Proof. Multiplying equation (14) by ${\varphi }_p$ , summing over p and using that $V \geq 0$ , we obtain

(20) $$ \begin{align} 2 \|p{\varphi}\|_2^2 = - \sum_{p} \hat V_N(p) {\varphi}_p - \sum_{p,q} \hat V_N(p-q) {\varphi}_p {\varphi}_q \leq - \sum_{p} \hat V_N(p) {\varphi}_p. \end{align} $$

On one hand, this implies that $\| p {\varphi } \|_2 \lesssim \| V_N \|_2 \| {\varphi } \|_2 < \infty $ (the last bound is not uniform in N; it follows from equation (15)). On the other hand, the estimate (20) leads to

$$\begin{align*}2 \| p {\varphi} \|^2_2 \leq \| V_N / |p| \|_2 \| p {\varphi} \|_2. \end{align*}$$

Dividing by $\| p {\varphi } \|_2$ and squaring, we obtain

(21) $$ \begin{align} \|p{\varphi}\|_2^2 \lesssim \sum_p \frac{|\hat V_N(p)|^2}{p^2} \lesssim \| \hat{V}_N \|_{\infty}^2 \| |p|^{-2} \chi_{|p| < N} \|_1 + \| \hat{V}_N \|_{\infty} \| \hat{V}_N \|_2 \| |p|^{-2} \chi_{|p|> N} \|_2 \lesssim N^{-1}. \end{align} $$

Using equation (14) again, we obtain the pointwise bound

(22) $$ \begin{align} | p^2 {\varphi}_p| \leq |\hat{V}_N (p)| + \Big[ \sum_q \frac{|\hat{V}_N (p-q)|^2}{q^2} \Big]^{1/2} \| q {\varphi} \|_2 \lesssim N^{-1}, \end{align} $$

where we proceeded as in (21) to bound $\| |\hat {V}_N|^2 * |q|^{-2} \|_{\infty }$ . This proves the bound (18) and immediately implies the bounds for $\| \tilde {\varphi } \|_2 , \| \tilde {\varphi } \|_{\infty }$ . To obtain the bound on $ \| \tilde {\varphi } \|_1$ , we divide equation (14) by $|p|^2$ . Proceeding as in (21), we obtain

$$ \begin{align*} \sum_{p\neq 0} \frac{|\hat{V}_N (p)|}{|p|^2} \lesssim 1 \, , \end{align*} $$

and hence we only have to bound $\||p|^{-2} (\hat {V}_N \ast {\varphi }) \|_{1}$ . Iterating equation (14) and using the regularising estimate $\||p|^{-2} \hat {V}_N \ast g\|_{6p/(6+p) + \varepsilon } \leq C_{\varepsilon } \|\hat {V}_N\|_2 \|g\|_{p}$ for all $\varepsilon>0$ , $p \geq 6/5$ , $g \in \ell ^{p}(\Lambda ^*)$ and some $C_{\varepsilon }> 0$ , we obtain that $\|{\varphi }\|_1 < \infty $ . Separating high and low momenta, we obtain for $A \geq 1$ and $\varepsilon>0$

$$ \begin{align*} \|\varphi\|_{1} &\lesssim 1+ \| \chi_{|p|> A N} |p|^{-2} \|_{2} \|V_N \check {\varphi}\|_{2} + \|\chi_{|p| \leq A N}|p|^{-2}\|_{1} \| \hat{V}_N \ast {\varphi} \|_{\infty} \\ &\lesssim 1 + A^{-\tfrac{1}{2}} \|{\varphi}\|_{1} + A \, , \end{align*} $$

where we used that $\|\check {\varphi }\|_{\infty } \leq \|{\varphi }\|_1$ and the Hölder inequality as in (22) to estimate $\| \hat {V}_N \ast {\varphi } \|_{\infty }$ . Taking A sufficiently large but fixed, we obtain $\|\varphi \|_{1} \lesssim 1$ .

The estimate (19) follows by noticing that, from the definition given in equation (16),

$$\begin{align*}\Big| 8\pi \mathfrak{a}_N - \hat{V} (0) - N \sum_p V_N (p) \tilde{\varphi}_p \Big| \leq N \sum_{|p| \leq N^{\alpha}} |\hat{V}_N (0)| |{\varphi}_p| \lesssim \frac{1}{N} \sum_{|p| \leq N^{\alpha}}\frac{1}{|p|^2} \lesssim N^{-1+\alpha}, \end{align*}$$

where we use equation (18) and $\| \hat {V}_N \|_{\infty } \lesssim N^{-1}$ .

Proof. We have

(30) $$ \begin{align} e^{-{\mathcal{B}}_2}H_2e^{{\mathcal{B}}_2}=N \sum_{p} \hat V_N(p) &\left(a^{\dagger}_p a_p + \int_0^1 e^{-s{\mathcal{B}}_2} [ a^{\dagger}_p a_p,{\mathcal{B}}_2] e^{s{\mathcal{B}}_2} \mathrm{d}s\right) \notag \\ &\qquad - e^{-{\mathcal{B}}_2} \left( \sum_{p} \hat V_N(p) a^{\dagger}_p a_p\mathcal{N}_+ + \frac {\hat V_N(0)}{2} \mathcal{N}_+(\mathcal{N}_+ -1) \right) e^{{\mathcal{B}}_2}. \end{align} $$

The term on the second line is controlled using Lemma 3 by $N^{-1} \mathcal {N}_+^2$ . For the second term in parenthesis in the first line, we use Lemma 3 to estimate

$$ \begin{align*} \pm \sum_{p} N \hat V_N(p) [ a^{\dagger}_p a_p,{\mathcal{B}}_2] &= \pm \sum_{p} N \hat V_N(p) \tilde {\varphi}_p a^{\dagger}_p a^{\dagger}_{-p} a_0 a_0 +\text{h.c.} \\ &\leq C \|\tilde {\varphi} \|_{2} (\mathcal{N}_0+1) (\mathcal{N}_++1) \leq N^{- \alpha/2} (\mathcal{N}_+ + 1), \end{align*} $$

where we used that $\mathcal {N}_0\lesssim N$ and Lemma 2. Finally, since $V_N$ is even, we obtain

$$ \begin{align*} N |\hat V_N (p) - \hat V_N(0)| \leq \|x^2V\|_{1} N^{-2} p^2 \leq C N^{-2} p^2, \end{align*} $$

which gives

$$ \begin{align*} \pm \left(N \sum_{p} \hat V_N(p) a^{\dagger}_p a_p - \hat V(0) \mathcal{N}_+\right) \leq N^{-2} H_1.\\[-50pt] \end{align*} $$

Proof. We can rewrite

(31) $$ \begin{align} e^{-{\mathcal{B}}_2} Q_3 e^{{\mathcal{B}}_2} = \sum_{q,r} \hat{V}_N(r)\left[e^{-{\mathcal{B}}_2}a_{q+r}^{\dagger} a_{-r}^{\dagger} e^{{\mathcal{B}}_2} e^{-{\mathcal{B}}_2} a_{q} a_0 e^{{\mathcal{B}}_2} + \text{h.c.} \right]. \end{align} $$

Via Duhamel’s formula, we have

(32) $$ \begin{align} e^{-{\mathcal{B}}_2}a_{q+r}^{\dagger} a_{-r}^{\dagger} e^{{\mathcal{B}}_2} = a^{\dagger}_{q+r} a^{\dagger}_{-r} + \int_0^1 e^{-s{\mathcal{B}}_2} [a^{\dagger}_{q+r} a^{\dagger}_{-r}, {\mathcal{B}}_2] e^{s{\mathcal{B}}_2} \mathrm{d}s \end{align} $$

and

$$ \begin{align*} e^{-{\mathcal{B}}_2} a_{q} a_0 e^{{\mathcal{B}}_2} = a_q a_0 + \int_0^1 e^{-s{\mathcal{B}}_2} (\tilde {\varphi}_q a^{\dagger}_{-q} a_0 a_0 a_0 - a_q \sum_l \tilde {\varphi}_l a_{-l} a_l a_0^{\dagger}) e^{s{\mathcal{B}}_2} \mathrm{d}s. \end{align*} $$

The product of the first terms, combined with its hermitian conjugate, corresponds to $Q_3$ in the statement of the lemma. All other terms will be estimated in three steps.

Step 1. Passing to x-space, we find

(33) $$ \begin{align} \begin{aligned} \Big| \sum_{q,r}& \tilde {\varphi}_q \hat{V}_N(r) \int_0^1 \big\langle \xi , a_{q+r}^{\dagger} a_{-r}^{\dagger} e^{-s{\mathcal{B}}_2} a^{\dagger}_{-q} a_0 a_0 a_0 e^{s{\mathcal{B}}_2} \xi \rangle \, \mathrm{ d}s \Big| \\ &= \int_0^1 \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y V_N(x-y) \big\langle \xi, {\check a}^{\dagger}_x {\check a}_y^{\dagger} e^{-s{\mathcal{B}}_2} {\check a}^{\dagger}(\check{\tilde{\varphi}}_x) a_0 a_0 a_0 e^{s{\mathcal{B}}_2} \xi \big\rangle \, \mathrm{d}s \\ &\leq \Big( \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y V_N(x-y) \| {\check a}_x {\check a}_y \xi \|^2 \Big)^{1/2} \\ &\hspace{2cm} \times \Big( \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y V_N (x-y) \int_0^1 \| {\check a}^{\dagger}(\check{\tilde{\varphi}}_x) a_0 a_0 a_0 e^{s{\mathcal{B}}_2} \xi \|^2 \mathrm{d}s \Big)^{1/2} \\ &\leq C N^{3/2} \| V_N \|_1^{1/2} \| \tilde{\varphi} \|_2 \| Q_4^{1/2} \xi \| \| (\mathcal{N}_+ + 1)^{1/2} \xi \| \\ &\leq C N^{-\alpha/2} \langle \xi , (Q_4 + \mathcal{N}_+ + 1) \xi \rangle, \end{aligned} \end{align} $$

where we used Lemma 2, Lemma 3 and the bound $\mathcal {N}_0 \leq N$ .

Step 2. Similarly, we have

(34) $$ \begin{align} \begin{aligned} \Big| \sum_{q,r} &\hat{V}_N(r) \sum_l \tilde {\varphi}_l \int_0^1 \mathrm{d}s \, \big\langle \xi, a_{q+r}^{\dagger} a_{-r}^{\dagger} e^{-s{\mathcal{B}}_2} a_{q} a_{-l} a_l a^{\dagger}_0 e^{s{\mathcal{B}}_2} \xi \big\rangle \Big| \\ &= \Big| \sum_l \tilde {\varphi}_l \int_0^1 \mathrm{d}s \int_{\Lambda^2} \mathrm{ d}x \mathrm{d}y \, V_N(x-y) \langle \xi , a^{\dagger}_x a^{\dagger}_y e^{-s{\mathcal{B}}_2} a_x a_{-l} a_l a^{\dagger}_0 e^{s{\mathcal{B}}_2} \xi \rangle \Big| \\ &\leq C \| \tilde {\varphi} \|_2 \|V_N \|^{1/2}_1 \| Q_4^{1/2} \xi \| \| (\mathcal{N}_++1)^{2} \xi \| \leq C N^{-\alpha/2} \langle \xi , Q_4 + \mathcal{N}_+ + 1) \xi \rangle. \end{aligned} \end{align} $$

Step 3. The remaining term has the form

$$ \begin{align*} \sum_{q,r} \hat{V}_N(r)\int_0^1e^{-s{\mathcal{B}}_2}[a_{q+r}^{\dagger} a_{-r}^{\dagger},{\mathcal{B}}_2] e^{(s-1){\mathcal{B}}_2} a_{q} a_0 e^{{\mathcal{B}}_2} \mathrm{ d}s. \end{align*} $$

A straightforward computation gives

$$ \begin{align*} [a_{q+r}^{\dagger} a_{-r}^{\dagger},{\mathcal{B}}_2] = - \left( \tilde{\varphi}_r a^{\dagger}_{q+r} a_r + \tilde{\varphi}_{-r -q} a^{\dagger}_{-r} a_{-q-r} \right)a^{\dagger}_0a^{\dagger}_0. \end{align*} $$

The contributions of the two terms in parenthesis can be handled similarly. Let us consider, for example, the expectation

(35) $$ \begin{align} \begin{aligned} \Big| \sum_{q,r} \hat{V}_N(r) \tilde {\varphi}_r \int_0^1& \langle \xi, e^{-s{\mathcal{B}}_2}a_{q+r}^{\dagger} a_{r} a^{\dagger}_0 a^{\dagger}_0e^{(s-1){\mathcal{B}}_2} a_{q} a_0 e^{{\mathcal{B}}_2} \xi \rangle \, \mathrm{d}s \Big| \\ &\lesssim \frac{1}{N} \sum_{r,q} |\tilde{\varphi}_r| \int_0^1 \| a_{q+r} a_0 a_0 e^{(s-1){\mathcal{B}}_2} \xi\| \| a_r e^{s {\mathcal{B}}_2} a_q a_0 e^{{\mathcal{B}}_2} \xi \| \mathrm{d}s \\ &\lesssim \| \tilde {\varphi} \|_2 \| \mathcal{N}_+^{1/2} \xi \| \int_0^1 \Big( \sum_{r,q} \| a_r e^{(s-1) {\mathcal{B}}_2} a_q a_0 e^{{\mathcal{B}}_2} \xi \|^2 \Big)^{1/2} \mathrm{d}s \\ &\lesssim \| \tilde {\varphi}\|_2 \| \mathcal{N}_+^{1/2} \xi \| \| (\mathcal{N}_+ +1)^{3/2} \xi \| \leq N^{-\alpha/2} \langle \xi , (\mathcal{N}_+ + 1) \xi \rangle, \end{aligned} \end{align} $$

where we used Lemma 2 and Lemma 3.

