$\displaystyle\leq\int_{0}^{|t|}ds\,|\langle\Xi,\mathbb{U}_{u,\alpha,1}^{\infty}(s)^{*}(\mathbb{H}_{u,\alpha,1}^{\infty}(s)-\mathbb{H}_{u,\alpha,1}^{\Lambda}(s))\mathbb{U}^{\Lambda}_{u,\alpha,1}(s)\Psi\rangle|$
$\displaystyle\leq\varepsilon_{\Lambda}\int_{0}^{|t|}ds\,e^{CR^{\rm
D}(s)}\|(\mathbb{T}+\mathcal{N}+1)^{\frac{1}{2}}\mathbb{U}_{u,\alpha,1}^{\infty}(s)\Xi\|\,\|(\mathbb{T}+\mathcal{N}+1)^{\frac{1}{2}}\mathbb{U}^{\Lambda}_{u,\alpha,1}(s)\Psi\|$

where $R^{\rm
D}(s)=\|u_{s}\|_{H^{3}}^{2}(1+\|\alpha_{s}\|_{\mathfrak{h}_{3/2}})^{2}$ and
$\varepsilon_{\Lambda}\to 0$ as $\Lambda\to 0$. The right hand side thus
converges to zero by Property (i). This implies strong convergence of
$\mathbb{U}_{u,\alpha,1}^{\Lambda}$ to $\mathbb{U}_{u,\alpha,1}^{\infty}$ by
unitarity of $\mathbb{U}_{u,\alpha,1}^{\Lambda}$,
$\Lambda\in\mathbb{R}_{+}\cup\\{\infty\\}$, as argued in the proof of
Proposition 4.3.

In Appendix B, we provide a proof that $\mathbb{U}_{u,\alpha,1}^{\Lambda}(t)$
is a Bogoliubov transformation. For finite $\Lambda$, this is essentially due
to the square-integrability of the kernels in (5.10) appearing in the terms
with $b^{*}b^{*}$, $b^{*}a^{*}$ and $a^{*}a^{*}$. For $\Lambda=\infty$,
however, the kernel corresponding to $a^{*}b^{*}$ fails to meet the
Hilbert–Schmidt criterion. In this case, we establish the statement by proving
a suitable approximation argument and utilizing the fact that
$\mathbb{U}_{u,\alpha,1}^{\Lambda}(t)\to\mathbb{U}_{u,\alpha,1}^{\infty}(t)$
converges strongly.

The final statement can be derived using the same logic as in the proof of
Proposition 4.3, in combination with Proposition 5.2(i). ∎

###  5.2. Dressing identity and proof of Theorem 1.4

We can now prepare for the proof of Theorem 1.4. We will start by making
precise how the unitaries $\mathbb{W}_{u,\alpha}^{\Lambda}(\theta)$
interpolate between the Bogoliubov dynamics
$\mathbb{U}_{u,\alpha,\theta}^{\Lambda}(t)$ for different $\theta$. For
$\theta=1$ we can then take the limit $\Lambda\to\infty$ of
$\mathbb{U}_{u,\alpha,1}^{\Lambda}(t)$ to obtain information on the behavior
of $\mathbb{U}^{\Lambda}_{u,\alpha,0}(t)$ as $\Lambda\to\infty$.

Let $(u,\alpha)\in H^{3}(\mathbb{R}^{3})\oplus\mathfrak{h}_{5/2}$ with
$\|u\|_{L^{2}}=1$, and consider the evolutions

$\mathbb{U}^{\Lambda}_{\mathfrak{D}[\theta](u,\alpha),\theta}(t)\qquad\text{and}\qquad\mathbb{W}_{\mathfrak{s}_{0}[t](u,\alpha)}^{\Lambda}(\theta),$
(5.13)

with generators
$\mathbb{H}^{\Lambda}_{\mathfrak{D}[\theta](u,\alpha),\theta}(t)$ and
$\mathbb{D}_{\mathfrak{s}_{0}[t](u,\alpha)}^{\Lambda}(\theta)$, respectively
(see Propositions 5.2 and 4.3). It is important to keep in mind that the
subscripts refer to the initial condition of the mean-field flow, which is
used to define the generator of each evolution.

The next result shows that the two flows in (5.13) commute, up to a global
phase. The additional phase is due to the fact that we wrote
$\mathbb{H}_{\mathfrak{D}[\theta](u,\alpha),\theta}^{\Lambda}(t)$ in normal
order, which is not preserved by the transformations. As discussed below, this
identity is the key ingredient for the renormalization of the
Nelson–Bogoliubov Hamiltonian.

######  Proposition 5.3.

Let $(u,\alpha)\in H^{3}(\mathbb{R}^{3})\oplus\mathfrak{h}_{5/2}$ with
$\|u\|_{L^{2}}=1$. If we denote

$E^{\Lambda}_{\theta}=(2\theta-\theta^{2})\langle
G^{\Lambda}_{0},B^{\Lambda}_{0}\rangle+\tfrac{1}{2}\langle
u,V_{\theta}^{\Lambda}\ast|u|^{2}u\rangle$

then for all $t,\theta\in\mathbb{R}$ and $\Lambda\in(0,\infty)$ we have the
identity

$\displaystyle\mathbb{U}^{\Lambda}_{u,\alpha,0}(t)e^{-itE^{\Lambda}_{\theta}}=\mathbb{W}_{\mathfrak{s}_{0}[t](u,\alpha)}^{\Lambda}(\theta)^{*}\,\mathbb{U}^{\Lambda}_{\mathfrak{D}[\theta](u,\alpha),\theta}(t)\,\mathbb{W}^{\Lambda}_{u,\alpha}(\theta).$

This proposition follows from the uniqueness of both sides by comparing their
derivatives. The lengthy calculation is given in Section 5.3. Assuming this
for now, we can prove Theorem 1.4 and Proposition 4.6.

###### Proofs of Theorem 1.4 and Proposition 4.6.

By strong continuity of $\mathbb{W}_{u,\alpha}^{\Lambda}(\theta)$ in $\Lambda$
(Lemma 4.3) and of $\mathbb{U}^{\Lambda}_{u,\alpha,1}(t)$ (Proposition 5.2),
we have

$\operatorname*{\mathrm{s--
lim}}_{\Lambda\to\infty}\mathbb{W}_{\mathfrak{s}_{0}[t](u,\alpha)}^{\Lambda}(1)^{*}\,\mathbb{U}^{\Lambda}_{\mathfrak{D}(u,\alpha),1}(t)\,\mathbb{W}^{\Lambda}_{u,\alpha}(0)\\\
=\mathbb{W}_{\mathfrak{s}_{0}[t](u,\alpha)}^{\infty}(1)^{*}\,\mathbb{U}^{\infty}_{\mathfrak{D}(u,\alpha),1}(t)\,\mathbb{W}^{\infty}_{u,\alpha}(0)$
(5.14)

By Proposition 5.3 with $\theta=1$ and
$\mathbb{U}_{u,\alpha}^{\Lambda}(t)=\mathbb{U}_{u,\alpha,0}^{\Lambda}(t)$ this
shows that

$\mathbb{U}_{u,\alpha}(t)=\operatorname*{\mathrm{s--
lim}}_{\Lambda\to\infty}\mathbb{U}^{\Lambda}_{u,\alpha,0}(t)e^{-itE^{\Lambda}_{1}}=\mathbb{W}_{\mathfrak{s}_{0}[t](u,\alpha)}^{\infty}(1)^{*}\,\mathbb{U}^{\infty}_{\mathfrak{D}(u,\alpha),1}(t)\,\mathbb{W}^{\infty}_{u,\alpha}(0).$
(5.15)

Strong continuity of $t\mapsto\mathbb{U}_{u,\alpha}(t)$ follows since the
right hand side is strongly continuous, as for
$\Psi\in\mathcal{F}\otimes\mathcal{F}$ the maps

$t\mapsto\mathbb{W}_{\mathfrak{s}_{0}[t](u,\alpha)}^{\infty}(1)^{*}\Psi,\,\qquad
t\mapsto\mathbb{U}^{\infty}_{\mathfrak{D}(u,\alpha),1}(t)\Psi\,\qquad
t\mapsto\mathbb{W}^{\infty}_{u,\alpha}(0)\Psi.$ (5.16)

are continuous.

To show the mapping property for $\mathbb{U}_{u,\alpha}(t)$, consider
$\Gamma(q_{u_{t}})$ as in the proof of Proposition 4.3 for
$(u_{t},\alpha_{t})=\mathfrak{s}[t](u,\alpha)$. By the same argument as in
(4.13) one shows that
$\|\Gamma(q_{u_{t}})\mathbb{U}^{\Lambda}_{u,\alpha,0}(t)\Psi\|=0$ for every
$\Psi\in\mathcal{F}\otimes\mathcal{F}$ and all $\Lambda\in\mathbb{R}_{+}$,
$t\in\mathbb{R}$ . The desired result then follows from
$\|\Gamma(q_{u_{t}})\mathbb{U}_{u,\alpha,0}^{\Lambda}(t)\Psi\|\rightarrow\|\Gamma(q_{u_{t}})\mathbb{U}_{u,\alpha}(t)\Psi\|^{2}$
as $\Lambda\to\infty$.

The property that $\mathbb{U}_{u,\alpha}(t)$ is a Bogoliubov transformation is
a direct consequence of the result that it is a composition of three
Bogoliubov transformations. ∎

######  Remark 5.2 (On the renormalized Nelson–Bogoliubov Hamiltonian).

If the evolution $\mathbb{U}_{u,\alpha}(t)$ was a semi-group, we could deduce
from Theorem 1.4 the existence of a generator $\mathbb{H}_{u,\alpha}$ that
would represent the renormalization of the Nelson–Bogoliubov Hamiltonian
(1.22). But $\mathbb{U}_{u,\alpha}(t)$ is associated with a non-autonomous
evolution equation and $\mathbb{H}_{u,\alpha}$ should depend on the time $t$.
In this setting, only the following, weaker theory is available (see [81] for
details). Consider the extension of $\mathbb{U}_{u,\alpha}(t)$ to a two-
parameter family $\mathbb{U}_{u,\alpha}(t,s)$ with
$\mathbb{U}_{u,\alpha}(t,0)=\mathbb{U}_{u,\alpha}(t)$. On the Banach space
$C_{\infty}(\mathbb{R},\mathcal{F}\otimes\mathcal{F})$ of continuous
$\mathcal{F}\otimes\mathcal{F}$-valued functions tending to zero at infinity,
we can define the corresponding evolution semi-group of isometries by

$(T(t)\chi)(s)=\mathbb{U}_{u,\alpha}(s,s-t)\chi(s-t).$ (5.17)

Then $T$ has a generator $A$, $D(A)\subset
C_{\infty}(\mathbb{R},\mathcal{F}\otimes\mathcal{F})$, which corresponds
formally to

$(A\chi)(s)=\tfrac{d}{dt}T(t)\big{|}_{t=0}\chi(s)=\big{(}-i\mathbb{H}_{u,\alpha}(s)-\tfrac{d}{ds}\big{)}\chi(s).$
(5.18)

However, we do not have any information concerning the domain or self-
adjointness of $\mathbb{H}_{u,\alpha}(t)$ (one could consider using [81, Thm.
2.9], but this does not apply since the solutions provided by Proposition 5.2
need not be differentiable in the norm of $\mathcal{F}\otimes\mathcal{F}$).

###  5.3. Proof of the dressing identity

In this section, we derive the dressing identity for the Nelson–Bogoliubov
evolution, that is, we prove Proposition 5.3. To enhance the clarity of the
presentation, we will use the shorthand notation

$\displaystyle\mathbb{U}^{\Lambda}_{\theta}(t)$
$\displaystyle:=\mathbb{U}^{\Lambda}_{\mathfrak{D}[\theta](u,\alpha),\theta}(t)\quad\qquad$
$\displaystyle\mathbb{W}_{t}^{\Lambda}(\theta):=\mathbb{W}^{\Lambda}_{\mathfrak{s}_{0}[t](u,\alpha)}(\theta)$
(5.19) $\displaystyle\mathbb{H}^{\Lambda}_{\theta}(t)$
$\displaystyle:=\mathbb{H}^{\Lambda}_{\mathfrak{D}[\theta](u,\alpha),\theta}(t)\qquad$
$\displaystyle\
\mathbb{D}^{\Lambda}_{t}(\theta):=\mathbb{D}^{\Lambda}_{\mathfrak{s}_{0}[t](u,\alpha)}(\theta)\qquad$
.

###### Proof of Proposition 5.3.

Adopting (5.19), the identity we aim to prove becomes
$\mathbb{U}^{\Lambda}_{0}(t)e^{-itE^{\Lambda}_{1}}=\mathbb{W}_{t}^{\Lambda}(\theta)^{*}\,\mathbb{U}^{\Lambda}_{\theta}(t)\,\mathbb{W}^{\Lambda}_{0}(\theta)$.
The idea is to prove a differential version of this identity after taking
derivatives in $t$ and $\theta$. To make this precise, first note that the
coefficients of $\mathbb{D}_{t}^{\Lambda}(\theta)$ given in (4.10) depend on
$u,\alpha$ in a differentiable way. Since for $(u,\alpha)\in
H^{2}(\mathbb{R}^{3})\oplus\mathfrak{h}_{1}$ the flow
$\mathfrak{s}_{\theta}[t](u,\alpha)$ is differentiable in $t$ in the
$L^{2}$-sense, this implies that for any $\Psi\in D(\mathcal{N})$ the mapping
$t\mapsto\mathbb{D}^{\Lambda}_{t}(\theta)\Psi$ is differentiable in
$\mathcal{F}\otimes\mathcal{F}$. Since $\mathbb{W}_{t}^{\Lambda}(\theta)$
preserves $D(\mathcal{N})$, we deduce from Duhamel’s formula (4.2) that
$\mathbb{W}_{t}^{\Lambda}(\theta)\Psi$ is differentiable in $t$ for $\Psi\in
D(\mathcal{N})$ and

$\big{(}i\partial_{t}\mathbb{W}_{t}^{\Lambda}(\theta)^{*}\big{)}\mathbb{W}_{t}^{\Lambda}(\theta)\Psi=-\int_{0}^{\theta}d\eta\,\mathbb{W}_{t}^{\Lambda}(\eta)^{*}\big{(}\partial_{t}\mathbb{D}_{t}^{\Lambda}\big{)}(\eta)\mathbb{W}_{t}^{\Lambda}(\eta)\Psi.$
(5.20)

Denote by
$\mathbb{V}_{\theta}^{\Lambda}(t):=\mathbb{W}_{t}^{\Lambda}(\theta)^{*}\mathbb{U}^{\Lambda}_{\theta}(t)\mathbb{W}^{\Lambda}_{0}(\theta)$
the right hand side of the identity we want to prove. It follows from our
previous considerations and Lemma 4.4 that for $\Psi\in
D((\mathcal{N}+\mathbb{T})^{1/2})$ we have

$\displaystyle i\partial_{t}\mathbb{V}_{\theta}^{\Lambda}(t)\Psi$
$\displaystyle=\big{(}i\partial_{t}\mathbb{W}_{t}^{\Lambda}(\theta)^{*}\big{)}\mathbb{U}^{\Lambda}_{\theta}(t)\mathbb{W}^{\Lambda}_{0}(\theta)\Psi+\mathbb{W}_{t}^{\Lambda}(\theta)^{*}\mathbb{H}_{\theta}^{\Lambda}(t)\mathbb{U}^{\Lambda}_{\theta}(t)\mathbb{W}^{\Lambda}_{0}(\theta)\Psi$
$\displaystyle=:\mathbb{B}_{\theta}(t)\mathbb{V}_{\theta}^{\Lambda}(t)\Psi$
(5.21)

in $D((\mathcal{N}+\mathbb{T})^{-1/2})$. By uniqueness of the solutions proved
in Proposition 5.2 our claim will follow if we can show that the generators of
$\mathbb{U}_{0}^{\Lambda}(t)e^{-iE_{\theta}^{\Lambda}t}$ and
$\mathbb{V}_{\theta}^{\Lambda}(t)$ are equal, that is for every
$t\in\mathbb{R}$,

$\displaystyle\mathbb{H}_{0}^{\Lambda}(t)+E_{\theta}^{\Lambda}$
$\displaystyle=\mathbb{B}_{\theta}(t)$ (5.22)
$\displaystyle=-\int_{0}^{\theta}d\eta\,\mathbb{W}_{t}^{\Lambda}(\eta)^{*}\big{(}\partial_{t}\mathbb{D}_{t}^{\Lambda}\big{)}(\eta)\mathbb{W}_{t}^{\Lambda}(\eta)+\mathbb{W}_{t}^{\Lambda}(\theta)^{*}\mathbb{H}_{\theta}^{\Lambda}(t)\mathbb{W}_{t}^{\Lambda}(\theta).$

Equality holds for $\theta=0$ since $E_{0}^{\Lambda}=0$ and
$\mathbb{W}_{t}^{\Lambda}(0)=1$, so it is sufficient to prove that for all
$\Psi,\Xi\in D((\mathcal{N}+\mathbb{T})^{1/2})$

$\displaystyle 0$
$\displaystyle=i\partial_{\theta}\Big{\langle}\Xi,\Big{(}\mathbb{H}_{0}^{\Lambda}(t)+E_{\theta}^{\Lambda}-\mathbb{B}_{\theta}(t)\Big{)}\Psi\Big{\rangle}$
(5.23)
$\displaystyle=\Big{\langle}\mathbb{W}_{t}^{\Lambda}(\theta)\Xi,\Big{(}i\partial_{\theta}E_{\theta}^{\Lambda}+i\partial_{t}\mathbb{D}_{t}^{\Lambda}(\theta)-i\partial_{\theta}\mathbb{H}_{\theta}^{\Lambda}(t)-[\mathbb{H}_{\theta}^{\Lambda}(t),\mathbb{D}_{t}^{\Lambda}(\theta)]\Big{)}\mathbb{W}_{t}^{\Lambda}(\theta)\Psi\Big{\rangle}.$

where we anticipated differentiability of $\mathbb{H}_{\theta}^{\Lambda}$
which follows easily from the explicit calculation of its derivative below.

The remainder of the proof is an explicit calculation of this quadratic form.
For ease of presentation, we set, using that the flows commute,

$(u_{t}^{\theta},\alpha_{t}^{\theta}):=\mathfrak{s}_{\theta}[t]\circ\mathfrak{D}[\theta](u,\alpha)=\mathfrak{D}[\theta]\circ\mathfrak{s}_{0}[t](u,\alpha).$
(5.24)

Moreover, we do not make the dependence of the different objects on
$t,\theta,\Lambda$ explicit everywhere and adopt the following shorthand
notation

$\displaystyle h_{\theta}$
$\displaystyle=h_{u_{t}^{\theta},\alpha_{t}^{\theta},\theta}$ (5.25a)
$\displaystyle\tau_{t}$ $\displaystyle=\tau_{u_{t},\alpha_{t}}$ (5.25b)
$\displaystyle q$ $\displaystyle=q_{u^{\theta}_{t}}=1-|u_{t}\rangle\langle
u^{\theta}_{t}|$ (5.25c) $\displaystyle\kappa^{\Lambda}_{t}(k,x)$
$\displaystyle=\big{(}qiB_{(\cdot)}^{\Lambda}(k)u_{t}^{\theta}\big{)}(x)$
(5.25d) $\displaystyle L^{\Lambda}_{\theta}(k)$
$\displaystyle=(1-\theta)G^{\Lambda}_{(\cdot)}+2B^{\Lambda}_{(\cdot)}(k)k(-i\theta\nabla+\theta^{2}F_{\alpha_{t}^{\theta}}(\cdot))$
(5.25e) $\displaystyle M^{\Lambda}(k,l)$ $\displaystyle=\langle
u^{\theta}_{t},kB^{\Lambda}_{(\cdot)}(k)lB^{\Lambda}_{(\cdot)}(l)u^{\theta}_{t}\rangle$
(5.25f) $\displaystyle K^{(1),\Lambda}_{\theta}$
$\displaystyle=K^{(1),\Lambda}_{\theta,u_{t}^{\theta}}$ (5.25g) $\displaystyle
K^{(2),\Lambda}_{\theta}$
$\displaystyle=K^{(2),\Lambda}_{\theta,u_{t}^{\theta}}$ (5.25h) $\displaystyle
V^{\Lambda}_{\theta}(x)$
$\displaystyle=-4\theta\mathrm{Re}\big{\langle}G^{\Lambda}_{x},B_{0}\big{\rangle}+2\theta^{2}\mathrm{Re}\big{\langle}B_{x}^{\Lambda},\omega
B_{0}\big{\rangle}$ (5.25i)

Note that $M^{\Lambda}(k,l)$ is independent of $\theta$ as the flow
$\mathfrak{D}$ preserves the modulus of $u$.

After commuting $\mathbb{H}_{\theta}$ and $\mathbb{D}_{t}$, the expression
from (5.23) takes the form

$\displaystyle-i\partial_{t}\mathbb{D}_{t}^{\Lambda}(\theta)+i\partial_{\theta}\mathbb{H}^{\Lambda}_{\theta}(t)+[\mathbb{H}^{\Lambda}_{\theta}(t),\mathbb{D}_{t}^{\Lambda}(\theta)]$
$\displaystyle\ =\int
dx\,b_{x}^{*}\big{(}-i\partial_{t}\tau_{t}+i\partial_{\theta}h_{\theta}+[h_{\theta},\tau_{t}]\big{)}b_{x}\,$
(5.26a) $\displaystyle\quad+\int
dxdy\,D(x,y)b_{x}^{*}b_{y}+iE_{b}+\bigg{(}\int
dxdy\,\widetilde{D}(x,y)\,b_{x}^{*}b_{y}^{*}-\text{h.c.}\bigg{)}$ (5.26b)
$\displaystyle\quad+\int
dkdx\Big{(}X(k,x)\,a_{k}^{*}b_{x}^{*}+\widetilde{X}(k,x)\,a_{k}b_{x}^{*}\Big{)}-\text{h.c.}$
(5.26c) $\displaystyle\quad+\int
dkdl\,A(k,l)a_{k}^{*}a_{l}+iE_{a}+\bigg{(}\int
dkdl\,\widetilde{A}(k,l)a_{k}^{*}a_{l}^{*}-\text{h.c.}\bigg{)},$ (5.26d)

in the sense of sesquilinear forms on $D((\mathcal{N}+\mathbb{T})^{1/2})$. We
will now show that all the coefficients of creation and annihilation operators
vanish, and that $E_{a}+E_{b}=\partial_{\theta}E^{\Lambda}_{\theta}$.

Mean-field part (5.26a): By (3.15), Lemma 5.1 and (5.24) we have
$\partial_{\theta}u^{\theta}_{t}=-i\tau_{t}u^{\theta}_{t}$ and
$\partial_{t}u_{t}^{\theta}=-ih_{\theta}u_{t}^{\theta}$, and since the
derivatives commute

$\displaystyle
0=(\partial_{t}\partial_{\theta}-\partial_{\theta}\partial_{t})u_{t}^{\theta}=\big{(}-(i\partial_{t}\tau_{t})+i(\partial_{\theta}h_{\theta})+[h_{\theta},\tau_{t}]\big{)}u_{t}^{\theta}.$
(5.27)

Since (5.24) is a bijection, this implies that (5.26a) vanishes.

