Numerical quadrature in the Brillouin zone for periodic Schrödinger operators

Abstract

As a consequence of Bloch’s theorem, the numerical computation of the fermionic ground state density matrices and energies of periodic Schrödinger operators involves integrals over the Brillouin zone. These integrals are difficult to compute numerically in metals due to discontinuities in the integrand. We perform an error analysis of several widely-used quadrature rules and smearing methods for Brillouin zone integration. We precisely identify the assumptions implicit in these methods and rigorously prove error bounds. Numerical results for two-dimensional periodic systems are also provided. Our results shed light on the properties of these numerical schemes, and provide guidance as to the appropriate choice of numerical parameters.

Appendix: Analytic properties of the integrand

The goal of this appendix is to study the analytic properties of the functions $F_{\varepsilon ,\sigma }^{{\mathcal {N}}}$, $F_{\varepsilon ,\sigma }^{E}$ and $F_{\varepsilon ,\sigma }^{\rho , W}$ defined respectively by

$$\begin{aligned} F_{\varepsilon ,\sigma }^{{\mathcal {N}}}(\mathbf{k})&:= \mathrm{Tr}_{L^2_\mathrm {per}} \left[ f^{\sigma } \left( H_\mathbf{k}- \varepsilon \right) \right] , \\ F_{\varepsilon ,\sigma }^{E}(\mathbf{k})&:= \mathrm{Tr}_{L^2_\mathrm {per}} \left[ H_\mathbf{k}f^{\sigma } \left( H_\mathbf{k}- \varepsilon \right) \right] , \\ F_{\varepsilon ,\sigma }^{\rho , W}(\mathbf{k})&:= \mathrm{Tr}_{L^2_\mathrm {per}} \left[ W f^{\sigma } \left( H_\mathbf{k}- \varepsilon \right) \right] . \end{aligned}$$

We prove the following two results. The first one studies the analytic properties of these functions near the real line.

Lemma A.1

(Analyticity near the real line) Let $f^1$ be any of the smearing functions (5.6)–(5.6$'''$). Then, there exist $C \in {\mathbb R}_+$ and ${Y} > 0$ such that, for all $\varepsilon \in [\varepsilon _F - \delta _0, \varepsilon _F + \delta _0]$ and all $0 < \sigma \leqslant \sigma _0$, the maps $F_{\varepsilon ,\sigma }^{X}$ admits an analytic continuation on $S_{\sigma Y}$, and it holds that

$$\begin{aligned} \sup _{\mathbf{z}\in S_{\sigma Y}} \left| F_{\varepsilon ,\sigma }^{{\mathcal {N}}}(\mathbf{z}) \right| + \sup _{\mathbf{z}\in S_{\sigma Y}} \left| F_{\varepsilon ,\sigma }^{E}(\mathbf{z}) \right| + \sup _{{\mathbf{z}\in S_{\sigma Y}, \; \Vert W\Vert _{H^{-(s+2)}_\mathrm {per}=1}}} \left| F_{\varepsilon ,\sigma }^{\rho , W}(\mathbf{z}) \right| \leqslant C \sigma ^{-1}. \end{aligned}$$

Our second result studies the analytic properties on the entire complex plane.

Lemma A.2

(Analyticity on ${\mathbb C}$) Let $f^1$ be one of the Gaussian-type smearing functions (5.6$'$)–(5.6$'''$). Then, the maps $F_{\varepsilon ,\sigma }^{X}$ are entire, and there exists $C \in {\mathbb R}_+$ and $\eta > 0$ such that, for all $Y \geqslant 1$, all $\varepsilon \in [\varepsilon _F - \delta _0, \varepsilon _F + \delta _0]$ and all $0 < \sigma \leqslant \sigma _0$,

$$\begin{aligned} \sup _{\mathbf{z}\in S_{\sigma Y}} \left| F_{\varepsilon ,\sigma }^{{\mathcal {N}}}(\mathbf{z}) \right| + \sup _{\mathbf{z}\in S_{\sigma Y}} \left| F_{\varepsilon ,\sigma }^{E}(\mathbf{z}) \right| + \sup _{{\mathbf{z}\in S_{\sigma Y}, \;\Vert W\Vert _{H^{-(s+2)}_\mathrm {per}} = 1}} \left| F_{\varepsilon ,\sigma }^{\rho , W}(\mathbf{z}) \right| \leqslant C {\mathrm {e}}^{\eta \frac{Y^4}{\sigma ^{2}}}. \end{aligned}$$

Let us first highlight the idea of the proofs of these lemmas. First, we will obtain bounds that only depend on $\Vert V_{\mathrm {per}}\Vert _{L^\infty }$, so that we can absorb $\varepsilon \in (\varepsilon _{F}-\delta _{0},\varepsilon _{F}+\delta _{0})$ in $V_{\mathrm {per}}$. Without loss of generality, we take $\varepsilon = 0$ and drop the subscript $\varepsilon$. In addition, to shorten the presentation, we only do the proof for the energy. In the following, $F_{\sigma }$ denotes $F_{\varepsilon = 0, \sigma }^{E}$.