Proof. Straightforward calculations yield

$$ \begin{align*} [H_1,{\mathcal{B}}_2] = \sum_{p } p^2 \tilde {\varphi}_p [a_p^{\dagger} a_{-p}^{\dagger} a_0 a_0 + \text{h.c.}] \end{align*} $$

and

(38) $$ \begin{align} [Q_4, {\mathcal{B}}_2] = \frac{1}{2} \sum_{p,q} \hat{V}_N(p-q) \tilde {\varphi}_q a_{p}^{\dagger} a_{-p}^{\dagger} a_0a_0 + \sum_{r,p,q} \hat{V}_N(r) \tilde {\varphi}_p a_{p+r}^{\dagger} a_q^{\dagger} a_{-p}^{\dagger} a_{q+r}a_0 a_0 + \text{h.c.} \end{align} $$

Hence,

$$\begin{align*}\begin{aligned} [H_1 + Q_4,{\mathcal{B}}_2] + Q_2 = \; &\sum_{|p|> N^{\alpha}} \Big(p^2 {\varphi}_p + \frac{1}{2} \sum_{q} \hat{V}_N(p-q) {\varphi}_q + \frac{1}{2} \hat{V}_N (p)\Big) (a_p^{\dagger} a_{-p}^{\dagger} a_0 a_0+ \text{h.c.} ) \\ &- \frac{1}{2} \sum_{|p| > N^{\alpha} , |q| \leq N^{\alpha}} \hat{V}_N (p-q) {\varphi}_q (a_p^{\dagger} a_{-p}^{\dagger} a_0 a_0+ \text{h.c.} ) \\ &+ \frac{1}{2} \sum_{|p| \leq N^{\alpha}} \Big( \sum_{|q| > N^{\alpha}} \hat{V}_N (p-q) {\varphi}_q + \hat{V}_N (p) \Big) (a_p^{\dagger} a_{-p}^{\dagger} a_0 a_0+ \text{h.c.} ) \\ &+ \sum_{r,p,q} \hat{V}_N(r) \tilde {\varphi}_p a_{p+r}^{\dagger} a_q^{\dagger} a_{-p}^{\dagger} a_{q+r}a_0 a_0 \\ = \: &\tilde Q^{\prime}_2 + \sum_{r,p,q} \hat{V}_N(r) \tilde {\varphi}_p a_{p+r}^{\dagger} a_q^{\dagger} a_{-p}^{\dagger} a_{q+r}a_0 a_0, \end{aligned} \end{align*}$$

where we used the scattering equation (14) and the definition of $\tilde Q'_2$ given in equation (27). Comparing with equation (26), we find equation (36).

To prove the estimate (37), we write

$$\begin{align*}\int_0^1 e^{-t \mathcal{B}_2} \Gamma_2 e^{t \mathcal{B}_2} \mathrm{d}t = \sum_{p,q,r} \hat{V}_{N}(r)\tilde{\varphi}_p \int_0^1 e^{-t{\mathcal{B}}_2} a_{p+r}^{\dagger} a_q^{\dagger} e^{t{\mathcal{B}}_2}e^{-t{\mathcal{B}}_2} a_{-p}^{\dagger} a_{q+r} a_0 a_0 e^{t{\mathcal{B}}_2} \mathrm{d}t, \end{align*}$$

and we proceed similarly as in Lemma 6. We omit further details.

Proof. First, we claim that

(41) $$ \begin{align} \int_0^1 \int_s^1 e^{-t{\mathcal{B}}_2}[Q_2,{\mathcal{B}}_2]e^{t{\mathcal{B}}_2} \mathrm{d}t \mathrm{d}s &= \frac { N(N-1)}{2} \sum_p \hat{V}_N(p) \tilde {\varphi}_p -N \mathcal{N}_+ \sum_p \hat{V}_N(p) \tilde {\varphi}_p + \mathcal{E}^{\prime}_{[Q_2,{\mathcal{B}}_2]} \end{align} $$

with

(42) $$ \begin{align} \mathcal{E}^{\prime}_{[Q_2,{\mathcal{B}}_2]} = - 2N &\sum_p \hat{V}_N (p) \tilde{\varphi}_p \int_0^1 \int_s^1 \left[ e^{-t \mathcal{B}_2} \mathcal{N}_+ e^{t \mathcal{B}_2} - \mathcal{N}_+ \right] \mathrm{d}t \mathrm{ d}s \nonumber \\ &\quad + \sum_p \hat{V}_N(p) \tilde{\varphi}_p \int_0^1 \int_s^1 e^{-t{\mathcal{B}}_2} \mathcal{N}_+( \mathcal{N}_+ + 1 )e^{t{\mathcal{B}}_2}\mathrm{d}t \mathrm{d}s \nonumber \\ &\quad +2\sum_p \hat{V}_N(p) \tilde {\varphi}_p \int_0^1 \int_s^1 e^{-t{\mathcal{B}}_2} \mathcal{N}_0(\mathcal{N}_0 -1) a^{\dagger}_p a_pe^{t{\mathcal{B}}_2} \mathrm{d}t \mathrm{d}s \nonumber \\ &\quad -\sum_{p,q}\hat{V}_N(p) \tilde {\varphi}_q \int_0^1 \int_s^1 e^{-t{\mathcal{B}}_2} a^{\dagger}_{p} a^{\dagger}_{-p} a_{-q} a_{q} (2 \mathcal{N}_0+1 ) e^{t{\mathcal{B}}_2} \mathrm{d}t \mathrm{d}s. \end{align} $$

To prove equation (42), we calculate

(43) $$ \begin{align}\nonumber [Q_2,{\mathcal{B}}_2] &= \frac{1}{4} \sum_{p,q} \hat{V}_N(p) \tilde{\varphi}_q [a_p^{\dagger} a_{-p}^{\dagger} a_0a_0 + a_{-p}a_pa_0^{\dagger} a_0^{\dagger} , a_q^{\dagger} a_{-q}^{\dagger} a_0a_0 - a_{-q}a_q a_0^{\dagger} a_0^{\dagger}] \\ &= \frac{1}{4} \sum_{p,q} \hat{V}_N(p) \tilde{\varphi}_q \left( [a_{-p}a_pa_0^{\dagger} a_0^{\dagger} , a_q^{\dagger} a_{-q}^{\dagger} a_0a_0] - [a_p^{\dagger} a_{-p}^{\dagger} a_0a_0,a_{-q}a_q a_0^{\dagger} a_0^{\dagger}] \right). \end{align} $$

The two terms in brackets are hermitian conjugates. Hence, it suffices to compute the second one

$$ \begin{align*} - [a_p^{\dagger} a_{-p}^{\dagger} a_0a_0,a_{-q}a_q a_0^{\dagger} a_0^{\dagger}] &= [a_{-q}a_q, a_p^{\dagger} a_{-p}^{\dagger}] a_0^{\dagger} a_0^{\dagger} a_0a_0 - a_p^{\dagger} a_{-p}^{\dagger} a_{-q} a_q [a_0a_0,a_0^{\dagger} a_0^{\dagger}], \end{align*} $$

where

(44) $$ \begin{align} [a_{-p}a_p,a_q^{\dagger} a_{-q}^{\dagger} ] = (\delta_{p,q} + \delta_{p,-q})( 1 + a^{\dagger}_p a_p + a^{\dagger}_{-p} a_{-p} ), \end{align} $$

and

(45) $$ \begin{align} \begin{aligned} a_0^{\dagger} a_0^{\dagger} a_0 a_0 &= \mathcal{N}_0(\mathcal{N}_0-1)= N(N -1) - 2 N \mathcal{N}_+ + \mathcal{N}_+(\mathcal{N}_+ +1), \\ [a_0a_0,a^{\dagger}_0a^{\dagger}_0] &= 2(2 \mathcal{N}_0 + 1). \end{aligned} \end{align} $$

Inserting these identities on the right-hand side of equation (43), conjugating with $e^{t\mathcal {B}_2}$ and integrating over $t\ \mathrm{and}\ s$ , we obtain equation (42).

With the definition given in equation (16), we write

$$\begin{align*}\begin{aligned} \frac{N(N-1)}{2} \sum_p \hat{V}_N (p) \tilde{{\varphi}}_p &= \frac{(N-1)}{2} (\hat{V} (0) - 8 \pi \mathfrak{a}_N) - \frac{N(N-1)}{2} \sum_{|p| \leq N^{\alpha}} \hat{V}_N (p) {\varphi}_p , \end{aligned} \end{align*}$$

and we use the bound (19) to estimate

$$\begin{align*}\pm \Big[ - N \mathcal{N}_+ \sum_p \hat{V}_N (p) \tilde{\varphi}_p + (8\pi \mathfrak{a}_N - \hat{V} (0)) \mathcal{N}_+ \Big] \lesssim N^{\alpha-1} \mathcal{N}_+. \end{align*}$$

Thus, Lemma 8 follows from equation (41), if we can prove that $\mathcal {E}^{\prime }_{[Q_2,{\mathcal {B}}_2]}$ satisfies the estimate given in (40).

Using the bound

$$\begin{align*}\Big| \sum_p \hat{V}_N (p) \tilde{\varphi}_p \Big| \lesssim N^{-1} \end{align*}$$

and Lemma 3, we can bound the first term on the right-hand side of equation (42) by

$$\begin{align*}\pm 2N \sum_p \hat{V}_N (p) \tilde{\varphi}_p \int_0^1 \int_s^1 \left[ e^{-t \mathcal{B}_2} \mathcal{N}_+ e^{t \mathcal{B}_2} - \mathcal{N}_+ \right] \mathrm{d}t \mathrm{d}s \lesssim N^{-\alpha/2} (\mathcal{N}_+ + 1). \end{align*}$$

Also, the second term on the right-hand side of equation (42) can be bounded with Lemma 3; we find

$$\begin{align*}\pm \sum_p \hat{V}_N (p) \tilde{\varphi}_p \int_0^1 \int_s^1 e^{-t \mathcal{B}_2} \mathcal{N}_+ ( \mathcal{N}_+ + 1) e^{t \mathcal{B}_2} \mathrm{d}t \mathrm{d}s \lesssim N^{-1} (\mathcal{N}_+ + 1)^2. \end{align*}$$

For the third term, we use $\| \hat {V}_N \|_{\infty } \lesssim N^{-1}$ , $\| \tilde {\varphi } \|_{\infty } \lesssim N^{-1-2\alpha }$ together with $\mathcal {N}_0 (\mathcal {N}_0 -1) \leq N^2$ and Lemma 3 again to conclude that

$$\begin{align*}\pm 2 \sum_p \hat{V}_N (p) \tilde{\varphi}_p \int_0^1 \int_s^1 e^{-t \mathcal{B}_2} \mathcal{N}_0 ( \mathcal{N}_0 + 1) e^{t \mathcal{B}_2} \mathrm{d}t \mathrm{d}s \lesssim N^{-2\alpha} \mathcal{N}_+. \end{align*}$$

To control the last term on the right-hand side of equation (42), we write

(46) $$ \begin{align} \sum_{p,q} \hat{V}_N(p) \tilde {\varphi}_q \int_0^1 \int_s^1 e^{-t{\mathcal{B}}_2} & a^{\dagger}_{p} a^{\dagger}_{-p} a_{-q} a_{q} (2 \mathcal{N}_0+1) e^{t{\mathcal{B}}_2} \mathrm{d}t \mathrm{d}s \notag \\ &\quad =\sum_{p} \hat{V}_N(p) \int_0^1 \int_s^1 e^{-t{\mathcal{B}}_2} a^{\dagger}_{p} a^{\dagger}_{-p} e^{t{\mathcal{B}}_2} e^{-t{\mathcal{B}}_2} \Phi (2 \mathcal{N}_0+1) e^{t{\mathcal{B}}_2} \mathrm{d}t \mathrm{ d}s, \end{align} $$

where we define $\Phi = \sum _q \tilde {\varphi }_q a_{-q} a_q$ so that, by Lemma 2,

(47) $$ \begin{align} \| \Phi \xi \| \lesssim \| \tilde{\varphi} \|_2 \| \mathcal{N}_+ \xi \| \lesssim N^{-1-\alpha/2} \| \mathcal{N}_+ \xi \|. \end{align} $$

Next, we expand

(48) $$ \begin{align} e^{- t{\mathcal{B}}_2} a^{\dagger}_{p} a^{\dagger}_{-p} e^{t{\mathcal{B}}_2} = a^{\dagger}_{p} a^{\dagger}_{-p} + \int_0^t e^{-\tau {\mathcal{B}}_2} [a^{\dagger}_{p} a^{\dagger}_{-p} ,{\mathcal{B}}_2]e^{\tau {\mathcal{B}}_2}d\tau. \end{align} $$

Inserting this identity into equation (46), we obtain two contributions. The first contribution can be controlled by passing to position space. We find

$$\begin{align*}\begin{aligned} \pm \sum_{p} \hat{V}_N(p) &\int_0^1 \int_s^1 a^{\dagger}_{p} a^{\dagger}_{-p} e^{-t{\mathcal{B}}_2} \Phi (2 \mathcal{N}_0+1) e^{t{\mathcal{B}}_2} \mathrm{d}t \mathrm{d}s \\ &= \pm \int_0^1 \int_s^1 \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y \kappa V_N(x-y) {\check a}_x^{\dagger} {\check a}_y^{\dagger} e^{-t{\mathcal{B}}} \Phi (2 \mathcal{N}_0 +1) e^{t{\mathcal{B}}} \mathrm{d}t \mathrm{d}s + \text{h.c.} \\ &\lesssim \delta Q_4 + \delta^{-1} N^2 \|\varphi\|_{L^2}^2 \|V_N\|_{1} (\mathcal{N}_++1)^2 \lesssim N^{-\alpha/2} Q_4 + N^{-1-\alpha/2} (\mathcal{N}_++1)^2 .\end{aligned} \end{align*}$$

On the other hand, the contribution arising from the second term on the right-hand side of equation (48) can be controlled by

(49) $$ \begin{align} \pm \sum_{p} \hat{V}_N(p) \tilde {\varphi}_p \int_0^1 \int_s^1\int_0^t e^{-\tau {\mathcal{B}}_2} a^{\dagger}_0 a^{\dagger}_0 (2 a^{\dagger}_p a_p + 1 )e^{\tau {\mathcal{B}}_2} e^{-t{\mathcal{B}}} \Phi &(2 \mathcal{N}_0 +1) e^{t{\mathcal{B}}} \mathrm{d}t \mathrm{d}s + \text{h.c.} \notag \\ &\qquad \lesssim N^{-1 -\alpha/2} (\mathcal{N}_++1)^2. \end{align} $$

This concludes the proof of (39) and (40).

