proof will be spit into two parts. In Lemma 5.6 we derive an explicit
representation of the residuum $R_{J}$ by sorting the operator $H_{J}$ in
terms of powers in $p$ and $b_{>d}$, and in Theorem 5.7 we will make sure that
this residuum is indeed small compared to the operator $\mathbb{T}_{N}$
defined in Eq. (4.11), which is quadratic in the operators $p$ and $b_{>d}$.

In order to illustrate the emergence of the additional constants $c_{J}$ in
the residuum $R_{J}$ in Eq. (5.11), let us first investigate the following toy
problem.

Example. Consider the toy Hamiltonian $H_{\mathrm{toy}}:=b_{1}^{\dagger}b_{1}$
and the corresponding Hartree functional
$\mathcal{E}_{\mathrm{toy}}:\mathbb{C}\longrightarrow\mathbb{C}$ given by
$\mathcal{E}_{\mathrm{toy}}[z]:=|z|^{2}$. Using $b_{1}=q_{1}+ip_{1}$ and the
commutation relation $[ip_{1},q_{1}]=\frac{1}{2N}$, we obtain

$\displaystyle
H_{\mathrm{toy}}=q_{1}^{2}\\!+\\!p_{1}^{2}\\!-\\!\frac{1}{2N}=q_{1}^{2}\\!-\\!\frac{1}{4}\left(b_{1}\\!-\\!b_{1}^{\dagger}\right)^{2}\\!-\\!\frac{1}{2N}=q_{1}^{2}\\!+\\!\frac{1}{2}b_{1}^{\dagger}b_{1}\\!-\\!\frac{1}{4}b_{1}^{2}\\!-\\!\frac{1}{4}\left(b_{1}^{\dagger}\right)^{2}\\!-\\!\frac{1}{4N}.$
(5.13)

Let $D_{\mathcal{V}}$ be the derivative with respect to the imaginary part and
$z=t+is\in\mathbb{C}$, then

$\displaystyle\frac{1}{2}D^{2}_{\mathcal{V}}|_{0}\,\mathcal{E}_{\mathrm{toy}}(z)=\frac{1}{2}D^{2}|_{0}\,\mathcal{E}_{\mathrm{toy}}(is)=s^{2}=\frac{1}{2}|z|^{2}-\frac{1}{4}z^{2}-\frac{1}{4}\bar{z}^{2}.$

With the definition
$c_{\mathrm{toy}}:=-\frac{1}{8}\partial_{t}^{2}\big{|}_{t=0}\mathcal{E}_{\mathrm{toy}}[t]=-\frac{1}{4}$
we can therefore rewrite Eq. (5.13) as

$\displaystyle
H_{\mathrm{toy}}=\mathcal{E}_{\mathrm{toy}}[q_{1}]+\frac{1}{2}D^{2}_{\mathcal{V}}|_{0}\mathcal{E}_{\mathrm{toy}}(b_{1})+\frac{c_{\mathrm{toy}}}{N}.$

######  Definition 5.5 (Taylor approximation of the square root).

Let $\eta_{m}$ be the function defined in Eq. (5.7) and let us define the
constant $c_{m}:=\frac{m}{8}d$. We then define the residuum corresponding to
the operator Taylor approximation of
$\Big{(}1-\mathbb{L}^{\prime}\Big{)}^{\frac{m}{2}}$, for different degrees of
accuracy, as

$\displaystyle E_{m}^{0}:$
$\displaystyle=\Big{(}1-\mathbb{L}^{\prime}\Big{)}^{\frac{m}{2}}-\eta_{m}\left(q\right),$
$\displaystyle E_{m}^{1}:$
$\displaystyle=\Big{(}1-\mathbb{L}^{\prime}\Big{)}^{\frac{m}{2}}-\eta_{m}\left(q\right)-D_{\mathcal{V}}\big{|}_{q}\eta_{m}\big{(}b_{\geq
1}\big{)},$ $\displaystyle E_{m}^{2}:$
$\displaystyle=\Big{(}1-\mathbb{L}^{\prime}\Big{)}^{\frac{m}{2}}-\eta_{m}\left(q\right)-D_{\mathcal{V}}\big{|}_{q}\eta_{m}\big{(}b_{\geq
1}\big{)}-\frac{1}{2}D^{2}_{\mathcal{V}}\big{|}_{0}\eta_{m}\big{(}b_{\geq
1}\big{)}-\frac{c_{m}}{N}.$

######  Lemma 5.6.

Let $J=(i,j)\in\\{0,\dots,4\\}^{2}$ be such that $\lambda_{J}\neq 0$, where
$\lambda_{J}$ is defined in Lemma 5.3, and let $R_{J}$ be the residuum defined
in Eq. (5.11). By distinguishing different cases with the help of the index
$e_{J}:=\left|\\{\ell\in J:\ell\in\\{3,4\\}\\}\right|$ and the index
$m_{J}:=\left|\\{\ell\in J:\ell=0\\}\right|$, we can explicitly express
$R_{J}$ as

  * •

In the case $m_{J}=2$, i.e. $J=(0,0)$: $R_{(0,0)}=\left(u_{0}^{\dagger}\cdot
T\cdot u_{0}\right)E_{2}^{2}$.

  * •

In the case $e_{J}=0$ and $m_{J}<2$: $R_{J}=\left(h_{i}^{\dagger}\cdot T\cdot
h_{j}\right)E_{m_{J}}^{1}$.

  * •

In the case $e_{J}=1$, there exists a constant $C$ and functions
$F_{J}:\mathbb{R}^{d}\longrightarrow\mathbb{R}$ with $|F_{J}(t)|\leq C|t|$,
such that

$\displaystyle R_{J}=\left(h_{i}^{\dagger}\cdot T\cdot
h_{j}\right)E_{m_{J}}^{0}+\frac{F_{J}(q)}{N}.$ (5.14)

  * •

For $e_{J}=2$ we distinguish further between the individual cases and obtain

$\displaystyle\ \ \ \ R_{(3,3)}$
$\displaystyle=\left(ip^{\prime}-ip\right)^{\dagger}\cdot T\cdot
ip^{\prime}+(ip)^{\dagger}\cdot T\cdot\left(ip^{\prime}-ip\right),$
$\displaystyle R_{(3,4)}$
$\displaystyle=\left(ip^{\prime}-ip\right)^{\dagger}\cdot T\cdot
b_{>d}=R_{(4,3)}^{\dagger},$ $\displaystyle R_{(4,4)}$ $\displaystyle=0.$

###### Proof.

The Lemma can be verified by straightforward computations for the different
individual cases. For the purpose of illustration, we will explicitly carry
out the computations for the case $J=(3,j)$ with $j\in\\{0,1,2\\}$, i.e. we
are going to verify Eq. (5.14) for this special case. Using the definition of
$E^{0}_{m}$ in Definition 5.5, the observation $h_{j}=e_{j}(q)$ and the fact
that $\left(ip^{\prime}_{\ell}\right)^{\dagger}=b_{\geq
1}^{\dagger}\cdot\big{(}u_{\ell}-\partial_{u_{\ell}}f(q)\big{)}-\big{(}u_{\ell}-\partial_{u_{\ell}}f(q)\big{)}^{\dagger}\cdot
b_{\geq 1}$, we obtain

$\displaystyle H_{J}=\left(ip^{\prime}\right)^{\dagger}\cdot T\cdot
e_{j}(q)\left(1-\mathbb{L}^{\prime}\right)^{\frac{m}{2}}=\left(ip^{\prime}\right)^{\dagger}\cdot
T\cdot e_{j}(q)\eta_{m}(q)+\left(ip^{\prime}\right)^{\dagger}\cdot T\cdot
e_{j}(q)E^{0}_{m}$ $\displaystyle\ \ =\frac{1}{2}\\!\sum_{\ell=1}^{d}b_{\geq
1}^{\dagger}\\!\cdot\\!\big{(}u_{\ell}\\!-\\!\partial_{u_{\ell}}f(q)\\!\big{)}u_{\ell}^{\dagger}\\!\cdot\\!T\\!\cdot\\!e_{j}(q)\,\eta_{m}(q)-\frac{1}{2}\\!\sum_{\ell=1}^{d}\big{(}u_{\ell}\\!-\\!\partial_{u_{\ell}}f(q)\\!\big{)}^{\dagger}\\!\cdot\\!b_{\geq
1}u_{\ell}^{\dagger}\\!\cdot\\!T\\!\cdot\\!e_{j}(q)\,\eta_{m}(q)$
$\displaystyle\ \ \ \ \ \ \ \ \ +\left(ip^{\prime}\right)^{\dagger}\cdot
T\cdot e_{j}(q)E^{0}_{m},$ (5.15)

where $m:=m_{J}$. Our goal is to commute $b_{\geq 1}$ in
$\big{(}u_{\ell}-\partial_{u_{\ell}}f(q)\big{)}^{\dagger}\cdot b_{\geq
1}u_{\ell}^{\dagger}\cdot T\cdot e_{j}(q)\,\eta_{m}(q)$ to the right side, in
order to obtain an expression which is of the same form as (5.10). We define
the corresponding functions $w$ and $\widetilde{w}$ as

$\displaystyle w:=-\frac{1}{2}\sum_{\ell=1}^{d}\left(u_{\ell}^{\dagger}\cdot
T\cdot e_{j}\left(\vec{t}\ \right)\\!\eta_{m}\left(\vec{t}\
\right)\right)\big{(}u_{\ell}-\partial_{t_{\ell}}f(t)\big{)}$

and $\widetilde{w}(t):=-w(t)$. The commutation law
$\left[g(q),\big{(}u_{\ell}\\!-\\!\partial_{u_{\ell}}f(q)\\!\big{)}^{\dagger}\\!\cdot\\!b_{\geq
1}\right]=\left[g(q),ip_{\ell}\right]=-\frac{1}{2N}\partial_{\ell}g(q)$, for
$C^{1}$ functions $g:\mathbb{R}^{d}\longrightarrow\mathbb{R}$ then yields

$\displaystyle-\frac{1}{2}\sum_{\ell=1}^{d}\big{(}u_{\ell}-\partial_{u_{\ell}}f(q)\big{)}^{\dagger}\\!\cdot\\!b_{\geq
1}\,u_{\ell}^{\dagger}\cdot T\cdot e_{j}(q)\,\eta_{m}(q)=w(q)^{\dagger}\cdot
b_{\geq 1}+\frac{1}{N}y(q),$

where $y:\mathbb{R}^{d}\longrightarrow\mathbb{R}$ is defined as
$y(t):=-\frac{1}{4}\sum_{\ell}\partial_{\ell}\left(u_{\ell}\cdot T\cdot
e_{j}(\vec{t}\ )\eta_{m}(\vec{t}\ )\right)$. Furthermore

$\displaystyle D_{\mathcal{V}}\big{|}_{\vec{t}}\ \mathcal{E}_{J}(z)\\!$
$\displaystyle=\\!e_{3}(z)^{\dagger}\\!\cdot\\!T\\!\cdot\\!e_{j}\left(\vec{t}\
\right)\\!\eta_{m}\\!\left(\vec{t}\
\right)\\!=\\!\sum_{\ell=1}^{d}\\!\left(i\mathfrak{Im}\left[\\!\big{(}u_{\ell}\\!-\\!\partial_{t_{\ell}}f(t)\big{)}^{\dagger}\\!\cdot\\!z\\!\right]u_{\ell}\right)^{\dagger}\\!\cdot\\!T\\!\cdot\\!e_{j}\left(\vec{t}\
\right)\\!\eta_{m}\\!\left(\vec{t}\ \right)$
$\displaystyle=w(t)^{\dagger}\cdot z+z^{\dagger}\cdot\widetilde{w}(t).$

Consequently we can rewrite Eq. (5.1) as

$\displaystyle\left(ip^{\prime}\right)^{\dagger}\cdot T\cdot
e_{j}(q)\left(1-\mathbb{L}^{\prime}\right)^{\frac{m}{2}}$
$\displaystyle=D_{\mathcal{V}}\big{|}_{q}\ \mathcal{E}_{J}\left(b_{\geq
1}\right)+\frac{1}{N}y(q)+\left(ip^{\prime}\right)^{\dagger}\cdot T\cdot
e_{j}(q)E^{0}_{m}.$

Note that $\mathcal{E}_{J}\left(\vec{t}\,\right)=0$ and
$D^{2}_{\mathcal{V}}\big{|}_{0}\mathcal{E}_{J}=0$. Therefore Eq. (5.14)
follows from the fact that $F(t):=y(t)-c_{J}$ is Lipschitz and $F(0)=0$, which
implies that there exists a constant $C$ such that $|F(t)|\leq C|t|$. ∎

For the proof of the following Theorem, we will use various operator estimates
derived in Appendices C and D.

######  Theorem 5.7.

Let $J\in\\{0,\dots,4\\}^{2}$ be such that $\lambda_{J}\neq 0$ and let $R_{J}$
be the residuum defined in Eq. (5.11). Then,

$\displaystyle R_{J}=o_{*}\left(\mathbb{T}_{N}\right),$

with $\mathbb{T}_{N}$ defined in Eq. (4.11) and the $o_{*}(\cdot)$ notation
from Definition 4.10.

###### Proof.

Recall the definitions in Lemma 5.6 of $e_{J}:=\left|\\{l\in
J:l\in\\{3,4\\}\\}\right|$, which counts how many of the indices in $J=(i,j)$
are equal to $3$ or $4$, $m_{J}:=\left|\\{l\in J:l=0\\}\right|$, which counts
how many of the indices are zero, and the residuum $R_{J}$ defined in Eq.
(5.11). In order to prove the statement of the Theorem, we are going to verify
$R_{J}=o_{*}\left(\mathbb{T}_{N}\right)$ for all $J$ with $\lambda_{J}\neq 0$.

The case $J=(0,0)$: In this case we have the identity
$R_{(0,0)}=\left(u_{0}^{\dagger}\cdot T\cdot u_{0}\right)E_{2}^{2}$, hence we
have to verify $E_{2}^{2}=o_{*}\left(\mathbb{T}_{N}\right)$. In order to do
this, recall the function $\eta_{2}(x)=1-\|F(x)\|^{2}$ from Eq. (5.7) and let
us compute using Lemma 4.9

$\displaystyle 1-\mathbb{L}^{\prime}$
$\displaystyle=1-\left(q+f(q)+ip^{\prime}+b_{>d}\right)^{\dagger}\cdot\left(q+f(q)+ip^{\prime}+b_{>d}\right)$
$\displaystyle=1-q^{\dagger}\cdot q-f(q)^{\dagger}\cdot
f(q)-f(q)^{\dagger}\cdot b_{>d}-b_{>d}^{\dagger}\cdot f(q)$ $\displaystyle\ \
\ \ -b_{>d}^{\dagger}\cdot b_{>d}-\left(p^{\dagger}\cdot
p\right)+\frac{d}{2N}-p^{\dagger}\cdot\left(p^{\prime}-p\right)-\left(p^{\prime}-p\right)^{\dagger}\cdot
p^{\prime}$
$\displaystyle=\eta_{2}(q)+D_{\mathcal{V}}\big{|}_{q}\eta_{2}\big{(}b_{\geq
1}\big{)}+D^{2}_{\mathcal{V}}\big{|}_{0}\eta_{2}\big{(}b_{\geq
1}\big{)}+\frac{d}{4N}-p^{\dagger}\cdot\left(p^{\prime}-p\right)-\left(p^{\prime}-p\right)^{\dagger}\cdot
p^{\prime},$

where we used $\eta_{2}(q)=1-q^{\dagger}\cdot q-f(q)^{\dagger}\cdot f(q)$ and
$p^{\dagger}\cdot
p=\frac{1}{4}\sum_{j=1}^{d}\left(2b_{j}^{\dagger}b_{j}-b_{j}^{2}-\left(b_{j}^{\dagger}\right)^{2}\right)+\frac{d}{4N}$.
Note that $\frac{c_{2}}{N}=\frac{d}{4N}$, where $c_{2}$ is the constant from
Definition 5.5. Since $p^{2}\leq\mathbb{T}_{N}$, it is clear that
$p^{2}=O_{*}\left(\mathbb{T}_{N}\right)$. In Lemmata C.6 and C.5, we will
verify that $\left(p^{\prime}\right)^{2}=O_{*}\left(\mathbb{T}_{N}\right)$ and
$\left(p^{\prime}-p\right)^{2}=o_{*}\left(\mathbb{T}_{N}\right)$. Therefore we
obtain by the operator Cauchy–Schwarz inequality in the auxiliary Lemma C.1
that $p^{\dagger}\cdot\left(p^{\prime}-p\right)$ as well as
$\left(p^{\prime}-p\right)^{\dagger}\cdot p^{\prime}$ are of order
$o_{*}\left(\mathbb{T}_{N}\right)$. We conclude
$E_{2}^{2}=o_{*}\left(\mathbb{T}_{N}\right)$.

The case $e_{J}=0$, with $J\neq(0,0)$: In this case $m_{J}\in\\{0,1\\}$ and
the error is given by

$\displaystyle R_{J}=h_{i}^{\dagger}\cdot T\cdot
h_{j}\,E^{1}_{m_{J}}=e_{i}\left(q\right)^{\dagger}\cdot T\cdot
e_{j}\left(q\right)\,E^{1}_{m_{J}}.$

We clearly have $E_{0}^{1}=0$. For $m_{J}=1$, let us define the function
$V(t):=e_{i}\left(\vec{t}\ \right)^{\dagger}\cdot T\cdot e_{j}\left(\vec{t}\
\right)$, which satisfies $V(t)\leq C|t|$ for a constant $C$. In Lemma D.2 we
will then verify that $V(q)E^{1}_{1}=o_{*}\left(\mathbb{T}_{N}\right)$.

The case $e_{J}=1$: In this case the error reads
$R_{J}=\left(h^{\dagger}_{i}\cdot T\cdot
h_{j}\right)\,E^{0}_{m_{J}}+\frac{F_{J}(q)}{N}$, where $F_{J}(t)\leq C|t|$ for
some constant $C$. Using Lemma D.2 and Lemma C.4 from the Appendix, we obtain
that $\left(E^{0}_{1}\right)^{\dagger}E^{0}_{1}=o_{*}\left(T_{N}\right)$ and
$\frac{F_{J}(q)}{N}=o_{*}\left(\frac{1}{N}\right)$. Regarding the first term,
note that $E^{0}_{0}=0$. Hence, we assume w.l.o.g. $m_{J}=1$. We are done once
we can verify

$\displaystyle\left(h^{\dagger}_{i}\cdot T\cdot
h_{j}\right)\cdot\left(h^{\dagger}_{i}\cdot T\cdot
h_{j}\right)^{\dagger}=O_{*}\left(\mathbb{T}_{N}\right)$ (5.16)

in case one of the indices $i,j$ is in $\\{3,4\\}$ and the other is zero. Let
us first assume $i\in\\{3,4\\}$. Then $h^{\dagger}_{i}\cdot T\cdot
h_{j}=h_{i}^{\dagger}\cdot w$, with $w:=T\cdot u_{0}\in\mathcal{H}$, and
therefore Eq. (5.16) follows from Lemma C.6 in the case $i=3$ and from Lemma
C.5 in the case $i=4$. The proof of the case $j\in\\{3,4\\}$ follows
analogously.

The case $e_{J}=2$: In this case, the error is a linear combination of
$\left(ip^{\prime}-ip\right)^{\dagger}\cdot T\cdot h_{j}$ and
$h_{i}^{\dagger}\cdot T\cdot\left(ip^{\prime}-ip\right)$ with
$h_{i},h_{j}\in\\{p^{\prime},b_{>d}\\}$. Note that
$A:=\sqrt{T}\left(1_{\mathcal{H}}-\pi_{>d}\right)$ is bounded, and therefore

$\displaystyle\left(ip^{\prime}-ip\right)^{\dagger}\cdot
T\cdot\left(ip^{\prime}-ip\right)=\left(ip^{\prime}-ip\right)^{\dagger}\cdot
A^{\dagger}A\cdot\left(ip^{\prime}-ip\right)$ $\displaystyle\ \ \ \
\leq\|A\|^{2}\
\left(ip^{\prime}-ip\right)^{\dagger}\cdot\left(ip^{\prime}-ip\right)=o_{*}\left(\mathbb{T}_{N}\right)$

by Lemma C.5. Similarly, we have $\left(p^{\prime}\right)^{\dagger}\cdot
T\cdot p^{\prime}\leq\|A\|^{2}(p^{\prime})^{\dagger}\cdot
p^{\prime}=O_{*}\left(\mathbb{T}_{N}\right)$ by Lemma C.6. Hence Lemma C.1
tells us that $\left(ip^{\prime}-ip\right)^{\dagger}\cdot T\cdot h_{j}$ and
$h_{i}^{\dagger}\cdot T\cdot\left(ip^{\prime}-ip\right)$ are of order
$o_{*}\left(\mathbb{T}_{N}\right)$. ∎

######  Corollary 5.8.

Recall the functional $\mathcal{E}_{A}$ defined in Eq. (5.9) and let us define
the constant $c:=\sum_{J\in\\{0,\dots,4\\}^{2}}\lambda_{J}c_{J}$. Then

$\displaystyle\mathcal{W}_{N}\widetilde{A}_{N}\mathcal{W}_{N}^{-1}$
$\displaystyle=\mathcal{E}_{A}(q)+D_{\mathcal{V}}\big{|}_{q}\mathcal{E}_{A}\left(b_{\geq
1}\right)+\frac{1}{2}D^{2}_{\mathcal{V}}\big{|}_{0}\mathcal{E}_{A}\left(b_{\geq
1}\right)+\frac{c}{N}+o_{*}\left(\mathbb{T}_{N}\right).$

###### Proof.

The statement follows from combining Lemma 5.3 and Theorem 5.7. ∎

###  5.2 Taylor Expansion of
$\mathcal{W}_{N}\widetilde{B}_{N}\mathcal{W}_{N}^{-1}$

Similar to the previous subsection, we introduce atoms $H_{J}$ in Definition
5.9 as well as their classical counterparts $\mathcal{E}_{J}$ in Definition
5.10. In Lemma 5.11 we explain how
$\mathcal{W}_{N}\widetilde{B}_{N}\mathcal{W}_{N}^{-1}$ and $\mathcal{E}_{B}$
can be written in terms of $H_{J}$ and $\mathcal{E}_{J}$, respectively.

######  Definition 5.9.

Recall the definition of
$h_{i}:\mathrm{dom}[\mathcal{N}]\longrightarrow\mathcal{F}_{0}\otimes\mathcal{H}$
from Definition 5.1. For a multi-index $J=(i,j,k,\ell)$ with
$i,j,k,\ell\in\\{0,\dots,4\\}$, we define an operator $H_{J}$ on
$\mathcal{W}_{N}\mathcal{F}_{\leq N}$ as

$\displaystyle H_{J}:$ $\displaystyle=\left(h_{i}\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot\hat{v}\cdot h_{k}\ \underline{\otimes}\
h_{\ell}\left(1-\mathbb{L}^{\prime}\right)^{\frac{m_{J}}{2}},$

where $m_{J}$ counts how many of the indices $i,j,k,\ell$ are zero.