We wish to study the analytic properties of the function

$$\begin{aligned} F_{\sigma }(\mathbf{k}) = \mathrm{Tr}_{L^2_\mathrm {per}} \left[ H_{\mathbf{k}} f^1 \left( H_\mathbf{k}/\sigma \right) \right] , \end{aligned}$$

where $f^1$ is one of the smearing functions (5.6)–(5.6$'''$). Formally, $H_{\mathbf{k}}$ admits an analytic continuation to the whole complex plane: when $\mathbf{z}= \mathbf{k}+{\mathrm {i}}\mathbf{y}\in {\mathbb C}^{d}$, we set

$$\begin{aligned} H_{\mathbf{z}} = - \frac{1}{2} \Delta _{\mathbf{z}} + V_{\mathrm {per}}, \end{aligned}$$

where

$$\begin{aligned} - \Delta _{\mathbf{z}}&= (-{\mathrm {i}}\nabla + \mathbf{z})^{2} = (-{\mathrm {i}}\nabla + \mathbf{k})^{2} + 2 {\mathrm {i}}\mathbf{y}\cdot (-{\mathrm {i}}\nabla + \mathbf{k}) - {|\mathbf{y}|^{2}}. \end{aligned}$$

(A.1)

The operator $H_{\mathbf{z}}$ is not self-adjoint (it is not even a normal operator), so it is difficult to compute the analytical extension $F_\sigma (\mathbf{z})$ of $F_{\sigma }(\mathbf{k})$. We will use a representation in terms of contour integrals. Formally, it holds that

$$\begin{aligned} \mathrm{Tr}_{L^2_\mathrm {per}} \left[ H_{\mathbf{z}} f^{1} \left( H_\mathbf{z}/\sigma \right) \right]&= \mathrm{Tr}_{L^{2}_{\mathrm {per}}} \left[ \dfrac{1}{2 \pi {\mathrm {i}}} \oint _{\mathscr {C}} \lambda f^{1}(\lambda /\sigma ) \left( \dfrac{1}{\lambda - H_\mathbf{z}} \right) {\mathrm {d}}\lambda \right] \\&= \dfrac{1}{2 \pi {\mathrm {i}}} \oint _{\mathscr {C}} \lambda f^{1}(\lambda /\sigma ) \mathrm{Tr}_{L^2_\mathrm {per}} \left[ \dfrac{1}{\lambda - H_\mathbf{z}} \right] {\mathrm {d}}\lambda \end{aligned}$$

for some (infinite) contour $\mathscr {C}$ enclosing the spectrum of $H_{\mathbf{z}}$. Unfortunately, we cannot commute the trace and the integral in the last line whenever the dimension d is greater than 1. The reason is that the operator $(\lambda - H_\mathbf{z})^{-1}$ is not trace-class when $d \geqslant 2$ (see (2.2)). Instead, we consider the contour integral¹

$$\begin{aligned} G_{\sigma }(\mathbf{z}) = \dfrac{1}{2 \pi {\mathrm {i}}} \int _{\mathscr {C}} \left[ \lambda f^{1}(\lambda /\sigma )(\lambda + \Sigma )^2 \right] \mathrm{Tr}_{L^2_\mathrm {per}} \left( \dfrac{1}{(H_\mathbf{z}+ \Sigma )^2} \dfrac{1}{\lambda - H_\mathbf{z}} \right) {\mathrm {d}}\lambda , \end{aligned}$$

(A.2)

where $\Sigma \in {\mathbb R}$ is a well-chosen shift, and prove that $G_{\sigma }$ is an analytic continuation of $F_{\sigma }$.

We prove Lemmas A.1 and A.2 in the following two sections. In both cases we prove that there exists appropriate contours such that $G_{\sigma } = F_\sigma$ using a perturbation argument. When $\mathbf{z}= \mathbf{k}+ {\mathrm {i}}\mathbf{y}\in {\mathbb C}^3$ with $| \mathbf{y}|$ small, we see $H_\mathbf{z}$ as a perturbation of $H_\mathbf{k}$, while when $\mathbf{y}$ is large, we see it as a perturbation of the free operator $-\frac{1}{2} \Delta _\mathbf{z}$.

1.1 Proof of Lemma A.1

We introduce $\mathscr {C}$ the parabolic contour in the complex plane defined by

$$\begin{aligned} \mathscr {C}:= \left\{ \lambda \in {\mathbb C}, | \mathrm{Im}\ \lambda |^2 = \sigma ^{2}\left( 1 + \frac{\mathrm{Re}\ \lambda }{\Sigma }\right) \right\} , \quad \text {where we set} \quad \Sigma := \Vert V\Vert _{L^\infty }+1. \end{aligned}$$

(A.3)

It is the (unique) parabola that passes through the points $-\Sigma$ and $\pm {\mathrm {i}}\sigma$. In particular, it does not encounter the poles of the Fermi–Dirac function at $(2{\mathbb Z}+ 1)\pi \sigma {\mathrm {i}}$. Let us prove that $\mathscr {C}$ encloses the spectrum of $H_{\mathbf{k}+ {\mathrm {i}}\mathbf{y}}$ for $\mathbf{y}$ small enough (see Fig. 10).