Terms quadratic in $b,b^{*}$ and $E_{b}$ (5.26b): In the commutator
$[\mathbb{H}^{\Lambda}_{\theta}(t),\mathbb{D}_{t}^{\Lambda}(\theta)]$ a
quadratic term in $b,b^{*}$ can arise either by commuting two terms with
$b^{\\#}a^{\\#}$, or terms with $b^{\\#}b^{\\#}$. In the first case, the
coefficients are combinations of $L^{\Lambda}_{\theta}(k)$ and
$\kappa^{\Lambda}_{t}(k,\cdot)$ integrated over $k$. In the latter case, some
terms have already been taken into account in (5.26a) and only the commutators
of $K^{(1),\Lambda}_{\theta}$, $K^{(2),\Lambda}_{\theta}$ with $\tau_{t}$
remain. Combining these with the derivatives of $K^{(1),\Lambda}_{\theta}$,
$K_{\theta}^{(2),\Lambda}$ and putting them in normal order yields

$\displaystyle D(x,y)$
$\displaystyle=i\partial_{\theta}K^{\Lambda,(1)}_{\theta}(x,y)-\big{(}\tau_{t}(x)-\tau_{t}(y)\big{)}K^{\Lambda,(1)}_{\theta}(x,y)$
$\displaystyle\quad-\int
dk\,\big{(}q(L^{\Lambda}_{\theta}(k)+L^{\Lambda}_{\theta}(-k)^{*})u^{\theta}_{t}\big{)}(x)\overline{\kappa^{\Lambda}_{t}(k,y)}$
$\displaystyle\quad+\int
dk\,\kappa^{\Lambda}_{t}(k,x)\overline{\big{(}q(L_{\theta}^{\Lambda}(k)+L^{\Lambda}_{\theta}(-k)^{*})u^{\theta}_{t}\big{)}(y)},$
(5.28a) $\displaystyle E_{b}$ $\displaystyle=i\int
dxdk\,\overline{\kappa^{\Lambda}_{t}(k,x)}\big{(}q(L^{\Lambda}_{\theta}(k)+L^{\Lambda}_{\theta}(-k)^{*})\big{)}u^{\theta}_{t}(x),$
(5.28b) $\displaystyle\widetilde{D}(x,y)$
$\displaystyle=\frac{i}{2}\partial_{\theta}K_{\theta}^{(2),\Lambda}(x,y)-\frac{1}{2}(\tau_{t}(x)+\tau_{t}(y))K_{\theta}^{(2),\Lambda}(x,y)$
$\displaystyle\quad+\int
dk\,\big{(}q(L^{\Lambda}_{\theta}(k)^{*}+L^{\Lambda}_{\theta}(-k))u^{\theta}_{t}\big{)}(x)\kappa^{\Lambda}_{t}(k,y).$
(5.28c)

We now show that $D(x,y)=0$. First, we may observe that
$i\partial_{\theta}u^{\theta}_{t}=\tau_{t}u^{\theta}_{t}$ and
$q=1-|u^{\theta}_{t}\rangle\langle u^{\theta}_{t}|$ imply for any operator
$T_{\theta}$ the identity

$i\partial_{\theta}(qT_{\theta}u^{\theta}_{t})=\tau_{t}qT_{\theta}u^{\theta}_{t}+q[T_{\theta},\tau_{t}]u^{\theta}_{t}+q(i\partial_{\theta}T_{\theta})u^{\theta}_{t}.$
(5.29)

Since $[V^{\Lambda}_{\theta},\tau_{t}]=0$, this gives us

$\displaystyle
i\partial_{\theta}K^{(1),\Lambda}_{\theta}(x,y)-\big{(}\tau_{t}(x)-\tau_{t}(y)\big{)}K^{(1),\Lambda}_{\theta}(x,y)$
$\displaystyle\qquad=\int
dzdz^{\prime}q(x,z)u^{\theta}_{t}(z)(i\partial_{\theta}V^{\Lambda}_{\theta}(z-z^{\prime}))\bar{u}^{\theta}_{t}(z^{\prime})q(z^{\prime},y).$
(5.30)

To evaluate the terms involving
$L^{\Lambda}_{\theta}(k)+L^{\Lambda}_{\theta}(-k)^{*}$, we first calculate
using $\overline{B^{\Lambda}_{x}}(-k)=B^{\Lambda}_{x}(k)$

$\displaystyle(L^{\Lambda}_{\theta}(k)+L^{\Lambda}_{\theta}(-k)^{*})u^{\theta}_{t}(x)$
$\displaystyle=\big{(}(1-\theta)2G^{\Lambda}_{x}(k)+2\theta
k^{2}B^{\Lambda}_{x}(k)\big{)}u^{\theta}_{t}(x).$ (5.31)

This gives

$\displaystyle-\int
dk\,\big{(}q(L^{\Lambda}_{\theta}(k)+L^{\Lambda}_{\theta}(-k)^{*})u\big{)}(x)\overline{\kappa^{\Lambda}_{t}(k,y)}$
(5.32) $\displaystyle=\int
dzdz^{\prime}q(x,z)u^{\theta}_{t}(z)\Big{(}2i(1-\theta)\langle
B^{\Lambda}_{z^{\prime}},G^{\Lambda}_{z}\rangle+2i\theta\langle
k^{2}B^{\Lambda}_{z^{\prime}},B^{\Lambda}_{z}\rangle\Big{)}\bar{u}^{\theta}_{t}(z^{\prime})q(z^{\prime},y).$

Adding this and minus its complex conjugate with $x,y$ exchanged (which leads
to an exchange of $z,z^{\prime}$) gives with
$(k^{2}+\omega(k))B_{x}(k)=G_{x}(k)$

$\displaystyle\int
dzdz^{\prime}q(x,z)u^{\theta}_{t}(z)\big{(}\underbrace{4i(1-\theta)\mathrm{Re}\langle
B^{\Lambda}_{z^{\prime}},G^{\Lambda}_{z}\rangle+4i\theta\mathrm{Re}\langle
k^{2}B^{\Lambda}_{z^{\prime}},B^{\Lambda}_{z}\rangle}_{=4i\mathrm{Re}\langle
B^{\Lambda}_{z^{\prime}},G^{\Lambda}_{z}\rangle-4\theta
i\mathrm{Re}\langle\omega
B^{\Lambda}_{z^{\prime}},B^{\Lambda}_{z}\rangle=-i\partial_{\theta}V^{\Lambda}_{\theta}}\big{)}\bar{u}^{\theta}_{t}(z^{\prime})q(z^{\prime},y).$
(5.33)

Combined with (5.3), this shows that $D\equiv 0$.

For $\widetilde{D}$ this follows from the same calculation, using that
$\langle
B_{x}^{\Lambda},G_{y}^{\Lambda}\rangle=\widehat{B_{0}^{\Lambda}G_{0}^{\Lambda}}(x-y)$
is real-valued since $G,B$ are even functions. By the same reasoning, the
value of the constant is

$\displaystyle E_{b}$ $\displaystyle=\begin{aligned} -i\int
dxdzdz^{\prime}q(x,z)u^{\theta}_{t}(z)&\Big{(}2i(1-\theta)\langle
B^{\Lambda}_{z^{\prime}},G^{\Lambda}_{z}\rangle\\\ &+2i\theta\langle
k^{2}B^{\Lambda}_{z^{\prime}},B^{\Lambda}_{z}\rangle\Big{)}\bar{u}^{\theta}_{t}(z^{\prime})q(z^{\prime},x)\end{aligned}$
$\displaystyle=-\frac{i}{2}\int
dzdz^{\prime}q(z^{\prime},z)(-i\partial_{\theta}V^{\Lambda}_{\theta}(z-z^{\prime})u^{\theta}_{t}(z)\bar{u}^{\theta}_{t}(z^{\prime}))$
$\displaystyle=2\mathrm{Re}\langle
G_{0}^{\Lambda},B_{0}^{\Lambda}\rangle-2\theta\mathrm{Re}\langle\omega
B^{\Lambda}_{0},B^{\Lambda}_{0}\rangle+\partial_{\theta}\tfrac{1}{2}\langle
u,V^{\Lambda}_{\theta}\ast|u|^{2}u\rangle,$ (5.34)

where we used that $|u^{\theta}_{t}|^{2}$ is independent of $\theta$ and
integrates to one.

Mixed terms in $a,a^{*}$, $b,b^{*}$ (5.26c). Mixed terms with, say,
$a^{*}_{k}b^{*}_{x}$ arise from the derivatives of the respective terms in
$\mathbb{H}^{\Lambda}_{\theta}$, $\mathbb{D}_{t}^{\Lambda}$, and from the
commutator $[\mathbb{H}^{\Lambda}_{\theta},\mathbb{D}_{t}^{\Lambda}]$ if one
commutes a term with one $a^{\\#}$ and one $b^{\\#}$ with terms with two
$a^{\\#}$s or two $b^{\\#}$s. The commutator

$\displaystyle\bigg{[}$ $\displaystyle\int
dkdl\Big{(}2M^{\Lambda}(k,-l)a^{*}_{k}a^{*}_{l}+M^{\Lambda}(k,l)a^{*}_{k}a^{*}_{l}+\text{h.c.}\Big{)},$
$\displaystyle\int
dxdm\,\Big{(}\kappa^{\Lambda}_{t}(m,x)a^{*}_{m}b^{*}_{x}-\kappa^{\Lambda}_{t}(-m,x)a_{m}b^{*}_{x}\Big{)}+\text{h.c}\bigg{]}$
(5.35)

vanishes identically, as one easily checks.

We group the remaining terms into two parts, $X=X^{(1)}+X^{(2)}$, which vanish
separately. Spelling things out, we set

$\displaystyle X^{(1)}(k,x)$
$\displaystyle=(i\partial_{\theta}-\tau_{t}(x))(qL^{\Lambda}_{\theta}(k)u^{\theta}_{t})(x)+(-i\partial_{t}+h_{\theta}+\omega(k))\kappa^{\Lambda}_{t}(k,x),$
$\displaystyle X^{(2)}(k,x)$ $\displaystyle=\int
dy\Big{(}K_{\theta}^{(1),\Lambda}(x,y)\kappa_{t}^{\Lambda}(k,y)+K_{\theta}^{(2),\Lambda}(x,y)\overline{\kappa^{\Lambda}_{t}(-k,y)}\Big{)},$
$\displaystyle\widetilde{X}^{(1)}(k,x)$
$\displaystyle=(i\partial_{\theta}-\tau_{t}(x))(qL^{\Lambda}_{\theta}(k)^{*}u^{\theta}_{t})(x)+(i\partial_{t}-h_{\theta}+\omega(k))\kappa^{\Lambda}_{t}(-k,x),$
$\displaystyle\widetilde{X}^{(2)}(k,x)$ $\displaystyle=-X^{(2)}(-k,x).$

To see that $X^{(1)}(k,x)=0$, we use (5.29) with
$T_{\theta}=L^{\Lambda}_{\theta}(k)$ and the identity

$\nabla\tau_{t}(x)=\nabla 2\mathrm{Re}\langle
iB_{x},\alpha_{t}^{\theta}\rangle=2\mathrm{Re}\langle
kB_{x},\alpha_{t}^{\theta}\rangle=F_{\alpha_{t}^{\theta}}(x)$ (5.36)

to obtain

$\displaystyle(i\partial_{\theta}-\tau_{t}(x))(qL^{\Lambda}_{\theta}(k)u^{\theta}_{t})=q[L^{\Lambda}_{\theta}(k),\tau_{t}]u_{t}^{\theta}+q(i\partial_{\theta}L^{\Lambda}_{\theta}(k))u_{t}^{\theta}$
$\displaystyle\quad=q\Big{(}2\theta
B^{\Lambda}_{(\cdot)}(k)k[-i\nabla,\tau_{t}]-iG^{\Lambda}_{(\cdot)}(k)+2iB^{\Lambda}_{(\cdot)}(k)k(-i\nabla+2\theta
F_{\alpha_{t}^{\theta}})\Big{)}u^{\theta}_{t}$
$\displaystyle\quad=q\Big{(}-iG^{\Lambda}_{(\cdot)}(k)+2B^{\Lambda}_{(\cdot)}(k)k\nabla+2i\theta
B^{\Lambda}_{(\cdot)}(k)kF_{\alpha_{t}^{\theta}}\Big{)}u^{\theta}_{t}.$ (5.37)

Using

$\displaystyle[-\Delta,B_{x}^{\Lambda}(k)]$
$\displaystyle=k^{2}B_{x}^{\Lambda}(k)+2kB_{x}^{\Lambda}(k)i\nabla$
$\displaystyle[A_{\alpha},B^{\Lambda}_{x}(k)]$
$\displaystyle=[2(-i\nabla)\langle
kB_{x},\alpha\rangle+\text{h.c},B^{\Lambda}_{x}(k)]=-2kB^{\Lambda}_{x}(k)\underbrace{2\mathrm{Re}\langle
kB_{x},\alpha\rangle}_{=F_{\alpha}(x)}$

we find in the same way

$\displaystyle(-i\partial_{t}+h_{\theta}+\omega(k))\kappa^{\Lambda}_{t}(k,x)$
$\displaystyle=q\Big{(}i\omega(k)B^{\Lambda}_{(\cdot)}+[-\Delta+\theta
A_{\alpha_{t}^{\theta}},iB^{\Lambda}_{(\cdot)}(k)]\Big{)}u^{\theta}_{t}$
$\displaystyle=q\Big{(}iG^{\Lambda}_{(\cdot)}-2B^{\Lambda}_{(\cdot)}(k)k\nabla-2i\theta
B^{\Lambda}_{(\cdot)}(k)kF_{\alpha_{t}^{\theta}}\Big{)}u^{\theta}_{t}.$ (5.38)

This equals the negative of (5.37), so $X^{(1)}\equiv 0$.

The equality $X^{(2)}\equiv 0$ follows simply by expanding the expressions:

$\displaystyle\int
dy\,K_{\theta}^{(2),\Lambda}(x,y)\overline{\kappa_{t}^{\Lambda}(-k,y)}$
$\displaystyle=\int
dydzdz^{\prime}dz^{\prime\prime}\,q(x,z)q(y,z^{\prime})u^{\theta}_{t}(z)u^{\theta}_{t}(z^{\prime})V^{\Lambda}_{\theta}(z-z^{\prime})\overline{iB^{\Lambda}_{z^{\prime\prime}}(-k)u^{\theta}_{t}(z^{\prime\prime})q(y,z^{\prime\prime})}$
$\displaystyle=\int
dzdz^{\prime}dz^{\prime\prime}\,q(x,z)q(z^{\prime\prime},z^{\prime})u^{\theta}_{t}(z)u^{\theta}_{t}(z^{\prime})V^{\Lambda}_{\theta}(z-z^{\prime})\bar{u}^{\theta}_{t}(z^{\prime\prime})(-iB^{\Lambda}_{z^{\prime\prime}}(k))$
$\displaystyle=-\int
dzdz^{\prime}\,q(x,z)u^{\theta}_{t}(z)|u^{\theta}_{t}(z^{\prime})|^{2}V^{\Lambda}_{\theta}(z-z^{\prime})iB^{\Lambda}_{z^{\prime}}(k)$
$\displaystyle\quad+i\int
dz\,q(x,z)(V^{\Lambda}_{\theta}\ast|u^{\theta}_{t}|^{2})(z)\langle
u^{\theta}_{t},B^{\Lambda}_{\cdot}(k)u^{\theta}_{t}\rangle$
$\displaystyle=-\int
dy\,K_{\theta}^{(1),\Lambda}(x,y)\kappa_{t}^{\Lambda}(k,y),$ (5.39)

where the last equality is obtained by performing the same calculation for
$K^{(1)}\kappa$, which just changes the location of some complex conjugates.
This implies vanishing of $\widetilde{X}^{(2)}$ and the argument for
$\widetilde{X}^{(1)}$ is completely analogous to that for $X^{(1)}$.

Terms quadratic in $a,a^{*}$ (5.26d). The only way to obtain a term with two
$a^{\\#}$s from the commutator
$[\mathbb{H}^{\Lambda}_{\theta},\mathbb{D}_{t}^{\Lambda}]$ is to commute two
terms with an $a^{\\#}$ and one $b^{\\#}$ each. Since the coefficient
$M^{\Lambda}(k,l)$ (5.11b) of the terms with two $a^{\\#}$s in
$\mathbb{H}^{\Lambda}_{\theta}(t)$ is independent of $\theta$, we obtain for
the coefficients in (5.26d)

$\displaystyle A(k,l)$ $\displaystyle=-4i\theta M^{\Lambda}(k,-l)$ (5.40a)
$\displaystyle\quad-\int
dx\Big{(}(qL^{\Lambda}_{\theta}(k)u^{\theta}_{t})(x)\overline{\kappa^{\Lambda}_{t}(l,x)}+\overline{(qL^{\Lambda}_{\theta}(k)^{*}u^{\theta}_{t})(x)}\kappa^{\Lambda}_{t}(-l,x)\Big{)}$
$\displaystyle\quad+\int
dx\Big{(}\overline{(qL^{\Lambda}_{\theta}(l)u^{\theta}_{t})(x)}\kappa^{\Lambda}_{t}(k,x)+(qL^{\Lambda}_{\theta}(l)^{*}u^{\theta}_{t})(x)\overline{\kappa^{\Lambda}_{t}(-k,x)}\Big{)}$
$\displaystyle E_{a}$ $\displaystyle=-i\int
dkdx\Big{(}\overline{(qL^{\Lambda}_{\theta}(k)u^{\theta}_{t})(x)}\kappa^{\Lambda}_{t}(k,x)+(qL^{\Lambda}_{\theta}(k)^{*}u^{\theta}_{t})(x)\overline{\kappa^{\Lambda}_{t}(-k,x)}\Big{)}$
(5.40b) $\displaystyle\widetilde{A}(k,l)$ $\displaystyle=\begin{aligned}
2i\theta M^{\Lambda}(k,l)+\int
dx\,\Big{(}&\overline{(qL^{\Lambda}_{\theta}(k)^{*}u^{\theta}_{t})(x)}\kappa^{\Lambda}_{t}(l,x)\\\
&+(qL^{\Lambda}_{\theta}(k)u^{\theta}_{t})(x)\overline{\kappa^{\Lambda}_{t}(-l,x)}\Big{)}.\end{aligned}$
(5.40c)

To see that $A(k,l)=0$, we first calculate using that
$q^{2}=q=1-|u^{\theta}_{t}\rangle\langle u^{\theta}_{t}|$ and

$\displaystyle\int
dx\Big{(}(qL^{\Lambda}_{\theta}(k)u^{\theta}_{t})(x)\overline{\kappa^{\Lambda}_{t}(l,x)}+\overline{qL^{\Lambda}_{\theta}(k)^{*}u^{\theta}_{t}(x)}\kappa^{\Lambda}_{t}(-l,x)\Big{)}$
$\displaystyle\quad=\begin{aligned} \int
dzdz^{\prime}\,q(z^{\prime},z)\Big{(}&(L^{\Lambda}_{\theta}(k)u^{\theta}_{t})(z)(-iB^{\Lambda}_{z^{\prime}}(-l))\bar{u}^{\theta}_{t}(z^{\prime})\\\
&+\overline{(L^{\Lambda}_{\theta}(k)^{*}u^{\theta}_{t})(z^{\prime})}iB^{\Lambda}_{z}(-l)u^{\theta}_{t}(z)\Big{)}\end{aligned}$
$\displaystyle\quad=\langle
u^{\theta}_{t},i[L^{\Lambda}_{\theta}(k),B_{(\cdot)}^{\Lambda}(-l)]u^{\theta}_{t}\rangle$
$\displaystyle\quad=2i\theta\langle
u^{\theta}_{t},B_{(\cdot)}^{\Lambda}(-l)lkB_{(\cdot)}^{\Lambda}(k)u_{t}^{\theta}\rangle=-2i\theta
M^{\Lambda}(k,-l).$ (5.41)

The second line in (5.40a) is the complex conjugate of this with $k,l$
exchanged, so it equals $2i\theta M^{\Lambda}(k,-l)$. This implies that
$A\equiv 0$. The argument for $\widetilde{A}\equiv 0$ is essentially the same.

It remains to evaluate $E_{a}$. We have by the calculation of $A(k,l)$

$\displaystyle E_{a}=2\theta\int dkM^{\Lambda}(k,-k)=-2\theta\langle
k^{2}B_{0}^{\Lambda},B_{0}^{\Lambda}\rangle.$ (5.42)

Consequently with (5.34)

$\displaystyle E_{a}+E_{b}$ $\displaystyle=2\mathrm{Re}\langle
G_{0}^{\Lambda},B_{0}^{\Lambda}\rangle-2\theta\mathrm{Re}\langle(k^{2}+\omega)B^{\Lambda}_{0},B^{\Lambda}_{0}\rangle+\partial_{\theta}\tfrac{1}{2}\langle
u,V^{\Lambda}_{\theta}\ast|u|^{2}u\rangle$
$\displaystyle=(2-2\theta)\mathrm{Re}\langle
G_{0}^{\Lambda},B_{0}^{\Lambda}\rangle+\partial_{\theta}\tfrac{1}{2}\langle
u,V^{\Lambda}_{\theta}\ast|u|^{2}u\rangle$
$\displaystyle=\partial_{\theta}E^{\Lambda}_{\theta}.$ (5.43)

This completes the proof of the proposition. ∎

##  6\. Estimates for the generators

In this section we establish the inequalities on the different generators of
the dynamics considered in the previous sections. This includes the generators
of the fluctuation dynamics for $e^{-itH_{N}^{\mathrm{D}}}$ and its Bogoliubov
approximation used in the proofs of Theorems 3.2 and 4.2, given in Sections
6.4, 6.3, respectively. Similar bounds for the generators associated with the
dressing flow and its Bogoliubov approximation are given in Section 6.5.

###  6.1. Fock space operator bounds

We start by proving a general bound for operators on Fock space that will
prove very useful.

######  Lemma 6.1.

Let $n_{a},n_{b},m_{a},m_{b}\in\mathbb{N}_{0}$, $M=n_{a}+n_{b}+m_{a}+m_{b}$,
$s,t\in\mathbb{R}$ and

$T:L^{2}(\mathbb{R}^{3})^{\otimes m_{b}}\otimes\mathfrak{h}_{t}^{\otimes
m_{a}}\to L^{2}(\mathbb{R}^{3})^{\otimes
n_{b}}\otimes\mathfrak{h}_{-s}^{\otimes n_{a}}$

be a bounded operator of norm $\tau$ with an integral kernel
$T((K,X),(L,Y))\in\mathscr{S}^{\prime}(\mathbb{R}^{3M})$. Set

$A_{n}(K)=\prod_{i=1}^{n}a_{k_{i}}\text{ and
}B_{n}(X)=\prod_{i=1}^{n}b_{x_{i}}.$

Then for all $0\leq r_{b}\leq n_{b}+m_{b}$, $0\leq r_{a}\leq n_{a}+m_{a}$,

$\displaystyle\bigg{|}\Big{\langle}\chi,\int_{\mathbb{R}^{3M}}T(X,K,Y,L)B^{*}_{n_{b}}(X)A^{*}_{n_{a}}(K)A_{m_{a}}(L)B_{m_{b}}(Y)dXdKdYdL\xi\Big{\rangle}\bigg{|}$
$\displaystyle\leq\tau\|(\mathcal{N}_{b}+M)^{\frac{n_{b}+m_{b}-r_{b}}{2}}(\mathcal{N}_{a}+M+1)^{r_{a}}\textnormal{d}\Gamma_{a}(\omega^{2s})^{\frac{n_{a}}{2}}\chi\|$
$\displaystyle\qquad\times\|(\mathcal{N}_{b}+M)^{\frac{r_{b}}{2}}(\mathcal{N}_{a}+1)^{-r_{a}}\textnormal{d}\Gamma_{a}(\omega^{2t})^{\frac{m_{a}}{2}}\xi\|.$

###### Proof.

To keep the notation manageable we give the proof in the case $n_{b}=m_{b}=0$,
the generalization is straightforward. Set $n=n_{a}$, $m=m_{a}$, and let
$\chi\in D(\textnormal{d}\Gamma(\omega^{2s})^{n/2})$, then

$\displaystyle\int\Big{\|}\prod_{i=1}^{n}\omega^{s}(k_{i})a_{k_{i}}\chi\Big{\|}^{2}dK$
$\displaystyle=\int\Big{(}\prod_{i=1}^{n}\omega^{2s}(k_{i})\Big{)}\Big{\langle}\chi,a_{k_{1}}^{*}\cdots
a_{k_{n}}^{*}a_{k_{n}}\cdots a_{k_{1}}\chi\Big{\rangle}dK$
$\displaystyle=\|\textnormal{d}\Gamma_{a}(\omega^{2s})^{n/2}\chi\|^{2},$ (6.1)

so

$(k_{1},\dots,k_{n})\mapsto\Big{(}\prod_{i=1}^{n}\omega^{s}(k_{i})a_{k_{i}}\Big{)}\chi\in
L^{2}(\mathbb{R}^{3n},\mathcal{F}\otimes\mathcal{F}),$

and analogously for $\xi$. Hence

$\displaystyle\bigg{|}\Big{\langle}\chi,\int
T(K,L)A_{n}^{*}(K)A_{m}(L)dKdL\,\xi\Big{\rangle}\bigg{|}$
$\displaystyle=\begin{aligned} \bigg{|}\int
dKdL\Big{\langle}(\mathcal{N}_{a}+1)^{r_{a}}&\Big{(}\prod_{i=1}^{n}\omega^{s}(k_{i})a_{k_{i}}\Big{)}\chi,\\\
&T(K,L)(\mathcal{N}_{a}+1)^{-r_{a}}\Big{(}\prod_{i=1}^{n}\omega^{-s}(k_{i})\Big{)}A_{m}(L)\xi\Big{\rangle}\bigg{|}\end{aligned}$
$\displaystyle\leq\|\textnormal{d}\Gamma_{a}(\omega^{2s})^{\frac{n}{2}}(\mathcal{N}_{a}+n+1)^{r_{a}}\chi\|$
$\displaystyle\qquad\times\Big{\|}\int\Big{(}\prod_{i=1}^{n}\omega^{-s}(k_{i})\Big{)}T(K,L)A_{m}(L)(\mathcal{N}_{a}+m+1)^{-r_{a}}\xi
dL\Big{\|}_{L^{2}(\mathbb{R}^{3n}_{K},\mathcal{F}\otimes\mathcal{F})}$
$\displaystyle\leq\|T\|_{\mathfrak{h}_{t}^{\otimes
m}\to\mathfrak{h}_{-s}^{\otimes
n}}\|\textnormal{d}\Gamma_{a}(\omega^{2s})^{\frac{n}{2}}(\mathcal{N}_{a}+M+1)^{r_{a}}\chi\|\|\textnormal{d}\Gamma_{a}(\omega^{2t})^{\frac{m}{2}}(\mathcal{N}_{a}+1)^{-r_{a}}\xi\|.$
(6.2)

This proves the claim. ∎

Two special cases of the previous lemma we use frequently are given separately
below.

######  Lemma 6.2.

For any $s\in\mathbb{R}$ and $\chi,\xi\in\mathcal{F}\otimes\mathcal{F}$, we
have

$\displaystyle\big{|}\big{\langle}\chi,\textnormal{d}\Gamma_{a}(\omega^{s/2})\xi\big{\rangle}\big{|}$
$\displaystyle\leq\|\textnormal{d}\Gamma_{a}(\omega^{s})^{1/2}\chi\|\,\|\mathcal{N}_{a}^{1/2}\xi\|,$
(6.3a) $\displaystyle\|\textnormal{d}\Gamma_{a}(\omega^{s/2})\chi\|$
$\displaystyle\leq\|\mathcal{N}_{a}^{1/2}\textnormal{d}\Gamma_{a}(\omega^{s})^{1/2}\chi\|.$
(6.3b)

###### Proof.

The first inequality is a special case of Lemma 6.1 ($t=0$,
$T\alpha=\omega^{s}\alpha$, $m_{a}=n_{a}=1$, $r_{a}=0$). The second inequality
follows from the first by taking the supremum over $\|\xi\|=1$. ∎

###  6.2. Preliminary estimates

Here we provide some bounds on the terms appearing in the mean-field
Hamiltonians as well as the kernels of the Bogoliubov Hamiltonians. These will
later be combined with the operator bounds from the previous section to prove
the estimates for the generators.

######  Lemma 6.3.

Let $G$, $B$ and $V$ be defined as in (2.17), (2.18) and (3.2b). For every
$s>0$ there exists a constant $C>0$ such that

$\displaystyle\|\widehat{V}\|_{L^{1+s}}$ $\displaystyle\leq C,$ (6.4a)
$\displaystyle\|kB_{0}\|_{\mathfrak{h}_{-s}}$ $\displaystyle\leq C,$ (6.4b)
$\displaystyle\forall
n\in\mathbb{N}_{0}\quad\big{\langle}|k|^{n}B_{0},|\alpha|\big{\rangle}$
$\displaystyle\leq C\|\alpha\|_{\mathfrak{h}_{(n-1)+s}},$ (6.4c)
$\displaystyle\forall n\in\\{1,2,3\\}\quad\|\langle
k^{n}B_{(\cdot)},\alpha\rangle u\|_{{L^{2}}}$ $\displaystyle\leq
C\|\alpha\|_{\mathfrak{h}_{(n-2)/2+s}}\|u\|_{H^{n/2}}$ (6.4d) for all $u\in
H^{n}(\mathbb{R}^{3})$ and $\alpha\in\mathfrak{h}_{n-1+s}$. Moreover, for
every $\varepsilon>0$ there exists $C>0$ so that for $u\in
H^{1}(\mathbb{R}^{3})$, $\alpha\in\mathfrak{h}_{1/2}$ $\displaystyle\|\langle
G_{(\cdot)},\alpha\rangle u\|_{L^{2}}$
$\displaystyle\leq\varepsilon(\|u\|_{H^{1}}^{2}+\|\alpha\|_{\mathfrak{h}_{1/2}}^{2})+C\|u\|_{L^{2}}^{2}.$
(6.4e)

###### Proof of Lemma 6.3.

In view of the formula for $V$, we have
$(2\pi)^{3/2}\widehat{V}(k)=-4G_{0}(k)B_{0}(k)+2\omega(k)B_{0}^{2}(k)$, and
the first three inequalities then follow immediately from the integrability
properties of $G_{0}(k)=\omega(k)^{-1/2}$ and
$B_{0}(k)=(k^{2}+\omega(k))^{-1}\omega(k)^{-1/2}$.