######  Definition 5.10.

Recall the definition of $e_{i}:\mathcal{H}_{0}\longrightarrow\mathcal{H}$ and
$\eta_{m}$ from Definition 5.2. For a multi-index $J=(i,j,k,\ell)$ with
$i,j,k,\ell\in\\{0,\dots,4\\}$, we define
$\mathcal{E}_{J}:\mathcal{H}_{0}\cap\mathrm{dom}[T]\longrightarrow\mathbb{C}$

$\displaystyle\mathcal{E}_{J}(z):$ $\displaystyle=\Big{[}e_{i}(z)\otimes
e_{j}(z)\Big{]}^{\dagger}\cdot\hat{v}\cdot e_{k}(z)\otimes e_{\ell}(z)\
\eta_{m_{J}}(z),$

where $m_{J}$ counts how many of the indices $i,j,k,\ell$ are zero and
$\eta_{m}$ are the functions defined in Eq. (5.7).

######  Lemma 5.11.

Let us define for all $i,j,k,\ell\in\\{1,\dots,4\\}$ the coefficients
$\lambda_{(0,0,0,0)}:=\frac{1}{2}$, $\lambda_{(i,0,0,0)}:=2$,
$\lambda_{(i,j,0,0)}:=\lambda_{(i,0,k,0)}:=\lambda_{(0,j,k,0)}:=1$,
$\lambda_{(i,j,k,0)}:=2$, $\lambda_{(i,j,k,\ell)}:=\frac{1}{2}$ and all other
coefficients are defined as $\lambda_{J}:=0$. Then

$\displaystyle\mathcal{W}_{N}\,\widetilde{B}_{N}\,\mathcal{W}_{N}^{-1}$
$\displaystyle=\sum_{J\in\\{0,\dots,4\\}^{4}}\lambda_{J}\
\mathfrak{Re}\left[H_{J}\right].$

Furthermore, the functional $\mathcal{E}_{B}$ defined as

$\displaystyle\mathcal{E}_{B}(z)$
$\displaystyle:=\sum_{J\in\\{0,\dots,4\\}^{4}}\lambda_{J}\
\mathfrak{Re}\left[\mathcal{E}_{J}(z)\right],$ (5.17)

is an extension of $\mathcal{E}^{\prime}_{B}\big{|}_{B_{r}}$ defined in Eq.
(5.6), where
$B_{r}:=\\{z\in\mathcal{H}_{0}\cap\mathrm{dom}\left[T\right]:\|z\|<r\\}$ and
$r>0$ is a constant such that $\|F(z)\|<\frac{1}{2}$ for all
$z\in\mathcal{H}_{0}$ with $\|z\|<r$.

The proof of Lemma 5.11 works analogously to the proof of Lemma 5.3. Following
the strategy from Subsection 5.1 we are going to verify that the residuum
$R_{J}$

$\displaystyle R_{J}:$
$\displaystyle=H_{J}-\mathcal{E}_{J}(q)-D_{\mathcal{V}}\big{|}_{q}\mathcal{E}_{J}\left(b_{\geq
1}\right)-\frac{1}{2}D^{2}_{\mathcal{V}}\big{|}_{0}\mathcal{E}_{J}\left(b_{\geq
1}\right)-\frac{c_{J}}{N}$ (5.18)

is small, where the constant $c_{J}$ are given by
$c_{(0,0,0,0)}:=-\frac{1}{8}\sum_{j=1}^{d}\partial^{2}_{t_{j}}\big{|}_{t=0}\mathcal{E}_{(0,0,0,0)}\left(\vec{t}\
\right)$ and

$\displaystyle c_{(3,3,0,0)}:$
$\displaystyle=-\frac{1}{8}\sum_{j=1}^{d}\partial_{t_{j}}^{2}\big{|}_{t=0}\mathcal{E}_{(1,1,0,0)}\left(\vec{t}\
\right),\ \ c_{(3,1,0,0)}:=-c_{(3,3,0,0)},\ \ \ c_{(1,3,0,0)}:=c_{(3,3,0,0)},$
$\displaystyle c_{(3,0,3,0)}:$
$\displaystyle=\frac{1}{8}\sum_{j=1}^{d}\partial_{t_{j}}^{2}\big{|}_{t=0}\mathcal{E}_{(1,0,1,0)}\left(\vec{t}\
\right),\ \ \ \ c_{(1,0,3,0)}:=-c_{(3,0,3,0)},\ \ \
c_{(3,0,1,0)}:=-c_{(3,0,3,0)},$ $\displaystyle c_{(0,3,3,0)}:$
$\displaystyle=\frac{1}{8}\sum_{j=1}^{d}\partial_{t_{j}}^{2}\big{|}_{t=0}\mathcal{E}_{(0,1,1,0)}\left(\vec{t}\
\right),\ \ \ \ c_{(0,1,3,0)}:=-c_{(0,3,3,0)},\ \ \
c_{(0,3,1,0)}:=-c_{(0,3,3,0)},$ (5.19)

and all other constants are defined as $c_{J}:=0$. The proof will be split
into two parts. In Lemma 5.12 we derive an explicit representation of the
residuum $R_{J}$ by sorting the operator $H_{J}$ in terms of powers in $p$ and
$b_{>d}$, and in Theorem 5.16 we will make sure that this residuum is indeed
small compared to the operator $\mathbb{T}_{N}$.

######  Lemma 5.12.

Let $J=(i,j,k,\ell)\in\\{0,\dots,4\\}^{4}$ be such that $\lambda_{J}\neq 0$,
where $\lambda_{J}$ is defined in Lemma 5.11, and let $R_{J}$ be the residuum
defined in Eq. (5.18). By distinguishing different cases with the help of the
indices $e_{J}:=\left|\\{\ell\in J:\ell\in\\{3,4\\}\\}\right|$ and
$m_{J}:=\left|\\{\ell\in J:\ell=0\\}\right|$, we can explicitly express
$R_{J}$ as:

  * •

In the case $m_{J}=4$, i.e. $J=(0,0,0,0)$: $R_{J}=\left(h_{i}\
\underline{\otimes}\ h_{j}\right)^{\dagger}\cdot\hat{v}\cdot h_{k}\
\underline{\otimes}\ h_{\ell}\,E_{4}^{2}$.

  * •

In the case $e_{J}=0$ and $m_{J}<4$: $R_{J}=\left(h_{i}\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot\hat{v}\cdot h_{k}\ \underline{\otimes}\
h_{\ell}\,E_{m_{J}}^{1}$.

  * •

In the case $e_{J}=1$, there exists a constant $C$ and functions
$F_{J}:\mathbb{R}^{d}\longrightarrow\mathbb{R}$ with $|F_{J}(t)|\leq C|t|$,
such that $R_{J}=\left(h_{i}\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot\hat{v}\cdot h_{k}\ \underline{\otimes}\
h_{\ell}\,E_{m_{J}}^{0}+\frac{F_{J}(q)}{N}$.

  * •

In the case $e_{J}=2$ and $m_{J}=2$ when two of the indices are $4$:

$\displaystyle R_{J}=-\left(h_{i}\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot\hat{v}\cdot h_{k}\ \underline{\otimes}\
h_{\ell}\,\mathbb{L}^{\prime}.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ $

  * •

In the case $e_{J}=2$ and $m_{J}=2$ when one of the indices is $3$ and another
one is $4$, let us define $\widetilde{h}_{3}:=p^{\prime}-p$ and
$\widetilde{h}_{r}:=h_{r}$ for $r\in\\{0,1,2,4\\}$. Then,

$\displaystyle\ \ \ \ \ \ R_{J}=-\left(h_{i}\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot\hat{v}\cdot h_{k}\ \underline{\otimes}\
h_{\ell}\,\mathbb{L}^{\prime}+\left(\widetilde{h}_{i}\ \underline{\otimes}\
\widetilde{h}_{j}\right)^{\dagger}\cdot\hat{v}\cdot\widetilde{h}_{k}\
\underline{\otimes}\ \widetilde{h}_{\ell}.$

  * •

In the case $e_{J}=2$ and $m_{J}=2$ when two of the indices are $3$, let us
define the coefficients
$\Lambda_{(3,3,0,0)}^{r,r^{\prime}}:=-\left(u_{r}\otimes
u_{r^{\prime}}\right)^{\dagger}\cdot\hat{v}\cdot u_{0}\otimes u_{0}$,
$\Lambda_{(3,0,3,0)}^{r,r^{\prime}}:=\left(u_{r}\otimes
u_{0}\right)^{\dagger}\cdot\hat{v}\cdot u_{r^{\prime}}\otimes u_{0}$ and
$\Lambda_{(0,3,3,0)}^{r,r^{\prime}}:=\left(u_{0}\otimes
u_{r}\right)^{\dagger}\cdot\hat{v}\cdot u_{r^{\prime}}\otimes u_{0}$. Then,

$\displaystyle R_{J}\\!=$
$\displaystyle\\!-\\!\left(h_{i}\,\underline{\otimes}\,h_{j}\right)^{\dagger}\cdot\hat{v}\cdot
h_{k}\,\underline{\otimes}\,h_{\ell}\,\mathbb{L}^{\prime}\\!+\\!\sum_{r,r^{\prime}=1}^{d}\\!\Lambda_{J}^{r,r^{\prime}}\left[(p^{\prime}_{r}\\!-\\!p_{r})\cdot
p^{\prime}_{r^{\prime}}\\!\\!+p_{r}\cdot(p^{\prime}_{r^{\prime}}\\!-\\!p_{r^{\prime}})\right].$
(5.20)

  * •

In the cases $e_{J}=2$ and $m_{J}<2$, respectively $e_{J}>2$: $R_{J}=H_{J}$.

###### Proof.

Similar to the proof of Lemma 5.6, the proof of Lemma 5.12 follows from a
straightforward computation for the individual cases. For the purpose of
illustration, we will explicitly carry out the computations for the case
$J=(3,0,3,0)$, i.e. we are going to verify Eq. (5.20). Since
$\mathcal{E}_{(3,0,3,0)}(z)$ is quadratic in $\pi(z)$, we immediately obtain
$\mathcal{E}_{(3,0,3,0)}\left(\vec{t}\ \right)=0$ and
$D_{\mathcal{V}}\big{|}_{\vec{t}}\ \mathcal{E}_{(3,0,3,0)}=0$. Let us define
the coefficients $\lambda_{\alpha,\gamma}:=\left(u_{\alpha}\otimes
u_{0}\right)^{\dagger}\cdot\hat{v}\cdot u_{\gamma}\otimes u_{0}$, the operator
$Q=\frac{1}{2}\sum_{\alpha,\gamma=1}^{d}\hat{v}_{\alpha 0,\gamma 0}\
u_{\alpha}\cdot u_{\gamma}^{\dagger}$ and
$G\in\mathcal{H}_{0}\otimes\mathcal{H}_{0}$ by
$G=-\frac{1}{4}\sum_{\alpha,\gamma=1}^{d}\hat{v}_{\alpha 0,\gamma 0}\
u_{\alpha}\otimes u_{\gamma}$. Then

$\displaystyle D^{2}_{\mathcal{V}}\big{|}_{0}\
\mathcal{E}_{(3,0,3,0)}(z)=z^{\dagger}\cdot Q\cdot z+G^{\dagger}\cdot z\otimes
z+\left(z\otimes z\right)\cdot G$

and therefore $D^{2}_{\mathcal{V}}\big{|}_{0}\
\mathcal{E}_{(3,0,3,0)}\left(b_{\geq 1}\right)=b_{\geq 1}^{\dagger}\cdot
Q\cdot b_{\geq 1}+G^{\dagger}\cdot b_{\geq 1}\otimes b_{\geq 1}+\left(b_{\geq
1}\otimes b_{\geq 1}\right)\cdot G$. This concludes the proof of Eq. (5.20),
since

$\displaystyle H_{J}-\left(ip^{\prime}\ \underline{\otimes}\
u_{0}\right)^{\dagger}\cdot\hat{v}\cdot ip^{\prime}\ \underline{\otimes}\
u_{0}\left(-\mathbb{L}^{\prime}\right)-\sum_{r,r^{\prime}=1}^{d}\Lambda_{J}^{r,r^{\prime}}\left[(p^{\prime}_{r}-p_{r})\cdot
p^{\prime}_{r^{\prime}}+p_{r}\cdot(p^{\prime}_{r^{\prime}}-p_{r^{\prime}})\right]$
$\displaystyle\ =\left(ip\ \underline{\otimes}\
u_{0}\right)^{\dagger}\cdot\hat{v}\cdot ip\ \underline{\otimes}\
u_{0}\\!=\\!\sum_{\alpha,\gamma=1}^{d}\\!\left(u_{\alpha}\otimes
u_{0}\right)^{\dagger}\cdot\hat{v}\cdot u_{\gamma}\otimes u_{0}\
\frac{1}{2}\left(b_{\alpha}-b_{\alpha}^{\dagger}\right)^{\dagger}\cdot\frac{1}{2}\left(b_{\gamma}-b_{\gamma}^{\dagger}\right)$
$\displaystyle\ =b_{\geq 1}^{\dagger}\cdot Q\cdot b_{\geq 1}+\left(b_{\geq 1}\
\underline{\otimes}\ b_{\geq 1}\right)^{\dagger}\cdot G+G^{\dagger}\cdot
b_{\geq 1}\ \underline{\otimes}\ b_{\geq 1}+\frac{c_{(3,0,3,0)}}{N}.$

∎

In the remainder of this subsection, we are going to verify that the residuum
$R_{J}$ is small compared to the quadratic operator $\mathbb{T}_{N}$. Note
that the error term in the last case of Lemma 5.12 is quite different from the
other cases, since it simply corresponds to the whole operator $H_{J}$. This
is not surprising, however, since the second order Taylor approximation of an
object that is already of an higher order than two is zero, i.e. the residuum
coincides with the object itself. With the help of the following three results
in Lemma 5.13, Lemma 5.14 and Theorem 5.15, we will systematically verify that
$H_{J}$ is small compared to the quadratic operator $\mathbb{T}_{N}$ in the
cases $e_{J}=2$ and $m_{J}<2$, respectively $e_{J}>2$. Regarding all other
cases, we will verify the smallness of the residuum in Theorem 5.16. In order
to do this, we will repeatedly use results derived in Appendices C and D.

######  Lemma 5.13.

For indices $i,j\in\\{0,\dots,4\\}$, we have the following estimates:

  * •

In case one of the indices is contained in $\\{3,4\\}$, we have

$\displaystyle\left(h_{i}\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot|\hat{v}|\cdot h_{i}\ \underline{\otimes}\
h_{j}=O_{*}\left(\mathbb{T}_{N}\right).$

  * •

In case one of the indices is contained in $\\{3,4\\}$ and the other one is
contained in $\\{1,\dots,4\\}$, we have

$\displaystyle\left(h_{i}\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot|\hat{v}|\cdot h_{i}\ \underline{\otimes}\
h_{j}=o_{*}\left(\mathbb{T}_{N}\right).$

###### Proof.

We will repeatedly use the inequality $|v|\leq\Lambda\left(T+1\right)=:S$ from
Assumption 1.1, which implies together with the translation-invariance of $T$
the inequalities $|\hat{v}|\leq S\otimes 1_{\mathcal{H}}$ and $|\hat{v}|\leq\
1_{\mathcal{H}}\otimes S$.

The case $i\in\\{1,2\\}$ and $j\in\\{3,4\\}$: Recall that $h_{k}=e_{k}(q)$ for
$k\in\\{0,1,2\\}$ and let us define the function $\phi(t):=e_{i}\left(\vec{t}\
\right)^{\dagger}\cdot S\cdot e_{i}\left(\vec{t}\ \right)$. Using the
inequality $|\hat{v}|\leq S\otimes 1_{\mathcal{H}}$ we obtain

$\displaystyle\left(e_{i}(q)\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot|\hat{v}|\cdot e_{i}(q)\ \underline{\otimes}\
h_{j}\leq h_{j}^{\dagger}\cdot\phi(q)\cdot h_{j}.$

Since $|\phi(t)|\leq C\left(|t|+|t|^{2}\right)$ for a constant $C$, we obtain
$h_{3}^{\dagger}\cdot\phi(q)\cdot h_{3}=o_{*}\left(\mathbb{T}_{N}\right)$ and
$h_{4}^{\dagger}\cdot\phi(q)\cdot h_{4}=o_{*}\left(\mathbb{T}_{N}\right)$ by
Lemmata C.4 and C.6.

The case $i\in\\{3,4\\}$ and $j\in\\{1,2\\}$: Making use of the commutation
laws $[b_{\alpha},q_{\beta}]=0$ and
$[ip_{\alpha},q_{\beta}]=\frac{1}{2N}\delta_{\alpha,\beta}$, this case follows
from the previous one.

The case $i=3,j=3$: Let $\pi_{\leq d}:=\sum_{r=1}^{d}u_{r}\cdot
u_{r}^{\dagger}$. Since $\left((p^{\prime})^{\dagger}\cdot
p^{\prime}\right)^{2}=o_{*}\left(\mathbb{T}_{N}\right)$, we obtain

$\displaystyle\left(p^{\prime}\ \underline{\otimes}\
p^{\prime}\right)^{\dagger}\cdot|\hat{v}|\cdot p^{\prime}\
\underline{\otimes}\
p^{\prime}\leq(p^{\prime})^{\dagger}\cdot\left((p^{\prime})^{\dagger}\cdot
p^{\prime}\right)\otimes S\cdot p^{\prime}\leq\big{\|}\pi_{\leq
d}\,S\,\pi_{\leq d}\big{\|}\left((p^{\prime})^{\dagger}\cdot
p^{\prime}\right)^{2}=o_{*}\left(\mathbb{T}_{N}\right).$

The case $i=4$ and $j\in\\{3,4\\}$: Note that

$\displaystyle\left(b_{>d}\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot|\hat{v}|$ $\displaystyle\cdot b_{>d}\
\underline{\otimes}\ h_{j}\leq 2\left(f(q)\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot|\hat{v}|\cdot f(q)\ \underline{\otimes}\ h_{j}$
$\displaystyle+2\left((b_{>d}+f(q))\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot|\hat{v}|\cdot(b_{>d}+f(q))\ \underline{\otimes}\
h_{j}$

By the previous case $i\in\\{1,2\\}$ and $j\in\\{3,4\\}$, we know that
$\left(f(q)\ \underline{\otimes}\ h_{j}\right)^{\dagger}\cdot|\hat{v}|\cdot
f(q)\ \underline{\otimes}\ h_{j}=o_{*}\left(\mathbb{T}_{N}\right)$. For the
second contribution, recall the definition of $\pi_{M,N}$ from Remark 4.11 and
let $\hat{\pi}_{M,N}$ be the orthogonal projection onto the subspace
$\mathcal{W}_{N}\mathcal{F}^{+}_{\leq M}\subset\mathcal{F}_{0}$, where

$\displaystyle\mathcal{F}^{+}_{\leq
M}:=\mathds{1}_{[0,M]}\left(\sum_{j>d}^{\infty}a_{j}^{\dagger}\cdot
a_{j}\right).$ (5.21)

Since we have $b_{k}\,\hat{\pi}_{M,N}=\hat{\pi}_{M,N}\,b_{k}\,\hat{\pi}_{M,N}$
for $k>d$ and $[p^{\prime}_{j},\hat{\pi}_{M,N}]=0$ by Lemma C.3, we obtain
using $\pi_{M,N}=\hat{\pi}_{M,N}\,\pi_{M,N}$ (which follows from
$\mathcal{W}_{N}\mathcal{F}_{\leq
M}\subset\mathcal{W}_{N}\mathcal{F}^{+}_{\leq M}$)

$\displaystyle\pi_{M,N}\left((b_{>d}+f(q))\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot|\hat{v}|\cdot(b_{>d}+f(q))\ \underline{\otimes}\
h_{j}\,\pi_{M,N}$ $\displaystyle\ \ \
\leq\pi_{M,N}\,h_{j}^{\dagger}\cdot\left(\hat{\pi}_{M,N}\,(b_{>d}+f(q))^{\dagger}\cdot(b_{>d}+f(q))\,\hat{\pi}_{M,N}\right)\otimes
S\cdot h_{j}\,\pi_{M,N}$ $\displaystyle\ \ \ \leq\frac{M}{N}\
\pi_{M,N}\,h_{j}^{\dagger}\cdot S\cdot h_{j}\,\pi_{M,N}\leq C\ \frac{M}{N}\
\pi_{M,N}\,\mathbb{T}_{N}\,\pi_{M,N}$

for a constant $0<C<\infty$, where we used $h_{j}^{\dagger}\cdot S\cdot
h_{j}=O_{*}\left(\mathbb{T}_{N}\right)$ and and the characterization of the
$O_{*}(\cdot)$ notation in Remark 4.11 for the last inequality. Using this
characterization for the inequality above yields $\left((b_{>d}+f(q))\
\underline{\otimes}\ h_{j}\right)^{\dagger}\cdot|\hat{v}|\cdot(b_{>d}+f(q))\
\underline{\otimes}\ h_{j}=o_{*}\left(\mathbb{T}_{N}\right)$.

The case $i=3$ and $j=4$: Making use of the commutation laws
$[ip^{\prime}_{\alpha},b]=-\frac{1}{2N}\partial_{\alpha}f(q)$, this case
follows from the previous one.

The case $i=0$ and $j\in\\{3,4\\}$, respectively $i\in\\{3,4\\}$ and $j=0$:
Since $h_{0}=1_{\mathcal{F}_{0}}\otimes u_{0}$ commutes with
$h_{3}=ip^{\prime}$ and $h_{4}=b_{>d}$, we assume w.l.o.g. $i=0$ and
$j\in\\{3,4\\}$. With $\lambda:=u_{0}^{\dagger}\cdot S\cdot u_{0}$, we obtain

$\displaystyle\left(u_{0}\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot|\hat{v}|\cdot u_{0}\ \underline{\otimes}\
h_{j}\leq\lambda\ h_{j}^{\dagger}\cdot
h_{j}=O_{*}\left(\mathbb{T}_{N}\right).$

∎

######  Lemma 5.14.

In the following, let
$G:\mathbb{R}^{d}\longrightarrow\mathcal{H}\otimes\mathcal{H}$ be a
differentiable function and let us define the operators
$X,Y:\mathrm{dom}[\mathcal{N}]\longrightarrow{\mathcal{F}_{0}}\otimes\mathcal{H}$
as

$\displaystyle X$ $\displaystyle:=\left(ip^{\prime}\right)^{\dagger}\otimes
1_{\mathcal{H}}\cdot
G(q)=-\sum_{k=1}^{\infty}\left(\sum_{j=1}^{d}iG_{j,k}p^{\prime}_{j}\right)\otimes
u_{k},$ $\displaystyle Y$ $\displaystyle:=b_{>d}^{\dagger}\otimes
1_{\mathcal{H}}\cdot
G(q)=\sum_{k=1}^{\infty}\left(\sum_{j>d}iG_{j,k}b^{\dagger}_{j}\right)\otimes
u_{k}.$

Then we have the estimates

$\displaystyle X^{\dagger}\cdot X$ $\displaystyle\leq
2\left[\left(p^{\prime}\right)^{\dagger}\cdot\|G\|^{2}(q)\cdot
p^{\prime}+\frac{d}{N^{2}}\sum_{\alpha=1}^{d}\|\partial_{\alpha}G\|^{2}(q)\right],$
$\displaystyle Y^{\dagger}\cdot Y$ $\displaystyle\leq
b_{>d}^{\dagger}\cdot\|G\|^{2}(q)\cdot b_{>d}+\frac{1}{N}\|G\|^{2}(q).$

The proof of Lemma 5.14 is based on the commutation relations
$[b_{\alpha},b_{\beta}^{\dagger}]=\frac{1}{N}\delta_{\alpha,\beta}$ and
$[p_{\alpha},q_{\beta}]=\frac{1}{2iN}\delta_{\alpha,\beta}$, and is left to
the reader.

