external electric potential are not difficult to bound because $W$ is a
bounded function. To obtain bounds of the correct size in $h$ for the terms
involving the periodic vector potential $A_{h}$, we need to use that $A(0)=0$,
which is guaranteed by Assumption 1.1. This is relevant for example when we
estimate our equivalent of the term on the left side of [16, Eq. (6.24)], that
is,

$\|[\pi_{\mathbf{A}_{h}}^{2}+W_{h}(r)-p_{r}^{2}]\sigma_{0}\|_{2}$

with $p_{r}=-\mathrm{i}\nabla_{r}$ and $\sigma_{0}$ in (5.7). We write the
operator multiplying $\sigma_{0}$ as

$\pi_{r}^{2}-p_{r}^{2}+W(r)+A(r)\cdot\pi_{r}+\pi_{r}\cdot A(r)+|A(r)|^{2},$
(6.4)

where $\pi_{r}=-\mathrm{i}\nabla_{r}+\mathbf{A}_{\mathbf{B}}(r)$. When we use
(5.6), we see that the terms involving $|A_{h}|^{2}$ and $W_{h}$ are bounded
by

$\displaystyle\bigl{(}\|A_{h}\|_{\infty}^{2}+\|W_{h}\|_{\infty}\bigr{)}\
\|\sigma_{0}\|_{2}\leqslant
C\varepsilon^{-\nicefrac{{1}}{{2}}}h^{4}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{h})}.$
(6.5)

Moreover, from [16, Eq. (6.24)] we know that

$\|[\pi_{r}^{2}-p_{r}^{2}]\sigma_{0}\|_{2}\leqslant
C\varepsilon^{-\nicefrac{{1}}{{2}}}h^{4}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{h})}.$

To obtain a bound for the contribution from the fourth and the fifth term on
the right side of (6.4), we write

$A_{h}(r)=h^{2}\int_{0}^{1}\mathrm{d}t\ (DA)(hrt)\cdot r,$

where $DA$ denotes the Jacobi matrix of $A$. Hence,

$\|A_{h}(r)\cdot\pi_{r}\,\sigma_{0}\|_{2}\leqslant\ h^{2}\|DA\|_{\infty}\,\|\
|\cdot|\pi\alpha_{*}\|_{2}\,\|\Psi_{>}\|_{2}\leqslant
Ch^{4}\varepsilon^{-\nicefrac{{1}}{{2}}}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{h})}.$

The term involving $\pi_{r}\cdot A_{h}(r)$ can be treated similarly when we
commute $\pi_{r}$ to the right. In combination, the above considerations show

$\|[\pi_{\mathbf{A}_{h}}^{2}+W_{h}(r)-p_{r}^{2}]\sigma_{0}\|_{2}\leqslant
Ch^{4}\varepsilon^{-\nicefrac{{1}}{{2}}}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{h})}.$

All other bounds in the proof of the equivalent of [16, Lemma 6.2] in our
setting that involve $W_{h}$ or $A_{h}$ can be estimated with similar ideas.
We therefore omit further details.

— Appendix —

##  Appendix A Gauge-Invariant Perturbation Theory for
$K_{{T_{\mathrm{c}}},\mathbf{A}}-V$

In this appendix, we discuss the behavior of the eigenvalues below the
essential spectrum of the operator $K_{{T_{\mathrm{c}}},\mathbf{A}}-V$ for
small $h>0$, where $K_{{T_{\mathrm{c}}},\mathbf{A}}\coloneqq
K_{{T_{\mathrm{c}}},\mathbf{A},0}$ with $K_{T,\mathbf{A},0}$ defined in
(5.13). We recall that the full magnetic vector potential is given by
$\mathbf{A}\coloneqq\mathbf{A}_{e_{3}}+A$ with the magnetic vector potential
$\mathbf{A}_{e_{3}}(x)=\frac{1}{2}e_{3}\wedge x$ of a constant magnetic field
and a periodic vector potential $A$ that satisfies $A(0)=0$. We also recall
the notation $\mathbf{A}_{h}(x)\coloneqq h\mathbf{A}(hx)$. The goal of this
appendix is to prove the following proposition.

######  Proposition A.1.

Assume $(1+|\cdot|^{2})V\in L^{2}(\mathbb{R}^{3})\cap
L^{\infty}(\mathbb{R}^{3})$ and let $A\in
W^{3,\infty}(\mathbb{R}^{3};\mathbb{R}^{3})$ be a periodic function. Then
there is $h_{0}>0$ such that for $0<h\leqslant h_{0}$ the following statements
hold:

  1. (a)

Let $\lambda$ be an isolated eigenvalue of multiplicity $m\in\mathbb{N}$ of
the operator $K_{T_{\mathrm{c}}}-V$ with spectral projection $P$. Then there
are $m$ eigenvalues $\lambda_{1}(h),\ldots,\lambda_{m}(h)$ of the operator
$K_{{T_{\mathrm{c}}},\mathbf{A}}-V$ with spectral projection $P(h)$ such that

$\displaystyle\max_{i=1,\ldots,m}|\lambda_{i}(h)-\lambda|$
$\displaystyle\leqslant Ch^{2},$ $\displaystyle\|P(h)-P\|_{\infty}$
$\displaystyle\leqslant Ch^{2}.$ (A.1)

  2. (b)

Assume that $K_{T_{\mathrm{c}}}-V$ has a simple lowest eigenvalue with
eigenfunction $\alpha_{*}$ and denote by $\alpha_{*}^{h}$ the eigenfunction to
the lowest eigenvalue of $K_{{T_{\mathrm{c}}},\mathbf{A}}-V$, which is
normalized such that $\langle\alpha_{*},\alpha_{*}^{h}\rangle\geqslant 0$
holds. Then we have the bound

$\displaystyle\|(1+\pi^{2})(\alpha_{*}^{h}-\alpha_{*})\|_{2}\leqslant Ch^{2},$
(A.2)

where $\pi^{2}=(-\mathrm{i}\nabla+\mathbf{A}_{\mathbf{B}})^{2}$ denotes the
magnetic Laplacian.