Fig. 10

Spectrum of the operator $H_{\mathbf{z}}$ for $\mathbf{z}= \mathbf{k}+ {\mathrm {i}}\mathbf{y}$, and the contour $\mathscr {C}$ that encloses it, while avoiding the poles of the Fermi–Dirac function at $(2{\mathbb Z}+1)\sigma {\mathrm {i}}$

Lemma A.3

There exist $C \in {\mathbb R}^+$ and $Y > 0$ such that, for all $0<\sigma \leqslant \sigma _{0}$, all $\mathbf{z}= \mathbf{k}+ {\mathrm {i}}\mathbf{y}\in S_{\sigma Y}$, and all $\lambda \in \mathscr {C}$, we have the following estimates:

$$\begin{aligned} \Vert (H_{\mathbf{z}}-\lambda )^{-1}\Vert _{\mathscr {B}}&\leqslant C \sigma ^{-1},\end{aligned}$$

(A.4)

$$\begin{aligned} \Vert (H_{\mathbf{z}}+\Sigma )^{-1}\Vert _{{\mathfrak {S}}_{2}}&\leqslant C,\end{aligned}$$

(A.5)

$$\begin{aligned} \mathrm{Tr}_{L^2_\mathrm {per}} \left( \dfrac{1}{(H_\mathbf{z}+ \Sigma )^2} \dfrac{1}{\lambda - H_\mathbf{z}} \right)&\leqslant C \sigma ^{-1},\end{aligned}$$

(A.6)

$$\begin{aligned} \partial _{\mathbf{z}} \mathrm{Tr}_{L^2_\mathrm {per}} \left( \dfrac{1}{(H_\mathbf{z}+ \Sigma )^2} \dfrac{1}{\lambda - H_\mathbf{z}} \right)&\leqslant C \sigma ^{-1}. \end{aligned}$$

(A.7)

Proof

From (A.1), it holds that

$$\begin{aligned} H_{\mathbf{z}} = \frac{1}{2} (-{\mathrm {i}}\nabla + \mathbf{k})^{2} + {\mathrm {i}}\mathbf{y}\cdot (-{\mathrm {i}}\nabla + \mathbf{k}) - \frac{1}{2}{|\mathbf{y}|^{2}} + V_\mathrm {per}= \left( H_{\mathbf{k}} - \tfrac{1}{2} |\mathbf{y}|^{2} \right) + {\mathrm {i}}\mathbf{y}\cdot (-{\mathrm {i}}\nabla + \mathbf{k}). \end{aligned}$$

For $Y \leqslant 1$, the spectrum of the self-adjoint operator $H_{\mathbf{k}} - \frac{1}{2} |\mathbf{y}|^{2}$ is contained in $[-\Sigma + \tfrac{1}{2}, +\infty )$, which is disjoint from $\mathscr {C}$. In particular, for $\lambda \in \mathscr {C}$, we have

$$\begin{aligned} \left( \lambda - H_\mathbf{z}\right) = \left( \lambda - (H_{\mathbf{k}} - \tfrac{1}{2} |\mathbf{y}|^{2}) \right) \left( 1 + {\mathrm {i}}\mathbf{y}\cdot \left( \lambda - (H_{\mathbf{k}} - \tfrac{1}{2} |\mathbf{y}|^{2}) \right) ^{-1} ({\mathrm {i}}\nabla + \mathbf{k}) \right) . \end{aligned}$$

(A.8)

Let us evaluate the norm of the normal operator $\left( \lambda - (H_{\mathbf{k}} - \tfrac{1}{2} |\mathbf{y}|^{2}) \right) ^{-1}$. From (A.3), we have

$$\begin{aligned} \left\| \left( \lambda - (H_{\mathbf{k}} - \tfrac{1}{2} |\mathbf{y}|^{2})\right) ^{-1}\right\| _{{\mathcal {B}}}&= {{\,\mathrm{dist}\,}}\left( \lambda , {{\,\mathrm{Spec}\,}}(H_{\mathbf{k}} - \tfrac{1}{2} |\mathbf{y}|^{2})\right) ^{-1} \\&\leqslant \left[ {(\mathrm{Im}\lambda )^{2} + (-\Sigma + \tfrac{1}{2} - \mathrm{Re}\lambda )_{+}^{2}} \right] ^{-1/2}\\&= \left[ {\sigma ^{2}(1+ \tfrac{\mathrm{Re}\lambda }{\Sigma }) + (-\Sigma + \tfrac{1}{2} - \mathrm{Re}\lambda )_{+}^{2}}\right] ^{-1/2} \leqslant C \sigma ^{-1}, \end{aligned}$$

where $x_{+} := \max (x,0)$ and where $C \in {\mathbb R}_{+}$ is a constant that depends only on $\Sigma$ and $\sigma _{0}$. The last inequality is obtained by optimizing over all $\mathrm{Re}\lambda \geqslant -\Sigma$.

From the fact that $H_{\mathbf{k}}$ is a bounded perturbation of $\tfrac{1}{2} (-{\mathrm {i}}\nabla + \mathbf{k})^{2}$, and using similar calculations, we easily get that there exists $C \in {\mathbb R}_+$ that depends only on $\Sigma$ and $\sigma _0$ such that

$$\begin{aligned} \Vert {\mathrm {i}}\mathbf{y}\cdot (\lambda - (H_{\mathbf{k}} - \tfrac{1}{2} |\mathbf{y}|^{2}))^{-1} \cdot (-{\mathrm {i}}\nabla + \mathbf{k})\Vert _{{\mathcal {B}}}\leqslant C \sigma ^{-1} | \mathbf{y}|. \end{aligned}$$

(A.9)

As a consequence, for ${Y} \leqslant 1/(2C)$ and $| \mathbf{y}| \leqslant \sigma Y$, the operator on the right parenthesis of (A.8) is invertible, and its inverse is bounded in norm by 2. Inverting (A.8) leads to (A.4).

Inequality (A.5) is proved in a similar way (notice that the operator $(\Sigma - (H_{\mathbf{k}} - \frac{1}{2} |\mathbf{y}|^{2}))^{-1}$ is Hilbert–Schmidt by (2.2)). Inequality (A.6) is a consequence of (A.4) and (A.5), together with the operator inequality $\left| \mathrm{Tr}(B^2A) \right| \leqslant \Vert B \Vert _{{\mathfrak {S}}_2}^2 \Vert A \Vert _{\mathscr {B}}$.

