by direct computation, we see
$\partial_{h}^{K}({\mathscr{G}}^{\theta}-\sqrt{h}z)\in S^{K-\frac{1}{2}}$
while $\partial_{h}^{K}R_{n}^{\pm}=0$, for $K>0$.

Then consider $\mathcal{E}_{n}=\mathcal{P}_{n}^{-1}$. First of all, by the
proof of Lemma 7.6, we have $\mathcal{E}_{n}(x,D_{x_{1}},\xi_{2})\in
S({\mathbb{R}}_{x_{2},\xi_{2}}^{2};\mathcal{L}(B_{x_{1}}^{k}\times{\mathbb{C}}^{2};B_{x_{1}}^{k+1}\times{\mathbb{C}}^{2}))$.
By differentiating
$\mathcal{E}_{n}=\mathcal{E}_{n}\\#\mathcal{P}_{n}\\#\mathcal{E}_{n}$ w.r.t.
$h$ and using (7.36) and (7.37), we have

$\partial_{h}\mathcal{E}_{n}=-\mathcal{E}_{n}\\#\partial_{h}\mathcal{P}_{n}\\#\mathcal{E}_{n}+\sum\limits_{|\alpha|=|\beta|=1}C_{\alpha,\beta}\left(\partial_{x_{2},\xi_{2}}^{\alpha}\mathcal{E}_{n}\\#\partial_{x_{2},\xi_{2}}^{\beta}\mathcal{P}_{n}\\#\mathcal{E}_{n}\right).$
(7.40)

Since $\partial_{h}\mathcal{P}_{n}\in S^{\frac{1}{2}}$, thus
$\partial_{h}\mathcal{E}_{n}\in S^{\frac{1}{2}}$ above. By differentiating
(7.40) w.r.t. $h$ and using (7.36) and (7.37), we see that
$\partial_{h}^{2}\mathcal{E}_{n}\in S^{\frac{3}{2}}$. An iterative argument
shows that $\partial_{h}^{K}\mathcal{E}_{n}\in S^{K-\frac{1}{2}}$. In
particular, $\partial_{h}^{K}E_{n,\pm}\in S^{K-\frac{1}{2}}$. Furthermore, by
differentiating $E_{n,\pm}^{-1}=E_{n,\pm}^{-1}\\#E_{n,\pm}\\#E_{n,\pm}^{-1}$
w.r.t. $h$ and using (7.37) and (7.36), we have

$\begin{split}\partial_{h}E_{n,\pm}^{-1}=-E_{n,\pm}^{-1}\\#\partial_{h}E_{n,\pm}\\#E_{n,\pm}^{-1}-\sum\limits_{|\alpha|=|\beta|=1}C_{\alpha,\beta}\partial_{x_{2},\xi_{2}}^{\alpha}E_{n,\pm}^{-1}\\#\partial_{x_{2},\xi_{2}}^{\beta}E_{n,\pm}\\#E_{n,\pm}^{-1}.\end{split}$
(7.41)

When $|\operatorname{Im}z|\geq h^{\delta}$, by (7.23) and [Zw12, Theorem
4.23(ii)], we see that

$\|\partial_{h}E_{n,\pm}^{-1}\|=\mathcal{O}(h^{-\frac{3}{2}}|\operatorname{Im}z|^{-2})+\mathcal{O}(h^{-\frac{1}{2}}|\operatorname{Im}z|^{-3})=\mathcal{O}(h^{-\frac{3}{2}}|\operatorname{Im}z|^{-2}).$

Furthermore, since $[D_{x_{j}},A^{W}]=(D_{x_{j}}A)^{W}$ and
$-[x_{j},A^{W}]=(hD_{\xi_{j}}A)^{W}$, we see that

$\begin{split}\|\operatorname{ad}_{L_{j_{1}}}\circ\cdots\operatorname{ad}_{L_{j_{N}}}(\partial_{h}E_{n,\pm}^{-1})^{W}\|=\mathcal{O}\left(\frac{h^{-\frac{3}{2}}}{|\operatorname{Im}z|^{2}}\
\frac{h^{N}}{|\operatorname{Im}z|^{N}}\right).\end{split}$

By [DS99, Prop. 8.4], we get

$\|\partial_{x_{2}}^{\alpha}\partial_{\xi_{2}}^{\beta}\partial_{h}E_{n,\pm}^{-1}(x_{2},\xi_{2};z,h)\|_{{\mathbb{C}}_{2\times
2}}\leq
C_{\alpha,\beta}\max\left(1,\frac{h^{\frac{3}{2}}}{|\operatorname{Im}z|^{3}}\right)h^{-\frac{3}{2}}|\operatorname{Im}z|^{-2-|\alpha|-|\beta|}.$
(7.42)

Iterating this process by taking $\partial_{h}$ of (7.41), expanding it and
using (7.36), (7.37) and (7.42), we see that every time we differentiate, we
derive an extra order of $1/(h|\operatorname{Im}z|^{2})$. Thus we obtain
(7.38) for $K>0$. Then

$\partial_{h}r_{n}=\partial_{h}\partial_{z}E_{n,\pm}\\#E_{n,\pm}^{-1}+\partial_{z}E_{n,\pm}\\#\partial_{h}E_{n,\pm}^{-1}+\sum\limits_{|\alpha|=|\beta|=1}C_{\alpha,\beta}(\partial_{x_{2},\xi_{2}}^{\alpha}\partial_{z}E_{n,\pm})\\#(\partial_{x_{2},\xi_{2}}^{\beta}E_{n,\pm}^{-1}).$

By (7.23), (7.38) and [Zw12, Theorem 4.23(ii)], we see that
$\|\partial_{h}r_{n}^{W}\|=\mathcal{O}(h^{-1}|\operatorname{Im}z|^{-2})$. By
the same argument as for $E_{n,\pm}^{-1}$, we get (7.39). ∎

We shall now focus on $K=1$, for simplicity, and study the asymptotic
expansion of $\partial_{h}r_{n}$.

######  Lemma 7.10 (Asymptotic expansion).

Let $0<\delta<1/2$ and $|\operatorname{Im}z|\geq h^{\delta}$, then
$\partial_{h}r_{n}$ has an asymptotic expansion in $S_{\delta}^{1+2\delta}$:

$\partial_{h}r_{n}\sim\sum\limits_{j=1}^{\infty}\ \frac{j}{2}\
h^{\frac{j}{2}-1}r_{n,j}=:\sum\limits_{j=1}^{\infty}h^{\frac{j}{2}-1}q_{n,j},\text{~{}where~{}}r_{n,j}\text{~{}are~{}given~{}in~{}Lemma~{}\ref{lemma:
explicit_asymptotic_expansion_needed}.}$ (7.43)

Then $h^{\frac{j}{2}-1}q_{n,j}\in S_{\delta}^{(j+1)\delta+1-\frac{j}{2}}$. Let
$Q_{n,J}:=\partial_{h}r_{n}-\sum\limits_{j=1}^{J-1}h^{\frac{j}{2}-1}q_{n,j}\in
S_{\delta}^{(J+1)\delta+1-\frac{J}{2}}$, i.e., $\text{ for all
}\alpha,\beta>0$, there is $C_{\alpha,\beta,n}^{\prime\prime}$ s.t.

$|\partial_{x_{2},}^{\alpha}\partial_{\xi_{2}}^{\beta}Q_{n,J}|\leq
C_{\alpha,\beta,n}^{\prime\prime}h^{\frac{J}{2}-1-(J+1)\delta-\delta(|\alpha|+|\beta|)}.$
(7.44)

Furthermore, for the expansion of
$\operatorname{Tr}_{{\mathbb{C}}^{2}}(\partial_{h}r_{n})$, we have for
$\eta=x_{2}+i\xi_{2}$,

$\begin{split}\operatorname{Chiral}\
{\mathscr{H}}_{{\operatorname{c}},n}(J=3):&\operatorname{Tr}_{{\mathbb{C}}^{2}}(h^{-\frac{1}{2}}q_{n,1}+q_{n,2})=\frac{\lambda_{n}^{2}}{z^{3}}\mathfrak{U}(\eta),\\\
\operatorname{Anti-Chiral}\
{\mathscr{H}}_{{\operatorname{ac}},n}^{\theta}(J=2):&\operatorname{Tr}_{{\mathbb{C}}^{2}}(h^{-\frac{1}{2}}q_{n,1})=\frac{s_{n}^{2}(z^{2}+c_{n}^{2})}{(z^{2}-c_{n}^{2})^{2}\sqrt{h}}.\end{split}$
(7.45)

We will prove that the termwise differentiation of the asymptotic expansion of
$r_{n}$ in (7.25) is indeed an asymptotic expansion of $\partial_{h}r_{n}$ in
$S_{\delta}^{2\delta+1}$.

###### Proof.

Let $g=\sqrt{h}$ and consider
$r_{n}\sim\sum\limits_{j=0}^{\infty}g^{j}r_{n,j}$. By Borel’s theorem (c.f.
[Zw12, Theorem 4.15]) with a varification of $g$ smoothness (c.f. [Ho03,
Theorem 1.2.6]), we see that for such $r_{n,j}\in
C^{\infty}({\mathbb{R}}_{x_{2},\xi_{2}}^{2})$, there is $\tilde{r}_{n}\in
C^{\infty}({\mathbb{R}}_{g}^{+}\times{\mathbb{R}}_{x_{2},\xi_{2}})$ s.t.
$\tilde{r}_{n}=\sum\limits_{j=0}^{\infty}g^{j}r_{n,j}$. Thus

$\partial_{g}\tilde{r}_{n}=\sum\limits_{j=1}^{\infty}jg^{j-1}r_{n,j}.$ (7.46)

On the other hand, by uniqueness in Borel’s theorem, we see that
$\tilde{r}_{n}-r_{n}=\mathcal{O}(h^{\infty})$. Thus
$\partial_{g}\tilde{r}_{n}-\partial_{g}r_{n}=\mathcal{O}(g^{\infty})$. Thus
(7.46) is also an asymptotic expansion of $\partial_{g}r_{n}$. Furthermore,
since $\partial_{h}r_{n}=\frac{1}{2\sqrt{h}}\partial_{g}r_{n}$, thus we proved
$\partial_{h}r_{n}$ has the following asymptotic expansion in
$S_{\delta}^{1+2\delta}$:

$\partial_{h}r_{n}\sim\frac{1}{2\sqrt{h}}\sum\limits_{j=1}^{\infty}jh^{\frac{j-1}{2}}r_{n,j}=\sum\limits_{j=1}^{\infty}\
\frac{j}{2}h^{\frac{j}{2}-1}r_{n,j}.$

The rest of the Lemma follows from Lemma 7.7. ∎

###### Proof of Prop. 3.1 .

Recall that $f_{0}(z)=f(z+\sqrt{2|n|/h})$ also depends on $h$. By
differentiating (7.29) w.r.t. $h$, we get

$\begin{split}\partial_{h}&\tilde{\operatorname{Tr}}(f({\mathscr{H}}^{\theta}_{c,n}))=\frac{i}{4\pi^{2}h^{2}|E|}\int_{{\mathbb{C}}}\int_{E}\partial_{\bar{z}}\tilde{f}_{0}(z)\operatorname{Tr}_{{\mathbb{C}}^{2}}(r_{n})\
dx_{2}\ d\xi_{2}~{}dz\wedge d\bar{z}\\\
&+\frac{i\sqrt{2|n|/h}}{8\pi^{2}h^{2}|E|}\int_{{\mathbb{C}}}\int_{E}\partial_{\bar{z}}\tilde{f^{\prime}}_{0}(z+\sqrt{2n/h})\operatorname{Tr}_{{\mathbb{C}}^{2}}(r_{n})\
dx_{2}\ d\xi_{2}~{}dz\wedge d\bar{z}\\\
&-\frac{i}{4\pi^{2}h|E|}\int_{{\mathbb{C}}}\int_{E}\partial_{\bar{z}}\tilde{f}_{0}(z)\operatorname{Tr}_{{\mathbb{C}}^{2}}(\partial_{h}r_{n})\
dx_{2}\ d\xi_{2}~{}dz\wedge d\bar{z}:=B_{1}+B_{2}-B_{3}.\end{split}$ (7.47)

The asymptotic expansion of
$B_{1}=-\frac{1}{h}\tilde{\operatorname{Tr}}(f({\mathscr{H}}^{\theta}))$ and
$B_{2}=-\sqrt{\frac{|n|}{2h^{3}}}\tilde{\operatorname{Tr}}(f^{\prime}({\mathscr{H}}^{\theta}))$
follows directly from (7.34). While $B_{3}$ can be computed by splitting the
integral as in Subsection 7.5:

$\begin{split}\left[\int_{{\mathbb{C}}}\partial_{\bar{z}}\tilde{f}_{0}\operatorname{Tr}_{{\mathbb{C}}^{2}}(\partial_{h}r_{n})dz\wedge
d\bar{z}\right](x_{2},\xi_{2};h)=&\int_{{\mathbb{C}}}\partial_{\bar{z}}\tilde{f}_{0}\sum\limits_{j=1}^{J-1}h^{\frac{j}{2}-1}\operatorname{Tr}_{{\mathbb{C}}^{2}}(q_{n,j})dz\wedge
d\bar{z}\\\ +&\int_{|\operatorname{Im}z|\geq
h^{\delta}}\partial_{\bar{z}}\tilde{f}_{0}\operatorname{Tr}_{{\mathbb{C}}^{2}}(Q_{n,J})dz\wedge
d\bar{z}\\\ +&\int_{|\operatorname{Im}z|\leq
h^{\delta}}\partial_{\bar{z}}\tilde{f}_{0}\operatorname{Tr}_{{\mathbb{C}}^{2}}(Q_{n,J})dz\wedge
d\bar{z}:=A_{1}^{\prime}+A_{2}^{\prime}+A_{3}^{\prime},\end{split}$

and we imitate the estimates of $A_{1}$, $A_{2}$, $A_{3}$ in the Subsection
7.5 with $\partial_{h}r_{n}$ instead of $r_{n}$, and we use Lemma 7.9 and 7.10
instead of Lemma 7.6 and 7.7. In short, we need (7.32) and (7.45) for
$A_{1}^{\prime}$, (7.44) for $A_{2}^{\prime}$, (7.39) and Lemma 7.10 for
$A_{3}^{\prime}$ and we derive that

$\begin{cases}A_{1,{\operatorname{c}}}^{\prime}=\pi
if^{\prime\prime}(\lambda_{n,B})\lambda_{n}^{2}\mathfrak{U}(\eta),\quad
A_{1,{\operatorname{ac}}}^{\prime}=\frac{\pi
i}{\sqrt{h}}\left(s_{n}^{2}f^{\prime}(\lambda_{n,B}-c_{n})+s_{n}^{2}f^{\prime}(\lambda_{n,B}+c_{n})\right),\\\
|A_{2}^{\prime}|\leq
C_{n,K,f,{\mathscr{V}}}h^{\frac{J}{2}-1-(J+1)\delta},|A_{3}^{\prime}|\leq
C_{n,K,f,{\mathscr{V}}}h^{(K-2)\delta-1},\end{cases}$

from which we can find $B_{3}$. And we summarize $B_{1}$, $B_{2}$, $B_{3}$
below:

For the chiral model where $J=3$, we have

$\begin{split}&B_{1,{\operatorname{c}}}=-\frac{1}{\pi
h^{2}}f(\lambda_{n,B})-\frac{|n|}{2\pi
h}\operatorname{Ave}(\mathfrak{U})f^{\prime\prime}(\lambda_{n,B})-\mathcal{O}_{n,K,f,{\mathscr{V}}}(h^{-\frac{1}{2}-4\delta}+h^{(K-1)\delta-2}),\\\
&B_{2,{\operatorname{c}}}=-\frac{\sqrt{2|n|}}{2\pi
h^{\frac{5}{2}}}f^{\prime}(\lambda_{n,B})-\frac{(2|n|)^{\frac{3}{2}}}{8\pi
h^{\frac{3}{2}}}\operatorname{Ave}(\mathfrak{U})f^{\prime\prime\prime}(\lambda_{n,B})-\mathcal{O}_{n,K,f,{\mathscr{V}}}(h^{-1-4\delta}+h^{(K-1)\delta-\frac{5}{2}})\\\
&B_{3,{\operatorname{c}}}=\frac{|n|}{2\pi
h}f^{\prime\prime}(\lambda_{n,B})\operatorname{Ave}(\mathfrak{U})+\mathcal{O}_{n,K,f,{\mathscr{V}}}(h^{-\frac{1}{2}-4\delta}+h^{(K-2)\delta-2})\end{split}$

and for the anti-chiral model where $J=2$, we have

$\begin{split}&B_{1,{\operatorname{ac}}}=-\frac{1}{2\pi
h^{2}}t_{n,0}(f)-\frac{1}{2\pi
h^{\frac{3}{2}}}t_{n,1}(f)-\mathcal{O}_{n,K,f,{\mathscr{V}}}(h^{-1-3\delta}+h^{(K-1)\delta-2}),\\\
&B_{2,{\operatorname{ac}}}=-\frac{\sqrt{2|n|}}{4\pi
h^{\frac{5}{2}}}t_{n,0}(f^{\prime})-\frac{\sqrt{2|n|}}{4\pi
h^{2}}t_{n,1}(f^{\prime})-\mathcal{O}_{n,K,f,{\mathscr{V}}}(h^{-\frac{3}{2}-3\delta}+h^{(K-1)\delta-\frac{5}{2}}),\\\
&B_{3,{\operatorname{ac}}}=\frac{1}{4\pi
h^{\frac{3}{2}}}t_{n,1}(f)+\mathcal{O}_{n,K,f,{\mathscr{V}}}(h^{-1-3\delta}+h^{(K-2)\delta-2}).\end{split}$

Combining the estimates of $B_{1}$, $B_{2}$, $B_{3}$, rearranging in the
ascending order in $h$, we get (3.6) and (3.7). ∎

##  8\. Magnetic response quantities

This section discusses applications of the regularized trace expansions
derived in the previous section, cf. Theorems 4 and 5 as well as Proposition
3.1. They form the rigorous foundation of our analysis in this section and we
shall focus on qualitative features rather here, instead. We study magnetic
response quantities by thoroughly analyzing the following cases:

  * •

The free or non-interacting case, corresponds to two non-interacting sheets of
graphene modeled by the direct sum of two magnetic Dirac operators, see also
[BZ19, BHJZ21] for similar results in a quantum graph model and [SGB04] for a
thorough analysis of the magnetic Dirac operator, directly.

  * •

The chiral case, which corresponds to pure AB/BA interaction.

  * •

The anti-chiral case, which corresponds to pure AA’/BB’ interaction.

For our analysis of the de Haas-van Alphen effect, we shall employ a cut-off
function $\eta_{N}\in C_{c}^{\infty}({\mathbb{R}})$ that is one on the
interval $[0,\sqrt{2BN}]$ and smoothly decays to zero outside of that
interval, enclosing precisely $N+1$ Landau levels and
$\eta_{N}^{\operatorname{sym}}$ which is equal to one on
$[-\sqrt{2BN},\sqrt{2BN}].$ The choice of cut-off function mainly plays the
role of a reference frame. In particular, for the study of magnetic
oscillations it seems more natural to consider $\eta_{N}$ instead of
$\eta_{N}^{\operatorname{sym}}$ as the former cut-off function singles out the
effect of individual Landau levels moving past a fixed chemical potential
$\mu.$ We shall employ the leading order terms for the regularized trace in
this section, as specified in Theorems 4 and 5 and Proposition 3.1. For this
reason, we write functionals $\rho(f),$ where $f\in C^{\infty}({\mathbb{R}}),$
as $\rho(f)\sim g,$ to indicate that $g$ are the first terms in the asymptotic
expansion of $\rho(f)$ and analogously for derivaives of $\rho(f)$ with
respect to the magnetic field.

Figure 9. Smoothed out longitudinal conductivity
$\sigma_{xx}\propto-\rho(\lambda n^{\prime}_{\beta}(\lambda-\mu))$ with
$n_{\beta}$, the Fermi-Dirac distribution, showing Shubnikov-de Haas
oscillations. On the left, $B=30$ and on the right $B=50$ for $\beta=1.5.$
with $\alpha_{i}=\frac{3}{5}.$

###  8.1. Shubnikov-de Haas oscillations

We shall start by discussing _Shubnikov - de Haas (SdH)_ oscillations in the
density of states. A common method of measuring SdH oscillations is by
measuring longitudinal conductivity and resistivity, see also [W11, Tan11]. In
the following, let $\sigma\in\mathbb{R}^{2\times 2}$ be the conductivity
matrix, such that the current density $j=\sigma E$, where $E$ is an external
electric field, then the resistivity matrix is just $\rho=\sigma^{-1}.$ Hence,
we shall focus on conductivities in the sequel.

The SdH oscillations are most strongly pronounced at low temperatures in the
regime of strong magnetic fields and describe oscillations in the longitudinal
conductivity $\sigma_{xx}$ of the material.

The expression for the longitudinal conductivity goes back to Ando et al [A70,
A82] who derived the following relation, see also [FS14],

$\sigma_{xx}(\beta,\mu,B)=-\int_{0}^{\infty}n_{\beta}^{\prime}(\lambda-\mu)\lambda\eta^{\text{sym}}_{N}(\lambda)\
d\rho(\lambda),$

where $n_{\beta}(x)=\frac{1}{e^{\beta x}+1}$ is the Fermi-Dirac statistics. In
the free case, i.e. without any tunnelling potential, the oscillations happen
precisely at the relativistic Landau levels. For the chiral model,
oscillations caused by higher Landau levels are enhanced compared to the free
case, whereas oscillations in the anti-chiral case are much more smoothed out.

The oscillatory behaviour of the longitudinal conductivity is visible both as
a function of chemical potential, for a fixed magnetic field strength, as
shown in Fig. 9 as well as function of inverse magnetic field in Fig. 10 for
fixed chemical potential.

Figure 10. Smoothed out longitudinal conductivity
$\sigma_{xx}\propto-\rho(\lambda n^{\prime}_{\beta}(\lambda-\mu))$ with
$n_{\beta}$, the Fermi-Dirac distribution, showing Shubnikov-de Haas
oscillations. On the left, $B=30$ and on the right $B=50$, both for
$\beta=2.5.$ with $\alpha_{i}=0.35.$

###  8.2. De Haas-van Alphen oscillations

In 1930, de Haas and van Alphen who discovered that both the magnetization and
the magnetic susceptibility of metals show an oscillatory profile as a
function of $B^{-1}.$ This effect is called the de Haas-van Alphen (dHvA)
effect. Even in the simpler case of graphene, both the experimental as well as
theoretical foundations of that effect are not yet well-understood [L11, KH14,
SGB04]. One problem in understanding the dHvA effect [SGB04], lies in the
dependence of the chemical potential on the external magnetic field. To
simplify mathematical analysis, it is more convenient to work in the grand-
canonical ensemble, which is also discussed in [CM01, SGB04, KF17]. The
comparison with the canonical ensemble is made in this subsection as well.

The grand thermodynamic potential for a DOS measure $\rho$, at inverse
temperature $\beta,$ and field-independent chemical potential $\mu$ is defined
as

$\Omega(\beta,\mu,B):=(f_{\beta}*(\eta_{N}\rho))(\mu),$

where $f_{\beta}(x):=-\beta^{-1}\log(e^{\beta x}+1).$ The magnetization $M$
and susceptibility $\chi$ are then in the grand-canonical ensemble defined as

$M(\beta,\mu,B)=-\frac{\partial\Omega_{\beta}(\mu,B)}{\partial B}\text{ and
}\chi(\beta,\mu,B)=\frac{\partial M_{\beta}(\mu,B)}{\partial B}.$

The susceptibility describes the response of a material to an external
magnetic field. When $\chi>0$ the material is paramagnetic, when $\chi<0$
diamagnetic, and strongly enhanced $\chi\gg 1$ for ferromagnets.

While the approximation of computing the magnetization in the grand canonical
ensemble is common, one should strictly speaking compute it in the canonical
ensemble, instead.

Figure 11. Magnetization and susceptibility for $\beta=4$, $\alpha_{i}=3/5,$
and chemical potential $\mu=5$.

In this case, the charge density $\varrho$ given by the Fermi-Dirac
statistics, with $n_{\beta}(x):=\frac{1}{e^{\beta x}+1}$, according to

$\varrho=-\frac{\Omega_{\beta}(\mu,B)}{\partial\mu}=\rho(n_{\beta}(\cdot-\mu))$

is fixed and the chemical potential becomes a function of $\rho$ and $B$.

To see that this uniquely defines $\mu$ as a function of $\varrho$ and $B$
large enough, it is sufficient to observe that

$\mu\mapsto\sum_{n\in\mathbb{Z}}(\eta_{N}n_{\mu})(\lambda_{n}\sqrt{B})$

is a monotonically increasing function.

Figure 12.  Charge density with respect to chemical potential. Magnetic field
$B=30$ for $\beta=1/2$ and $\beta=2$. We consider $100$ Landau levels around
zero and an anti-chiral model with $\theta=0.$

The Helmholtz free energy is then given as

$F_{\beta}(\varrho,B)=\Omega_{\beta}(\mu(\varrho,B),B)+\mu(\rho,B)\varrho$

with the magnetization given as the derivative

$M(\beta,\varrho,B)=-\frac{\partial F_{\beta}(\varrho,B)}{\partial B}.$

Hence, the magnetization in the canonical ensemble is also given by

$M(\beta,\varrho,B)=-\frac{\partial\Omega_{\beta}(\mu,B)}{\partial
B}\Big{|}_{\mu=\mu(\varrho,B)},$

where the difference to the grand-canonical ensemble lies in the $B$-dependent
chemical potential. The dHvA oscillations are shown in Figures 7 and 11, with
the AB’/BA’ interaction leading to enhanced oscillations and the AA’/BB’
interaction damping the oscillations, compared to the non-interacting case.

###  8.3. Quantum Hall effect

Figure 13. Full quantum Hall conductivity (8.1) on the left with $\beta=2$,
$B=40$ and on the right the high temperature conductivity (8.3) with
$\beta=5$,$B=50$.

The (transversal) quantum Hall conductivity $\sigma_{xy}$ is, by the Streda
formula [MK12, (16)], for a Fermi energy $\mu$ given by

$\sigma_{xy}(\beta,\mu,B)=\sum_{n=-N}^{N}\frac{\partial\rho(\eta_{N}n_{\beta}(\bullet-\mu))}{\partial
B}.$ (8.1)

In case of the chiral Hamiltonian, the Gibbs factor
$\gamma_{\beta,n}(\mu)=e^{\beta(\lambda_{n}\sqrt{B}-\mu)}$ allows us to write

$\begin{split}&\sigma_{xy,c}(\beta,\mu,B)=\Bigg{(}\sum_{n=-N}^{N}\frac{n_{\beta}(\lambda_{n}\sqrt{B}-\mu)}{\pi}\left(1-\frac{\beta\lambda_{n}\sqrt{B}}{2}\gamma_{\beta,n}(\mu)n_{\beta}(\lambda_{n}\sqrt{B}-\mu)\right)\\\
&+\sum_{n=-N}^{N}-\frac{\lambda_{n}|\lambda_{n}|^{2}\beta^{3}\operatorname{Ave}(\mathfrak{U})}{4\pi\sqrt{B}}n^{4}_{\beta}(\lambda_{n}\sqrt{B}-\mu)\left(\gamma_{\beta,n}(\mu)-4\gamma_{\beta,n}(\mu)^{2}+\gamma_{\beta,n}(\mu)^{3}\right)\Bigg{)}\left(1+o(1)\right)\end{split}$
(8.2)

At very low temperatures, and $\mu$ well between two Landau levels, the
contribution of the derivative of the Landau levels with respect to $B$ can be
discarded.

We then obtain the high-temperature limiting expression

$\hat{\sigma}_{xy,c}(\beta,\mu,B):=\sum_{n=-N}^{N}\frac{\eta_{\beta}(\lambda_{n,B}-\mu)}{\pi}\xrightarrow[\beta\to\infty]{}|\\{n;|\lambda_{n,B}|\leq\mu\\}|$
(8.3)

as $n_{\beta}(\lambda_{n,B}-\mu)\to 1-H(\lambda_{n}\sqrt{B}-\mu)$ for
$\beta\uparrow\infty,$ where $H$ is the Heaviside function.

This expression reveals the well-known staircase profile of the Hall
conductivity which can already be concluded in this model in the
$\beta\to\infty$ limit from Proposition 6.2.

For the AA’/BB’ interaction, the situation is rather different. Due to the
broadening and splitting of the Landau levels, the staircase profile is less
pronounced at non-zero temperature. Setting

$\hat{\sigma}_{xy,ac}(\beta,\mu,B):=t_{n,0}(n_{\beta}(\bullet-\mu))-\frac{t_{n,1}(n_{\beta}(\bullet-\mu))}{2\sqrt{B}},$

where in the limit $\beta\to\infty,$ the second term vanishes, for $\mu$ away
from the spectrum as $n_{\beta}^{\prime}$ is a $\delta_{0}$ approximating
sequence such that also in case of the AA’/BB’ interaction
$\lim_{\beta\to\infty}\hat{\sigma}_{xy}(\beta,\mu,B)=|\\{n;|\lambda_{n,B}|\leq\mu\\}|.$

##  Appendix A Asymptotic expansion

In this appendix, we shall prove Prop. A.1 which, in particular, includes the
proof of Lemma 7.7. The quantization is as in Subsection 7.3.

######  Proposition A.1.

Let $h_{0}$, $E_{n,\pm}$ be as in Lemma 7.5. For $h\in[0,h_{0})$, $|z|\leq
2\|\mathscr{W}\|_{\infty}$, we have

  1. (1)

The symbol $\frac{1}{\sqrt{h}}E_{n,\pm}$ has an asymptotic expansion in $S$:
There are $a_{n,j,k}\in S$ s.t.

$\frac{1}{\sqrt{h}}E_{n,\pm}(x_{2},\xi_{2};z,h)\sim\sum\limits_{j=0}^{\infty}h^{\frac{j}{2}}E_{n,j}(x_{2},\xi_{2};z)\
\text{with}\
E_{n,j}=\sum\limits_{k=0}^{j-1}a_{n,j,k}(x_{2},\xi_{2})z^{k},j\geq 1.$ (A.1)

In particular, $E_{n,0}=z-z_{n,0},\quad E_{n,1}=-z_{n,1},\quad
E_{n,2}=-z_{n,2},$ where $z_{n,j}$ are given in Lemma A.2.