######  Remark A.2.

The above Proposition should be compared to [16, Proposition A.1], where a
similar statement is proved in the special case, where $V\geqslant 0$ and
where $\mathbf{A}(x)=\frac{1}{2}e_{3}\wedge x$. In the reference the
assumption for $V$ is needed to assure that the Birman–Schwinger operator
related to $K_{T,\mathbf{A}}-V$ is self-adjoint. To prove the above
proposition we investigate the resolvent kernel of $K_{T,\mathbf{A}}-V$ rather
than the Birman–Schwinger operator, which is the reason why we do not need the
positivity assumption for $V$. The main technical ingredient of our proof is
the estimate for the integral kernel of the resolvent of $K_{T}-V$ in Lemma
A.4 below. Afterwards, we apply a standard version of gauge-invariant
perturbation theory. As has been shown in [40, Chapter 6] one can also obtain
exponential decay estimates for the resolvent kernel of $K_{T}-V$ via
Combes–Thomas estimates. The proof, however, requires more effort than for the
case of the Laplacian because $K_{T}$ has a more complicated structure. Since
these estimates are not needed for the proof of Proposition A.1,we refrain
from presenting them here.

We denote the resolvent of $K_{T,\mathbf{A}}-V$ at
$z\in\rho(K_{T,\mathbf{A}}-V)$ by
$\mathcal{R}_{\mathbf{A}}^{z,V}\coloneqq(z-(K_{{T_{\mathrm{c}}},\mathbf{A}}-V))^{-1}$.
The integral kernel of $\mathcal{R}_{\mathbf{A}}^{z,V}$ is denoted by
$\mathcal{G}_{\mathbf{A}}^{z,V}(x,y)$. We highlight that we use another symbol
for the resolvent and for its integral kernel in this section. Since we
frequently work with $L^{p}$-norms of integral kernels this simplifies our
notation. If $\mathbf{A}=0$, we write $\mathcal{R}^{z,V}$ for
$\mathcal{R}_{0}^{z,V}$ and $\mathcal{G}^{z,V}$ for $\mathcal{G}_{0}^{z,V}$.
Similarly, $\mathcal{G}^{z}$ stands for $\mathcal{G}^{z,0}$. Before we give
the proof of the above proposition, we state and prove four preparatory
lemmas.

###  A.1 Preparatory lemmas

The first lemma concerns the regularity of the kernel $\mathcal{G}^{z}$.

######  Lemma A.3.

There is a continuous function
$a\colon\mathbb{C}\backslash[2{T_{\mathrm{c}}},\infty)\to\mathbb{R}_{+}$ such
that

$\displaystyle\int_{\mathbb{R}^{3}}\mathrm{d}x\
(1+x^{2})|\nabla\mathcal{G}^{z}(x)|\leqslant a(z).$ (A.3)

###### Proof.

We use the resolvent identity
$\mathcal{R}^{z}=\mathcal{R}^{0}+z\mathcal{R}^{0}\mathcal{R}^{z}$ to write
$\mathcal{G}^{z}$ as

$\displaystyle\mathcal{G}^{z}(x)=\mathcal{G}^{0}(x)+z\int_{\mathbb{R}^{3}}\mathrm{d}v\;\mathcal{G}^{0}(x-v)\,\mathcal{G}^{z}(v),$
(A.4)

which implies

$\displaystyle\|(1+|\cdot|^{2})\nabla\mathcal{G}^{z}\|_{1}\leqslant\|(1+2|\cdot|^{2})\nabla\mathcal{G}^{0}\|_{1}\left(1+|z|\,\|(1+|\cdot|^{2})\mathcal{G}^{z}\|_{1}\right).$
(A.5)

The second $L^{1}(\mathbb{R}^{3})$-norm on the right side of (A.5) is bounded
by

$\displaystyle\|(1+|\cdot|^{2})\mathcal{G}^{z}\|_{1}\leqslant C\
\|(1+|\cdot|^{2})^{2}\mathcal{G}^{z}\|_{2}=C\,\Bigl{(}\int_{\mathbb{R}^{3}}\mathrm{d}p\
\Bigl{|}(1-\Delta_{p})^{2}\frac{1}{z-K_{T_{\mathrm{c}}}(p)}\Bigr{|}^{2}\Bigr{)}^{\nicefrac{{1}}{{2}}},$
(A.6)

which, when multiplied with $z$, meets the requirements of the lemma. It
therefore remains to consider $\|(1+|\cdot|^{2})\nabla\mathcal{G}^{0}\|_{1}$.

From [21, Eq. (A.6)] and [39, Theorem 6.23] we know that $\mathcal{G}^{0}(x)$
can be written as

$\displaystyle\mathcal{G}^{0}(x)$
$\displaystyle=\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{1}{n-\frac{1}{2}}\text{Im}\
g_{0}^{\mathrm{i}(n-\frac{1}{2})2\pi{T_{\mathrm{c}}}}(x)$ (A.7)
$\displaystyle=\frac{1}{2\pi^{2}|x|}\sum_{n=1}^{\infty}\frac{1}{n-\frac{1}{2}}\text{Im}\exp\Bigl{(}\mathrm{i}\sqrt{\mu+\mathrm{i}\left(n-\frac{1}{2}\right)2\pi{T_{\mathrm{c}}}}\,|x|\Bigr{)},$

where $g_{0}^{z}$ is the resolvent kernel of the Laplacian in (4.3) and
$\sqrt{\cdot}$ denotes the principal square root. We use
$\operatorname{\mathrm{Im}}\mathrm{e}^{\mathrm{i}z}=\sin(\operatorname{\mathrm{Re}}z)\exp(-\operatorname{\mathrm{Im}}z)$
for $z\in\mathbb{C}$ and