For (6.4d), we use the Fourier representation in $x$ together with Parseval to
write

$\displaystyle\|\big{\langle}k^{n}B_{(\cdot)},\alpha\big{\rangle}u\|_{{L^{2}}}^{2}=\frac{1}{(2\pi)^{3}}\Big{\|}\int
dk\,k^{n}B_{0}(k)\alpha(k)\int
dp\,e^{ip(\cdot)}\,\hat{u}(p-k)\Big{\|}_{L^{2}}^{2}$ (6.5)
$\displaystyle=\Big{\|}\int
dk\,k^{n}B_{0}(k)\alpha(k)\,\hat{u}(\cdot-k)\Big{\|}_{L^{2}}^{2}$
$\displaystyle=\int
dpdkd\ell\,\frac{k^{n}B_{0}(k)\overline{\alpha(k)}}{\omega(p-k)^{n/2}}\frac{\ell^{n}B_{0}(\ell)\alpha(\ell)}{\omega(p-\ell)^{n/2}}\omega(p-k)^{\frac{n}{2}}\widehat{u}(p-k)\omega(p-\ell)^{\frac{n}{2}}\overline{\widehat{u}(p-\ell)}.$

Since $k\mapsto|k|^{n}B_{0}(k)^{2}\omega(k)^{2-2s}$ and
$k\mapsto\omega(k)^{-n}$ are radial and decreasing functions for $s>0$ and
$n\in\\{1,2,3\\}$, it follows by symmetric rearrangement that

$\displaystyle\sup_{p\in\mathbb{R}^{3}}\int
dk\frac{|k|^{n}B_{0}(k)^{2}\omega(k)^{2-2s}}{\omega(p-k)^{n}}\leq\int
dk\,B_{0}(k)^{2}\omega(k)^{2-2s}\leq C.$ (6.6)

With Cauchy–Schwarz we thus find

$\displaystyle\|\big{\langle}k^{n}B_{(\cdot)},\alpha\big{\rangle}u\|_{{L^{2}}}^{2}$
(6.7) $\displaystyle\leq\int
dpdkd\ell\,\frac{|k|^{n}B_{0}(k)^{2}\omega(k)^{2-2s}}{\omega(p-k)^{n}}|\ell|^{n}\omega(\ell)^{2s-2}|\alpha(\ell)|^{2}\omega(p-\ell)^{n}|\widehat{u}(p-\ell)|^{2}$
$\displaystyle\leq C\int d\ell\,\omega(\ell)^{n+2s-2}|\alpha(\ell)|^{2}\int
dp\,(|p|^{2}+1)^{\frac{n}{2}}|\widehat{u}(p)|^{2}\leq
C\|\alpha\|_{\mathfrak{h}_{n/2+s-1}}^{2}\|u\|_{H^{n/2}}^{2}.$

The final inequality (6.4e) is proved in a similar way, but to obtain the
small constant in front of the energy norm we start by introducing a cutoff
$\Lambda<\infty$. We then bound

$\|\langle G_{(\cdot)},\alpha\rangle
u\|_{L^{2}}\leq\|\langle\mathds{1}_{|\cdot|>\Lambda}G_{(\cdot)},\alpha\rangle
u\|_{L^{2}}+\tfrac{\varepsilon}{2}\|\alpha\|_{\mathfrak{h}_{1/2}}^{2}+\tfrac{1}{2\varepsilon}\|\mathds{1}_{|\cdot|\leq\Lambda}G_{0}\|^{2}_{\mathfrak{h}_{-1/2}}\|u\|_{L^{2}}.$
(6.8)

Now the first term is treated exactly as for (6.4d), which yields

$\displaystyle\|\langle\mathds{1}_{|\cdot|>\Lambda}G_{(\cdot)},\alpha\rangle
u\|_{L^{2}}^{2}\leq\|\alpha\|_{\mathfrak{h}_{1/2}}^{2}\|u\|_{H^{1}}^{2}\int_{|k|>\Lambda}dk\frac{|G_{0}(k)|^{2}}{\omega(k)^{3}}.$
(6.9)

Choosing $\Lambda$ so that the final integral is less or equal to
$\varepsilon^{2}$ proves the claim. ∎

The next lemma collects bounds for the different potentials that appear in the
mean-field equations introduced in Section 3.2.

######  Lemma 6.4.

Let $V$ be defined by (3.2b) and $f_{u}$ by (3.4e). There exists a constant
$C>0$ such that for all $u\in H^{1}(\mathbb{R}^{3})$ with $\|u\|_{L^{2}}=1$

$\displaystyle\|V\ast|u|^{2}\|_{L^{\infty}}+\|V^{2}\ast|u|^{2}\|_{L^{\infty}}+\|\nabla(V\ast|u|^{2})\|_{L^{\infty}}$
$\displaystyle\leq C\|u\|_{H^{1}}^{3/2},$ (6.10a)
$\displaystyle\|f_{u}\|_{L^{\infty}}+\|f_{u}\|_{\mathfrak{h}_{1/2}}$
$\displaystyle\leq C\|u\|_{H^{1}}^{2}.$ (6.10b)

Moreover, for every $s>0$ there exists a constant $C>0$ such that for all
$u\in H^{1}(\mathbb{R}^{3})$, $\|u\|_{L^{2}}=1$ and
$\alpha\in\mathfrak{h}_{1+s}$, the objects $F_{\alpha}$, $g_{u,\alpha}$,
$\mu_{u,\alpha}$ defined in (3.4c)–(3.4f) satisfy

$\displaystyle\|F_{\alpha}\|_{L^{\infty}}$ $\displaystyle\leq
C\|\alpha\|_{\mathfrak{h}_{s}}$ (6.11a) $\displaystyle\|\nabla
F_{\alpha}\|_{L^{\infty}}$ $\displaystyle\leq
C\|\alpha\|_{\mathfrak{h}_{1+s}}$ (6.11b)
$\displaystyle\|g_{u,\alpha}\|_{L^{\infty}}+\|g_{u,\alpha}\|_{\mathfrak{h}_{1/2}}$
$\displaystyle\leq C\|u\|_{H^{1}}\|\alpha\|_{\mathfrak{h}_{s}}$ (6.11c)
$\displaystyle|\mu_{u,\alpha}|$ $\displaystyle\leq
C\|u\|_{H^{1}}^{2}\|\alpha\|_{\mathfrak{h}_{s}}^{2}.$ (6.11d)

###### Proof.

For the first term in (6.10a), applying Young’s inequality and Parseval,

$\displaystyle\|V\ast|u|^{2}\|_{L^{\infty}}\leq\|V\|_{L^{2}}\||u|^{2}\|_{L^{2}}=\|\widehat{V}\|_{L^{2}}\|u\|_{L^{4}}^{2}\stackrel{{\scriptstyle\eqref{eq:bound:FT:V}}}{{\leq}}C\|u\|_{L^{4}}^{2},$
(6.12)

and then the Cauchy-Schwarz and Sobolev inequalities yield

$\displaystyle\|u\|_{L^{4}}^{2}\leq\|u\|_{L^{2}}^{1/2}\|u\|_{L^{6}}^{3/2}\leq
C\|u\|_{H^{1}}^{3/2}.$ (6.13)

For the convolution involving $V^{2}$, we proceed similarly, and obtain

$\displaystyle\|V^{2}\ast|u|^{2}\|_{L^{\infty}}\leq\|V^{2}\|_{L^{2}}\||u|^{2}\|_{L^{2}}=\|V\|_{L^{4}}^{2}\|u\|_{L^{4}}^{2}\leq
C\|V\|_{L^{4}}^{2}\|u\|_{H^{1}}^{3/2}.$ (6.14)

By Hausdorff–Young, the Fourier transform is bounded from $L^{4/3}$ to
$L^{4}$, so by (6.4a)

$\displaystyle\|V\|_{L^{4}}\leq C\|\widehat{V}\|_{L^{4/3}}\leq C.$ (6.15)

For the convolution involving the gradient, we use the same inequalities to
estimate

$\displaystyle\|\nabla(V\ast|u|^{2})\|_{L^{\infty}}\leq
2\|V\|_{L^{4}}\|\overline{u}\nabla u\|_{L^{4/3}}\leq C\|\overline{u}\nabla
u\|_{L^{4/3}}.$ (6.16)

With Hölder and (6.13) we conclude that

$\displaystyle\|\overline{u}\nabla u\|_{L^{4/3}}\leq C\|u\|_{L^{4}}\|\nabla
u\|_{{L^{2}}}\leq C\|u\|_{H^{1}}^{3/2}$ (6.17)

and thus obtain (6.10a).

In view of $f_{u}(k)=\big{\langle}u,kB_{(\cdot)}(k)(-i\nabla)u\big{\rangle}$,
the bound on $\|f_{u}\|_{L^{\infty}}$ is obvious. To bound the
$\mathfrak{h}_{1/2}$-norm of $f_{u}$, we use that

$\|f_{u}\sqrt{\omega}\|_{{L^{2}}}\leq
2\|kB_{0}\sqrt{\omega}\|_{L^{4}}\|\mathscr{F}[\overline{u}\nabla
u]\|_{L^{4}}.$ (6.18)

This implies the bound by Hausdorff-Young inequality and (6.17).

The bounds on
$F_{\alpha}(x)=2\mathrm{Re}\big{\langle}kB_{x},\alpha\big{\rangle}$ and
$\nabla F_{\alpha}$ follow directly from the from the fact that
$|k|B(k)\in\mathfrak{h}_{-s}$. The bound on $F_{\alpha}$ implies that on the
$L^{\infty}$-norm of
$g_{u,\alpha}=2\big{\langle}u,kB_{(\cdot)}(k)F_{\alpha}u\big{\rangle}$. For
the $\mathfrak{h}_{1/2}$-norm, writing
$g_{u,\alpha}=2kB_{0}\cdot\mathscr{F}[F_{\alpha}|u|^{2}]$ gives

$\displaystyle\|g_{u,\alpha}\sqrt{\omega}\|_{{L^{2}}}$ $\displaystyle\leq
2\|kB_{0}\sqrt{\omega}\|_{L^{4}}\|\mathscr{F}[F_{\alpha}|u|^{2}]\|_{L^{4}}$
$\displaystyle\leq C\|F_{\alpha}|u|^{2}\|_{L^{4/3}}\leq
C\|F_{\alpha}\|_{L^{\infty}}\||u|^{2}\|_{L^{4/3}}.$ (6.19)

The claimed bound then follows from (6.13) by Hölder’s inequality, since
$\|u\|_{H^{1}}\geq 1$.

The estimate for
$\mu_{u,\alpha}=\tfrac{1}{2}\big{\langle}u,V\ast|u|^{2}u\big{\rangle}+\mathrm{Re}\big{\langle}\alpha,f_{u}\big{\rangle}+\mathrm{Re}\big{\langle}\alpha,g_{u,\alpha}\big{\rangle}$
follows from the previous bounds. ∎

We have similar bounds on the mean-field Hamiltonian.

######  Lemma 6.5.

Let $h_{u,\alpha}$ be defined by (3.4a). For every $s>0$ there is a constant
$C>0$ such that for all $(u,\alpha)\in H^{3}\oplus\mathfrak{h}_{1+s}$ with
$\|u\|_{L^{2}}=1$

$\displaystyle\|h_{u,\alpha}u\|_{{L^{2}}}$ $\displaystyle\leq
C\big{(}\|u\|_{H^{2}}+\|u\|_{H^{1}}^{2}\big{)}\big{(}1+\|\alpha\|_{\mathfrak{h}_{s}}^{2}\big{)},$
$\displaystyle\|\nabla h_{u,\alpha}u\|_{{L^{2}}}$ $\displaystyle\leq
C\big{(}\|u\|_{H^{3}}+\|u\|_{H^{1}}^{5/2}\big{)}\big{(}1+\|\alpha\|_{\mathfrak{h}_{1+s}}\|\alpha\|_{\mathfrak{h}_{s}}^{2}\big{)}.$

###### Proof of Lemma 6.5.

The proof follows from Lemmas 6.3 and 6.4 in combination with (we use
$\|\cdot\|_{H^{s}}\geq 1$ and
$\|\cdot\|_{\mathfrak{h}_{s}}\geq\|\cdot\|_{\mathfrak{h}_{r}}$ for $s\geq r$)

$\displaystyle\|h_{u,\alpha}u\|_{{L^{2}}}$ $\displaystyle\leq\|\Delta
u\|_{L^{2}}+2\|\big{\langle}k^{2}B_{(\cdot)},\alpha\big{\rangle}u\|_{{L^{2}}}+4\big{\langle}|k|B_{0},|\alpha|\big{\rangle}\|\nabla
u\|_{{L^{2}}}$
$\displaystyle\quad+\|F_{\alpha}\|_{L^{\infty}}^{2}+\|V\ast|u|^{2}\|_{L^{\infty}}+|\mu_{u,\alpha}|$
$\displaystyle\leq
C\big{(}\|u\|_{H^{2}}+\|u\|_{H^{1}}^{2}\big{)}\big{(}1+\|\alpha\|_{\mathfrak{h}_{s}}^{2},\big{)}$
(6.20)

and

$\displaystyle\|\nabla h_{u,\alpha}u\|_{{L^{2}}}$
$\displaystyle\leq\|\nabla\Delta
u\|_{L^{2}}+2\|\big{\langle}k^{3}B_{(\cdot)},\alpha\big{\rangle}u\|_{{L^{2}}}+6\|\big{\langle}k^{2}B_{(\cdot)},\alpha\big{\rangle}\nabla
u\|_{{L^{2}}}$ $\displaystyle\
+4\big{\langle}|k|B_{0},|\alpha|\big{\rangle}\|\Delta
u\|_{{L^{2}}}+2\|F_{\alpha}\|_{L^{\infty}}\|\nabla
F_{\alpha}\|_{L^{\infty}}+\|F_{\alpha}\|_{L^{\infty}}^{2}\|\nabla u\|_{L^{2}}$
$\displaystyle\
+\|\nabla(V\ast|u|^{2})\|_{L^{\infty}}+\|V\ast|u|^{2}\|_{L^{\infty}}\|\nabla
u\|_{{L^{2}}}+|\mu_{u,\alpha}|\|\nabla u\|_{L^{2}}$ $\displaystyle\leq
C\big{(}\|u\|_{H^{3}}+\|u\|_{H^{1}}^{5/2}\big{)}\big{(}1+\|\alpha\|_{\mathfrak{h}_{1+s}}\|\alpha\|_{\mathfrak{h}_{s}}^{2}\big{)}.$
(6.21)

∎

Next, we state suitable bounds for the time-derivatives of $u_{t}$,
$\alpha_{t}$ and $\mu_{u_{t},\alpha_{t}}$. Note that the constant $C$ in the
bound is uniform in $t$ but not in $(u,\alpha)$, as it depends on the energy
of the initial condition.

######  Lemma 6.6.

Let $(u,\alpha)\in H^{3}(\mathbb{R}^{3})\oplus\mathfrak{h}_{5/2}$ with
$\|u\|_{L^{2}}=1$ and $(u_{t},\alpha_{t})=\mathfrak{s}^{\rm D}[t](u,\alpha)$
denote the solution to (3.3). Let $\mu_{u_{t},\alpha_{t}}$ be defined as in
(3.4d). There exists a constant $C>0$ such that for all $|t|\geq 0$

$\displaystyle\|\dot{u}_{t}\|_{H^{1}}+\|\dot{\alpha}_{t}\|_{\mathfrak{h}_{1/2}}+|\dot{\mu}_{u_{t},\alpha_{t}}|$
$\displaystyle\leq
C\,\|u_{t}\|_{H^{3}}\,\big{(}1+\|\alpha_{t}\|_{\mathfrak{h}_{3/2}}\big{)}.$

###### Proof.

We use that
$\|\dot{u}_{t}\|_{H^{1}}^{2}=\|\dot{u}_{t}\|_{L^{2}}^{2}+\|\nabla\dot{u}_{t}\|_{{L^{2}}}^{2}$
and $i\dot{u}_{t}=h_{u_{t},\alpha_{t}}u_{t}$ . Since
$\|u_{t}\|_{H^{1}}+\|\alpha_{t}\|_{\mathfrak{h}_{1/2}}\leq C$ (by Proposition
2.2 and Lemma 3.3), we obtain from Lemma 6.5 for $s=\tfrac{1}{2}$

$\displaystyle\|h_{u_{t},\alpha_{t}}u_{t}\|_{{L^{2}}}$ $\displaystyle\leq
C\|u_{t}\|_{H^{2}},$ (6.22a) $\displaystyle\|\nabla
h_{u_{t},\alpha_{t}}u_{t}\|_{{L^{2}}}$ $\displaystyle\leq
C\|u_{t}\|_{H^{3}}\big{(}1+\|\alpha_{t}\|_{\mathfrak{h}_{3/2}}\big{)}.$
(6.22b)

With the aid of Lemma 6.4, one easily verifies

$\displaystyle\|\dot{\alpha}_{t}\|_{\mathfrak{h}_{1/2}}\leq\|\alpha_{t}\|_{\mathfrak{h}_{3/2}}+\|u_{t}\|_{H^{1}}^{2}\|\alpha_{t}\|_{\mathfrak{h}_{1/2}}.$
(6.23)

Recall that
$\mu_{u,\alpha}=\tfrac{1}{2}\big{\langle}u,V\ast|u|^{2}u\big{\rangle}+\mathrm{Re}\big{\langle}\alpha,f_{u}\big{\rangle}+\mathrm{Re}\big{\langle}\alpha,g_{u,\alpha}\big{\rangle}$.
Since $V$ is an even function,

$\displaystyle\big{|}\tfrac{d}{dt}\big{\langle}u,V\ast|u|^{2}u\big{\rangle}\big{|}$
$\displaystyle=4|\mathrm{Re}\big{\langle}\dot{u}_{t},V\ast|u_{t}|^{2}u_{t}\big{\rangle}|$
$\displaystyle\leq
4\|V\ast|u_{t}|^{2}\|_{L^{\infty}}\|\dot{u}_{t}\|_{L^{2}}\leq
C\|\dot{u}_{t}\|_{L^{2}}.$ (6.24)

We further estimate

$\displaystyle\|\dot{f}_{u_{t}}\|_{\mathfrak{h}_{-1/2}}$ $\displaystyle\leq
2\|kB_{0}\|_{\mathfrak{h}_{-1/2}}\left(\|\dot{u}_{t}\|_{L^{2}}\|u_{t}\|_{H^{1}}+\|\dot{u}_{t}\|_{H^{1}}\|u_{t}\|_{L^{2}}\right)\leq
C\|\dot{u}_{t}\|_{H^{1}},$
$\displaystyle\|\dot{g}_{u_{t},\alpha_{t}}\|_{\mathfrak{h}_{-1/2}}$
$\displaystyle\leq 4\|\big{\langle}\dot{u}_{t},kB_{(\cdot)}\cdot
F_{\alpha_{t}}u_{t}\big{\rangle}\|_{\mathfrak{h}_{-1/2}}+2\|\big{\langle}u_{t},kB_{(\cdot)}\cdot
F_{\dot{\alpha}_{t}}u_{t}\big{\rangle}\|_{\mathfrak{h}_{-1/2}}$
$\displaystyle\leq\big{(}2\|\dot{u}_{t}\|_{L^{2}}\|F_{\alpha_{t}}\|_{L^{\infty}}+\|F_{\dot{\alpha}_{t}}\|_{L^{\infty}}\big{)}\|kB_{0}\|_{\mathfrak{h}_{-1/2}}.$
(6.25)

With (6.4c) we have $\|F_{\dot{\alpha}_{t}}\|_{L^{\infty}}\leq
C\|\dot{\alpha}_{t}\|_{\mathfrak{h}_{1/2}}$, and hence

$\displaystyle|\dot{\mu}_{u_{t},\alpha_{t}}|$ $\displaystyle\leq
C\|\dot{u}_{t}\|_{L^{2}}+\|\dot{\alpha}_{t}\|_{L^{2}}\|f_{u_{t}}+g_{u_{t},\alpha_{t}}\|_{L^{2}}+\|\alpha_{t}\|_{\mathfrak{h}_{1/2}}\|\dot{f}_{u_{t}}+\dot{g}_{u_{t},\alpha_{t}}\|_{\mathfrak{h}_{-1/2}}$
$\displaystyle\leq
C(\|\dot{\alpha}_{t}\|_{\mathfrak{h}_{1/2}}+\|\dot{u}_{t}\|_{H^{1}}).$ (6.26)

This completes the proof of the lemma. ∎

The next lemma summarizes estimates for the different kernels (and their time-
derivatives) that appear in the (dressed) Bogoliubov Hamiltonian (4.1) and the
fluctuation Hamiltonian introduced in (3.8), given explicitly by (6.68).

We introduce the integral kernels

$\displaystyle N_{u}(x,k,l)$ $\displaystyle=(q_{u}kB_{(\cdot)}(k)\cdot
lB_{(\cdot)}(l)u)(x),$ (6.27a) $\displaystyle Q_{u}(x,y,k,l)$
$\displaystyle=(q_{u}kB_{(\cdot)}(k)\cdot lB_{(\cdot)}(l)q_{u})(x,y),$ (6.27b)

where we note that $kB_{x}(k)\cdot lB_{x}(l)$ acts as a multiplication
operator in $x$. Recalling the definition
$\big{(}L_{\alpha}(k)f\big{)}(x)=2kB_{x}(k)\big{(}(-i\nabla+F_{\alpha}(x))f\big{)}(x)$
from (4.2a), we set

$\displaystyle\ell_{t}^{(1)}(x,k)$
$\displaystyle=(q_{u_{t}}L_{\alpha_{t}}(k)u_{t})(x),\quad\ell_{t}^{(2)}(x,k)=(q_{u_{t}}L_{\alpha_{t}}(k)^{*}u_{t})(x).$
(6.28)

Moreover, for $u\in L^{2}(\mathbb{R}^{3})$ and $\\{u\\}^{\perp}\subset
L^{2}(\mathbb{R}^{3})$, we consider the operators

$\displaystyle K_{u}^{(3)}$
$\displaystyle:\\{u\\}^{\perp}\rightarrow\\{u\\}^{\perp}\otimes\\{u\\}^{\perp}$
$\displaystyle\psi$
$\displaystyle\mapsto(K_{u}^{(3)}\psi)(x_{1},x_{2})=(q_{u})_{1}(q_{u})_{2}W_{u}(x_{1},x_{2})u(x_{1})(q_{u}\psi)(x_{2}),$
(6.29a) $\displaystyle K_{u}^{(4)}$
$\displaystyle:\\{u\\}^{\perp}\otimes\\{u\\}^{\perp}\rightarrow\\{u\\}^{\perp}\otimes\\{u\\}^{\perp}$
$\displaystyle\psi$
$\displaystyle\mapsto(K_{u}^{(4)}\psi)(x_{1},x_{2})=(q_{u})_{1}(q_{u})_{2}W_{u}(x_{1},x_{2})(q_{u}\otimes
q_{u}\psi)(x_{1},x_{2}).$ (6.29b)

with, for $V$ defined in (3.2b),

$\displaystyle
W_{u}(x_{1},x_{2})=V(x_{1}-x_{2})-V\ast|u|^{2}(x_{1})-V\ast|u|^{2}(x_{2})+\big{\langle}u,V\ast|u|^{2}u\big{\rangle}.$
(6.30)

######  Lemma 6.7.

Let $(u,\alpha)\in H^{3}(\mathbb{R}^{3})\oplus\mathfrak{h}_{5/2}$ with
$\|u\|_{L^{2}}=1$, denote by $(u_{t},\alpha_{t})=\mathfrak{s}^{\rm
D}[t](u,\alpha)$ the solution to (3.3), and
$\rho(t)=\|u_{t}\|_{H^{3}}^{2}\,(1+\|\alpha_{t}\|_{\mathfrak{h}_{3/2}})^{2}$.
There exists a constant $C>0$ such that for all $|t|\geq 0$

$\displaystyle\|K_{u_{t}}^{(2)}\|_{L^{2}(\mathbb{R}^{6})}$ $\displaystyle\leq
C\hskip 56.9055pt\|\dot{K}_{u_{t}}^{(2)}\|_{L^{2}(\mathbb{R}^{6})}$
$\displaystyle\leq C\sqrt{\rho(t)}$ (6.31a)
$\displaystyle\|K_{u_{t}}^{(1)}\|_{L^{2}\to L^{2}}$ $\displaystyle\leq C\hskip
56.9055pt\|\dot{K}_{u_{t}}^{(1)}\|_{L^{2}\to L^{2}}$ $\displaystyle\leq
C\sqrt{\rho(t)}$ (6.31b)
$\displaystyle\|\ell_{t}^{(1)}\|_{L^{2}\otimes\mathfrak{h}_{-1/4}}$
$\displaystyle\leq C\hskip
56.9055pt\|\dot{\ell}_{t}^{(1)}\|_{L^{2}\otimes\mathfrak{h}_{-1/4}}$
$\displaystyle\leq C\sqrt{\rho(t)}$ (6.31c)
$\displaystyle\|\ell_{t}^{(2)}\|_{\mathfrak{h}_{1/4}\to L^{2}}^{2}$
$\displaystyle\leq C\hskip
56.9055pt\|\dot{\ell}_{t}^{(2)}\|_{\mathfrak{h}_{1/4}\to L^{2}}$
$\displaystyle\leq C\sqrt{\rho(t)}$ (6.31d)
$\displaystyle\|M_{u_{t}}\|_{(L^{2})^{\otimes 2}}$ $\displaystyle\leq C\hskip
56.9055pt\|\dot{M}_{u_{t}}\|_{(L^{2})^{\otimes 2}}$ $\displaystyle\leq
C\sqrt{\rho(t)}$ (6.31e) $\displaystyle\|K_{u_{t}}^{(3)}\|_{L^{2}\to
L^{2}\otimes L^{2}}$ $\displaystyle\leq C\hskip
56.9055pt\|\dot{K}_{u_{t}}^{(3)}\|_{L^{2}\to L^{2}\otimes L^{2}}$
$\displaystyle\leq C\sqrt{\rho(t)}$ (6.31f)
$\displaystyle\|N_{u_{t}}\|_{L^{2}\otimes\mathfrak{h}_{-1/8}^{\otimes 2}}$
$\displaystyle\leq C\hskip
56.9055pt\|\dot{N}_{u_{t}}\|_{L^{2}\otimes\mathfrak{h}_{-1/8}^{\otimes 2}}$
$\displaystyle\leq C\sqrt{\rho(t)}$ (6.31g)
$\displaystyle\|Q_{u_{t}}\|_{L^{2}\otimes\mathfrak{h}_{1/8}^{\otimes 2}\to
L^{2}}$ $\displaystyle\leq C\hskip
56.9055pt\|\dot{Q}_{u_{t}}\|_{L^{2}\otimes\mathfrak{h}_{1/8}^{\otimes 2}\to
L^{2}}$ $\displaystyle\leq C\sqrt{\rho(t)}$ (6.31h)

where $L^{2}$ stands for $L^{2}(\mathbb{R}^{3})$.