We finally prove (A.7). For all $\mu$ in the resolvent set of $H_{\mathbf{z}}$, we have

$$\begin{aligned} \partial _{\mathbf{z}} \left( \frac{1}{H_{\mathbf{z}} - \mu } \right)&= - \left( \frac{1}{H_{\mathbf{z}} - \mu } \right) \partial _{\mathbf{z}} H_{\mathbf{z}} \left( \frac{1}{H_{\mathbf{z}} - \mu } \right) = - \frac{1}{H_{\mathbf{z}} - \mu } (-{\mathrm {i}}\nabla + \mathbf{z}) \frac{1}{H_{\mathbf{z}} - \mu }. \end{aligned}$$

We then use similar arguments and the fact that the operator $(-{\mathrm {i}}\nabla + \mathbf{z}) \left( H_{\mathbf{z}} - \lambda \right) ^{-1}$ is bounded. $\square$

We can now prove the analyticity of $G_\sigma$ defined in (A.2).

Lemma A.4

There exist $C \in {\mathbb R}_+$ and $Y > 0$ such that, for all $0<\sigma \leqslant \sigma _{0}$, the function $G_{\sigma }$ is analytic on $S_{\sigma Y}$, and

$$\begin{aligned} \sup _{\mathbf{z}\in S_{\sigma Y }} |G_{\sigma }(\mathbf{z})| \leqslant C \sigma ^{-1}. \end{aligned}$$

Proof

From the previous Lemma A.3, we already see that $G_\sigma$ is analytic. Let us prove the bound. It holds that

$$\begin{aligned} |G_{\sigma }(\mathbf{z})|&\leqslant C \sigma ^{-1} \int _{\mathscr {C}} |f^{1}(\sigma ^{-1} \lambda ) \lambda (\lambda + \Sigma )^2| |{\mathrm {d}}\lambda |. \end{aligned}$$

To evaluate the last integral, we parametrize the contour $\mathscr {C}$ with $\lambda (t) := \Sigma (t^{2} - 1) + {\mathrm {i}}\sigma t$ for $t \in {\mathbb R}$, so that

$$\begin{aligned} | \lambda | = \left( \Sigma ^2 (t^2 - 1)^2 + \sigma ^2 t^2 \right) ^{1/2} \leqslant C(t^2 + 1) \quad \text {and} \quad |{\mathrm {d}}\lambda | = \sqrt{4 t^{2} \Sigma ^{2} + \sigma ^{2}} {\mathrm {d}}t \leqslant C (| t | + 1) {\mathrm {d}}t, \end{aligned}$$

(A.10)

for some $C \in {\mathbb R}_{+}$ that depends only on $\Sigma$ and $\sigma _0$. We obtain

$$\begin{aligned} \left| G_{\sigma }(\mathbf{z}) \right|&\leqslant C \sigma ^{-1} \left( \int _{{\mathbb R}} \left| f^{1}\left( \sigma ^{-1} \Sigma (t^{2} - 1) + {\mathrm {i}}t \right) \right| (1+| t |^{7}){\mathrm {d}}t \right) . \end{aligned}$$

Let us prove that the last integral is uniformly bounded for $0<\sigma \leqslant \sigma _{0}$. We prove this result in full details for the Fermi–Dirac smearing $f^{1}(x) = (1+{\mathrm {e}}^{x})^{-1}$, the other cases being similar. We split the integral in the regions $|t| \leqslant \pi /2$ and $|t| > \pi /2$. For $| t | \leqslant \pi /2$, it holds that $\cos t \geqslant 0$, so that

$$\begin{aligned} \left| f^1 \left( \sigma ^{-1} \Sigma (t^{2} - 1) + {\mathrm {i}}t \right) \right|&= \left| 1 + {\mathrm {e}}^{\sigma ^{-1} \Sigma (t^{2} - 1)} {\mathrm {e}}^{{\mathrm {i}}t} \right| ^{-1} \leqslant \left| \mathrm{Re}\left( 1 + {\mathrm {e}}^{\sigma ^{-1} \Sigma (t^{2} - 1)} {\mathrm {e}}^{{\mathrm {i}}t} \right) \right| ^{-1} \\&= \left| 1 + {\mathrm {e}}^{\sigma ^{-1} \Sigma (t^{2} - 1)} \cos t \right| ^{-1} \leqslant 1. \end{aligned}$$

We deduce that the integral over $| t | \leqslant \pi /2$ is uniformly bounded for $0<\sigma \leqslant \sigma _0$. For $t \geqslant \pi /2 > 1$, it holds that $(t^2 - 1) > 0$, so that

$$\begin{aligned} \left| 1 + {\mathrm {e}}^{\sigma ^{-1} \Sigma (t^{2} - 1)} {\mathrm {e}}^{{\mathrm {i}}t} \right| ^{-1} \leqslant \left| {\mathrm {e}}^{\sigma ^{-1} \Sigma (t^{2} - 1)} -1 \right| ^{-1} \leqslant \left| {\mathrm {e}}^{\sigma _0^{-1} \Sigma ( t^2 - 1)} -1 \right| ^{-1}, \end{aligned}$$

where the right-hand side no longer depends on $0<\sigma \leqslant \sigma _0$. Finally, we check that the integral

$$\begin{aligned} \int _{| t | \geqslant \frac{\pi }{2}}\left| {\mathrm {e}}^{\sigma _0^{-1} \Sigma ( t^2 - 1)} -1 \right| ^{-1} (1 + | t |^7) {\mathrm {d}}t \end{aligned}$$

is absolutely convergent, and independent of $0<\sigma \leqslant \sigma _0$. This ends the proof of Lemma A.4. $\square$

We now prove, as claimed, that $G_{\sigma }$ is an analytic extension of $F_{\sigma }$.