$\displaystyle\sqrt{a+\mathrm{i}b}=\frac{1}{\sqrt{2}}\sqrt{\sqrt{a^{2}+b^{2}}+a}+\frac{\mathrm{i}\operatorname{sgn}(b)}{\sqrt{2}}\sqrt{\sqrt{a^{2}+b^{2}}-a}$
(A.8)

for $a,b\in\mathbb{R}$ to see that

$\displaystyle\text{Im}\exp\Bigl{(}\mathrm{i}\sqrt{\mu+\mathrm{i}\left(n-\frac{1}{2}\right)2\pi{T_{\mathrm{c}}}}\,|x|\Bigr{)}=\sin\left(|x|c_{n}^{+}\right)\exp\left(-|x|c_{n}^{-}\right),$
(A.9)

where

$\displaystyle
c_{n}^{\pm}=\frac{1}{\sqrt{2}}\sqrt{\sqrt{\mu^{2}+((n-1/2)2\pi{T_{\mathrm{c}}})^{2}}\pm\mu}.$
(A.10)

In particular,

$\displaystyle\nabla\mathcal{G}^{0}(x)$
$\displaystyle=-\frac{x}{2\pi^{2}|x|^{3}}\sum_{n=1}^{\infty}\frac{1}{n-\frac{1}{2}}\sin\left(|x|c_{n}^{+}\right)\exp\left(-|x|c_{n}^{-}\right)$
$\displaystyle\hskip
50.0pt+\frac{x}{2\pi^{2}|x|^{2}}\sum_{n=1}^{\infty}\frac{c_{n}^{+}}{n-\frac{1}{2}}\cos\left(|x|c_{n}^{+}\right)\exp\left(-|x|c_{n}^{-}\right)$
$\displaystyle\hskip
50.0pt-\frac{x}{2\pi^{2}|x|^{2}}\sum_{n=1}^{\infty}\frac{c_{n}^{-}}{n-\frac{1}{2}}\sin\left(|x|c_{n}^{+}\right)\exp\left(-|x|c_{n}^{-}\right).$
(A.11)

The above formula implies the bound

$\displaystyle\|(1+|\cdot|^{2})\nabla\mathcal{G}^{0}\|_{1}$
$\displaystyle\leqslant\sum_{n=1}^{\infty}\frac{C}{n}\int_{0}^{\infty}\mathrm{d}r\left(1+r^{2}+(r+r^{3})\sqrt{1+n}\right)\exp\left(-rc_{n}^{-}\right)\leqslant
C\sum_{n=1}^{\infty}n^{-\frac{3}{2}}.$ (A.12)

This proves the claim of the lemma. ∎

The second lemma concerns bounds for the operator norm of
$(z-(K_{T_{\mathrm{c}}}-V))^{-1}$ and commutators of this operator with $x$,
when viewed as maps from $L^{2}(\mathbb{R}^{3})$ to
$L^{\infty}(\mathbb{R}^{3})$. Here and in the following we denote by
$\|A\|_{2;\infty}$ the norm of a bounded operator $A$ from
$L^{2}(\mathbb{R}^{3})\rightarrow L^{\infty}(\mathbb{R}^{3})$. From [47,
Corollary A.1.2] we know that if such an operator is also a bounded from
$L^{2}(\mathbb{R}^{3})$ to $L^{2}(\mathbb{R}^{3})$ then it has an integral
kernel given by a measurable function $A(x,y)$, which obeys

$\displaystyle\operatorname*{ess\,sup}_{x\in\mathbb{R}^{3}}\left(\int_{\mathbb{R}^{3}}\mathrm{d}y\;|A(x,y)|^{2}\right)^{\nicefrac{{1}}{{2}}}<\infty.$
(A.13)

Moreover, the norm $\|A\|_{2;\infty}$ equals the norm of the integral kernel
of $A$ in (A.13).

The following Lemma A.4 implies that the resolvent kernel $\mathcal{G}^{z,V}$
satisfies

$\displaystyle\operatorname*{ess\,sup}_{x\in\mathbb{R}^{3}}\Bigl{(}\int_{\mathbb{R}^{3}}\mathrm{d}y\;\bigl{(}1+|x-y|^{4}\bigr{)}|\mathcal{G}^{z,V}(x,y)|^{2}\Bigr{)}^{\nicefrac{{1}}{{2}}}<\infty.$
(A.14)

In passing we note that our assumptions on $V$ would allow for more: it can be
shown that $\mathcal{G}^{z,V}$ is exponentially decaying in the sense that
(A.14) holds with $1+|x-y|^{4}$ replaced by $\mathrm{e}^{\delta|x-y|}$ for
some $\delta>0$ depending on the distance of $z$ to the spectrum of
$K_{T_{\mathrm{c}}}-V$. For more information we refer to Remark A.2 above.