######  Theorem 5.15.

Let $\mathbb{L}^{\prime}$ be the operator from Definition 4.8. Then we have
the following estimates:

  * •

In case at least two of the indices $i,j,k,\ell\in\\{1,\dots,4\\}$ are
contained in $\\{3,4\\}$, we have

$\displaystyle\left(h_{i}\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot\hat{v}\cdot h_{k}\ \underline{\otimes}\
h_{\ell}=o_{*}\left(\mathbb{T}_{N}\right),$

  * •

In case at least two of the indices $i,j,k\in\\{0,\dots,4\\}$ are contained in
$\\{3,4\\}$, we have

$\displaystyle\left(h_{i}\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot\hat{v}\cdot h_{k}\ \underline{\otimes}\
u_{0}\,\mathbb{L}^{\prime}=o_{*}\left(\mathbb{T}_{N}\right),$

  * •

In case at least two of the indices $i,j,k\in\\{1,\dots,4\\}$ are contained in
$\\{3,4\\}$, we have

$\displaystyle\left(h_{i}\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot\hat{v}\cdot h_{k}\ \underline{\otimes}\
u_{0}\,\sqrt{1-\mathbb{L}^{\prime}}=o_{*}\left(\mathbb{T}_{N}\right).$

###### Proof.

Let us denote with $e_{(a,b)}$ the number of indices in $(a,b)$ that are
elements of $\\{3,4\\}$. In the following, we will verify the theorem
separately for the case $e_{(i,j)}\geq 1$ and $e_{(k,\ell)}\geq 1$, and the
case $e_{(k,\ell)}=0$. Note that the case $e_{(i,j)}=0$ is only possible for
the first bullet point, and the proof of the statement follows from the case
$e_{(k,\ell)}=0$, since

$\displaystyle\left[\left(h_{i}\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot\hat{v}\cdot h_{k}\ \underline{\otimes}\
h_{\ell}\right]^{\dagger}=\left(h_{k}\ \underline{\otimes}\
h_{\ell}\right)^{\dagger}\cdot\hat{v}\cdot h_{i}\ \underline{\otimes}\ h_{j}.$

The case $e_{(i,j)}\geq 1$ and $e_{(k,\ell)}\geq 1$: Let us define the
operators $A:=h_{i}\ \underline{\otimes}\ h_{j}$ and $Q:=\hat{v}$, and
depending on the concrete bullet point let us define $B$ as $h_{k}\
\underline{\otimes}\ h_{\ell}$, $h_{k}\ \underline{\otimes}\
u_{0}\,\mathbb{L}^{\prime}$ or $h_{k}\ \underline{\otimes}\
u_{0}\,\sqrt{1-\mathbb{L}^{\prime}}$. In any case we have to verify

$\displaystyle A^{\dagger}\cdot Q\cdot B=o_{*}\left(\mathbb{T}_{N}\right).$

By Lemma C.1, it is enough to verify that one of the operators
$A^{\dagger}\cdot|Q|\cdot A$ and $B^{\dagger}\cdot|Q|\cdot B$ is of order
$o_{*}\left(\mathbb{T}_{N}\right)$, and the other one is of order
$O_{*}\left(\mathbb{T}_{N}\right)$, which follows from Lemma 5.13 and the
auxiliary Corollary D.4.

The case $e_{(k,\ell)}=0$: In this case we have $i,j\in\\{3,4\\}$ for any of
the bullet points. Let us define the function
$G:\mathbb{R}^{d}\longrightarrow\mathcal{H}\otimes\mathcal{H}$ by
$G(t):=1_{\mathcal{H}}\otimes\left(T+1\right)^{-\frac{1}{2}}\cdot\hat{v}\cdot
e_{k}\left(\vec{t}\ \right)\otimes e_{\ell}\left(\vec{t}\ \right)$. Note that
$G(t)\in\mathcal{H}\otimes\mathcal{H}$ follows from Assumption 1.1. We define
the operator $X:=\left(T+1\right)^{\frac{1}{2}}\cdot h_{i}$ and depending on
the concrete bullet point let us define $Y:=h^{\dagger}_{j}\otimes
1_{\mathcal{H}}\cdot G(q)\,Z$ with $Z:=1_{\mathcal{F}_{0}}$,
$Z:=\mathbb{L}^{\prime}$ or $Z:=\sqrt{1-\mathbb{L}^{\prime}}$. In the
following, we have to verify $X^{\dagger}\cdot
Y=o_{*}\left(\mathbb{T}_{N}\right)$. Since $i\in\\{3,4\\}$, we know that
$X^{\dagger}\cdot X=O_{*}\left(\mathbb{T}_{N}\right)$. By the Cauchy–Schwarz
like result in Lemma C.1, it is therefore enough to verify $Y^{\dagger}\cdot
Y=o_{*}\left(\mathbb{T}_{N}\right)$. Applying Lemma 5.14 yields in any case

$\displaystyle Y^{\dagger}\cdot Y$
$\displaystyle=Z^{\dagger}\left(h^{\dagger}_{j}\otimes 1_{\mathcal{H}}\cdot
G(q)\right)^{\dagger}\cdot\left(h^{\dagger}_{j}\otimes 1_{\mathcal{H}}\cdot
G(q)\right)Z$ $\displaystyle\leq 2\
Z^{\dagger}\left[h_{j}^{\dagger}\cdot\|G(q)\|^{2}\cdot
h_{j}+\frac{1}{N}\|G(q)\|^{2}+\frac{d}{N^{2}}\sum_{r=1}^{d}\|\partial_{r}G(q)\|^{2}\right]Z,$

and Corollary D.4 then yields that $Z^{\dagger}\,\|G(q)\|^{2}\,Z$ and
$Z^{\dagger}\,\frac{1}{N}\left(\sum_{r=1}^{d}\|\partial_{r}G(q)\|^{2}\right)Z$
are of order $o_{*}(1)$. Therefore, $Z^{\dagger}\,\frac{1}{N}\|G(q)\|^{2}Z$
and
$Z^{\dagger}\,\frac{d}{N^{2}}\left(\sum_{r=1}^{d}\|\partial_{r}G(q)\|^{2}\right)Z$
are both of order $o_{*}\left(\mathbb{T}_{N}\right)$. Finally,
$Z^{\dagger}\,h_{j}^{\dagger}\cdot\|G(q)\|^{2}\cdot
h_{j}\,Z=o_{*}\left(\mathbb{T}_{N}\right)$ follows from the auxiliary Lemmata
C.4 and C.6, and the auxiliary Corollary D.4. ∎

######  Theorem 5.16.

Let $J\in\\{0,\dots,4\\}^{4}$ be such that $\lambda_{J}\neq 0$ and let $R_{J}$
be the residuum defined in Eq. (5.18). Then,

$\displaystyle R_{J}$ $\displaystyle=o_{*}\left(\mathbb{T}_{N}\right).$

###### Proof.

Let $J=(i,j,k,\ell)$ be a multi index with $\lambda_{J}\neq 0$, and recall the
index $e_{J}:=\left|\\{l\in J:l\in\\{3,4\\}\\}\right|$ and the index
$m_{J}:=\left|\\{l\in J:l=0\\}\right|$ from Lemma 5.12 as well as the residuum
defined in Eq. (5.18). In order to prove the statement of the Theorem, we have
to verify $R_{J}=o_{*}\left(\mathbb{T}_{N}\right)$ for all
$J\in\\{0,\dots,4\\}^{4}$.

The case $e_{J}=0$ and $m_{J}=0$: In this case we have a trivial residuum
$R_{J}=0$.

The case $e_{J}=0$ and $m_{J}=1$: In this case, $R_{J}=V(q)E_{1}^{1}$, with
$V(t):=\big{(}e_{i}\left(\vec{t}\ \right)\otimes e_{j}\left(\vec{t}\
\right)\big{)}^{\dagger}\cdot\hat{v}\cdot e_{k}\left(\vec{t}\ \right)\otimes
e_{\ell}\left(\vec{t}\ \right)$. Since the $C^{1}$ function $V$ satisfies
$F(0)=0$, we obtain $V(q)E_{1}^{1}=o_{*}\left(\mathbb{T}_{N}\right)$ by Lemma
D.2.

The case $e_{J}=0$ and $m_{J}=2$: In this case $R_{J}=V(q)E_{2}^{1}$, with
$V(t):=\big{(}e_{i}\left(\vec{t}\ \right)\otimes e_{j}\left(\vec{t}\
\right)\big{)}^{\dagger}\cdot\hat{v}\cdot e_{k}\left(\vec{t}\ \right)\otimes
e_{\ell}\left(\vec{t}\ \right)$. We compute

$\displaystyle E_{2}^{1}$
$\displaystyle=\left(1-\mathbb{L}^{\prime}\right)-\eta_{2}(q)-D_{\mathcal{V}}\big{|}_{q}\eta_{2}\big{(}b_{\geq
1}\big{)}=-\left(b_{>d}^{\dagger}\cdot b_{>d}+(p^{\prime})^{\dagger}\cdot
p^{\prime}-\frac{d}{2N}\right).$

By Lemmata C.4 and C.6 we know that $V(q)\,b_{>d}^{\dagger}\cdot
b_{>d}=V(q)\,b_{>d}^{\dagger}\cdot b_{>d}$ and
$V(q)\left(p^{\prime}\right)^{\dagger}\cdot p^{\prime}$ are of order
$o_{*}\left(\mathbb{T}_{N}\right)$, and consequently
$V(q)E_{2}^{1}=o_{*}\left(\mathbb{T}_{N}\right)$.

The case $e_{J}=0$ and $m_{J}=3$: In this case $R_{J}=V(q)E_{3}^{1}$, with
$V(t):=\big{(}e_{i}\left(\vec{t}\ \right)\otimes e_{j}\left(\vec{t}\
\right)\big{)}^{\dagger}\cdot\hat{v}\cdot e_{k}\left(\vec{t}\ \right)\otimes
e_{\ell}\left(\vec{t}\ \right)$. We compute

$\displaystyle E_{3}^{1}:$
$\displaystyle=\left(1-\mathbb{L}^{\prime}\right)\sqrt{1-\mathbb{L}^{\prime}}-\eta_{3}(q)-D_{\mathcal{V}}\big{|}_{q}\eta_{3}\big{(}b_{\geq
q}\big{)}$
$\displaystyle=(1\\!-\\!\eta_{2}(q))E_{1}^{1}\\!-\\!\left(f(q)^{\dagger}\\!\cdot\\!b_{>d}\\!+\\!b_{>d}\\!\cdot\\!f(q)\right)E_{1}^{0}\\!-\\!\left(\\!b_{>d}^{\dagger}\\!\cdot\\!b_{>d}\\!+\\!(p^{\prime})^{\dagger}\\!\cdot\\!p^{\prime}\\!-\\!\frac{d}{2N}\\!\right)\sqrt{1\\!-\\!\mathbb{L}^{\prime}}.$

By Lemma D.2, we know that
$V(q)(1-\eta_{2}(q))E_{1}^{1}=o_{*}\left(\mathbb{T}_{N}\right)$ and
$\left(E_{1}^{0}\right)^{2}=o_{*}\left(\mathbb{T}_{N}\right)$. Note that we
further have $\left[V(q)\left(f(q)^{\dagger}\cdot b_{>d}+b_{>d}\cdot
f(q)\right)\right]^{2}=o_{*}\left(\mathbb{T}_{N}\right)$, and therefore the
product $V(q)\left(f(q)^{\dagger}\cdot b_{>d}+b_{>d}\cdot
f(q)\right)E_{1}^{0}$ is of order $o_{*}\left(\mathbb{T}_{N}\right)$ as well.
By making use of Lemmata C.4 and C.6, and Corollary D.4, we obtain

$\displaystyle
V(q)\left(\\!b_{>d}^{\dagger}\\!\cdot\\!b_{>d}\\!+\\!(p^{\prime})^{\dagger}\\!\cdot\\!p^{\prime}\\!-\\!\frac{d}{2N}\\!\right)\sqrt{1\\!-\\!\mathbb{L}^{\prime}}=o_{*}\left(\mathbb{T}_{N}\right).$

The case $e_{J}=0$ and $m_{J}=4$: In this case $R_{J}=\left(u_{0}\otimes
u_{0}\right)\cdot\hat{v}\cdot u_{0}\otimes u_{0}\ E_{4}^{2}$. We compute

$\displaystyle E_{4}^{2}:$
$\displaystyle=\left(1-\mathbb{L}^{\prime}\right)^{2}-\eta_{4}(q)-D_{\mathcal{V}}\big{|}_{q}\eta_{4}\big{(}b_{\geq
1}\big{)}-D^{2}_{\mathcal{V}}\big{|}_{0}\eta_{4}\big{(}b_{\geq
1}\big{)}-\frac{c_{4}}{N}$ $\displaystyle=\left(f(q)^{\dagger}\cdot
b_{>d}+b_{>d}\cdot f(q)\right)^{2}+\Big{\\{}f(q)^{\dagger}\cdot
b_{>d}+b_{>d}\cdot f(q),\ \left(p^{\prime}\right)^{\dagger}\cdot
p^{\prime}+b_{>d}^{\dagger}\cdot b_{>d}\Big{\\}}$ $\displaystyle\ \
+\left(\left(p^{\prime}\right)^{\dagger}\cdot p^{\prime}+b_{>d}^{\dagger}\cdot
b_{>d}\right)^{2}+\Big{\\{}\eta_{2}(q),\
\left(p^{\prime}\right)^{\dagger}\cdot p^{\prime}+b_{>d}^{\dagger}\cdot
b_{>d}\Big{\\}}$ $\displaystyle\ \
+2p^{\dagger}\cdot\left(p^{\prime}-p\right)+2\left(p^{\prime}-p\right)^{\dagger}\cdot
p^{\prime}-\frac{d}{N}\mathbb{L}^{\prime}+\frac{d^{2}}{4N^{2}},$

with the notation $\\{A,B\\}:=AB+BA$. Clearly
$\frac{d}{N}\mathbb{L}^{\prime}=o_{*}\left(\mathbb{T}_{N}\right)$. From
Lemmata C.4, C.5 and C.6, we know that all the operators
$p^{\dagger}\cdot\left(p^{\prime}-p\right)$,
$\left(p^{\prime}-p\right)^{\dagger}\cdot p^{\prime}$,
$\left((p^{\prime})^{\dagger}\cdot p^{\prime}\right)^{2}$,
$\left(b_{>d}^{\dagger}\cdot b_{>d}\right)^{2}$, $\left(b_{>d}^{\dagger}\cdot
f(q)+f(q)^{\dagger}\cdot b_{>d}\right)^{2}$,
$\eta_{2}(q)\,b_{>d}^{\dagger}\cdot b_{>d}$ and
$\eta_{2}(q)\,\left(p^{\prime}\right)^{\dagger}\cdot p^{\prime}\,\eta_{2}(q)$
are of order $o_{*}\left(\mathbb{T}_{N}\right)$. Consequently,
$\\{\eta_{2}(q),\left(p^{\prime}\right)^{\dagger}\cdot
p^{\prime}+b_{>d}^{\dagger}\cdot b_{>d}\\}$ and $\Big{\\{}f(q)^{\dagger}\cdot
b_{>d}+b_{>d}\cdot f(q),\ \left(p^{\prime}\right)^{\dagger}\cdot
p^{\prime}+b_{>d}^{\dagger}\cdot b_{>d}\Big{\\}}$ are of order
$o_{*}\left(\mathbb{T}_{N}\right)$ as well.

The case $e_{J}=1$: In this case, we have $R_{J}=\left(h_{i}\
\underline{\otimes}\ h_{j}\right)^{\dagger}\cdot\hat{v}\cdot h_{k}\
\underline{\otimes}\ h_{\ell}\,E_{m_{J}}^{0}+\frac{F_{J}(q)}{N}$. By Lemma
Lemma C.4, we know that $\frac{F_{J}(q)}{N}=o_{*}\left(\mathbb{T}_{N}\right)$.
Since we know that
$\left(E_{m_{J}}^{0}\right)^{2}=o_{*}\left(\mathbb{T}_{N}\right)$ by Corollary
D.5, we are done once we can verify that $X_{J}\cdot
X_{J}^{\dagger}=O_{*}\left(\mathbb{T}_{N}\right)$, where $X_{J}:=\left(h_{i}\
\underline{\otimes}\ h_{j}\right)^{\dagger}\cdot\hat{v}\cdot h_{k}\
\underline{\otimes}\ h_{\ell}$.

With $3\in\\{i,j,k,\ell\\}$: Let us first assume $j=3$, and define
$w(t):=e_{i}\left(\vec{t}\ \right)^{\dagger}\otimes
1_{\mathcal{H}}\cdot\hat{v}\cdot e_{k}\left(\vec{t}\ \right)\otimes
e_{\ell}\left(\vec{t}\ \right)$. Clearly,
$X_{J}=\left(ip^{\prime}\right)^{\dagger}\cdot w(q)$ and therefore
$X_{J}X_{J}^{\dagger}=O_{*}\left(\mathbb{T}_{N}\right)$ follows from C.6. The
other cases $i=3,k=3$ and $\ell=3$ follows from the commutation relation
$[ip^{\prime}_{\alpha},q_{\beta}]=\frac{1}{2N}\delta_{\alpha,\beta}$.

With $4\in\\{i,j,k,\ell\\}$: In any case, $X_{J}$ is either equal to
$w(q)^{\dagger}\cdot b_{>d}$ or $b^{\dagger}_{>d}\cdot w(q)$, where
$w:\mathbb{R}^{d}\longrightarrow\mathcal{H}$ with $\|w(t)\|\leq c|t|^{j}$ and
$j\geq 0$. Note that we use the commutativity of $q_{j}$ and $b_{>d}$ here.
Therefore, Lemma C.5 implies $X_{J}\cdot
X_{J}^{\dagger}=O_{*}\left(\mathbb{T}_{N}\right)$.

The case $e_{J}=2$ and $m_{J}=2$: In any case, we know by the second bullet
point of Theorem 5.15, that $\left(h_{i}\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot\hat{v}\cdot h_{k}\ \underline{\otimes}\
h_{\ell}\left(-\mathbb{L}^{\prime}\right)=o_{*}\left(\mathbb{T}_{N}\right)$.
In case $\\{i,j,k,\ell\\}=\\{0,4\\}$, this is the whole residuum $R_{J}$. In
case $\\{i,j,k,\ell\\}=\\{0,3\\}$, the residuum reads

$\displaystyle R_{J}=\left(h_{i}\ \underline{\otimes}\
h_{j}\right)^{\dagger}\cdot\hat{v}\cdot h_{k}\ \underline{\otimes}\
h_{\ell}\left(-\mathbb{L}^{\prime}\right)+\sum_{r,r^{\prime}=1}^{d}\Lambda_{J}^{r,r^{\prime}}\left[(p^{\prime}_{r}-p_{r})\cdot
p^{\prime}_{r^{\prime}}+p_{r}\cdot(p^{\prime}_{r^{\prime}}-p_{r^{\prime}})\right].$

Since any of the products $(p^{\prime}_{r}-p_{r})\cdot
p^{\prime}_{r^{\prime}}$ and
$p_{r}\cdot(p^{\prime}_{r^{\prime}}-p_{r^{\prime}})$ are of order
$o_{*}\left(\mathbb{T}_{N}\right)$, we conclude
$R_{J}=o_{*}\left(\mathbb{T}_{N}\right)$. The case
$\\{i,j,k,\ell\\}=\\{0,3,4\\}$ works similarly, and is left to the reader.