Proof. Using that $ \chi _{|p| \leq N^{\alpha }} \tilde \varphi _p = 0$ , similar computations as in the proof of Lemma 41 yield

(51) $$ \begin{align} \begin{aligned} [\tilde Q_2 &,{\mathcal{B}}_2] \\ = \; &\frac{1}{4} \sum_{|p|>N^{\alpha}, |q|\leq N^{\alpha},r} \hat V_N(p-q) {\varphi}_q \tilde {\varphi}_r \\ &\times (a^{\dagger}_p a^{\dagger}_{-p} a_r a_{-r}[a_0 a_0,a^{\dagger}_0 a^{\dagger}_0] +[a^{\dagger}_p a^{\dagger}_{-p},a_r a_{-r}]a^{\dagger}_0 a^{\dagger}_0 a_0 a_0)+\text{h.c.} \\ = \; &\sum_{|p|>N^{\alpha}, |q| \leq N^{\alpha}} \frac{1}{4}\hat V_N(p-q) {\varphi}_q \\ & \times \big( (2 a^{\dagger}_p a^{\dagger}_{-p} \Phi (2\mathcal{N}_0+1) + \text{h.c.}) - 4 \tilde {\varphi}_p (a^{\dagger}_p a_p +a^{\dagger}_{-p} a_{-p}) \mathcal{N}_0 (\mathcal{N}_0-1) -4\tilde{\varphi}_p \mathcal{N}_0 (\mathcal{N}_0-1)\big), \end{aligned} \end{align} $$

with the notation $\Phi = \sum _r \tilde {\varphi }_r a_r a_{-r}$ . To bound the contribution arising from the first term in parentheses, we decompose

(52) $$ \begin{align} \begin{aligned} \frac{1}{2} &\sum_{|p|>N^{\alpha}, |q|\leq N^{\alpha}} \hat V_N(p-q) {\varphi}_q a^{\dagger}_p a^{\dagger}_{-p} \Phi (2\mathcal{N}_0+1)\\ = \; &\frac{1}{2} \sum_{p,|q|\leq N^{\alpha}} \hat V_N(p-q) {\varphi}_q a^{\dagger}_p a^{\dagger}_{-p} \Phi (2\mathcal{N}_0+1) - \frac{1}{2} \sum_{|p|, |q|\leq N^{\alpha}} \hat V_N(p-q) {\varphi}_q a^{\dagger}_p a^{\dagger}_{-p} \Phi (2\mathcal{N}_0+1). \end{aligned} \end{align} $$

The first term can be controlled by switching to position space. With the notation $\check {{\varphi }}^<$ for the Fourier series of $\chi _{|q| \leq N^{\alpha }} {\varphi }_q$ , we find

$$\begin{align*}\begin{aligned} \pm \frac{1}{2} &\sum_{p,|q|\leq N^{\alpha}} \hat V_N(p-q) {\varphi}_q a^{\dagger}_p a^{\dagger}_{-p} \Phi (2\mathcal{N}_0+1) \\ &= \pm \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y V_N (x-y) \check {\varphi}^< (x-y) a^{\dagger}_x a^{\dagger}_y \Phi (2\mathcal{N}_0+1) + \text{h.c.} \nonumber \\ &\lesssim \delta Q_4 + \delta^{-1} N^{-\alpha} \|V_N^{1/2} \check{\varphi}^< \|_2^2 (\mathcal{N}_++1)^2, \end{aligned} \end{align*}$$

where we used the bound (47) for $\Phi $ and $\mathcal {N}_0 \leq N$ . With

(53) $$ \begin{align} \| V_N^{1/2} \check{\varphi}^< \|_2^2 = \sum_{|p|, |q| \leq N^{\alpha}} \hat{V}_N (p-q) {\varphi}_p {\varphi}_q \lesssim \frac{1}{N^3} \Big[ \sum_{|p| \leq N^{\alpha}} \frac{1}{|p|^2} \Big]^2 \lesssim N^{2\alpha - 3} \end{align} $$

and choosing $\delta = N^{-\alpha /2}$ , we conclude (since $\alpha < 1$ ) that

$$\begin{align*}\pm \frac{1}{2} \sum_{p,|q|\leq N^{\alpha}} \hat V_N(p-q) {\varphi}_q a^{\dagger}_p a^{\dagger}_{-p} \Phi (2\mathcal{N}_0+1) \lesssim N^{-\alpha/2} Q_4 + N^{-1-\alpha/2} (\mathcal{N}_+ + 1)^2. \end{align*}$$

For the second term on the right-hand side of equation (52), we estimate

$$\begin{align*}\pm \frac{1}{2} \sum_{|p|, |q|\leq N^{\alpha}} \hat V_N(p-q) {\varphi}_q a^{\dagger}_p a^{\dagger}_{-p} \Phi (2\mathcal{N}_0+1) + \text{h.c.} \lesssim N^{-\alpha/2} \| \chi_{|p|\leq N^{\alpha}} \hat V_N \ast {\varphi}^{<} \|_{2} \| (\mathcal{N}_+ + 1)^2, \end{align*}$$

where, again, we used the bound (47) and $\mathcal {N}_0 \leq N$ . With

$$\begin{align*}\| \chi_{|p|\leq N^{\alpha}} \hat V_N \ast {\varphi}^{<}\|_{2} \leq \|\chi_{|p|\leq N^{\alpha}}\|_{L^2} \|V_N \check{\varphi}^{<}\|_{1} \lesssim N^{3\alpha/2}\| V_N^{1/2}\|_{2} \| V_N^{1/2} \check {\varphi}^{<}\|_{2} \lesssim N^{-2+5\alpha/2} \end{align*}$$

we conclude that

$$\begin{align*}\pm \frac{1}{2} \sum_{|p|, |q|\leq N^{\alpha}} \hat V_N(p-q) {\varphi}_q a^{\dagger}_p a^{\dagger}_{-p} \Phi (2\mathcal{N}_0+1) + \text{h.c.} \lesssim N^{-2-2\alpha} (\mathcal{N}_+ + 1)^2. \end{align*}$$

The contribution arising from the second term in parentheses on the right-hand side of equation (51) can be bounded by

$$ \begin{align*} \pm \sum_{|p|>N^{\alpha}, |q| \leq N^{\alpha}} \hat V_N(p-q) {\varphi}_q \tilde {\varphi}_p a^{\dagger}_p a_p \mathcal{N}_0 (\mathcal{N}_0-1) \lesssim N^{-1- \alpha} \mathcal{N}_+, \end{align*} $$

using that $ \|\tilde {\varphi } (\hat V_N \ast {\varphi }^{<})\|_{L^{\infty }} \lesssim \|\tilde {\varphi } \|_{\infty } \| V_N \check {\varphi }^{<}\|_1 \lesssim N^{-3-\alpha }$ . For the contribution arising from the last term on the right-hand side of equation (51), we write $\mathcal {N}_0(\mathcal {N}_0-1)= N(N -1) - 2 N \mathcal {N}_+ + \mathcal {N}_+ (\mathcal {N}_+ +1)$ . The contribution proportional to $N(N-1)$ produces the main term on the right-hand side of equation (50). The other contributions can be bounded, noticing that

$$ \begin{align*} \Big| \sum_{|p|> N^{\alpha}, |q|\leq N^{\alpha}} \hat{V}_N(p-q) {\varphi}_p {\varphi}_q \Big| \leq \|\check {\tilde {\varphi}} V_N \check {\varphi}^{<}\|_{1} \leq \|\check {\tilde {\varphi}}\|_{\infty} \|V_N \check{\varphi}^< \|_1 \lesssim N^{-2+\alpha}, \end{align*} $$

where we used $\| \check {\tilde {\varphi }} \|_{\infty } \lesssim \| \tilde {\varphi } \|_1 \lesssim 1$ , by Lemma 2.

Proof of Proposition 4.

Combining equation (28) with the bounds proven in Lemma 5, Lemma 6, Lemma 7, Lemma 8 and Lemma 9, we conclude that

(54) $$ \begin{align} e^{-\mathcal{B}_2} H_N e^{\mathcal{B}_2} = 4 \pi \mathfrak{a}_N (N-1) + A_{\alpha} + H_1 + \tilde{H}_2 + \tilde{Q}^{\prime}_2 + Q_3 + Q_4 + \mathcal{E}, \end{align} $$

where

(55) $$ \begin{align} \pm \mathcal{E} \lesssim N^{-\alpha/2} Q_4 + \big [ N^{-\alpha/2} + N^{\alpha-1} \big] (\mathcal{N}_+ + 1) + N^{-1+\alpha} \mathcal{N}_+^2 + N^{-2} H_1 \end{align} $$

and where we defined

$$\begin{align*}\begin{aligned} A_{\alpha} &= - \frac{N(N-1)}{2} \Big[ \sum_{|p| \leq N^{\alpha}} \hat{V}_N (p) {\varphi}_p + \sum_{|p|> N^{\alpha}, |q| \leq N^{\alpha}} \hat{V}_N (p-q) {\varphi}_p {\varphi}_q \Big] \\ &= - \frac{N(N-1)}{2} \Big[ \sum_{|p| \leq N^{\alpha}} (\hat{V}_N (p) + \hat{V}_N * {\varphi}) {\varphi}_p - \sum_{|p|, |q| \leq N^{\alpha}} \hat{V}_N (p-q) {\varphi}_p {\varphi}_q \Big]. \end{aligned} \end{align*}$$

The second term in parentheses can be estimated as in (53). Setting, in position space, $f = 1 + \check {{\varphi }}$ , we find

(56) $$ \begin{align} A_{\alpha} = - \frac{N(N-1)}{2} \sum_{|p| \leq N^{\alpha}} (\hat{V}_N * \hat{f}) (p) {\varphi}_p + \mathcal{O} (N^{2\alpha-1}). \end{align} $$

From equation (16), we have $\widehat {V_N f} (0) = 8\pi \mathfrak {a}_N$ . Hence

(57) $$ \begin{align} | (\hat{V}_N * \hat{f}) (p) - 8 \pi \mathfrak{a}_N / N | \leq \int_{\Lambda} V_N (x) f (x) |e^{-ip \cdot x} -1| \mathrm{d}x \leq C |p| / N^2. \end{align} $$

Moreover, from the scattering equation (14), we find

$$\begin{align*}{\varphi}_p = - \frac{1}{2p^2} (\hat{V}_N * \hat{f}) (p), \end{align*}$$

which implies, by the bound (57),

$$\begin{align*}\Big| {\varphi}_p + \frac{4\pi \mathfrak{a}_N}{N p^2} \Big| \leq \frac{1}{2p^2} \big| (\hat{V}_N * \hat{f}) (p) - 8 \pi \mathfrak{a}_N \big| \leq \frac{C}{|p|N^2}. \end{align*}$$

Inserting in equation (56), we obtain

$$\begin{align*}A_{\alpha} = \sum_{|p| \leq N^{\alpha}} \frac{(4\pi \mathfrak{a}_N)^2}{p^2} + \mathcal{O} (N^{2\alpha-1}). \end{align*}$$

To conclude the proof of Proposition 4, we still have to compare the operator $\tilde Q^{\prime }_2$ appearing on the right-hand side of equation (54) with the operator $\tilde Q_2$ defined in equation (25). From equation (27), we can write, using again the notation $f = 1+ \check {{\varphi }}$ ,

(58) $$ \begin{align} \tilde Q^{\prime}_2 - \tilde Q_2 = \frac{1}{2} \sum_{|p| \leq N^{\alpha}} \Big[ (\hat{V}_N * \hat{f})(p) - \frac{8\pi \mathfrak{a}_N}{N} \Big] \, a_p^{\dagger} a_{-p}^{\dagger} a_0 a_0 - \frac{1}{2} \sum_{p, |q| \leq N^{\alpha}} \hat{V}_N (p-q) {\varphi}_q a_p^{\dagger} a_{-p}^{\dagger} a_0 a_0 + \text{h.c.} \end{align} $$

The first term on the right-hand side of equation (58) can be bounded using the bound (57) by

$$\begin{align*}\begin{aligned} \pm \sum_{|p| \leq N^{\alpha}} \Big[ (\hat{V}_N * \hat{f})(p) &- \frac{8\pi \mathfrak{a}_N}{N} \Big] \, \big( a_p^{\dagger} a_{-p}^{\dagger} a_0 a_0 +\text{h.c.}\big) \\ &\lesssim \frac{1}{N} \| |p| \chi_{|p| \leq N^{\alpha}} \|_2 (\mathcal{N}_+ + 1) \lesssim N^{-1+5\alpha/2} (\mathcal{N}_+ + 1). \end{aligned} \end{align*}$$

For the second term on the right-hand side of equation (58), we set ${\varphi }^<_p = {\varphi }_p \chi _{|p| \leq N^{\alpha }}$ and estimate, switching to position space,

$$\begin{align*}\begin{aligned} \pm \sum_{p} &(\hat{V}_N * {\varphi}^< )(p) \, a_p^{\dagger} a_{-p}^{\dagger} a_0 a_0 + \text{h.c.} \\ &= \pm \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y \, V_N (x-y) \check{{\varphi}}^< (x-y) a_x^{\dagger} a_y^{\dagger} a_0 a_0 + \text{h.c.} \\ &\lesssim \delta Q_4 + \delta^{-1} N^2 \| V_N \|_1 \| \check{{\varphi}}^< \|_{\infty}^2 \lesssim \delta Q_4 + \delta^{-1} N^{-1+2\alpha} \leq N^{-\alpha/2} Q_4 + N^{-1+5\alpha/2}, \end{aligned} \end{align*}$$

since $\| \check {{\varphi }}^< \|_{\infty } \lesssim \| {\varphi }^< \|_1 \lesssim N^{\alpha }$ , from Lemma 2 (in the last step, we chose $\delta = N^{-\alpha /2}$ ). The last two estimates show that the difference $\tilde Q^{\prime }_2 - \tilde Q_2$ can be added to the error (55) and therefore conclude the proof of Proposition 4.