######  Lemma A.4.

Assume that $V\in L^{\infty}(\mathbb{R}^{3})$ vanishes at infinity. Then there
is a continuous function $a\colon\rho(K_{T_{\mathrm{c}}}-V)\to\mathbb{R}_{+}$
such that

$\displaystyle\|\mathcal{R}^{z,V}\|_{2;\infty}+\bigl{\|}[x,\mathcal{R}^{z,V}]\bigr{\|}_{2;\infty}+\bigl{\|}[x,[x,\mathcal{R}^{z,V}]]\bigr{\|}_{2;\infty}\leqslant
a(z).$ (A.15)

###### Proof.

We start by proving the bound for the first term on the right side of (A.15).
We use the fact that $(1-\Delta)^{-1}$ is a bounded linear map from
$L^{2}(\mathbb{R}^{3})$ to $L^{\infty}(\mathbb{R}^{3})$ and the resolvent
identity to estimate

$\displaystyle\|\mathcal{R}^{z,V}\|_{2;\infty}\leqslant
C\|(1-\Delta)\mathcal{R}^{z,V}\|_{\infty}\leqslant
C\|(1-\Delta)\mathcal{R}^{z}\|_{\infty}\left(1+\|V\|_{\infty}\|\mathcal{R}^{z,V}\|_{\infty}\right).$
(A.16)

Our assumptions on $V$ imply
$\rho(K_{T_{\mathrm{c}}}-V)\subseteq\rho(K_{T_{\mathrm{c}}})$, whence both
$z$-dependent terms on the right side meet the requirements of the lemma.

To obtain a bound for the second term on the right side of (A.15), we note
that

$\displaystyle[x,\mathcal{R}^{z,V}]$
$\displaystyle=\mathcal{R}^{z,V}[K_{T_{\mathrm{c}}},x]\mathcal{R}^{z,V},$
$\displaystyle[K_{T_{\mathrm{c}}},x]$ $\displaystyle=-\mathrm{i}(\nabla
f)(-\mathrm{i}\nabla),$ (A.17)

where $f(p)\coloneqq K_{T_{\mathrm{c}}}(p)$ is the symbol in (1.14). Using
this, we estimate

$\displaystyle\|[x,\mathcal{R}^{z,V}]\|_{2;\infty}\leqslant\|\mathcal{R}^{z,V}\|_{2;\infty}\,\|(\nabla
f)(-\mathrm{i}\nabla)\mathcal{R}^{z,V}\|_{\infty}.$ (A.18)

A bound for the first factor on the right side was obtained in (A.16). Using
the resolvent identity again, we bound the second factor by

$\displaystyle\|(\nabla
f)(-\mathrm{i}\nabla)\mathcal{R}^{z,V}\|_{\infty}\leqslant\|(\nabla
f)(-\mathrm{i}\nabla)\mathcal{R}^{z}\|_{\infty}\left(1+\|V\|_{\infty}\|\mathcal{R}^{z,V}\|_{\infty}\right),$
(A.19)

which proves the claim for the second term on the right side of (A.15). A
bound for the third term can be derived similarly, and we therefore leave the
remaining details to the reader. This proves the claim. ∎

######  Lemma A.5.

Assume $(1+|\cdot|^{2})V\in L^{2}(\mathbb{R}^{3})\cap
L^{\infty}(\mathbb{R}^{3})$ and let $\alpha$ be an eigenfunction of the
operator $K_{T_{\mathrm{c}}}-V$ with eigenvalue $\lambda<2{T_{\mathrm{c}}}$.
Then we have

$\displaystyle\|\,|\cdot|\nabla\alpha\|_{2}+\|\,|\cdot|^{2}\alpha\|_{2}<\infty.$
(A.20)

###### Proof.

We use the eigenvalue equation to write the Fourier transform of $\alpha$ as

$\displaystyle\widehat{\alpha}(p)=-\frac{1}{\lambda-
K_{T_{\mathrm{c}}}(p)}\,(\widehat{V}\ast\widehat{\alpha})(p).$ (A.21)

Using Young’s inequality, we see that this implies

$\displaystyle\|\,|\cdot|^{2}\alpha\|$
$\displaystyle=\Bigl{(}\int_{\mathbb{R}^{3}}\mathrm{d}p\;\Bigl{|}\Delta_{p}\frac{1}{\lambda-
K_{T_{\mathrm{c}}}(p)}\,(\widehat{V}\ast\widehat{\alpha})(p)\Bigr{|}^{2}\Bigr{)}^{\nicefrac{{1}}{{2}}}$
$\displaystyle\leqslant
C\left(\|\widehat{V}\ast\widehat{\alpha}\|_{\infty}+\|\Delta(\widehat{V}\ast\widehat{\alpha})\|_{\infty}\right)\leqslant
C\,\|(1+|\cdot|^{2})V\|_{\infty}.$ (A.22)

To prove the other bound, we use the resolvent identity to write (A.21) as

$\displaystyle\widehat{\alpha}(p)=-\frac{1}{K_{T_{\mathrm{c}}}(p)}\,(\widehat{V}\ast\widehat{\alpha})(p)+\frac{\lambda}{K_{T_{\mathrm{c}}}(p)}\frac{1}{\lambda-
K_{T_{\mathrm{c}}}(p)}\,(\widehat{V}\ast\widehat{\alpha})(p).$ (A.23)

We argue as in (A.22) to see that the $L^{2}(\mathbb{R}^{3})$-norm of
$\nabla_{p}p\frac{\lambda}{K_{T_{\mathrm{c}}}(p)}\frac{1}{\lambda-
K_{T_{\mathrm{c}}}(p)}(\widehat{V}\ast\widehat{\alpha})(p)$ is bounded by a
constant times $\|\,|\cdot|V\|_{\infty}$. To treat the other term, we go back
to position space and note that

$\displaystyle\Bigl{(}\int_{\mathbb{R}^{3}}\mathrm{d}x\left|\int_{\mathbb{R}^{3}}\mathrm{d}y\;|x|\,|\nabla\mathcal{G}^{0}(x-y)|\,|V\alpha(y)|\right|^{2}\Bigr{)}^{\nicefrac{{1}}{{2}}}\leqslant\|(1+|\cdot|)\nabla\mathcal{G}^{0}\|_{1}\,\|(1+|\cdot|)V\|_{\infty}.$
(A.24)

In combination with Lemma A.3, these considerations prove the claim. ∎

The last lemma provides us with a convenient representation for the operator
$K_{T,\mathbf{A}}$. Its proof can be found in [16, Lemma 6.4]. Before we can
state the lemma we need the following definition.

