to be vanishing without relying on the assumptions of Proposition 2.1, as shown in “Appendix B.1”.

To begin with, we note that

$$\begin{aligned}&\int _{-T}^0 ds\; s e^{\eta s}e^{i(E_{m',k'}-E_{m,k})s} \nonumber \\&\quad =\left[ \frac{iT}{E_{m,k}-E_{m',k'}+i\eta }-\frac{1}{(E_{m,k}-E_{m',k'}+i\eta )^2} \right] \exp [-\eta T-i(E_{m',k'}-E_{m,k})T] \nonumber \\&\quad \quad +\frac{1}{(E_{m,k}-E_{m',k'}+i\eta )^2}. \end{aligned}$$

(D.1)

Since the first term in the right-hand side is vanishing in the limit $T\nearrow \infty$, we will treat only the second term in the following. The corresponding contribution is written

$$\begin{aligned}&\sum _{m,k:E_{m,k}\le E_{\mathrm{F}}}\int _{-\infty }^0 ds\; \eta s e^{\eta s}\langle \Phi _{m,k},[J_j,\chi _{[j,j+\ell ]}(s)]\Phi _{m,k}\rangle \\&\quad =\eta \sum _{m,k:E_{m,k}\le E_{\mathrm{F}}}\; \sum _{m',k':E_{m',k'}>E_{\mathrm{F}}}\Biggl \{ \langle \Phi _{m,k},J_j\Phi _{m',k'}\rangle \langle \Phi _{m',k'},\chi _{[j,j+\ell ]}\Phi _{m,k} \rangle \frac{1}{[E_{m,k}-E_{m',k'}+i\eta ]^2}\\&\qquad -\langle \Phi _{m,k},\chi _{[j,j+\ell ]}\Phi _{m',k'}\rangle \langle \Phi _{m',k'},J_j\Phi _{m,k} \rangle \frac{1}{[E_{m,k}-E_{m',k'}-i\eta ]^2}\Biggr \}. \end{aligned}$$

The matrix element of $\chi _{[j,j+\ell ]}$ in the right-hand side can be calculated as

$$\begin{aligned} \langle \Phi _{m',k'},\chi _{[j,j+\ell ]}\Phi _{m,k} \rangle =\frac{1}{M}e^{ij(k-k')}\frac{e^{i({\hat{\ell }}+1)(k-k')}-1}{e^{i(k-k')}-1}\langle u_{m',k'},u_{m,k}\rangle , \end{aligned}$$

where we have taken $\ell =p{\hat{\ell }}$ with a large positive integer ${\hat{\ell }}$.

In the same way as in “Appendix C”, it is sufficient to treat Cases (i) and (ii). In the following, we consider only Case (i) because both of Cases can be treated in the same way. In the former case, the above matrix element can be written as

$$\begin{aligned} \langle \Phi _{m,k'},\chi _{[j,j+\ell ]}\Phi _{m,k} \rangle \sim \frac{i}{M}\frac{e^{-i({\hat{\ell }}+1)(\kappa +\kappa ')}-1}{\kappa +\kappa '}. \end{aligned}$$

Therefore, the corresponding contribution is

$$\begin{aligned} {\tilde{g}}_j(\eta ,{\hat{\ell }})=\frac{v_{\mathrm{F}}}{4\pi ^2}\int _0^\delta d\kappa \int _0^\delta d\kappa ' \frac{\eta }{\kappa +\kappa '}\left\{ \frac{e^{-i({\hat{\ell }}+1)(\kappa +\kappa ')}-1}{[v_\mathrm{F}(\kappa +\kappa ')-i\eta ]^2} +\frac{e^{i({\hat{\ell }}+1)(\kappa +\kappa ')}-1}{[v_\mathrm{F}(\kappa +\kappa ')+i\eta ]^2}\right\} . \end{aligned}$$

In the double limit ${\hat{\ell }}\nearrow \infty$ and $\eta \searrow 0$, we have

$$\begin{aligned} \lim _{\eta \searrow 0}\lim _{{\hat{\ell }}\nearrow \infty } {\tilde{g}}_j(\eta ,{\hat{\ell }})=-\frac{1}{4\pi ^2}\int _0^\infty dw\left\{ \frac{1}{(w-i)^2}+\frac{1}{(w+i)^2}\right\} =0, \end{aligned}$$

where we have used Riemann–Lebesgue argument for ${\hat{\ell }}\nearrow \infty$.

Finally, we remark the following: We can expect that the uniform bound (2.11) in Proposition 2.1 holds in the case of periodic potentials because the oscillatory integrals which appear in the contributions from the first term in the right-hand side of (B.1) and the first term in the right-hand side of (D.1) cancel the factor T which appears in their contributions for a large T. In fact, one can prove this statement under certain assumptions about the spectrum of the Hamiltonian $H_0$ and its Fourier transform.