Proof. We proceed similarly as in the proof of [Reference Boccato, Brennecke, Cenatiempo and Schlein7, Prop. 5.1]. For $\xi \in \mathcal {F}$ , we set $f(s) = \langle \xi , e^{-s{\mathcal {B}}_3} (\mathcal {N}_++1) e^{s{\mathcal {B}}_3} \xi \rangle $ . For $s \in (0;1)$ , we find

(62) $$ \begin{align} \begin{aligned} f'(s) &= \langle \xi, e^{-s{\mathcal{B}}_3} [(\mathcal{N}_++1),{\mathcal{B}}_3] e^{s{\mathcal{B}}_3} \xi \rangle \\ &= \sum_{p,q} \tilde {\varphi}_p \chi_{|q| \leq N^{\alpha}} \, \langle \xi, e^{-s{\mathcal{B}}_3} a^{\dagger}_{p+q} a^{\dagger}_{-p} a_q a_0 e^{s\mathcal{B}_3} \xi \rangle + \text{h.c.} \\ &\lesssim \delta \sum_{p,q} \langle \xi, e^{-s{\mathcal{B}}_3} a^{\dagger}_{p+q} a^{\dagger}_{-p} a_{-p} a_{p+q} e^{s{\mathcal{B}}_3} \xi \rangle + \delta^{-1} \sum_{p,q} |\tilde {\varphi}_p|^2 \, \langle \xi, e^{-s{\mathcal{B}}_3} a^{\dagger}_q a_q a^{\dagger}_0 a_0 e^{s{\mathcal{B}}_3} \xi \rangle \\ &\lesssim \delta \langle e^{-s{\mathcal{B}}_3} \mathcal{N}_+^2 e^{s{\mathcal{B}}_3} \xi \rangle + C \delta^{-1} N \| \tilde {\varphi} \|_2^2 \langle \xi , e^{-s{\mathcal{B}}_3} \mathcal{N}_+ e^{s {\mathcal{B}}_3} \xi \rangle \lesssim N^{-\alpha/2} f(s), \end{aligned} \end{align} $$

where we put $\delta = N^{-1-\alpha /2}$ and used that $\mathcal {N}_+, \mathcal {N}_0 \leq N$ , $\| {\varphi } \|_2 \lesssim N^{-1-\alpha /2}$ , by Lemma 2. With Gronwall’s lemma [Reference Pachpatte26, Theorem 1.2.2], we obtain $f(s) \lesssim \langle \xi , (\mathcal {N}_+ + 1) \xi \rangle $ for all $s \in [-1;1]$ , proving the bound (61). Inserting this estimate on the right-hand side of (62) and integrating over s, we obtain the desired bound (60). For $k> 1$ , the bound given in (61) can be shown similarly.

Proof. A simple computation shows that

$$ \begin{align*} [H_1,{\mathcal{B}}_3] &= \sum_{p,q} \left[(p+q)^2 + p^2 - q^2 \right] \tilde {\varphi}_p \chi_{|q| \leq N^{\alpha}} a^{\dagger}_{p+q} a^{\dagger}_{-p} a_q a_0 + \text{h.c.}, \\ &= 2 \sum_{p,q} p^2 \tilde {\varphi}_p \chi_{|q| \leq N^{\alpha}} a^{\dagger}_{p+q} a^{\dagger}_{-p} a_q a_0 +\text{h.c.}+ \mathcal{E}_{[H_1,{\mathcal{B}}_3]}, \end{align*} $$

with

$$ \begin{align*} \mathcal{E}_{[H_1,{\mathcal{B}}_3]}& = 2 \sum_{p,q} p\cdot q \tilde {\varphi}_p \chi_{|q| \leq N^{\alpha}} a^{\dagger}_{p+q} a^{\dagger}_{-p} a_q a_0 +\text{h.c.} \end{align*} $$

Using the scattering equation (14) yields (67). We now estimate $\mathcal {E}_{[H_1,{\mathcal {B}}_3]}$ . Using $|q| \leq N^{\alpha }$ , we find, for any $\delta>0$ ,

$$ \begin{align*} \pm \mathcal{E}_{[H_1,{\mathcal{B}}_3]} &= \pm 2 \sum_{p,q} p\cdot q \tilde {\varphi}_p \chi_{|q| \leq N^{\alpha}} a^{\dagger}_{p+q} a^{\dagger}_{-p} (\mathcal{N}_++1)^{-1/2} (\mathcal{N}_++1)^{1/2} a_q a_0 + \text{h.c.} \\ &\lesssim \delta \sum_{p,q} p^2 a^{\dagger}_{p+q} a^{\dagger}_{-p} (\mathcal{N}_++1)^{-1} a_{-p} a_{p+q} + \delta^{-1} \sum_{p,|q| \leq N^{\alpha}} q^2 |\tilde {\varphi}_p |^2 \, a^{\dagger}_{q} (\mathcal{N}_++1) a_{q}(a^{\dagger}_{0} a_{0}) \\ &\lesssim \delta H_1 + \delta^{-1} N^{1+2\alpha} \|\tilde {\varphi}\|_2^2 (\mathcal{N}_++1)^2. \end{align*} $$

Choosing $\delta = N^{-3\alpha /2}$ , we conclude that

$$\begin{align*}\pm \mathcal{E}_{[H_1,{\mathcal{B}}_3]} \lesssim N^{-3\alpha/2} H_1 + N^{-1+5\alpha/2} (\mathcal{N}_++1)^2. \end{align*}$$

Let us now turn to (68). Recalling (10) and (59), we find

$$ \begin{align*} [Q_4,{\mathcal{B}}_3] &= \frac{1}{2} \sum_{r,p,q} \sum_{m,n} \hat{V}_N (r) \tilde {\varphi}_m \chi_{|n| \leq N^{\alpha}} \big[ a_{p+r}^{\dagger} a_q^{\dagger} a_p a_{q+r}, a_{m+n}^{\dagger} a_{-m}^{\dagger} a_n a_0 \big] + \text{h.c.} \\ &= \frac{1}{2} \sum_{r,p,q} \sum_{m,n} \hat{V}_N (r) \tilde {\varphi}_m \chi_{|n| \leq N^{\alpha}} \\ &\hspace{1.5cm} \times \Big\{ a_{p+r}^{\dagger} a_q^{\dagger} \big[ a_p a_{q+r} , a_{m+n}^{\dagger} a_{-m}^{\dagger} \big] a_n a_0 + a_{m+n}^{\dagger} a_{-m}^{\dagger} \big[ a_{p+r}^{\dagger} a_q^{\dagger} , a_n a_0 \big] a_p a_{q+r} \Big\}. \end{align*} $$

Using equation (6) and rearranging all terms in normal order, we arrive at

$$ \begin{align*} [Q_4,{\mathcal{B}}_3] &= \sum_{r,p,q} \hat V_N(p-r)\tilde {\varphi}_r \chi_{|q| \leq N^{\alpha}} (a^{\dagger}_{p+q} a^{\dagger}_{-p} a_q a_0 + \text{h.c.})+ \mathcal{E}_{[Q_4,{\mathcal{B}}_3]}, \end{align*} $$

where

$$\begin{align*}\begin{aligned} 2 \mathcal{E}_{[Q_4,{\mathcal{B}}_3]}= \; &- \sum_{p,q,m, r} \hat V_N(r) \tilde {\varphi}_m \chi_{|p+r| \leq N^{\alpha}} \, a^{\dagger}_{m+p+r} a^{\dagger}_{-m} a^{\dagger}_q a_{q+r} a_p a_0 \\[4pt] & - \sum_{p,q,m, r} \hat V_N(r) \tilde {\varphi}_m \chi_{|q| \leq N^{\alpha}} \, a^{\dagger}_{m+q} a^{\dagger}_{-m} a^{\dagger}_{p+r} a_{q+r} a_p a_0 \\[4pt] & + \sum_{p,q,m, r} \hat V_N(r) \tilde {\varphi}_m \chi_{|q+r-m| \leq N^{\alpha}} \, a^{\dagger}_{p+r} a^{\dagger}_{-m} a^{\dagger}_{q} a_{q+r-m} a_p a_0 \\[4pt] & + \sum_{p,q,m, r} \hat V_N(r) \tilde {\varphi}_m \chi_{|p-m| \leq N^{\alpha}} \, a^{\dagger}_{p+r} a^{\dagger}_{-m} a^{\dagger}_{q} a_{q+r} a_{p-m} a_0 \\[4pt] & + \sum_{p,q,m, r} \hat V_N(r) \tilde {\varphi}_{-q-r} \chi_{|m| \leq N^{\alpha}} \, a^{\dagger}_{p+r} a^{\dagger}_{q} a^{\dagger}_{-q-r+m} a_{p} a_{m} a_0 \\[4pt] & + \sum_{p,q,m, r} \hat V_N(r) \tilde {\varphi}_{-p} \chi_{|p-m| \leq N^{\alpha}} \, a^{\dagger}_{p+r} a^{\dagger}_{q} a^{\dagger}_{-m} a_{q+r} a_{p-m} a_0 + \text{h.c.} =: \; \sum_{i=1}^6 \mathcal{E}_i. \end{aligned} \end{align*}$$

For a parameter $\delta> 0$ , we find

$$ \begin{align*} \pm \mathcal{E}_1 &\lesssim \delta \|\hat V_N\|_{\infty}^2 \|\chi_{|\cdot|\leq N^{\alpha}}\|_{1} \mathcal{N}_+^{3} + \delta^{-1} \|\tilde {\varphi}\|_2^2 \| \chi_{|\cdot|\leq N^{\alpha}}\|_{1} \mathcal{N}^2_+ \mathcal{N}_0 \, , \\[4pt] &\lesssim N^{-1+5\alpha/2} \mathcal{N}_+^2, \end{align*} $$

where, in the last step, we chose $\delta = N^{-\alpha /2}$ and used $\mathcal {N}_0 \leq N$ (and Lemma 2). To estimate $\mathcal {E}_2, \dots , \mathcal {E}_6$ , we switch to position space. For arbitrary $\delta> 0$ , we find

$$ \begin{align*} \pm \mathcal{E}_2 &= \pm \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y V_N(x-y) \check{\chi}_{|\cdot|\leq N^{\alpha}}(z-y) a^{\dagger}_z a^{\dagger}(\check{\tilde {\varphi}}^z) a^{\dagger}_x a_x a_y a_0 + \text{h.c.} \\[4pt] &\leq \delta \|\chi_{|\cdot|\leq N^{\alpha}}\|_2^2 \, Q_4 a^{\dagger}_0 a_0 + \delta^{-1} \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y V_N(x-y) a^{\dagger}_z a^{\dagger}(\check{\tilde {\varphi}}^z) a^{\dagger}_x a_x a(\check{\tilde {\varphi}}^z) a_z \\[4pt] &\lesssim \delta N^{1+3\alpha} Q_4 + \delta^{-1} N^{-3-\alpha} \mathcal{N}_+^3 \\[4pt] &\lesssim N^{-\alpha/2} Q_4 + N^{-2+5\alpha/2} \mathcal{N}_+^3, \end{align*} $$

where, in the last line, we fixed $\delta = N^{-1-7\alpha /2}$ . Similarly, we find

$$ \begin{align*} \pm \mathcal{E}_3 &= \pm \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y V_N(x-y) a^{\dagger}_xa^{\dagger}_y a^{\dagger}(\check{\tilde {\varphi}}^z) a(\check{\chi}_{|\cdot|\leq N^{\alpha}}^y) a_x a_0 + \text{h.c.} \\[4pt] &\lesssim \delta Q_4 + \delta^{-1} N \|V_N\|_{1} \|\check \chi_{|\cdot|\leq N^{\alpha}}\|_{2}^2 \|\tilde {\varphi}\|_2^2 (\mathcal{N}_++1)^3 \\[4pt] &\lesssim N^{-\alpha/2} Q_4 + N^{-2+5\alpha/2} (\mathcal{N}_+ + 1)^3, \end{align*} $$

taking $\delta = N^{-\alpha /2}$ . Furthermore, for an arbitrary $\xi \in \mathcal {F}$ , we have

$$ \begin{align*} |\langle \xi, \mathcal{E}_4 \xi \rangle| &= \Big| \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y V_N(x-y) \check{\chi}_{|\cdot|\leq N^{\alpha}}(x-z) \langle \xi, a^{\dagger}_xa^{\dagger}_z a^{\dagger}(\check{\tilde {\varphi}}^x) a_y a_z a_0 \xi \rangle \Big| \\ &\leq \| \check{\chi}_{|.| \leq N^{\alpha}} \|_{\infty} \int _{\Lambda^2} \mathrm{d}x \mathrm{d}y V_N (x-y) \| a_x a_z a (\check{\tilde{\varphi}}^x) \xi \| \| a_y a_z a_0 \xi \| \\ &\lesssim \| \chi_{|.| \leq N^{\alpha}} \|_1 \| \tilde{\varphi} \|_2 \| V_N \|_1 \left[ \langle \xi , \mathcal{N}^3_+ \xi \rangle + \langle \xi, \mathcal{N}_+^2 \mathcal{N}_0 \xi \rangle \right] \\ &\lesssim N^{-1+5\alpha/2} \langle \xi, \mathcal{N}_+^2 \xi \rangle. \end{align*} $$

For $\mathcal {E}_5$ , we estimate

$$ \begin{align*} \pm \mathcal{E}_5 &= \pm \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y V_N(x-y) \check{\tilde {\varphi}}(y-z) a^{\dagger}_xa^{\dagger}_y a^{\dagger}_z (\mathcal{N}_++1)^{-1/2}(\mathcal{N}_++1)^{1/2} a_x a(\check{\chi}_{|\cdot|\leq N^{\alpha}}^z) a_0 + \text{h.c.} \\ &\leq \delta Q_4 + \delta^{-1} \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y V_N(x-y) |\check{\tilde {\varphi}}(z-y)|^2 a_x^{\dagger} a^{\dagger} (\check{\chi}^z_{|.| \leq N^{\alpha}}) a_0^{\dagger} (\mathcal{N}_+ + 1) a_0 a (\check{\chi}^z_{|.| \leq N^{\alpha}}) a_x \\ &\lesssim \delta Q_4 + \delta^{-1} N \|V_N\|_{1} \|\check \chi_{|\cdot|\leq N^{\alpha}}\|_{2}^2 \|\tilde {\varphi}\|_2^2 (\mathcal{N}_++1)^3 \\ &\lesssim N^{-\alpha/2} Q_4 + N^{-2+5\alpha/2} \mathcal{N}_+^3, \end{align*} $$

again choosing $\delta = N^{-\alpha /2}$ . By a simple change of variable, it is easy to check that $\mathcal {E}_6 = \mathcal {E}_5$ . This concludes the proof of the lemma.