Let us note the evident fact that the bounds (6.31a) – (6.31e) hold uniformly
also for the $\Lambda$-dependent kernels introduced in (5.10). For instance,
$\|K^{(2),\Lambda}_{1,u_{t}}\|_{L^{2}(\mathbb{R}^{6})}\leq C$ for all
$\Lambda\in\mathbb{R}_{+}\cup\\{\infty\\}$ with $K^{(2),\Lambda}_{1,u_{t}}$
defined by (5.11e). While we do not state them explicitly here, such uniform
bounds will be used in the proofs of Lemmas 6.8 and 6.9.

###### Proof.

Recall that $\|u_{t}\|_{L^{2}}=1$ and
$\|u_{t}\|_{H_{1}}+\|\alpha_{t}\|_{\mathfrak{h}_{1/2}}\leq C$ for all $|t|\geq
0$ by Proposition 2.2 and Lemma 3.3. We go through the claimed bounds line-by-
line.

Line (6.31a). We use that $\|q_{u_{t}}\|=1$, so
$\|K_{u_{t}}^{(2)}\|^{2}_{L^{2}(\mathbb{R}^{6})}\leq\|\widetilde{K}_{u_{t}}^{(2)}\|_{L^{2}(\mathbb{R}^{6})}$
and

$\displaystyle\|\widetilde{K}_{u_{t}}^{(2)}\|^{2}_{L^{2}(\mathbb{R}^{6})}$
$\displaystyle=\int dxdy\,|u_{t}(x)|^{2}V^{2}(x-y)|u_{t}(y)|^{2}$
$\displaystyle\leq\|V^{2}\ast|u_{t}|^{2}\|_{L^{\infty}}\|u_{t}\|_{L^{2}}^{2}\leq
C$ (6.32)

by Lemma 6.4. Invoking

$\displaystyle\dot{K}_{u_{t}}^{(2)}$ $\displaystyle=\dot{q}_{u_{t}}\otimes
q_{u_{t}}\widetilde{K}_{u_{t}}^{(2)}+q_{u_{t}}\otimes\dot{q}_{u_{t}}\widetilde{K}_{u_{t}}^{(2)}+q_{u_{t}}\otimes
q_{u_{t}}\tfrac{d}{dt}{\widetilde{K}}_{u_{t}}^{(2)}$ (6.33)

together with $\dot{q}_{u_{t}}=-\dot{p}_{u_{t}}$,
$\|\dot{p}_{u_{t}}\|_{L^{2}\rightarrow L^{2}}\leq 2\|\dot{u}_{t}\|_{{L^{2}}}$
and $\|\tfrac{d}{dt}\widetilde{K}_{u_{t}}^{(2)}\|_{L^{2}(\mathbb{R}^{6})}\leq
C\ \|\dot{u}_{t}\|_{{L^{2}}}$ as in (6.32), we obtain

$\displaystyle\|\dot{K}_{u_{t}}^{(2)}\|_{L^{2}(\mathbb{R}^{6})}\leq
C\|\dot{u}_{t}\|_{{L^{2}}}\leq C\sqrt{\rho(t)},$ (6.34)

where we employed Lemma 6.6 in the last step.

Line (6.31b). Since $|K_{u_{t}}^{(1)}(x,y)|=|K^{(2)}_{u_{t}}(x,y)|$ we can use
that the operator norm is bounded by the Hilbert–Schmidt norm,
$\|K_{u_{t}}^{(1)}\|_{L^{2}\to
L^{2}}\leq\|K_{u_{t}}^{(2)}\|_{L^{2}(\mathbb{R}^{6})}$, so that we can apply
the previous bounds. The time-derivative is bounded analogously.

Line (6.31c). We recall
$\ell_{t}^{(1)}(x,k)=(q_{u_{t}}kB_{(\cdot)}(k)\cdot(-i\nabla+F_{\alpha_{t}})u_{t})(x)$
and estimate

$\displaystyle\|\ell_{t}^{(1)}\|_{L^{2}\otimes\mathfrak{h}_{-1/4}}$
$\displaystyle\leq
2\|kB_{0}\|_{\mathfrak{h}_{-1/4}}\|(-i\nabla+F_{\alpha_{t}})u_{t}\|_{{L^{2}}}$
$\displaystyle\leq C(\|u_{t}\|_{H^{1}}+\|F_{\alpha_{t}}\|_{L^{\infty}})\leq
C,$ (6.35)

and similarly for

$\displaystyle\|\dot{\ell}_{t}^{(1)}\|_{L^{2}\otimes\mathfrak{h}_{-1/4}}$
$\displaystyle\leq
2\|kB_{0}\|_{\mathfrak{h}_{-1/4}}\|\tfrac{d}{dt}q_{u_{t}}e^{-ik(\cdot)}(-i\nabla+F_{\alpha_{t}})u_{t}\|_{L^{2}}$
(6.36) $\displaystyle\leq
C(\|\dot{u}_{t}\|_{H^{1}}+\|F_{\dot{\alpha}_{t}}\|_{L^{\infty}})\leq
C(\|\dot{u}_{t}\|_{H^{1}}+\|\dot{\alpha}_{t}\|_{\mathfrak{h}_{1/2}})\leq
C\sqrt{\rho(t)},$

where we used $\|\dot{q}_{u_{t}}\|_{L^{2}\to L^{2}}\leq
2\|\dot{u}_{t}\|_{L^{2}}\leq 2\|\dot{u}_{t}\|_{H^{1}}$ and
$\|F_{\dot{\alpha}_{t}}\|_{L^{\infty}}\leq
C\|\dot{\alpha}_{t}\|_{\mathfrak{h}_{1/2}}$.

Line (6.31d). We have

$\ell_{t}^{(2)}(x,k)=\ell^{(1)}_{t}(x,k)+2(q_{u_{t}}k^{2}B_{(\cdot)}(k)u_{t}\big{)}(x),$
(6.37)

and thus

$\displaystyle\|\ell^{(2)}_{t}\|_{\mathfrak{h}_{1/4}\to L^{2}}$
$\displaystyle\leq\|\ell^{(1)}\|_{\mathfrak{h}_{1/4}\to
L^{2}}+2\sup_{\|\eta\|_{\mathfrak{h}_{1/4}}=1}\Big{\|}\int
dk\,k^{2}B_{(\cdot)}(k)\eta(k)u_{t}\Big{\|}_{L^{2}}$
$\displaystyle\leq\|\ell^{(1)}\|_{L^{2}\otimes\mathfrak{h}_{-1/4}}+C\|u_{t}\|_{H^{1}}$
(6.38)

by (6.4d) (with $\alpha=\eta$, $u=u_{t}$, $n=2$ and $s=1/4$). Similarly, for
the time-derivative

$\displaystyle\|\dot{\ell}_{t}^{(2)}\|_{\mathfrak{h}_{1/4}\to L^{2}}$
$\displaystyle\leq\|\dot{\ell}^{(1)}\|_{L^{2}\otimes\mathfrak{h}_{-1/4}}+C\|\dot{u}_{t}\|_{H^{1}}\leq
C\sqrt{\rho(t)}.$ (6.39)

Line (6.31e). The estimate

$\displaystyle\left|M_{u_{t}}(k,l)\right|$
$\displaystyle=\left|\frac{kB_{0}(k)\cdot
lB_{0}(l)(2\pi)^{3/2}(k+l)\cdot\mathcal{F}[i\nabla\left|u_{t}\right|^{2}](k+l)}{\left|k+l\right|^{2}}\right|$
$\displaystyle\leq C\|u_{t}\nabla
u_{t}\|_{L^{1}}\frac{\left|k\right|B_{0}(k)\left|l\right|B_{0}(l)}{\left|k+l\right|}$
(6.40)

and the Hardy–Littlewood–Sobolev inequality imply

$\displaystyle\|M_{u_{t}}\|_{L^{2}(\mathbb{R}^{6})}$ $\displaystyle\leq
C\|u_{t}\|_{H^{1}}^{2}\|\left|\cdot\right|B_{0}\|_{L^{3}}^{2}\leq
C\|u_{t}\|_{H^{1}}^{2}.$ (6.41)

For $\dot{M}_{u_{t}}$ we obtain by the same argument

$\|\dot{M}_{u_{t}}\|_{L^{2}(\mathbb{R}^{6})}\leq
C\|u_{t}\|_{H^{1}}\|\dot{u}_{t}\|_{H^{1}},$ (6.42)

which implies the claim by Lemma 6.6.

Line (6.31f). Recall the definition of $W_{u}$ in (6.30), and denote
$q_{i}=(q_{u_{t}})_{i}$. Using $\|u_{t}\|_{H^{1}}\leq C$ we get

$\displaystyle\|K^{(3)}_{u_{t}}\|_{L^{2}\to L^{2}\otimes L^{2}}$
$\displaystyle=\sup_{\|\psi\|=1}\|q_{1}q_{2}W_{u_{t}}(x_{1},x_{2})u_{t}(x_{1})(q_{u_{t}}\psi)(x_{2})\|_{L^{2}\otimes
L^{2}}$ $\displaystyle\leq C\|q_{u_{t}}\|_{L^{2}\to
L^{2}}^{3}\big{(}\|V^{2}\ast|u_{t}|^{2}\|_{L^{\infty}}^{1/2}+\|V\ast|u_{t}|^{2}\|_{L^{\infty}}\big{)}\leq
C$ (6.43)

by Lemma 6.4. For the norm of the time-derivative, one computes

$\displaystyle(\dot{K}^{(3)}_{u_{t}}\psi)(x_{1},x_{2})=q_{1}q_{2}\dot{W}_{u_{t}}(x_{1},x_{2})u_{t}(x_{1})(q\psi)(x_{2})$
$\displaystyle\qquad\qquad+\dot{q}_{1}q_{2}W_{u_{t}}(x_{1},x_{2})u_{t}(x_{1})(q\psi)(x_{2})+q_{1}\dot{q}_{2}W_{u_{t}}(x_{1},x_{2})u_{t}(x_{1})(q\psi)(x_{2})$
$\displaystyle\qquad\qquad+q_{1}q_{2}W_{u_{t}}(x_{1},x_{2})\dot{u}_{t}(x_{1})(q\psi)(x_{2})+q_{1}q_{2}W_{u_{t}}(x_{1},x_{2})u_{t}(x_{1})(\dot{q}\psi)(x_{2})$
(6.44)

where each term can be estimated similarly as in (6.2). Using Lemma 6.4 in

$\displaystyle\|\tfrac{d}{dt}V\ast\left|u_{t}\right|^{2}\|_{L^{\infty}}\leq
2\|\dot{u}_{t}\|_{L^{2}}\|V^{2}\ast|u_{t}|^{2}\|_{L^{\infty}}^{1/2}\leq
C\|\dot{u}_{t}\|_{{L^{2}}},$ (6.45)

one obtains $\|\dot{W}_{u_{t}}\|_{L^{\infty}(\mathbb{R}^{6})}\leq
C\|\dot{u}_{t}\|_{{L^{2}}}$. Together with $\|\dot{q}_{u_{t}}\|_{L^{2}\to
L^{2}}\leq 2\|\dot{u}_{t}\|_{L^{2}}$, this leads to
$\|\dot{K}^{(3)}_{u_{t}}\|_{L^{2}\to L^{2}\otimes L^{2}}\leq C\sqrt{\rho(t)}$.

Lines (6.31g) and (6.31h). Recalling the definitions of $N_{t},Q_{t}$ in
(6.27a), (6.27b) it follows readily that

$\displaystyle\|N_{u_{t}}\|_{L^{2}\otimes\mathfrak{h}_{-1/8}^{\otimes
2}}+\|Q_{u_{t}}\|_{L^{2}\otimes\mathfrak{h}_{1/8}^{\otimes 2}\to L^{2}}$
$\displaystyle\leq C\|kB_{0}\|_{\mathfrak{h}_{-1/8}}^{2}\leq C$ (6.46a)
$\displaystyle\|\dot{N}_{u_{t}}\|_{L^{2}\otimes\mathfrak{h}_{-1/8}^{\otimes
2}}+\|\dot{Q}_{u_{t}}\|_{L^{2}\otimes\mathfrak{h}_{1/8}^{\otimes 2}\to L^{2}}$
$\displaystyle\leq
C\|kB_{0}\|_{\mathfrak{h}_{-1/8}}^{2}\|\dot{u}_{t}\|_{L^{2}}\leq
C\sqrt{\rho(t)}.$ (6.46b)

This completes the proof of the lemma. ∎

###  6.3. Estimates for the Bogoliubov Hamiltonians

The first Lemma provides bounds on the dressed Bogoliubov Hamiltonian (4.1)
and its time-derivative, its difference to the operator
$\mathbb{T}=\text{d}\Gamma_{b}(-\Delta)+\text{d}\Gamma_{a}(\omega)$, and its
commutator with the total number operator
$\mathcal{N}=\mathcal{N}_{b}+\mathcal{N}_{a}$. These imply existence and
uniqueness of the associated dynamics, as explained in Proposition 5.2.
Similar bounds also hold for the family of interpolating Bogoliubov
Hamiltonians (5.10) (recall that $\mathbb{H}_{u,\alpha}^{\rm
D}(t)=\mathbb{H}_{u,\alpha,1}^{\infty}(t)$). Note that for $\theta=1$ the
bounds in part (b) are uniform in $\Lambda$.

######  Lemma 6.8.

(a) Let $(u,\alpha)\in H^{3}(\mathbb{R}^{3})\oplus\mathfrak{h}_{5/2}$ with
$\|u\|_{L^{2}}=1$ and $(u_{t},\alpha_{t})=\mathfrak{s}^{\rm D}[t](u,\alpha)$
denote the solution to (3.3). There exists a constant $C>0$ such that for all
$t\in\mathbb{R}$

$\displaystyle\pm\big{(}\mathbb{H}_{u,\alpha}^{\rm D}(t)-\mathbb{T}\big{)}$
$\displaystyle\leq\tfrac{1}{2}\mathbb{T}+C(\mathcal{N}+1)$ (6.47a)
$\displaystyle\pm i[\mathcal{N},\mathbb{H}_{u,\alpha}^{\rm D}(t)]$
$\displaystyle\leq\tfrac{1}{2}\mathbb{T}+C(\mathcal{N}+1)$ (6.47b)
$\displaystyle\pm\tfrac{d}{dt}\mathbb{H}_{u,\alpha}^{\rm D}(t)$
$\displaystyle\leq\tfrac{1}{2}\mathbb{T}+C\rho(t)(\mathcal{N}+1)$ (6.47c)

as quadratic forms on $\mathcal{F}\otimes\mathcal{F}$, where
$\rho(t)=\|u_{t}\|_{H^{3}}^{2}(1+\|\alpha_{t}\|_{\mathfrak{h}_{3/2}})^{2}$.

(b) Let $(u,\alpha)\in H^{3}(\mathbb{R}^{3})\oplus\mathfrak{h}_{5/2}$ with
$\|u\|_{L^{2}}=1$ and $(u_{t},\alpha_{t})=\mathfrak{s}_{\theta}[t](u,\alpha)$
as defined in (5.1). There exists a constant $C>0$ such that for all
$t\in\mathbb{R}$, $|\theta|\leq 1$ and $\Lambda\in\mathbb{R}_{+}$

$\displaystyle\pm\big{(}\mathbb{H}_{u,\alpha,\theta}^{\Lambda}(t)-\mathbb{T}\big{)}$
$\displaystyle\leq\tfrac{1}{2}\mathbb{T}+C(1+|1-\theta|\Lambda)(\mathcal{N}+1)$
(6.48a) $\displaystyle\pm
i[\mathcal{N},\mathbb{H}_{u,\alpha,\theta}^{\Lambda}(t)]$
$\displaystyle\leq\tfrac{1}{2}\mathbb{T}+C(1+|1-\theta|\Lambda)(\mathcal{N}+1)$
(6.48b)
$\displaystyle\pm\tfrac{d}{dt}\mathbb{H}_{u,\alpha,\theta}^{\Lambda}(t)$
$\displaystyle\leq\tfrac{1}{2}\mathbb{T}+C(1+|1-\theta|\Lambda)\rho(t)(\mathcal{N}+1)$
(6.48c)

as quadratic forms on $\mathcal{F}\otimes\mathcal{F}$, where
$\rho(t)=\|u_{t}\|_{H^{3}}^{2}(1+\|\alpha_{t}\|_{\mathfrak{h}_{3/2}})^{2}$.

###### Proof.

The proof follows essentially by combining the operator bounds of Lemma 6.1
with the kernel bounds of Lemma 6.7. We give the details below.

Proof of (6.47a). Recall that

$\displaystyle\mathbb{H}^{\rm D}_{u,\alpha}(t)-\mathbb{T}$
$\displaystyle=\textnormal{d}\Gamma_{b}\big{(}A_{\alpha_{t}}+F^{2}_{\alpha_{t}}-\mu_{u_{t},\alpha_{t}}\big{)}+\mathbb{K}_{u_{t}}^{(1)}+\big{(}\mathbb{K}_{u_{t}}^{(2)}+\text{h.c.}\big{)}$
$\displaystyle\quad+\bigg{(}\int
dkdx\Big{(}\ell_{t}^{(1)}(x,k)b_{x}^{*}a_{k}^{*}+\ell_{t}^{(2)}(x,k)b_{x}^{*}a_{k}\Big{)}+\text{h.c.}\bigg{)}$
$\displaystyle\quad+\int dkdl\,M_{u_{t}}(k,l)\,\mathcal{A}_{kl}.$ (6.49)

Since
$A_{\alpha_{t}}=2(-i\nabla_{x})\cdot\big{\langle}kB_{x},\alpha_{t}\big{\rangle}+\text{h.c.}$
the first term in the first line is bounded, using the Cauchy–Schwarz
inequality, by

$\displaystyle\pm\textnormal{d}\Gamma_{b}\big{(}A_{\alpha_{t}}+F^{2}_{\alpha_{t}}-\mu_{u_{t},\alpha_{t}}\big{)}$
$\displaystyle\hskip
14.22636pt\leq\varepsilon\text{d}\Gamma_{b}(-\Delta)+\Big{(}\frac{4}{\varepsilon}\big{\langle}|k|B_{0},|\alpha_{t}|\big{\rangle}^{2}+\|F_{\alpha_{t}}\|_{L^{\infty}}^{2}+|\mu_{u_{t},\alpha_{t}}|\Big{)}\mathcal{N}_{b}$
$\displaystyle\hskip
14.22636pt\leq\varepsilon\text{d}\Gamma_{b}(-\Delta)+\frac{C}{\varepsilon}\mathcal{N}_{b},$
(6.50)

where the last bound follows from Lemmas 6.3 and 6.4, and
$\|u_{t}\|_{H^{1}}+\|\alpha_{t}\|_{\mathfrak{h}_{1/2}}\leq C$.

For the second and third term in the first line, we apply Lemmas 6.1 and 6.7
to get

$\displaystyle\pm\mathbb{K}_{u_{t}}^{(1)}$
$\displaystyle\leq\|K_{u_{t}}^{(1)}\|_{L^{2}\to L^{2}}(\mathcal{N}_{b}+1)\leq
C(\mathcal{N}_{b}+1),$ (6.51a)
$\displaystyle\pm(\mathbb{K}_{u_{t}}^{(2)}+\text{h.c.})$ $\displaystyle\leq
2\|K_{u_{t}}^{(2)}\|_{L^{2}(\mathbb{R}^{6})}(\mathcal{N}_{b}+1)\leq
C(\mathcal{N}_{b}+1).$ (6.51b)

For the mixed quadratic terms, we use the Lemma 6.1 choosing $m_{a}=m_{b}=0$,
$n_{a}=n_{b}=1$, $s=1/2$ and $r_{a}=0$, $r_{b}=1$ for the term involving
$b_{x}^{*}a_{k}^{*}$, and $m_{a}=1=n_{b}$, $n_{a}=m_{b}=0$, $t=1/2$ and
$r_{a}=r_{b}=0$ for the term involving $b_{x}^{*}a_{k}$. Thus gives

$\displaystyle\pm\bigg{(}\int
dkdx\Big{(}\ell^{(1)}_{t}(x,k)b_{x}^{*}a_{k}^{*}+\ell^{(2)}_{t}(x,k)b_{x}^{*}a_{k}\Big{)}+\text{h.c.}\bigg{)}$
$\displaystyle\quad\quad\quad\leq\varepsilon\text{d}\Gamma_{a}(\omega)+C\varepsilon^{-1}\Big{(}\|\ell_{t}^{(1)}\|^{2}_{L^{2}\otimes\mathfrak{h}_{-1/2}}+\|\ell_{t}^{(2)}\|^{2}_{\mathfrak{h}_{1/2}\to
L^{2}}\Big{)}(\mathcal{N}_{b}+1)$
$\displaystyle\quad\quad\quad\leq\varepsilon\text{d}\Gamma_{a}(\omega)+C\varepsilon^{-1}(\mathcal{N}_{b}+1),$
(6.52)

where we used again Lemma 6.7 and monotonicity of the
$\mathfrak{h}_{s}$-norms.

In the last term in (6.49) we have
$\mathcal{A}_{kl}=-2a_{k}^{*}a_{-l}+a_{k}^{*}a_{l}^{*}+a_{-l}a_{-k}$, so using
that the operator norm is bounded by the Hilbert-Schmidt norm, we have with
Lemma 6.1 and (6.31e)

$\displaystyle\pm\int dkdl\ M_{u_{t}}(k,l)\mathcal{A}_{k,l}$
$\displaystyle\leq C(\mathcal{N}_{a}+1).$ (6.53)

Proof of (6.47b). The commutator is easily found to be

$[\mathcal{N},\mathbb{H}^{\rm D}_{u,\alpha}(t)]=2\bigg{(}\int
dkdx\,\ell_{t}^{(1)}(x,k)b_{x}^{*}a_{k}^{*}-\text{h.c.}\bigg{)}\\\ +2\int
dkdl\,M_{u_{t}}(k,l)\,(a^{*}_{k}a_{l}^{*}-a_{-k}a_{-l}),$ (6.54)

which can be estimated exactly as in (6.3) and (6.53).

Proof of (6.47c). We compute

$\displaystyle\frac{d}{dt}\mathbb{H}^{\rm D}_{u,\alpha}(t)$
$\displaystyle=\textnormal{d}\Gamma_{b}(\dot{h}_{u_{t},\alpha_{t}})+\dot{\mathbb{K}}_{u_{t}}^{(1)}+(\dot{\mathbb{K}}_{u_{t}}^{(2)}+\text{h.c.})$
$\displaystyle\quad+\bigg{(}\int
dkdx\Big{(}\dot{\ell}^{(1)}_{t}(x,k)b_{x}^{*}a_{k}^{*}+\dot{\ell}^{(2)}_{t}(x,k)b_{x}^{*}a_{k}\Big{)}+\text{h.c.}\bigg{)}$
$\displaystyle\quad+\int dkdl\,\dot{M}_{u_{t}}(k,l)\,\mathcal{A}_{kl},$ (6.55)

with

$\displaystyle\dot{h}_{u_{t},\alpha_{t}}=A_{\dot{\alpha}_{t}}+2F_{\alpha_{t}}\cdot
F_{\dot{\alpha}_{t}}+\tfrac{d}{dt}V\ast\left|u_{t}\right|^{2}-\dot{\mu}_{u_{t},\alpha_{t}}.$
(6.56)

Since
$A_{\dot{\alpha}_{t}}=2(-i\nabla_{x})\cdot\big{\langle}kB_{x},\dot{\alpha}_{t}\big{\rangle}+\text{h.c.}$
we can use Cauchy–Schwarz and Lemma 6.3 to obtain

$\displaystyle\pm\textnormal{d}\Gamma_{b}(A_{\dot{\alpha}_{t}})\leq\varepsilon\text{d}\Gamma_{b}(-\Delta)+\frac{C}{\varepsilon}\|\dot{\alpha}_{t}\|_{\mathfrak{h}_{1/2}}^{2}\,\mathcal{N}_{b}.$
(6.57)

Recalling
$F_{\dot{\alpha}_{t}}(x)=2\mathrm{Re}\big{\langle}kB_{x},\dot{\alpha}_{t}\big{\rangle}$
we estimate

$\displaystyle\pm\textnormal{d}\Gamma_{b}(F_{\alpha_{t}}\cdot
F_{\dot{\alpha}_{t}})$
$\displaystyle\leq\|F_{\alpha_{t}}\|_{L^{\infty}}\|F_{\dot{\alpha}_{t}}\|_{L^{\infty}}\,\mathcal{N}_{b}\leq
C\|\dot{\alpha}_{t}\|_{\mathfrak{h}_{1/2}}\,\mathcal{N}_{b}.$ (6.58)

The time-derivative of the convolution is estimated in (6.45), and thus

$\displaystyle\pm\textnormal{d}\Gamma_{b}(\tfrac{d}{dt}V\ast|u_{t}|^{2})\leq
C\|\dot{u}_{t}\|_{L^{2}}\,\mathcal{N}_{b}.$ (6.59)

Combining the above estimates, we arrive at

$\displaystyle\pm\textnormal{d}\Gamma_{b}(\dot{h}_{u_{t},\alpha_{t}})$
$\displaystyle\leq\varepsilon\text{d}\Gamma_{b}(-\Delta)+\frac{C}{\varepsilon}\Big{(}\|\dot{\alpha}_{t}\|_{\mathfrak{h}_{1/2}}^{2}+\|\dot{u}_{t}\|_{{L^{2}}}\Big{)}\mathcal{N}_{b}.$
(6.60)

Similarly as in (6.51a) and (6.51b) we bound the remaining terms in the first
line in (6.3)

$\displaystyle\pm\dot{\mathbb{K}}_{u_{t}}^{(1)}$
$\displaystyle\leq\|\dot{K}_{u_{t}}^{(1)}\|_{L^{2}\rightarrow
L^{2}}\left(\mathcal{N}_{b}+1\right)\leq
C\sqrt{\rho(t)}\left(\mathcal{N}_{b}+1\right),$ (6.61a)
$\displaystyle\pm(\dot{\mathbb{K}}_{u_{t}}^{(2)}+\text{h.c.})$
$\displaystyle\leq
2\|\dot{K}_{u_{t}}^{(2)}\|_{L^{2}(\mathbb{R}^{6})}\left(\mathcal{N}_{b}+1\right)\leq
C\sqrt{\rho(t)}\left(\mathcal{N}_{b}+1\right).$ (6.61b)

By the reasoning of (6.3) we obtain

$\pm\bigg{(}\int
dkdx\Big{(}{\dot{\ell}_{t}^{(1)}}(x,k)b_{x}^{*}a_{k}^{*}+{{\dot{\ell}_{t}^{(2)}}(x,k)}b_{x}^{*}a_{k}\Big{)}+\text{h.c.}\bigg{)}\\\
\leq\varepsilon\text{d}\Gamma_{a}(\omega)+\frac{C\rho(t)}{\varepsilon}\left(\mathcal{N}_{b}+1\right),$
(6.62)

and analogously to (6.53) one also verifies that

$\displaystyle\pm\int dkdl\,\dot{M}_{u_{t}}(k,l)\,\mathcal{A}_{kl}$
$\displaystyle\leq C\rho(t)\,\left(\mathcal{N}_{a}+1\right).$ (6.63)

The above estimates prove (6.47c) and thus complete the proof of part (a) of
the lemma.