The cases $e_{J}=2$ and $m_{J}<2$, respectively $e_{J}>2$: We obtain for
$m_{J}=0$ by the first bullet point of Theorem 5.15, and for $m_{J}=1$ by the
third bullet point, that

$\displaystyle R_{J}=H_{J}=o_{*}\left(\mathbb{T}_{N}\right).$

∎

######  Corollary 5.17.

Recall the functional $\mathcal{E}_{B}$ defined in Eq. (5.17) and the constant
$c$ from Corollary 5.8. Then,

$\displaystyle\mathcal{W}_{N}\widetilde{B}_{N}\mathcal{W}_{N}^{-1}$
$\displaystyle=\mathcal{E}_{B}(q)+D_{\mathcal{V}}\big{|}_{q}\mathcal{E}_{B}\left(b_{\geq
1}\right)+\frac{1}{2}D^{2}_{\mathcal{V}}\big{|}_{0}\
\mathcal{E}_{B}\left(b_{\geq
1}\right)-\frac{c}{N}+o_{*}\left(\mathbb{T}_{N}\right).$

###### Proof.

Let us define $\tilde{c}:=\sum_{J\in\\{0,\dots,4\\}^{4}}\lambda_{J}c_{J}$.
Combining Lemma 5.11 and Theorem 5.16 immediately yields

$\displaystyle\mathcal{W}_{N}\widetilde{B}_{N}\mathcal{W}_{N}^{-1}$
$\displaystyle=\mathcal{E}_{B}(q)+D_{\mathcal{V}}\big{|}_{q}\mathcal{E}_{B}\left(b_{\geq
1}\right)+\frac{1}{2}D^{2}_{\mathcal{V}}\big{|}_{0}\
\mathcal{E}_{B}\left(b_{\geq
1}\right)+\frac{\tilde{c}}{N}+o_{*}\left(\mathbb{T}_{N}\right).$

Recall the definition of $c_{J}$ in Eq. (5.12) for $J\in\\{0,\dots,4\\}^{2}$,
respectively Eq. (5.19) for $J\in\\{0,\dots,4\\}^{4}$. Making use of the
observation that most of the $c_{J}$ are zero, we obtain

$\displaystyle c+\tilde{c}$
$\displaystyle=\sum_{J\in\\{0,\dots,4\\}^{2}}\lambda_{J}c_{J}+\sum_{J\in\\{0,\dots,4\\}^{4}}\lambda_{J}c_{J}=\lambda_{(0,0)}c_{(0,0)}+\lambda_{(1,1)}\Big{[}c_{(1,3)}+c_{(3,1)}+c_{(3,3)}\Big{]}$
$\displaystyle\
+\lambda_{(0,0,0,0)}c_{(0,0,0,0)}+\lambda_{(1,1,0,0)}\Big{[}c_{(3,3,0,0)}+c_{(3,1,0,0)}+c_{(1,3,0,0)}\Big{]}$
$\displaystyle\
+\lambda_{(1,0,1,0)}\Big{[}c_{(3,0,3,0)}+c_{(3,0,1,0)}+c_{(1,0,3,0)}\Big{]}+\lambda_{(0,1,1,0)}\Big{[}c_{(0,3,3,0)}+c_{(0,1,3,0)}+c_{(0,3,1,0)}\Big{]}$
$\displaystyle=-\frac{1}{8}\partial^{2}_{t_{j}}\big{|}_{t=0}\sum_{j=1}^{d}\Big{(}\lambda_{(0,0)}\mathcal{E}_{(0,0)}\left(\vec{t}\
\right)+\lambda_{(1,1)}\mathcal{E}_{(1,1)}\left(\vec{t}\
\right)+\lambda_{(0,0,0,0)}\mathcal{E}_{(0,0,0,0)}\left(\vec{t}\ \right)$
$\displaystyle\ +\lambda_{(1,1,0,0)}\mathcal{E}_{(1,1,0,0)}\left(\vec{t}\
\right)+\lambda_{(1,0,1,0)}\mathcal{E}_{(1,0,1,0)}\left(\vec{t}\
\right)+\lambda_{(0,1,1,0)}\mathcal{E}_{(0,1,1,0)}\left(\vec{t}\
\right)\Big{)}$
$\displaystyle=-\frac{1}{8}\sum_{j=1}^{d}\partial^{2}_{t_{j}}\big{|}_{t=0}\Big{(}\mathcal{E}_{A}\left(\vec{t}\,\right)+\mathcal{E}_{B}\left(\vec{t}\,\right)\Big{)}=0,$

where we have used in the first equality of the last line that
$\partial^{2}_{t_{j}}\big{|}_{t=0}\lambda_{J}\mathcal{E}_{J}\left(\vec{t}\,\right)=0$
for

$\displaystyle J\notin\\{(0,0),(1,1),(1,1,0,0),(1,0,1,0),(0,1,1,0)\\}$

and in the second equality of the last line that
$\mathcal{E}_{A}\left(\vec{t}\,\right)+\mathcal{E}_{B}\left(\vec{t}\,\right)=\mathcal{E}^{\prime}\left(\vec{t}\,\right)=e_{\mathrm{H}}$
for $t$ small enough, where $\mathcal{E}^{\prime}$ is defined in Eq. (5.1). ∎

###### Proof of Theorem 4.12..

Making use of Eq. (3.8), we obtain

$\displaystyle\left(\mathcal{W}_{N}U_{N}\right)N^{-1}H_{N}\left(\mathcal{W}_{N}U_{N}\right)^{-1}=\mathcal{W}_{N}\widetilde{A}_{N}\mathcal{W}_{N}^{-1}+\mathcal{W}_{N}\widetilde{B}_{N}\mathcal{W}_{N}^{-1}+o_{*}\left(\mathbb{T}_{N}\right),$
(5.22)

where we have used that $\mathcal{W}_{N}\,b_{\geq 1}^{\dagger}\cdot T\cdot
b_{\geq 1}\,\mathcal{W}_{N}^{-1}\leq 2\left(X_{1}+X_{2}\right)$ with

$\displaystyle X_{1}:$ $\displaystyle=\left(q+f(q)\right)^{\dagger}\cdot
T\cdot\left(q+f(q)\right)=o_{*}(1),$ $\displaystyle X_{2}:$
$\displaystyle=\left(ip^{\prime}+b_{>d}\right)^{\dagger}\cdot
T\cdot\left(ip^{\prime}+b_{>d}\right)=O_{*}\left(\mathbb{T}_{N}\right),$

see Lemmata C.4 and C.6. Combining Corollaries 5.8 and 5.17 yields

$\displaystyle\mathcal{W}_{N}\widetilde{A}_{N}\mathcal{W}_{N}^{-1}+\mathcal{W}_{N}\widetilde{B}_{N}\mathcal{W}_{N}^{-1}$
$\displaystyle=\mathcal{E}\left(q\right)\\!+D_{\mathcal{V}}\big{|}_{q}\mathcal{E}\Big{(}b_{\geq
1}\Big{)}+\\!\frac{1}{2}D_{\mathcal{V}}^{2}\big{|}_{0}\mathcal{E}\Big{(}b_{\geq
1}\Big{)}\\!+o_{*}\left(\mathbb{T}_{N}\right),$

with $\mathcal{E}:=\mathcal{E}_{A}+\mathcal{E}_{B}$. Furthermore, note that
$\mathcal{E}(z)=\mathcal{E}^{\prime}(z)$ for $\|z\|<r$ where
$\mathcal{E}^{\prime}$ is defined in Eq. (5.1), see Lemmata 5.3 and 5.11.
Therefore, $\mathcal{E}\left(\vec{t}\,\right)=e_{\mathrm{H}}$ and
$D_{\mathcal{V}}\big{|}_{t}\mathcal{E}=0$ for $t$ small enough. As we will
show in Lemma C.2, this implies
$\mathcal{E}\left(q\right)=e_{\mathrm{H}}+o_{*}\left(\mathbb{T}_{N}\right)$
and $D_{\mathcal{V}}\big{|}_{q}\mathcal{E}\Big{(}b_{\geq
1}\Big{)}=o_{*}\left(\mathbb{T}_{N}\right)$. Furthermore, we have
$\frac{1}{2}D_{\mathcal{V}}^{2}\big{|}_{0}\
\mathcal{E}=\mathrm{Hess}|_{u_{0}}\mathcal{E}_{\mathrm{H}}$ and therefore

$\displaystyle\frac{1}{2}D_{\mathcal{V}}^{2}\big{|}_{0}\
\mathcal{E}\Big{(}b_{\geq 1}\Big{)}=N^{-1}\mathbb{H}.$

In combination with Eq. (5.22) this concludes the proof. ∎

##  Appendix A Coercivity of the Hessian in Example (II)

In the following we are going to verify that the Hartree energy of a system of
pseudo-relativistic bosons in $\mathbb{R}^{3}$ interacting via a Newtonian
potential, given by

$\displaystyle\mathcal{E}_{g}\left[u\right]:=\big{\langle}\sqrt{m^{2}-\Delta}-m\big{\rangle}_{u}-(u\otimes
u)^{\dagger}\cdot\frac{g}{2|x-y|}\cdot u\otimes u,$

satisfies the coercivity assumption in Eq. (1.7) for $g$ small enough, see
Example (II) in the introduction. Note that we are using the notation
introduced in Section 3. Let us denote with $u_{g,\beta}$ the unique radial
minimizer of the functional $\mathcal{E}_{g}$ subject to the rescaled
condition $\|u\|=1+\beta$, i.e. $u_{g,\beta}$ is radial, and satisfies
$\|u_{g,\beta}\|=1+\beta$ and
$\mathcal{E}_{g}[u_{g,\beta}]=\underset{\|u\|=1+\beta}{\inf}\mathcal{E}_{g}[u]$.
Let us further denote the normed minimizers by $u_{g}:=u_{g,0}$. By a scaling
argument it is easy to see that $u_{g,\beta}=(1+\beta)u_{g(1+\beta)^{2}}$. For
real-valued functions $f$ and $h$ in $\\{u_{g,\beta}\\}^{\perp}$ we can
express the Hessian as
$\frac{1}{2}\mathrm{Hess}|_{u_{g,\beta}}\mathcal{E}_{g}[f+ih]=\braket{L^{+}_{g,\beta}}_{f}+\braket{L^{-}_{g,\beta}}_{h}$,
where $L^{+}_{g,\beta}$ and $L^{-}_{g,\beta}$ are selfadjoint operators given
by

$\displaystyle L^{-}_{g,\beta}$
$\displaystyle:=\sqrt{m^{2}-\Delta}-m-\mu_{g,\beta}-\left(1\otimes
u_{g,\beta}\right)^{\dagger}\cdot\frac{g}{|x-y|}\cdot 1\otimes u_{g,\beta},$
$\displaystyle L^{+}_{g,\beta}$ $\displaystyle:=L^{-}_{g,\beta}-\left(1\otimes
u_{g,\beta}\right)^{\dagger}\cdot\frac{2g}{|x-y|}\cdot u_{g,\beta}\otimes 1,$

with
$\mu_{g,\beta}:=\big{\langle}\sqrt{m^{2}-\Delta}-m\big{\rangle}_{u_{g,\beta}}-(u_{g,\beta}\otimes
u_{g,\beta})^{\dagger}\cdot\frac{g}{|x-y|}\cdot u_{g,\beta}\otimes
u_{g,\beta}$. Furthermore we denote the operators associated to the normed
minimizers $u_{g}$ by $L^{\pm}_{g}:=L^{\pm}_{g,0}$. Note that

$\displaystyle\braket{L^{-}_{g}-L^{+}_{g}}_{f}=\left(f\otimes
u_{g}\right)^{\dagger}\cdot\frac{2g}{|x-y|}\cdot u_{g}\otimes f>0$

for all $f\neq 0$, and consequently it is enough to verify the following
Theorem A.1 in order to prove Eq. (1.7).

######  Theorem A.1.

There exist constants $g_{0}$ and $\eta>0$ such that for all $0<g<g_{0}$ and
$f\in L^{2}\left(\mathbb{R}^{d}\right)$ with
$f\perp\\{u_{g},\partial_{x_{1}}u_{g},\partial_{x_{2}}u_{g},\partial_{x_{3}}u_{g}\\}$

$\displaystyle\braket{L^{+}_{g}}_{f}\geq\eta\|f\|^{2}.$

In order to prove Theorem A.1, we first need some auxiliary results regarding
the minimizers $u_{g,\beta}$ subject to the rescaled condition
$\|u_{g,\beta}\|=1+\beta$.

######  Lemma A.2.

Let us define $R_{g,\beta}:=u_{g,\beta}-u_{g}$ for $\beta\in[0,1)$ (where $1$
can be replaced by any other positive number). Then there exist constants
$g_{0},C>0$ such that

$\displaystyle L^{+}_{g}R_{g,\beta}=\delta_{g,\beta}u_{g}+\epsilon_{g,\beta},$

with $|\delta_{g,\beta}|\leq C\beta$ and $\|\epsilon_{g,\beta}\|\leq
C\|R_{g,\beta}\|^{2}$ for $g\in(0,g_{0})$ and $\beta\in[0,1)$.

###### Proof.

Since the elements $u_{g,\beta}$ are minimizers of $\mathcal{E}_{g}$, they
satisfy the corresponding Euler-Lagrange equations
$L^{-}_{g,\beta}u_{g,\beta}=0$. A straightforward computation yields

$\displaystyle 0=$ $\displaystyle
L^{-}_{g,\beta}u_{g,\beta}-L^{-}_{g}u_{g}=L^{+}_{g}R_{g,\beta}-\delta_{g,\beta}u_{g}-\epsilon_{g,\beta}$

with

$\displaystyle\delta_{g,\beta}$ $\displaystyle:=\mu_{g,\beta}-\mu_{g},$
$\displaystyle\epsilon_{g,\beta}$
$\displaystyle:=\left(\mu_{g,\beta}-\mu_{g}\right)R_{g,\beta}+\left(1\otimes
u_{g,\beta}\right)^{\dagger}\cdot\frac{g}{|x-y|}\cdot R_{g,\beta}\otimes
R_{g,\beta}$ $\displaystyle\ \ \ \ +\left(1\otimes
R_{g,\beta}\right)^{\dagger}\cdot\frac{g}{|x-y|}\cdot R_{g,\beta}\otimes
u_{g}+\left(1\otimes R_{g,\beta}\right)^{\dagger}\cdot\frac{g}{|x-y|}\cdot
u_{g}\otimes R_{g,\beta}.$ (A.1)

Let us first investigate the contributions involving $\frac{g}{|x-y|}$. From
[16, Proposition 1] it is clear that there exists a constant $C$ such that
$\|u_{g,\beta}\|_{H^{1}\left(\mathbb{R}^{3}\right)}\leq C<\infty$ for all $g$
small enough and $\beta\in[0,1)$. With the notation $S:=\sqrt{1-\Delta}$ we
obtain

$\displaystyle\left\|\left(1\otimes
u_{g,\beta}\right)^{\dagger}\cdot\frac{g}{|x-y|}\cdot R_{g,\beta}\otimes
R_{g,\beta}\right\|=\left\|\left(1\otimes Su_{g,\beta}\right)^{\dagger}\cdot
1\otimes S^{-1}\frac{g}{|x-y|}\cdot R_{g,\beta}\otimes R_{g,\beta}\right\|$
$\displaystyle\ \ \leq g\|Su_{g,\beta}\|\,\left\|1\otimes
S^{-1}\frac{1}{|x-y|}\right\|\,\|R_{g,\beta}\|^{2}\leq
Cg\left\|S^{-1}\frac{1}{|x|}\right\|\,\|R_{g,\beta}\|^{2},$

where $\left\|S^{-1}\frac{1}{|x|}\right\|$ is the operator norm of the bounded
one-particle operator $S^{-1}\frac{1}{|x|}$. Similarly, the other
contributions involving $\frac{g}{|x-y|}$ in Eq. (A.1) can be estimated by
$Cg\left\|S^{-1}\frac{1}{|x|}\right\|\,\|R_{g,\beta}\|^{2}$ as well. The
uniform control of the norm
$\|u_{g,\beta}\|_{H^{1}\left(\mathbb{R}^{3}\right)}\leq C<\infty$ furthermore
implies $|\delta_{g,\beta}|=|\mu_{g,\beta}-\mu_{g}|\leq\tilde{C}\beta$ for
some constant $\tilde{C}$. Note that
$\|R_{g,\beta}\|\geq\|u_{g,\beta}\|-\|u_{g}\|=\beta$, and consequently
$\|\left(\mu_{g,\beta}-\mu_{g}\right)R_{g,\beta}\|\leq\tilde{C}\beta\|R_{g,\beta}\|\leq\tilde{C}\|R_{g,\beta}\|^{2}$.
We conclude that

$\displaystyle\|\epsilon_{g,\beta}\|\leq\left(\tilde{C}+3Cg\left\|S^{-1}\frac{1}{|x|}\right\|\right)\|R_{g,\beta}\|^{2}.$

∎

######  Lemma A.3.

Let $R_{g,\beta}$ and $\epsilon_{g,\beta}$ be as in Lemma A.2. Then there
exists a constant $g_{0}>0$ such that $\underset{\beta\rightarrow
0}{\lim}\,\frac{\|\epsilon_{g,\beta}\|}{\braket{u_{g},R_{g,\beta}}}=0$ and
$\underset{\beta\rightarrow
0}{\limsup}\,\frac{|\delta_{g,\beta}|}{\braket{u_{g},R_{g,\beta}}}\leq C$ for
a suitable constant $C>0$ and $g\in(0,g_{0})$.

###### Proof.

By the results in [16] we know that $0$ is an isolated eigenvalue of
$L^{+}_{g}$, i.e. there exists a constant $\delta>0$ such that
$\sigma\left(L^{+}_{g}\right)\cap(-\delta,\delta)=\\{0\\}$, with corresponding
eigenvectors
$\partial_{x_{1}}u_{g},\partial_{x_{2}}u_{g},\partial_{x_{3}}u_{g}$. Since
$u_{g,\beta}$ is radial, we know that $R_{g,\beta}\perp\partial_{x_{j}}u_{g}$,
and therefore we obtain by Lemma A.2

$\displaystyle\delta\|R_{g,\beta}\|\leq\|L^{+}_{g}R_{g,\beta}\|\leq
C\left(\beta+\|R_{g,\beta}\|^{2}\right).$ (A.2)

Using [16, Proposition 1] again, it is clear that $\underset{g\rightarrow
0,\beta\rightarrow 0}{\lim}\,\|R_{g,\beta}\|=0$ and therefore there exists a
constant $g_{0}$ such that $\|R_{g,\beta}\|\leq\frac{\delta}{2C}$ for all
$g\in(0,g_{0})$ and $\beta$ small enough. Consequently Eq. (A.2) yields
$\|R_{g,\beta}\|\leq\frac{2C}{\delta}\beta$. Using the fact that
$\|u_{g,\beta}\|=1+\beta$, we further obtain

$\displaystyle
1+2\beta\leq(1+\beta)^{2}=1+2\braket{u_{g},R_{g,\beta}}+\|R_{g,\beta}\|^{2}\leq
1+2\braket{u_{g},R_{g,\beta}}+\left(\frac{2C}{\delta}\right)^{2}\beta^{2},$

and therefore

$\displaystyle\frac{\|\epsilon_{g,\beta}\|}{\braket{u_{g},R_{g,\beta}}}\leq\frac{C\left(\frac{2C}{\delta}\right)^{2}\beta^{2}}{\beta-2\left(\frac{C}{\delta}\right)^{2}\beta^{2}}\underset{\beta\rightarrow
0}{\longrightarrow}0,$
$\displaystyle\frac{|\delta_{g,\beta}|}{\braket{u_{g},R_{g,\beta}}}\leq\frac{C\beta}{\beta-2\left(\frac{C}{\delta}\right)^{2}\beta^{2}}\underset{\beta\rightarrow
0}{\longrightarrow}C.$

∎

###### Proof of Theorem A.1.

Let $Q$ denote the projection onto the space $\\{u_{g}\\}^{\perp}$. Clearly
there exists a $w\in L^{2}(\mathbb{R}^{3})$ such that

$\displaystyle L^{+}_{g}f=QL^{+}_{g}f+\braket{w,f}u_{g}$

for all $f\in L^{2}(\mathbb{R}^{3})$. With $R_{g,\beta},\delta_{g,\beta}$ and
$\epsilon_{g,\beta}$ from Lemma A.2 at hand, we obtain

$\displaystyle
L^{+}_{g}\left(\delta_{g,\beta}f-\braket{w,f}R_{g,\beta}\right)=\delta_{g,\beta}QL^{+}_{g}f-\braket{w,f}\epsilon_{g,\beta},$

and therefore
$\|L^{+}_{g}\left(\delta_{g,\beta}f-\braket{w,f}R_{g,\beta}\right)\|\leq|\delta_{g,\beta}|\,\|QL^{+}_{g}f\|+\|w\|\,\|\epsilon_{g,\beta}\|\,\|f\|$.
Using again that there exists a constant $\delta>0$ such that
$\sigma\left(L^{+}_{g}\right)\cap(-\delta,\delta)=\\{0\\}$ with corresponding
eigenvectors
$\partial_{x_{1}}u_{g},\partial_{x_{2}}u_{g},\partial_{x_{3}}u_{g}$, see [16],
and that $R_{g,\beta}$ as a radial function is orthogonal to them, we obtain
for all
$f\in\\{u_{g},\partial_{x_{1}}u_{g},\partial_{x_{2}}u_{g},\partial_{x_{3}}u_{g}\\}^{\perp}$

$\displaystyle\|L^{+}_{g}\left(\delta_{g,\beta}f-\braket{w,f}R_{g,\beta}\right)\|\geq\delta\|\delta_{g,\beta}f-\braket{w,f}R_{g,\beta}\|\geq\delta|\braket{u_{g},R_{g,\beta}}|\,|\braket{w,f}|.$

Combining the estimates we have so far yields

$\displaystyle|\braket{w,f}|\leq\frac{|\delta_{g,\beta}|}{\delta|\braket{u_{g},R_{g,\beta}}|}\|QL^{+}_{g}f\|+\frac{\|w\|\,\|\epsilon_{g,\beta}\|}{\delta|\braket{u_{g},R_{g,\beta}}|}\|f\|=:x_{\beta}\|QL^{+}_{g}f\|+y_{\beta}\|f\|.$

By Lemma A.3 we know that $\underset{\beta\rightarrow
0}{\limsup}|x_{\beta}|\leq C$ for some constant $C>0$ and
$y_{\beta}\underset{\beta\rightarrow 0}{\longrightarrow}0$. Using again that
$\sigma\left(L^{+}_{g}\right)\cap(-\delta,\delta)=\\{0\\}$, we obtain

$\displaystyle\delta\|f\|\leq\|L^{+}_{g}f\|\leq\|QL^{+}_{g}f\|+|\braket{w,f}|\leq\left(1+x_{\beta}\right)\|QL^{+}_{g}f\|+y_{\beta}\|f\|$

and consequently $\|QL^{+}_{g}f\|\geq\frac{\delta-
y_{\beta}}{1+x_{\beta}}\|f\|$. This holds for all (small) $\beta$, hence
$\beta\to 0$ gives $\|QL^{+}_{g}f\|\geq\frac{\delta}{1+C}\|f\|$. Finally note
that $QL^{+}_{g}Q\geq 0$ since
$\braket{L^{+}_{g}}_{f}=\frac{1}{2}\mathrm{Hess}|_{u_{g}}\mathcal{E}_{g}[f]\geq
0$ for real-valued $f\perp u_{g}$, which concludes the proof. ∎

##  Appendix B The Bogoliubov Operator

In the following we will prove Theorem 4.4, i.e. we are going to verify that
the Bogoliubov operator $\mathbb{H}$ constructed in Definition 4.3 is bounded
from below and that its ground state energy can be approximated by
$\Psi\in\bigcup_{M\in\mathbb{N}}\mathrm{dom}[a_{\geq
1}^{\dagger}\cdot(T+1)\cdot a_{\geq 1}]\cap\mathcal{F}_{\leq M}$ with
$\|\Psi\|=1$. Our strategy is to decouple the degenerate modes from the non-
degenerate ones and to apply the general framework for non-degenerate
Bogoliubov operators in [27].