  2. (2)

Let $0<\delta<1/2$, if $|\operatorname{Im}z|\geq h^{\delta}$, then
$\sqrt{h}E_{n,\pm}^{-1}$ has an asymptotic expansions in
$S_{\delta}^{\delta}$: There are $b_{n,j,k,l},c_{n,j,k}\in S$ s.t. in terms of
$\prod\limits_{l=0}^{k}b_{n,j,k,l}(x_{2},\xi_{2};z)=\sum\limits_{\alpha=0}^{j+k-2}z^{\alpha}c_{n,j,k}(x_{2},\xi_{2})$

$\begin{split}&\sqrt{h}E_{n,\pm}^{-1}\sim\sum\limits_{j=0}^{\infty}h^{\frac{j}{2}}F_{n,j}(x_{2},\xi_{2};z),\text{
with
}F_{n,j}=\sum_{k=0}^{j}(z-z_{n,0})^{-1}\prod\limits_{l=0}^{k}\left(b_{n,j,k,l}(x_{2},\xi_{2};z)(z-z_{n,0})^{-1}\right).\end{split}$
(A.2)

Thus $h^{\frac{j}{2}}F_{n,j}\in S^{j(\delta-\frac{1}{2})+\delta}$. In
particular, we have

$\begin{split}&F_{n,0}=(z-z_{n,0})^{-1},\ F_{n,1}=F_{n,0}z_{n,1}F_{n,0},\
F_{n,2}=F_{n,0}\left(z_{n,1}F_{n,1}+z_{n,2}F_{n,0}-\frac{\\{F_{n,0},z-z_{n,0}\\}}{2i}\right),\end{split}$
(A.3)

where $\\{\cdot,\cdot\\}$ is the Poisson bracket.

  3. (3)

Let $0<\delta<1/2$, if $|\operatorname{Im}z|\geq h^{\delta}$, then $r_{n}$ has
an asymptotic expansions in $S_{\delta}^{\delta}$: There are
$d_{n,j,k,l}(x_{2},\xi_{2};z),e_{n,j,k,\alpha}(x_{2},\xi_{2})\in S$, s.t. in
terms of
$\prod\limits_{l=0}^{k}d_{n,j,k,l}(x_{2},\xi_{2};z)=\sum\limits_{\alpha=0}^{j+k-2}z^{\alpha}e_{n,j,k,\alpha}(x_{2},\xi_{2})$

$\begin{split}&\quad
r_{n}(x_{2},\xi_{2};z,h)\sim\sum\limits_{j=0}^{\infty}h^{\frac{j}{2}}r_{n,j}(x_{2},\xi_{2};z,h),\text{with~{}}r_{n,j}=\sum\limits_{k=0}^{j}(z-z_{n,0})^{-1}\prod\limits_{l=0}^{k}\left(d_{n,j,k,l}(x_{2},\xi_{2};z)(z-z_{n,0})^{-1}\right).\end{split}$
(A.4)

Thus $h^{\frac{j}{2}}r_{n,j}\in S_{0}^{(J+1)\delta-\frac{J}{2}}$. In
particular,

$r_{n,0}=F_{n,0},\quad r_{n,1}=F_{n,1},\quad
r_{n,2}=F_{n,2}-(\partial_{z}z_{n,2})F_{n,0}.$ (A.5)

  4. (4)

In particular, denote $\eta=x_{2}+i\xi_{2}$, then the leading terms of
$\operatorname{Tr}_{{\mathbb{C}}^{2}}(r_{n})$ are:

$\begin{split}\operatorname{Chiral}\
{\mathscr{H}}_{{\operatorname{c}},n}:&\operatorname{Tr}_{{\mathbb{C}}^{2}}(r_{{\operatorname{c}},n,0}+h^{\frac{1}{2}}r_{{\operatorname{c}},n,1}+hr_{{\operatorname{c}},n,2})=\frac{2}{z}+0+\frac{\lambda_{n}^{2}}{z^{3}}\mathfrak{U}(\eta)h,\\\
\operatorname{Anti-Chiral}\
{\mathscr{H}}_{{\operatorname{ac}},n}^{\theta}:&\operatorname{Tr}_{{\mathbb{C}}^{2}}(r_{{\operatorname{ac}},n,0}+h^{\frac{1}{2}}r_{{\operatorname{ac}},n,1})=\frac{2z}{z^{2}-c_{n}^{2}}+\frac{2s_{n}^{2}(z^{2}+c_{n}^{2})}{(z^{2}-c_{n}^{2})^{2}}\sqrt{h},\end{split}$
(A.6)

where
$\mathfrak{U}(\eta)=\frac{\alpha_{1}^{2}}{8}\left[\alpha_{1}^{2}(|U_{-}(\eta)|^{2}-|U(\eta)|^{2})^{2}+4|\partial_{\bar{\eta}}\overline{U_{-}(\eta)}-\partial_{\eta}U(\eta)|^{2}\right]$,
$\partial_{\eta}=\frac{1}{2}(\partial_{x_{2}}-i\partial_{\xi_{2}})$,
$s_{n}(\eta)=\begin{cases}\alpha_{0}\sin({\frac{\theta}{2}})|V(\eta)|&n\neq
0\\\ \alpha_{0}|V(\eta)|&n=0,\end{cases}$ and
$c_{n}(\eta)=\begin{cases}\alpha_{0}\cos({\frac{\theta}{2}})|V(\eta)|&n\neq
0\\\ \alpha_{0}|V(\eta)|&n=0.\end{cases}$

We will prove Proposition A.1 in the rest of this appendix in two steps:
First, we compute explicitly the leading terms (three terms for the chiral
model, two for anti-chiral model) in the expansion of
$Z_{n}(x_{2},\xi_{2};z,h)$, the symbol of $Z_{n}^{W}$, where
$E_{n,\pm}=\sqrt{h}(z-Z_{n}^{W})$ by (7.16). Then, we derive in general the
$z$ dependence for each term in the expansion of $E_{n,\pm}$, from which we
build up both the legitimacy of the existence of asymptotic expansions of
$E_{n,\pm}^{-1}$ and $r_{n}$, and the $z$ dependence for each terms in the
expansions.

Explicit leading terms. Recall that by (7.16) and (7.20),
$E_{n,\pm}=\sqrt{h}(z-Z_{n}^{W})$ with

$\begin{split}Z_{n}^{W}&(x_{2},hD_{x_{2}};h)=R_{n}^{+}\tilde{{\mathscr{V}}}^{W}(I+\sqrt{h}E_{0,n}^{\theta}\tilde{{\mathscr{V}}}^{W})^{-1}R_{n}^{-}\\\
&=\sum\limits_{k=0}^{\infty}h^{\frac{k}{2}}(-1)^{k}R_{n}^{+}\tilde{{\mathscr{V}}}^{W}(E_{0,n}^{\theta}\tilde{{\mathscr{V}}}^{W})^{k}R_{n}^{-}=:\sum\limits_{k=0}^{\infty}h^{\frac{k}{2}}Q_{n,k}^{W}(x_{2},hD_{x_{2}};h),\end{split}$
(A.7)

where $R_{n}^{\pm}$, $E_{0,n}^{\theta}$, $\tilde{{\mathscr{V}}}^{W}$ are given
in (7.9), (7.13) and (7.20). Then we can express the asymptotic expansion of
$Z_{n}(x_{2},\xi_{2})$ in terms of $Q_{n,k}(x_{2},\xi_{2})$:

######  Proposition A.2.

Let
$Q_{n,k}^{W}(x_{2},hD_{x_{2}};h)=(-1)^{k}R_{n}^{+}\tilde{{\mathscr{V}}}^{W}(E_{0,n}^{\theta}\tilde{{\mathscr{V}}}^{W})^{k}R_{n}^{-}$.
Then symbols $Q_{n,0}$, $Q_{n,1}$, $Q_{n,2}$ have the following asymptotic
expansions

$\begin{split}&Q_{n,0}(x_{2},\xi_{2};h)=Q_{n,0}^{(0)}(x_{2},\xi_{2})+\sqrt{h}Q_{n,0}^{(1)}(x_{2},\xi_{2})+hQ_{n,0}^{(2)}(x_{2},\xi_{2})+\mathcal{O}_{S}(h^{\frac{3}{2}}),\\\
&Q_{n,1}(x_{2},\xi_{2};h)=Q_{n,1}^{(0)}(x_{2},\xi_{2})+\sqrt{h}Q_{n,1}^{(1)}(x_{2},\xi_{2})+\mathcal{O}_{S}(h),\\\
&Q_{n,2}(x_{2},\xi_{2};h)=Q_{n,2}^{(0)}(x_{2},\xi_{2})+\mathcal{O}_{S}(\sqrt{h}).\end{split}$

In the chiral model, for $\eta=x_{2}\ +i\xi_{2}$,
$D_{\eta}=\frac{1}{2}(D_{x_{2}}-iD_{\xi_{2}})$,

$\begin{split}&Q_{{\operatorname{c}},n,0}^{{(0)}}=Q_{{\operatorname{c}},n,0}^{{(2)}}=Q_{{\operatorname{c}},n,2}^{{(0)}}=0,\quad
Q_{{\operatorname{c}},n,1}^{{(0)}}=-\frac{\alpha_{1}^{2}\lambda_{n}}{4}\left[|U|^{2}-|U_{-}|^{2}\right]\sigma_{3},\\\
&Q_{{\operatorname{c}},n,0}^{{(1)}}=\frac{\lambda_{n}\alpha_{1}}{2}\begin{pmatrix}0&D_{\eta}U-D_{\bar{\eta}}\overline{U_{-}}\\\
D_{\eta}U_{-}-D_{\bar{\eta}}\overline{U}&0\end{pmatrix},\\\
&Q_{{\operatorname{c}},n,1}^{{(1)}}=\begin{cases}-\frac{\alpha_{1}^{2}z}{4}\left[2|n|(|U|^{2}+|U_{-}|^{2})\operatorname{1\hskip-2.75ptl}_{2\times
2}+(|U|^{2}-|U_{-}|^{2})\sigma_{3}\right]&n\neq 0,\\\
-\frac{\alpha_{1}^{2}z}{2}\begin{pmatrix}|U|^{2}&0\\\
0&|U_{-}|^{2}\end{pmatrix}&n=0.\end{cases}\end{split}$

While in the anti-chiral model, when
${\mathscr{H}}^{\theta}={\mathscr{H}}_{\operatorname{ac}}$, we have, when
$n=0$,

$\begin{split}&Q_{{\operatorname{ac}},0,0}^{(1)}=Q_{{\operatorname{ac}},0,1}^{(0)}=Q_{{\operatorname{ac}},0,1}^{(1)}=Q_{{\operatorname{ac}},0,2}^{(0)}=0,\\\
&Q_{{\operatorname{ac}},0,0}^{(0)}=\alpha_{0}\begin{pmatrix}0&e^{-\frac{\theta}{2}i}V\\\
e^{\frac{\theta}{2}}V^{*}&0\end{pmatrix},Q_{{\operatorname{ac}},0,0}^{(2)}=\frac{\alpha_{0}}{4}\begin{pmatrix}0&e^{-\frac{\theta}{2}i}\Delta_{x_{2},\xi_{2}}V\\\
e^{\frac{\theta}{2}i}\Delta_{x_{2},\xi_{2}}\bar{V}\end{pmatrix},\end{split}$

when $n\neq 0$,

$\begin{split}&Q_{{\operatorname{ac}},n,0}^{(1)}=0,\
Q_{{\operatorname{ac}},n,1}^{(0)}=\frac{\alpha_{0}^{2}|V|^{2}\sin^{2}(\tfrac{\theta}{2})}{2\lambda_{n}}\operatorname{1\hskip-2.75ptl}_{2\times
2},\
Q_{{\operatorname{ac}},n,1}^{(1)}=-\frac{z\alpha_{0}^{2}|V|^{2}\sin^{2}(\tfrac{\theta}{2})}{4\lambda_{n}^{2}}\operatorname{1\hskip-2.75ptl}_{2\times
2},\\\
&Q_{{\operatorname{ac}},n,0}^{(0)}=\alpha_{0}\cos(\tfrac{\theta}{2})\begin{pmatrix}0&V^{*}\\\
V&0\end{pmatrix},Q_{{\operatorname{ac}},n,2}^{(0)}=-\frac{\alpha_{0}^{3}|V|^{2}\sin^{2}(\tfrac{\theta}{2})\cos(\tfrac{\theta}{2})}{4\lambda_{n}^{2}}\begin{pmatrix}0&V\\\
V^{*}&0\end{pmatrix},\\\
&Q_{{\operatorname{ac}},n,0}^{(2)}=\frac{\alpha_{0}}{4}\left(2|n|\cos(\tfrac{\theta}{2})-i\sigma_{3}\sin(\tfrac{\theta}{2})\right)\begin{pmatrix}0&\Delta_{x_{2},\xi_{2}}V\\\
\Delta_{x_{2},\xi_{2}}\bar{V}&0\end{pmatrix}.\end{split}$

In particular, $Z_{n}$ has an asymptotic expansion
$Z_{n}\sim\sum\limits_{k=0}^{\infty}h^{\frac{k}{2}}z_{n,k}$ in $S$ with

$z_{n,0}=Q_{n,0}^{{(0)}},\ z_{n,1}=Q_{n,1}^{{(0)}}+Q_{n,0}^{{(1)}},\
z_{n,2}=Q_{n,2}^{{(0)}}+Q_{n,1}^{{(1)}}+Q_{n,0}^{{(2)}}.$

###### Proof.

$Q_{n,k}$ has the symbol
$Q_{n,k}(x_{2},\xi_{2})=(-1)^{k}\int_{{\mathbb{R}}_{x_{1}}}(K_{n}^{\theta}(x_{1}))^{*}\tilde{{\mathscr{V}}}^{w}\\#(E_{0,n}^{\theta}\tilde{{\mathscr{V}}}^{w})^{\\#k}K_{n}^{\theta}(x_{1})dx_{1}.$
Recall that by (7.10), (2.4), and (7.13), we have

$K_{n}^{\theta}=\begin{pmatrix}u_{n}^{\theta}&0\\\
0&u_{n}^{-\theta}\end{pmatrix},\quad{\mathscr{V}}=\begin{pmatrix}0&T\\\
T^{*}&0\end{pmatrix},\quad
T=\begin{pmatrix}\alpha_{0}V&\alpha_{1}\overline{U_{-}}\\\
\alpha_{1}U&\alpha_{0}V\end{pmatrix},\quad{E_{0,n}^{\theta}}=\begin{pmatrix}{e_{0,n}^{\theta}}&0\\\
0&{e_{0,n}^{-\theta}}\end{pmatrix}.$

Thus, inserting the above expressions into the definition of $Q_{n,k}$, we
find for its symbol

$Q_{n,k}=\int\begin{pmatrix}{u_{n}^{\theta}}^{*}&0\\\
0&({u_{n}^{-\theta}})^{*}\end{pmatrix}\begin{pmatrix}0&\tilde{T}^{w}\\\
(\tilde{T}^{w})^{*}&0\end{pmatrix}\left(\begin{pmatrix}{e_{0,n}^{\theta}}&0\\\
0&{e_{0,n}^{-\theta}}\end{pmatrix}\begin{pmatrix}0&\tilde{T}^{w}\\\
(\tilde{T}^{w})^{*}&0\end{pmatrix}\right)^{k}\begin{pmatrix}{u_{n}^{\theta}}^{*}&0\\\
0&{u_{n}^{-\theta}}\end{pmatrix}\frac{dx_{1}}{(-1)^{k}},$

where
$\tilde{T}^{w}=T^{w}({x_{2}+h^{\frac{1}{2}}x_{1},\xi_{2}-h^{\frac{1}{2}}D_{x_{1}}})$.
In particular,

$\begin{split}&Q_{n,0}=\begin{pmatrix}0&\int({u_{n}^{\theta}})^{*}\tilde{T}^{w}{u_{n}^{-\theta}}dx_{1}\\\
\int({u_{n}^{-\theta}})^{*}(\tilde{T}^{w})^{*}{u_{n}^{\theta}}dx_{1}\end{pmatrix},\\\
&Q_{n,1}=\begin{pmatrix}-\int({u_{n}^{\theta}})^{*}\tilde{T}^{w}{e_{0,n}^{-\theta}}(\tilde{T}^{w})^{*}{u_{n}^{\theta}}dx_{1}&0\\\
0&-\int({u_{n}^{-\theta}})^{*}(\tilde{T}^{w})^{*}{e_{0,n}^{\theta}}\tilde{T}^{w}{u_{n}^{-\theta}}dx_{1}\end{pmatrix},\text{
and }\\\
&Q_{n,2}=\begin{pmatrix}0&\int({u_{n}^{\theta}})^{*}\tilde{T}^{w}{e_{0,n}^{-\theta}}(\tilde{T}^{w})^{*}{e_{0,n}^{\theta}}\tilde{T}^{w}{u_{n}^{-\theta}}dx_{1}\\\
\int({u_{n}^{-\theta}})^{*}(\tilde{T}^{w})^{*}{e_{0,n}^{\theta}}\tilde{T}^{w}{e_{0,n}^{-\theta}}(\tilde{T}^{w})^{*}{u_{n}^{\theta}}dx_{1}&0\end{pmatrix}.\end{split}$
(A.8)

Notice that since both $\tilde{T}^{w}$ and ${e_{0,n}^{\theta}}$ depend on $h$,
we need to further expand them in order to obtain asymptotic expansions of
$Q_{n,k}$. Thus the proof of Proposition A.2 rests now on the following two
lemmas.