Turning to part (b), first note that for $\theta=1$ the proof is verbatim the
same as the one for part (a). To see this, recall that all kernels that appear
in $\mathbb{H}_{u,\alpha,1}^{\Lambda}(t)$ satisfy the same bounds as in Lemma
6.7, as explained thereafter (in particular, they are uniformly bounded in
$\Lambda$).

The crucial difference for $\theta\neq 1$, apart from the trivial
$\theta$-dependence of the kernels, is the appearance of the term

$\displaystyle(1-\theta)\int
dkdx\,\bigg{(}(q_{u_{t}}G_{x}^{\Lambda}(k)u_{t})a_{k}^{*}b_{x}^{*}+(q_{u_{t}}\overline{G_{x}^{\Lambda}(k)}u_{t})a_{k}^{*}b_{x}\bigg{)}+\text{h.c.}$
(6.64)

in $\mathbb{H}_{u,\alpha,\theta}^{\Lambda}$ which is not present when
$\theta=1$ (hence this term did not appear in the proof of part (a)). Since
$\|q_{u_{t}}G_{(\cdot)}^{\Lambda}u_{t}\|_{L^{2}\otimes
L^{2}}\leq\|G_{0}^{\Lambda}\|_{L^{2}}\leq C\Lambda$, we can bound the above
term by $C\Lambda(\mathcal{N}+1)$. All other contributions in
$\mathbb{H}_{u,\alpha,\theta}^{\Lambda}(t)$ are estimated as for $\theta=1$,
i.e. they are uniformly bounded in $\Lambda$. This explains the
$\Lambda$-dependent upper bound in (6.47a). The bounds for the commutator and
the time-derivative are obtained in the same way. ∎

The next lemma was used in the proof of Proposition 5.2 to show that
$\mathbb{U}_{1}^{\Lambda}(t)\to\mathbb{U}_{1}^{\infty}(t)$ strongly as
$\Lambda\to\infty$.

######  Lemma 6.9.

Let $(u,\alpha)\in H^{3}(\mathbb{R}^{3})\oplus\mathfrak{h}_{5/2}$ with
$\|u\|_{L^{2}}=1$, $(u_{t},\alpha_{t})=\mathfrak{s}_{1}[t](u,\alpha)$ as
defined in (5.1) and
$\rho(t)=\|u_{t}\|_{H^{3}}^{2}(1+\|\alpha_{t}\|_{\mathfrak{h}_{3/2}})^{2}$.
There is a family $\varepsilon_{\Lambda}>0$ with
$\varepsilon_{\Lambda}\xrightarrow{\Lambda\to\infty}0$ such that

$\big{|}\big{\langle}\chi,(\mathbb{H}_{u,\alpha,1}^{\infty}(t)-\mathbb{H}_{u,\alpha,1}^{\Lambda}(t))\phi\big{\rangle}\big{|}\\\
\leq\varepsilon_{\Lambda}e^{C\int_{0}^{|t|}\rho(s)ds}\|(\mathbb{T}+\mathcal{N}+1)^{1/2}\chi\|\|(\mathbb{T}+\mathcal{N}+1)^{1/2}\phi\|$

for all $t\in\mathbb{R}$ and $\chi,\phi\in D((\mathbb{T}+\mathcal{N})^{1/2})$.

###### Proof.

The difference
$\mathbb{H}_{u,\alpha,1}^{\infty}(t)-\mathbb{H}_{u,\alpha,1}^{\Lambda}(t)$ is
up to the term
$\textnormal{d}\Gamma_{b}(h_{u_{t},\alpha_{t}})+\textnormal{d}\Gamma_{a}(\omega)$
precisely of the same form as (5.10) with all kernels replaced by kernels of
the form $K_{u_{t},1}^{(1),\infty}-K_{u_{t},1}^{(1),\Lambda}$ and
$M^{\infty}_{u_{t},1}(k,l)-M_{u_{t},1}^{\Lambda}(k,l)$, and analogously for
the other terms. The claimed bound is now obtained following the same steps as
in the proof of (6.47a) and taking into account that, by Lemma 6.7 and
continuity in $\Lambda$, the norms of the kernel differences all vanish as
$\Lambda\to\infty$. ∎

###  6.4. Fluctuation generator for the dressed dynamics

We start by stating the precise form of the fluctuation generator $H^{\rm
D}_{u,\alpha}(t)$ introduced in (3.8). For $[a]_{+}=\max\\{0,a\\}$ and
$\mathfrak{s}^{\rm D}[t](u,\alpha)$ set

$\displaystyle H_{0}(t)$
$\displaystyle=\text{d}\Gamma_{a}(\omega)+\text{d}\Gamma_{b}(h_{u_{t},\alpha_{t}})+\big{[}1-\tfrac{\mathcal{N}_{b}}{N}\big{]}_{+}\mathbb{K}^{(1)}_{u_{t}}$
$\displaystyle\quad+\Big{(}\mathbb{K}^{(2)}_{u_{t}}\frac{\sqrt{[(N-\mathcal{N}_{b})(N-\mathcal{N}_{b}-1)]_{+}}}{N}+\text{h.c.}\Big{)}$
$\displaystyle\quad+\int
dkdx(q_{u_{t}}L_{\alpha_{t}}(k)u_{t})(x)\,b_{x}^{*}a_{k}^{*}\sqrt{[1-\tfrac{\mathcal{N}_{b}}{N}]_{+}}+\textnormal{h.c.}$
$\displaystyle\quad+\int
dkdx(q_{u_{t}}L_{\alpha_{t}}(k)^{*}u_{t})(x)b_{x}^{*}a_{k}\sqrt{[1-\tfrac{\mathcal{N}_{b}}{N}]_{+}}+\textnormal{h.c.}$
$\displaystyle\quad+\int
dkdl\,M_{u_{t}}(k,l)\big{[}1-\tfrac{\mathcal{N}_{b}}{N}\big{]}_{+}\,\big{(}-2a^{*}_{k}a_{-l}+a_{k}^{*}a_{l}^{*}+a_{-k}a_{-l}\big{)}$
(6.65)

with $L_{\alpha_{t}}(k)$, $M_{u_{t}}(k,l)$ and $\mathbb{K}^{(j)}_{u_{t}}$
defined in (4.2a), (3.4a) and (4.2c), respectively. Note that up to the
$N$-dependent factors $H_{0}(t)$ coincides with $\mathbb{H}_{u,\alpha}^{\rm
D}(t)$. We further introduce the operators

$\displaystyle\mathbb{K}^{(3)}_{u_{t}}$ $\displaystyle=\int
dx_{1}dx_{2}dx_{3}\,K_{u_{t}}^{(3)}(x_{1},x_{2},x_{3})b^{*}_{x_{1}}b^{*}_{x_{2}}b_{x_{3}}$
(6.66a) $\displaystyle\mathbb{K}^{(4)}_{u_{t}}$ $\displaystyle=\frac{1}{2}\int
dx_{1}dx_{2}dx_{3}dx_{4}\,K_{u_{t}}^{(4)}(x_{1},x_{2},x_{3},x_{4})b^{*}_{x_{1}}b^{*}_{x_{2}}b_{x_{3}}b_{x_{4}},$
(6.66b)

where for $u\in L^{2}(\mathbb{R}^{3})$ and $\\{u\\}^{\perp}\subset
L^{2}(\mathbb{R}^{3})$ we used the kernels of the operators $K_{u}^{(3)}$,
$K_{u}^{(4)}$ introduced in (6.29a), (6.29b). Lastly, recall (6.27a) and
(6.27b) and let

$\displaystyle J_{u,\alpha}(x,k,y)$
$\displaystyle=2(q_{u}kB_{(\cdot)}(k)\cdot(-i\nabla+F_{\alpha})q_{u})(x,y).$
(6.67)

The proof of the next lemma follows from a straightforward computation, which
is postponed to Appendix C.

######  Lemma 6.10.

For $(u,\alpha)\in H^{3}(\mathbb{R}^{3})\oplus\mathfrak{h}_{5/2}$ with
$\|u\|_{L^{2}}=1$ let
$(u_{t},\alpha_{t})=\mathfrak{s}^{\mathrm{D}}[t](u,\alpha)$ be the solution to
(3.3). The operator $H_{u,\alpha}^{{\rm D},\leq N}(t):\mathcal{F}^{\leq
N}_{\perp u_{t}}\otimes\mathcal{F}\rightarrow\mathcal{F}\otimes\mathcal{F}$
defined by (3.7) satisfies the identity $H^{{\rm D},\leq
N}_{u,\alpha}(t)=H_{u,\alpha}^{\rm D}(t)\restriction\mathcal{F}^{\leq
N}_{\perp u_{t}}\otimes\mathcal{F}$ where $H^{\rm
D}_{u,\alpha}(t):\mathcal{F}\otimes\mathcal{F}\rightarrow\mathcal{F}\otimes\mathcal{F}$
is given by

$\displaystyle H_{u,\alpha}^{\rm D}(t)$ $\displaystyle=\sum_{j=0}^{5}H_{j}(t)$
(6.68)

with $H_{0}(t)$ defined by (6.65), and

$\displaystyle H_{1}(t)$ $\displaystyle=\begin{aligned}
&-\frac{1}{N}\mathrm{d}\Gamma(V\ast|u_{t}|^{2}-\mu_{u_{t},\alpha_{t}})\\\
&\quad+\Big{(}\mathbb{K}^{(3)}_{u_{t}}\frac{\sqrt{[N-\mathcal{N}_{b}]_{+}}}{N}+\textnormal{h.c.}\Big{)}+\frac{1}{N}\mathbb{K}^{(4)}_{u_{t}},\end{aligned}$
(6.69a) $\displaystyle H_{2}(t)$
$\displaystyle=-\frac{1}{\sqrt{N}}\mathcal{N}_{b}\hat{\Phi}(f_{u_{t}}+g_{u_{t},\alpha_{t}}),$
(6.69b) $\displaystyle H_{3}(t)$ $\displaystyle=\frac{1}{\sqrt{N}}\int
dkdxdy\,J_{u_{t},\alpha_{t}}(x,k,y)\,b_{x}^{*}b_{y}\,a_{k}^{*}+\textnormal{h.c.},$
(6.69c) $\displaystyle H_{4}(t)$ $\displaystyle=\begin{aligned}
\frac{1}{\sqrt{N}}\int
dkdldx\Big{(}&N_{u_{t}}(x,k,l)\,b_{x}^{*}\big{[}1-\tfrac{\mathcal{N}_{b}}{N}\big{]}_{+}^{1/2}\\\
&+\overline{N_{u_{t}}(x,k,l)}\big{[}1-\tfrac{\mathcal{N}_{b}}{N}\big{]}_{+}^{1/2}b_{x}\Big{)}\mathcal{A}_{kl},\end{aligned}$
(6.69d) $\displaystyle H_{5}(t)$ $\displaystyle=\frac{1}{N}\int
dkdldxdy\,Q_{u_{t}}(x,y,k,l)\,b_{x}^{*}b_{y}\,\mathcal{A}_{kl},$ (6.69e)

where $\mathcal{A}_{kl}=-2a^{*}_{k}a_{-l}+a_{k}^{*}a_{l}^{*}+a_{-k}a_{-l}$.

The next lemma provides estimates for the fluctuation Hamiltonian that are an
important ingredient of the proof of Theorem 3.2.

######  Lemma 6.11.

Let $(u,\alpha)\in H^{3}(\mathbb{R}^{3})\oplus\mathfrak{h}_{5/2}$ with
$\|u\|_{L^{2}}=1$, $(u_{t},\alpha_{t})=\mathfrak{s}^{\rm D}[t](u,\alpha)$
denote the solution to (3.3). There exists a constant $C>0$ such that for all
$t\in\mathbb{R}$

$\displaystyle\pm\big{(}H^{\rm D}_{u,\alpha}(t)-\mathbb{T}\big{)}$
$\displaystyle\leq\tfrac{1}{2}\mathbb{T}+C(\mathcal{N}+1)(1+\tfrac{1}{N}\mathcal{N}_{b})^{2}$
(6.70a) $\displaystyle\pm i[\mathcal{N},H^{\rm D}_{u,\alpha}(t)]$
$\displaystyle\leq\tfrac{1}{2}\mathbb{T}+C(\mathcal{N}+1)(1+\tfrac{1}{N}\mathcal{N}_{b})^{2}$
(6.70b) $\displaystyle\pm\tfrac{d}{dt}H^{\rm D}_{u,\alpha}(t)$
$\displaystyle\leq\tfrac{1}{2}\mathbb{T}+C\rho(t)(\mathcal{N}+1)(1+\tfrac{1}{N}\mathcal{N}_{b})^{2}.$
(6.70c)

as quadratic forms on $\mathcal{F}\otimes\mathcal{F}$, where
$\rho(t)=\|u_{t}\|^{2}_{H^{3}}(1+\|\alpha_{t}\|_{\mathfrak{h}_{3/2}})^{2}$.

###### Proof of (6.70a).

Comparing (6.65) with (6.49) we see that $H_{0}(t)$ differs from
$\mathbb{H}^{\rm D}_{u,\alpha}(t)$ only by the factors

$\displaystyle 0\leq\big{[}1-\tfrac{\mathcal{N}_{b}}{N}\big{]}_{+}\leq 1,\quad
0\leq\tfrac{\sqrt{[(N-\mathcal{N}_{b})(N-\mathcal{N}_{b}-1)]_{+}}}{N}\leq 1$
(6.71)

and is thus estimated in analogy to the proof of Lemma 6.8.

For later purpose we consider $\big{\langle}\chi,H_{1}(t)\phi\big{\rangle}$
for $\chi,\phi\in\mathcal{F}\otimes\mathcal{F}$. The first two terms in
$H_{1}(t)$ are estimated by

$\displaystyle\big{|}\big{\langle}\chi,\frac{1}{N}\textnormal{d}\Gamma_{b}\big{(}V\ast|u_{t}|^{2}-\mu_{u_{t},\alpha_{t}}\big{)}\phi\big{\rangle}\big{|}\leq\frac{C}{N}\|\mathcal{N}_{b}^{1/2}\chi\|\,\|\mathcal{N}_{b}^{1/2}\phi\|,$
(6.72)

where we used Lemma 6.4, and

$\displaystyle\Big{|}\big{\langle}\chi,\Big{(}\mathbb{K}^{(3)}_{u_{t}}\tfrac{\sqrt{[N-\mathcal{N}_{b}]_{+}}}{N}+\text{h.c.}\Big{)}\phi\big{\rangle}\Big{|}$
$\displaystyle\quad\leq\tfrac{1}{\sqrt{N}}\|K^{(3)}_{u_{t}}\|_{L^{2}\to
L^{2}\otimes
L^{2}}\|(\mathcal{N}_{b}+1)^{1/2}\chi\|\,\|(\mathcal{N}_{b}+1)\phi\|$
$\displaystyle\quad\leq\tfrac{C}{\sqrt{N}}\|(\mathcal{N}_{b}+1)^{1/2}\chi\|\,\|(\mathcal{N}_{b}+1)\phi\|,$
(6.73)

by Lemmas 6.1 and 6.7.

To estimate the term involving $\mathbb{K}_{u_{t}}^{(4)}$ we recall (3.2b) and
write $W_{u_{t}}(x,y)=W^{{\rm
b}}_{u_{t}}(x,y)-4\mathrm{Re}\big{\langle}G_{x},B_{y}\big{\rangle}$, where
$W_{u_{t}}^{{\rm b}}(x,y)$ is point-wise bounded. Using the symmetry of
$\chi^{(n)},\phi^{(n)}\in\mathcal{F}^{(n)}\otimes\mathcal{F}$ in the $n$
particle coordinates, we find

$\big{\langle}\chi,\mathbb{K}_{u_{t}}^{(4)}\phi\big{\rangle}\\\
=\sum_{n=2}^{\infty}\frac{n(n-1)}{2}\big{\langle}q_{1}q_{2}\chi^{(n)},\big{(}W_{u_{t}}^{\rm{b}}(x_{1},x_{2})-4\mathrm{Re}\big{\langle}G_{x_{1}},B_{x_{2}}\big{\rangle}\big{)}q_{1}q_{2}\phi^{(n)}\big{\rangle}$
(6.74)

with

$\displaystyle\Big{|}\sum_{n=2}^{\infty}\frac{n(n-1)}{2}\big{\langle}q_{1}q_{2}\chi^{(n)},W_{u_{t}}^{\rm{b}}(x_{1},x_{2})q_{1}q_{2}\phi^{(n)}\big{\rangle}\Big{|}$
$\displaystyle\leq
C\|\mathcal{N}_{b}^{\frac{1}{2}}\chi\|\|\mathcal{N}_{b}^{\frac{3}{2}}\phi\|.$
(6.75)

We use the Cauchy–Schwarz inequality and the fact that

$e^{ikx}=(1+(-i\nabla_{1}-k)^{2})^{-1/2}e^{ikx}(1-\Delta_{1})^{1/2}$ (6.76)

to bound the remaining term by

$\displaystyle\Big{|}\sum_{n=2}^{\infty}\frac{n(n-1)}{2}\big{\langle}q_{1}q_{2}\chi^{(n)},\mathrm{Re}\big{\langle}G_{x_{1}},B_{x_{2}}\big{\rangle}q_{1}q_{2}\phi^{(n)}\big{\rangle}\Big{|}$
(6.77) $\displaystyle\leq
C\Big{(}\sum_{n=0}^{\infty}n^{3}\|\phi^{(n)}\|^{2}\Big{)}^{1/2}\Big{(}\sum_{n=0}^{\infty}n\|\mathrm{Re}\big{\langle}G_{x_{1}},B_{x_{2}}\big{\rangle}q_{1}q_{2}\chi^{(n)}\|^{2}\Big{)}^{1/2}$
$\displaystyle\leq C\|\mathcal{N}_{b}^{3/2}\phi\|$ $\displaystyle\
\times\Big{(}\sum_{n=0}^{\infty}n\Big{\|}\int
dk(1+(-i\nabla_{1}-k)^{2})^{-\frac{1}{2}}\overline{G_{x_{1}}(k)}B_{x_{2}}(k)(1-\Delta_{1})^{\frac{1}{2}}q_{1}q_{2}\chi^{(n)}\Big{\|}^{2}\Big{)}^{\frac{1}{2}}.$

The last factor is bounded by

$\sup_{p\in\mathbb{R}^{3}}\left\\{\int
dk\,\frac{G_{0}(k)B_{0}(k)}{(1+(p-k)^{2})^{1/2}}\right\\}\left(\sum_{n=0}^{\infty}n\|\left(1-\Delta_{1}\right)^{1/2}q_{1}\phi^{(n)}\|^{2}\right)^{1/2}\\\
\leq C\|\textnormal{d}\Gamma_{b}(1-\Delta)^{1/2}\chi\|,$ (6.78)

where the supremum over $p\in\mathbb{R}^{3}$ is finite by the same argument as
in (6.6), and where we further used

$\displaystyle\|\left(1-\Delta_{1}\right)^{1/2}q_{1}\chi^{(n)}(t)\|\leq
C\|u_{t}\|_{H^{1}}\|\left(1-\Delta_{1}\right)^{1/2}\chi^{(n)}(t)\|.$ (6.79)

Adding up the relevant terms, we arrive at the desired estimate

$\big{|}\big{\langle}\chi,H_{1}(t)\phi\big{\rangle}\big{|}\\\ \leq
C\|\left(\mathcal{N}+\mathbb{T}+1\right)^{1/2}\chi\|\Big{(}\tfrac{1}{\sqrt{N}}\|(\mathcal{N}_{b}+1)\phi\|+\tfrac{1}{N}\|(\mathcal{N}_{b}+1)\mathcal{N}_{b}^{1/2}\phi\|\Big{)}.$
(6.80)

The bound for $H_{2}(t)$ is straightforward,

$\displaystyle\big{|}\big{\langle}\chi,H_{2}(t)\phi\big{\rangle}\big{|}\leq
C\tfrac{1}{\sqrt{N}}\|(\mathcal{N}+1)^{1/2}\chi\|\,\|(\mathcal{N}+1)\phi\|,$
(6.81)

where we used that $\|f_{u_{t}}+g_{u_{t},\alpha_{t}}\|_{L^{2}}\leq C$ by Lemma
6.7.

With the definition of the kernel of $H_{3}(t)$ in (6.67), we write

$\displaystyle\big{\langle}\chi,H_{3}(t)\chi\big{\rangle}$ (6.82)
$\displaystyle\ =\sum_{n=1}^{\infty}\frac{2n}{\sqrt{N}}\int dk\,kB_{0}(k)\cdot
2\mathrm{Re}\big{\langle}q_{1}\chi^{(n)},e^{-ikx_{1}}(-i\nabla_{1}+F_{\alpha_{t}}(x_{1}))q_{1}\,a_{k}^{*}\chi^{(n)}\big{\rangle}$

with $\chi^{(n)}\in\mathcal{F}^{(n)}\otimes\mathcal{F}$. With
$\|F_{\alpha_{t}}\|_{L^{\infty}}\leq C$ and
$\|kB_{0}\|_{\mathfrak{h}_{-1/4}}\leq C$, we obtain by Cauchy–Schwarz

$\displaystyle\big{|}\big{\langle}\chi,H_{3}(t)\chi\big{\rangle}\big{|}\leq\sum_{n=1}^{\infty}\frac{4n}{\sqrt{N}}\|(-i\nabla_{1}+F_{\alpha_{t}}(x_{1}))q_{1}\chi^{(n)}\|\int
dk\,|kB_{0}(k)|\,\|a_{k}\chi^{(n)}\|$
$\displaystyle\quad\stackrel{{\scriptstyle\eqref{eq:nabla:q:Psi}}}{{\leq}}C\sum_{n=1}^{\infty}\frac{4n}{\sqrt{N}}\|(1-\Delta_{1})^{1/2}\chi^{(n)}\|\|\text{d}\Gamma_{a}(\sqrt{\omega})^{1/2}\chi^{(n)}\|$
$\displaystyle\quad\stackrel{{\scriptstyle\hphantom{\eqref{eq:bound:dG:sqrt:omega}}}}{{\leq}}\frac{C}{\sqrt{N}}\|\textnormal{d}\Gamma_{b}(1-\Delta)^{1/2}\chi\|\,\|\mathcal{N}_{b}^{1/2}\textnormal{d}\Gamma_{a}(\sqrt{\omega})^{1/2}\chi\|$
$\displaystyle\quad\stackrel{{\scriptstyle\eqref{eq:bound:dG:sqrt:omega}}}{{\leq}}\frac{C}{\sqrt{N}}\|\textnormal{d}\Gamma_{b}(1-\Delta)^{1/2}\chi\|\,\|d\Gamma_{a}(\omega)^{1/2}\chi\|^{1/2}\,\|\mathcal{N}_{a}^{1/2}\mathcal{N}_{b}\chi\|^{1/2},$
(6.83)

which implies the desired bound.