######  Definition B.1.

Let $Q_{\mathrm{H}}=\sum_{i,j\geq 1}Q_{i,j}\ u_{i}\cdot u_{j}^{\dagger}$ and
$G_{\mathrm{H}}=\sum_{i,j\geq 1}G_{ij}\ u_{i}\otimes u_{j}$ be as in Lemma
4.1, and let us denote the operator $Q_{\perp}:=\sum_{i,j>d}Q_{i,j}\
u_{i}\cdot u_{j}^{\dagger}$ on
$\mathcal{H}_{\perp}:=\braket{u_{0},u_{1},\dots,u_{d}}^{\perp}$ as well as
$G_{\perp}:=\sum_{i,j>d}G_{ij}u_{i}\otimes u_{j}$. Then we define the operator
$\mathbb{H}_{\perp}$ as

$\displaystyle\mathbb{H}_{\perp}:=a_{>d}^{\dagger}\cdot Q_{\perp}\cdot
a_{>d}+2\mathfrak{Re}\left[G_{\perp}^{\dagger}\cdot a_{>d}\
\underline{\otimes}\ a_{>d}\right].$

######  Lemma B.2.

The operator $\mathbb{H}_{\perp}$ is semi-bounded from below, i.e.
$\inf\sigma\left(\mathbb{H}_{\perp}\right)>-\infty$. Furthermore, there exists
a constant $R>0$ such that

$\displaystyle\mathbb{H}_{\perp}\leq R\left(a_{>d}^{\dagger}\cdot
Q_{\perp}\cdot a_{>d}+1\right).$ (B.1)

###### Proof.

Let us define the operator $G_{\mathrm{op}}$ on $\mathcal{H}_{\perp}$ by the
condition $z^{\dagger}\cdot G_{\mathrm{op}}\cdot
z=2G_{\perp}^{\dagger}\cdot\overline{z}\otimes z$, with $\overline{z}$ being
the usual complex conjugation in $L^{2}\left(\mathbb{R}^{d}\right)$. Then,
$z^{\dagger}\cdot Q_{\perp}\cdot
z+2\mathfrak{Re}\left[\overline{z}^{\dagger}\cdot G_{\mathrm{op}}\cdot
z\right]=\mathrm{Hess}|_{u_{0}}\mathcal{E}_{\mathrm{H}}[z]\geq\eta\|z\|^{2}$
for all $z\in\mathcal{H}_{\perp}$ with $\eta>0$ by Assumption 1.3. As pointed
out in Section 2.1 in [21], this implies $Q_{\perp}\geq r>0$ as well as

$\displaystyle\left(\begin{array}[]{rr}Q_{\perp}&G_{\mathrm{op}}^{\dagger}\\\
G_{\mathrm{op}}&Q_{\perp}\end{array}\right)\geq 0,$

where we have used that $Q_{\perp}$ is a real operator. Since $Q_{\perp}>0$,
this is further equivalent to
$G_{\mathrm{op}}\,Q_{\perp}^{-1}\,G_{\mathrm{op}}^{\dagger}\leq Q_{\perp}$.
Since
$G_{\mathrm{H}}\in\overline{\mathcal{H}_{0}\otimes\mathcal{H}_{0}}^{\|.\|_{*}}$,
where the $\|.\|_{*}$ norm is defined in Lemma 4.1, and since
$z^{\dagger}\cdot Q_{\perp}^{-\frac{1}{2}}\cdot z\leq
c\,z^{\dagger}\cdot\left(T+1\right)^{-\frac{1}{2}}\cdot z$ for a suitable
constant $c$ and $z\in\mathcal{H}_{\perp}$, which is an easy consequence of
the operator inequality in Lemma 4.1 and the fact that $Q_{\perp}\geq r>0$, we
obtain

$\displaystyle\mathrm{Tr}\left[G_{\mathrm{op}}\,Q_{\perp}^{-1}\,G_{\mathrm{op}}^{\dagger}\right]=\|1_{\mathcal{H}_{\perp}}\otimes
Q_{\perp}^{-\frac{1}{2}}\cdot
G_{\perp}\|^{2}_{\mathcal{H}\otimes\mathcal{H}}\leq
c^{2}\,\|G_{\mathrm{H}}\|^{2}_{*}<\infty,$ (B.2)

i.e. $G_{\mathrm{op}}\,Q_{\perp}^{-\frac{1}{2}}$ is a Hilbert-Schmidt
operator. By the general results in [27], this implies that
$\mathbb{H}_{\perp}$ is semi-bounded as well as the existence of a constant
$R>0$ such that Eq. (B.1) holds. ∎

######  Lemma B.3.

Let us define
$P_{j}:=\frac{1}{2i}\left(a_{j}-a_{j}^{\dagger}\right)=\sqrt{N}p_{j}$ for
$j\in\\{1,\dots,d\\}$, the constant $c_{0}:=\sum_{j=1}^{d}G_{j,j}$, the
quadratic function $\nu(y):=-4\sum_{j,k=1}^{d}G_{j,k}y_{j}y_{k}$ for
$y\in\mathbb{R}^{d}$ and the linear $\mathcal{H}_{\perp}$ valued function

$\displaystyle
u(y_{1},\dots,y_{d}):=4i\sum_{j=1}^{d}y_{j}\sum_{k>d}G_{j,k}u_{k}\in\mathcal{H}_{\perp}.$

Then we can rewrite the Bogoliubov operator $\mathbb{H}$ from Definition (4.3)
as

$\displaystyle\mathbb{H}=c_{0}+\nu\left(P_{1},\dots,P_{d}\right)+u\left(P_{1},\dots,P_{d}\right)^{\dagger}\cdot
a_{>d}+a_{>d}^{\dagger}\cdot
u\left(P_{1},\dots,P_{d}\right)+\mathbb{H}_{\perp}.$ (B.3)

###### Proof.

Since $\mathrm{Hess}|_{u_{0}}\mathcal{E}_{\mathrm{H}}[z]=z^{\dagger}\cdot
Q_{\mathrm{H}}\cdot z+2\mathfrak{Re}\left[G_{\mathrm{H}}^{\dagger}\cdot
z\otimes z\right]$ is degenerate in the directions $u_{1},\dots,u_{d}$, we
obtain $Q_{j,k}=-2G_{j,k}$ in case $j$ or $k$ is in $\\{1,\dots,d\\}$. Writing
$\mathbb{H}$ in coordinates therefore yields

$\displaystyle\mathbb{H}\\!$
$\displaystyle=\\!\\!\\!\sum_{j,k=1}^{d}\\!G_{j,k}\left(\\!-\\!2a_{j}^{\dagger}a_{k}\\!+\\!a_{j}a_{k}\\!+\\!a_{j}^{\dagger}a_{k}^{\dagger}\right)\\!+\\!2\sum_{j=1}^{d}\sum_{k>d}\\!G_{j,k}\left(a_{j}\\!-\\!a_{j}^{\dagger}\right)a_{k}\\!-\\!2\sum_{j=1}^{d}\sum_{k>d}G_{j,k}\left(a_{j}\\!-\\!a_{j}^{\dagger}\right)a_{k}^{\dagger}$
$\displaystyle\ \ \ \
+\sum_{j,k>d}\\!\left(Q_{j,k}a_{j}^{\dagger}a_{k}+G_{j,k}a_{j}a_{k}+G_{j,k}a_{j}^{\dagger}a_{k}^{\dagger}\right)$
$\displaystyle=\big{\\{}c_{0}+\nu\left(P_{1},\dots,P_{d}\right)\big{\\}}+u\left(P_{1},\dots,P_{d}\right)^{\dagger}\cdot
a_{>d}+a_{>d}^{\dagger}\cdot
u\left(P_{1},\dots,P_{d}\right)+\mathbb{H}_{\perp},$

where we have used
$-2a_{j}^{\dagger}a_{k}+a_{j}a_{k}+a_{j}^{\dagger}a_{k}^{\dagger}=-4P_{j}P_{k}+\delta_{j,k}$
in the second identity. ∎

######  Remark B.4.

In the subsequent Lemma B.5, we want to get rid of the term
$u\left(P_{1},\dots,P_{d}\right)^{\dagger}\cdot a_{>d}+a_{>d}^{\dagger}\cdot
u\left(P_{1},\dots,P_{d}\right)$ in Eq. (B.3) by completing the square, i.e.
by applying a shift $a_{>d}\mapsto a_{>d}+w\left(P_{1},\dots,P_{d}\right)$
where $w(y_{1},\dots,y_{d})\in\mathcal{H}_{\perp}$ is a suitable vector. In
the following we are going to construct such a $w(y)$. Let us first define the
$\mathbb{R}$-linear map
$L:\mathcal{H}_{\perp}\longrightarrow\mathcal{H}_{\perp}$

$\displaystyle z^{\dagger}\cdot L(w):=z^{\dagger}\cdot Q_{\perp}\cdot
w+2\left(w\otimes z\right)^{\dagger}\cdot G_{\perp},$

for all $z\in\mathcal{H}_{\perp}$. Furthermore, let us define the real inner
product $\braket{z,w}_{\mathbb{R}}:=\mathfrak{Re}\left[z^{\dagger}\cdot
w\right]$ on $\mathcal{H}_{\perp}$. Clearly, $L$ is symmetric with respect to
this inner product. By Assumption 1.3 we have for all
$w\in\mathcal{H}_{\perp}$

$\displaystyle\braket{w,L(w)}_{\mathbb{R}}=\mathrm{Hess}|_{u_{0}}\mathcal{E}_{\mathrm{H}}[w]\geq\eta\,\|w\|^{2},$
(B.4)

and consequently we can define $w(y)\in\mathcal{H}_{\perp}$ for all
$y\in\mathbb{R}^{d}$ as the solution of the equation

$\displaystyle L\cdot w(y)=-u(y).$ (B.5)

We note that $w(y)\in\mathrm{dom}[Q_{\perp}]$ due to the improved coercivity

$\displaystyle\braket{w,L(w)}_{\mathbb{R}}\geq\tilde{c}\,w^{\dagger}\cdot
Q_{\perp}\cdot w$ (B.6)

where $\tilde{c}$ is a suitable constant, which follows from the fact that

$\displaystyle 2\left|\left(w\otimes
w\right)^{\dagger}\\!\cdot\\!G_{\perp}\right|\leq
w^{\dagger}\\!\cdot\\!\left(\epsilon\,Q_{\perp}\\!+\\!\epsilon^{-1}c^{2}\|G_{\mathrm{H}}\|_{*}^{2}\right)\\!\cdot\\!w\leq\epsilon\,w^{\dagger}\\!\cdot\\!Q_{\perp}\\!\cdot\\!w\\!+\\!\frac{c^{2}\|G_{\mathrm{H}}\|_{*}^{2}}{\epsilon\eta}\braket{w,L(w)}_{\mathbb{R}}$

for all $\epsilon>0$, where $c$ is the constant in Eq. (B.2).

######  Lemma B.5.

Let $w:\mathbb{R}^{d}\rightarrow\mathcal{H}_{\perp}$ be the function defined
by Eq. (B.5) and let us define the unitary transformation
$\mathcal{R}:\mathcal{F}_{0}\longrightarrow\mathcal{F}_{0}$

$\displaystyle\mathcal{R}:$
$\displaystyle=\exp\left[w\left(P_{1},\dots,P_{d}\right)^{\dagger}\cdot
a_{>d}-a_{>d}^{\dagger}\cdot w\left(P_{1},\dots,P_{d}\right)\right].$

Then there exists a non-negative quadratic function
$\eta:\mathbb{R}^{d}\longrightarrow\mathbb{R}$, s.t.

$\displaystyle\mathcal{R}\,\mathbb{H}\,\mathcal{R}^{-1}=c_{0}+\eta\left(P_{1},\dots,P_{d}\right)+\mathbb{H}_{\perp},$
(B.7)

where $c_{0}$ and $\mathbb{H}_{\perp}$ are as in Lemma B.3 and Definition B.1.

###### Proof.

Let us define
$\eta(y_{1},\dots,y_{d}):=\nu(y_{1},\dots,y_{d})+\braket{w(y_{1},\dots,y_{d}),u(y_{1},\dots,y_{d})}_{\mathbb{R}}$.
With $\eta$ and the vector valued function $w$ at hand, we can rewrite Eq.
(B.3) as

$\displaystyle\mathbb{H}$
$\displaystyle=c_{0}+\eta\left(P_{1},\dots,P_{d}\right)+\Big{(}a_{>d}-w\left(P_{1},\dots,P_{d}\right)\Big{)}^{\dagger}\cdot
Q_{\perp}\cdot\Big{(}a_{>d}-w\left(P_{1},\dots,P_{d}\right)\Big{)}$
$\displaystyle\ \ \
+2\mathfrak{Re}\left[G_{\perp}^{\dagger}\cdot\Big{(}a_{>d}-w\left(P_{1},\dots,P_{d}\right)\Big{)}\
\underline{\otimes}\
\Big{(}a_{>d}-w\left(P_{1},\dots,P_{d}\right)\Big{)}\right].$ (B.8)

Eq. (B.7) follows now from the representation of $\mathbb{H}$ in Eq. (B.8) and
the fact that

$\displaystyle\mathcal{R}\,a_{>d}\,\mathcal{R}^{-1}=a_{>d}+w\left(P_{1},\dots,P_{d}\right).$

In order to see that $\eta$ is indeed non-negative, note that we can use
$\eta$ and $w$ to complete the square in
$\mathrm{Hess}|_{u_{0}}\mathcal{E}_{\mathrm{H}}[z]$ as well, i.e. for
$z=\sum_{j=1}^{d}\left(t_{j}+is_{j}\right)u_{j}+z_{>d}$ with
$t,s\in\mathbb{R}^{d}$ and $z_{>d}\in\mathcal{H}_{\perp}$ we can write
$\mathrm{Hess}|_{u_{0}}\mathcal{E}_{\mathrm{H}}[z]$ as

$\displaystyle\eta(s)\\!+\\!\left(z_{>d}\\!-\\!w(s)\right)^{\dagger}\\!\cdot\\!Q_{\perp}\\!\cdot\\!\left(z_{>d}\\!-\\!w(s)\right)\\!+\\!2\mathfrak{Re}\left[G_{\perp}^{\dagger}\cdot\left(z_{>d}\\!-\\!w(s)\right)\\!\otimes\\!\left(z_{>d}\\!-\\!w(s)\right)\right].$

Therefore, $\mathrm{Hess}|_{u_{0}}\mathcal{E}_{\mathrm{H}}[z]\geq 0$ for all
$z$ implies $\eta(s)\geq 0$ for all $s\in\mathbb{R}^{d}$. ∎

###### Proof of Theorem 4.4.

Since the function $\eta$ in Lemma B.5 is non-negative, we immediately obtain
the lower bound

$\displaystyle\inf\sigma\left(\mathbb{H}\right)\geq
c_{0}+\inf\sigma\left(\mathbb{H}_{\perp}\right)>-\infty.$

In order to verify the bound from below for the operator
$\mathbb{H}-r\mathbb{A}$, where $\mathbb{A}$ is defined in Eq. (4.3), we will
make use of the improved coercivity

$\displaystyle\mathrm{Hess}|_{u_{0}}\mathcal{E}_{\mathrm{H}}[z]\geq
r_{*}\left(\sum_{j=1}^{d}s_{j}^{2}+z_{>d}^{\dagger}\cdot(T+1)\cdot
z_{>d}\right),$ (B.9)

where $r_{*}$ is a suitable constant and
$z=\sum_{j=1}^{d}(t_{j}+is_{j})u_{j}+z_{>d}$ with
$z_{>d}\in\mathcal{H}_{\perp}$, which can be verified analogously to Eq. (B.6)
in Remark B.4. With the definition $\eta_{r}:=r_{*}-r$ for $r<r_{*}$ we
obtain, in analogy to Assumption 1.3,

$\displaystyle\mathrm{Hess}|_{u_{0}}\mathcal{E}_{\mathrm{H}}[z]-r\left(\sum_{j=1}^{d}s_{j}^{2}+z_{>d}^{\dagger}\cdot(T+1)\cdot
z_{>d}\right)\geq\eta_{r}\|z\|^{2}$

for all $z$ of the form $z=i\sum_{j=1}^{d}s_{j}u_{j}+z_{>d}$ with
$s_{j}\in\mathbb{R}$ and $z_{>d}\in\mathcal{H}_{\perp}$. Therefore we can
repeat the proof of the lower bound for the operator $\mathbb{H}-r\mathbb{A}$,
which yields

$\displaystyle\mathbb{H}-r\mathbb{A}\geq\inf\sigma\left(\mathbb{H}-r\mathbb{A}\right)>-\infty.$
(B.10)

Note that this further implies that the Friedrichs extension of the quadratic
form $\mathbb{H}$ is well-defined, i.e. $\mathbb{H}$ is semi-bounded and
closeable, since $\mathbb{H}$ is comparable to the non-negative selfadjoint
operator $\mathbb{A}$, i.e. there exist constants
$\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}>0$ with

$\displaystyle\alpha_{1}\mathbb{A}-\beta_{1}\leq\mathbb{H}\leq\alpha_{2}\mathbb{A}+\beta_{2}.$

In order to verify that there exists an approximate sequence of ground states
$\Psi_{M}$ with $\Psi_{M}\in\mathcal{F}_{\leq M}$ and
$\Psi_{M}\in\mathrm{dom}\left[a_{\geq 1}^{\dagger}\cdot(T+1)\cdot a_{\geq
1}\right]$, it is enough to prove that such states are dense in
$\mathrm{dom}\left[a_{\geq 1}^{\dagger}\cdot\left(T+1\right)\cdot a_{\geq
1}\right]$, the domain of the quadratic form which defines the Bogoliubov
operator $\mathbb{H}$ by Friedrichs extension, with respect to the norm
$\|\Psi\|^{2}_{\mathbb{H}}:=\braket{\mathbb{H}+C}_{\Psi}$ where
$C>-\inf\sigma\left(\mathbb{H}\right)$. The lower bound follows from Eq.
(B.10), while the upper bound follows from Eq. (B.3) and Inequality (B.1).
Furthermore, we have

$\displaystyle\|\Psi\|^{2}_{\mathbb{H}}\leq\alpha_{2}\braket{\mathbb{A}}_{\Psi}+(\beta_{2}+C)\|\Psi\|^{2}\leq\|\Psi\|^{2}_{\diamond},$

for all $\Psi\in\mathcal{F}_{0}$, where
$\|\Psi\|_{\diamond}^{2}:=\alpha_{2}\braket{a_{\geq
1}^{\dagger}\cdot(T+1)\cdot a_{\geq
1}}_{\Psi}+(\beta_{2}+C+\frac{d}{4})\|\Psi\|^{2}$. Clearly,
$\bigcup_{M}\mathcal{F}_{\leq M}\cap\mathrm{dom}\left[a_{\geq
1}^{\dagger}\cdot(T+1)\cdot a_{\geq 1}\right]$ is dense in the domain
$\mathrm{dom}\left[a_{\geq 1}^{\dagger}\cdot\left(T+1\right)\cdot a_{\geq
1}\right]$ with respect to the norm $\|.\|_{\diamond}$ and therefore it is
also dense with respect to $\|.\|_{\mathbb{H}}$. ∎

##  Appendix C Auxiliary Lemmata

In the following section we will derive various operator estimates involving
powers of the operators $p,p^{\prime},b_{>d}$ and functions of $q$, with an
emphasis on asymptotic results of the form $A_{N}=o_{*}(B_{N})$, where the
$o_{*}(\cdot)$ notation is introduced in Definition 4.10. It is a crucial
observation that all of our basic variables $q_{i},p_{j}$ and $b_{k}$ are of
order $o_{*}(1)$, and therefore the product of a basic variable with an
operator $A_{N}$ should be of order $o_{*}(A_{N})$, which we will verify for
specific examples $A_{N}$. Let us first discuss an important tool, which we
will repeatedly use, given by the following Cauchy–Schwarz inequality for
operators.

######  Lemma C.1.

For any $\lambda\in\mathbb{C}$ with $|\lambda|=1$, $t>0$, linear operators
$A:\mathcal{H}_{1}\longrightarrow\mathcal{H}_{2}$ and
$B:\mathcal{H}_{1}\longrightarrow\mathcal{H}_{2}$, and selfadjoint operator
$Q:\mathcal{H}_{2}\longrightarrow\mathcal{H}_{2}$, we have the operator
inequality

$\displaystyle\mathfrak{Re}\left[\lambda\ A^{\dagger}\cdot Q\cdot B\right]\leq
t\ A^{\dagger}\cdot\left|Q\right|\cdot A+t^{-1}\
B^{\dagger}\cdot\left|Q\right|\cdot B.$ (C.1)

Furthermore, let $A_{N},B_{N}$ be sequences of linear operators
$\mathcal{H}_{1}\longrightarrow\mathcal{H}_{2}$, $Q$ a selfadjoint operator on
$\mathcal{H}_{2}$ and $C_{N}:\mathcal{H}_{1}\longrightarrow\mathcal{H}_{1}$ a
sequence of non-negative operators, which satisfy
$A_{N}^{\dagger}\cdot\left|Q\right|\cdot A_{N}=O_{*}\left(C_{N}\right)$ and
$B_{N}^{\dagger}\cdot\left|Q\right|\cdot B_{N}=o_{*}\left(C_{N}\right)$. Then,

$\displaystyle A_{N}^{\dagger}\cdot Q\cdot B_{N}=o_{*}\left(C_{N}\right),$
$\displaystyle B_{N}^{\dagger}\cdot Q\cdot A_{N}=o_{*}\left(C_{N}\right).$

###### Proof.

Let $Q=U|Q|$ be the polar decomposition of $Q$. Inequality (C.1) immediately
follows from the inequality

$\displaystyle 0\leq\left(\sqrt{t}\ A-\sqrt{\frac{1}{t}}\lambda\
UB\right)^{\dagger}\cdot|Q|\cdot\left(\sqrt{t}\ A-\sqrt{\frac{1}{t}}\lambda\
UB\right).$

By our assumption $A_{N}^{\dagger}\cdot\left|Q\right|\cdot
A_{N}=O_{*}\left(C_{N}\right)$ we know that there exist constants
$c,\delta_{0}>0$, such that
$\pi_{M,N}\,A_{N}^{\dagger}\cdot\left|Q\right|\cdot A_{N}\,\pi_{M,N}\leq
c\braket{C_{N}}_{\Psi}$ for all $\frac{M}{N}\leq\delta_{0}$. Furthermore, by
our assumption $B_{N}^{\dagger}\cdot\left|Q\right|\cdot
B_{N}=o_{*}\left(C_{N}\right)$, there exists a function
$\epsilon:\mathbb{R}^{+}\longrightarrow\mathbb{R}^{+}$ with
$\underset{\delta\rightarrow 0}{\lim}\epsilon(\delta)$, such that
$\pi_{M,N}\,B_{N}^{\dagger}\cdot\left|Q\right|\cdot
B_{N}\,\pi_{M,N}\leq\epsilon\left(\frac{M}{N}\right)C_{N}$. Applying
Inequality (C.1) with $t:=\sqrt{\epsilon\left(\frac{M}{N}\right)}$ yields for
all $\lambda\in\mathbb{C}$ with $|\lambda|=1$ and $\frac{M}{N}\leq\delta_{0}$

$\displaystyle\pi_{M,N}\,\mathfrak{Re}\left[\lambda\ A_{N}^{\dagger}\cdot
Q\cdot B_{N}\right]\,\pi_{M,N}\leq\sqrt{\epsilon\left(\frac{M}{N}\right)}\
C_{N}.$

∎

Consider a function $g:\mathbb{R}^{d}\longrightarrow\mathbb{R}$. The following
Lemma states that the operator $g(q)$ depends, up to an exponentially small
error, only on the local data of $g$ in an arbitrary small neighborhood
$[-\epsilon,\epsilon]^{d}$ of the origin, i.e.
$g(q)=\widetilde{g}(q)+O_{*}\left(e^{-\delta N}\right)$ in case
$g|_{[-\epsilon,\epsilon]^{d}}=\widetilde{g}|_{[-\epsilon,\epsilon]^{d}}$.
This property plays a key role in the proof of the main technical Theorem
4.12, since the involved functions are (somewhat arbitrary) extensions of
locally constructed functions with specific properties, which the extensions
no longer have, see for example the definition of
$f:\mathbb{R}^{d}\longrightarrow\mathcal{H}_{0}$ in Definition 4.7.