######  Lemma A.3 (Expansion of $\tilde{T}^{w}$ and ${e_{0,n}^{\theta}}$).

  1. (1)

Let $T\in C^{\infty}_{b}({\mathbb{R}}_{x}^{2})$. Recall that
$\tilde{T}(x,\xi):=T(x_{2}+h^{\frac{1}{2}}x_{1},\xi_{2}-h^{\frac{1}{2}}\xi_{1})\in
S({\mathbb{R}}_{x,\xi}^{4})$. Then

$\begin{split}\tilde{T}^{w}(x,D_{x_{1}},\xi_{2})=&T(x_{2},\xi_{2})+\sqrt{h}\langle\nabla_{x_{2},\xi_{2}}T(x_{2},\xi_{2}),(x_{1},-D_{x_{1}})\rangle\\\
+&\frac{h}{2}\langle(x_{1},-D_{x_{1}}),\operatorname{Hess}T(x_{2},\xi_{2})(x_{1},-D_{x_{1}})^{T}\rangle+\mathcal{O}_{S({\mathbb{R}}_{x_{2},\xi_{2}}^{2};\mathcal{L}(B^{3}_{x_{1}};B^{0}_{x_{1}}))}(h^{\frac{3}{2}})\end{split}$
(A.9)

  2. (2)

Let ${e_{0,n}^{\theta}}$ be as in (7.13). Then
${e_{0,n}^{\theta}}(x,D_{x_{1}},\xi_{2})$ has asymptotic expansion
${e_{0,n}^{\theta}}\sim\sum\limits_{k=0}^{\infty}h^{\frac{k}{2}}\sigma_{k}(e_{0,n}^{\theta})$
where $\sigma_{k}({e_{0,n}^{\theta}})=\sum\limits_{m\neq
n}\frac{z^{k}u_{m}^{\theta}(u_{m}^{\theta})^{*}}{(\lambda_{m}-\lambda_{n})^{k+1}}.$

######  Lemma A.4 (Projections).

Let $S_{n}^{\theta}=\operatorname{span}\\{u_{n}^{\theta},u_{-n}^{\theta}\\}$
with $S_{n}:=S_{n}^{0}.$ The following properties hold:

  1. (1)

We have $S_{n}^{\theta}=S_{n}^{-\theta}$, in particular

${u_{n}^{\theta}}=\cos\left({\frac{\theta}{2}}\right){u_{n}^{-\theta}}+i\sin\left({\frac{\theta}{2}}\right){u_{-n}^{-\theta}}.$
(A.10)

  2. (2)

Let $M=\begin{pmatrix}0&\alpha\\\ \beta&0\end{pmatrix}\in\mathbb{C}^{2\times
2}$ then $Mu_{n}\in S_{n-1}\cup S_{n+1}$, for any $n\geq 0$. More specifically
for $\theta=0$

$\begin{split}Mu_{\pm n}&=\frac{\alpha
i}{2}(u_{n+1}-u_{-(n+1)}){\mp}\frac{\beta i}{2}(u_{n-1}+u_{-(n-1)}),\text{ for
}n\geq 2\\\ Mu_{\pm 1}&=\frac{\alpha
i}{2}(u_{2}-u_{-2})\pm\frac{\beta}{\sqrt{2}}u_{0},\text{ and
}Mu_{0}=\frac{\alpha}{\sqrt{2}}(u_{1}-u_{-1}).\end{split}$ (A.11)

  3. (3)

We have $x_{1}{u_{n}^{\theta}}\in S_{n-1}^{\theta}\cup S_{n+1}^{\theta}$,
$D_{x_{1}}{u_{n}^{\theta}}\in S_{n-1}^{\theta}\cup S_{n+1}^{\theta}$. More
specifically

$\begin{split}x_{1}u_{\pm
n}^{\theta}&=\frac{\sqrt{2}}{4}[u_{n-1}^{\theta}(\sqrt{n}\pm+\sqrt{n-1})+u_{-(n-1)}^{\theta}(\sqrt{n}\mp\sqrt{n-1})\\\
&+u_{n+1}^{\theta}(\sqrt{n+1}+\sqrt{n})\pm
u_{-(n+1)}^{\theta}(\sqrt{n+1}\mp\sqrt{n})],\text{ for }|n|\geq 2\\\
x_{1}u_{\pm
1}^{\theta}&=\frac{i}{2}u_{0}^{\theta}+\frac{\sqrt{2}}{4}[u_{2}^{\theta}(\sqrt{2}\pm\sqrt{1})+u_{-2}^{\theta}(\sqrt{2}\mp\sqrt{1})]\text{
and
}x_{1}u_{0}^{\theta}=\frac{\sqrt{2}i}{4}(u_{1}^{\theta}+u_{-1}^{\theta}).\end{split}$
(A.12)

###### Proof.

We omit the proof of this Lemma here as it follows from straightforward but
lengthy basis expansions and the simple observation that $\langle
u_{m}^{-\theta},u_{n}^{\theta}\rangle=\cos\left({\frac{\theta}{2}}\right)\delta_{m,n}+i\sin\left({\frac{\theta}{2}}\right)\delta_{m,-n}.$
∎

From the preceding Lemmas A.3 and A.4, we can compute the asymptotic expansion
of each term of $Q_{n,k}$ in (A.8) and therefore prove Prop.A.2.

For the $(1,2)$-entry of $Q_{n,0}$, by Lemma A.3, we have

$\begin{split}\int({u_{n}^{\theta}})^{*}\tilde{T}^{w}{u_{n}^{-\theta}}dx_{1}=&\int({u_{n}^{\theta}})^{*}T{u_{n}^{-\theta}}dx_{1}+\sqrt{h}\int({u_{n}^{\theta}})^{*}\langle\nabla_{x_{2},\xi_{2}}T,(x_{1},-D_{x_{1}})\rangle{u_{n}^{-\theta}}dx_{1}\\\
&+\frac{h}{2}\int({u_{n}^{\theta}})^{*}\langle(x_{1},-D_{x_{1}}),\operatorname{Hess}T(x_{2},\xi_{2})(x_{1},-D_{x_{1}})^{T}\rangle{u_{n}^{-\theta}}dx_{1}\\\
=:&t_{n,0}^{(0)}+\sqrt{h}t_{n,0}^{(1)}+ht_{n,0}^{(2)}+\mathcal{O}_{S({\mathbb{R}}_{x_{2},\xi_{2}}^{2};{\mathbb{C}}_{2\times
2})}(h^{\frac{3}{2}}).\end{split}$

Specializing now to the chiral case, in which case $\theta=0,$ we choose

$T(x_{2},\xi_{2})=\begin{pmatrix}0&\alpha_{1}\overline{U(x_{2},\xi_{2})}\\\
\alpha_{1}U_{-}(x_{2},\xi_{2})&0\end{pmatrix}$

where in the chiral case, by Lemmas A.3 and A.4, we see that

$\begin{split}&t_{{\operatorname{c}},n,0}^{{(0)}}=0,\
t_{{\operatorname{c}},n,0}^{{(1)}}=\frac{\lambda_{n}\alpha_{1}i}{2}(\partial_{\bar{w}}\overline{U_{-}}-\partial_{w}U),\text{
and }t_{{\operatorname{c}},n,0}^{{(2)}}=0,\end{split}$

while in the anti-chiral case, choosing
$T(x_{2},\xi_{2})=\alpha_{0}V(x_{2},\xi_{2})\operatorname{id}_{{\mathbb{C}}_{2\times
2}}$

$\begin{split}&t_{{\operatorname{ac}},n,0}^{{(0)}}=\begin{cases}\alpha_{0}\cos({\frac{\theta}{2}})V&n\neq
0,\\\ \alpha_{0}e^{-\frac{\theta}{2}i}V&n=0,\end{cases},\quad
t_{{\operatorname{ac}},n,0}^{{(1)}}=0,\text{ and
}t_{{\operatorname{ac}},n,0}^{{(2)}}=\begin{cases}\frac{\alpha_{0}}{4}(2|n|\cos({\frac{\theta}{2}})-i\sigma_{3}\sin({\frac{\theta}{2}}))\Delta_{x_{2},\xi_{2}}V,&n\neq
0\\\
\frac{\alpha_{0}}{4}e^{-i\frac{\theta}{2}}\Delta_{x_{2},\xi_{2}}V&n=0.\end{cases}\end{split}$

Due to the conjugacy relation
$\int({u_{n}^{\theta}})^{*}(\tilde{T}^{w})^{*}{u_{n}^{-\theta}}dx_{1}=(\int({u_{n}^{-\theta}})^{*}\tilde{T}^{w}{u_{n}^{\theta}}dx_{1})^{*}$,
the expansion of $Q_{n,0}^{\theta}$ follows by (A.8).

Similarly for the $(1,1)$-entry $Q_{n,1}^{\theta}$, denote

$-\int({u_{n}^{\theta}})^{*}\tilde{T}^{w}{e_{0,n}^{-\theta}}(\tilde{T}^{w})^{*}{u_{n}^{\theta}}dx_{1}=:t_{n,1}^{{(0)}}+t_{n,1}^{(1)}\sqrt{h}+\mathcal{O}_{S({\mathbb{R}}_{x_{2},\xi_{2}}^{2};{\mathbb{C}}_{2\times
2})}(h)$

where, using Lemma A.9, in the chiral case,

$\begin{split}&t_{{\operatorname{c}},n,1}^{(0)}=-\frac{\alpha_{1}^{2}\lambda_{n}}{4}(|U|^{2}-|U_{-}|^{2})\text{
and
}t_{{\operatorname{c}},n,1}^{{(1)}}=\begin{cases}-\frac{\alpha_{1}^{2}z}{4}[2|n|(|U|^{2}+|U_{-}|^{2})+(|U|^{2}-|U_{-}|^{2})],&n\neq
0\\\ -\frac{\alpha_{1}^{2}z}{2}|U|^{2},&n=0\end{cases}\end{split}$

and in the anti-chiral case

$\begin{split}&t_{{\operatorname{ac}},n,1}^{{(0)}}=\begin{cases}\frac{\alpha_{0}^{2}|V|^{2}\sin^{2}({\frac{\theta}{2}})}{2\lambda_{n}},&n\neq
0\\\ 0,&n=0\end{cases}\text{ and
}t_{{\operatorname{ac}},n,1}^{{(1)}}=\begin{cases}-\frac{\alpha_{0}^{2}|V|^{2}\sin^{2}({\frac{\theta}{2}})z}{4\lambda_{n}^{2}},&n\neq
0\\\ 0,&n=0.\end{cases}\end{split}$

In a similar fashion, the $(2,2)$-entry of $Q_{n,1}$, defined in (A.8), can be
obtained by precisely the same computations after only replacing $\theta$ by
$-\theta$ and $T^{*}$ by $T$, i.e. $U$ switching with $U_{-}$ and using
$V^{*}$ instead of $V$. Thus the asymptotic expansion of $Q_{n,1}^{\theta}$
follows.

Similarly for $Q_{n,2}^{\theta}$ we restrict us to the $(1,2)$ entry in (A.8).
Then, we denote

$\int({u_{n}^{\theta}})^{*}\tilde{T}^{w}{e_{0,n}^{-\theta}}(\tilde{T}^{w})^{*}{e_{0,n}^{\theta}}\tilde{T}^{w}{u_{n}^{-\theta}}dx_{1}=:t_{n,2}^{{(0)}}+\mathcal{O}_{S({\mathbb{R}}_{x_{2},\xi_{2}}^{2};{\mathbb{C}}_{2\times
2})}(\sqrt{h}).$

It follows then by Lemma A.9, that in the chiral model,
$t_{{\operatorname{c}},n,2}^{{(0)}}=0$ while in the anti-chiral model,
$t_{n,2}^{{(0)}}=-\frac{\alpha_{0}^{3}|V|^{2}\sin^{2}({\frac{\theta}{2}})\cos({\frac{\theta}{2}})}{4\lambda_{n}^{2}}V.$
By the conjugacy relation

$\int({u_{n}^{-\theta}})^{*}(\tilde{T}^{w})^{*}{e_{0,n}^{\theta}}\tilde{T}^{w}{e_{0,n}^{-\theta}}(\tilde{T}^{w})^{*}{u_{n}^{\theta}}dx_{1}=[\int({u_{n}^{\theta}})^{*}\tilde{T}^{w}{e_{0,n}^{-\theta}}(\tilde{T}^{w})^{*}{e_{0,n}^{\theta}}\tilde{T}^{w}{u_{n}^{-\theta}}dx_{1}]^{*},$

this also yields directly the expansion of
$Q_{{\operatorname{ac}},n,2}^{\theta}$.

∎

Existence, derivation and $z$ dependence. Now we prove the rest of Prop. A.1,
which includes the existence and derivation of asymptotic expansion of
$E_{n,\pm}^{-1}$ and $r_{n}$ and the $z$ dependence of each terms in the
expansions of $E_{n,\pm}$, $E_{n,\pm}^{-1}$ and $r_{n}$.