To bound $H_{4}(t)$, we apply Lemma 6.1 (for instance for the term involving
$b_{x}^{*}a_{k}^{*}a_{-l}$ we choose $m_{b}=0,n_{b}=m_{a}=n_{a}=1$, $s=t=1/4$
and $r_{a}=r_{b}=0$) and use from Lemma 6.7 that
$\|N_{t}\|_{\mathfrak{h}_{1/4}\to
L^{2}\otimes\mathfrak{h}_{-1/4}}+\|N_{t}\|_{L^{2}\otimes\mathfrak{h}_{1/4}^{\otimes
2}\to\mathbb{C}}\leq 2\|N_{t}\|_{L^{2}\to\mathfrak{h}_{-1/4}^{\otimes 2}}\leq
2\|N_{t}\|_{L^{2}\otimes\mathfrak{h}_{-1/8}^{\otimes 2}}\leq C$. This implies

$\displaystyle\big{|}\big{\langle}\chi,H_{4}(t)\chi\big{\rangle}\big{|}$
$\displaystyle\leq
C\Big{[}\|(\mathcal{N}_{b}+1)^{1/2}\text{d}\Gamma_{a}(\sqrt{\omega})^{1/2}\chi\|\|\text{d}\Gamma_{a}(\sqrt{\omega})^{1/2}\chi\|$
(6.84)
$\displaystyle\qquad+\|(1+\mathcal{N}_{b})^{\tfrac{1}{2}}(1+\mathcal{N}_{a})^{\tfrac{1}{2}}\chi\|\|(1+\mathcal{N}_{a})^{-\tfrac{1}{2}}\text{d}\Gamma_{a}(\sqrt{\omega})\chi\|$
$\displaystyle\qquad+\|(\mathcal{N}_{b}+1)^{1/2}(\mathcal{N}_{a}+1)^{1/2}\chi\|\|(\mathcal{N}_{a}+1)^{-1/2}\text{d}\Gamma_{a}(\sqrt{\omega})\chi\|\Big{]}.$

The desired bound now follows easily from (6.3a). For $H_{5}(t)$, we use that
$\|Q_{t}\|_{L^{2}\otimes\mathfrak{h}_{1/4}\to
L^{2}\otimes\mathfrak{h}_{-1/4}}+\|Q_{t}\|_{L^{2}\otimes\mathfrak{h}_{1/4}^{\otimes
2}\to L^{2}}\leq C\|Q_{t}\|_{L^{2}\otimes\mathfrak{h}_{1/8}^{\otimes 2}\to
L^{2}}\leq C$ by Lemma 6.7, and Lemma 6.1 imply

$\displaystyle\big{|}\big{\langle}\chi,H_{5}(t)\chi\big{\rangle}\big{|}$
$\displaystyle\leq\frac{C}{N}\Big{[}\|(\mathcal{N}_{b}+1)^{1/2}\text{d}\Gamma_{a}(\sqrt{\omega})^{1/2}\chi\|^{2}$
(6.85)
$\displaystyle\qquad+\|(\mathcal{N}_{b}+1)(\mathcal{N}_{a}+1)^{1/2}\chi\|\|(\mathcal{N}_{a}+1)^{-1/2}\text{d}\Gamma_{a}(\sqrt{\omega})\chi\|\Big{]}.$

The desired bound now follows from Lemma 6.2.

Proof of (6.70b). The commutation relations imply, e.g.,
$[\mathcal{N},a_{k}]=-a_{k}$, so the non-zero terms in the commutator
$[\mathcal{N},H_{u,\alpha}^{\mathrm{D}}(t)]$ have the same kernels, up to
signs, as those in $H_{u,\alpha}^{\mathrm{D}}(t)-\mathbb{T}$. They can thus be
estimated as in the proof of (6.70a), and we omit the details.

Proof of (6.70c). This inequality is obtained following similar steps as in
the proof of (6.70a) with some obvious modifications, like the use of the
bounds for the time-derivatives in Lemma 6.7 and the use of
$\|\dot{q}_{u_{t}}\|_{H^{1}}\leq C\sqrt{\rho(t)}$, cf. Lemma. 6.6. ∎

The next lemma shows that the fluctuation generator can be approximated by the
Bogoliubov Hamiltonian $\mathbb{H}^{\rm D}_{u,\alpha}(t)$ for large $N$, when
tested on suitable states.

######  Lemma 6.12.

Let $(u,\alpha)\in H^{3}(\mathbb{R}^{3})\oplus\mathfrak{h}_{5/2}$ with
$\|u\|_{L^{2}}=1$ and $(u_{t},\alpha_{t})=\mathfrak{s}^{\rm D}[t](u,\alpha)$
denote the solution to (3.3). There exists a constant $C>0$ such that

$\big{|}\big{\langle}\chi,\left(H^{\rm
D}_{u,\alpha}(t)-\mathbb{H}_{u,\alpha}^{\rm
D}(t)\right)\phi\big{\rangle}\big{|}\\\ \leq C\rho(t)N^{-1/2}\ln
N\|\left(\mathcal{N}+\mathbb{T}+1\right)^{1/2}\chi\|\|\left(\mathcal{N}^{3}+\mathbb{T}+1\right)^{1/2}\phi\|$
(6.86)

for all $\chi\in\mathcal{F}^{\leq N}\otimes\mathcal{F}$ and
$\phi\in\mathcal{F}\otimes\mathcal{F}$, where
$\rho(t)=\|u_{t}\|^{2}_{H^{3}}(1+\|\alpha_{t}\|_{\mathfrak{h}_{3/2}})^{2}$.

###### Proof.

Recalling the definitions of the fluctuation generator (6.68) and the
Bogoliubov Hamiltonian (6.49), we write

$\displaystyle\big{\langle}\chi,(H^{\rm
D}_{u,\alpha}(t)-\mathbb{H}_{u,\alpha}^{\rm D}(t))\phi\big{\rangle}$
$\displaystyle=\big{\langle}\chi,\mathbb{K}^{(1)}_{u_{t}}\left(\big{[}1-\tfrac{\mathcal{N}_{b}}{N}\big{]}_{+}-1\right)\phi\big{\rangle}$
(6.87a) $\displaystyle\
+\big{\langle}\chi,\left(\mathbb{K}^{(2)}_{u_{t}}\big{(}N^{-1}\sqrt{[(N-\mathcal{N}_{b})(N-\mathcal{N}_{b}-1)]_{+}}-1\big{)}+\text{h.c.}\right)\phi\big{\rangle}$
(6.87b) $\displaystyle\ +\big{\langle}\chi,\Big{(}\int
dxdk\,\big{(}q_{u_{t}}L_{\alpha_{t}}(k)u_{t}\big{)}(x)a^{*}_{k}b_{x}^{*}\Big{(}\big{[}1-\tfrac{\mathcal{N}_{b}}{N}\big{]}_{+}^{\frac{1}{2}}-1\Big{)}+\text{h.c.}\Big{)}\phi\big{\rangle}$
(6.87c) $\displaystyle\ +\big{\langle}\chi,\Big{(}\int
dxdk\,\big{(}q_{u_{t}}L_{\alpha_{t}}(k)^{*}u_{t}\big{)}(x)a_{k}b_{x}^{*}\Big{(}\big{[}1-\tfrac{\mathcal{N}_{b}}{N}\big{]}_{+}^{\frac{1}{2}}-1\Big{)}+\text{h.c.}\Big{)}\phi\big{\rangle}$
(6.87d) $\displaystyle\ +\big{\langle}\chi,\int
dkdl\,M_{u_{t}}(k,l)\mathcal{A}_{kl}\Big{(}\big{[}1-\tfrac{\mathcal{N}_{b}}{N}\big{]}_{+}-1\Big{)}\phi\big{\rangle}$
(6.87e) $\displaystyle\
+\big{\langle}\chi,\left(H_{1}(t)+H_{2}(t)+H_{3}(t)+H_{4}(t)+H_{5}(t)\right)\phi\big{\rangle}$
(6.87f)

When bounding these terms, we have to take care to put any powers of the
number operator exceeding one-half to the right, i.e., on $\phi$. At the same
time, the power of $\mathcal{N}+\mathbb{T}$ acting on $\phi$ cannot exceed
one-half, either. For the terms in (6.87a)–(6.87e) the estimates are rather
straightforward and given at the end of the proof. The most difficult estimate
is that for the term coming from $H_{3}(t)$, which is also responsible for the
presence of the factor $\ln N$ in the statement.

Term $H_{3}(t)$. Recall the expression for the kernel
$J_{u,\alpha}=2(q_{u}kB_{(\cdot)}(k)\cdot(-i\nabla+F_{\alpha})q_{u})$, which
multiplies $b_{x}^{*}b_{y}a_{k}^{*}$ in $H_{3}(t)$. This term is problematic,
for when the gradient acts on $\phi$ we cannot put further powers of
$\mathcal{N}$ on $\phi$ while keeping control by
$\|\left(\mathcal{N}^{3}+\mathbb{T}\right)^{1/2}\phi\|$. We deal with this
problem by using the identity

$\displaystyle k\cdot
B_{x_{1}}(k)\left(-i\nabla_{1}+F_{\alpha_{t}}(x_{1})\right)a^{*}_{k}$
$\displaystyle=\left(-i\nabla_{1}+F_{\alpha_{t}}(x_{1})\right)\cdot
kB_{x_{1}}(k)+a^{*}_{k}k^{2}B_{x_{1}}(k)$ (6.88)

and splitting the momentum integration into $|k|\leq\Lambda$ and
$|k|>\Lambda$. Together with the adjoint expression, which is less of a
problem, this gives

$\displaystyle\tfrac{1}{2}\big{\langle}\chi,H_{3}(t)\phi\big{\rangle}$
$\displaystyle=\frac{1}{\sqrt{N}}\sum_{n=0}^{\infty}n\big{\langle}\left(-i\nabla_{1}+F_{\alpha_{t}}(x_{1})\right)q_{1}\chi^{(n)},\int
dk\,k\overline{B_{x_{1}}(k)}a_{k}q_{1}\phi^{(n)}\big{\rangle}$ (6.89a)
$\displaystyle\
+\frac{1}{\sqrt{N}}\sum_{n=0}^{\infty}n\big{\langle}\chi^{(n)},q_{1}\int\limits_{\left|k\right|\geq\Lambda}dk\,kB_{x_{1}}(k)a_{k}^{*}\left(-i\nabla_{1}+F_{\alpha_{t}}(x_{1})\right)q_{1}\phi^{(n)}\big{\rangle}$
(6.89b) $\displaystyle\
+\frac{1}{\sqrt{N}}\sum_{n=0}^{\infty}n\big{\langle}\left(-i\nabla_{1}+F_{\alpha_{t}}(x_{1})\right)q_{1}\chi^{(n)},\int\limits_{\left|k\right|\leq\Lambda}dk\,kB_{x_{1}}(k)a_{k}^{*}q_{1}\phi^{(n)}\big{\rangle}$
(6.89c) $\displaystyle\
+\frac{1}{\sqrt{N}}\sum_{n=0}^{\infty}n\big{\langle}\int_{\left|k\right|\leq\Lambda}dk\,k^{2}\overline{B_{x_{1}}(k)}a_{k}q_{1}\chi^{(n)},q_{1}\phi^{(n)}\big{\rangle}.$
(6.89d)

In the first line, the gradient acts on $\chi$, so we can simply bound it as
in (6.83),

$\displaystyle|\eqref{eq:Norm approximation Gross transformed dynamics 2f1}|$
$\displaystyle\leq
CN^{-1/2}\|\textnormal{d}\Gamma_{b}(1-\Delta)^{1/2}\chi\|\|\textnormal{d}\Gamma_{a}(\omega)^{1/2}\phi\|^{1/2}\|(\mathcal{N}+1)^{3/2}\phi\|^{1/2}.$
(6.90)

In the second line, we can use that $\chi^{(n)}=0$ for $n>N$ to remove a
factor of $(n/N)^{1/2}$, since the lower cutoff $\Lambda$ will give us a small
pre-factor. With the Cauchy-Schwarz inequality and Lemma 6.1 this gives

$\displaystyle|\eqref{eq:Norm approximation Gross transformed dynamics
2f2}|\leq\sum_{n=0}^{\infty}n^{1/2}\Big{|}\big{\langle}\chi^{(n)},q_{1}\int_{\left|k\right|\geq\Lambda}dk\,kB_{x_{1}}(k)a_{k}^{*}\left(-i\nabla_{1}+F_{\alpha_{t}}(x_{1})\right)q_{1}\phi^{(n)}\big{\rangle}\Big{|}$
$\displaystyle\ \leq
C\Big{(}\sum_{n=0}^{\infty}\Big{\|}\int_{\left|k\right|\geq\Lambda}dk\,k\overline{B_{x_{1}}(k)}a_{k}q_{1}\chi^{(n)}\Big{\|}^{2}\Big{)}^{\frac{1}{2}}\Big{(}\sum_{n=0}^{\infty}n\|\left(1-\Delta_{1}\right)^{\frac{1}{2}}\phi^{(n)}\|^{2}\Big{)}^{\frac{1}{2}}$
$\displaystyle\ \leq
C\|\mathds{1}_{\left|\cdot\right|\geq\Lambda}\omega^{-1/2}kB_{0}\|_{L^{2}}\|\textnormal{d}\Gamma_{a}(\omega)^{1/2}\chi\|\|\textnormal{d}\Gamma_{b}(1-\Delta)^{1/2}\phi\|$
$\displaystyle\ \leq
C\Lambda^{-1/2}\|\mathbb{T}^{1/2}\chi\|\|\left(\mathcal{N}+\mathbb{T}\right)^{1/2}\phi\|,$
(6.91)

where we used
$\|\mathds{1}_{\left|\cdot\right|\geq\Lambda}\omega^{-1/2}kB_{0}\|_{L^{2}}\leq\sqrt{4\pi/\Lambda}$
in the last step. Lemma 6.1 together with
$\|\mathds{1}_{\left|\cdot\right|\leq\Lambda}kB_{0}\|_{2}\leq\sqrt{4\pi\ln\Lambda}$
yields for the third line

$\displaystyle|\eqref{eq:Norm approximation Gross transformed dynamics 2f3}|$
$\displaystyle\leq
CN^{-1/2}CN^{-1/2}\sqrt{\ln\Lambda}\|\textnormal{d}\Gamma_{b}(1-\Delta)^{1/2}\chi\|\|\left(\mathcal{N}+1\right)\phi\|.$
(6.92)

It remains to bound the last line, (6.89d). Here, we will need to use the
regularity of $\chi$ in $x$ to improve the integrability of $k^{2}B_{x}(k)$.
Using the identity (6.76) to this end, we obtain

$\displaystyle\Big{\|}\int\limits_{\left|k\right|\leq\Lambda}dk\,k^{2}\overline{B_{x_{1}}(k)}a_{k}\chi\Big{\|}^{2}=\int\limits_{\left|k\right|\leq\Lambda}dk\int\limits_{\left|l\right|\leq\Lambda}dl\,k^{2}B_{0}(k)l^{2}B_{0}(l)\big{\langle}e^{ikx_{1}}a_{k}\chi,e^{ilx_{1}}a_{l}\chi\big{\rangle}$
$\displaystyle=\int\limits_{\left|k\right|\leq\Lambda}dk\int\limits_{\left|l\right|\leq\Lambda}dl\,\begin{aligned}
k^{2}B_{0}(k)l^{2}B_{0}(l)\Big{\langle}&((-i\nabla_{1}-l)^{2}+1)^{-\frac{1}{2}}e^{ikx_{1}}(1-\Delta_{1})^{\frac{1}{2}}a_{k}\chi,\\\
&((-i\nabla_{1}-k)^{2}+1)^{-\frac{1}{2}}e^{ilx_{1}}(1-\Delta_{1})^{\frac{1}{2}}a_{l}\chi\Big{\rangle}\end{aligned}$
$\displaystyle\
\leq\int_{\left|k\right|\leq\Lambda}dk\,\int_{\left|l\right|\leq\Lambda}dl\,k^{4}B_{0}(k)^{2}\|((-i\nabla_{1}-k)^{2}+1)^{-1/2}e^{ilx_{1}}(1-\Delta_{1})^{1/2}a_{l}\chi\|^{2}$
$\displaystyle\
\leq\sup_{p\in\mathbb{R}^{3}}\left\\{\int_{\left|k\right|\leq\Lambda}dk\,\frac{k^{4}B_{0}(k)^{2}}{1+(p-k)^{2}}\right\\}\|\mathcal{N}_{a}^{1/2}\left(1-\Delta_{1}\right)^{1/2}\chi\|^{2}.$
(6.93)

By symmetric rearrangement (similarly as in (6.6)) the supremum over
$p\in\mathbb{R}^{3}$ is bounded by a constant times $\ln\Lambda$. With this
inequality, we can estimate the remaining term by

$\displaystyle|\eqref{eq:Norm approximation Gross transformed dynamics
2f4}|=\frac{1}{\sqrt{N}}\sum_{n=1}^{\infty}n\Big{|}\big{\langle}\int\limits_{\left|k\right|\leq\Lambda}dk\,k^{2}\overline{B_{x_{1}}(k)}a_{k}\mathcal{N}_{a}^{-\frac{1}{2}}q_{1}\chi^{(n)},q_{1}\left(\mathcal{N}_{a}+1\right)^{\frac{1}{2}}\phi^{(n)}\big{\rangle}\Big{|}$
$\displaystyle\quad\leq
C\frac{\sqrt{\ln\Lambda}}{\sqrt{N}}\Big{(}\sum_{n=0}^{\infty}n\|\left(1-\Delta_{1}\right)^{\frac{1}{2}}q_{1}\chi^{(n)}\|^{2}\Big{)}^{\frac{1}{2}}\Big{(}\sum_{n=0}^{\infty}n\|\left(\mathcal{N}_{a}+1\right)^{\frac{1}{2}}\phi^{(n)}\|^{2}\Big{)}^{\frac{1}{2}}$
$\displaystyle\quad\leq
C\frac{\sqrt{\ln\Lambda}}{\sqrt{N}}\|\left(\mathcal{N}+\mathbb{T}\right)^{\frac{1}{2}}\chi\|\|\left(\mathcal{N}+1\right)\phi\|.$
(6.94)

If we choose the cutoff parameter $\Lambda=N$ we thus arrive at

$\big{|}\big{\langle}\chi,H_{3}(t)\phi\big{\rangle}\big{|}\\\ \leq
CN^{-\frac{1}{2}}\|\left(\mathcal{N}+\mathbb{T}+1\right)^{\frac{1}{2}}\chi\|\big{(}\|\left(\mathcal{N}^{3}+\mathbb{T}+1\right)^{\frac{1}{2}}\chi\|+\sqrt{\ln
N}\|\left(\mathcal{N}+1\right)\phi\|\big{)}.$ (6.95)

The terms $H_{j}(t)$ for $j\neq 3$ are somewhat easier to treat, since if
there are any gradients (as in $H_{2}(t)$, via $f_{u}$) they act on $u_{t}$
and not $\chi,\phi$. The powers of $\omega$ needed to render the kernels
integrable are strictly less than one, so the possibility of distributing
factors of $\mathcal{N}$ given by Lemma 6.1 is sufficient to treat theses
terms, as we now show.

Term $H_{1}(t)+H_{2}(t)$. For this contribution we can use the already
established bounds from the proof of Lemma 6.11, that is, (6.80) and (6.81),
respectively.

Term $H_{4}(t)$. We use $\|N_{t}\|_{\mathfrak{h}_{1/8}\rightarrow
L^{2}\otimes\mathfrak{h}_{-1/8}}+\|N_{t}\|_{\mathfrak{h}_{1/8}^{\otimes
2}\rightarrow L^{2}}\leq\|N_{t}\|_{L^{2}\otimes\mathfrak{h}_{-1/8}^{\otimes
2}}\leq C$ (see Lemma 6.7) and Lemma 6.1 to get

$\displaystyle\big{|}\big{\langle}\chi,H_{4}(t)\phi\big{\rangle}\big{|}$
$\displaystyle\leq\frac{C}{\sqrt{N}}\begin{aligned}
\Big{[}&\|\left(\mathcal{N}+1\right)^{\frac{1}{2}}\chi\|\|\textnormal{d}\Gamma_{a}(\omega^{\frac{1}{4}})\phi\|\\\
&+\|\left(\mathcal{N}+1\right)^{-\frac{1}{2}}\textnormal{d}\Gamma_{a}(\sqrt{\omega})\chi\|\|\left(\mathcal{N}+1\right)\phi\|\\\
&+\|\textnormal{d}\Gamma_{a}(\sqrt{\omega})^{\frac{1}{2}}\chi\|\|\textnormal{d}\Gamma_{a}(\sqrt{\omega})^{\frac{1}{2}}\left(\mathcal{N}+1\right)^{\frac{1}{2}}\phi\|\Big{]}.\end{aligned}$
(6.96)

By means of (6.3b) we then obtain

$\displaystyle\big{|}\big{\langle}\chi,H_{4}(t)\phi\big{\rangle}\big{|}$
$\displaystyle\leq
CN^{-1/2}\|\left(\mathcal{N}+\mathbb{T}+1\right)^{1/2}\chi\|\|\left(\mathcal{N}^{3}+\mathbb{T}+1\right)^{1/2}\phi\|.$
(6.97)

Term $H_{5}(t)$. Recalling that $|k|B_{0}(k)\in\mathfrak{h}_{-s}$ for $s>0$,
Lemma 6.1 gives

$\displaystyle\big{|}\big{\langle}\chi,H_{5}(t)\phi\big{\rangle}\big{|}$
$\displaystyle\leq
CN^{-1}\Big{[}\|\mathcal{N}_{b}^{1/2}\textnormal{d}\Gamma_{a}(\sqrt{\omega})^{1/2}\chi\|\|\textnormal{d}\Gamma_{a}(\sqrt{\omega})^{1/2}\mathcal{N}_{b}^{1/2}\phi\|$
$\displaystyle\hskip
56.9055pt+\|\mathcal{N}_{b}^{1/2}\left(\mathcal{N}_{a}+1\right)^{-1/2}\textnormal{d}\Gamma_{a}(\sqrt{\omega})\chi\|\|\left(\mathcal{N}+1\right)\phi\|$
$\displaystyle\hskip
56.9055pt+\|\mathcal{N}_{b}\chi\|\|\textnormal{d}\Gamma_{a}(\omega^{1/4})\phi\|\Big{]}.$
(6.98)

Since $\chi=\mathds{1}_{\mathcal{N}_{b}\leq N}\chi$, Equation (6.3b) leads to

$\displaystyle\big{|}\big{\langle}\chi,H_{5}(t)\phi\big{\rangle}\big{|}$
$\displaystyle\leq
CN^{-1/2}\|\left(\mathcal{N}+\mathbb{T}+1\right)^{1/2}\chi\|\|\left(\mathcal{N}^{3}+\mathbb{T}+1\right)^{1/2}\phi\|.$
(6.99)

We conclude by estimating the terms from (6.87a)–(6.87e) by using that

$\displaystyle\pm\big{(}\big{[}1-\tfrac{\mathcal{N}_{b}}{N}\big{]}_{+}-1\big{)}$
$\displaystyle\leq N^{-1}\mathcal{N}_{b},$ (6.100a)
$\displaystyle\pm\big{(}N^{-1}\sqrt{[(N-\mathcal{N}_{b})(N-\mathcal{N}_{b}-1)]_{+}}-1\big{)}$
$\displaystyle\leq CN^{-1}\mathcal{N}_{b}.$ (6.100b)

Terms (6.87a) and (6.87b). Using (6.72), (6.73) and
$\mathds{1}_{\mathcal{N}_{b}\leq N}\chi=\chi$, we can estimate the first two
lines by

$\displaystyle|\eqref{eq:Norm approximation Gross transformed dynamics
2b}|+|\eqref{eq:Norm approximation Gross transformed dynamics 2bb}|$
$\displaystyle\leq
CN^{-1/2}\|\left(\mathcal{N}+1\right)^{1/2}\chi\|\|\left(\mathcal{N}+1\right)\phi\|.$
(6.101)

Term (6.87c). Using Lemmas 6.7 and 6.1 we arrive at

$\displaystyle|\eqref{eq:Norm approximation Gross transformed dynamics
2c}|=2\Big{|}\mathrm{Re}\big{\langle}\chi,\int
dxdk\,\ell^{(1)}(x,k)a^{*}_{k}b_{x}^{*}\Big{(}\big{[}1-\tfrac{\mathcal{N}_{b}}{N}\big{]}_{+}^{1/2}-1\Big{)}\phi\big{\rangle}\Big{|}$
$\displaystyle\stackrel{{\scriptstyle\hphantom{\eqref{eq:expansion of the
number
operators}}}}{{\leq}}\|\ell^{(1)}\|_{L^{2}\otimes\mathfrak{h}_{-1/2}}\|\text{d}\Gamma_{a}(\omega)^{1/2}\chi\|\|\left(\mathcal{N}+1\right)^{1/2}([1-\tfrac{\mathcal{N}_{b}}{N}]_{+}^{1/2}-1)\phi\|$
$\displaystyle\qquad\qquad+\|\ell^{(1)}\|_{L^{2}\otimes\mathfrak{h}_{-1/4}}\|([1-\tfrac{\mathcal{N}_{b}}{N}]_{+}^{1/2}-1)\chi\|\|\text{d}\Gamma_{a}(\sqrt{\omega})^{1/2}\mathcal{N}_{b}^{1/2}\phi\|$
$\displaystyle\stackrel{{\scriptstyle\eqref{eq:expansion of the number
operators}}}{{\leq}}CN^{-1}\|\text{d}\Gamma_{a}(\omega)^{\frac{1}{2}}\chi\|\|\left(\mathcal{N}+1\right)^{\frac{3}{2}}\phi\|+CN^{-1}\|\mathcal{N}_{b}\chi\|\|\text{d}\Gamma_{a}(\sqrt{\omega})^{\frac{1}{2}}\mathcal{N}_{b}^{\frac{1}{2}}\phi\|$
$\displaystyle\stackrel{{\scriptstyle\eqref{eq:normbound:dGamma(omega^s/2)}}}{{\leq}}CN^{-\frac{1}{2}}(\|\text{d}\Gamma_{a}(\omega)^{\frac{1}{2}}\chi\|+\|N^{-\frac{1}{2}}\mathcal{N}_{b}\chi\|)(\|\left(\mathcal{N}+1\right)^{3/2}\phi\|+\|\mathbb{T}^{\frac{1}{2}}\phi\|).$
(6.102)

This implies the claimed bound since $\mathds{1}_{\mathcal{N}_{b}\leq
N}\chi=\chi$.

Term (6.87d). This term is treated in close analogy to the previous one,
leading to

$\displaystyle|\eqref{eq:Norm approximation Gross transformed dynamics 2d}|$
$\displaystyle\leq
N^{-1/2}\|\left(\mathcal{N}+\mathbb{T}+1\right)^{1/2}\chi\|\|\left(\mathcal{N}^{3}+\mathbb{T}+1\right)^{1/2}\phi\|.$
(6.103)

Term (6.87e). By means of Lemmas 6.7 and 6.1, and (6.100a) we get

$\displaystyle|\eqref{eq:Norm approximation Gross transformed dynamics 2e}|$
$\displaystyle\leq
CN^{-1}\|\left(\mathcal{N}+1\right)^{1/2}\chi\|\|\left(\mathcal{N}+1\right)^{3/2}\phi\|.$
(6.104)

This completes the proof of the lemma. ∎

###  6.5. Estimates for the dressing transformation

In this section, we will derive estimates for the fluctuation generator
associated with the dressing transformation $D_{u,\alpha}$ defined in (3.3),
and its quadratic approximation $\mathbb{D}^{\Lambda}_{u,\alpha}$ defined in
(4.10). We also give the proof of Lemma 3.6.