######  Lemma C.2.

Let $q_{1},\dots,q_{d}$ be the operators defined in Eq. (4.4) and let
$g:\mathbb{R}^{d}\rightarrow\mathbb{R}$ be a function such that
$g|_{[-\epsilon,\epsilon]^{d}}=0$ for some $\epsilon>0$. Furthermore, assume
that $g$ satisfies the growth condition $|g(t)|\leq C|t|^{2j}$, with $C>0$ and
$j\in\mathbb{N}$. Then

$\displaystyle g\left(q\right)=O_{*}\left(e^{-\delta N}\right)$

for some $\delta>0$.

###### Proof.

Using the elementary estimate
$|t|^{2j}=\left(\sum_{r=1}^{d}t_{r}^{2}\right)^{j}\leq d^{j}\max_{1\leq r\leq
d}t_{r}^{2j}$ yields

$\displaystyle|g(t)|\leq d^{j}C\sum_{r=1}^{d}t_{r}^{2j}\
\mathds{1}_{(\epsilon,\infty)}\left(|t_{r}|\right).$

In the following we want to verify that there exist constants $C$, $\delta>0$
and $\delta_{0}>0$ such that
$\braket{q_{r}^{2j}\mathds{1}_{(\epsilon,\infty)}(}{q_{r}}{)}_{\Psi}\leq
Ce^{-\delta N}$ for all states $\Psi\in\mathcal{W}_{N}\mathcal{F}_{\leq M}$,
$\|\Psi\|=1$, with $\frac{M}{N}\leq\delta_{0}$ and $r\in\\{1,\dots,d\\}$.
Since $\mathcal{W}_{N}\,q_{r}\,\mathcal{W}_{N}^{-1}=q_{r}$, it is equivalent
to verify this for $\Psi\in\mathcal{F}_{\leq M}$ instead. Due to the
reflection symmetry $q_{r}\mapsto-q_{r}$ of $q_{r}^{2j}$, it is furthermore
enough to verify that
$\braket{\mathds{1}_{(\epsilon,\infty)}(q_{r})q_{r}^{2j}}_{\Psi}\leq
Ce^{-\delta N}$ for all states $\Psi\in\mathcal{F}_{\leq M}$ with
$\frac{M}{N}\leq\delta_{0}$. Note that the operators
$q_{r}:=\frac{1}{\sqrt{2N}}\frac{a_{r}+a_{r}^{\dagger}}{\sqrt{2}}$ depend on
$N$. In the following we will make use of the description of the Fock space
$\mathcal{F}_{\leq M}$ in terms of Hermite polynomials $h_{n}$, i.e. for
$r\in\\{1,\dots,d\\}$ and $\Psi\in\mathcal{F}_{\leq M}$ there exist states
$\Psi_{n}\in\mathcal{F}_{\leq M-n}$ with $a_{r}\Psi_{n}=0$, such that
$\Psi=\sum_{n=0}^{M}h_{n}\\!\left(\frac{a_{r}+a_{r}^{\dagger}}{\sqrt{2}}\right)\Psi_{n}$,
see for example Eq. (1.26), respectively Exercise 1(ii), in [25]. Furthermore
we define the density matrix
$\gamma_{r}(x,y):=\sum_{n_{1},n_{2}=0}^{M}\braket{\Psi_{n_{1}},\Psi_{n_{2}}}h_{n_{2}}(x)h_{n_{1}}(y)\frac{1}{\sqrt{\pi}}e^{-\frac{x^{2}+y^{2}}{2}}$
on $L^{2}\left(\mathbb{R}\right)$. With $\gamma_{r}$ at hand we have

$\displaystyle\braket{\mathds{1}_{(\epsilon,\infty)}(q_{r})q_{r}^{2j}}_{\Psi}=\int_{\sqrt{2N}\epsilon}^{\infty}\left(\frac{x}{\sqrt{2N}}\right)^{2j}\gamma_{r}(x,x)\mathrm{d}x.$

In order to estimate this quantity, let us define the harmonic oscillator
Hamiltonian $H:=-\frac{\mathrm{d}^{2}}{\mathrm{d}x^{2}}+x^{2}$ on
$L^{2}(\mathbb{R})$. Since $\gamma_{r}$ involves only eigenfunctions
$h_{n}(x)e^{-\frac{x^{2}}{2}}$ of $H$ with $n\leq M$, we have the operator
inequality $\gamma_{r}\leq e^{2M+1-H}$. Using the Mehler kernel for $e^{-H}$
therefore yields for $c:=\frac{1}{\sqrt{2\pi\,\mathrm{sinh}(2)}}$ and
$\lambda:=\mathrm{coth}(2)-\mathrm{cosech(2)}>0$, and all
$M\leq\epsilon^{2}\lambda N/2$

$\displaystyle\int_{\sqrt{2N}\epsilon}^{\infty}\left(\frac{x}{\sqrt{2N}}\right)^{2j}\gamma_{r}(x,x)\mathrm{d}x\leq
c\,e^{\epsilon^{2}\lambda
N+1}\int_{\sqrt{2N}\epsilon}^{\infty}\left(\frac{x}{\sqrt{2N}}\right)^{2j}e^{-\lambda
x^{2}}\mathrm{d}x=O_{N\rightarrow\infty}\left(e^{-\epsilon^{2}\lambda
N}\right).$

∎

The following Lemma is an auxiliary result, which will be useful for the
verification of various asymptotic results involving the operator $b_{>d}$.

######  Lemma C.3.

Recall the operators $\mathcal{W}_{N},\mathbb{L}^{\prime},p_{j}^{\prime}$ and
$f(q)$ from Definition 4.8 and the definition of $\hat{\pi}_{M,N}$ above Eq.
(5.21). Then, $q_{j}$ and $p^{\prime}_{j}$ commute with $\hat{\pi}_{M,N}$ for
$j\in\\{1,\dots,d\\}$,
$\hat{\pi}_{M,N}\mathbb{L}^{\prime}\Psi=\mathbb{L}^{\prime}\hat{\pi}_{M,N}\Psi$
for all $\Psi\in\mathcal{W}_{N}\mathcal{F}_{\leq N}$, and
$b_{k}\,\hat{\pi}_{M,N}=\hat{\pi}_{M,N}\,b_{k}\,\hat{\pi}_{M,N}$ for all
$k>d$. Furthermore, we have for all $M\leq N$ the estimate

$\displaystyle\hat{\pi}_{M,N}\left(b_{>d}+f(q)\right)^{\dagger}\cdot\left(b_{>d}+f(q)\right)\hat{\pi}_{M,N}\leq\frac{M}{N},$
(C.2) $\displaystyle\hat{\pi}_{M,N}\,b_{>d}^{\dagger}\cdot
b_{>d}\,\hat{\pi}_{M,N}\leq 4.$ (C.3)

###### Proof.

Recall $\mathcal{N}:=\sum_{j=1}^{\infty}a_{j}^{\dagger}a_{j}$ and let us
define $\mathcal{N}_{+}:=\sum_{j>d}a_{j}^{\dagger}\cdot a_{j}$. Since
$\mathcal{N}_{+}$ commutes with $\mathbb{L}$ and $q_{j},p_{j}$ for
$j\in\\{1,\dots,d\\}$, we obtain that
$\hat{\pi}_{M,N}=\mathcal{W}_{N}\,\mathds{1}_{[0,M]}\left(\mathcal{N}_{+}\right)\mathcal{W}_{N}^{-1}$
commutes with $q_{j}=\mathcal{W}_{N}\,q_{j}\,\mathcal{W}_{N}^{-1}$ and
$p_{j}^{\prime}=\mathcal{W}_{N}\,p_{j}\,\mathcal{W}_{N}^{-1}$. Similarly
$\hat{\pi}_{M,N}\mathbb{L}^{\prime}\Psi=\mathbb{L}^{\prime}\hat{\pi}_{M,N}\Psi$
for all $\Psi\in\mathcal{W}_{N}\mathcal{F}_{\leq N}$. Making use of the fact
that
$b_{j}=\mathcal{W}_{N}\,\left(b_{j}-f_{j}(q)\right)\,\mathcal{W}_{N}^{-1}$
yields

$\displaystyle b_{j}\,\hat{\pi}_{M,N}$
$\displaystyle=\mathcal{W}_{N}\left(b_{j}-f_{j}(q)\right)\mathds{1}_{[0,M]}\left(\mathcal{N}_{+}\right)\mathcal{W}_{N}^{-1}$
$\displaystyle=\mathcal{W}_{N}\,\mathds{1}_{[0,M]}\left(\mathcal{N}_{+}\right)\left(b_{j}-f_{j}(q)\right)\mathds{1}_{[0,M]}\left(\mathcal{N}_{+}\right)\mathcal{W}_{N}^{-1}=\hat{\pi}_{M,N}\,b_{j}\,\hat{\pi}_{M,N}.$

Inequality (C.2) follows from
$\big{(}b_{>d}+f(q)\big{)}^{\dagger}\cdot\big{(}b_{>d}+f(q)\big{)}=\frac{1}{N}\mathcal{W}_{N}\,\mathcal{N}_{+}\,\mathcal{W}_{N}^{-1}$
and

$\displaystyle\hat{\pi}_{M,N}\,\left(b_{>d}+f(q)\right)^{\dagger}\cdot\left(b_{>d}+f(q)\right)\,\hat{\pi}_{M,N}=\frac{1}{N}\mathcal{W}_{N}\,\mathcal{N}_{+}\mathds{1}_{[0,M]}\left(\mathcal{N}_{+}\right)\,\mathcal{W}_{N}^{-1}\leq\frac{M}{N}.$

In order to verify Inequality (C.3), note that $f(t)^{\dagger}\cdot f(t)\leq
1$ for all $t$. Applying the Cauchy–Schwarz inequality as in C.1 yields

$\displaystyle\hat{\pi}_{M,N}\,b_{>d}^{\dagger}\cdot
b_{>d}\,\hat{\pi}_{M,N}\\!$
$\displaystyle\leq\\!2\hat{\pi}_{M,N}\left(b_{>d}+f(q)\right)^{\dagger}\cdot\left(b_{>d}+f(q)\right)\hat{\pi}_{M,N}\\!+\\!2\hat{\pi}_{M,N}\,f(q)^{\dagger}\\!\cdot
f(q)\,\hat{\pi}_{M,N}$ $\displaystyle\leq 2\frac{M}{N}+2\leq 4.$

∎

The proof of the main technical Theorem 4.12 consists of two steps: First one
has to identify the residuum $R_{J}$, which is carried out in the Lemmata 5.6
and 5.12, and in the second step one has to derive asymptotic results for
these residua $R_{J}$, which is carried out in the Theorems 5.7 and 5.16. The
following three Lemmata provide asymptotic results for the types of operators
most frequently encountered during our analysis of $R_{J}$.

######  Lemma C.4.

Let $\phi,\Phi:\mathbb{R}^{d}\longrightarrow\mathbb{R}$ be functions with
$|\phi(t)|\leq C|t|^{k}$ and $|\Phi(t)|\leq C(1+|t|^{k})$ for some $k\geq 1$.
Then, $\phi(q)=o_{*}(1)$ and $\Phi(q)=O_{*}(1)$. Furthermore,

$\displaystyle\phi(q)\,b_{>d}^{\dagger}\cdot b_{>d}$
$\displaystyle=o_{*}\left(b_{>d}^{\dagger}\cdot b_{>d}+\frac{1}{N}\right),$
(C.4) $\displaystyle\Phi(q)\,b_{>d}^{\dagger}\cdot b_{>d}$
$\displaystyle=O_{*}\left(b_{>d}^{\dagger}\cdot b_{>d}+\frac{1}{N}\right).$
(C.5)

###### Proof.

In the following, let $0\leq\tau\leq 1$ be a smooth function with
$\mathrm{supp}\left(\tau\right)\subset B_{1}(0)$ and $\tau(t)=1$ for all $t\in
B_{\frac{1}{2}}(0)$, and let $\tau_{r}(t):=\tau\left(\frac{t}{r}\right)$ for
$r>0$. Clearly $\phi(q)=\tau_{r}(q)\phi(q)+(1-\tau_{r}(q))\phi(q)$. By our
assumptions we know that $|\tau_{r}\phi|\leq\epsilon_{r}$ with
$\epsilon_{r}\underset{r\rightarrow 0}{\longrightarrow}0$ and
$(1-\tau_{r})\phi$ is zero in a neighborhood of zero, hence
$(1-\tau_{r}(q))\phi(q)=O_{*}\left(e^{-\delta N}\right)$ by Lemma C.2. We
conclude that $|\phi(q)|\leq\epsilon_{r}+O_{*}\left(e^{-\delta N}\right)$ for
all $r>0$, and consequently $\phi(q)=o_{*}(1)$. The corresponding statement
for $\Phi(q)$ follows from the fact that $\Phi(q)\leq C+\phi(q)$ with
$\phi(t):=|t|^{k}$ and $\phi(q)=o_{*}(1)$.

Let us write similar to before $\phi(q)\,b_{>d}^{\dagger}\cdot
b_{>d}=\tau_{r}(q)\phi(q)\,b_{>d}^{\dagger}\cdot
b_{>d}+(1-\tau_{r}(q))\phi(q)\,b_{>d}^{\dagger}\cdot b_{>d}$. In order to
verify Eq. (C.5). First of all $\tau_{r}(q)\phi(q)\,b_{>d}^{\dagger}\cdot
b_{>d}\\!\leq\epsilon_{r}b_{>d}^{\dagger}\cdot b_{>d}$, where we use that $q$
commutes with $b_{>d}$. For the treatment of the second term, recall
Inequality (C.3) and
$\pi_{M,N}(1-\tau_{r}(q))^{2}|\phi_{r}(q)|^{2}\pi_{M,N}\leq C^{2}e^{-2\delta
N}$ for $\frac{M}{N}\leq\delta$ with $C,\delta>0$, which follows from Lemma
C.2. Hence,

$\displaystyle\pi_{M,N}(1\\!-\\!\tau_{r}(q))|\phi(q)|\,b_{>d}^{\dagger}\cdot
b_{>d}\,\pi_{M,N}\\!=\\!\pi_{M,N}(1\\!-\\!\tau_{r}(q))|\phi(q)|\hat{\pi}_{M,N}\,b_{>d}^{\dagger}\cdot
b_{>d}\,\pi_{M,N}$ $\displaystyle\
\leq\Big{\|}\pi_{M,N}(1\\!-\\!\tau_{r}(q))|\phi(q)|\Big{\|}\
\Big{\|}\hat{\pi}_{M,N}\,b_{>d}^{\dagger}\cdot
b_{>d}\,\hat{\pi}_{M,N}\Big{\|}\leq 4Ce^{-\delta N}.$

We conclude that $\pi_{M,N}\phi(q)\pi_{M,N}\leq 4Ce^{-\delta
N}+\epsilon_{r}b_{>d}^{\dagger}\cdot b_{>d}$ for $\frac{M}{N}\leq\delta$, and
therefore $\phi(q)=o_{*}\left(b_{>d}^{\dagger}\cdot
b_{>d}+\frac{1}{N}\right)$. The corresponding statement for
$\Phi(q)\,b_{>d}^{\dagger}\cdot b_{>d}$ follows as above. ∎

######  Lemma C.5.

Given $w:\mathbb{R}^{d}\longrightarrow\mathcal{H}$ with $\|w(t)\|\leq c\
|t|^{k}$ and $W:\mathbb{R}^{d}\longrightarrow\mathcal{H}$ with $\|W(t)\|\leq
c\left(1+|t|^{k}\right)$ for some $c>0$ and $k\geq 1$, we define
$X:=W(q)^{\dagger}\cdot b_{>d}$ and $Y:=w(q)^{\dagger}\cdot b_{>d}$. Then,
$X^{\dagger}X$ and $XX^{\dagger}$ are of order
$O_{*}\left(b_{>d}^{\dagger}\cdot b_{>d}+\frac{1}{N}\right)$, and
$Y^{\dagger}Y$ and $YY^{\dagger}$ are of order
$o_{*}\left(b_{>d}^{\dagger}\cdot b_{>d}+\frac{1}{N}\right)$. Furthermore, for
$\Phi:\mathbb{R}^{d}\longrightarrow\mathbb{R}$ with $|\Phi(t)|\leq
c\left(1+|t|^{j}\right)$, we obtain

$\displaystyle\Phi(q)\left(b_{>d}^{\dagger}\cdot
b_{>d}\right)^{2}=o_{*}\left(b_{>d}^{\dagger}\cdot b_{>d}+\frac{1}{N}\right).$

Recall the operator $p^{\prime}$ from Definition 4.8. We have

$\displaystyle\left(p^{\prime}-p\right)^{\dagger}\cdot\left(p^{\prime}-p\right)=o_{*}\left(b_{>d}^{\dagger}\cdot
b_{>d}+\frac{1}{N}\right).$

###### Proof.

Let us define $G(t):=W(t)^{\dagger}\cdot W(t)$ and $g(t):=w(t)^{\dagger}\cdot
w(t)$. Then we obtain by Lemma C.4 together with the inequality $W(t)\cdot
W(t)^{\dagger}\leq G(t)\ 1_{\mathcal{H}}$ the estimate

$\displaystyle X^{\dagger}X$ $\displaystyle=b_{>d}^{\dagger}\cdot W(q)\cdot
W(q)^{\dagger}\cdot b_{>d}\leq G(q)\,b_{>d}^{\dagger}\cdot
b_{>d}=O_{*}\left(b_{>d}^{\dagger}\cdot b_{>d}+\frac{1}{N}\right).$

Similarly, $Y^{\dagger}Y\leq g(q)\,b_{>d}^{\dagger}\cdot
b_{>d}=o_{*}\left(b_{>d}^{\dagger}\cdot b_{>d}+\frac{1}{N}\right)$. For the
reversed order, we use the fact that $\|G(q)\|^{2}=O_{*}\left(1\right)$ and
$\|g(q)\|^{2}=o_{*}\left(1\right)$

$\displaystyle XX^{\dagger}$
$\displaystyle=X^{\dagger}X+\frac{1}{N}\|G(q)\|^{2}=O_{*}\left(b_{>d}^{\dagger}\cdot
b_{>d}+\frac{1}{N}\right),$ $\displaystyle YY^{\dagger}$
$\displaystyle=Y^{\dagger}Y+\frac{1}{N}\|g(q)\|^{2}=o_{*}\left(b_{>d}^{\dagger}\cdot
b_{>d}+\frac{1}{N}\right).$

For the next statement, note that we have $\left(b_{>d}^{\dagger}\cdot
b_{>d}\right)^{2}=b_{>d}^{\dagger}\cdot\left(b_{>d}^{\dagger}\cdot
b_{>d}+\frac{1}{N}\right)\cdot b_{>d}$ and $b_{>d}^{\dagger}\cdot b_{>d}\leq
2(b_{>d}\\!+\\!f(q))^{\dagger}\\!\cdot\\!(b_{>d}\\!+\\!f(q))+2f(q)^{\dagger}\\!\cdot\\!f(q)$,
and consequently

$\displaystyle\Phi(q)\left(b_{>d}^{\dagger}\cdot
b_{>d}\right)^{2}=\Phi(q)\,b_{>d}^{\dagger}\cdot\left(b_{>d}^{\dagger}\cdot
b_{>d}+\frac{1}{N}\right)\cdot b_{>d}$ $\displaystyle\leq
2b_{>d}^{\dagger}\\!\cdot\\!(b_{>d}\\!+\\!f(q))^{\dagger}\\!\cdot\\!\Phi(q)\\!\cdot\\!(b_{>d}\\!+\\!f(q))\\!\cdot\\!b_{>d}\\!+\\!2\Phi(q)f(q)^{\dagger}\\!\cdot\\!f(q)\,b_{>d}^{\dagger}\\!\cdot\\!b_{>d}\\!+\\!\frac{\Phi(q)}{N}\,b_{>d}^{\dagger}\\!\cdot\\!b_{>d}.$

Note that
$2\Phi(q)f(q)^{\dagger}\\!\cdot\\!f(q)\,b_{>d}^{\dagger}\\!\cdot\\!b_{>d}$ and
$\frac{\Phi(q)}{N}\,b_{>d}^{\dagger}\\!\cdot\\!b_{>d}$ are of order
$o_{*}\left(b_{>d}^{\dagger}\cdot b_{>d}+\frac{1}{N}\right)$ by Lemma C.4. For
the other term in the inequality above, note that we have the estimate

$\displaystyle\ \ \ \
\pi_{M,N}\,b_{>d}^{\dagger}\cdot\left[\Phi(q)(b_{>d}+f(q))^{\dagger}\cdot(b_{>d}+f(q))\right]\cdot
b_{>d}\,\pi_{M,N}$
$\displaystyle=\pi_{M,N}\,b_{>d}^{\dagger}\cdot\left[\Phi(q)\hat{\pi}_{M,N}(b_{>d}+f(q))^{\dagger}\cdot(b_{>d}+f(q))\hat{\pi}_{M,N}\right]\cdot
b_{>d}\,\pi_{M,N}$ $\displaystyle\leq\frac{M}{N}\
\pi_{M,N}\Phi(q)\,b_{>d}^{\dagger}\cdot b_{>d}\,\pi_{M,N}\leq C\
\frac{M}{N}\pi_{M,N}\left(b_{>d}^{\dagger}\cdot
b_{>d}+\frac{1}{N}\right)\,\pi_{M,N},$

where we have used that $\pi_{M,N}\Phi(q)\,b_{>d}^{\dagger}\cdot
b_{>d}\,\pi_{M,N}\leq C\ \pi_{M,N}\left(b_{>d}^{\dagger}\cdot
b_{>d}+\frac{1}{N}\right)\pi_{M,N}$ for $\frac{M}{N}\leq\delta_{0}<1$, see
Lemma C.4. In order to verify the last part of the Lemma, let us define the
operators $Y_{\ell}:=\partial_{\ell}f(q)^{\dagger}\cdot b_{>d}$. From the
previous part of this Lemma we know

$\displaystyle\left(p^{\prime}-p\right)^{\dagger}\cdot\left(p^{\prime}-p\right)=\sum_{\ell=1}^{d}\mathfrak{Im}\left[Y_{\ell}\right]^{2}\leq\frac{1}{2}\sum_{\ell=1}^{d}\left(Y_{\ell}^{\dagger}Y_{\ell}+Y_{\ell}Y_{\ell}^{\dagger}\right)=o_{*}\left(b_{>d}^{\dagger}\cdot
b_{>d}+\frac{1}{N}\right).$

∎

For the following Lemma C.6 as well as for the results in Appendix D, it is
convenient to define the operator

$\displaystyle\mathbb{Q}_{N}:=p^{\dagger}\cdot p+b_{>d}^{\dagger}\cdot
b_{>d}+\frac{1}{N}.$ (C.6)

Since $\mathbb{Q}_{N}\leq\mathbb{T}_{N}$, where $\mathbb{T}_{N}$ is defined in
Eq. (4.11), any sequence with $X_{N}=O_{*}\left(\mathbb{Q}_{N}\right)$,
respectively $X_{N}=o_{*}\left(\mathbb{Q}_{N}\right)$, satisfies
$X_{N}=O_{*}\left(\mathbb{T}_{N}\right)$, respectively
$X_{N}=o_{*}\left(\mathbb{T}_{N}\right)$, as well.