######  Definition A.6 (Speaker path).

Let $R>0$, assume that $\mu\geqslant-1$ and define the following complex paths

$\displaystyle\begin{split}u_{1}(t)&\coloneqq\frac{\pi\mathrm{i}}{2{\beta_{\mathrm{c}}}}+(1+\mathrm{i})t\\\
u_{2}(t)&\coloneqq\frac{\pi\mathrm{i}}{2{\beta_{\mathrm{c}}}}-(\mu+1)t\\\
u_{3}(t)&\coloneqq-\frac{\pi\mathrm{i}}{2{\beta_{\mathrm{c}}}}t-(\mu+1)\\\
u_{4}(t)&\coloneqq-\frac{\pi\mathrm{i}}{2{\beta_{\mathrm{c}}}}-(\mu+1)(1-t)\\\
u_{5}(t)&\coloneqq-\frac{\pi\mathrm{i}}{2{\beta_{\mathrm{c}}}}+(1-\mathrm{i})t\end{split}$
$\displaystyle\begin{split}\phantom{\frac{\mathrm{i}}{{\beta_{\mathrm{c}}}}}t&\in[0,R],\\\
\phantom{\frac{\mathrm{i}}{{\beta_{\mathrm{c}}}}}t&\in[0,1],\\\
\phantom{\frac{\mathrm{i}}{{\beta_{\mathrm{c}}}}}t&\in[-1,1],\\\
\phantom{\frac{\mathrm{i}}{{\beta_{\mathrm{c}}}}}t&\in[0,1],\\\
\phantom{\frac{\mathrm{i}}{{\beta_{\mathrm{c}}}}}t&\in[0,R].\end{split}$

The speaker path is defined as the union of paths $u_{i}$, $i=1,\ldots,5$,
with $u_{1}$ taken in reverse direction, i.e.,

$\displaystyle{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}_{R}\coloneqq\mathop{\dot{-}}u_{1}\mathop{\dot{+}}u_{2}\mathop{\dot{+}}u_{3}\mathop{\dot{+}}u_{4}\mathop{\dot{+}}u_{5}.$

If $\mu<-1$ we choose the same path as in the case $\mu=-1$.

######  Lemma A.7.

Let $H\colon\mathcal{D}(H)\rightarrow\mathcal{H}$ be a self-adjoint operator
on a separable Hilbert space $\mathcal{H}$ with $H\geqslant-\mu$ and let
$\beta>0$. Then, we have

$\displaystyle\frac{H}{\tanh(\frac{\beta}{2}H)}=H+\lim_{R\to\infty}\int_{{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}_{R}}\frac{\mathrm{d}z}{2\pi\mathrm{i}}\Bigl{(}\frac{z}{\tanh(\frac{\beta}{2}z)}-z\Bigr{)}\frac{1}{z-H},$
(A.25)

with the speaker path ${\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip 0.7pt}}_{R}$
in Definition A.6. The above integral including the limit is understood as an
improper Riemann integral with respect to the uniform operator topology.

In the following, we use the symbol $\int_{{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip 0.7pt}}}$ to
denote the integral on the right side of (A.25) including the limit and we
denote ${\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}=\bigcup_{R>0}{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip 0.7pt}}_{R}$.

###  A.2 Proof of Proposition A.1

The proof of Proposition A.1 is based on an adaption of gauge-invariant
perturbation theory for Schrödinger operators as introduced in [42] to our
setting. The core of the argument is contained in the following lemma.

######  Lemma A.8.

Assume that $(1+|\cdot|^{2})V\in L^{2}(\mathbb{R}^{3})\cap
L^{\infty}(\mathbb{R}^{3})$. There is a continuous function
$a\colon\rho(K_{T_{\mathrm{c}}}-V)\to\mathbb{R}_{+}$ such that the following
holds: For every compact set $K\subset\rho(K_{T_{\mathrm{c}}}-V)$ there is a
constant $h(K)>0$ such that for $0<h<h(K)$ we have

$\displaystyle\mathcal{R}_{\mathbf{A}}^{z,V}=\mathcal{S}_{\mathbf{A}_{h}}^{z,V}+h^{2}\,\eta_{h}(z)\quad\text{
with }\quad\|(1+\pi^{2})\eta_{h}(z)\|_{\infty}\leqslant a(z).$ (A.26)

Here $\mathcal{S}_{\mathbf{A}}^{z,V}$ denotes the operator defined by the
kernel

$\displaystyle\mathcal{S}_{\mathbf{A}}^{z,V}(x,y)$
$\displaystyle\coloneqq\mathrm{e}^{\mathrm{i}\Phi_{\mathbf{A}}(x,y)}\,\mathcal{G}^{z,V}(x,y)$
(A.27)

with the phase factor $\Phi_{\mathbf{A}}(x,y)$ in (4.2).