Lemma A.5

For all $0 < \sigma \leqslant \sigma _{0}$ and all $\mathbf{k}\in {\mathbb R}^d$, it holds that $G_{\sigma }(\mathbf{k}) = F_{\sigma }(\mathbf{k})$.

Proof

We first approximate $H_{\mathbf{k}}$ by a finite-rank operator, then apply the Cauchy residual formula, and pass to the limit. Recall that $H_\mathbf{k}$ is a self-adjoint operator with spectral decomposition (2.1). For $Q \in {\mathbb N}^*$, we introduce the truncated operator

$$\begin{aligned} H_\mathbf{k}^Q := \sum _{n =1}^Q \varepsilon _{n \mathbf{k}} | u_{n\mathbf{k}} \rangle \langle u_{n\mathbf{k}} |. \end{aligned}$$

We have

$$\begin{aligned} \left| G_\sigma (\mathbf{k}) - F_\sigma (\mathbf{k}) \right| \leqslant&\left| G_\sigma (\mathbf{k}) - \mathrm{Tr}_{L^2_\mathrm {per}} \left( H_\mathbf{k}^Q f^\sigma (H_\mathbf{k}^Q) \right) \right| \end{aligned}$$

(A.11)

$$\begin{aligned}&+ \left| \mathrm{Tr}_{L^2_\mathrm {per}} \left( H_\mathbf{k}^Q f^\sigma (H_\mathbf{k}^Q) \right) - \mathrm{Tr}_{L^2_\mathrm {per}} \left( H_\mathbf{k}f^\sigma (H_\mathbf{k}) \right) \right| . \end{aligned}$$

(A.12)

We first focus on (A.12). Using the asymptotic (2.2) and the decay properties of $f^\sigma$, it holds that

$$\begin{aligned} \mathrm{Tr}_{L^2_\mathrm {per}} \left( H_\mathbf{k}f^\sigma (H_\mathbf{k}) \right) - \mathrm{Tr}_{L^2_\mathrm {per}} \left( H_\mathbf{k}^Q f^\sigma (H_\mathbf{k}^Q) \right) = \sum _{n \geqslant Q} \varepsilon _{n \mathbf{k}} f^{\sigma }(\varepsilon _{n\mathbf{k}}) \xrightarrow [Q \rightarrow \infty ]{} 0. \end{aligned}$$

(A.13)

We now focus on the right-hand side of (A.11). For $M \geqslant \varepsilon _{Q \mathbf{k}} + 1$ we denote by $\mathscr {C}_{M}$ the positively oriented closed contour defined by

$$\begin{aligned} \mathscr {C}_M := \left\{ \lambda \in \mathscr {C}, \ \mathrm{Re}\lambda \leqslant M \right\} \bigcup \left[ M + {\mathrm {i}}\sigma \left( 1 + \frac{M}{\Sigma } \right) , M - {\mathrm {i}}\sigma \left( 1 + \frac{ M}{\Sigma } \right) \right] . \end{aligned}$$

The contour $\mathscr {C}_M$ is obtained by truncating the parabola $\mathscr {C}$ to the region $\mathrm{Re}\lambda \leqslant M$ and closing the contour by a segment. For all $M \geqslant \varepsilon _{Q \mathbf{k}} + 1$, this contour encloses the spectrum of $H_\mathbf{k}^Q$, so that, from the Cauchy residual formula,

$$\begin{aligned} \mathrm{Tr}_{L^2_\mathrm {per}} \left( H_\mathbf{k}^Q f^\sigma (H_\mathbf{k}^Q) \right) = \frac{1}{2 \pi {\mathrm {i}}} \oint _{\mathscr {C}_M} \left[ f^\sigma (\lambda ) \lambda (\lambda + \Sigma )^2 \right] \mathrm{Tr}_{L^2_\mathrm {per}} \left( \dfrac{1}{(H_\mathbf{k}^Q + \Sigma )^2} \dfrac{1}{\lambda - H_\mathbf{k}^Q} \right) {\mathrm {d}}\lambda . \end{aligned}$$

As $M \rightarrow \infty$, and using the same arguments as in the proofs of Lemmas A.3 and A.4, we see that the right-hand side converges to the integral over the full contour $\mathscr {C}$. Moreover, we have the point-wise convergence

$$\begin{aligned} \forall \lambda \in \mathscr {C}, \quad&\lim _{Q \rightarrow \infty } \mathrm{Tr}_{L^2_\mathrm {per}} \left( \dfrac{1}{(H_\mathbf{k}^Q + \Sigma )^2} \dfrac{1}{\lambda - H_\mathbf{k}^Q} \right) = \lim _{Q \rightarrow \infty } \sum _{n =1}^{Q} \frac{1}{(\varepsilon _{n\mathbf{k}} + \Sigma )^{2}(\lambda -\varepsilon _{n\mathbf{k}})} \\&\qquad =\sum _{n=1}^{\infty } \frac{1}{(\varepsilon _{n\mathbf{k}} + \Sigma )^{2}(\lambda -\varepsilon _{n\mathbf{k}})} = \mathrm{Tr}_{L^2_\mathrm {per}} \left( \dfrac{1}{(H_\mathbf{k}+ \Sigma )^2} \dfrac{1}{\lambda - H_\mathbf{k}} \right) . \end{aligned}$$

We conclude from the dominated convergence theorem that $\mathrm{Tr}_{L^2_\mathrm {per}} \left( H_\mathbf{k}^Q f^\sigma (H_\mathbf{k}^Q) \right)$ converges to $G_{\sigma }(\mathbf{k})$ as $Q \rightarrow \infty$. The proof of Lemma A.5 follows. $\square$

Combining Lemma A.5 together with Lemma A.4 ends the proof of Lemma A.1.