######  Lemma C.6.

Let $\phi:\mathbb{R}^{d}\longrightarrow\mathbb{R}$ be a function with
$|\phi(t)|\leq c\ |t|^{k}$ and $\Phi:\mathbb{R}^{d}\longrightarrow\mathbb{R}$
with $|\Phi(t)|\leq c\left(1+|t|^{k}\right)$ for some constant $c$ and $k\geq
1$. Then

$\displaystyle\left(p^{\prime}\right)^{\dagger}\cdot\phi(q)\cdot
p^{\prime}=o_{*}\left(\mathbb{Q}_{N}\right),$
$\displaystyle\left(p^{\prime}\right)^{\dagger}\cdot\Phi(q)\cdot
p^{\prime}=O_{*}\left(\mathbb{Q}_{N}\right).$

In case the partial derivatives $\partial_{i}\phi(t)$, $\partial_{j}\Phi(t)$
and $\partial_{i}\partial_{j}\Phi(t)$ are bounded by
$c\left(1+|t|^{j}\right)$, we also have

$\displaystyle\phi(q)\left(p^{\prime}\right)^{\dagger}\cdot
p^{\prime}\,\phi(q)=o_{*}\left(\mathbb{Q}_{N}\right),$
$\displaystyle\Phi(q)\left[\left(p^{\prime}\right)^{\dagger}\cdot
p^{\prime}\right]^{2}\Phi(q)=o_{*}\left(\mathbb{Q}_{N}\right).$

###### Proof.

Since $p^{\dagger}\cdot p\leq\mathbb{Q}_{N}$ and
$\left(p^{\prime}-p\right)^{\dagger}\cdot\left(p^{\prime}-p\right)=O_{*}\left(\mathbb{Q}_{N}\right)$
by Lemma C.5, we obtain $(p^{\prime})^{\dagger}\cdot
p^{\prime}=O_{*}\left(\mathbb{Q}_{N}\right)$ as well, i.e.

$\displaystyle\pi_{M,N}(p^{\prime})^{\dagger}\cdot p^{\prime}\pi_{M,N}\leq
C_{1}\ \pi_{M,N}\mathbb{Q}_{N}\pi_{M,N}$

for all $M,N$ with $\frac{M}{N}\leq\delta_{1}<1$ where $\delta_{1}$ and
$C_{1}$ are suitable constants. By Lemma C.4, we know that
$\pi_{M,N}\Phi(q)\pi_{M,N}\leq C_{2}$ for all $\frac{M}{N}\leq\delta_{2}<1$
where $\delta_{2}$ and $C_{2}$ are suitable constants, and
$\pi_{M,N}\phi(q)\pi_{M,N}\leq\epsilon\left(\frac{M}{N}\right)$ with
$\lim_{\delta\rightarrow 0}\ \epsilon(\delta)=0$. Based on the observation
that $p^{\prime}\,\pi_{M,N}=\pi_{M+1,N}\,p^{\prime}\,\pi_{M,N}$, we obtain for
all $M,N$ that satisfy
$\frac{M}{N}\leq\delta:=2\mathrm{min}\\{\delta_{1},\delta_{2}\\}$

$\displaystyle\pi_{M,N}\left(p^{\prime}\right)^{\dagger}\cdot\Phi(q)\cdot
p^{\prime}\,\pi_{M,N}\\!=\\!\pi_{M,N}\left(p^{\prime}\right)^{\dagger}\cdot\pi_{M\\!+\\!1,N}\Phi(q)\pi_{M\\!+\\!1,N}\cdot
p^{\prime}\,\pi_{M,N}$ $\displaystyle\ \ \ \ \leq C_{1}C_{2}\
\pi_{M,N}\cdot\mathbb{Q}_{N}\cdot\pi_{M,N}.$

Similarly, we have
$\pi_{M,N}\left(p^{\prime}\right)^{\dagger}\cdot\phi(q)\cdot
p^{\prime}\,\pi_{M,N}\leq C_{1}\epsilon\left(\frac{M}{N}\right)\
\pi_{M,N}\,\mathbb{Q}_{N}\,\pi_{M,N}$. Hence,
$\left(p^{\prime}\right)^{\dagger}\cdot\Phi(q)\cdot
p^{\prime}=O_{*}\left(\mathbb{Q}_{N}\right)$ and
$\left(p^{\prime}\right)^{\dagger}\cdot\phi(q)\cdot
p^{\prime}=o_{*}\left(\mathbb{Q}_{N}\right)$. In case we have a polynomial
bound on the partial derivatives as well, let us define
$w(t):=\frac{1}{2}\sum_{\ell=1}^{d}\partial_{\ell}\phi(t)\otimes u_{\ell}$ and
$W(t):=\frac{1}{2}\sum_{\ell=1}^{d}\partial_{\ell}\Phi(t)\otimes u_{\ell}$.
Using the commutation relation
$[ip^{\prime}_{j},q_{k}]=\frac{\delta_{j,k}}{2N}$, we compute

$\displaystyle\phi(q)\,\left(p^{\prime}\right)^{\dagger}\cdot
p^{\prime}\,\phi(q)=\left(\phi(q)\cdot
ip^{\prime}+\frac{1}{N}w(q)\right)^{\dagger}\cdot\left(\phi(q)\cdot
ip^{\prime}+\frac{1}{N}w(q)\right).$

From the previous part, we know that $\left(\phi(q)\cdot
ip^{\prime}\right)^{\dagger}\cdot\phi(q)\cdot
ip^{\prime}=o_{*}\left(\mathbb{Q}_{N}\right)$. Furthermore, Lemma C.4 tells us
that $w(q)^{\dagger}\cdot w(q)=O_{*}\left(1\right)$, and therefore
$\frac{1}{N}w(q)^{\dagger}\cdot\frac{1}{N}w(q)=o_{*}\left(\mathbb{Q}_{N}\right)$.
Hence, $\phi(q)\left(p^{\prime}\right)^{\dagger}\cdot p^{\prime}\,\phi(q)$ is
of order $o_{*}\left(\mathbb{Q}_{N}\right)$ as well. The last estimate in the
Lemma can be verified analogously. ∎

##  Appendix D Analysis of the Operator Square Root

In the following section we derive asymptotic results for operators involving
the square root $\sqrt{1-\mathbb{L}^{\prime}}$, where $\mathbb{L}^{\prime}$ is
defined in Definition 4.8, allowing us to prove a Taylor approximation for the
operators $\left(1-\mathbb{L}^{\prime}\right)^{\frac{m}{2}}$, see Definition
5.5. The easiest case $m=2$ will be discussed in the following Lemma D.1, the
case $m=1$ is the content of Lemma D.2 and the case $m=3$ is covered by
Corollary D.5.

######  Lemma D.1.

Recall the operator $\mathbb{Q}_{N}$ from Eq. (C.6) and the function $f$ from
Definition 4.7, and let us define
$g(t):=\sum_{j=1}^{d}t_{j}^{2}+f(t)^{\dagger}\cdot f(t)$. Then,

$\displaystyle\left[\mathbb{L}^{\prime}-g(q)\right]^{2}=o_{*}\left(\mathbb{Q}_{N}\right).$
(D.1)

###### Proof.

Using the transformation laws in Lemma 4.9 we obtain

$\displaystyle\mathbb{L}^{\prime}-g(q)=f(q)^{\dagger}\cdot
b_{>d}+b_{>d}^{\dagger}\cdot f(q)+\left(p^{\prime}\right)^{\dagger}\cdot
p^{\prime}+b_{>d}^{\dagger}\cdot b_{>d}-\frac{d}{2N}.$

By Lemma C.5, we know that $\left[f(q)^{\dagger}\cdot
b_{>d}+b_{>d}^{\dagger}\cdot f(q)\right]^{2}=o_{*}\left(\mathbb{Q}_{N}\right)$
and $\left[b_{>d}^{\dagger}\cdot
b_{>d}\right]^{2}=o_{*}\left(\mathbb{Q}_{N}\right)$, and by Lemma C.6 we know
that $\left[\left(p^{\prime}\right)^{\dagger}\cdot
p^{\prime}\right]^{2}=o_{*}\left(\mathbb{Q}_{N}\right)$. ∎

######  Lemma D.2.

Let Assumption 1.3 hold and recall the function $\eta_{1}$ from Eq. (5.7).
Then,

$\displaystyle\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right]^{2}=o_{*}\left(\mathbb{Q}_{N}\right).$
(D.2)

Furthermore for any function $V:\mathbb{R}^{d}\longrightarrow\mathbb{R}$ with
$|V(t)|\leq c\left(|t|+|t|^{k}\right)$ and bounded derivatives
$|\partial_{t_{i}}V(t)|+|\partial_{t_{i}}\partial_{t_{j}}V(t)|\leq
c\left(1+|t|^{k}\right)$ for some $k\geq 1$, we have

$\displaystyle V(q)$
$\displaystyle\Big{[}\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)-D_{\mathcal{V}}\big{|}_{q}\eta_{1}\big{(}b_{\geq
1}\big{)}\Big{]}=o_{*}\left(\mathbb{Q}_{N}\right).$ (D.3)

###### Proof.

Let us define $h(x):=\chi\left(x\right)\sqrt{1-x}$, where
$\chi:[0,\infty)\longrightarrow[0,1]$ is the function from the definition of
$\eta_{1}$ in Eq. (5.7), as well as the operator $Q:=q^{\dagger}\cdot
q+f(q)^{\dagger}\cdot f(q)$. By the support properties of $\chi$ we have for
all $\frac{M}{N}<\frac{1}{2}$ and $\Psi\in\mathcal{W}_{N}\mathcal{F}_{\leq M}$

$\displaystyle\sqrt{1-\mathbb{L}^{\prime}}\,\Psi=h\left(\mathbb{L}^{\prime}\right)\Psi,$

and therefore it is enough to verify the statements of this Lemma for
$h\left(\mathbb{L}^{\prime}\right)$ instead of $\sqrt{1-\mathbb{L}^{\prime}}$.
With $h$ at hand, we have
$\eta_{1}(q)=h\left(\|F(q)\|^{2}\right)=h\left(Q\right)$ and

$\displaystyle D_{\mathcal{V}}\big{|}_{q}\eta_{1}(v)=w(q)^{\dagger}\cdot
v+v^{\dagger}\cdot w(q)$

with $w(t):=h^{\prime}\left(\sum_{j=1}^{d}t_{j}^{2}+f(t)^{\dagger}\cdot
f(t)\right)f(t)$, for all $v\in\mathcal{H}_{0}$. Hence
$w(q)=h^{\prime}\left(Q\right)f(q)$. In the following, let $\hat{h}$ be the
Fourier transform of the smooth function $h$, normalized such that
$h(x)=\int\hat{h}(z)\ e^{izx}\ \mathrm{d}z$. Then,

$\displaystyle
h\left(\mathbb{L}^{\prime}\right)-h\left(Q\right)=\int\hat{h}(z)\left(e^{iz\mathbb{L}^{\prime}}-e^{izQ}\right)\
\mathrm{d}z.$

In order to investigate the integrand, we use the following integral
representation

$\displaystyle e^{iz\mathbb{L}^{\prime}}-e^{izQ}$
$\displaystyle=i\int_{0}^{z}e^{iy\mathbb{L}^{\prime}}\left(\mathbb{L}^{\prime}-Q\right)e^{i(z-y)Q}\
\mathrm{d}y$ $\displaystyle=i\int_{0}^{z}e^{iy\mathbb{L}^{\prime}}e^{i(z-y)Q}\
\mathrm{d}y\left(\mathbb{L}^{\prime}-Q\right)+i\int_{0}^{z}e^{iy\mathbb{L}^{\prime}}\left[\mathbb{L}^{\prime},e^{i(z-y)Q}\right]\
\mathrm{d}y.$

Let us define the operators
$B_{z}:=i\int_{0}^{z}e^{iy\mathbb{L}^{\prime}}e^{i(z-y)Q}\ \mathrm{d}y$ and
$R_{z}:=i\int_{0}^{z}e^{iy\mathbb{L}^{\prime}}\left[\mathbb{L}^{\prime},e^{i(z\\!-\\!y)Q}\right]\mathrm{d}y$.
Clearly, $\|B_{z}\|\leq|z|$. Regarding $R_{z}$, note that every term in the
definition of $\mathbb{L}^{\prime}$ commutes with $Q$, except
$\left(p^{\prime}\right)\cdot p^{\prime}$, which satisfies the relation
$\left[p_{j}^{\prime},\phi(q)\right]=\left[p_{j},\phi(q)\right]=\frac{1}{i2N}\left(\partial_{j}\phi\right)(q)$.
We define the family of functions

$\displaystyle\phi_{x}(t):=e^{ix\left(\sum_{j=1}^{d}t_{j}^{2}+f(t)^{\dagger}\cdot
f(t)\right)}$ (D.4)

and compute

$\displaystyle\left[\mathbb{L}^{\prime},e^{ixQ}\right]$
$\displaystyle=\left[\left(p^{\prime}\right)\cdot
p^{\prime},\phi_{x}(q)\right]=\sum_{j=1}^{d}\left[\left(p_{j}^{\prime}\right)^{2},\phi_{x}(q)\right]$
$\displaystyle=\frac{1}{iN}\sum_{j=1}^{d}\partial_{j}\phi_{x}(q)p_{j}^{\prime}-\frac{1}{4N^{2}}\sum_{j=1}^{d}\partial_{j}^{2}\phi_{x}(q).$

We have the estimates $\left|\partial_{j}\phi_{x}(t)\right|\leq c|x|\ |t|$ and
$\left|\partial^{2}_{j}\phi_{x}(t)\right|\leq
c\left(1+|x|^{2}\right)\left(1+|t|^{2}\right)$ for some $c>0$, where we use
the fact that $t\mapsto f(t)$ is a
$C^{2}\left(\mathbb{R}^{d},\mathcal{H}_{0}\right)$ function, see Definition
4.7. As before, let $\pi_{M,N}$ be the orthogonal projection onto
$\mathcal{W}_{N}\left(\mathcal{F}_{\leq M}\right)$. By Lemma C.4

$\displaystyle\|\partial_{j}\phi_{x}(q)\pi_{M,N}\|$ $\displaystyle\leq c|x|\
\|\ |q|\,\pi_{M,N}\|\leq\tilde{c}|x|,$
$\displaystyle\|\partial_{j}^{2}\phi_{x}(q)\pi_{M,N}\|$ $\displaystyle\leq
c\left(1+|x|^{2}\right)\|\left(1+|q|^{2}\right)\pi_{M,N}\|\leq\tilde{c}\left(1+|x|^{2}\right),$

for some constant $\tilde{c}$ and all $x\in\mathbb{R}$ and all $M\leq N$. Note
that $p_{j}^{\prime}\mathcal{W}_{N}\mathcal{F}_{\leq
M}\subset\mathcal{W}_{N}\mathcal{F}_{\leq M+1}$ and
$\|p_{j}^{\prime}\,\pi_{M,N}\|\leq\sqrt{\frac{M+1}{N}}$, and consequently we
have for all $M\leq N-1$

$\displaystyle\|\\!\left[\mathbb{L}^{\prime},e^{ixQ}\right]\pi_{M,N}\|$
$\displaystyle\leq\frac{2}{N}\sum_{j=1}^{d}\\!\|\partial_{j}\phi_{x}(q)\,\pi_{M\\!+\\!1,N}\|\
\|p_{j}^{\prime}\,\pi_{M,N}\|\\!+\\!\frac{1}{N^{2}}\sum_{j=1}^{d}\\!\|\partial_{j}^{2}\phi_{x}(q)\,\pi_{M,N}\|$
$\displaystyle\leq\frac{d\tilde{c}}{N}|x|+\frac{d\tilde{c}}{4N^{2}}\left(1+|x|^{2}\right)\leq\frac{2d\tilde{c}}{N}\left(1+|x|^{2}\right).$

Therefore, $\|R_{z}\pi_{M,N}\|\leq\frac{C}{N}\left(1+|z|^{3}\right)$ for some
constant $C$.

Let us define $B:=\int\hat{h}(z)B_{z}\ \mathrm{d}z$ and
$R:=\int\hat{h}(z)R_{z}\ \mathrm{d}z$. From our estimates on $B_{z},R_{z}$, we
deduce $\|B\|\leq\int|\hat{h}(z)|\ |z|\ \mathrm{d}z:=C_{1}<\infty$ and
$\|R\,\pi_{M,N}\|\leq\frac{C}{N}\int|\hat{h}(z)|\left(1+|z|^{3}\right)\
\mathrm{d}z:=\frac{C_{2}}{N}<\infty$. Hence,
$R^{\dagger}R=o_{*}\left(\mathbb{Q}_{N}\right)$. Since
$h\left(\mathbb{L}^{\prime}\right)-h\left(Q\right)=B\,\left(\mathbb{L}^{\prime}-Q\right)+R$,
we obtain the estimate

$\displaystyle\left[h\left(\mathbb{L}^{\prime}\right)-h\left(Q\right)\right]^{2}=[B\left(\mathbb{L}^{\prime}-Q\right)+R]^{\dagger}[B\left(\mathbb{L}^{\prime}-Q\right)+R]$
$\displaystyle\ \ \ \leq
2\left(\mathbb{L}^{\prime}-Q\right)B^{\dagger}B\left(\mathbb{L}^{\prime}-Q\right)+2R^{\dagger}R$
$\displaystyle\ \ \ \leq
2(C_{1})^{2}\left(\mathbb{L}^{\prime}-Q\right)^{2}+2R^{\dagger}R=o_{*}\left(\mathbb{Q}_{N}\right),$

where we have used that $\left(\mathbb{L}^{\prime}-Q\right)^{2}$ is of order
$o_{*}\left(\mathbb{Q}_{N}\right)$, see Lemma D.1. This proves Eq. (D.2).

In order to verify Eq. (D.3) let us compute

$\displaystyle\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)-D_{\mathcal{V}}\big{|}_{q}\eta_{1}\big{(}b_{\geq
1}\big{)}=\\!h\left(\mathbb{L}^{\prime}\right)\\!-\\!h\left(Q\right)-h^{\prime}\left(Q\right)\left(b_{>d}^{\dagger}\cdot
f(q)\\!+\\!f(q)^{\dagger}\cdot b_{>d}\right)$ $\displaystyle\ \
=\int\hat{h}(z)\left[\\!i\\!\int_{0}^{z}e^{iy\mathbb{L}^{\prime}}\left(\mathbb{L}^{\prime}\\!-\\!Q\right)e^{i(z-y)Q}\
\mathrm{d}y-iz\ e^{izQ}\left(f(q)^{\dagger}\cdot
b_{>d}\\!+\\!b_{>d}^{\dagger}\cdot f(q)\right)\right]\ \mathrm{d}z$
$\displaystyle\ \
=R\\!+\\!\int\hat{h}(z)\left[\\!i\\!\int_{0}^{z}e^{iy\mathbb{L}^{\prime}}e^{i(z-y)Q}\mathrm{d}y\left(\mathbb{L}^{\prime}\\!-\\!Q\right)\\!-\\!iz\
e^{izQ}\left(f(q)^{\dagger}\cdot b_{>d}\\!+\\!b_{>d}^{\dagger}\cdot
f(q)\right)\right]\mathrm{d}z$ $\displaystyle\ \
=R+i\int\hat{h}(z)\int_{0}^{z}\left(e^{iy\mathbb{L}^{\prime}}-e^{iyQ}\right)e^{i(z-y)Q}\
\mathrm{d}y\ \mathrm{d}z\left(\mathbb{L}^{\prime}-Q\right)$ (D.5)
$\displaystyle\ \ \ \ \ +i\int\hat{h}(z)z\ e^{izQ}\
\mathrm{d}z\left(\mathbb{L}^{\prime}-Q-f(q)^{\dagger}\cdot
b_{>d}-b_{>d}^{\dagger}\cdot f(q)\right).$

Let $V$ be a function that satisfies the assumptions of the Lemma. To complete
the proof, we need to verify that
$V(q)\left[h\left(\mathbb{L}^{\prime}\right)-h\left(Q\right)-h^{\prime}\left(Q\right)\left(f(q)^{\dagger}\cdot
b_{>d}+b_{>d}^{\dagger}\cdot f(q)\right)\right]$ is of order
$o_{*}\left(\mathbb{Q}_{N}\right)$. By Lemma C.4, we know that
$|V|^{2}(q)=o_{*}(1)$ and from the previous part it is clear that
$\pi_{M,N}\,R^{\dagger}R\,\pi_{M,N}=O_{*}(\frac{1}{N^{2}})$. Hence,
$V(q)R=o_{*}(\frac{1}{N})$ and especially $V(q)R=o_{*}(\mathbb{Q}_{N})$.
Regarding the second term in Eq. (D.5), recall that
$e^{iy\mathbb{L}^{\prime}}-e^{iyQ}=\left(\mathbb{L}^{\prime}-Q\right)\,B_{-y}^{\dagger}+R_{-y}^{\dagger}$.
Therefore,