Proof. With Lemma 12, we find, using the scattering equation (14),

$$\begin{align*}\Gamma_3 = \big[ H_1 + Q_4 , {\mathcal{B}}_3 \big] + Q_3 = \tilde{Q}_{3,1} + \tilde{Q}_{3,2} + Q_3^> + \mathcal{E}_{[H_1,{\mathcal{B}}_3]} + \mathcal{E}_{[Q_4,{\mathcal{B}}_3]} \end{align*}$$

with

(71) $$ \begin{align} \tilde Q_{3,1} & = - \sum_{p,q} p^2 {\varphi}_p \chi_{|p| \leq N^{\alpha}} \chi_{|q| \leq N^{\alpha}} \, a^{\dagger}_{p+q} a^{\dagger}_{-p} a_q a_0 + \text{h.c.}, \end{align} $$

(72) $$ \begin{align} \tilde Q_{3,2} &= - \sum_{p,q, r} \hat{V}_N (p-r) {\varphi}_r \chi_{|r| \leq N^{\alpha}} \chi_{|q| \leq N^{\alpha}} \, a^{\dagger}_{p+q} a^{\dagger}_{-p} a_q a_0 + \text{h.c.}, \end{align} $$

(73) $$ \begin{align} Q_3^{>} & = \sum_{p,q} \hat V_N(p) \chi_{|q| > N^{\alpha}} a^{\dagger}_{p+q} a^{\dagger}_{-p} a_q a_0 +\text{h.c.} \end{align} $$

It follows easily from Lemma 2 that $\| \chi _{|p| \leq N^{\alpha }} p^2 {\varphi }_p\|_{2}\lesssim N^{-1+3\alpha /2}$ ; thus

$$\begin{align*}\pm \tilde Q_{3,1} \lesssim N^{-1/2 + 3\alpha/2} (\mathcal{N}_++1)^{3/2}. \end{align*}$$

Denoting ${\varphi }^{<}_p = {\varphi }_p \chi _{|p| \leq N^{\alpha }}$ , we write $\tilde {Q}_{3,2}$ in position space as

$$ \begin{align*} \pm \tilde Q_{3,2} &= \pm \int_{\Lambda^3} \mathrm{d}x \mathrm{d}y \mathrm{d}z \, V_N (x-y) \check{{\varphi}}^{<} (x-y) \check{\chi}_{|\cdot|\leq N^{\alpha}}(x-z) a^{\dagger}_x a^{\dagger}_y a_z a_0 + \text{h.c.}\\ &\leq \delta Q_4 + \delta^{-1} N \| V_N^{1/2} \check{\varphi}^< \|_2^2 \|\chi_{|\cdot|\leq N^{\alpha}}\|_{\infty}^2 (\mathcal{N}_++1) \\ &\lesssim N^{-\alpha/2} Q_4 + N^{-2+5\alpha/2} (\mathcal{N}_+ + 1) \, , \end{align*} $$

where we chose $\delta = N^{-\alpha /2}$ and used the estimate

(74) $$ \begin{align} \int_{\Lambda} a^{\dagger} (\check{\chi}_{|.| \leq N^{\alpha}}^x) \, a (\check{\chi}_{|.| \leq N^{\alpha}}^x) \, \mathrm{d}x = d\Gamma (\chi^2_{|p| \leq N^{\alpha}}) \leq \| \chi_{|.| \leq N^{\alpha}} \|_{\infty}^2 \, \mathcal{N}_+ \leq \mathcal{N}_+. \end{align} $$

Proceeding similarly, we find

$$ \begin{align*} \pm Q_3^{>} &= \pm \int_{\Lambda^3} \mathrm{d}x \mathrm{d}y \mathrm{d}z \, V_N (x-y) \check{\chi}_{|\cdot|> N^{\alpha}}(x-z) a^{\dagger}_x a^{\dagger}_y a_z a_0 + \text{h.c.}\\ &\lesssim \delta Q_4 + \delta^{-1} N^{1-2\alpha} \| V_N \|_1 H_1 \lesssim N^{-\alpha/2} Q_4 + N^{-3\alpha/2} H_1, \end{align*} $$

where we took $\delta = N^{-\alpha /2}$ and used that

$$\begin{align*}\int_{\Lambda} a^{\dagger}(\check{\chi}_{|\cdot|> N^{\alpha}}^{x}) a(\check{\chi}_{|\cdot|> N^{\alpha}}^x) \mathrm{d}x = \sum_{|p| > N^{\alpha}} a_p^{\dagger} a_p \leq N^{-2\alpha} H_1. \end{align*}$$

Combining the bounds for $\tilde {Q}_{3,1} , \tilde {Q}_{3,2}, Q_3^>$ with the estimates for $\mathcal {E}_{[H_1,{\mathcal {B}}_3]}, \mathcal {E}_{[Q_4,{\mathcal {B}}_3]}$ from Lemma 12, we obtain the bound (70).

Proof. For arbitrary $\xi \in \mathcal {F}$ , we define $f(s) = \langle \xi , e^{-s{\mathcal {B}}_3} Q_4 e^{s {\mathcal {B}}_3} \xi \rangle $ so that

$$\begin{align*}f' (s) = \langle \xi, e^{-s {\mathcal{B}}_3} [ Q_4 , \mathcal{B}_3 ] e^{s {\mathcal{B}}_3} \xi \rangle. \end{align*}$$

From Lemma 12, we find

$$\begin{align*}[Q_4,{\mathcal{B}}_3] = \sum_{r,p,q} \hat V_N(p-r)\tilde {\varphi}_r \chi_{|q| \leq N^{\alpha}} (a^{\dagger}_{p+q} a^{\dagger}_{-p} a_q a_0 + \text{h.c.})+ \mathcal{E}_{[Q_4,{\mathcal{B}}_3]}, \end{align*}$$

where

$$\begin{align*}\pm \mathcal{E}_{[Q_4,{\mathcal{B}}_3]} \lesssim N^{-\alpha/2} Q_4 + N^{-1+5\alpha/2} (\mathcal{N}_+ + 1)^2. \end{align*}$$

Switching to position space, we have

$$\begin{align*}\begin{aligned} \sum_{r,p,q} \hat V_N(p-r) &\tilde {\varphi}_r \chi_{|q| \leq N^{\alpha}} (a^{\dagger}_{p+q} a^{\dagger}_{-p} a_q a_0 + \text{h.c.}) \\ &= \sum_{r,p,q} \hat V_N(p-r)\tilde {\varphi}_r \chi_{|q| \leq N^{\alpha}} (a^{\dagger}_{p+q} a^{\dagger}_{-p} a_q a_0 + \text{h.c.})\\ &= \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y \, V_N (x-y) \check{\tilde{\varphi}} (x-y) a^{\dagger}_x a^{\dagger}_y a( \check{\chi}_{|\cdot| \leq N^{\alpha}}^x) a_0 + \text{h.c.} \\ &\lesssim Q_4 + \int \mathrm{d}x \mathrm{d}y \, V_N(x-y) |\check{\tilde{\varphi}} (x-y)|^2 a_0^{\dagger} a^{\dagger} (\check{\chi}^x_{|.| \leq N^{\alpha}}) a (\check{\chi}^x_{|.| \leq N^{\alpha}}) a_0 \\ &\lesssim Q_4 + N \| V_N \|_1 \| \tilde {\varphi} \|_{\infty}^2 \| \chi_{|.| \leq N^{\alpha}}^2 \|_{\infty} \lesssim Q_4 + \mathcal{N}_+ , \end{aligned} \end{align*}$$

where we used Lemma 2 and argued as in (74). We conclude that

$$\begin{align*}\pm \big[ Q_4 , {\mathcal{B}}_3 \big] \lesssim Q_4 + \mathcal{N}_+ + N^{-1+5\alpha/2} (\mathcal{N}_+ + 1)^2. \end{align*}$$

Therefore, using Lemma 3, we find

$$\begin{align*}f' (s) \lesssim f(s) + \langle \xi, \mathcal{N}_+ \xi \rangle + N^{-1+5\alpha/2} \langle \xi , (\mathcal{N}_+ + 1)^2 \xi \rangle. \end{align*}$$

By the Gronwall lemma, we obtain the bound (75).

To prove the estimate (76), we proceed similarly. For $\xi \in \mathcal {F}$ , we define $g(s) = \langle \xi , e^{-s{\mathcal {B}}_3} H_1 e^{s {\mathcal {B}}_3} \xi \rangle $ for any $|s| \leq 1$ , which leads to

$$\begin{align*}g' (s) = \langle \xi , e^{-\mathcal{B}_3} H_1 e^{s\mathcal{B}_3} \xi \rangle. \end{align*}$$

From Lemma 12, we have

$$\begin{align*}\big[ H_1, {\mathcal{B}}_3 \big] = \sum_{p,q,r} \hat{V}_N (p-r) (\delta_{0,r} + {\varphi}_r) \chi_{|p|> N^{\alpha}} \chi_{|q| \leq N^{\alpha}} a_{p+q}^{\dagger} a_{-p}^{\dagger} a_q a_0 + \text{h.c.} + \mathcal{E}_{[H_1,{\mathcal{B}}_3]}, \end{align*}$$

where

$$\begin{align*}\pm \mathcal{E}_{[H_1,{\mathcal{B}}_3]} \leq N^{-3\alpha/2} H_1 + N^{-1+5\alpha/2} (\mathcal{N}_+ + 1)^2. \end{align*}$$

Writing $\chi _{|p|> N^{\alpha }} = 1 - \chi _{|p| \leq N^{\alpha }}$ , we decompose

$$\begin{align*}\sum_{p,q,r} \hat{V}_N (p-r) (\delta_{0,r} + {\varphi}_r) \chi_{|p|> N^{\alpha}} \chi_{|q| \leq N^{\alpha}} a_{p+q}^{\dagger} a_{-p}^{\dagger} a_q a_0 + \text{h.c.} = \mathcal{E}_1 + \mathcal{E}_2, \end{align*}$$

where

$$\begin{align*}\begin{aligned} \mathcal{E}_1 &= \pm \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y V_N (x-y) (1 + \check{{\varphi}}) (x-y) a_x^{\dagger} a_y^{\dagger} a(\check{\chi}_{|.| \leq N^{\alpha}}^x) a_0 + \text{h.c.} \\ &\lesssim Q_4 + N \|V_N\|_{1} \|(1+\check {\varphi})\|_{\infty}^2 \|\chi_{|\cdot|\leq N^{\alpha}}\|_{\infty}^2 \mathcal{N}_+ \lesssim Q_4 + \mathcal{N}_+ \end{aligned} \end{align*}$$

and

$$\begin{align*}\begin{aligned} \pm \mathcal{E}_2 &= \pm \sum_{p,q,r} \hat{V}_N (p-r) (\delta_{0,r} + {\varphi}_r ) \chi_{|p| \leq N^{\alpha}} \chi_{|q| \leq N^{\alpha}} a_{p+q}^{\dagger} a_{-p}^{\dagger} a_q a_0 + \text{h.c.} \\ & \lesssim \| V_N \|_1 \| (1+ \check{{\varphi}}) \|_{\infty} \Big[ \delta \mathcal{N}_+^2 + \delta^{-1} N \| \chi_{|.| \leq N^{\alpha}} \|_2^2 \mathcal{N}_+ \Big] \\ &\lesssim N^{-1} \Big[ \delta \mathcal{N}_+^2 + \delta^{-1} N^{1+3\alpha} \mathcal{N}_+ \Big] \lesssim \mathcal{N}_+ + N^{-1+3\alpha} \mathcal{N}_+^2\, , \end{aligned} \end{align*}$$

choosing in the last step $\delta = N^{3\alpha }$ . Thus, with Lemma 3, we find

$$\begin{align*}g' (s) \lesssim f(s) + g(s) + \langle \xi, \mathcal{N}_+ \xi \rangle + N^{-1+3\alpha} \langle \xi, (\mathcal{N}_+ + 1)^2 \xi \rangle. \end{align*}$$

With the estimate (75) and applying Gronwall’s lemma, we obtain the desired estimate (76).

Proof. We compute

(77) $$ \begin{align} \begin{aligned} [Q_3,{\mathcal{B}}_3] &= \sum_{p,q,r,s } \hat V_N(p) \tilde {\varphi}_r \chi_{|s|\leq N^{\alpha}} [a^{\dagger}_{p+q} a^{\dagger}_{-p}a_q a_0 + a^{\dagger}_0 a^{\dagger}_q a_{-p} a_{p+q}, a^{\dagger}_{r+s}a^{\dagger}_{-r} a_sa:0 - a^{\dagger}_0 a^{\dagger}_sa:{-r} a_{r+s}] \\ &=\sum_{p,q,r,s } \hat V_N(p) \tilde {\varphi}_r \chi_{|s|\leq N^{\alpha}} [a^{\dagger}_{p+q} a^{\dagger}_{-p}a_q a_0, a^{\dagger}_{r+s}a^{\dagger}_{-r} a_sa:0] + \text{h.c.} \\ &\quad + \sum_{p,q,r,s } \hat V_N(p) \tilde {\varphi}_r \chi_{|s|\leq N^{\alpha}} [a^{\dagger}_0 a^{\dagger}_sa:{-r} a_{r+s},a^{\dagger}_{p+q} a^{\dagger}_{-p}a_q a_0] + \text{h.c.} =: (\textrm{I}) + (\textrm{II}). \end{aligned} \end{align} $$

We start by estimating term $(\textrm {I})$ . With the canonical commutation relations, we obtain

(78) $$ \begin{align} \begin{aligned} (\textrm{I}) = \; &\sum_{p,q,r}\hat V_N(p) \left[ \tilde {\varphi}_r \chi_{|q-r|\leq N^{\alpha}} a^{\dagger}_{p+q} a^{\dagger}_{-p}a^{\dagger}_{-r} a_{q-r} + \tilde {\varphi}_q \chi_{|r|\leq N^{\alpha}} a^{\dagger}_{p+q} a^{\dagger}_{-p}a^{\dagger}_{-q+r} a_{r} \right. \\ &\hspace{1cm} - \left. \tilde {\varphi}_r \chi_{|q+p|\leq N^{\alpha}} a^{\dagger}_{r+p+q} a^{\dagger}_{-r}a^{\dagger}_{-p} a_{q} - \tilde {\varphi}_r \chi_{|p|\leq N^{\alpha}} a^{\dagger}_{r-p} a^{\dagger}_{-r}a^{\dagger}_{p+q} a_{q}\right] a_0a_0 + \text{h.c.} \\ =: \; &(\textrm{I})_a + (\textrm{I})_b + (\textrm{I})_c + (\textrm{I})_d. \end{aligned} \end{align} $$