###### Proof of Prop. A.1.

By (A.7) and Prop. A.2, $E_{n,\pm}=\sqrt{h}(z-Z_{n})$, and $Z_{n}$ has an
asymptotic expansion in $S$. Thus, $\frac{1}{\sqrt{h}}E_{n,\pm}$ also has an
asymptotic expansion in $S$:
$\frac{1}{\sqrt{h}}E_{n,\pm}\sim\sum\limits_{j}h^{\frac{j}{2}}E_{n,j}$ with
$E_{n,j}\in S$. To exhibit the $z$-dependence, we notice that only $E_{0,n}$
depends on $z$ in (A.7). Thus, by (7.24), we have

$\begin{split}Z_{n}^{W}&=R_{n}^{+}\tilde{{\mathscr{V}}}^{W}(\operatorname{1\hskip-2.75ptl}+\sqrt{h}E_{0,n}\tilde{{\mathscr{V}}}^{W})^{-1}R_{n}^{-}=R_{n}^{+}\tilde{{\mathscr{V}}}^{W}R_{n}^{-}+\sum\limits_{\alpha=1}^{\infty}R_{n}^{+}\tilde{{\mathscr{V}}}^{W}(\sqrt{h}E_{0,n}\tilde{{\mathscr{V}}}^{W})^{\alpha}R_{n}^{-}\\\
&=R_{n}^{+}\tilde{{\mathscr{V}}}^{W}R_{n}^{-}+\sum\limits_{\alpha=1}^{\infty}h^{\frac{\alpha}{2}}R_{n}^{+}\tilde{{\mathscr{V}}}^{W}\left[\sum\limits_{m\neq
n}\frac{K_{m}^{\theta}(K_{m}^{\theta})^{*}}{\lambda_{m}-\lambda_{n}}\sum\limits_{\beta=0}^{\infty}\left(\frac{\sqrt{h}z}{\lambda_{m}-\lambda_{n}}\right)^{\beta}\right]^{\alpha}R_{n}^{-}\\\
&=R_{n}^{+}\tilde{{\mathscr{V}}}^{W}R_{n}^{-}+\sum\limits_{\alpha=1}^{\infty}\sum\limits_{\gamma=0}^{\infty}h^{\frac{\alpha+\gamma}{2}}z^{\gamma}A_{n,\alpha,\gamma}^{W}(x_{2},hD_{x_{2}})=R_{n}^{+}\tilde{\mathscr{V}}^{W}R_{n}^{-}-\sum\limits_{j=1}^{\infty}h^{\frac{j}{2}}\left(\sum\limits_{k=0}^{j-1}z^{k}a_{n,j,k}^{W}(x_{2},hD_{x_{2}})\right)\end{split}$

for some appropriate $A_{n,\alpha,\gamma}(x_{2},\xi_{2})\in S$ and
$a_{n,j,k}(x_{2},\xi_{2})\in S$. Thus we proved part (1).

For $\sqrt{h}E_{n,\pm}^{-1}$, first of all, by a parametrix construction using
the formal expansion of the sharp product

$a\widetilde{\\#}b\sim\sum_{k}\frac{1}{k!}\left.\left(\left(\frac{ih}{2}\sigma(D_{x_{2}},D_{\xi_{2}};D_{y},D_{\eta})\right)^{k}\left(a(x_{2},\xi_{2})b(y,\eta)\right)\right)\right|_{x_{2}=y,\
\xi_{2}=\eta},$ (A.13)

we can formally derive (A.2) and (A.3). More specifically, there is a formal
expansion of $\sqrt{h}E_{n,\pm}^{-1}$, which is denoted by
$\sqrt{h}F_{n}\sim\sum\limits_{j}h^{\frac{j}{2}}F_{n,j}$, s.t.
$\frac{1}{\sqrt{h}}E_{n,\pm}\widetilde{\\#}\sqrt{h}F_{n}=\operatorname{1\hskip-2.75ptl}_{2\times
2}$. Denote $\sigma(D_{x_{2}},D_{\xi_{2}};D_{y},D_{\eta})$ in (A.13) by
$\sigma$, we can solve for $F_{n,j}$ by considering

$\begin{split}\operatorname{1\hskip-2.75ptl}_{2\times
2}&=E_{n,\pm}\widetilde{\\#}F_{n}^{-1}\sim\sum\limits_{\alpha=0}^{\infty}\sum\limits_{\beta=0}^{\infty}h^{\frac{\alpha+\beta}{2}}E_{n,\alpha}\widetilde{\\#}F_{n,\beta}\\\
&=\sum\limits_{\alpha=0}^{\infty}\sum\limits_{\beta=0}^{\infty}h^{\frac{\alpha+\beta}{2}}\sum\limits_{\gamma=0}^{\infty}h^{\gamma}\left.\left(\left(\frac{i\sigma}{2}\right)^{\gamma}(E_{n,\alpha}(x_{2},\xi_{2})F_{n,\beta}(y,\eta))\right)\right|_{x_{2}=y,\xi_{2}=\eta}\\\
&=\sum\limits_{j=0}^{\infty}\sum\limits_{\beta=0}^{j}\sum\limits_{\alpha=0}^{j-\beta}h^{\frac{j}{2}}\left.\left(\left(\frac{i\sigma}{2}\right)^{\frac{j-\alpha-\beta}{2}}(E_{n,\alpha}(x_{2},\xi_{2})F_{n,\beta}(y,\eta))\right)\right|_{x_{2}=y,\xi_{2}=\eta}.\\\
\end{split}$

Then we compare the parameter of the term of $h^{\frac{j}{2}}$ on both sides
and get

$-E_{n,0}F_{n,j}=\sum\limits_{\beta=0}^{j-1}\sum\limits_{\alpha=0}^{j-\beta}\left.\left(\left(\frac{i\sigma}{2}\right)^{\frac{j-\alpha-\beta}{2}}(E_{n,\alpha}(x_{2},\xi_{2})F_{n,\beta}(y,\eta))\right)\right|_{x_{2}=y,\xi_{2}=\eta},$

from which we can solve for $F_{n,j}$. Furthermore, by (A.1) and
$E_{n,0}=z-z_{n,0}$, we can check inductively that for $j\geq 0$, there are
$b_{n,j,k,l}$, $c_{n,j,k}$ s.t.

$\begin{split}&F_{n,j}=\sum_{k=0}^{j}(z-z_{n,0})^{-1}\prod\limits_{l=0}^{k}\left(b_{n,j,k,l}(x_{2},\xi_{2};z)(z-z_{n,0})^{-1}\right),\\\
&\text{with~{}}\prod\limits_{l=0}^{k}b_{n,j,k,l}(x_{2},\xi_{2};z)=\sum\limits_{\alpha=0}^{j+k-2}z^{\alpha}c_{n,j,k}(x_{2},\xi_{2}),\text{~{}for~{}appropriate~{}}c_{n,j,k}\in
S.\end{split}$

Notice that $\widetilde{\\#}$ differs from the actural sharp product $\\#$:

$a\\#b=e^{\frac{ih}{2}\sigma(D_{x_{2}},D_{\xi_{2}};D_{y},D_{\eta})}\left(a(x_{2},\xi_{2})b(y,\eta)\right)|_{x_{2}=y,\
\xi_{2}=\eta}.$ (A.14)

Now we claim that this formal expansion for $\sqrt{h}F_{n}$ is legitimate as
an asymptotic expansion in $S_{\delta}^{\delta}$ and in fact, it is exactly
the asymptotic expansion of $\sqrt{h}E_{n,\pm}$ when $|z|\leq
2\|{\mathscr{V}}\|_{\infty}$ and $|\operatorname{Im}z|\geq h^{\delta}$. In
fact, $\sqrt{h}(E_{n,\pm}^{-1}-F_{n})\in S^{-\infty}$.

In fact, since $|z|$ is bounded and $|\operatorname{Im}z|\geq h^{\delta}$ and
$F_{n,j}$ is a rational function in $z$, thus $h^{\frac{j}{2}}F_{n,j}\in
S_{\delta}^{j(\delta-\frac{1}{2})+\delta}$. Since
$j(\delta-\frac{1}{2})+\delta\to-\infty$, (A.2) is not only a formal expansion
but is indeed an asymptotic expansion of $F_{n}$ in the symbol class
$S_{\delta}^{\delta}$.

Furthermore, comparing (A.13) with (A.14), we see that
$F_{n}\\#E_{n,\pm}=1-R_{n}$ with $R_{n}\in S^{-\infty}$. By Beal’s lemma,
there is $\tilde{R}_{n}\in S^{-\infty}$ s.t.
$(1-R_{n}^{W})^{-1}=1-\tilde{R}_{n}^{W}$. Thus
$\sqrt{h}E_{n,\pm}^{-1}=F\\#(1-\tilde{R}_{n}^{W})\in S_{\delta}^{\delta}$ and
have exactly the same asymptotic expansion as $F_{n}$ in (A.2) since
$\tilde{R}_{n}\in S_{\delta}^{-\infty}$. Thus part (2) is proved.

It follows that $r_{n}:=\partial_{z}E_{n,\pm}\\#E_{n,\pm}^{-1}$ is also well-
defined with an asymptotic expansion in $S_{\delta}^{\delta}$. Since

$\begin{split}r_{n}&\sim\sum\limits_{\alpha=0}^{\infty}h^{\frac{\alpha}{2}}\partial_{z}E_{n,\alpha}\\#\sum\limits_{\beta=0}^{\infty}h^{\frac{\beta}{2}}F_{n,j}=\sum\limits_{\alpha=0}^{\infty}\sum\limits_{\beta=0}^{\infty}h^{\frac{\alpha+\beta}{2}}\sum\limits_{\gamma=0}^{\infty}h^{\gamma}\left.\left(\left(\frac{i\sigma}{2}\right)^{\gamma}(E_{n,\alpha}(x_{2},\xi_{2};z)F_{n,\beta}(y,\eta;z))\right)\right|_{x_{2}=y,\xi_{2}=\eta}\\\
&=\sum_{j=0}^{\infty}\sum\limits_{\alpha=0}^{j}\sum\limits_{\beta=0}^{j-\alpha}h^{\frac{j}{2}}r_{n,j,\alpha,\beta}\left.\left(\left(\frac{i\sigma}{2}\right)^{\frac{j-\alpha-\beta}{2}}(E_{n,\alpha}(x_{2},\xi_{2};z)F_{n,\beta}(y,\eta;z))\right)\right|_{x_{2}=y,\xi_{2}=\eta}.\end{split}$

Combining it with part (1) and (2) and the fact that $\sigma$ is linear in
$D_{x_{2}}$, $D_{\xi_{2}}$, we get part (3). Part (4) follows directly from
part (1), (2), (3) with Prop. A.2. ∎

##  Appendix B Lemmas for the proof of 7.8

In this subsection, we provide several lemmas that together complete the proof
of Lemma 7.8. We start with a proposition that expresses the Hilbert-Schmidt
norm of the quantization in terms of its operator-valued symbol.

######  Proposition B.1.

Let ${\mathscr{H}}_{1}$, ${\mathscr{H}}_{2}$ be two Hilbert spaces. Let
$P:{\mathbb{R}}^{2}\to\mathcal{L}({\mathscr{H}}_{1};{\mathscr{H}}_{2})$ be an
operator-valued symbol in the symbol class
$S({\mathbb{R}}^{2}_{y,\eta};\mathcal{L}({\mathscr{H}}_{1};{\mathscr{H}}_{2}))$.
Furthermore, let $\|\cdot\|_{\operatorname{HS}}$ denote the Hilbert-Schmidt
norm of maps ${\mathscr{H}}_{1}$ to ${\mathscr{H}}_{2}$ or
$L^{2}({\mathbb{R}}_{y};{\mathscr{H}}_{1})$ to
$L^{2}({\mathbb{R}}_{y};{\mathscr{H}}_{2})$. Then

$\|P^{W}(y,hD_{y})\|_{\operatorname{HS}}^{2}=\frac{1}{2\pi
h}\int_{{\mathbb{R}}^{2}}\|P(y,\eta)\|_{\operatorname{HS}}^{2}\ dy\ d\eta.$

In particular, if ${\mathscr{H}}_{1}={\mathscr{H}}_{2}={\mathbb{R}}$, for the
scalar-valued symbol $P$, we have

$\|P^{W}(y,hD_{y})\|_{\operatorname{HS}}^{2}=\frac{\|P(y,\eta)\|_{L^{2}({\mathbb{R}}^{2};{\mathbb{R}})}^{2}}{2\pi
h}.$ (B.1)

The next Lemma allows us to interchange the order of trace and integration.

######  Lemma B.2.

Let $E_{n,-}$, $E_{n,+}$ be as in (7.15). Let
$\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}$,
$\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W}$ be as in the proof of Lemma
7.7. Then, there exists a constant $C>0$ such that

$\begin{split}&\|\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W}E_{n,-}\|_{\operatorname{HS}(L^{2}({\mathbb{R}}^{2}_{x}),L^{2}({\mathbb{R}}_{x_{2}}))}\leq
Ch^{-1/2}R\text{ and
}\|E_{n,-}\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}\|_{\operatorname{HS}(L^{2}({\mathbb{R}}^{2}_{x}),L^{2}({\mathbb{R}}_{x_{2}}))}\leq
Ch^{-1/2}R.\end{split}$

###### Proof.

The first equation follows from (B.1). For the second equation, we first
recall that

######  Claim 1.

If $a\in S({\mathbb{R}}^{2n};\mathcal{L}(X,Y);m_{1})$, $b\in
S({\mathbb{R}}^{2n};\operatorname{HS}(Y,Z);m_{2})$ and $m_{1}m_{2}(x,\xi)\in
L^{2}({\mathbb{R}}^{2n}_{x,\xi})$, where $m_{1},m_{2}$ are order functions,
then

$b\\#a\in S({\mathbb{R}}^{2n};\operatorname{HS}(X,Z);m_{1}m_{2})\text{ and
}(b\\#a)^{W}=b^{W}a^{W}\in\operatorname{HS}(L^{2}({\mathbb{R}}^{n}_{x};X);L^{2}({\mathbb{R}}^{n}_{x};Y)).$

Furthermore, we claim that

######  Claim 2.

For any $k^{\prime}$ s.t. $1<k^{\prime}$, we have

  1. (1)

$E_{n,-}(x_{2},\xi_{2})\in
S({\mathbb{R}}^{2}_{x_{2},\xi_{2}};\mathcal{L}(B^{-k^{\prime}}_{x_{1}};{\mathbb{C}}^{2}))$,

  2. (2)

$\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{w}(x,D_{x_{1}},\xi_{2})\in
S({\mathbb{R}}^{2}_{x_{2},\xi_{2}};\operatorname{HS}(L^{2}_{x_{1}};B^{-k^{\prime}}_{x_{1}});m)$,
where $m(x_{2},\xi_{2})=(1+(|(x_{2},\xi_{2})|-R)_{+})^{-k^{\prime}}$ is the
order function.

Then it follows that, by Claim 1, we have
$E_{n,-}\\#\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{w}\in
S({\mathbb{R}}^{2}_{x_{2},\xi_{2}};\operatorname{HS}(L^{2}_{x_{1}});m)$, i.e.

$\|E_{n,-}\\#\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{w}(x_{2},\xi_{2})\|_{\operatorname{HS}(L^{2}_{x_{1}})}\leq
m(x_{2},\xi_{2})=(1+(|(x_{2},\xi_{2})|-R)_{+})^{-k^{\prime}}.$

Thus by Prop. B.1, since for all $k>0$,

$\int_{{\mathbb{R}}^{2}}[1+(|(x_{2},\xi_{2})|-R)_{+}]^{-2k}dxd\xi=\pi
R^{2}+\mathcal{O}(R^{\max(1,-2k+2)})=\mathcal{O}(R^{2}),$

we get
$\|E_{n,-}\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}\|_{\operatorname{HS}(L^{2}({\mathbb{R}}_{x_{2}};L^{2}({\mathbb{R}}_{x_{1}};{\mathbb{C}}^{4}));L^{2}({\mathbb{R}}_{x_{2}};{\mathbb{C}}^{2}))}\leq
Ch^{-1/2}R$ and the Lemma is proved. ∎

###### Proof of Claim 2.

For part $(1)$, note that $I:B^{m}\rightarrow B^{n}$ for $m>n+\frac{1}{2}$ is
Hilbert-Schmidt: Let $(e_{l})_{l\in{\mathbb{N}}}$ be orthonormal eigenbasis of
$B^{m}$ with $(1+y^{2}+D_{y}^{2})e_{l}=\lambda_{l}e_{l}$. Then
$\|e_{l}\|_{B^{m}}^{2}=\|(1+y^{2}+D_{y}^{2})^{m}e_{l}\|_{L^{2}}^{2}=1$, so
$\|e_{l}\|_{B^{n}}^{2}=\|(1+y^{2}+D_{y}^{2})^{n}e_{l}\|_{L^{2}}^{2}=(\lambda_{l}^{n-m})^{2}$.
Since $2(n-m)<-1$ and $\lambda_{l}=\mathcal{O}(l)$ as $l\rightarrow\infty$,
$\|I\|_{\operatorname{HS}}^{2}\leq
C\sum_{l=1}^{\infty}\lambda_{l}^{-2(n-m)}<\infty$ i.e. $I$ is Hilbert-Schmidt.