$\displaystyle\int\hat{h}(z)i\int_{0}^{z}\left(e^{iy\mathbb{L}^{\prime}}-e^{iyQ}\right)e^{i(z-y)Q}\
\mathrm{d}y\ \mathrm{d}z\left(\mathbb{L}^{\prime}-Q\right)$ $\displaystyle\ \
\
=\int\hat{h}(z)i\int_{0}^{z}\left[\left(\mathbb{L}^{\prime}-Q\right)B_{-y}^{\dagger}+R^{\dagger}_{-y}\right]e^{i(z-y)Q}\
\mathrm{d}y\ \mathrm{d}z\left(\mathbb{L}^{\prime}-Q\right)$ $\displaystyle\ \
\
=\left[\left(\mathbb{L}^{\prime}-Q\right)\widetilde{B}^{\dagger}+\widetilde{R}^{\dagger}\right]\left(\mathbb{L}^{\prime}-Q\right),$

with $\widetilde{B}:=-i\int\hat{h}(z)\int_{0}^{z}e^{i(y-z)Q}B_{-y}\
\mathrm{d}y\ \mathrm{d}z$ and
$\widetilde{R}:=-i\int\hat{h}(z)\int_{0}^{z}e^{i(y-z)Q}R_{-y}\ \mathrm{d}y\
\mathrm{d}z$. In the following we want to verify that
$V(q)\left[\left(\mathbb{L}^{\prime}-Q\right)\widetilde{B}^{\dagger}+\widetilde{R}^{\dagger}\right]\left(\mathbb{L}^{\prime}-Q\right)=o_{*}\left(\mathbb{Q}_{N}\right)$.
Since
$\left(\mathbb{L}^{\prime}-Q\right)^{2}=o_{*}\left(\mathbb{Q}_{N}\right)$ by
Lemma D.1, it is enough to verify that
$V(q)\widetilde{R}^{\dagger}\widetilde{R}V(q)$ and
$V(q)\left(\mathbb{L}^{\prime}-Q\right)\widetilde{B}^{\dagger}\widetilde{B}\left(\mathbb{L}^{\prime}-Q\right)V(q)$
are of order $o_{*}\left(\mathbb{Q}_{N}\right)$. Recall that we have the
identity
$R_{y}=i\int_{0}^{y}e^{ix\mathbb{L}^{\prime}}\left[\mathbb{L}^{\prime}\\!-\\!Q,e^{i(y\\!-\\!x)Q}\right]\mathrm{d}x=i\int_{0}^{y}e^{ix\mathbb{L}^{\prime}}\left[(p^{\prime})^{\dagger}\cdot
p^{\prime},\phi_{x}(q)\right]\mathrm{d}x$ with the function $\phi_{x}$ from
Eq. (D.4). We can further express $[(p^{\prime})^{\dagger}\cdot
p^{\prime},\phi_{x}(q)]V(q)$ as

$\displaystyle\sum_{j=1}^{d}\left(\frac{1}{iN}\partial_{j}\phi_{x}(q)V(q)p_{j}^{\prime}-\frac{1}{2N^{2}}\partial_{j}\phi_{x}(q)\partial_{j}V(q)-\frac{1}{4N^{2}}\partial_{j}^{2}\phi_{x}(q)V(q)\right).$

Similar to before, this leads to the estimate
$\|R_{z}V(q)\pi_{M,N}\|\leq\frac{\widetilde{C}}{N}\left(1+|z|^{3}\right)$ for
some constant $\widetilde{C}$, and consequently
$\|\widetilde{R}V(q)\pi_{M,N}\|\leq\frac{\widetilde{C}_{1}}{N}$ for some
constant $\widetilde{C}_{1}$. Hence we have
$V(q)\widetilde{R}^{\dagger}\widetilde{R}V(q)=o_{*}\left(\mathbb{Q}_{N}\right)$.
Regarding the term
$V(q)\left(\mathbb{L}^{\prime}-Q\right)\widetilde{B}^{\dagger}\widetilde{B}\left(\mathbb{L}^{\prime}-Q\right)V(q)$,
note that $\|\widetilde{B}\|^{2}=:\tilde{C}_{2}<\infty$. Applying the
Cauchy–Schwarz yields

$\displaystyle
V(q)\left(\mathbb{L}^{\prime}-Q\right)\widetilde{B}^{\dagger}\widetilde{B}\left(\mathbb{L}^{\prime}-Q\right)V(q)\leq\tilde{C}_{2}\
V(q)\left(\mathbb{L}^{\prime}-Q\right)^{2}V(q)$
$\displaystyle\\!\leq\\!5\tilde{C}_{2}V(q)\\!\left[\\!f(q)^{\dagger}\\!\cdot\\!b_{>d}\,b_{>d}^{\dagger}\\!\cdot\\!f(q)\\!+\\!b_{>d}^{\dagger}\\!\cdot\\!f(q)f(q)^{\dagger}\\!\cdot\\!b_{>d}\\!+\\!\left(\\!(p^{\prime})^{\dagger}\\!\cdot\\!p^{\prime}\\!\right)^{2}\\!+\\!\left(\\!b_{>d}^{\dagger}\\!\cdot\\!b_{>d}\\!\right)^{2}\\!+\\!\frac{d^{2}}{N^{2}}\\!\right]\\!V(q).$

Let us define the function $w(t):=V(t)f(t)$. By Lemma C.5 we obtain that

$\displaystyle V(q)\,f(q)^{\dagger}\cdot b_{>d}\ b_{>d}^{\dagger}\cdot
f(q)\,V(q)=w(q)^{\dagger}\cdot b_{>d}\ b_{>d}^{\dagger}\cdot w(q)$
$\displaystyle=o_{*}\left(\mathbb{Q}_{N}\right),$ $\displaystyle
V(q)\,b_{>d}^{\dagger}\cdot f(q)\ f(q)^{\dagger}\cdot
b_{>d}\,V(q)=b_{>d}^{\dagger}\cdot w(q)\,w(q)^{\dagger}\cdot b_{>d}$
$\displaystyle=o_{*}\left(\mathbb{Q}_{N}\right),$

and $V(q)\left(b_{>d}^{\dagger}\cdot
b_{>d}\right)^{2}V(q)=o_{*}\left(\mathbb{Q}_{N}\right)$. Furthermore,
$V(q)\left[\left(p^{\prime}\right)^{\dagger}\cdot
p^{\prime}\right]^{2}V(q)=o_{*}\left(\mathbb{Q}_{N}\right)$ by Lemma C.6. We
conclude that
$V(q)\left(\mathbb{L}^{\prime}-Q\right)\widetilde{B}^{\dagger}\widetilde{B}\left(\mathbb{L}^{\prime}-Q\right)V(q)=o_{*}\left(\mathbb{Q}_{N}\right)$.

Let us now verify that the final term
$\widetilde{V}(q)\left(\mathbb{L}^{\prime}-Q-b_{>d}^{\dagger}\cdot
f(q)-f(q)^{\dagger}\cdot b_{>d}\right)$ in Eq. (D.5) is of order
$o_{*}\left(\mathbb{Q}_{N}\right)$, where
$\widetilde{V}(t):=V(t)\int\hat{h}(z)iz\
e^{iz\left(\sum_{j=1}^{d}t_{j}^{2}+f(t)^{\dagger}\cdot f(t)\right)}\
\mathrm{d}z$. By the definition of $\mathbb{L}^{\prime}$ and $Q$, we have the
identity

$\displaystyle\widetilde{V}(q)\left(\mathbb{L}^{\prime}-Q-f(q)^{\dagger}\cdot
b_{>d}-b_{>d}^{\dagger}\cdot f(q)\right)=\widetilde{V}(q)b_{>d}^{\dagger}\cdot
b_{>d}+\widetilde{V}(q)\left(p^{\prime}\right)^{\dagger}\cdot
p^{\prime}-\frac{d}{2N}V(q).$

The first term is of order $o_{*}\left(\mathbb{Q}_{N}\right)$ by Lemma C.4,
the second term is by Lemma C.6 and regarding the last term we know that
$\frac{d}{2N}V(q)=o_{*}\left(\mathbb{Q}_{N}\right)$ by Lemma C.4. ∎

Before we can verify the Taylor approximation for the operator
$\left(1-\mathbb{L}\right)^{\frac{3}{2}}$ in Corollary D.5, we need the
following two results, which are of independent relevance for the proof of
Theorem 5.16.

######  Lemma D.3.

We have $\left(\mathbb{L}^{\prime}\right)^{2}=o_{*}\left(1\right)$, and
furthermore

$\displaystyle\sqrt{1-\mathbb{L}^{\prime}}\,$
$\displaystyle\mathbb{Q}_{N}\,\sqrt{1-\mathbb{L}^{\prime}}=O_{*}\left(\mathbb{Q}_{N}\right),$
(D.6) $\displaystyle\mathbb{L}^{\prime}\,$
$\displaystyle\mathbb{Q}_{N}\,\mathbb{L}^{\prime}=o_{*}\left(\mathbb{Q}_{N}\right).$
(D.7)

###### Proof.

Note that $\|\mathbb{L}^{\prime}\pi_{M,N}\|=\frac{M}{N}$ for all $M\leq N$,
and therefore we immediately obtain
$\left(\mathbb{L}^{\prime}\right)^{2}=o_{*}\left(1\right)$. In order to verify
Equations (D.6) and (D.7), it is enough to prove that
$\sqrt{1-\mathbb{L}^{\prime}}\left(\xi^{\dagger}\cdot\xi\right)\sqrt{1-\mathbb{L}^{\prime}}=O_{*}\left(\mathbb{Q}_{N}\right)$
and
$\mathbb{L}^{\prime}\left(\xi^{\dagger}\cdot\xi\right)\mathbb{L}^{\prime}=o_{*}\left(\mathbb{Q}_{N}\right)$
for $\xi\in\\{p^{\prime},b_{>d}\\}$.

The case $\xi=p^{\prime}$: In order to verify
$\sqrt{1-\mathbb{L}^{\prime}}\left(\xi^{\dagger}\cdot\xi\right)\sqrt{1-\mathbb{L}^{\prime}}=O_{*}\left(\mathbb{Q}_{N}\right)$,
observe that we have for all $\Psi\in\mathcal{W}_{N}\mathcal{F}_{\leq N-1}$
the commutation law

$\displaystyle
p_{j}^{\prime}\sqrt{1-\mathbb{L}^{\prime}}\Psi=\frac{\sqrt{1-\mathbb{L}^{\prime}-\frac{1}{N}}+\sqrt{1-\mathbb{L}^{\prime}+\frac{1}{N}}}{2}p_{j}^{\prime}\Psi+\frac{\sqrt{1-\mathbb{L}^{\prime}-\frac{1}{N}}-\sqrt{1-\mathbb{L}^{\prime}+\frac{1}{N}}}{2}q_{j}\Psi.$

For $M\leq N-2$, let us define the operators
$B_{M,N}:=\frac{\sqrt{1-\mathbb{L}^{\prime}-\frac{1}{N}}+\sqrt{1-\mathbb{L}^{\prime}+\frac{1}{N}}}{2}\pi_{M+1,N}$
and
$\widetilde{B}_{M,N}:=\frac{\sqrt{1-\mathbb{L}^{\prime}-\frac{1}{N}}-\sqrt{1-\mathbb{L}^{\prime}+\frac{1}{N}}}{2}\pi_{M+1,N}$.
Note that $\|B_{M,N}\|\leq 1$ and
$\|\widetilde{B}_{M,N}\|^{2}\leq\frac{C}{N^{2}}$ for all
$\frac{M}{N}\leq\delta_{0}$, where $C$ and $0<\delta<1$ are suitable
constants. Consequently

$\displaystyle\pi_{M,N}\sqrt{1-\mathbb{L}^{\prime}}\,\left(p^{\prime}\right)^{\dagger}\\!\cdot\\!p^{\prime}\,\sqrt{1-\mathbb{L}^{\prime}}\pi_{M,N}=\left|\left(B_{M,N}\\!\otimes\\!1_{\mathcal{H}}\\!\cdot\\!p^{\prime}\\!+\\!\widetilde{B}_{M,N}\\!\otimes\\!1_{\mathcal{H}}\\!\cdot\\!q\right)\pi_{M,N}\right|^{2}$
$\displaystyle\ \ \leq 2\left|B_{M,N}\otimes 1_{\mathcal{H}}\cdot
p^{\prime}\,\pi_{M,N}\right|^{2}+2\left|\widetilde{B}_{M,N}\otimes
1_{\mathcal{H}}\cdot q\,\pi_{M,N}\right|^{2}$ $\displaystyle\ \
\leq\pi_{M,N}\,\left(p^{\prime}\right)^{\dagger}\cdot
p^{\prime}\,\pi_{M,N}+\frac{C(d+1)}{N^{2}},$

which concludes the proof of
$\sqrt{1-\mathbb{L}^{\prime}}\left(p^{\prime}\right)^{\dagger}\cdot
p^{\prime}\,\sqrt{1-\mathbb{L}^{\prime}}=O_{*}\left(\mathbb{Q}_{N}\right)$.
The estimate $\mathbb{L}^{\prime}\left(p^{\prime}\right)^{\dagger}\cdot
p^{\prime}\,\mathbb{L}^{\prime}=o_{*}\left(\mathbb{Q}_{N}\right)$ follows from
an analogue commutation law.

The case $\xi=b_{>d}$: In order to verify
$\sqrt{1-\mathbb{L}^{\prime}}\,b_{>d}^{\dagger}\cdot
b_{>d}\,\sqrt{1-\mathbb{L}^{\prime}}=O_{*}\left(\mathbb{Q}_{N}\right)$, note
that
$\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right]^{2}=o_{*}\left(\mathbb{Q}_{N}\right)$
by D.2, i.e. there exists a function $\epsilon$ with
$\epsilon(\delta)\underset{\delta\rightarrow 0}{\longrightarrow}0$ and
$\pi_{M,N}\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right]^{2}\pi_{M,N}\leq\epsilon\left(\frac{M}{N}\right)\
\pi_{M,N}\,\mathbb{Q}_{N}\,\pi_{M,N}$. By Lemma C.3, we know that
$\hat{\pi}_{M,N}\,b_{>d}^{\dagger}\cdot b_{>d}\,\hat{\pi}_{M,N}\leq C$ for a
constant $C$. Furthermore
$\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right]\hat{\pi}_{M,N}=\hat{\pi}_{M,N}\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right]$.
Let us define
$S:=\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right]b_{>d}^{\dagger}\cdot
b_{>d}\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right]$, and estimate

$\displaystyle\pi_{M,N}S\pi_{M,N}$
$\displaystyle=\pi_{M,N}\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right]\hat{\pi}_{M,N}\,b_{>d}^{\dagger}\\!\cdot\\!b_{>d}\,\hat{\pi}_{M,N}\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right]\pi_{M,N}$
$\displaystyle\leq 4\
\pi_{M,N}\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right]^{2}\pi_{M,N}\leq
4\ \epsilon\left(\frac{M}{N}\right)\ \pi_{M,N}\,\mathbb{Q}_{N}\,\pi_{M,N}.$

Hence, $S=o_{*}\left(\mathbb{Q}_{N}\right)$ and therefore

$\displaystyle\sqrt{1-\mathbb{L}^{\prime}}\,b_{>d}^{\dagger}\cdot
b_{>d}\,\sqrt{1-\mathbb{L}^{\prime}}\leq
2\left(S+\eta_{1}(q)\,b_{>d}^{\dagger}\cdot
b_{>d}\,\eta_{1}(q)\right)=O_{*}\left(\mathbb{Q}_{N}\right).$

The proof of $\mathbb{L}^{\prime}\,b_{>d}^{\dagger}\cdot
b_{>d}\,\mathbb{L}^{\prime}=o_{*}\left(\mathbb{Q}_{N}\right)$ can be carried
out in a similar fashion. ∎

######  Corollary D.4.

Let $X_{N}$ be a sequence with $X_{N}=O_{*}\left(\mathbb{Q}_{N}\right)$ and
$Y_{N}$ a sequence with $Y_{N}=o_{*}\left(\mathbb{Q}_{N}\right)$. Then,

$\displaystyle\sqrt{1-\mathbb{L}^{\prime}}$ $\displaystyle
X_{N}\sqrt{1-\mathbb{L}^{\prime}}=O_{*}\left(\mathbb{Q}_{N}\right),$ (D.8)
$\displaystyle\sqrt{1-\mathbb{L}^{\prime}}$ $\displaystyle
Y_{N}\sqrt{1-\mathbb{L}^{\prime}}=o_{*}\left(\mathbb{Q}_{N}\right),$ (D.9)
$\displaystyle\mathbb{L}^{\prime}$ $\displaystyle
X_{N}\mathbb{L}^{\prime}=o_{*}\left(\mathbb{Q}_{N}\right).$ (D.10)

###### Proof.

The Corollary follows from Lemma D.3 and the fact that $\pi_{M,N}$ commutes
with $\sqrt{1-\mathbb{L}^{\prime}}$ and $\mathbb{L}^{\prime}$. For the purpose
of illustration, let us verify Eq. (D.8). By the assumptions of the Corollary
we know that there exist constants $C$ and $\delta>0$, such that
$\pi_{M,N}\,X_{N}\,\pi_{M,N}\leq C\pi_{M,N}\,\mathbb{Q}_{N}\,\pi_{M,N}$.
Consequently

$\displaystyle\pi_{M,N}$
$\displaystyle\,\sqrt{1-\mathbb{L}^{\prime}}X_{N}\sqrt{1-\mathbb{L}^{\prime}}\,\pi_{M,N}=\sqrt{1-\mathbb{L}^{\prime}}\pi_{M,N}\,X_{N}\,\pi_{M,N}\sqrt{1-\mathbb{L}^{\prime}}$
$\displaystyle\leq
C\pi_{M,N}\,\sqrt{1-\mathbb{L}^{\prime}}\mathbb{Q}_{N}\sqrt{1-\mathbb{L}^{\prime}}\,\pi_{M,N}=O_{*}\left(\mathbb{Q}_{N}\right),$

where we have used Eq. (D.6) from Lemma D.3 in the last equality. ∎

######  Corollary D.5.

Let Assumption 1.3 hold and let $\eta_{m}$ be the functions from Eq. (5.7),
with $m\in\\{0,\dots,3\\}$. Then

$\displaystyle\left[\left(1-\mathbb{L}^{\prime}\right)^{\frac{m}{2}}-\eta_{m}(q)\right]^{2}=o_{*}\left(\mathbb{Q}_{N}\right).$

###### Proof.

The case $m=0$ is trivial. The case $m=1$ is the content of Lemma D.2 and the
case $m=2$ follows from Lemma D.1. Let us now verify the statement in the case
$m=3$. Using the fact that $\eta_{3}(t)=\eta_{2}(t)\eta_{1}(t)$, we obtain

$\displaystyle\left(1-\mathbb{L}^{\prime}\right)\sqrt{1-\mathbb{L}^{\prime}}-\eta_{3}(q)=\left[\left(1-\mathbb{L}^{\prime}\right)-\eta_{2}(q)\right]\sqrt{1-\mathbb{L}^{\prime}}+\eta_{2}(q)\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right]$
$\displaystyle=-\left(f(q)^{\dagger}\cdot b_{>d}+b_{>d}^{\dagger}\cdot
f(q)+(p^{\prime})^{\dagger}\cdot p^{\prime}+b_{>d}^{\dagger}\cdot
b_{>d}-\frac{d}{2N}\right)\sqrt{1-\mathbb{L}^{\prime}}$ $\displaystyle\ \ \ \
+\tau_{r}(q)\eta_{2}(q)\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right]+\left[1-\tau_{r}(q)\right]\eta_{2}(q)\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right],$

where $\tau:\mathbb{R}^{d}\longrightarrow\mathbb{R}$ is a function with
$\tau|_{B_{1}(0)}=0$, $\tau|_{\mathbb{R}^{d}\setminus B_{2}(0)}=1$ and
$0\leq\tau\leq 1$. Since the function $\tilde{\eta}=(1-\tau)\eta_{2}$ is
bounded by a constant $c$, we obtain using Lemma D.2

$\displaystyle\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right]\tilde{\eta}^{2}(q)\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right]\leq
c\
\left[\sqrt{1-\mathbb{L}^{\prime}}-\eta_{1}(q)\right]^{2}=o_{*}\left(\mathbb{Q}_{N}\right).$

Note that $\eta^{\prime}:=\left(\tau\ \eta_{2}\right)^{2}$ is zero in a
neighborhood of zero. Therefore, $\eta^{\prime}(q)=O_{*}\left(e^{-\delta
N}\right)$ and $\eta^{2}_{1}(q)\eta^{\prime}(q)=O_{*}\left(e^{-\delta
N}\right)$ for some $\delta>0$ by Lemma C.2. By Corollary D.4, we obtain in
particular that
$\sqrt{1-\mathbb{L}^{\prime}}\eta^{\prime}(q)\sqrt{1-\mathbb{L}^{\prime}}=o_{*}\left(\mathbb{Q}_{N}\right)$.
Hence we have the estimate

$\displaystyle\left[\\!\sqrt{1\\!-\\!\mathbb{L}^{\prime}}\\!-\\!\eta_{1}(q)\\!\right]\eta^{\prime}(q)\left[\sqrt{1\\!-\\!\mathbb{L}^{\prime}}\\!-\\!\eta_{1}(q)\\!\right]\\!\leq\\!2\sqrt{1\\!-\\!\mathbb{L}^{\prime}}\eta^{\prime}(q)\sqrt{1\\!-\\!\mathbb{L}^{\prime}}\\!+\\!2\eta^{2}_{1}(q)\eta^{\prime}(q)\\!=\\!o_{*}\left(\mathbb{Q}_{N}\right).$

By Lemma C.5, Lemma C.6 and Corollary D.4, we know that the operators

$\displaystyle\sqrt{1-\mathbb{L}^{\prime}}\left(b_{>d}^{\dagger}\cdot
f(q)+f(q)^{\dagger}\cdot b_{>d}\right)^{2}\sqrt{1-\mathbb{L}^{\prime}}$

$\sqrt{1-\mathbb{L}^{\prime}}\left((p^{\prime})^{\dagger}\cdot
p^{\prime}\right)^{2}\sqrt{1-\mathbb{L}^{\prime}}$ as well as
$\sqrt{1-\mathbb{L}^{\prime}}\left(b_{>d}^{\dagger}\cdot
b_{>d}\right)^{2}\sqrt{1-\mathbb{L}^{\prime}}$ are of order
$o_{*}\left(\mathbb{Q}_{N}\right)$ as well, and therefore

$\displaystyle\sqrt{1-\mathbb{L}^{\prime}}\left(b_{>d}^{\dagger}\cdot
f(q)+f(q)^{\dagger}\cdot b_{>d}+(p^{\prime})^{\dagger}\cdot
p^{\prime}+b_{>d}^{\dagger}\cdot
b_{>d}-\frac{d}{2N}\right)^{2}\sqrt{1-\mathbb{L}^{\prime}}=o_{*}\left(\mathbb{Q}_{N}\right).$

We conclude that
$\left(1-\mathbb{L}^{\prime}\right)\sqrt{1-\mathbb{L}^{\prime}}-\eta_{3}(q)=T_{1}+T_{2}+T_{3}$
is a sum of terms with
$T_{i}^{\dagger}T_{i}=o_{*}\left(\mathbb{Q}_{N}\right)$, and therefore
$\left[\left(1-\mathbb{L}^{\prime}\right)\sqrt{1-\mathbb{L}^{\prime}}-\eta_{3}(q)\right]^{2}=o_{*}\left(\mathbb{Q}_{N}\right)$.
∎

Acknowledgments

We are grateful to Rupert Frank for helpful discussions at an early stage of
this project. Funding from the European Union’s Horizon 2020 research and
innovation programme under the ERC grant agreement No 694227 is acknowledged.

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