To estimate the first term, we rewrite it in position space. We find

$$ \begin{align*} \pm (\textrm{I})_a &= \pm \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y \, V_N(x-y) a^{\dagger}_x a^{\dagger}_y a^{\dagger}(\check {\tilde {\varphi}}^x) a(\check{ \chi}^x_{|\cdot| \leq N^{\alpha}}) a_0 a_0 + \text{h.c.} \\[3pt] &\leq \delta Q_4 + \delta^{-1} \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y \, V_N(x-y) a^{\dagger} (\check{ \chi}^x_{|\cdot| \leq N^{\alpha}}) a(\check {\tilde {\varphi}}^x)a^{\dagger}(\check {\tilde {\varphi}}^x) a(\check{ \chi}^x_{|\cdot| \leq N^{\alpha}}) a^{\dagger}_0a^{\dagger}_0 a_0 a_0 \\[3pt] &\lesssim \delta Q_4 + \delta^{-1} N^2 \|\tilde {\varphi}\|_2^2 \|V_N\|_1 \|\chi_{|\cdot| \leq N^{\alpha}}\|_{\infty}^2 (\mathcal{N}_++1)^2 \\[3pt] &\lesssim \delta Q_4 + \delta^{-1} N^{-\alpha-1} (\mathcal{N}_++1)^2 \leq N^{-\alpha/2} Q_4 + N^{-1-\alpha/2} (\mathcal{N}_+ + 1)^2, \end{align*} $$

where we used that $\mathcal {N}_0 \leq N$ and the bound (74) and, in the last step, set $\delta = N^{-\alpha /2}$ . The second term in equation (78) is dealt with similarly. We obtain

$$ \begin{align*} \pm (\textrm{I})_b &= \pm \int_{\Lambda^3} \mathrm{d}x \mathrm{d}y \mathrm{d}z \, V_N(x-y) \check {\tilde {\varphi}}(x-z) \, a^{\dagger}_x a^{\dagger}_y a^{\dagger}_z a(\check{ \chi}^z_{|\cdot| \leq N^{\alpha}}) a_0 a_0 + \text{h.c.} \\[3pt] &\leq \delta Q_4 + \delta^{-1}\int_{\Lambda^3} \mathrm{d}x \mathrm{d}y \mathrm{d}z \, V_N(x-y) |\check {\tilde {\varphi}}(x-z)|^2 a^{\dagger}(\check{ \chi}^z_{|\cdot| \leq N^{\alpha}}) (\mathcal{N}_++1)a(\check{ \chi}^z_{|\cdot| \leq N^{\alpha}}) a^{\dagger}_0a^{\dagger}_0a_0a_0 \\[3pt] & \lesssim \delta Q_4 + \delta^{-1} N^2 \|\tilde {\varphi}\|_2^2 \|V_N\|_{1} \|\chi_{|\cdot| \leq N^{\alpha}}\|_{\infty}^2 (\mathcal{N}_++1)^2 \\[3pt] &\lesssim N^{-\alpha/2} Q_4 + N^{-1-\alpha/2} (\mathcal{N}_+ + 1)^2, \end{align*} $$

choosing again $\delta = N^{-\alpha /2}$ . For the third term in equation (78), we bound it, for an arbitrary $\delta> 0$ , with Cauchy-Schwarz by

$$ \begin{align*} \pm (\textrm{I})_c&\lesssim \delta \sum_{p,q,r}|\hat V_N(p)| a^{\dagger}_{r+p+q} a^{\dagger}_{-r}a^{\dagger}_{-p} (\mathcal{N}_+ +1)^{-1}a_{-p} a_{-r} a_{r+p+q} \\ &\qquad + \delta^{-1} \sum_{p,q,r} |\hat V_N(p)| |\tilde {\varphi}_r|^2 \chi_{|q+p|\leq N^{\alpha}} a^{\dagger}_{q}(\mathcal{N}_++1)a_{q}a^{\dagger}_0a^{\dagger}_0a_0a_0 \\ &\lesssim (\delta \|\hat V_N\|_{\infty} + \delta^{-1} N^2 \|\tilde {\varphi}\|_{2}^2 \|\hat V_N \ast \chi_{|\cdot|\leq N^{\alpha}}\|_{\infty})(\mathcal{N}_++1)^2 \\ &\lesssim N^{- 1+5\alpha/2} (\mathcal{N}_+ + 1)^2, \end{align*} $$

where, at the end, we took $\delta = N^{-\alpha }$ and used $\|\hat V_N \ast \chi _{|\cdot |\leq N^{\alpha }}\|_{\infty } \leq \|\hat V_N\|_{\infty } \|\chi _{|\cdot |\leq N^{\alpha }}\|_{1} \lesssim N^{-1+3\alpha }$ . The last term in equation (78) can be bounded, again by Cauchy-Schwarz, by

$$ \begin{align*} \pm (\textrm{I})_d &\lesssim \delta (\mathcal{N}_+ + 1)^2 + \delta^{-1} N^2 \| \hat{V}_N \|_{\infty}^2 \| \tilde{\varphi} \|_2^2 \| \chi_{|.| \leq N^{\alpha}} \|_1 (\mathcal{N}_+ + 1)^2 \lesssim N^{-1+\alpha} (\mathcal{N}_+ + 1)^2, \end{align*} $$

where we used $\delta = N^{\alpha }$ .

Let us now consider term $(\textrm {II})$ in equation (77). We write

(79) $$ \begin{align} \begin{aligned} (\textrm{II}) &= \sum_{p,q,r,s} \hat{V}_N (p) \tilde{\varphi}_r \chi_{|s| \leq N^{\alpha}} \\ &\hspace{1cm} \times \Big\{ a_0^{\dagger} a_s^{\dagger} \big[ a_{-r} a_{r+s} , a^{\dagger}_{p+q} a^{\dagger}_{-p} \big] a_q a_0 + a_{p+q}^{\dagger} a_{-p}^{\dagger} \big[ a_0^{\dagger} a_s^{\dagger} , a_q a_0 \big] a_{-r} a_{r+s} \Big\} + \text{h.c.} \\ &=: (\textrm{II})_a + (\textrm{II})_b. \end{aligned} \end{align} $$

With

$$\begin{align*}a_{p+q}^{\dagger} a_{-p}^{\dagger} \big[ a_0^{\dagger} a_s^{\dagger} , a_q a_0 \big] a_{-r} a_{r+s} = - \delta_{sq} \, a_{p+q}^{\dagger} a_{-p}^{\dagger} a_0^{\dagger} a_0 a_{-r} a_{r+s} - a_{p+q}^{\dagger} a_{-p}^{\dagger} a_q a_s^{\dagger} a_{-r} a_{r+s}, \end{align*}$$

we obtain

$$\begin{align*}\begin{aligned} (\textrm{II})_b = \; &-\sum_{p,q,r} \hat{V}_N (p) \tilde{\varphi}_r \chi_{|q| \leq N^{\alpha}} \, a_{p+q}^{\dagger} a_{-p}^{\dagger} a_0^{\dagger} a_0 a_{-r} a_{r+q} \\ &- \sum_{p,q,r,s} \hat{V}_N (p) \tilde{\varphi}_r \chi_{|s| \leq N^{\alpha}} \, a_{p+q}^{\dagger} a_{-p}^{\dagger} a_s^{\dagger} a_q a_{-r} a_{r+s} \\ &- \sum_{p,r,q} \hat{V}_N (p) \tilde{\varphi}_r \chi_{|q| \leq N^{\alpha}} \, a_{p+q}^{\dagger} a^{\dagger}_{-p} a_{-r} a_{r+q} =: (\textrm{II})_{b1} + (\textrm{II})_{b2} + (\textrm{II})_{b3}. \end{aligned} \end{align*}$$

We can bound $(\textrm {II})_{b3}$ by switching to position space. We find

$$\begin{align*}\begin{aligned} \pm (\textrm{II})_{b3} &= \pm \int_{\Lambda^4} \mathrm{d}x \mathrm{d}y \mathrm{d}u \mathrm{d}v \, V_N (x-y) \check{\chi}_{|.| \leq N^{\alpha}} (x-u) \check{\tilde {\varphi}} (u-v) a_x^{\dagger} a_y^{\dagger} a_u a_v + \text{h.c.} \\ &\lesssim \delta \| \check{\tilde {\varphi}} \|_2^2 \, Q_4 + \delta^{-1} \| \check{\chi}_{|.| \leq N^{\alpha}} \|_2^2 \, (\mathcal{N}_+ + 1)^2 \lesssim N^{-\alpha/2} Q_4 + N^{-3 +5 \alpha/2} (\mathcal{N}_+ + 1)^2. \end{aligned} \end{align*}$$

Term $(\textrm {II})_{b1}$ can be bounded analogously, but it contains an additional factor $\mathcal {N}_0 = a_0^{\dagger } a_0 \leq N$ . Thus

$$\begin{align*}\pm (\textrm{II})_{b1} \leq N^{-\alpha/2} Q_4 + N^{-1 +5 \alpha/2} (\mathcal{N}_+ + 1)^2. \end{align*}$$

Term $(\textrm {II})_{b1}$ can also be bounded in position space. We obtain

$$\begin{align*}\begin{aligned} \pm (\textrm{II})_{b2} &= \pm \int_{\Lambda^3} \mathrm{d}x \mathrm{d}y \mathrm{d}u \, V_N (x-y) \, a_x^{\dagger} a_y^{\dagger} a^{\dagger} (\check{\chi}_{|.| \leq N^{\alpha}}^u) a_x a (\check{\tilde {\varphi}}^u) a_u + \text{h.c.} \\ &\leq \delta \| \check{\chi}_{|.| \leq N^{\alpha}} \|_2^2 \, Q_4 \mathcal{N}_+ + \delta^{-1} \| V_N \|_1 \| \check{\tilde {\varphi}} \|_2^2 (\mathcal{N}_+ + 1)^3 \lesssim N^{-\alpha/2} Q_4 + N^{-1+5\alpha/2} (\mathcal{N}_+ + 1)^2, \end{aligned} \end{align*}$$

where we chose $\delta = N^{-1-7\alpha /2}$ .

Let us now consider term $(\textrm {II})_a$ , defined on the right-hand side of equation (79). With

$$\begin{align*}\begin{aligned} a_0^{\dagger} &a_s^{\dagger} \big[ a_{-r} a_{r+s} , a^{\dagger}_{p+q} a^{\dagger}_{-p} \big] a_q a_0 \\ &= a_0^{\dagger} a_s^{\dagger} \Big\{ \delta_{-r,p+q} a_{-p}^{\dagger} a_{r+s} + \delta_{r,p} a^{\dagger}_{p+q} a_{r+s} + \delta_{r+s, p+q} a_{-r} a_{-p}^{\dagger} + \delta_{r+s,-p} a_{-r} a_{p+q}^{\dagger} \Big\} a_q a_0, \end{aligned} \end{align*}$$

we obtain, rearranging the terms in normal order (with appropriate changes of variables),

(80) $$ \begin{align} \begin{aligned} (\textrm{II})_a = \; &\sum_{p,q} \big( \hat{V}_N (p) + \hat{V}_N (p-q) \big) \tilde{\varphi}_p \chi_{|q| \leq N^{\alpha}} a_0^{\dagger} a_q^{\dagger} a_q a_0 \\ &+ 4 \sum_{p,q,s} \hat{V}_N (p) \tilde{\varphi}_{p+q} \chi_{|s| \leq N^{\alpha}} \, a_0^{\dagger} a_s^{\dagger} a_{-p}^{\dagger} a_{-p-q+s} a_q a_0 + \text{h.c.} \\ =: \; &2 \sum_{p,q} \big( \hat{V}_N (p) + \hat{V}_N (p-q) \big) \tilde{\varphi}_p \chi_{|q| \leq N^{\alpha}} a_0^{\dagger} a_q^{\dagger} a_q a_0 + (\textrm{II})_{a1}, \end{aligned} \end{align} $$

where we can bound, with $\mathcal {N}_0 = a_0^{\dagger } a_0 \leq N$ ,

$$\begin{align*}\begin{aligned} \pm (\textrm{II})_{a1} \lesssim N \| \hat{V}_N \|_{\infty} \Big[ \delta \| \tilde {\varphi} \|_2^2 + \delta^{-1} \| \chi_{|.| \leq N^{\alpha}} \|_2^2 \Big] \mathcal{N}_+^2 \lesssim N^{-1+\alpha} (\mathcal{N}_+ + 1)^2, \end{aligned} \end{align*}$$

choosing $\delta = N^{1-\alpha }$ . Collecting all the estimates we have proved so far, we conclude from equation (77) that

$$\begin{align*}\big[ Q_3, \mathcal{B}_3 \big] = 2 \sum_{p,q} \big( \hat{V}_N (p) + \hat{V}_N (p-q) \big) \tilde{\varphi}_p \chi_{|q| \leq N^{\alpha}} a_0^{\dagger} a_q^{\dagger} a_q a_0 + \mathcal{E}_1, \end{align*}$$

where

(81) $$ \begin{align} \pm \mathcal{E}_1 \lesssim N^{-\alpha/2} Q_4 + N^{-1+5\alpha/2} (\mathcal{N}_+ + 1)^2. \end{align} $$

At the expense of adding an additional small error to the right-hand side of the estimate (81), in the main term, we can replace $a_0^{\dagger } a_0 = N - \mathcal {N}_+$ by a factor of N, since

$$\begin{align*}\pm \sum_{p,q} (\hat{V}_N (p) + \hat{V}_N (p-q) ) \tilde{\varphi}_p \chi_{|q| \leq N^{\alpha}} a_q^{\dagger} \mathcal{N}_+ a_q \lesssim N^{-1} \| \tilde {\varphi} \|_1 (\mathcal{N}_++1)^2 \lesssim N^{-1} (\mathcal{N}_+ + 1)^2. \end{align*}$$

Moreover, from

(82) $$ \begin{align} \pm N\sum_{p,q} &\left( \hat V_N(p-q)- \hat V_N(p)\right)\tilde {\varphi}_p \chi_{|q| \leq N^{\alpha}} a^{\dagger}_q a_q \notag \\ &\qquad\qquad\lesssim \sum_{p,q} | q/N| \| \nabla \hat V\|_{\infty} |\tilde {\varphi}_p| \chi_{|q| \leq N^{\alpha}} a^{\dagger}_q a_q\lesssim \|\tilde {\varphi}\|_1 N^{-1+\alpha} \mathcal{N}_+ \lesssim N^{-1+\alpha} \mathcal{N}_+, \end{align} $$

we arrive at

$$\begin{align*}\big[ Q_3, \mathcal{B}_3 \big] = 4N \sum_{p} \hat{V}_N (p) \tilde{\varphi}_p \sum_{|q| \leq N^{\alpha}} a_q^{\dagger} a_q + \mathcal{E}_2, \end{align*}$$

where

$$\begin{align*}\pm \mathcal{E}_2 \lesssim N^{-\alpha/2} Q_4 + N^{-1+\alpha} \mathcal{N}_+ + N^{-1+5\alpha/2} (\mathcal{N}_+ + 1)^2. \end{align*}$$