###### Proof.

We employ (4.18) and apply the integral representation in Lemma A.7 to
$K_{{T_{\mathrm{c}}},\mathbf{A}}$ to see that

$\displaystyle
K_{{T_{\mathrm{c}}},\mathbf{A}}^{x}\,\mathrm{e}^{\mathrm{i}\Phi_{\mathbf{A}_{h}}(x,y)}=\mathrm{e}^{\mathrm{i}\Phi_{\mathbf{A}_{h}}(x,y)}\,K_{{T_{\mathrm{c}}},\mathbf{A}_{y}}^{x}$
(A.28)

holds. Here $K_{{T_{\mathrm{c}}},\mathbf{A}}^{x}$ is understood to act on the
$x$-coordinate and $\mathbf{A}_{y}(x)\coloneqq\widetilde{\mathbf{A}}(x,y)$
denotes the vector potential in transversal Poincaré gauge relative to the
point $y$, defined in (4.12). Using Lemma A.7 again, we also find the identity

$\displaystyle K_{{T_{\mathrm{c}}},\mathbf{A}_{y}}-K_{T_{\mathrm{c}}}$
$\displaystyle=\left(-2\mathrm{i}(\mathbf{A}_{h})_{y}(x)\nabla_{x}-\mathrm{i}\operatorname{div}(\mathbf{A}_{h})_{y}(x)+|(\mathbf{A}_{h})_{y}(x)|^{2}\right)$
(A.29) $\displaystyle\hskip 20.0pt+\int_{{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}}\frac{\mathrm{d}z}{2\pi\mathrm{i}}\;\varphi(z)\;\frac{1}{z+(-\mathrm{i}\nabla+(\mathbf{A}_{h})_{y})^{2}+\mu}$
$\displaystyle\hskip
40.0pt\times\left(-2\mathrm{i}(\mathbf{A}_{h})_{y}(x)\nabla_{x}-\mathrm{i}\operatorname{div}(\mathbf{A}_{h})_{y}(x)+|(\mathbf{A}_{h})_{y}(x)|^{2}\right)\frac{1}{z+\Delta+\mu},$

where $\varphi(z)=(z/\tanh(z/(2{T_{\mathrm{c}}}))-z)$. Let us define the
operator $\mathcal{T}_{\mathbf{A}}^{z,V}$ via the equation

$\displaystyle\left(z-(K_{{T_{\mathrm{c}}},\mathbf{A}}-V)\right)\mathcal{S}_{\mathbf{A}_{h}}^{z,V}=1-\mathcal{T}_{\mathbf{A}_{h}}^{z,V}.$
(A.30)

Using (A.28), (A.29), and (A.30), we write the integral kernel of
$\mathcal{T}_{\mathbf{A}}^{z,V}$ as

$\displaystyle\mathcal{T}_{\mathbf{A}}^{z,V}(x,y)$
$\displaystyle=\mathrm{e}^{\mathrm{i}\Phi_{\mathbf{A}}(x,y)}\Big{[}\left(-2\mathrm{i}\mathbf{A}_{y}(x)\nabla_{x}-\mathrm{i}\operatorname{div}\mathbf{A}_{y}(x)+|\mathbf{A}_{y}(x)|^{2}\right)\mathcal{G}^{z,V}(x,y)$
$\displaystyle\hskip 20.0pt+\int_{{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}}\frac{\mathrm{d}\zeta}{2\pi\mathrm{i}}\;\varphi(\zeta)\int_{\mathbb{R}^{6}}\mathrm{d}v\mathrm{d}w\;\frac{1}{\zeta-(-\mathrm{i}\nabla+\mathbf{A}_{y})^{2}+\mu}(x,v)$
$\displaystyle\hskip
30.0pt\times\left(-2\mathrm{i}\mathbf{A}_{y}(v)\nabla_{v}-\mathrm{i}\operatorname{div}\mathbf{A}_{y}(v)+|\mathbf{A}_{y}(v)|^{2}\right)g_{0}^{\zeta}(v-w)\,\mathcal{G}^{z,V}(v,y)\Big{]}.$
(A.31)

In the next step we use this formula to prove a bound for the operator norm of
$\mathcal{T}_{\mathbf{A}}^{z,V}$.

Let us denote the first and the second term on the right side of (A.31) by
$\mathcal{T}_{\mathbf{A}}^{(1)}(x,y)$ and
$\mathcal{T}_{\mathbf{A}}^{(2)}(x,y)$, respectively. Using (4.24) and (4.25),
we see that

$\displaystyle|\mathcal{T}_{\mathbf{A}_{h}}^{(1)}(x,y)|\leqslant
Ch^{2}\bigl{(}|x-y|\,|\nabla_{x}\mathcal{G}^{z,V}(x,y)|+\bigl{(}|x-y|+|x-y|^{2}\bigr{)}|\mathcal{G}^{z,V}(x,y)|\bigr{)}.$
(A.32)

Using the resolvent identity
$\mathcal{R}^{z,V}=\mathcal{R}^{z}+\mathcal{R}^{z}V\mathcal{R}^{z,V}$, we
estimate the first term on the right side of (A.32) by

$\displaystyle\left|x-y\right|\left|\nabla_{x}\mathcal{G}^{z,V}(x,y)\right|\leqslant$
$\displaystyle\left|x-y\right|\left|\nabla\mathcal{G}^{z}(x-y)\right|$
$\displaystyle+\int_{\mathbb{R}^{3}}\mathrm{d}w\
\left||x-w|\,\nabla\mathcal{G}^{z}(x-w)\,V(w)\,\mathcal{G}^{z,V}(w,y)\right|$
$\displaystyle+\int_{\mathbb{R}^{3}}\mathrm{d}w\
\left|\nabla\mathcal{G}^{z}(x-w)\,V(w)\,|w-y|\,\mathcal{G}^{z,V}(w,y)\right|.$
(A.33)

Eq. (A.33) allows us to obtain the following bound for the operator norm of
$\mathcal{T}_{\mathbf{A}}^{(1)}$:

$\displaystyle\|\mathcal{T}_{\mathbf{A}_{h}}^{(1)}\|_{\infty}$
$\displaystyle\leqslant
Ch^{2}\big{[}\left\||\cdot|\nabla\mathcal{G}^{z}\right\|_{1}\left(1+\left\|V\right\|_{2}\|\mathcal{R}^{z,V}\|_{2;\infty}\right)$
$\displaystyle\hskip
20.0pt+\left(1+\|\nabla\mathcal{G}^{z}\|_{1}\left\|V\right\|_{2}\right)\|[x,\mathcal{R}^{z,V}]\|_{2;\infty}+\|[x,[x,\mathcal{R}^{z,V}]]\|_{2;\infty}\big{]}.$
(A.34)

From Lemma A.3 and A.4, we know that there is a continuous
$a\colon\rho(K_{T_{\mathrm{c}}}-V)\to\mathbb{R}_{+}$ such that the right side
of (A.34) is bounded by $a(z)$. In the following we will denote by $a(z)$ a
generic function with these properties whose precise form may change from line
to line.