1.2 Proof of Lemma A.2

We now focus on Gaussian-type smearing functions. The idea of the proof is similar to previously. We need however to re-define $G_\sigma$ for large $| \mathbf{y}|$ (i.e. choose an appropriate contour). In the sequel, we fix $Y \geqslant 1$ and provide estimates uniform in $\mathbf{z}\in S_Y$.

Looking at the operator $-\frac{1}{2} \Delta _{\mathbf{z}}$ in Fourier basis, we see that its spectrum is

$$\begin{aligned} \sigma \left( -\tfrac{1}{2} \Delta _\mathbf{z}\right) := \left\{ \tfrac{1}{2}(\mathbf{K}+ \mathbf{z})^{2} \right\} _{\mathbf{K}\in \mathcal {R}^*} \left\{ \frac{1}{2} (|\mathbf{K}+ \mathbf{k}|^{2} - |\mathbf{y}|^{2}) + {\mathrm {i}}\mathbf{y}\cdot (\mathbf{K}+\mathbf{k}) \right\} _{\mathbf{K}\in \mathcal {R}^*}, \end{aligned}$$

hence is contained in the parabolic set $(\mathrm{Im}\lambda )^{2} \leqslant Y^{2}(2\mathrm{Re}\lambda + Y^{2})$. In the sequel, we take an parabolic integration contour that encloses this region. More specifically, for $\alpha > 1$ and $Y \geqslant 1$, we introduce the dilated contour

$$\begin{aligned} \mathscr {C}_{\alpha ,Y} := \left\{ \lambda \in {\mathbb C}, \ \mathrm{Im}\lambda ^{2} = \alpha ^{2} Y^{2}(2\mathrm{Re}\lambda + \alpha ^{2} Y^{2})\right\} . \end{aligned}$$

As $\alpha$ increases, the distance between $\mathscr {C}_{\alpha ,Y}$ and $\sigma (-\tfrac{1}{2} \Delta _\mathbf{z})$ goes to $+ \infty$. Since the operator $-\tfrac{1}{2} \Delta _\mathbf{z}$ is normal, this implies that there exists $\alpha _c \geqslant 1$ such that

$$\begin{aligned} \forall Y \geqslant 1, \quad \forall \mathbf{z}\in S_Y, \quad \forall \lambda \in \mathscr {C}_{\alpha _c, Y}, \quad \left\| (\lambda +\tfrac{1}{2} \Delta _\mathbf{z})^{-1} \right\| _\mathscr {B}\leqslant \left( 2 \Vert V \Vert _{L^\infty } \right) ^{-1}. \end{aligned}$$

The contour $\mathscr {C}:= \mathscr {C}_{\alpha _c, Y}$ is our integration contour for $\mathbf{z}\in S_Y$ (see Fig. 11).

Fig. 11

The spectra of the operator $-\tfrac{1}{2} \Delta _\mathbf{z}$ and $H_{\mathbf{z}}$, and the contour $\mathscr {C}$ that encloses them

For $\lambda \in \mathscr {C}$, it holds that (compare with (A.8))

$$\begin{aligned} \left( \lambda - H_{\mathbf{z}}\right) = \left( \lambda + \tfrac{1}{2} \Delta _{\mathbf{z}} - V \right) = \left( \lambda + \tfrac{1}{2} \Delta _\mathbf{z}\right) \left( 1 - (\lambda + \tfrac{1}{2} \Delta _\mathbf{z})^{-1} V \right) . \end{aligned}$$

As in Lemma A.3 we deduce the following inequalities. We do not repeat the proof, as it is similar. We denote by $\Sigma := \frac{\alpha _c^{2}Y^{2}}{2}$ for the sake of clarity.

Lemma A.6

There exists $C \in {\mathbb R}_+$ such that, for all $Y \geqslant 1$, all $\mathbf{z}\in S_Y$, and all $\lambda \in \mathscr {C}$, it holds that

$$\begin{aligned}&\Vert (H_{\mathbf{z}}-\lambda )^{-1}\Vert _{\mathscr {B}} \leqslant C, \quad \Vert (H_{\mathbf{z}}+\Sigma )^{-1}\Vert _{{\mathfrak {S}}_{2}} \leqslant C,\\&\mathrm{Tr}_{L^2_\mathrm {per}} \left( \dfrac{1}{(H_\mathbf{z}+ \Sigma )^2} \dfrac{1}{\lambda - H_\mathbf{z}} \right) \leqslant C, \quad \text {and} \quad \partial _{\mathbf{z}} \mathrm{Tr}_{L^2_\mathrm {per}} \left( \dfrac{1}{(H_\mathbf{z}+ \Sigma )^2} \dfrac{1}{\lambda - H_\mathbf{z}} \right) \leqslant C. \end{aligned}$$