Conjugating with $e^{t {\mathcal {B}}_3}$ and integrating over $t\ \mathrm{and}\ s$ , we obtain, with the help of Lemma 10 and Lemma 14 (and the observation that the first estimate in Lemma 10 also holds, if we replace $\mathcal {N}_+$ on the left-hand side by $\sum _{|p| \leq N^{\alpha }} a_p^{\dagger } a_p$ ),

$$\begin{align*}\int_0^1 \int_s^1 e^{-t{\mathcal{B}}_3} [Q_3, {\mathcal{B}}_3] e^{t {\mathcal{B}}_3} \mathrm{d}t \mathrm{ d}s = 2N \sum_q \hat{V}_N (q) \tilde{\varphi}_q \sum_{|p| \leq N^{\alpha}} a_p^{\dagger} a_p + \mathcal{E}_3, \end{align*}$$

where

$$\begin{align*}\pm \mathcal{E}_3 \lesssim N^{-\alpha/2} (Q_4 + \mathcal{N}_+ + 1) + N^{-1+\alpha} \mathcal{N}_+ + N^{-1+5\alpha/2} (\mathcal{N}_+ + 1)^2. \end{align*}$$

The claim now follows from equation (19) and the observation that

$$\begin{align*}\mathcal{N}_+ - \sum_{|p| \leq N^{\alpha}} a_p^{\dagger} a_p = \sum_{|p| \geq N^{\alpha}} a_p^{\dagger} a_p \leq N^{-2\alpha} H_1.\\[-47pt] \end{align*}$$

Proof. We have

$$\begin{align*}\begin{aligned} e^{-{\mathcal{B}}_3} \tilde Q_2 e^{{\mathcal{B}}_3} - \tilde Q_2 &= \int_0^1 e^{-s{\mathcal{B}}_3}[ \tilde Q_2,{\mathcal{B}}_3] e^{s{\mathcal{B}}_3}\mathrm{d}s \\ &= \frac{4\pi \mathfrak{a}_N}{N} \sum_{|r| \leq N^{\alpha}} \int_0^1 e^{-s{\mathcal{B}}_3}[ a^{\dagger}_r a^{\dagger}_{-r} a_0 a_0 + \text{h.c.}, {\mathcal{B}}_3] e^{s{\mathcal{B}}_3}\mathrm{d}s. \end{aligned} \end{align*}$$

We compute the commutator

(84) $$ \begin{align} [a^{\dagger}_r a^{\dagger}_{-r} a_0 a_0 &+ \text{h.c.} , a_{p+q}^{\dagger} a_{-p}^{\dagger} a_q a_0 - \text{h.c.} ] \end{align} $$

(85) $$ \begin{align} &=[a^{\dagger}_r a^{\dagger}_{-r}a_0a_0,a^{\dagger}_{p+q} a^{\dagger}_{-p} a_q a_0]+[a^{\dagger}_0 a^{\dagger}_0 a_r a_{-r},a^{\dagger}_{p+q} a^{\dagger}_{-p} a_q a_0]+\text{h.c.}\nonumber \\ &= 2\Big\{ a^{\dagger}_0 a^{\dagger}_0 \big( \delta_{p+q,r} a^{\dagger}_{-p} a_{-p-q} +\delta_{p,r} a^{\dagger}_{p+q} a_{p} \big) a_q a_0 \nonumber\\ &\qquad\qquad\qquad -\delta_{r,q}a^{\dagger}_{p+q} a^{\dagger}_{-p} a^{\dagger}_{-q} a_0 a_0 a_0- a^{\dagger}_{p+q} a^{\dagger}_{-p} a^{\dagger}_{0} a_{q} a_r a_{-r}\Big\} +\text{h.c.} \end{align} $$

Hence, we obtain

$$\begin{align*}\begin{aligned} \frac{4\pi \mathfrak{a}_N}{N} \sum_{|r| \leq N^{\alpha}} \big[ a_r^{\dagger} &a_{-r}^{\dagger} a_0 a_0 + \text{h.c.} , {\mathcal{B}}_3 \big] \\ =\; &\frac{8\pi \mathfrak{a}_N}{N} \sum_{p,q : |p+q| \leq N^{\alpha}} \tilde{\varphi}_p \chi_{|q| \leq N^{\alpha}} a_0^{\dagger} a_0^{\dagger} a_{-p}^{\dagger} a_{-p-q} a_q a_0 \\ &- \frac{4\pi \mathfrak{a}_N}{N} \sum_{p, |q| \leq N^{\alpha}} \tilde {\varphi}_p \chi_{|q| \leq N^{\alpha}} a_{p+q}^{\dagger} a_{-p}^{\dagger} a_{-q}^{\dagger} a_0 a_0 a_0 \\ &- \frac{4\pi \mathfrak{a}_N}{N} \sum_{p, q, |r| \leq N^{\alpha}} \tilde {\varphi}_p \chi_{|q| \leq N^{\alpha}} a_{p+q}^{\dagger} a_{-p}^{\dagger} a_{0}^{\dagger} a_q a_r a_{-r} =: (\textrm{I}) + (\textrm{II}) + (\textrm{III}). \end{aligned} \end{align*}$$

With Cauchy-Schwarz and using the bounds from Lemma 2, we can bound

$$\begin{align*}\begin{aligned} \pm (\textrm{I}) &\lesssim N^{-1/2-\alpha/2} (\mathcal{N}_+ + 1)^{3/2} , \\ \pm (\textrm{II}) &\lesssim N^{-1/2+\alpha} (\mathcal{N}_+ + 1)^{3/2} , \\ \pm (\textrm{III}) &\lesssim N^{-3/2+\alpha} (\mathcal{N}_+ + 1)^{5/2}. \end{aligned} \end{align*}$$

The claim now follows with Lemma 10.

Proof of Proposition 11.

Recall from equation (66) that

$$\begin{align*}\begin{aligned} e^{-{\mathcal{B}}_3} e^{-{\mathcal{B}}_2} H_N e^{{\mathcal{B}}_2} e^{{\mathcal{B}}_3} = \; & 4 \pi \mathfrak{a}_N (N-1) + \sum_{|p|\leq N^{\alpha}} \frac{(4\pi \mathfrak{a}_N)^2}{p^2} + H_1 + Q_4 + e^{-{\mathcal{B}}_3}(\tilde H_2 + \tilde Q_2 + \mathcal E_{{\mathcal{B}}_2})e^{{\mathcal{B}}_3} \\ &+ \int_0^1 e^{-t{\mathcal{B}}_3} \Gamma_3 e^{t{\mathcal{B}}_3} \mathrm{d}t + \int_0^1 \int_s^1 e^{-t{{\mathcal{B}}_3}}[Q_3,{{\mathcal{B}}_3}]e^{t{{\mathcal{B}}_3}} \mathrm{d}t \mathrm{d}s, \end{aligned} \end{align*}$$

where

$$\begin{align*}\pm \mathcal{E}_{{\mathcal{B}}_2} \leq N^{-\alpha/2} Q_4 + \big[ N^{-\alpha/2} + N^{-1+5\alpha/2} \big] (\mathcal{N}_+ + 1) + N^{-1+\alpha} \mathcal{N}_+^2 + N^{-2} H_1. \end{align*}$$

With Lemma 10 and Lemma 14, this also implies that

$$\begin{align*}\begin{aligned} \pm e^{-{\mathcal{B}}_3} \mathcal{E}_{{\mathcal{B}}_2} e^{{\mathcal{B}}_3} \lesssim \; &N^{-\alpha/2} Q_4 + N^{-2} H_1 + \big[ N^{-\alpha/2} + N^{-1+5\alpha/2} \big] (\mathcal{N}_+ + 1) \\ &+ N^{-1+2\alpha} (\mathcal{N}_+ + 1)^2. \end{aligned} \end{align*}$$

Applying the first bound in Lemma 10 to the operator $\tilde H_2 = (2\hat {V} (0) - 8 \pi \mathfrak {a}_N) \mathcal {N}_+$ , defined in equation (24), we obtain

$$\begin{align*}\pm \Big[ e^{-{\mathcal{B}}_3} \tilde H_2 e^{{\mathcal{B}}_3} - \tilde H_2 \Big] \lesssim N^{-\alpha/2} (\mathcal{N}_+ + 1). \end{align*}$$

Combining Lemma 13 with Lemma 14, we obtain

$$\begin{align*}\begin{aligned} \int_0^1 e^{-t {\mathcal{B}}_3} &\Gamma_3 e^{t {\mathcal{B}}_3} \mathrm{d}t \\ \lesssim \; & N^{-3\alpha/2} H_1 + N^{-\alpha/2} Q_4 + N^{-\alpha/2} (\mathcal{N}_+ + 1) \\ &+ N^{3\alpha/2-1/2} (\mathcal{N}_+ + 1)^{3/2} + N^{-1+5\alpha/2} (\mathcal{N}_+ + 1)^2. \end{aligned} \end{align*}$$

Together with the bounds in Lemmas 15 and 16, and with the observation that

$$\begin{align*}8\pi \mathfrak{a}_N \sum_{|p|> N^{\alpha}} a_p^{\dagger} a_p \leq N^{-2\alpha} H_1, \end{align*}$$

we conclude the proof of Proposition 11.

Proof. The proof of the bound (90) is standard (based on Gronwall’s lemma and the bounds in Lemma 17). To prove equation (91), we define $g(s) = e^{-s{\mathcal {B}}_4} H_1 e^{s{\mathcal {B}}_4}$ and compute (using the commutation relations after equation (89))

$$ \begin{align*} g'(s) &= e^{-s{\mathcal{B}}_4} [H_1,{\mathcal{B}}_4] e^{s{\mathcal{B}}_4} = \sum_{|p| \leq N^{\alpha}} p^2 \tau_p e^{-s{\mathcal{B}}_4} \big[ b^{\dagger}_p b^{\dagger}_{-p} + \text{h.c.} \big] e^{s{\mathcal{B}}_4} \lesssim e^{-s {\mathcal{B}}_4} H_1 e^{s {\mathcal{B}}_4} + \sum_{|p| \leq N^{\alpha}} p^2 \tau_p^2. \end{align*} $$

From Lemma 17, we have $\sum _p p^2 \tau _p^2 \lesssim N^{\alpha }$ ; with Gronwall’s lemma, we obtain the bound (91).

To show the bound (92), we set $h(s) = e^{-s{\mathcal {B}}_4} Q_4 e^{s{\mathcal {B}}_4}$ , and then

(93) $$ \begin{align} h' (s) = e^{-s{\mathcal{B}}_4} [Q_4, {\mathcal{B}}_4] e^{s{\mathcal{B}}_4}. \end{align} $$

Proceeding as in (38), we find (we use here the convention that $\tau _q = 0$ , for $|q|> N^{\alpha }$ )

$$\begin{align*}\begin{aligned} [Q_4, {\mathcal{B}}_4 ] = \; & \frac{1}{2} \sum_{q} (\hat{V}_N * \tau) (q) b_q^{\dagger} b_{-q}^{\dagger} + \sum_{p,q,s} \hat{V}_N (q-p) \tau_s b^{\dagger}_{p+s-q} b^{\dagger}_q a^{\dagger}_{-s} a_p +\text{h.c.} \end{aligned} \end{align*}$$

Switching to position space, we write

$$\begin{align*}[Q_4, {\mathcal{B}}_4 ] = \frac{1}{2} \int_{\Lambda^2} \mathrm{d}x \mathrm{d}y V_N (x-y) \check{\tau} (x-y) b_x^{\dagger} b_y^{\dagger} + \int_{\Lambda^3} \mathrm{d}x \mathrm{d}y \mathrm{d}z V_N (x-y) \check{\tau} (x-z) b_x^{\dagger} b_y^{\dagger} a_z^{\dagger} a_y + \text{h.c.} \end{align*}$$

With the bounds from Lemma 17, we conclude

$$\begin{align*}\pm [Q_4, {\mathcal{B}}_4] \lesssim Q_4 + N^{-1+2\alpha} + N^{-1} (\mathcal{N}_+ + 1)^2. \end{align*}$$

Inserting this in equation (93), applying the bound (90) and then Gronwall’s lemma, we obtain the desired bound (92).