######  Lemma 6.13.

Let $\mathbb{D}^{\Lambda}_{u,\alpha}(\theta)$ and $D_{u,\alpha}(\theta)$ be
defined by (4.10) and (3.3). There exists a constant $C>0$, such that for all
$(u,\alpha)\in H^{1}(\mathbb{R}^{3})\oplus\mathfrak{h}_{0}$ with
$\|u\|_{L^{2}}=1$, $|\theta|\leq 1$ and
$\Lambda\in\mathbb{R}_{+}\cup\\{\infty\\}$

$\displaystyle\pm\mathbb{D}^{\Lambda}_{u,\alpha}(\theta)$ $\displaystyle\leq
C(\|u\|_{H^{1}}^{2}+\|\alpha\|_{\mathfrak{h}_{0}})(\mathcal{N}+1)$
$\displaystyle\pm i[\mathcal{N},\mathbb{D}_{u,\alpha}^{\Lambda}(\theta)]$
$\displaystyle\leq
C(\|u\|_{H^{1}}^{2}+\|\alpha\|_{\mathfrak{h}_{0}})(\mathcal{N}+1)$
$\displaystyle\pm\tfrac{d}{d\theta}\mathbb{D}_{u,\alpha}^{\Lambda}(\theta)$
$\displaystyle\leq
C(\|u\|_{H^{1}}^{3/2}+\|\alpha\|_{\mathfrak{h}_{0}})(\mathcal{N}+1)$
$\displaystyle\pm i[\mathcal{N},D_{u,\alpha}(\theta)]$ $\displaystyle\leq
C(\|u\|_{H^{1}}^{2}+\|\alpha\|_{\mathfrak{h}_{0}})(\mathcal{N}+1)(1+(\tfrac{1}{N}\mathcal{N}_{b})^{1/2})$

in the sense of quadratic forms on $\mathcal{F}\otimes\mathcal{F}$ and

$\displaystyle\big{|}\langle\phi,(D_{u,\alpha}(\theta)-\mathbb{D}_{u,\alpha}^{\infty}(\theta))\chi\rangle\big{|}\leq
C\|\phi\|\,\|(\mathcal{N}+1)^{3/2}\chi\|N^{-1/2}$

for all $\phi,\chi\in\mathcal{F}\otimes\mathcal{F}$.

###### Proof.

Recall that $\mathbb{D}_{u,\alpha}^{\Lambda}(\theta)$ and
$D_{u,\alpha}(\theta)$ are defined w.r.t. the mean-field flow
$(u^{\theta},\alpha^{\theta})=\mathfrak{D}[\theta](u,\alpha)$ and that
$|u^{\theta}|=|u|$ by (3.20). One readily shows that
$\|\tau_{u,\alpha}\|_{L^{\infty}}\leq 3\|B_{0}\|_{L^{2}}\|\alpha\|_{L^{2}}$,
and
$\|\kappa^{\Lambda}_{u^{\theta}}\|_{L^{2}(\mathbb{R}^{6})}\leq\|B_{0}\|_{L^{2}}$.
By means of (3.20) and (6.13) one further obtains

$\displaystyle\|\partial_{\theta}\kappa^{\Lambda}_{u^{\theta}}\|_{L^{2}(\mathbb{R}^{6})}=\|\tau_{u,\alpha}\kappa_{u^{\theta}}^{\Lambda}\|_{L^{2}(\mathbb{R}^{6})}\leq
3\|B_{0}\|^{2}_{L^{2}}\|\alpha\|_{L^{2}}.$ (6.105)

Using this, the estimates involving $\mathbb{D}^{\Lambda}_{u,\alpha}(\theta)$
follow standard bounds for creation and annihilation operators that are
special cases of Lemma 6.1.

Since $0\leq[1-\tfrac{\mathcal{N}_{b}}{N}]_{+}\leq 1$ the commutator of
$\mathcal{N}$ with the first two terms in (3.3) can be bounded as before. Now
let us denote the last term in (3.3), which is cubic in the
creation/annihilation operators, by $D_{u,\alpha}^{(3)}(\theta)$. Using the
canonical commutation relations and again standard estimates for creation and
annihilation operators, one obtains

$\displaystyle\pm i[\mathcal{N},D_{u,\alpha}^{(3)}(\theta)]$
$\displaystyle\leq
4\|B_{0}\|_{L^{2}}N^{-1/2}\mathcal{N}_{b}\,\mathcal{N}_{a}^{1/2}.$ (6.106)

This proves the bound on the commutator $[\mathcal{N},D_{u,\alpha}^{(3)}]$.

To show the last inequality, write

$\displaystyle D_{u,\alpha}(\theta)-\mathbb{D}^{\infty}_{u,\alpha}(\theta)$
(6.107) $\displaystyle=\int
dxdk\Big{(}\kappa^{\infty}_{u^{\theta}}(k,x)a_{k}^{*}-\kappa^{\infty}_{u^{\theta}}(-k,x)a_{k}\Big{)}b_{x}^{*}\big{(}\big{[}1-\tfrac{\mathcal{N}_{b}}{N}]_{+}^{1/2}-1\big{)}$
$\displaystyle\quad+\int
dxdk\Big{(}\overline{\kappa^{\infty}_{u^{\theta}}(k,x)}a_{k}-\overline{\kappa^{\infty}_{u^{\theta}}(-k,x)}a_{k}^{*}\Big{)}b_{x}\big{(}\big{[}1-\tfrac{\mathcal{N}_{b}-1}{N}]_{+}^{1/2}-1\big{)}+D_{u,\alpha}^{(3)}(\theta).$

It is straightforward to show that

$|\langle\phi,D_{u,\alpha}^{(3)}(\theta)\chi\rangle|\leq
4\|B_{0}\|_{L^{2}}\|\phi\|\|\mathcal{N}_{b}\,\mathcal{N}_{a}^{1/2}\chi\|N^{-1/2}.$
(6.108)

For the first two terms, the bounds
$\|\kappa^{\infty}_{u^{\theta}}\|_{L^{2}(\mathbb{R}^{6})}\leq\|B_{0}\|_{{L^{2}}}$
and $([1-\tfrac{\mathcal{N}_{b}-j}{N}]_{+}^{1/2}-1)^{2}\leq
CN^{-1}(\mathcal{N}_{b}+1)$ for $j\in\\{0,1\\}$ imply that

$\Big{|}\langle\phi,\int
dxdk\,\Big{(}\kappa^{\infty}_{u^{\theta}}(k,x)a_{k}^{*}-\kappa^{\infty}_{u^{\theta}}(-k,x)a_{k}\Big{)}b_{x}^{\bullet}\big{(}\big{[}1-\tfrac{\mathcal{N}_{b}-j}{N}]_{+}^{1/2}-1\big{)}\chi\rangle\Big{|}\\\
\leq
C\|\phi\|\|(\mathcal{N}_{b}+1)^{1/2}\mathcal{N}_{b}^{1/2}\mathcal{N}_{a}^{1/2}\chi\|N^{-1/2}$
(6.109)

where $\bullet\in\\{\varnothing,*\\}$. This completes the proof of the lemma.
∎

We now turn to the proof of Lemma 3.6, which relates the energy of excitations
in $W^{\mathrm{D}}\Psi_{N}$ to the difference of the energy of $\Psi$ to its
mean-field energy. As explained below the statement of the lemma, it is not
possible to apply the strategy of the proof of Theorem 3.2 because
$\mathbb{T}$ is not dominated by the generator $D_{u,\alpha}(\theta)$ in
(3.3). Instead, our proof relies on comparing
$X_{\mathfrak{D}}^{*}\mathbb{T}X_{\mathfrak{D}}$ directly with the difference
between the many-body energy per particle and the dressed mean-field energy
$\mathcal{E}_{1}$, evaluated at $\mathfrak{D}(u,\alpha)$. Energy estimates of
this kind were previously used in a different context in [56].

###### Proof of Lemma 3.6.

We recall Lemma 3.1 and the fact that
$\mathcal{E}=\mathcal{E}_{1}\circ\mathfrak{D}$ as shown by Equation (5.5) for
$\mathcal{E}_{0}=\mathcal{E}$. With this at hand, we write the difference
between the many-body energy per particle and the mean-field energy as

$\displaystyle
N^{-1}\big{\langle}\Psi_{N},H_{N}\Psi_{N}\big{\rangle}-\mathcal{E}(u,\alpha)=N^{-1}\big{\langle}W^{\rm
D}\Psi_{N},H_{N}^{\rm D}W^{\rm
D}\Psi_{N}\big{\rangle}-\mathcal{E}_{1}\circ\mathfrak{D}(u,\alpha).$ (6.110)

Moreover, for $\zeta={X}_{\mathfrak{D}(u,\alpha)}W^{\rm D}\Psi$ , we can use
(2.15b) to write the relevant $\gamma$ functional as

$\displaystyle\gamma\big{[}W^{\rm
D}\Psi_{N},\mathfrak{D}(u,\alpha)\big{]}=N^{-1}\big{\langle}\zeta,\mathbb{T}\zeta\big{\rangle}.$
(6.111)

To relate the expressions on the right-hand side of (6.110) and (6.111), we
make use of the excitation map $X_{\mathfrak{D}(u,\alpha)}$. To do so, we
rewrite $H_{N}^{\rm D}$ in terms of the fluctuation generator
$H_{\mathfrak{D}(u,\alpha)}^{\rm D}(0)$ from (6.68). This will allow us to
employ previously established estimates.

To ease up the notation, we set from now on
$(u^{\mathrm{D}},\alpha^{\mathrm{D}})=\mathfrak{D}(u,\alpha)$ and the
shorthand $q=q_{u^{\mathrm{D}}}$, $h=h_{\mathfrak{D}(u,\alpha)}$,
$f=f_{u^{\mathrm{D}}}$, $g=g_{u^{\mathrm{D}},\alpha^{\mathrm{D}}}$ (see
(3.4a)–(3.4f) for the definitions of these objects). We can employ the results
from Appendix C to obtain

$\displaystyle X_{\mathfrak{D}(u,\alpha)}H_{N}^{\rm
D}X_{\mathfrak{D}(u,\alpha)}^{*}$ $\displaystyle=H^{\rm
D}_{\mathfrak{D}(u,\alpha)}(0)+\big{\langle}u^{\mathrm{D}},hu^{\mathrm{D}}\big{\rangle}\big{(}N-\mathcal{N}_{b}\big{)}+N\big{\langle}\alpha^{\mathrm{D}},\omega\alpha^{\mathrm{D}}\big{\rangle}$
$\displaystyle\quad+\sqrt{N}\Phi\big{(}\omega\alpha^{\mathrm{D}}+f+g\big{)}-b^{*}(u^{\mathrm{D}})b(qhu^{\mathrm{D}})$
$\displaystyle\quad+\sqrt{N-\mathcal{N}_{b}}b(qhu^{\mathrm{D}})+b^{*}(qhu^{\mathrm{D}})\sqrt{N-\mathcal{N}_{b}}.$
(6.112)

From the first inequality of Lemma 6.11, the fact that
$\mathds{1}_{\mathcal{N}_{b}\leq N}\zeta=\zeta$, and (6.100a), we get

$\displaystyle\langle\zeta,\mathbb{T}\zeta\rangle$ $\displaystyle\leq
2\langle\zeta,H_{\mathfrak{D}(u,\alpha)}^{\rm
D}(0)\zeta\rangle+C\langle\zeta,\mathcal{N}\zeta\rangle.$ (6.113)

With the formula (5.3) for the dressed mean-field energy $\mathcal{E}_{1}$, we
arrive at

$\displaystyle
N^{-1}\big{\langle}\zeta,\mathbb{T}\zeta\big{\rangle}-\left|N^{-1}\big{\langle}\Psi_{N},H_{N}\Psi_{N}\big{\rangle}-\mathcal{E}(u,\alpha)\right|$
(6.114) $\displaystyle\leq
CN^{-1}\langle\zeta,\mathcal{N}\zeta\rangle+\Big{|}-N^{-1}\big{\langle}u^{\mathrm{D}},hu^{\mathrm{D}}\big{\rangle}\big{\langle}\zeta,\mathcal{N}_{b}\zeta\big{\rangle}-N^{-1}\big{\langle}\zeta,b^{*}(u^{\mathrm{D}})b(qhu^{\mathrm{D}})\zeta\big{\rangle}$
$\displaystyle\qquad+N^{-\frac{1}{2}}\big{\langle}\zeta,\Phi\big{(}\omega\alpha^{\mathrm{D}}+f+g\big{)}\zeta\big{\rangle}+2N^{-\frac{1}{2}}\mathrm{Re}\big{\langle}\zeta,\sqrt{1-\mathcal{N}_{b}/N}b(qhu^{\mathrm{D}})\zeta\big{\rangle}\Big{|}.$

We bound the terms on the right hand side of in (6.114) by (using that
$\|\zeta\|=\|\Psi_{N}\|=1$)

$|\eqref{eq:energy difference rest}|\leq
C\big{\langle}\zeta,\mathcal{N}\zeta\big{\rangle}N^{-1}\big{(}1+\|hu^{\mathrm{D}}\|_{L^{2}}\big{)}\\\
+CN^{-1/2}\big{(}\|\omega\alpha^{\mathrm{D}}+f+g\|_{L^{2}}\|\mathcal{N}_{a}^{1/2}\zeta\|+C\|\mathcal{N}_{b}^{1/2}\zeta\|\big{)}.$
(6.115)

By Lemmas 6.4 and 6.5, the norms of $f$, $g$, $hu^{\mathrm{D}}$ are bounded in
terms of the $H^{2}\oplus\mathfrak{h}_{1/2}$-norm of
$(u^{\mathrm{D}},\alpha^{\mathrm{D}})$, which by Lemma 3.3 is controlled by
the norm of $(u,\alpha)\in H^{2}\oplus\mathfrak{h}_{3/2}$. Thus there exists a
constant $C$, depending on this norm, so that

$\displaystyle|\eqref{eq:energy difference rest}|\leq
C\big{(}N^{-1}\big{\langle}\zeta,\mathcal{N}\zeta\big{\rangle}+(N^{-1}\big{\langle}\zeta,\mathcal{N}\zeta\big{\rangle})^{1/2}\big{)}.$
(6.116)

Moreover by (2.15a) and Lemma 3.5 for $\theta=1$, we have

$\displaystyle N^{-1}\big{\langle}\zeta,\mathcal{N}\zeta\big{\rangle}$
$\displaystyle=\beta\big{[}W^{\rm D}\Psi_{N},\mathfrak{D}(u,\alpha)\big{]}\leq
C\big{(}\beta\big{[}\Psi_{N},(u,\alpha)\big{]}+N^{-1}\big{)}.$ (6.117)

Combined with (6.116) this proves the statement of the Lemma. ∎

##  Appendix A Initial states

###### Proof of Proposition 1.2.

Let $W^{\rm D}_{\geq K}=W(N^{-1/2}\sum_{j=1}^{N}B_{K,x_{j}})$ and $W^{\rm
D}_{\geq K,x_{j}}=W(N^{-1/2}B_{K,x_{j}})$. The first inequality of the
Proposition can be obtained similarly as [61, Prop. II.2]. More explicitly, we
use (2.8) and (2.9) to estimate

$\displaystyle N^{-1}\|\mathcal{N}_{a}^{1/2}W^{*}(\sqrt{N}\alpha)\Psi_{N,K}\|$
$\displaystyle=N^{-1}\int d^{3}k\,\|a_{k}(W^{\rm D}_{\geq K})^{*}(u^{\otimes
N}\otimes\Omega)\|^{2}$ $\displaystyle\leq\|B_{K,0}\|_{L^{2}}^{2}\leq
CK^{-2}.$ (A.1)

By means of

$\displaystyle W^{\rm D}_{\geq K}(q_{u})_{1}(W^{\rm D}_{\geq K})^{*}$
$\displaystyle=(q_{u})_{1}+|u\rangle\langle u|_{1}-W^{\rm D}_{\geq
K,x_{1}}|u\rangle\langle u|_{1}(W^{\rm D}_{\geq K,x_{1}})^{*}$ (A.2)

and

$\displaystyle\big{(}1-(W^{\rm D}_{\geq K,x_{1}})\big{)}\big{(}1-(W^{\rm
D}_{\geq K,x_{1}})^{*}\big{)}\leq N^{-1}\hat{\Phi}(iB_{K,x_{1}})^{2}$ (A.3)

we get

$\displaystyle\left|\big{\langle}\Psi_{N,K},(q_{u})_{1}\Psi_{N,K}\big{\rangle}\right|$
$\displaystyle\leq 2\|\left(1-(W^{\rm D}_{\geq K,x_{1}})^{*}\right)u^{\otimes
N}\otimes W(\sqrt{N}\alpha)\Omega\|$ $\displaystyle\leq
2N^{-1/2}\|B_{K,0}\|_{L^{2}}\|\big{(}\mathcal{N}_{a}+1\big{)}^{1/2}u^{\otimes
N}\otimes W(\sqrt{N}\alpha)\Omega\|$ $\displaystyle\leq
CK^{-1}\big{(}N^{-1/2}+\|\alpha\|_{L^{2}}\big{)},$ (A.4)

and thus $\beta[\Psi_{N,K},u,\alpha]\leq
CK^{-1}\left(1+\|\alpha\|_{L^{2}}\right)$. Since $W^{\rm D}_{\geq
K}=\prod_{j=1}^{N}W^{\rm D}_{\geq K,x_{j}}$, the transformation relations of
the dressing transformation from [46, Section II] lead to

$\displaystyle W^{\rm D}_{\geq K}H_{N}(W^{\rm D}_{\geq K})^{*}$
$\displaystyle=\sum_{j=1}^{N}\Big{[}-\Delta_{j}+N^{-1/2}\hat{\Phi}(\mathds{1}(\left|\cdot\right|\leq
K)G_{x_{j}})$
$\displaystyle\quad+N^{-1}\big{(}a(kB_{K,x_{j}})^{2}+\text{h.c.}+2a^{*}(kB_{K,x_{j}})a(kB_{K,x_{j}})\big{)}$
$\displaystyle\quad-2N^{-1/2}\big{(}i\nabla_{x_{j}}\cdot
a(kB_{K,x_{j}})+a^{*}(kB_{K,x_{j}})\cdot i\nabla_{x_{j}}\big{)}\Big{]}$
$\displaystyle\quad+N^{-1}\sum_{i<j}V_{K}(x_{i}-x_{j})+\textnormal{d}\Gamma_{a}(\omega)+E_{K}$
(A.5)

with

$\displaystyle V_{K}(x_{i}-x_{j})$
$\displaystyle=2\mathrm{Re}\big{\langle}B_{K,x_{i}},\omega
B_{K,x_{j}}\big{\rangle}-4\mathrm{Re}\big{\langle}G_{x_{i}},B_{K,x_{j}}\big{\rangle}$
(A.6)

and

$\displaystyle E_{K}=\int_{\left|k\right|\leq
K}\frac{dk}{\omega(k)\left(k^{2}+\omega(k)\right)}.$ (A.7)

The shifting property of the Weyl operator (2.9) then lets write the
expectation value of the energy per particle as

$N^{-1}\big{\langle}\Psi_{N,K},H_{N}\Psi_{N,K}\big{\rangle}=N^{-1}E_{K}\mathcal{E}(u,\alpha)\\\
+\Big{\langle}u,\big{(}2\mathrm{Re}\big{\langle}G_{(\cdot)},\alpha_{\geq
K}\big{\rangle}+A_{\alpha_{\geq K},(\cdot)}+F_{\alpha_{\geq
K}}^{2}+\frac{1}{2}V_{K}*\left|u\right|^{2}\big{)}u\Big{\rangle}$ (A.8)

with $\alpha_{\geq K}=\mathds{1}_{\left|\cdot\right|\geq K}\alpha$ and
$A_{\alpha}$, $F_{\alpha}$ as defined in (3.4b), (3.4c). Note that
$\left|E_{K}\right|\leq C\left(1+\ln K\right)$. By means of
$\sup_{x\in\mathbb{R}^{3}}\left|\big{\langle}kB_{x},\alpha_{\geq
K}\big{\rangle}\right|\leq CK^{-1}\|\alpha\|_{\mathfrak{h}_{1}}$,
$\|V_{K}\|\leq CK^{-3/2}$ and $\|u\|_{L^{2}}=1$ we get

$\displaystyle\left|\big{\langle}u,A_{\alpha_{\geq
K},(\cdot)}u\big{\rangle}\right|$ $\displaystyle\leq
CK^{-1}\|\alpha\|_{\mathfrak{h}_{1}}\|u\|_{H^{1}},\quad\|F_{\alpha_{\geq
K}}\|_{L^{\infty}}\leq CK^{-1}\|\alpha\|_{\mathfrak{h}_{1}}$ (A.9)

and

$\displaystyle\|V_{K}*\left|u\right|^{2}\|_{L^{\infty}}$
$\displaystyle\leq\|V_{K}\|_{L^{2}}\|\left|u\right|^{2}\|_{L^{2}}\leq
CK^{-3/2}\|u\|_{H^{1}}^{2}.$ (A.10)

Inequality (6.9) with $G_{(\cdot)}\mapsto\omega^{-1/2}G_{(\cdot)}$ and
$\alpha\mapsto\omega^{1/2}\alpha$ leads to

$\|\mathrm{Re}\big{\langle}G_{(\cdot)},\alpha_{\geq
K}\big{\rangle}u\|_{L^{2}}\leq
CK^{-1}\|\alpha\|_{\mathfrak{h}_{1}}\|u\|_{H^{1}}.$ (A.11)

In total, we obtain

$\left|N^{-1}\big{\langle}\Psi_{N,K},H_{N}\Psi_{N,K}\big{\rangle}-\mathcal{E}(u,\alpha)\right|\\\
\leq C\left(K^{-1}+N^{-1}(1+\ln
K)\right)\left(\|u\|_{H^{1}}^{2}+\|\alpha\|_{\mathfrak{h}_{1}}^{2}\right).$
(A.12)

∎

##  Appendix B Bogoliubov transformations

For a linear map $T$ on a complex Hilbert space, we denote by
$\bar{T}f=\overline{T\bar{f}}$ its complex conjugate.

######  Lemma B.1.

Let $\mathscr{H}$ be a Hilbert space and $\mathbb{U}_{n}$ with
$n\in\mathbb{N}$ be a family of unitary Bogoliubov transformations on the Fock
space over $\mathscr{H}$, that is, there exist $\mathfrak{u}_{n}$ linear,
bounded and $\mathfrak{v}_{n}$ Hilbert Schmidt, so that

$\mathbb{U}_{n}^{*}a^{*}(f)\mathbb{U}_{n}=a^{*}(\mathfrak{u}_{n}f)+a(\mathfrak{v}_{n}\bar{f}),\qquad\mathbb{U}_{n}^{*}a(f)\mathbb{U}_{n}=a(\overline{\mathfrak{u}}_{n}f)+a^{*}(\overline{\mathfrak{v}_{n}f}).$

Assume that

$\mathbb{U}_{\infty}:=\operatorname*{\mathrm{s--
lim}}_{n\to\infty}\mathbb{U}_{n}$

exists, and moreover there exists a self-adjoint $A$, $D(A)$ with $A\geq 1$
and $C>0$ so that for all $n\in\mathbb{N}\cup\\{\infty\\}$ and $\Psi\in
D(\textnormal{d}\Gamma(A)^{1/2})$ it holds

$\langle\mathbb{U}_{n}\Psi,\mathcal{N}\mathbb{U}_{n}\Psi\rangle\leq
C\langle\Psi,(1+\textnormal{d}\Gamma(A))\Psi\rangle.$

Then $\mathbb{U}_{\infty}$ is a Bogoliubov transformation and the
corresponding maps $\mathfrak{u},\mathfrak{v}$ satisfy
$\|\mathfrak{u}\|_{\mathscr{H}\to\mathscr{H}}\leq C+1$,
$\|\mathfrak{v}\|_{\mathfrak{S}_{2}(\mathscr{H})}\leq C$.

###### Proof.

We start by showing that for $\Psi,\Phi\in D(\mathrm{d}\Gamma(A)^{1/2})$,
$f\in\mathscr{H}$, and $\bullet\in\\{\varnothing,*\\}$

$\lim_{n\to\infty}\langle\Phi,\mathbb{U}_{n}^{*}a^{\bullet}(f)\mathbb{U}_{n}\Psi\rangle=\langle\Phi,\mathbb{U}_{\infty}^{*}a^{\bullet}(f)\mathbb{U}_{\infty}\Psi\rangle.$
(B.1)

To see this, note that

$\displaystyle\Big{|}\langle\Phi,\mathbb{U}_{n}^{*}a^{\bullet}(f)\mathbb{U}_{n}\Psi\rangle-\langle\Phi,\mathbb{U}_{\infty}^{*}a^{\bullet}(f)\mathbb{U}_{\infty}\Psi\rangle\Big{|}$
(B.2) $\displaystyle\quad\leq\begin{aligned}
C\|f\|_{\mathscr{H}}\Big{(}&\|(\mathbb{U}_{n}-\mathbb{U}_{\infty})\Phi\|_{\mathcal{F}}\|(1+\mathrm{d}\Gamma(A))^{1/2}\Psi\|_{\mathcal{F}}\\\
&+\|(\mathbb{U}_{n}-\mathbb{U}_{\infty})\Psi\|_{\mathcal{F}}\|(1+\mathrm{d}\Gamma(A))^{1/2}\Phi\|_{\mathcal{F}}\Big{)},\end{aligned}$

which tends to zero since $\mathbb{U}_{n}$ converges strongly to
$\mathbb{U}_{\infty}$.