For part $(2)$, let $k$ be s.t. $0<k+1<k^{\prime}$, we have

$\begin{split}\|\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{w}\|_{\operatorname{HS}(L^{2}_{x_{1}};B^{-k^{\prime}}_{x_{1}})}&\leq\|I\|_{\operatorname{HS}(B^{-k}_{x_{1}};B^{-k^{\prime}}_{x_{1}})}\|\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{w}\|_{\mathcal{L}(L^{2}_{x_{1}};B^{-k}_{x_{1}})}\leq
C_{k,k^{\prime}}\|\Lambda^{-k}\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{w}\|_{\mathcal{L}(L^{2}_{x_{1}};L^{2}_{x_{1}})}\end{split}$
(B.2)

where $\Lambda^{-k}=(1+D_{x_{1}}^{2}+x_{1}^{2})^{-k/2}$. To estimate
$\|\Lambda^{-k}\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{w}\|_{\mathcal{L}(L^{2}_{x_{1}};L^{2}_{x_{1}})}$,
we view it as quantization in $(x_{1},\xi_{1})$. The corresponding symbol is
$\Lambda^{-k}\\#\tilde{\operatorname{1\hskip-2.75ptl}}_{R}:=K_{x_{2},\xi_{2}}(x_{1},\xi_{1})$.
We will show that

$\begin{split}&|K_{x_{2},\xi_{2}}(x_{1},\xi_{1})|\leq
C_{k}m(x_{2},\xi_{2})=C_{k}[1+(|(x_{2},\xi_{2})|-R)_{+}]^{-k},\\\
&|\partial_{x_{1}}^{\alpha}\partial^{\beta}_{\xi_{1}}K_{x_{2},\xi_{2}}(x_{1},\xi_{1})|\leq
C_{k}[1+(|(x_{2},\xi_{2})|-R)_{+}]^{-k}.\end{split}$ (B.3)

So that

$\frac{1}{m(x_{2},\xi_{2})}(\Lambda^{-k}\\#\tilde{\operatorname{1\hskip-2.75ptl}}_{R})_{x_{2},\xi_{2}}(x_{1},\xi_{1})\in
S({\mathbb{R}}^{2}_{x_{1},\xi_{1}}).$

Thus

$\|\Lambda^{-k}\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{w}\|_{\mathcal{L}(L^{2}_{x_{1}};L^{2}_{x_{1}})}\leq
C_{k}[1+(|(x_{2},\xi_{2})|-R)_{+}]^{-k}.$

By (B.2),

$\|\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{w}\|_{\operatorname{HS}(L^{2}_{x_{1}};B^{-k^{\prime}}_{x_{1}})}\leq
C_{k,k^{\prime}}[1+(|(x_{2},\xi_{2})|-R)_{+}]^{-k}.$

Furthermore, notice that for multi-indices $\alpha,\beta$,

$|\partial_{x_{1}}^{\alpha}\partial_{\xi_{1}}^{\beta}\Lambda_{-k}(x_{1},\xi_{1})|\leq
C_{\alpha,\beta}(1+|(x_{1},\xi_{1})|^{-(2k+|\alpha|+|\beta|)}).$

Similarly, we can show that

$\|\partial_{x_{2}}^{\alpha}\partial_{\xi_{2}}^{\beta}\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{w}\|_{\operatorname{HS}(L^{2}_{x_{1}};B^{-k^{\prime}}_{x_{1}})}\leq
C_{k,k^{\prime}}[1+(|(x_{2},\xi_{2})|-R)_{+}]^{-k}.$

Thus $\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{w}(x,D_{x_{1}},\xi_{2})\in
S({\mathbb{R}}^{2}_{x_{2},\xi_{2}};\operatorname{HS}(L^{2}_{x_{1}};B^{-k^{\prime}}_{x_{1}});m)$
and the claim is proved. So we only need to show (B.3). Recall
$K_{x_{2},\xi_{2}}(x_{1},\xi_{1})=(\Lambda_{-k}\\#\tilde{\operatorname{1\hskip-2.75ptl}}_{R})(x_{1},\xi_{1})$.
Since

$a^{W}\\#_{1}b^{W}(x_{1},\xi_{1})=\frac{1}{4\pi^{2}}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}e^{i(y_{1}\eta_{2}-y_{2}\eta_{1})}a(x_{1}-y_{1},\xi_{1}-\eta_{1})b(x_{1}-y_{2},\xi_{1}-\eta_{2})dy_{1}dy_{2}d\eta_{2}d\eta_{2}.$

Hence, we find

$K_{x_{2},\xi_{2}}(x_{1},\xi_{1})=\frac{1}{4\pi^{2}}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}e^{iQ(Y)}\Lambda_{-k}(x_{1}-y_{1},\xi_{1}-\eta_{1})\operatorname{1\hskip-2.75ptl}_{R}(x_{2}+h^{1/2}(x_{1}-y_{2}),\xi_{2}-h^{1/2}(\xi_{1}-\eta_{2}))dY$

where $Y=(y_{1},\eta_{2},y_{2},\eta_{2})$ and
$Q(Y)=y_{1}\eta_{2}-y_{2}\eta_{1}$. Note that integrand is non-zero only if

$|(y-y_{2},\eta-\eta_{2})|\geq\frac{1}{Ch^{1/2}}(|(x_{2},\xi_{2})|-R)$

for some $C>0$. Define

$L^{T}(Y,D_{Y})=(\gamma_{Y}+(1-\gamma_{Y})|\nabla
Q(Y)|^{2})^{-1}(\gamma_{Y}+(1-\gamma_{Y})\nabla Q(Y)D_{Y})$ (B.4)

where $L^{T}$ denotes formal adjoint of $L$ and $\gamma_{Y}\in
C_{0}^{\infty}({\mathbb{R}}^{4})$ such that $\gamma_{Y}=1$ for $|Y|<1-\beta$
and $\gamma_{Y}=0$ for $|Y|<1$ for some $0<\beta<1$. Then
$L^{T}e^{iQ(Y)}=e^{iQ(Y)}$. Moreover,
$L(Y,D_{Y})=a_{0}(Y)+\sum_{j=1}^{4}a_{j}(Y)D_{Y_{j}}$ where

$\begin{split}|\partial^{\alpha}a_{0}|&\leq C(|Y|+1)^{-(2+|\alpha|)},\text{
and }|\partial^{\alpha}a_{j}|\leq C(|Y|+1)^{-(1+|\alpha|)}.\end{split}$

Taking adjoint operator $L=L^{T}$, $(N+5)$-times, with $N\geq k$,

$K_{x_{2},\xi_{2}}(x_{1},\xi_{1})=\frac{1}{4\pi^{2}}\int_{{\mathbb{R}}^{2}}\int_{{\mathbb{R}}^{2}}e^{iQ(Y)}\Gamma_{N,-k,R}^{(x_{2},\xi_{2})}(x_{1},\xi_{1};Y)dY$

where

$\Gamma_{N,-k,R}^{(x_{2},\xi_{2})}=L^{N+5}[\Lambda_{-k}(x_{1}-y_{1},\xi_{1}-\eta_{1})\operatorname{1\hskip-2.75ptl}_{R}(x_{2}+h^{1/2}(x_{1}-y_{2}),\xi_{2}-h^{1/2}(\xi_{1}-\eta_{2}))].$

Since

$\begin{split}|\Gamma_{N,-k,R}^{(x_{2},\xi_{2})}|\leq\sum_{j=1}^{N+5}C_{j}|Y|^{j}|D_{Y}[\Lambda_{-k}(x_{2}-y_{1},\xi_{2}-\eta_{1})\operatorname{1\hskip-2.75ptl}_{R}(x_{2}+h^{1/2}(x_{1}-y_{2}),\xi_{2}-h^{1/2}(\xi_{1}-\eta_{2}))]|\\\
\leq\tilde{C}(1+|Y|^{2})^{-\frac{N+5}{2}}(1+|(x_{2}-y_{1},\xi_{2}-\eta_{1})|)^{-2k}H\bigg{(}|(x_{2}-y_{2},\xi_{2}-\eta_{2})|-\frac{|(x_{2},\xi_{2})|-R}{Ch^{1/2}}\bigg{)},\end{split}$

where $H$ is the Heaviside step function. Then on the support of this
function,

$\text{either}\quad|(x_{1},\xi_{1})|\leq\frac{(|(x_{2},\xi_{2})|-R)_{+}}{2Ch^{1/2}}\quad\text{or}\quad|(x_{1},\xi_{1})|\geq\frac{(|(x_{2},\xi_{2})|-R)_{+}}{2Ch^{1/2}}.$

In the first case,

$|(y_{2},\eta_{2})|\geq\frac{(|(x_{2},\xi_{2})|-R)_{+}}{2Ch^{1/2}}.$

Thus

$|\Gamma_{N,-k,R}^{(x_{2},\xi_{2})}(x_{1},\xi_{1})|\leq\frac{\tilde{\tilde{C}}[(1+(|(x_{2},\xi_{2})|-R)_{+}]^{-N}}{(1+|Y|^{2})^{5/2}}.$

In the second case,

$\begin{split}|\Gamma_{N,-k,R}^{(x_{2},\xi_{2})}(x_{1},\xi_{1})|&\leq\frac{\tilde{\tilde{C}}[1+|(x_{2}-y_{1},\xi_{2}-\eta_{1})|]^{-2k}[1+|(y_{1},\eta_{1})|]^{-N}}{(1+|Y|^{2})^{5/2}}\\\
&\leq\frac{\tilde{\tilde{C}}[(1+(|(x_{2},\xi_{2})|-R)_{+}]^{-\min{(N,2k)}}}{(1+|Y|^{2})^{5/2}}.\end{split}$

Choose $N\geq k$ so that in either case,

$|\Gamma_{N,-k,R}^{(x_{2},\xi_{2})}(x_{1},\xi_{1})|\leq\frac{\tilde{\tilde{C}}[(1+(|(x_{2},\xi_{2})|-R)_{+}]^{-k}}{(1+|Y|^{2})^{5/2}}.$

Then

$|\Lambda^{-k}\\#\tilde{\operatorname{1\hskip-2.75ptl}}_{R}(x)|=|K_{x_{2},\xi_{2}}(x_{1},\xi_{1})|\leq\frac{1}{4\pi^{2}}\int_{{\mathbb{R}}^{4}}|\Gamma_{N,-k,R}^{(x_{2},\xi_{2})}(x_{1},\xi_{1})|dY\leq
C_{k}[1+(|(x_{2},\xi_{2})|-R)_{+}]^{-k}.$

We proceed as before to get the same bound on
$|\partial_{x_{1}}^{\alpha}\partial_{\xi_{1}}^{\beta}\Gamma_{N,-k,R}^{(x_{2},\xi_{2})}(x_{1},\xi_{1})|$.
Hence

$|\partial_{x_{1}}^{\alpha}\partial_{\xi_{1}}^{\beta}K_{x_{2},\xi_{2}}(x_{1},\xi_{1})|\leq
C_{k}[1+(|(x_{2},\xi_{2})|-R)_{+}]^{-k}.$

Thus (B.3) is proved. ∎

######  Lemma B.3.

Let $E_{n,-},E_{n,+},E_{n,\pm}$ be as in (7.15). For $\operatorname{Im}z\neq
0$, both operators

$\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}E_{n,+}E_{n,\pm}^{-1}E_{n,-}\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}\text{~{}and~{}}\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W}E_{n,-}E_{n,+}E_{n,\pm}^{-1}\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W}$

are trace class as bounded linear operators
$\mathcal{L}(L^{2}({\mathbb{R}}_{x_{2}};L^{2}({\mathbb{R}}_{x_{1}};{\mathbb{C}}^{4})))$
and $\mathcal{L}(L^{2}({\mathbb{R}}_{x_{2}};{\mathbb{C}}^{2}))$, respectively.

###### Proof.

By Lemma B.2, the fact that
$\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}E_{n,+}$ is the adjoint of
$E_{n,-}\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}$ and boundedness of
$E_{n,\pm}$ from (7.18), we have

$\begin{split}&\operatorname{Tr}_{1}(\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}E_{n,+}E_{n,\pm}^{-1}E_{n,-}\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W})\leq\frac{CR^{2}}{h^{\frac{3}{2}}|\operatorname{Im}z|}\text{
and
}\operatorname{Tr}_{2}(\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W}E_{n,-}E_{n,+}E_{n,\pm}^{-1}\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W})\leq\frac{CR^{2}}{h^{\frac{3}{2}}|\operatorname{Im}z|}.\end{split}$

∎

The second proposition allows us to change the position of $E_{n,-}$ in the
averaging and limiting process in the proof of Lemma 7.8.

######  Lemma B.4.

Let $E_{n,-},E_{n,+},E_{n,\pm}$ be as in (7.15), then

$\operatorname{Tr}_{L^{2}({\mathbb{R}}^{2}_{x};{\mathbb{C}}^{4})}(\operatorname{1\hskip-2.75ptl}_{R}^{W}E_{n,+}E_{n,\pm}^{-1}E_{n,-}\operatorname{1\hskip-2.75ptl}_{R}^{W})-\operatorname{Tr}_{L^{2}({\mathbb{R}}_{x_{2}};{\mathbb{C}}^{2})}(\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W}E_{n,-}E_{n,+}E_{n,\pm}^{-1}\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W})\leq\frac{CR^{\frac{3}{2}}}{h|\operatorname{Im}z|}.$

###### Proof.

Since $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$ when $AB$ and $BA$ are
both of trace class.

$\begin{split}&\operatorname{Tr}_{1}(\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}E_{n,+}E_{n,\pm}^{-1}E_{n,-}\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W})-\operatorname{Tr}_{2}(\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W}E_{n,-}E_{n,+}E_{n,\pm}^{-1}\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W})\\\
=&\operatorname{Tr}_{2}(E_{n,-}(\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W})^{2}E_{n,+}E_{n,\pm}^{-1})-\operatorname{Tr}_{2}((\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W})^{2}E_{n,-}E_{n,+}E_{n,\pm}^{-1})\\\
=&\operatorname{Tr}_{2}\left[(E_{n,-}\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}-\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W}E_{n,-})\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}E_{n,+}E_{n,\pm}^{-1}\right]+\operatorname{Tr}_{2}\left[\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W}(E_{n,-}\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}-\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W}E_{n,-})E_{n,+}E_{n,\pm}^{-1}\right]\\\
:=&\operatorname{Tr}_{2}\left[[E_{n,-},\operatorname{1\hskip-2.75ptl}_{R}]_{w}\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}E_{n,+}E_{n,\pm}^{-1}\right]+\operatorname{Tr}_{2}\left[\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W}[E_{n,-},\operatorname{1\hskip-2.75ptl}_{R}]_{w}E_{n,+}E_{n,\pm}^{-1}\right]\\\
:=&\operatorname{Tr}_{2}(A_{1})+\operatorname{Tr}_{2}(A_{2})\end{split}$

where
$[E_{n,-},\operatorname{1\hskip-2.75ptl}_{R}]_{W}:=E_{n,-}\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}-\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{W}E_{n,-}$.
Then the following claim completes the proof.