To obtain a bound for the operator norm of
$\mathcal{T}_{\mathbf{A}}^{(2)}(z)$, we first estimate its kernel by

$\displaystyle|\mathcal{T}_{\mathbf{A}_{h}}^{(2)}(x,y)|\leqslant
Ch^{2}\Bigl{(}\int_{{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}}\text{d}|\zeta|\;\left|\varphi(\zeta)\right|\Bigr{)}\sup_{\zeta\in{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}}\int_{\mathbb{R}^{6}}\mathrm{d}v\mathrm{d}w\;\Bigl{|}\frac{1}{\zeta-(-\mathrm{i}\nabla+(\mathbf{A}_{h})_{y})^{2}+\mu}(x,v)\Bigr{|}$
$\displaystyle\hskip 14.22636pt\times\big{[}|v-y|\ |\nabla
g_{0}^{\zeta}(v-w)|+\left(|v-y|+|v-y|^{2}\right)|g_{0}^{\zeta}(v-w)|\big{]}|\mathcal{G}^{z,V}(w,y)|.$
(A.35)

From Lemma 4.2 we know that the absolute value of the resolvent kernel of the
magnetic Laplacian is bounded from above by a function only depending on
$x-v$, whose $L^{1}(\mathbb{R}^{3})$-norm is bounded by a constant times
$f(\operatorname{\mathrm{Re}}\zeta,\operatorname{\mathrm{Im}}\zeta)$ with $f$
in (4.4). This, in particular, implies that this $L^{1}(\mathbb{R}^{3})$-norm
is uniformly bounded in $\zeta\in{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip 0.7pt}}$ and
$h$ as long as the latter is small enough, compare this to the bound in [16,
Eq. (6.19)]. We use this bound, $|v-y|\leqslant|v-w|+|w-y|$, and the resolvent
identity for $\mathcal{R}^{z,V}$ to bound the operator norm of
$\mathcal{T}_{\mathbf{A}}^{(2)}$ by

$\displaystyle\|\mathcal{T}_{\mathbf{A}_{h}}^{(2)}\|_{\infty}$
$\displaystyle\leqslant h^{2}C\big{(}\sup_{\zeta\in{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}}\|(1+|\cdot|)\nabla g_{0}^{\zeta}\|_{1}\;,\;\sup_{\zeta\in{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}}\|(1+|\cdot|^{2})g_{0}^{\zeta}\|_{1}\;,\;\|(1+|\cdot|^{2})\mathcal{G}^{z}\|_{1}\;,\;$
$\displaystyle\hskip
50.0pt\|V\|_{2}\;,\;\|\mathcal{R}^{z,V}\|_{2;\infty}\;,\;\|[x,\mathcal{R}^{z,V}]\|_{2,\infty}\;,\;\|[x,[x,\mathcal{R}^{z,V}]]\|_{2,\infty}\big{)}.$
(A.36)

The constant on the right side is an affine function of each of its arguments.
From Lemma 4.4 we know that the norms involving $g_{0}^{\zeta}$ are finite. In
combination with Lemmas A.3 and A.4, this implies that the right side of
(A.36) is bounded by $a(z)$. We conclude that

$\displaystyle\|\mathcal{T}_{\mathbf{A}_{h}}^{z,V}\|_{\infty}\leqslant
a(z)\,h^{2}$ (A.37)

holds.

Let $K\subset\rho(K_{T_{\mathrm{c}}}-V)$ be compact. The above bounds allow us
to find $h_{0}(K)>0$ such that for $z\in K$ and as long as $0<h<h_{0}(K)$ we
can write the resolvent of $K_{{T_{\mathrm{c}}},\mathbf{A}}-V$ as

$\displaystyle\frac{1}{z-(K_{{T_{\mathrm{c}}},\mathbf{A}}-V)}=\mathcal{S}_{\mathbf{A}_{h}}^{z,V}+h^{2}\,\eta_{h}(z)\quad\text{
with }\quad\eta_{h}(z)\coloneqq
h^{-2}\mathcal{S}_{\mathbf{A}_{h}}^{z,V}\sum_{n=1}^{\infty}(\mathcal{T}_{\mathbf{A}_{h}}^{z,V})^{n}.$
(A.38)

To show that the operator norm of $(1+\pi^{2})\eta_{h}(z)$ is bounded by
$a(z)$, we use

$\displaystyle\|(1+\pi^{2})\eta_{h}(z)\|_{\infty}\leqslant\|(1+\pi^{2})\mathcal{S}_{\mathbf{A}_{h}}^{z,V}\|_{\infty}\;\sum_{n=1}^{\infty}h^{2(n-1)}a(z)^{n}.$
(A.39)

With the resolvent identity for $\mathcal{R}^{z,V}$ and Lemma A.3, we easily
see that the operator norm of $(1+\pi^{2})\mathcal{S}_{\mathbf{A}_{h}}^{z,V}$
is bounded by $a(z)$. This proves the claim. ∎