We now define, for $0<\sigma \leqslant \sigma _0$ and $Y \geqslant 1$, the function defined on $S_Y$ by

$$\begin{aligned} G_{\sigma ,Y}(\mathbf{z}) := \dfrac{1}{2 \pi {\mathrm {i}}} \int _{\mathscr {C}} \left[ \lambda f^{1}(\lambda /\sigma )(\lambda + \Sigma )^2 \right] \mathrm{Tr}_{L^2_\mathrm {per}} \left( \dfrac{1}{(H_\mathbf{z}+ \Sigma )^2} \dfrac{1}{\lambda - H_\mathbf{z}} \right) {\mathrm {d}}\lambda . \end{aligned}$$

(A.14)

Before stating a bound on $G_{\sigma ,Y}$, we need the following technical lemma on the growth of $f^{1}$:

Lemma A.7

(Growth of $f^{1}$ in the complex plane) When $f^{1}$ is one of the Gaussian-type smearing functions (5.6$''$–5.6$'''$), then $f^1$ is entire, and there exists $C \in {\mathbb R}_{+}$ and $q \geqslant 0$ such that

$$\begin{aligned} \forall x,y \in {\mathbb R}, \quad \left| f^1(x + {\mathrm {i}}y) \right| \leqslant \left\{ \begin{aligned}&C (1+(x^2 + y^2)^q) \left( {\mathrm {e}}^{ y^2 - x^2} \right) \quad \text {if} \quad x \geqslant 0, \\&C (1+(x^2 + y^2)^q) \left( 1 + {\mathrm {e}}^{ y^2 - x^2} \right) \quad \text {if} \quad x < 0. \end{aligned} \right. \end{aligned}$$

(A.15)

Proof

We first handle the case when $f^{1}$ is the Gaussian smearing function (in which case we can choose $q = 0$). We have

$$\begin{aligned} f^1(x) = \frac{1}{2} \left( 1 - \mathrm{erf} \left( x \right) \right) = \frac{1}{\sqrt{\pi }} \int _{x}^\infty {\mathrm {e}}^{-t^2} {\mathrm {d}}t. \end{aligned}$$

First recall that for $x \in {\mathbb R}$, we have $0< f^1(x) < 1$. Moreover, for $x \geqslant 1$, it holds that

$$\begin{aligned} 0 \leqslant f^1(x) \leqslant \ \frac{1}{\sqrt{\pi }} \int _{x}^\infty t {\mathrm {e}}^{-t^2} {\mathrm {d}}t = \frac{1}{2 \sqrt{\pi }} {\mathrm {e}}^{-x^2}. \end{aligned}$$

This already proves (A.15) for $y = 0$. Then, we notice that an analytic continuation of $f^1$ is given by

$$\begin{aligned} f^1(x + {\mathrm {i}}y) = \dfrac{1}{\sqrt{\pi }} \left( \int _x^\infty {\mathrm {e}}^{-t^2} {\mathrm {d}}t + {\mathrm {i}}\int _{0}^{y} {\mathrm {e}}^{-(x + {\mathrm {i}}t)^2} {\mathrm {d}}t\right) = f^1(x) + {\mathrm {i}}\dfrac{{\mathrm {e}}^{-x^2}}{\sqrt{\pi }} \int _0^y {\mathrm {e}}^{- 2 {\mathrm {i}}x t} {\mathrm {e}}^{t^2} {\mathrm {d}}t. \end{aligned}$$

Lemma A.7 then follows from the inequalities

$$\begin{aligned} \left| f^1(x + {\mathrm {i}}y) \right| \leqslant \left| f^1(x) \right| + \dfrac{{\mathrm {e}}^{-x^2}}{\sqrt{\pi }} \int _{[0,y]} {\mathrm {e}}^{t^2} {\mathrm {d}}t, \end{aligned}$$

together with the fact that

$$\begin{aligned} \int _{[0,y]} {\mathrm {e}}^{t^2} {\mathrm {d}}t = \int _{[0,1]} {\mathrm {e}}^{t^2} {\mathrm {d}}t + \int _{[1,y]} {\mathrm {e}}^{t^2} {\mathrm {d}}t \leqslant \int _{[0,1]} {\mathrm {e}}^{t^2} {\mathrm {d}}t + \int _{[1,y]} t {\mathrm {e}}^{t^2} {\mathrm {d}}t \leqslant C {\mathrm {e}}^{y^2}. \end{aligned}$$

The case of the Methfessel–Paxton and cold smearing schemes follows immediately by noting that they differ from the Gaussian smearing function by terms of the form $x^n {\mathrm {e}}^{-x^{2}}$, and the fact that

$$\begin{aligned} \left| z^n {\mathrm {e}}^{-z^2} \right| = ( x^2 + y^2 )^{n/2} {\mathrm {e}}^{y^2 - x^2}. \end{aligned}$$

$\square$

We are now in position to prove estimates on $G_{\sigma , Y}$ defined on (A.14).