Now let $f\in D(A)$, so $a^{*}(f)\Omega\in D(\mathrm{d}\Gamma(A)^{1/2})$. Then
we have, using that $\mathbb{U}_{n}$ is a Bogoliubov transformation for
$n\in\mathbb{N}$,

$\displaystyle\langle
a^{*}(f)\Omega,\mathbb{U}_{\infty}a^{*}(g)\mathbb{U}_{\infty}^{*}\Omega\rangle$
$\displaystyle=\lim_{n\to\infty}\langle
a^{*}(f)\Omega,(a^{*}(\mathfrak{u}_{n}g)+a(\mathfrak{v}_{n}\bar{g}))\Omega\rangle$
$\displaystyle=\lim_{n\to\infty}\langle f,\mathfrak{u}_{n}g\rangle,$ (B.3a)
$\displaystyle\langle
a^{*}(f)\Omega,\mathbb{U}_{\infty}a(g)\mathbb{U}_{\infty}^{*}\Omega\rangle$
$\displaystyle=\lim_{n\to\infty}\langle
a^{*}(f)\Omega,(a(\overline{\mathfrak{u}}_{n}g)+a^{*}(\overline{\mathfrak{v}_{n}g}))\Omega\rangle$
$\displaystyle=\lim_{n\to\infty}\langle
f,\overline{\mathfrak{v}_{n}g}\rangle.$ (B.3b)

Since moreover

$|\langle
a^{*}(f)\Omega,\mathbb{U}_{\infty}a^{\bullet}(g)\mathbb{U}_{\infty}^{*}\Omega\rangle|\leq\|f\|_{\mathscr{H}}\|g\|_{\mathscr{H}}\overbrace{\|(1+\mathcal{N}^{1/2})\mathbb{U}_{\infty}\Omega\|_{\mathcal{F}}}^{\leq
C+1},$ (B.4)

the operators $\mathfrak{u}_{n},\mathfrak{v}_{n}$ converge weakly to operators
$\mathfrak{u},\mathfrak{v}$ with norm less than $C+1$. Weak convergence of
$\mathfrak{u}_{n}$, $\mathfrak{v}_{n}$, implies that for $\Phi\in
D(\mathcal{N}^{1/2})$, $\Psi\in\mathcal{F}$, and $f\in\mathscr{H}$

$\lim_{n\to\infty}\langle\Phi,a(\mathfrak{u}_{n}f)\Psi\rangle=\langle\Phi,a(\mathfrak{u}f)\Psi\rangle,\qquad\lim_{\Lambda\to\infty}\langle\Phi,a(\mathfrak{v}_{n}f)\Psi\rangle=\langle\Phi,a(\mathfrak{v}f)\Psi\rangle,$
(B.5)

and thus for $\Phi,\Psi\in D(\mathcal{N}^{1/2})$

$\langle\Phi,\mathbb{U}_{\infty}^{*}a^{*}(f)\mathbb{U}_{\infty}\Psi\rangle=\langle
a(\mathfrak{u}f)\Phi,\Psi\rangle+\langle\Phi,a(\mathfrak{v}\bar{f})\Psi\rangle=\langle\Phi,(a^{*}(\mathfrak{u}f)+a(\mathfrak{v}\bar{f}))\Psi\rangle,$
(B.6)

and similarly for $a(f)$. Moreover, we have

$\|\mathfrak{v}_{n}\|_{\mathfrak{S}_{2}}^{2}=\|\mathcal{N}^{1/2}\mathbb{U}_{n}\Omega\|^{2}\leq
C,$ (B.7)

so the sequence $\mathfrak{v}_{n}$ is bounded in $\mathfrak{S}_{2}$, whence it
has a subsequence that converges weakly in $\mathfrak{S}_{2}$. Since $\langle
f,\mathfrak{v}_{n}g\rangle=\mathrm{Tr}(|f\rangle\langle g|\mathfrak{v}_{n})$,
the limit must be $\mathfrak{v}$, so $\mathfrak{v}\in\mathfrak{S}_{2}$ with
norm less than $C$. This proves the claim. ∎

######  Lemma B.2.

The unitaries $\mathbb{W}^{\Lambda}_{u,\alpha}(\theta)$ and
$\mathbb{U}_{u,\alpha,1}^{\Lambda}(t)$ defined in Propositions 4.10 and 5.2
are Bogoliubov transformations under the hypothesis given there.

###### Proof.

In view of the strong convergence of $\mathbb{U}^{\Lambda}_{u,\alpha,1}(t)$ to
$\mathbb{U}^{\infty}_{u,\alpha,1}(t)$ and the bound from Proposition 5.2(i),
it is sufficient it prove the claim on $\mathbb{U}_{u,\alpha,1}$ for
$\Lambda<\infty$. For $\mathbb{W}^{\Lambda}_{u,\alpha}$ such a distinction is
not necessary. Note that the terms in the generators
$\mathbb{D}^{\infty}_{u,\alpha}$ and $\mathbb{H}_{u,\alpha,1}^{\Lambda}$,
$\Lambda<\infty$ of these unitaries with two creation operators ($b^{*}$
and/or $a^{*}$) have coefficients that are square integrable functions of
their arguments (compare Lemma 6.7). Since this is the only relevant property,
we give an exemplary proof in the case of $\mathbb{W}_{u,\alpha}^{\infty}$.

As a first step, consider the (tentative) equations for the operators
$\mathfrak{u}(t)$, $\mathfrak{v}(t)$ associated with a Bogoliubov
transformation. These are usually expressed in terms of the matrix

$\mathcal{V}=\begin{pmatrix}\mathfrak{u}&\mathfrak{v}\\\
\overline{\mathfrak{v}}&\overline{\mathfrak{u}}\end{pmatrix}\,$ (B.8)

where, in our case, $\mathfrak{u},\mathfrak{v}:L^{2}(\mathbb{R}^{3})\oplus
L^{2}(\mathbb{R}^{3})\to L^{2}(\mathbb{R}^{3})\oplus L^{2}(\mathbb{R}^{3})$
are linear and bounded. For the case of $\mathbb{W}_{u,\alpha}^{\infty}$,
asking that

$i\partial_{\theta}\big{(}c^{*}(\mathfrak{u}(\theta)(f\oplus
g))+c(\mathfrak{v}(\theta)(\bar{f}\oplus\bar{g}))\big{)}=[\mathbb{D}_{u,\alpha}^{\infty}(\theta),c^{*}(\mathfrak{u}(\theta)(f\oplus
g))+c(\mathfrak{v}(\theta)(\bar{f}\oplus\bar{g}))]$ (B.9)

yields the equations for $\mathcal{V}$

$i\partial_{\theta}\mathcal{V}(\theta)=\mathcal{A}(\theta)\mathcal{V}(\theta),$
(B.10)

with

$\displaystyle\mathcal{A}(\theta)$
$\displaystyle=\begin{pmatrix}A_{\mathrm{d}}&-A_{\mathrm{o}}\\\
\bar{A}_{\mathrm{o}}&-\bar{A}_{\mathrm{d}}\end{pmatrix},$ (B.11)
$\displaystyle
A_{\mathrm{o}}=\begin{pmatrix}0&\kappa_{u^{\theta}}^{\infty}(\cdot,x)\\\
\kappa_{u^{\theta}}^{\infty}(k,\cdot)&0\end{pmatrix},$ $\displaystyle\quad
A_{\mathrm{d}}=\begin{pmatrix}\tau_{u,\alpha}&-\kappa_{u^{\theta}}^{\infty}(-(\cdot),x)\\\
-\overline{\kappa}_{u^{\theta}}^{\infty}(-k,\cdot)&0\end{pmatrix},$

where $\kappa$ acts as an integral operator by integrating in the variable
denoted by $(\cdot)$. These equations admit a unique solution
$\mathcal{V}(\theta)$ with $\mathcal{V}(0)=1$ ($\mathcal{A}$ is a bounded
perturbation of the diagonal terms, which are generators).

Since $\kappa_{u^{\theta}}^{\infty}\in L^{2}(\mathbb{R}^{6})$, the off-
diagonal part $A_{\mathrm{o}}$ is a Hilbert-Schmidt operator, and consequently
the off-diagonal part $\mathfrak{v}(\theta)$ of $\mathcal{V}(\theta)$ is also
Hilbert-Schmidt [12, Lem.4.9]. By Shale-Stinespring criterion (see, e.g., [12,
Lem. 4.2]) there thus exists a Bogoliubov transformation $\mathbb{V}(\theta)$
associated to $\mathfrak{u}(\theta)$, $\mathfrak{v}(\theta)$, i.e.
$\mathcal{V}(\theta)$. This transformation is determined up to a
$\theta$-dependent phase. We can fix this phase by asking that
$\mathbb{V}(\theta)\Omega=\mathbb{W}_{0}^{\infty}(\theta)\Omega$, as we now
show.

The vacuum vector $\Omega$ spans the one-dimensional space on which $c(f\oplus
g)=b(f)+a(g)$ vanish for all $f,g\in L^{2}(\mathbb{R}^{3})$. Then
$\mathbb{V}(\theta)\Omega$ spans the joint kernel of
$\mathbb{V}(\theta)c(f\oplus g)\mathbb{V}(\theta)^{*}$. Using the equation
satisfied by $\mathcal{V}(\theta)^{-1}$, we find (since we do not know a
priori that $\mathbb{V}$ is differentiable,
$i\partial_{\theta}\mathbb{V}(\theta)$ denotes the weak derivative)

$\displaystyle 0$
$\displaystyle=i\partial_{\theta}\big{(}\mathbb{V}(\theta)c(f\oplus
g)\mathbb{V}(\theta)^{*}\big{)}\mathbb{V}(\theta)\Omega$
$\displaystyle=\big{(}\mathbb{V}(\theta)c(f\oplus
g)\mathbb{V}(\theta)^{*}\big{)}i\partial_{\theta}\mathbb{V}(\theta)\Omega+[\mathbb{V}(\theta)c(f\oplus
g)\mathbb{V}(\theta)^{*},\mathbb{D}_{0}^{\infty}(\theta)]\mathbb{V}(\theta)\Omega.$
(B.12)

It follows that

$i\partial_{\theta}\mathbb{V}(\theta)\Omega-\mathbb{D}_{0}^{\infty}(\theta)\mathbb{V}(\theta)\Omega$
(B.13)

is in the kernel of $\mathbb{V}(\theta)c(f\oplus g)\mathbb{V}(\theta)^{*}$,
and thus proportional to $\mathbb{V}(\theta)\Omega$. The constant of
proportionality must be real, since $\mathbb{V}(\theta)$ is unitary, and thus
we can set it to zero by adjusting the phase, i.e., setting
$\widetilde{\mathbb{V}}(\theta)=\mathbb{V}(\theta)e^{i\int_{0}^{\theta}\nu(s)ds}$,
which is also a Bogoliubov transformation associated with
$\mathcal{V}(\theta)$. By uniqueness of the solution proved in Proposition
4.3, we thus have
$\widetilde{\mathbb{V}}(\theta)\Omega=\mathbb{W}^{\infty}_{0}(\theta)\Omega$.
Using the explicit action of $\mathbb{V}(\theta)$ on the creation and
annihilation operators, one shows by induction that
$\widetilde{\mathbb{V}}(\theta)\Psi=\mathbb{W}_{0}^{\infty}(\theta)$ for any
state $\Psi$ obtained by application of a finite number of creation and
annihilation operators (see the proof of [12, Lem.4.8]). Since the span of
such $\Psi$ is dense, this proves equality and thus that
$\mathbb{W}_{0}^{\infty}(\theta)$ is a Bogoliubov transformation. ∎

##  Appendix C Fluctuation generator of the dressed dynamics

In this section we provide the derivation of the fluctuation generator of the
dressed Nelson dynamics.

In order to disentangle the calculation, we write the excitation map as

$X_{u,\alpha}=X_{u}\otimes W^{*}(\sqrt{N}\alpha),$ (C.1)

where $X_{u}:\bigotimes_{\rm
sym}^{N}L^{2}(\mathbb{R}^{3})\to\mathcal{F}_{\perp u}$ acts as
$\Psi_{N}\mapsto((X_{u}\Psi_{N})^{(k)})_{k=0}^{N}$ with

$(X_{u}\Psi_{N})^{(k)}=\binom{N}{k}^{1/2}\prod_{i=1}^{k}(q_{u})_{i}\big{\langle}u^{\otimes(N-k)},\Psi_{N}\big{\rangle}_{L^{2}(\mathbb{R}^{3(N-k)})}\in\mathcal{F}^{(k)}_{\perp
u}$ (C.2)

We first calculate the result of applying only the Weyl operator.

######  Lemma C.1.

For $(u,\alpha)\in H^{3}(\mathbb{R}^{3})\oplus\mathfrak{h}_{5/2}$ with
$\|u\|_{L^{2}}=1$ let
$(u_{t},\alpha_{t})=\mathfrak{s}^{\mathrm{D}}[t](u,\alpha)$ be the solution to
(3.3). Let

$H_{\alpha}^{\mathrm{D},\leq
N}(t)=i\dot{W}(\sqrt{N}\alpha_{t})^{*}W(\sqrt{N}\alpha_{t})+W(\sqrt{N}\alpha_{t})^{*}H_{N}^{\rm
D}W(\sqrt{N}\alpha_{t})$ (C.3)

then

$\displaystyle H_{\alpha}^{\mathrm{D},\leq N}(t)$
$\displaystyle=\sum_{j=1}^{N}\Big{(}-\Delta_{x_{j}}+A_{\alpha_{t},x_{j}}+F^{2}_{\alpha_{t}}(x_{j})-\mathrm{Re}\big{\langle}\alpha_{t},f_{u_{t}}+g_{u_{t},\alpha_{t}}\big{\rangle}\Big{)}$
(C.4a)
$\displaystyle\quad+\frac{1}{N}\sum_{i<j}V(x_{i}-x_{j})+\textnormal{d}\Gamma_{a}(\omega)-\sqrt{N}\hat{\Phi}(f_{u_{t}}+g_{u_{t},\alpha_{t}})$
(C.4b) $\displaystyle\quad+\frac{2}{\sqrt{N}}\sum_{j=1}^{N}\int
dk\,kB_{x_{j}}(k)\cdot\big{(}-i\nabla_{j}+F_{\alpha_{t}}(x_{j})\big{)}a_{k}^{*}+\mathrm{h.c.}$
(C.4c)
$\displaystyle\quad+\frac{1}{N}\sum_{j=1}^{N}\big{(}a(kB_{x_{j}})^{2}+\textnormal{h.c.}+2a^{*}(kB_{x_{j}})a(kB_{x_{j}})\big{)}$
(C.4d)

###### Proof.

We recall the form of $H^{\mathrm{D}}_{N}$ given in Lemma 3.1. By the shift
property of the Weyl operator (2.9), we have

$\displaystyle
W(\sqrt{N}\alpha_{t})^{*}\textnormal{d}\Gamma_{a}(\omega)W(\sqrt{N}\alpha_{t})$
$\displaystyle=\textnormal{d}\Gamma_{a}(\omega)+\hat{\Phi}(\sqrt{N}\omega\alpha_{t})+N\big{\langle}\alpha_{t},\omega\alpha_{t}\big{\rangle},$
(C.5a) $\displaystyle
W(\sqrt{N}\alpha_{t})^{*}\hat{A}_{x_{j}}W(\sqrt{N}\alpha_{t})$
$\displaystyle=\hat{A}_{x_{j}}+\sqrt{N}A_{\alpha_{t},x_{j}}.$ (C.5b) with
$A_{\alpha_{t},x}=-2i\nabla_{x}\cdot\big{\langle}kB_{x},\alpha_{t}\big{\rangle}+\text{h.c.}$,
and $\displaystyle
W(\sqrt{N}\alpha_{t})^{*}\big{(}a(kB_{x_{j}})^{2}+\textnormal{h.c.}+2a^{*}(kB_{x_{j}})a(kB_{x_{j}})\big{)}W(\sqrt{N}\alpha_{t})$
$\displaystyle\
=\big{(}a(kB_{x_{j}})^{2}+\textnormal{h.c.}+2a^{*}(kB_{x_{j}})a(kB_{x_{j}})\big{)}$
$\displaystyle\qquad+2\sqrt{N}F_{\alpha_{t}}(x_{j})\hat{\Phi}(kB_{x_{j}})+NF^{2}_{\alpha_{t}}(x_{j}),$
(C.5c)

where we inserted
$F_{\alpha_{t}}(x)=2\mathrm{Re}\big{\langle}kB_{x},\alpha_{t}\big{\rangle}$
and used

$\displaystyle
2\big{|}\big{\langle}\alpha_{t},kB_{x_{j}}\big{\rangle}\big{|}^{2}+\big{\langle}\alpha_{t},kB_{x_{j}}\big{\rangle}^{2}+\big{\langle}kB_{x_{j}},\alpha_{t}\big{\rangle}^{2}=\big{(}2\mathrm{Re}\big{\langle}kB_{x_{j}},\alpha_{t}\big{\rangle}\big{)}^{2}.$
(C.6)

The operators $-\Delta_{x_{j}}$ and $V(x_{j}-x_{i})$ are left unchanged by
$W(\sqrt{N}\alpha)$, so it remains to the term with the time-derivative. We
use the formula
$W(\sqrt{N}\alpha_{t})^{*}=e^{a(\sqrt{N}\alpha_{t})}e^{-a^{*}(\sqrt{N}\alpha_{t})}e^{\frac{N}{2}\|\alpha_{t}\|^{2}}$
to compute

$\displaystyle\tfrac{d}{dt}W(\sqrt{N}\alpha_{t})^{*}$
$\displaystyle=e^{a(\sqrt{N}\alpha_{t})}a(\sqrt{N}\dot{\alpha}_{t})e^{-a^{*}(\sqrt{N}\alpha_{t})}e^{\frac{N}{2}\|\alpha_{t}\|^{2}}$
$\displaystyle\quad-e^{a(\sqrt{N}\alpha_{t})}e^{-a^{*}(\sqrt{N}\alpha_{t})}e^{\frac{N}{2}\|\alpha_{t}\|^{2}}\big{(}a^{*}(\sqrt{N}\dot{\alpha}_{t})-N\mathrm{Re}\big{\langle}\dot{\alpha}_{t},\alpha_{t}\big{\rangle}\big{)}$
$\displaystyle=W(\sqrt{N}\alpha_{t})^{*}\Big{(}a(\sqrt{N}\dot{\alpha}_{t})-a^{*}(\sqrt{N}\dot{\alpha}_{t})-N\mathrm{Im}\big{\langle}\dot{\alpha}_{t},\alpha_{t}\big{\rangle}\Big{)}$
(C.7)

where the last step follows from

$\displaystyle
e^{a^{*}(\sqrt{N}\alpha_{t})}a(\sqrt{N}\dot{\alpha}_{t})e^{-a^{*}(\sqrt{N}\alpha_{t})}=a(\sqrt{N}\dot{\alpha}_{t})-N\big{\langle}\dot{\alpha}_{t},\alpha_{t}\big{\rangle}.$
(C.8)

Inserting the equation of motion (5.2) for $\alpha_{t}$, we find

$\displaystyle
i\tfrac{d}{dt}W(\sqrt{N}\alpha_{t})^{*}=W(\sqrt{N}\alpha_{t})^{*}\big{(}-\sqrt{N}\hat{\Phi}(i\dot{\alpha}_{t})+N\mathrm{Re}\big{\langle}\alpha_{t},i\dot{\alpha}_{t}\big{\rangle}\big{)}$
(C.9)
$\displaystyle=W(\sqrt{N}\alpha_{t})^{*}\big{(}-\sqrt{N}\hat{\Phi}(\omega\alpha_{t}+f_{u_{t}}+g_{u_{t},\alpha_{t}})+N\mathrm{Re}\big{\langle}\alpha_{t},\omega\alpha_{t}+f_{u_{t}}+g_{u_{t},\alpha_{t}}\big{\rangle}\big{)},$

and with the shift property of the Weyl operator (2.9), we get

$(i\tfrac{d}{dt}W(\sqrt{N}\alpha_{t})^{*})W(\sqrt{N}\alpha_{t})\\\
=-\sqrt{N}\hat{\Phi}(\omega\alpha_{t}+f_{u_{t}}+g_{u_{t},\alpha_{t}})-N\mathrm{Re}\big{\langle}\alpha_{t},\omega\alpha_{t}+f_{u_{t}}+g_{u_{t},\alpha_{t}}\big{\rangle}.$
(C.10)

Noting that the terms involving $\omega\alpha_{t}$ in (C.10) and (C.5a) cancel
gives the claim. ∎

The excitation map for the particles $X_{u}$ satisfies a general
transformation property.

######  Lemma C.2.

For any densely defined operator $B:D(B)\subseteq L^{2}(\mathbb{R}^{3})\to
L^{2}(\mathbb{R}^{3})$ and $u\in D(B)$ we have

$\displaystyle
X_{u}\bigg{(}\sum_{j=1}^{N}B_{j}\bigg{)}X_{u}^{*}=\big{\langle}u,Bu\big{\rangle}(N-\mathcal{N}_{b})$
$\displaystyle\quad+\int
dx\bigg{(}(q_{u}Bu)(x)b_{x}^{*}\sqrt{[N-\mathcal{N}_{b}]_{+}}+\overline{(q_{u}B^{*}u)}(x)\sqrt{[N-\mathcal{N}_{b}]_{+}}b_{x}\bigg{)}$
$\displaystyle\quad+\int dxdy\,b_{x}^{*}Bb_{y}$ (C.11)

as an operator identity on $\mathcal{F}^{(\leq N)}_{\perp u}$.

###### Proof.

This follows by writing $\sum_{j}B_{j}$ as the restriction to
$L^{2}(\mathbb{R}^{3})^{\otimes N}$ of $\mathrm{d}\Gamma(B)$ and using the
identities of Lemma 2.1. ∎

With this, we can now give the proof of the formula for the fluctuation
Hamiltonian.

###### Proof of Lemma 6.10.

We first calculate

$H_{u,\alpha}^{\mathrm{D},\leq N}(t)=X_{u_{t}}H_{\alpha}^{\mathrm{D},\leq
N}(t)X_{u_{t}}^{*}+i\dot{X}_{u_{t}}X_{u_{t}}^{*}$ (C.12)

and then add convenient terms that vanish on $\mathcal{F}_{\perp u_{t}}^{\leq
N}\otimes\mathcal{F}$ to obtain the symmetric expression for
$H_{u,\alpha}^{\mathrm{D}}(t)$.

The terms in $H_{\alpha}^{\mathrm{D},\leq N}(t)$ that are not either invariant
under $X_{u}$ or of the well-known form arising for many-boson systems are
those of lines (C.4c) and (C.4d). Using
$\overline{B_{x_{j}}(k)}=B_{x_{j}}(-k)$ we write

$\big{(}a(kB_{x_{j}})^{2}+\textnormal{h.c.}+2a^{*}(kB_{x_{j}})a(kB_{x_{j}})\big{)}\\\
=\int dkdl\,kB_{x_{j}}(k)\cdot
lB_{x_{j}}(l)\underbrace{\big{[}-2a_{k}^{*}a_{-l}+a_{k}^{*}a_{l}^{*}+a_{-k}a_{-l}\big{]}}_{=:\mathcal{A}_{kl}}$
(C.13)

and by Lemma C.2

$\displaystyle\frac{1}{N}X_{u_{t}}\Big{(}\int
dkdl\,\sum_{j=1}^{N}\big{(}kB_{x_{j}}(k)\cdot
lB_{x_{j}}(l)\big{)}\,\mathcal{A}_{kl}\Big{)}X_{u_{t}}^{*}$
$\displaystyle\quad=\frac{1}{N}\int
dkdl\underbrace{\big{\langle}u_{t},kB_{(\cdot)}(k)\cdot
lB_{(\cdot)}(l)u_{t}\big{\rangle}}_{=M_{u_{t}}(k,l)}(N-\mathcal{N}_{b})\,\mathcal{A}_{kl}$
$\displaystyle\quad\quad+\frac{1}{N}\int
dkdldx\bigg{(}\underbrace{(q_{u_{t}}kB_{(\cdot)}(k)\cdot
lB_{(\cdot)}(l)u_{t})(x)}_{=N_{u_{t}}(x,k,l)}b_{x}^{*}\sqrt{[N-\mathcal{N}_{b}]_{+}}+\text{h.c}$
$\displaystyle\quad\quad+\frac{1}{N}\int
dkdldxdy\underbrace{(q_{u_{t}}kB_{(\cdot)}(k)\cdot
lB_{(\cdot)}(l)q_{u_{t}})(x,y)}_{=Q_{u_{t}}(x,y,k,l)}b_{x}^{*}b_{y}\,\mathcal{A}_{kl}.$
(C.14)

Recalling the formulas (3.4e), (3.4f), and (4.2a) for $f_{u_{t}}(k)$,
$g_{u_{t},\alpha_{t}}(k)$, and $L_{\alpha}(k)$, we also find

$\displaystyle\frac{2}{\sqrt{N}}X_{u_{t}}\bigg{(}\sum_{j=1}^{N}\int
dk\,kB_{x_{j}}(k)\cdot\big{(}-i\nabla_{j}+F_{\alpha_{t}}(x_{j})\big{)}a_{k}^{*}+\text{h.c.}\bigg{)}X_{u_{t}}^{*}$
(C.15) $\displaystyle\quad=\frac{2}{\sqrt{N}}\int
dk\,X_{u_{t}}\bigg{(}\sum_{j=1}^{N}kB_{x_{j}}(k)\cdot\big{(}-i\nabla_{j}+F_{\alpha_{t}}(x_{j})\big{)}\bigg{)}X_{u_{t}}^{*}a_{k}^{*}+\text{h.c.}$
$\displaystyle\quad=\frac{1}{\sqrt{N}}\int
dk\big{(}f_{u_{t}}(k)+g_{u_{t},\alpha_{t}}(k)\big{)}(N-\mathcal{N}_{b})\,a_{k}^{*}+\text{h.c.}$
$\displaystyle\qquad+\frac{2}{\sqrt{N}}\int
dkdx(q_{u_{t}}L_{\alpha_{t}}(k)u_{t})(x)b_{x}^{*}a_{k}^{*}\sqrt{[N-\mathcal{N}_{b}]_{+}}+\text{h.c.}$
$\displaystyle\qquad+\frac{2}{\sqrt{N}}\int
dkdx(\overline{q_{u_{t}}L_{\alpha_{t}}(k)^{*}u_{t})}(x)\sqrt{[N-\mathcal{N}_{b}]_{+}}b_{x}a_{k}^{*}+\text{h.c.}$
$\displaystyle\quad\quad+\frac{2}{\sqrt{N}}\int
dkdxdy\underbrace{(q_{u_{t}}kB_{(\cdot)}(k)\cdot(-i\nabla+F_{\alpha_{t}})q_{u_{t}})(x,y)}_{J_{u_{t},\alpha_{t}}(x,k,y)}b_{x}^{*}b_{y}\,a_{k}^{*}+\text{h.c.}$

The term $\sqrt{N}\hat{\Phi}(f_{u_{t}}+g_{u_{t},\alpha_{t}})$ from the first
line cancels with the corresponding term in $H_{\alpha}^{\mathrm{D},\leq
N}(t)$, and the remaining term
$-N^{-1/2}\mathcal{N}_{b}\hat{\Phi}(f_{u_{t}}+g_{u_{t},\alpha_{t}})$ equals
$H_{2}(t)$ from the formula for $H^{\rm D}_{u,\alpha}(t)$. We thus have