######  Claim 3.

For $\operatorname{Im}z\neq 0$, $A_{1}$, $A_{2}$ are trace class operators and
there is a $C>0$ such that

$\operatorname{Tr}_{2}(A_{1}),\operatorname{Tr}_{2}(A_{2})\leq
Ch^{-1}|\operatorname{Im}z|^{-3/2}R^{3/2}.$

∎

###### proof of Claim 3.

From Lemma B.2, we already know

$\|[E_{n,-},\operatorname{1\hskip-2.75ptl}_{R}]_{W}\|_{\operatorname{HS}^{\operatorname{W}}}\leq
Ch^{-1/2}R,\quad$

where
$\operatorname{HS}^{\operatorname{W}}=\operatorname{HS}(L^{2}({\mathbb{R}}_{x_{2}};L^{2}({\mathbb{R}}_{x_{1}};{\mathbb{C}}^{4}));L^{2}({\mathbb{R}}_{x_{2}};{\mathbb{C}}^{2}))$.
We will improve the upper bound from $Ch^{-1/2}R$ to $Ch^{-1/2}R^{1/2}$.

Let $\bar{\chi}_{R}^{c}=1-\bar{\chi}_{R}$,
$\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{c}=1-\tilde{\operatorname{1\hskip-2.75ptl}}_{R}$.
First notice that from the proof of Lemma B.2, and replacing $\bar{\chi}_{R}$
by $\bar{\chi}_{R}^{c}$, we have

$\begin{split}&\|[E_{n,-},\operatorname{1\hskip-2.75ptl}_{R}]_{w}(x_{2},\xi_{2})\|_{\operatorname{HS}}\leq
C_{k}[1+(R-|(x_{2},\xi_{2})|)_{+}]^{-k},\\\
&\|[E_{n,-},\operatorname{1\hskip-2.75ptl}_{R}^{c}]_{w}(x_{2},\xi_{2})\|_{\operatorname{HS}}\leq
C_{k}[1+(|(x_{2},\xi_{2})|-R)_{+}]^{-k}\end{split}$

where
$[E_{n,-},\operatorname{1\hskip-2.75ptl}_{R}]_{w}(x_{2},\xi_{2})=E_{n,-}\\#\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{w}-\bar{\operatorname{1\hskip-2.75ptl}}_{R}\\#E_{n,-}$
is the symbol in $(x_{2},\xi_{2})$ of
$[E_{n,-},\operatorname{1\hskip-2.75ptl}_{R}]_{W}$ and
$\operatorname{HS}=\operatorname{HS}(L^{2}({\mathbb{R}}_{x_{1}};{\mathbb{C}}^{4});{\mathbb{C}}^{2})$.
Since
$[E_{n,-},\operatorname{1\hskip-2.75ptl}_{R}]_{w}=-[E_{n,-},\operatorname{1\hskip-2.75ptl}_{R}^{c}]_{w}$,
we have

$\|[E_{n,-},\operatorname{1\hskip-2.75ptl}_{R}]_{w}(x_{2},\xi_{2})\|_{\operatorname{HS}}\leq
C_{k}[1+||(x_{2},\xi_{2})|-R|]^{-k}.$

Thus by Prop. B.1 and a direct foward computation of the following integral

$\int_{{\mathbb{R}}^{2}_{x_{2},\xi_{2}}}[1+||(x_{2},\xi_{2})|-R|]^{-2k}dx_{2}d\xi_{2}=\frac{1}{(2k-2)(2k-1)}+\frac{R}{2k-1}=\mathcal{O}(R),$

we find that

$\|[E_{n,-},\operatorname{1\hskip-2.75ptl}_{R}]_{W}\|_{\operatorname{HS}^{\operatorname{W}}}\leq
Ch^{-1/2}R^{1/2}.$

Since $\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}E_{n,+}$ is the adjoint
of $E_{n,-}\tilde{\operatorname{1\hskip-2.75ptl}}_{R}^{W}$, this yields that

$\operatorname{Tr}(A_{1})\leq
Ch^{-3/2}R^{3/2},\quad\operatorname{Tr}(A_{2})\leq Ch^{-3/2}R^{3/2}.$

∎

The next Lemma exhibits the averaging property of the periodic symbols to
reduce the regularized trace to a fundamental cell.

######  Lemma B.5.

Let $E_{n,-},E_{n,+},E_{n,\pm}$, $\bar{\operatorname{1\hskip-2.75ptl}}_{R}$ be
as in (7.15). Then

$\lim\limits_{R\to\infty}\frac{1}{4R^{2}}\int_{{\mathbb{R}}^{2}}\operatorname{Tr}_{{\mathbb{C}}^{2}}(\bar{\operatorname{1\hskip-2.75ptl}}_{R}\\#\partial_{z}E_{n,\pm}\\#E_{n,\pm}^{-1}\\#\bar{\operatorname{1\hskip-2.75ptl}}_{R})~{}d{x_{2}}~{}d{\xi_{2}}=\frac{1}{|E|}\int_{E}\partial_{\bar{z}}\tilde{f}\operatorname{Tr}_{{\mathbb{C}}^{2}}(\partial_{z}E_{n,\pm}\\#E_{n,\pm}^{-1})~{}d{x_{2}}~{}d{\xi_{2}}.$

###### Proof.

Let $T=\partial_{z}E_{n,\pm}\\#E_{n,\pm}^{-1}$,
$\mathcal{T}=\bar{\operatorname{1\hskip-2.75ptl}}_{R}\\#\partial_{z}E_{n,\pm}\\#E_{n,\pm}^{-1}\\#\bar{\operatorname{1\hskip-2.75ptl}}_{R}=\bar{\operatorname{1\hskip-2.75ptl}}_{R}\\#\partial_{z}E_{n,\pm}\\#E_{n,\pm}^{-1}\\#\bar{\operatorname{1\hskip-2.75ptl}}_{R}\\#1$.
We add $1$ since formulae of compositions of $4$ symbols are more accessible
in this case. Thus for $X_{2}=(x_{2},\xi_{2})$,

$\mathcal{T}(X_{2})=\frac{1}{(2\pi h)^{4}}\int
e^{iQ_{4}(Y)/h}\bar{\operatorname{1\hskip-2.75ptl}}_{R}(X_{2}-Y_{1})T(X_{2}-Y_{2})\bar{\operatorname{1\hskip-2.75ptl}}_{R}(X_{2}-Y_{3})1(X_{2}-Y_{4})dY$

where $Y=(Y_{1},Y_{2},Y_{3},Y_{4})$, $Y_{i}=(y_{i},\eta_{i})$,
$dY=\prod_{i}dy_{i}d\eta_{i}$ and

$Q_{4}(Y)=\sigma(Y_{1},Y_{2})+\sigma(Y_{2},Y_{3})+\sigma(Y_{3},Y_{4})+\sigma(Y_{1},Y_{4})+\sigma(Y_{4},Y_{2})+\sigma(Y_{3},Y_{1})$

where $\sigma$ is the canonical symplectic form:
$\sigma(Y_{i},Y_{j})=\eta_{i}y_{j}-y_{i}\eta_{j}$.

######  Claim 4.

$\mathcal{T}$ satisfy

$\begin{cases}|\mathcal{T}(X_{2})|\leq
C_{N}(|X_{2}|-R)^{8-N}|\operatorname{Im}z|^{-(7+N)},&|X_{2}|>R+1\\\
|\mathcal{T}(X_{2})-T(X_{2})|\leq
3C_{N}(R-|X_{2}|)^{8-N}|\operatorname{Im}z|^{-(7+N)},&|X_{2}|<R-1\\\
|\mathcal{T}(X_{2})|\leq C_{N},&R-1\leq|X_{2}|\leq R+1\end{cases}$ (B.5)

Thus when $R\to\infty$, the tails all vanish.
$\operatorname{Tr}_{{\mathbb{C}}^{2}}(T(X_{2}))$ is periodic since $V$ is.
Thus integrating over $|X_{2}|<R-1$ and then devided by $4R^{2}$ gives us the
average

$\frac{1}{|E|}\int_{E}\partial_{\bar{z}}\tilde{f}\operatorname{Tr}_{{\mathbb{C}}^{2}}(T(X_{2}))dx_{2}d\xi_{2}.$

Thus Lemma B.5. ∎

###### Proof of Claim 4.

First consider when $|X_{2}|>R+1$. Note that $Q_{4}(Y)/h=Q_{4}(h^{-1/2}Y)$. We
can modify the definition of $L^{t}$ in (B.4), and still make

$L^{t}(h^{-{1/2}}Y,h^{1/2}D_{Y})e^{iQ_{4}/h}=e^{iQ_{4}/h}.$

Let $Y^{\prime}=\begin{pmatrix}A&0\\\ 0&-A\end{pmatrix}Y$, with
$Y=(y_{1},y_{2},y_{3},y_{4},\eta_{1},\eta_{2},\eta_{3},\eta_{4})^{T}$ and
invertible

$A=\begin{pmatrix}0&-1&1&-1\\\ 1&0&-1&1\\\ -1&1&0&-1\\\
1&-1&1&0\end{pmatrix}.$

Thus $\|Y\|\leq C\|Y^{\prime}\|\leq C^{\prime}\|Y\|$ for some $C,C^{\prime}>0$
and one can check

$\|\nabla Q_{4}(Y)\|^{2}=\|Y^{\prime}\|^{2}\sim\|Y\|^{2}.$

Thus denominator of

$L=b_{0}(h^{-{1/2}}Y)+\sum\limits_{i=1}^{8}b_{i}(h^{-1/2}Y)h^{1/2}D_{Y_{i}}$

with $|\partial^{\alpha}b_{i}|\leq(1+\|Y\|)^{-1-|\alpha|}$. And thus

$L^{N}=\sum_{|\beta|\leq N}b_{\beta}^{N}(h^{-1/2}Y)(h^{1/2}D_{Y})^{\beta}$
(B.6)

with $|\partial^{\alpha}b^{N}_{\beta}(Y)|\leq
C(\|Y\|+1)^{-2N+|\beta|-|\alpha|}$. Therefore, for any $N\in{\mathbb{N}}$,

$\mathcal{T}(X_{2})=\frac{1}{(2\pi h)^{4}}\int
e^{iQ_{4}(Y)}L^{N}[\bar{\operatorname{1\hskip-2.75ptl}}_{R}(X_{2}-Y_{1})T(X_{2}-Y_{2})\bar{\operatorname{1\hskip-2.75ptl}}_{R}(X_{2}-Y_{3})1(X_{2}-Y_{4})]dY.$
(B.7)

Since $|X_{2}|>R+1$, we have $|Y_{1}|,|Y_{3}|>|X|-R$. Since

$\|T\|\leq\frac{C}{|\operatorname{Im}z|}$

thus by [Sj89, Theorem 8.4],

$|\partial^{\alpha}_{x_{2}}\partial^{\beta}_{\xi_{2}}T|\leq
C_{\alpha,\beta}\max\left(1,\left(\frac{h}{|\operatorname{Im}z|^{2}}\right)^{3}\right)|\operatorname{Im}z|^{7+N}.$

Thus take $N>8$, by (B.7) and (B.6), we have

$\mathcal{T}(X_{2})\leq\frac{C_{N}}{(|X_{2}|-R)^{N-8}|\operatorname{Im}z|^{7+N}}.$

When $|X_{2}|<R-1$, we can rewrite

$\begin{split}\mathcal{T}(X_{2})=\bar{\operatorname{1\hskip-2.75ptl}}_{R}\\#T\\#\bar{\operatorname{1\hskip-2.75ptl}}_{R}\\#1=1\\#T\\#1\\#1-\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{c}\\#T\\#1\\#1-1\\#T\\#\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{c}\\#1+\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{c}\\#T\\#\bar{\operatorname{1\hskip-2.75ptl}}_{R}^{c}\\#1\end{split}$

where the last three terms can be computed in similar ways as the first case
and bounded by

$\frac{C_{N}}{(R-|X_{2}|)^{N-8}|\operatorname{Im}z|^{7+N}}.$

While the first term is precisely $T(X_{2})$.

Finally, when $R-1<|X_{2}|<R+1$, $\mathcal{T}(X_{2})$ is bounded if
$|\operatorname{Im}z|\neq 0$. Thus we proved (B.5). ∎

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$\displaystyle=\mathbb{E}_{U_{L}}\quantity[\Tr[O\,U_{L}P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}P_{i}U_{L}^{\dagger}]]$
(57)
$\displaystyle=\Tr\quantity[O\,\mathbb{E}_{U_{L}}\quantity[U_{L}P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}P_{i}U_{L}^{\dagger}]]$
(58)
$\displaystyle=\Tr\quantity[O\,\frac{\Tr[P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}P_{i}]\mathds{1}}{2^{n}}]=\frac{\Tr[O]}{2^{n}}=0\,,$
(59)

where in the first line we exchanged the trace and the expectation value since
both are linear operations, and in the second line we made use of Eq. (53) for
the first moment of the Haar distribution. Similarly, one can show that if
$\mathbb{U}_{R}$ forms a 1-design, then averaging over it yields the same
result, namely $\mathbb{E}_{U_{R}}[f(\bm{\theta}+\pi\bm{e}_{i})]=0$. The same
calculation for $f(\bm{\theta})$ shows that
$\mathbb{E}_{U_{R}}[f(\bm{\theta})]=\mathbb{E}_{U_{L}}[f(\bm{\theta})]=0$.

Thus, for every diagonal element of the Hessian, if either $\mathbb{U}_{L}$ or
$\mathbb{U_{R}}$ is a 1-design (that is Eq. (53) hold), then its expectation
value vanishes

$\mathbb{E}_{U_{R},U_{L}}[H_{ii}]=0\quad\forall i\quad\text{if either
{\hbox{\mathbb{U}_L}} or {\hbox{\mathbb{U_R}}} is a 1-design}.$ (60)

The variance of the diagonal elements can be calculated in a similar manner,
even though the calculation is more involved. Substituting Eq. (55) in the
defition of the variance, one obtains

$\displaystyle\text{Var}[H_{ii}]$
$\displaystyle:=\mathbb{E}[H_{ii}^{2}]-\mathbb{E}[H_{ii}]^{2}=\mathbb{E}[H_{ii}^{2}]$
$\displaystyle=\frac{1}{4}\quantity[\mathbb{E}[f(\bm{\theta}+\pi\bm{e}_{i})^{2}]+\mathbb{E}[f(\bm{\theta})^{2}]-2\mathbb{E}[f(\bm{\theta}+\pi\bm{e}_{i})f(\bm{\theta})]]\,.$
(61)

In order to use Eq. (54) for second moment integrals, we can rewrite these
expectation values as follow

$\displaystyle\mathbb{E}[f(\bm{\theta}+\pi\bm{e}_{i})^{2}]$
$\displaystyle=\mathbb{E}\quantity[\Tr[O\,U_{L}P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}P_{i}U_{L}^{\dagger}]^{2}]$
$\displaystyle=\mathbb{E}\quantity[\Tr[O\,U_{L}P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}P_{i}U_{L}^{\dagger}]\matrixelement{0}{U_{R}^{\dagger}P_{i}U_{L}^{\dagger}OU_{L}P_{i}U_{R}}{0}]$
$\displaystyle=\mathbb{E}\quantity[\Tr\quantity[O\,U_{L}P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}P_{i}U_{L}^{\dagger}OU_{L}P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}P_{i}U_{L}^{\dagger}]]$
$\displaystyle=\Tr\quantity[\mathbb{E}[O\,U_{L}P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}P_{i}U_{L}^{\dagger}OU_{L}P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}P_{i}U_{L}^{\dagger}]]\,,$
(62)

and similarly for the remaining two terms. Assuming that the set of unitaries
$\mathbb{U}_{L}$ generated by $U_{L}$ is a 2-design, then