Lemma A.8

There exists $C \in {\mathbb R}_+$ and $\eta > 0$ such that, for all $0 < \sigma \leqslant \sigma _{0}$ and all $Y \geqslant 1$, the function $G_{\sigma ,Y}$ is analytic on $S_Y$, and

$$\begin{aligned} \sup _{\mathbf{z}\in S_{Y}} |G_{\sigma ,Y}(\mathbf{z})| \leqslant C {\mathrm {e}}^{\eta \frac{Y^{4}}{\sigma ^{2}}}. \end{aligned}$$

Proof

For the sake of clarity, we do the proof when $f^1$ is the Gaussian smearing function (i.e. $q = 0$). From Lemma A.6, it holds that

$$\begin{aligned} |G_{\sigma ,Y}(\mathbf{z})|&\leqslant C \int _{\mathscr {C}} | \lambda f^{1}(\lambda /\sigma ) (\lambda + \Sigma )^2| |{\mathrm {d}}\lambda |. \end{aligned}$$

We parametrize the contour $\mathscr {C}$ with $\lambda (t) := \tfrac{\alpha _c^{2} Y^{2}}{2} ( (t^{2}-1) + {\mathrm {i}}2t)$, so that (compare with (A.10))

$$\begin{aligned} | \lambda | \leqslant C Y^2 (t^2 + 1 ) \quad \text {and} \quad | {\mathrm {d}}\lambda | \leqslant C Y^2 (| t | + 1) {\mathrm {d}}t. \end{aligned}$$

Then,

$$\begin{aligned} \left| G_{\sigma ,Y}(\mathbf{z}) \right|&\leqslant C Y^{8} \int _{{\mathbb R}} \left| f^{1} \left( \tfrac{\alpha _c^{2}Y^{2}}{2 \sigma } \left( t^{2}-1+{\mathrm {i}}2t\right) \right) \right| (1+|t|^{7}) {\mathrm {d}}t. \end{aligned}$$

We split this integral in regions where $| t | \geqslant 1$, and $| t | \leqslant 1$. When $| t | \leqslant 1$, it holds that $\mathrm{Re}\lambda (t) \leqslant 0$. Together with the second inequality of Lemma A.7 and the inequalities

$$\begin{aligned} \forall -1 \leqslant t \leqslant 1, \quad (1 + | t |^7) \leqslant 2, \quad \text {and} \quad 4t^2 - (t^2 - 1)^2 \leqslant 4, \end{aligned}$$

we get

$$\begin{aligned} \int _{[-1, 1]} \left| f^{1} \left( \tfrac{\alpha _c^{2}Y^{2}}{2 \sigma } \left( t^{2}-1+{\mathrm {i}}2 t\right) \right) \right| (1+|t|^{7}) {\mathrm {d}}t&\leqslant 2 C \int _{[-1,1]} \left[ 1 + \exp \left( \tfrac{\alpha _c^4 Y^4}{4 \sigma ^2}(4 t^2 - (t^2 - 1)^2) \right) \right] {\mathrm {d}}t \\&\leqslant 4 C \left[ 1 + {\mathrm {e}}^{\frac{\alpha _c^4 Y^4}{\sigma ^2} } \right] \leqslant 8C {\mathrm {e}}^{\frac{\alpha _c^4 Y^4}{\sigma ^2}}. \end{aligned}$$

For $| t | \geqslant 1$, we use the first inequality of Lemma A.7, and obtain

$$\begin{aligned} \int _{[-1, 1]^c} \left| f^{1} \left( \tfrac{\alpha _c^{2}Y^{2}}{2\sigma } \left( t^{2}-1+{\mathrm {i}}2 t\right) \right) \right| (1+|t|^{7}) {\mathrm {d}}t&\leqslant 2C \int _1^\infty \exp \left( \tfrac{\alpha _c^4 Y^4}{4 \sigma ^2}(4 t^2 - (t^2-1)^2) \right) \\&\quad (1 + | t |^7){\mathrm {d}}t. \end{aligned}$$

When $t \geqslant 1$, we have the inequalities

$$\begin{aligned} (1 + | t |^7) \leqslant 2 | t |^7 \quad \text {while} \quad 6 t^2 \leqslant \frac{1}{2}(t^4 + 6^2) \quad \text {so that} \quad 4 t^2 - (t^2 - 1)^2 \leqslant - \tfrac{t^4}{2} + 17. \end{aligned}$$

As a result,

$$\begin{aligned} \int _1^\infty \exp \left( \tfrac{\alpha _c^4 Y^4}{4 \sigma ^2}(4 t^2 - (t^2-1)^2) \right) (1 + | t |^7){\mathrm {d}}t \leqslant 2 \exp \left( \tfrac{17 \alpha _c^4 Y^4}{4 \sigma ^2 } \right) \int _0^\infty \exp \left( \tfrac{- \alpha _c^4 Y^4}{8\sigma ^2}t^4 \right) t^7 {\mathrm {d}}t. \end{aligned}$$

We finally perform the change of variable $u = \frac{\alpha _c Y}{8^{1/4} \sqrt{\sigma }}t$ and get

$$\begin{aligned} \int _0^\infty \exp \left( \tfrac{- \alpha _c^4 Y^4}{8\sigma ^2}t^4 \right) t^7 {\mathrm {d}}t = \left( \frac{8^{1/4} \sqrt{\sigma }}{\alpha _c Y} \right) ^8 \int _0^\infty {\mathrm {e}}^{-u^4} u^7 {\mathrm {d}}u, \end{aligned}$$

where the right-hand side is uniformly bounded for $0<\sigma \leqslant \sigma _0$ and $Y \geqslant 1$. Combining all the inequalities concludes the proof of Lemma A.8$\square$

Lemma A.2 is a consequence of the Lemma A.8 and the following one, whose proof is similar to the one of Lemma A.5.

Lemma A.9

For all $0 < \sigma \leqslant \sigma _{0}$, all $Y \geqslant 1$, and all $\mathbf{k}\in {\mathbb R}^d$, it holds that $G_{\sigma ,Y}(\mathbf{k}) = F_{\sigma }(\mathbf{k})$.