With the resolvent estimates in Lemma A.8 at hand, we turn to the proof of
Proposition A.1. Let $\lambda<2{T_{\mathrm{c}}}$ be an eigenvalue of the
operator $K_{T_{\mathrm{c}}}-V$. Our assumption on $V$ guarantees that it has
finite multiplicity $m\in\mathbb{N}$. We choose $\varepsilon>0$ such that the
ball $B_{\varepsilon}(\lambda)\subseteq\mathbb{C}$ contains no other point of
the spectrum of $K_{T_{\mathrm{c}}}-V$ than $\lambda$ and define

$\displaystyle P(h)\coloneqq\int_{\partial
B_{\varepsilon}(\lambda)}\frac{\mathrm{d}z}{2\pi\mathrm{i}}\;\mathcal{R}_{\mathbf{A}}^{z,V}=\int_{\partial
B_{\varepsilon}(\lambda)}\frac{\mathrm{d}z}{2\pi\mathrm{i}}\;\mathcal{S}_{\mathbf{A}_{h}}^{z,V}+h^{2}\int_{\partial
B_{\varepsilon}(\lambda)}\frac{\mathrm{d}z}{2\pi\mathrm{i}}\;\eta_{h}(z).$
(A.40)

From Lemma A.8 we know that the operator norm of the second term on the right
side is bounded by a constant times $h^{2}$ provided $h$ is small enough. The
integral kernel of the first term is given by

$\displaystyle\Bigl{(}\int_{\partial
B_{\varepsilon}(\lambda)}\frac{\mathrm{d}z}{2\pi\mathrm{i}}\;\mathcal{S}_{\mathbf{A}}^{z,V}\Bigr{)}(x,y)=\mathrm{e}^{\mathrm{i}\Phi_{\mathbf{A}}(x,y)}\sum_{i=1}^{m}u_{i}(x)\overline{u_{i}(y)},$
(A.41)

where the vectors $\\{u_{i}\\}_{i=1}^{m}$ span the eigenspace of $\lambda$.
Let us denote by $P$ the projection onto that linear space. Using (A.40),
(A.41), and $|\Phi_{\mathbf{A}_{h}}(x,y)|\leqslant Ch^{2}(|x|^{2}+|y|^{2})$,
which follows from the assumption $A(0)=0$, we obtain the bound

$\displaystyle\|P(h)-P\|_{\infty}\leqslant
Ch^{2}\,\max_{i=1,\ldots,m}\|\,|\cdot|^{2}u_{i}\|_{2}.$ (A.42)

In combination with Lemma A.5, this proves $\operatorname{rank}P(h)=m$ for $h$
small enough as well as the second bound in (A.1).

To prove the bounds for the eigenvalues we use the identity

$\displaystyle(K_{{T_{\mathrm{c}}},\mathbf{A}}-V)P(h)=\int_{\partial
B_{\varepsilon}(\lambda)}\frac{\mathrm{d}z}{2\pi\mathrm{i}}\;z\,\mathcal{R}_{\mathbf{A}}^{z,V}.$
(A.43)

As long as $h$ is small enough, the rank of this operator equals $m$ and its
eigenvalues are given by $\lambda_{1}(h),\ldots,\lambda_{m}(h)$. Similar
arguments to the above for the spectral projections allow us to conclude that

$\displaystyle\|(K_{{T_{\mathrm{c}}},\mathbf{A}}-V)P(h)-(K_{T_{\mathrm{c}}}-V)P\|_{\infty}\leqslant
Ch^{2}\,\max_{i=1,\ldots,m}\|\,|\cdot|^{2}u_{i}\|_{2}$ (A.44)

holds. This proves the claimed bound for the eigenvalues. It remains to prove
(A.2).

Let us write $\alpha_{*}^{h}=a(h)\alpha_{*}+b(h)\phi_{h}$ with
$\langle\alpha_{*},\phi_{h}\rangle=0$ and $|a(h)|^{2}+|b(h)|^{2}=1$. Our
assumptions imply $a(h)=\langle\alpha_{*},\alpha_{*}^{h}\rangle\geqslant 0$.
We rewrite the equation $P(h)\alpha_{*}^{h}=\alpha_{*}^{h}$ to see that
$b(h)\phi_{h}=(P(h)-P)\alpha_{*}^{h}$. An application of (A.1) thus implies
$|b(h)|\leqslant Ch^{2}$. Using this, $|a(h)|^{2}+|b(h)|^{2}=1$, and the fact
that $a(h)\geqslant 0$, we see that $|a(h)-1|\leqslant Ch^{2}$. This allows us
to conclude that

$\displaystyle\|(1+\pi^{2})(\alpha_{*}^{h}-\alpha_{*})\|_{2}\leqslant|a(h)-1|\
\|(1+\pi^{2})\alpha_{*}\|_{2}+|b(h)|\ \|(1+\pi^{2})\phi_{h}\|_{2}\leqslant
Ch^{2}.$ (A.45)

To obtain the result, we used Lemma A.5 to see that
$\|(1+\pi^{2})\alpha_{*}\|_{2}<\infty$, as well as
$\|(1+\pi^{2})\phi_{h}\|_{2}\leqslant\|(1+\pi^{2})\alpha_{*}\|_{2}+\|(1+\pi^{2})\alpha_{*}^{h}\|_{2}$,
and

$\displaystyle\|(1+\pi^{2})\alpha_{*}^{h}\|_{2}\leqslant
C\|K_{{T_{\mathrm{c}}},\mathbf{A}}\alpha_{*}^{h}\|_{2}\leqslant
C\left(|\lambda(h)|+\|V\alpha_{*}^{h}\|_{2}\right)\leqslant
C\left(|\lambda(h)|+\|V\|_{\infty}\right).$ (A.46)

This proves (A.2) and also finishes the proof of Proposition A.1.

Acknowledgements

A. D. gratefully acknowledges funding from the Swiss National Science
Foundation through the Ambizione grant PZ00P2 185851.

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(Christian Hainzl) Mathematisches Institut der Universität München

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