$\displaystyle\mathbb{E}_{U_{L}}[f(\bm{\theta}+\pi\bm{e}_{i})^{2}]$
$\displaystyle=\Tr\quantity[\mathbb{E}_{U_{L}}[O\,U_{L}\underbrace{P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}P_{i}}_{B}U_{L}^{\dagger}OU_{L}\underbrace{P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}P_{i}}_{B}U_{L}^{\dagger}]]$
(63)
$\displaystyle=\Tr\quantity[\frac{\Tr[B^{2}]\Tr[O]O+\Tr[B]^{2}O^{2}}{2^{2n}-1}-\frac{\Tr[B^{2}]O^{2}+\Tr[B]^{2}\Tr[O]O}{2^{n}(2^{2n}-1)}]$
(64)
$\displaystyle=\frac{\Tr[O]^{2}+\Tr[O^{2}]}{2^{2n}-1}-\frac{\Tr[O^{2}]+\Tr[O]^{2}}{2^{n}(2^{2n}-1)}=\frac{1}{2^{n}+1}\,,$
(65)

where in the second line we made use of Eq. (54), and the third line the used
that $\Tr[B]=\Tr[B^{2}]=1$ since
$B=P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}P_{i}$ is a projector, and that
$\Tr[O]=0$ and $\Tr[O^{2}]=2^{n}$. Similarly, one can show that integration
over $\mathbb{U}_{R}$ yields the same result. Also, the same calculation leads
to
$\mathbb{E}_{U_{L}}[f(\bm{\theta})^{2}]=\mathbb{E}_{U_{R}}[f(\bm{\theta})^{2}]=1/(2^{n}+1)$.
Thus, if either $\mathbb{U}_{L}$ or $\mathbb{U_{R}}$ is a 2-design then

$\mathbb{E}_{U_{R},U_{L}}[f(\bm{\theta})^{2}]=\mathbb{E}_{U_{R},U_{L}}[f(\bm{\theta}+\pi\bm{e}_{i})^{2}]=\frac{1}{2^{n}+1}\quad\forall
i\quad\text{if either {\hbox{\mathbb{U}_L}} or {\hbox{\mathbb{U_R}}} is a
2-design}.$ (66)

Now we evaluate the correlation term
$\mathbb{E}[f(\bm{\theta}+\pi\bm{e}_{i})f(\bm{\theta})]$. If $\mathbb{U}_{L}$
is a 2-design, then

$\displaystyle\mathbb{E}_{U_{L}}[f(\bm{\theta}+\pi\bm{e}_{i})f(\bm{\theta})]$
$\displaystyle=\Tr\quantity[\mathbb{E}_{U_{L}}\quantity[O\,U_{L}P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}U_{L}^{\dagger}OU_{L}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}P_{i}U_{L}^{\dagger}]]$
$\displaystyle=\Tr\quantity[\frac{\Tr[P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}]^{2}O^{2}}{2^{2n}-1}-\frac{O^{2}}{2^{n}(2^{2n}-1)}]$
$\displaystyle=\frac{1}{2^{2n}-1}\quantity[2^{n}\,\Tr[P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}]^{2}-1]\,.$
(67)

While if $\mathbb{U}_{R}$ is a 2-design instead it holds

$\displaystyle\mathbb{E}_{U_{R}}[f(\bm{\theta}+\pi\bm{e}_{i})f(\bm{\theta})]$
$\displaystyle=\Tr\quantity[O\,U_{L}P_{i}\,\mathbb{E}_{U_{R}}\quantity[U_{R}\outerproduct{0}{0}U_{R}^{\dagger}U_{L}^{\dagger}OU_{L}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}]P_{i}U_{L}^{\dagger}]$
$\displaystyle=\Tr\quantity[O\,U_{L}P_{i}\,\frac{(2^{n}-1)U_{L}^{\dagger}OU_{L}}{2^{n}(2^{2n}-1)}\,P_{i}U_{L}^{\dagger}]$
$\displaystyle=\frac{1}{2^{n}(2^{n}+1)}\Tr[O\,U_{L}P_{i}U_{L}^{\dagger}OU_{L}P_{i}U_{L}^{\dagger}]\,.$
(68)

If both of them are 2-designs, then continuing from Eq. (B.2.1), one obtains

$\displaystyle\mathbb{E}_{U_{L},U_{R}}[f(\bm{\theta}+\pi\bm{e}_{i})f(\bm{\theta})]$
$\displaystyle=\frac{1}{2^{n}(2^{n}+1)}\Tr\quantity[\mathbb{E}_{U_{L}}\quantity[O\,U_{L}P_{i}U_{L}^{\dagger}OU_{L}P_{i}U_{L}^{\dagger}]]$
$\displaystyle=\frac{1}{2^{n}(2^{n}+1)}\Tr\quantity[\frac{\Tr[P_{i}]^{2}O^{2}+\Tr[P_{i}^{2}]\Tr[O]O}{2^{2n}-1}-\frac{\Tr[P_{i}^{2}]O^{2}+\Tr[P_{i}]^{2}\Tr[O]O}{2^{n}(2^{2n}-1)}]$
$\displaystyle=-\frac{1}{2^{n}(2^{n}+1)}\frac{\Tr[P_{i}^{2}]\Tr[O^{2}]}{2^{n}(2^{2n}-1)}=-\frac{1}{(2^{n}+1)(2^{2n}-1)}\in\order{2^{-3n}}$
(69)

Finally, plugging Eqs. (B.2.1), (B.2.1) and (B.2.1) in Eq. (B.2.1), one has
$\forall i=1,\ldots,M$

$\displaystyle\text{Var}_{U_{L},U_{R}}[H_{ii}]$
$\displaystyle=\frac{1}{2}\mathbb{E}[f(\bm{\theta})^{2}]-\frac{1}{2}\mathbb{E}[f(\bm{\theta}+\pi\bm{e}_{i})f(\bm{\theta})]$
$\displaystyle=\frac{1}{2(2^{n}+1)}-\frac{1}{2}\begin{cases}\displaystyle{\frac{1}{2^{2n}-1}\quantity[2^{n}\,\Tr[P_{i}U_{R}\outerproduct{0}{0}U_{R}^{\dagger}]^{2}-1]}\quad&\forall
i\,,\text{if $\mathbb{U}_{L}$ 2-design}\\\\[10.0pt]
\displaystyle{\frac{1}{2^{n}(2^{n}+1)}\Tr[O\,U_{L}P_{i}U_{L}^{\dagger}OU_{L}P_{i}U_{L}^{\dagger}]}\quad&\forall
i\,,\text{if $\mathbb{U}_{R}$ 2-design}\\\\[10.0pt]
\displaystyle{-\frac{1}{(2^{n}+1)(2^{2n}-1)}}\quad&\forall i\,,\text{if
$\mathbb{U}_{L},\mathbb{U}_{R}$ 2-designs}\\\ \end{cases}$ (70)

where $\mathbb{U}_{R}=\mathbb{U}_{R}^{(i)}$ and
$\mathbb{U}_{L}=\mathbb{U}_{L}^{(i)}$ are defined as in Eq. (56) and actually
depend on the index $i$ of the parameter.

Not surprisingly, as it happens for first order derivatives, also second order
derivatives of PQCs are found to be exponentially vanishing [36, 56], as from
Eq. (70) one can check that $\text{Var}[H_{ii}]\in\order{2^{-n}}$.

####  B.2.2 Statistics of the trace of the Hessian

The average value of the trace of the Hessian is easily found to be zero using
Eq. (60), in fact

$\mathbb{E}_{U_{R},U_{L}}[\Tr[H]]=\sum_{i=1}^{M}\mathbb{E}_{U^{(i)}_{R},U^{(i)}_{L}}[H_{ii}]=0\,,$
(71)

where we assume that for every parameter $i$ either $\mathbb{U}_{R}^{(i)}$ or
$\mathbb{U}_{L}^{(i)}$ is a 1-design. The variance of the trace is instead

$\displaystyle\text{Var}_{U_{R},U_{L}}[\Tr[H]]$
$\displaystyle=\text{Var}\quantity[\sum_{i=1}^{M}H_{ii}]=\sum_{i=1}^{M}\text{Var}[H_{ii}]+2\sum_{i<j}^{M}\text{Cov}[H_{ii}H_{jj}]\,.$
(72)

We can upper bound this quantity using the covariance inequality [79]

$\absolutevalue{\text{Cov}[H_{ii},H_{jj}]}\leq\sqrt{\text{Var}[H_{ii}]\text{Var}[H_{jj}]}\approx\text{Var}[H_{ii}]\,,$

were we assumed that $\text{Var}[H_{ii}]\approx\text{Var}[H_{jj}]\,\forall
i,j$. Using that $\text{Var}[H_{ii}]\in\order{2^{-n}}$ one finally has

$\displaystyle\text{Var}_{U_{R},U_{L}}[\Tr[H]]\leq\sum_{i=1}^{M}\text{Var}[H_{ii}]+2\sum_{i<j}^{M}\text{Var}[H_{ii}]\in\order{\frac{M^{2}}{2^{n}}}\,.$
(73)

Alternatively, one can obtain a tighter yet qualitative approximation by
explicitly considering the nature of the sums in Eq. (72). First, by using Eq.
(55), the covariance term is explicitly

$\text{Cov}[H_{ii},H_{jj}]=\mathbb{E}[H_{ii}H_{jj}]=\frac{1}{4}\mathbb{E}[(f_{i}-f)(f_{j}-f)]=\frac{1}{4}\mathbb{E}[f^{2}]+\frac{1}{4}\mathbb{E}[f_{i}f_{j}]-\frac{1}{4}\mathbb{E}[f_{i}f]-\frac{1}{4}\mathbb{E}[f_{j}f]\,,$
(74)

where for ease of notation we defined $f_{i,j}=f(\bm{\theta}+\pi\bm{e}_{i,j})$
and $f=f(\bm{\theta})$. Note that except for the first term which is always
positive, all remaining correlations terms can be both positive and negative.
Also, all of these terms are bounded from above by the same quantity, as via
Cauchy-Schwarz it follows

$\absolutevalue{\mathbb{E}[f_{i}f_{j}]}\leq\sqrt{\mathbb{E}[f_{i}^{2}]\mathbb{E}[f_{j}^{2}]}=\frac{1}{2^{n}+1}\quad\text{and}\quad\absolutevalue{\mathbb{E}[f_{i}f]}\leq\sqrt{\mathbb{E}[f_{i}^{2}]\mathbb{E}[f^{2}]}=\frac{1}{2^{n}+1}\,,$
(75)

where we have used $E[f^{2}]=E[f_{i}^{2}]=1/(2^{n}+1)$ from Eq. (66). Then,
the variance can be written as

$\displaystyle\text{Var}_{U_{R},U_{L}}[\Tr[H]]$
$\displaystyle=\sum_{i=1}^{M}\text{Var}[H_{ii}]+2\sum_{i<j}^{M}\mathbb{E}[H_{ii}H_{jj}]$
$\displaystyle=\sum_{i=1}^{M}\frac{\mathbb{E}[f^{2}]-\mathbb{E}[f_{i}f]}{2}+2\sum_{i<j}^{M}\frac{\mathbb{E}[f^{2}]+\mathbb{E}[f_{i}f_{j}]-\mathbb{E}[f_{i}f]-\mathbb{E}[f_{j}f]}{4}$
$\displaystyle=\frac{1}{2}\quantity(\sum_{i=1}^{M}+\sum_{i<j}^{M})\mathbb{E}[f^{2}]-\frac{1}{2}\quantity(\sum_{i=1}^{M}\mathbb{E}[f_{i}f]+\sum_{i<j}^{M}\mathbb{E}[f_{i}f]+\sum_{i<j}^{M}\mathbb{E}[f_{j}f])+\frac{1}{2}\sum_{i<j}^{M}\mathbb{E}[f_{i}f_{j}]$
$\displaystyle=\frac{M(M+1)}{4}\mathbb{E}[f^{2}]-\underbrace{\frac{M}{2}\sum_{i=1}^{M}\mathbb{E}[f_{i}f]+\frac{1}{2}\sum_{i<j}^{M}\mathbb{E}[f_{i}f_{j}]}_{\Delta}\,.$
(76)

Figure 16: Simulation results of evaluating the trace of the Hessian matrix
for two different hardware-efficient ansatzes with random values of the
parameters. The plot on the left is obtained using the layer template shown in
the figure for $n=6$ qubits and $l=6$ layers. The plot on the right instead
with $n=5$ and $l=5$ layers of the template shown in the corresponding inset.
The simulations are performed by sampling $2000$ random parameter vectors
$\bm{\theta}_{m}$ with $\theta_{i}\sim\text{Unif}[0,2\pi[$, evaluating the
trace of the Hessian matrix $\Tr[H(\bm{\theta})]$, and then building the
histogram to show its frequency distribution. In both experiments the measured
observable is $Z^{\otimes n}$. The length of the arrows are respectively:
“Numerical $2\sigma$” (black solid line) twice the numerical standard
deviation, “Approximation” (dashed red) twice the square root of the
approximation in Eq. (77), “Bound” (dashed-dotted green) twice the square root
of the upper Bound in Eq. (73). These parametrized circuits correspond to the
templates BasicEntanglinLayer and Simplified2Design defined in Pennylane [80],
and used for example in [55] to study barren plateaus.

##### Numerical simulations—

In addition to Figure 6 in the main text, in Figure 16 we report numerical
evidence for the trace of the Hessian for two common hardware-efficient
parametrized quantum circuit ansatzes. The histograms represent the frequency
of obtaining a given value of the trace of the Hessian $\Tr[H(\bm{\theta})]$
upon random assignments of the parameters. The length of the arrows are,
respectively: “Numerical $2\sigma$” (black solid line) twice the statistical
standard deviation computed from the numerical results, “Approximation”
(dashed red) twice the square root of the Eq. (B.2.2) with $\Delta=0$, “Bound”
(dashed-dotted green) twice the square root of the upper Bound in Eq. (73).

All simulations confirm the bound (73), and, more interestingly, both the
circuit on the left of Fig. 16 and the one in Fig. 6 in the main text, have a
numerical variance which is very well approximated by Eq. (B.2.2) with
$\Delta=0$. We conjecture this is due to the fact that all correlation terms
in Eq. (B.2.2) are roughly of the same order of magnitude (see Eq. (75)), and
can be either positive and negative, depending on the parameter and the
specifics of the ansatz. Thus, one can expect the whole contribution to either
vanish $\Delta\approx 0$, or be negligible with respect to the leading term.
If this is the case, then substituting $\mathbb{E}[f^{2}]=1/(2^{n}+1)$, the
variance of the Hessian is approximately

$\text{Var}_{U_{R},U_{L}}[\Tr[H]]\approx\frac{M(M+1)}{4}\mathbb{E}[f^{2}]=\frac{M(M+1)}{4(2^{n}+1)}\approx\frac{1}{4}\frac{M^{2}}{2^{n}}\,,$
(77)

which is four times smaller then the upper bound Eq. (73), but clearly has the
same scaling. While we numerically verified it also at other number of qubits,
more investigations are needed to understand if and when this approximation
holds, and we leave a detailed study of this phenomenon for future work.

##  Appendix C Visualization of CartPole policies obtained with Q-learning

(a) $\sigma=0$

(b) $\sigma=0.2$

Figure 17: Visualization of the Q-functions learned in the noise-free (a) and
noisy (b) settings. The red surface shows Q-values for pole angle and cart
position, orange for pole angle and cart velocity, and magenta for pole angle
and pole velocity.


* [186] Chong Sun et al.  “Finite-temperature density matrix embedding theory”  In _Phys. Rev. B_ 101, 2020, pp. 075131  DOI: 10.1103/PhysRevB.101.075131

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