Proof. For $s \in [0;1]$ , we define

$$\begin{align*}E(s) = \sum_{|p| \leq N^{\alpha}} \sqrt{|p|^4 + 16 \pi \mathfrak{a}_N p^2} \, \big(\gamma^{s}_p b_p^{\dagger} + \nu_p^{s} b_{-p} \big) \big( \gamma_p^{s} b_p + \nu^{s}_p b_{-p}^{\dagger} \big) \end{align*}$$

with the operators $b_p, b^{\dagger }_p$ defined in equation (88) and with the notation $\gamma _p^{s} = \cosh (s \tau _p)$ and $\nu _p^{s} = \sinh (s \tau _p)$ . In particular, for $s=1$ , this is exactly the operator appearing on the third line in equation (87). For $\psi $ in the sector $\{ \mathcal {N} = N \}$ , we define $f_{\psi } :[0;1] \to \mathbb {R}$ by

$$\begin{align*}f_{\psi} (s) = \langle \psi , e^{-s{\mathcal{B}}_4} E(s) e^{s{\mathcal{B}}_4} \psi \rangle. \end{align*}$$

The idea is that the generalised Bogoliubov transformation $e^{s{\mathcal {B}}_4}$ approximately cancels (on states with few excitations) the symplectic rotations determined by the coefficients $\gamma _p^{s}$ , $\nu _p^{s}$ (it would precisely cancel them if the operators $b_p^{\dagger }, b_p$ satisfied canonical commutation relations); hence, on states with few excitations, we expect $f_{\psi }$ to be approximately constant in s. More precisely, we claim that

(96) $$ \begin{align} \big| f^{\prime}_{\psi} (s) \big| \lesssim N^{2\alpha-1} \langle \psi, (\mathcal{N}_+ + 1)^2 \psi \rangle. \end{align} $$

Assuming for a moment that the bound (96) holds true, we could conclude, integrating over $s \in [0;1]$ , that

$$\begin{align*}e^{-{\mathcal{B}}_4} E(1) e^{{\mathcal{B}}_4} = E(0) + \delta \end{align*}$$

with

$$\begin{align*}\pm \delta \lesssim N^{2\alpha-1} (\mathcal{N}_+ + 1)^2. \end{align*}$$

With the bounds from Lemma 18 (and noticing that the action of ${\mathcal {B}}_4$ on the high-momenta part of the kinetic energy is trivial), this would imply that

(97) $$ \begin{align} \begin{aligned} e^{-{\mathcal{B}}_4} &e^{-{\mathcal{B}}_3} e^{-{\mathcal{B}}_2} H_N e^{{\mathcal{B}}_2} e^{{\mathcal{B}}_3} e^{{\mathcal{B}}_4} \\ &= 4 \pi \mathfrak{a}_N (N-1) - \frac{1}{2} \sum_{|p| \leq N^{\alpha}} \Big[ p^2 + 8\pi \mathfrak{a}_N - \sqrt{|p|^4 + 16\pi \mathfrak{a}_N p^2} - \frac{(8 \pi \mathfrak{a}_N)^2}{2p^2} \Big] \\ &\quad + \sum_{|p| \leq N^{\alpha}} \sqrt{|p|^4 + 16 \pi \mathfrak{a}_N p^2} \, a_p^{\dagger} \frac{a_0 a_0^{\dagger}}{N} a_p + \sum_{|p|> N^{\alpha}} p^2 a_p^{\dagger} a_p + e^{-{\mathcal{B}}_4} Q_4 e^{{\mathcal{B}}_4} + \mathcal{E} , \end{aligned} \end{align} $$

where $\mathcal {E}$ satisfies the bound (95). Writing $a_0 a_0^{\dagger } = a_0^{\dagger } a_0 + 1 = N - \mathcal {N}_+ + 1$ , we could then replace

$$\begin{align*}\sum_{|p| \leq N^{\alpha}} \sqrt{|p|^4 + 16 \pi \mathfrak{a}_N p^2} \, a_p^{\dagger} \frac{a_0 a_0^{\dagger}}{N} a_p = \sum_{|p| \leq N^{\alpha}} \sqrt{|p|^4 + 16 \pi \mathfrak{a}_N p^2} \, a_p^{\dagger} a_p + \delta \end{align*}$$

with

$$\begin{align*}\pm \delta \leq \frac{1}{N} \sum_{|p| \leq N^{\alpha}} \sqrt{|p|^4 + 16 \pi \mathfrak{a}_N p^2} a_p^{\dagger} (\mathcal{N}_+ + 1) a_p \lesssim N^{-1+2\alpha} (\mathcal{N}_++1)^2. \end{align*}$$

Furthermore, since

$$ \begin{align*} \Big| \, p^2 + 8 \pi \mathfrak{a}_N - \sqrt{|p|^4 + 16 \pi \mathfrak{a}_N p^2} - \frac{(8\pi \mathfrak{a}_N)^2}{2p^2} \Big| \lesssim (1+p^2)^{-2}, \end{align*} $$

we could write

$$ \begin{align*} \sum_{|p|\leq N^{\alpha}} &\left[ p^2 + 8\pi \mathfrak{a}_N - \sqrt{|p|^4 + 16 \pi \mathfrak{a}_N p^2} - \frac{(8\pi \mathfrak{a}_N)^2}{2p^2}\right] \\ &\qquad\qquad\qquad\qquad = \sum_{p} \left[p^2 + 8\pi \mathfrak{a}_N - \sqrt{|p|^4 + 16 \pi \mathfrak{a}_N p^2} - \frac{(8\pi \mathfrak{a}_N)^2}{2p^2}\right] + {\mathcal O}(N^{-\alpha}). \end{align*} $$

Similarly, from $\big | p^2 - \sqrt {|p|^4 + 16 \pi \mathfrak {a}_N p^2} \big | \lesssim 1$ , we could bound

$$\begin{align*}\pm \sum_{|p|> N^{\alpha}} \big[ p^2 - \sqrt{|p|^4 + 16 \pi \mathfrak{a}_N p^2} \big] a_p^{\dagger} a_p \lesssim N^{-2\alpha} H_1. \end{align*}$$

Inserting all these estimates in equation (97), we would end up with (94) and (95).

It remains to show the bound (96). To this end, we observe that

(98) $$ \begin{align} f^{\prime}_{\psi} (s) = \frac{d}{ds} \big\langle \psi, e^{s {\mathcal{B}}_4} E(s) e^{-s{\mathcal{B}}_4} \psi \big\rangle = \big\langle \psi, e^{s {\mathcal{B}}_4} \big\{ \big[ {\mathcal{B}}_4, E(s) \big] + \frac{\partial E(s)}{\partial s} \big\} e^{-s{\mathcal{B}}_4} \psi \big\rangle. \end{align} $$

We have, denoting $\varepsilon _p = \sqrt {|p|^4 + 16 \pi \mathfrak {a}_N p^2}$ ,

$$\begin{align*}\begin{aligned} \big[ {\mathcal{B}}_4, E(s) \big] = \frac{1}{2} \sum_{|p| \leq N^{\alpha}} \varepsilon_p \sum_{|q| \leq N^{\alpha}} \tau_q \big\{ &\big[ b_q^{\dagger} b_{-q}^{\dagger}, (\gamma_p^s b_p^{\dagger} + \nu_p^s b_{-p}) \big] (\gamma_p^s b_p + \nu_p^s b_{-p}^{\dagger}) \\ &+ (\gamma_p^s b_p^{\dagger} + \nu_p^s b_{-p}) \big[ b_q^{\dagger} b_{-q}^{\dagger}, (\gamma_p^s b_p + \nu_p^s b_{-p}^{\dagger}) \big] \big\} + \text{h.c.} \end{aligned} \end{align*}$$

A long but straightforward computation, based on the commutation relations (89), leads to

(99) $$ \begin{align} \begin{aligned} \big[ &{\mathcal{B}}_4, E(s) \big] \\ = &\, -\frac{1}{2}\kern-1.2pt \sum_{|p| \leq N^{\alpha}} \varepsilon_p \tau_p \nu_p^s b_p^{\dagger} \big(1-\frac{\mathcal{N}_+}{N}\big) (\gamma_p^s b_p + \nu_p^s b_{-p}^{\dagger} ) - \frac{1}{2} \kern-1.2pt\sum_{|p| \leq N^{\alpha}} \kern-1.2pt\varepsilon_p \tau_p \nu_p^s \big(1-\frac{\mathcal{N}_+}{N} \big) b_p^{\dagger} (\gamma_p^s b_p + \nu_p^s b_{-p}^{\dagger} ) \\ &-\frac{1}{2} \sum_{|p| \leq N^{\alpha}} \varepsilon_p \tau_p \gamma_p^s (\gamma_p^s b^{\dagger}_p + \nu_p^s b_{-p} ) b_{-p}^{\dagger} \big(1-\frac{\mathcal{N}_+}{N}\big) -\frac{1}{2} \sum_{|p| \leq N^{\alpha}} \varepsilon_p \tau_p \gamma_p^s (\gamma_p^s b^{\dagger}_p + \nu_p^s b_{-p} )\big(1-\frac{\mathcal{N}_+}{N}\big) b_{-p}^{\dagger} \\ &+ \frac{1}{2N} \sum_{|p|, |q| \leq N^{\alpha}} \varepsilon_p \tau_q \nu_p^s \big[ b_q^{\dagger} a_{-q}^{\dagger} a_{-p} + a_q^{\dagger} a_{-p} b_{-q}^{\dagger} \big] (\gamma_p^s b_p + \nu_p^s b_{-p}^{\dagger} ) \\ &+ \frac{1}{2N} \sum_{|p|, |q| \leq N^{\alpha}} \varepsilon_p \tau_q \gamma_p^s (\gamma_p^s b^{\dagger}_p + \nu_p^s b_{-p}) \big[ b_q^{\dagger} a_{-q}^{\dagger} a_{p} + a_q^{\dagger} a_{p} b_{-q}^{\dagger} \big] + \text{h.c.} \end{aligned} \end{align} $$

To compute the explicit time derivative of the observable $E(s)$ , on the other hand, we notice that $ d\gamma _p^s /ds = \tau _p \nu _p^s$ and $d\nu _p^s / ds = \tau _p \gamma _p^s$ . Thus, we obtain

$$\begin{align*}\begin{aligned} \frac{\partial E(s)}{\partial s} = &\!\sum_{|p| \leq N^{\alpha}} \!\!\varepsilon_p \tau_p \big(\nu_p^s b_p^{\dagger} + \gamma_p^s b_{-p} \big) \big( \gamma_p^s b_p + \nu_p^s b_{-p}^{\dagger} \big) +\! \sum_{|p| \leq N^{\alpha}} \!\!\varepsilon_p \tau_p (\gamma_p^s b_p^{\dagger} + \nu_p^s b_{-p}) (\nu_p^s b_p + \gamma_p^s b_{-p}^{\dagger}). \end{aligned} \end{align*}$$

Combining the last equation with equation (99), we observe that (as expected) all large contributions cancel. We find

$$\begin{align*}\begin{aligned} \big[ {\mathcal{B}}_4, E(s) \big] + \frac{\partial E(s)}{\partial s} = \; &\frac{1}{N} \sum_{|p| \leq N^{\alpha}} \varepsilon_p \tau_p \nu_p^s \, \mathcal{N}_+ b_p^{\dagger} (\gamma_p^s b_p + \nu_p^s b_{-p}^{\dagger}) \\ &+\frac{1}{N} \sum_{|p| \leq N^{\alpha}} \varepsilon_p \tau_p \gamma_p^s \, (\gamma_p^s b^{\dagger}_p + \nu_p^s b_{-p}) \mathcal{N}_+ b_{-p}^{\dagger} \\ &+\frac{1}{N} \sum_{|p|, |q| \leq N^{\alpha}} \varepsilon_p \tau_q \nu_p^s \, b_q^{\dagger} a_{-q}^{\dagger} a_{-p} (\gamma_p^s b_p + \nu_p^s b_{-p}^{\dagger}) \\ &+ \frac{1}{N} \sum_{|p|, |q| \leq N^{\alpha}} \varepsilon_p \tau_q \gamma_p^s \, (\gamma_p^s b_p^{\dagger} + \nu_p^s b_{-p}) b_q^{\dagger} a_{-q}^{\dagger} a_p + \text{h.c.} \end{aligned} \end{align*}$$

Using the bounds in Lemma 17, with the estimate $\varepsilon _p \lesssim p^2$ and the restrictions $|p|, |q| \leq N^{\alpha }$ , we arrive at

$$\begin{align*}\pm \left\{ \big[ {\mathcal{B}}_4, E(s) \big] + \frac{\partial E(s)}{\partial s} \right\} \lesssim N^{2\alpha-1} (\mathcal{N}_+ + 1)^2. \end{align*}$$

From equation (98), this implies that

$$\begin{align*}|f^{\prime}_{\psi} (s)| \lesssim N^{2\alpha-1} \langle \psi, e^{-s{\mathcal{B}}_4} (\mathcal{N}_+ + 1)^2 e^{s{\mathcal{B}}_4} \psi \rangle. \end{align*}$$

Applying Lemma 18, we obtain the desired bound (96).

Proof. To take care of the terms on the second line of the bound (95), we use localisation in the number of particles, a tool developed in [Reference Lewin, Nam, Serfaty and Solovej17] and, in the present setting, in [Reference Boccato, Brennecke, Cenatiempo and Schlein7]. Here, we make use of the results of [Reference Lieb and Seiringer18, Reference Lieb and Seiringer19, Reference Nam, Rougerie and Seiringer24], which imply that, if $\psi _N \in L^2_s (\Lambda ^N)$ is a normalised sequence of approximate ground states of the Hamilton operator $H_N$ satisfying

$$\begin{align*}\Big| \frac{1}{N} \langle \psi_N, H_N \psi_N \rangle - 4\pi \mathfrak{a}_N \Big| \to 0 \end{align*}$$

as $N \to \infty $ , then $\psi _N$ exhibit condensation, in the sense that

(104) $$ \begin{align} \lim_{N \to \infty} \frac{1}{N} \langle \psi_N , \mathcal{N}_+ \psi_N \rangle = 0. \end{align} $$

Now let $f,g : \mathbb {R} \to [0,1]$ be smooth functions such that $f(s)^2 + g(s)^2 = 1$ for all $s \in \mathbb {R}$ , and $f(s) = 1$ for $s \leq 1/2$ , $f(s) =0$ for $s\geq 1$ . For $M_0 \geq 1$ , we define $f_{M_0}(\mathcal {N}_+) = f(\mathcal {N}_+ / M_0)$ and $g_{M_0}(\mathcal {N}_+) = g(\mathcal {N}_+/M_0)$ . Then we have

(105) $$ \begin{align} H_N &= f_{M_0} H_N f_{M_0} + g_{M_0} H_N g_{M_0} + \mathcal E_{M_0}, \end{align} $$

with

(106) $$ \begin{align} \mathcal E_{M_0} = \frac{1}{2} \left( [f_{M_0},[f_{M_0},H_N]] + [g_{M_0},[g_{M_0},H_N]]\right). \end{align} $$

In view of equation (9), we can write (with $h =f,g$ )

$$ \begin{align*} [h_{M_0} , [h_{M_0},H_N]] &= \big[ h (\mathcal{N}_+ / M_0) - h ((\mathcal{N}_+ -2)/M_0)\big]^2 \sum_{p \neq 0} \hat V_N(p) a_p^{\dagger} a_{-p}^{\dagger} a_0 a_0 + \text{h.c.}, \\ &\quad + \big[ h (\mathcal{N}_+ / M_0) - h ((\mathcal{N}_+ -1)/M_0)\big]^2 \sum_{q,r,q+r\neq 0} \hat V_N(r) a_{q+r}^{\dagger} a_{-r}^{\dagger} a_{q}a_0 + \text{h.c.} \end{align*} $$

This easily implies that

(107)

We choose $M_0 = \varepsilon N$ , for some $\varepsilon>0$ independent of N, to be fixed later. We introduce the notation $\mathcal {N}_{+}^{\, {\mathcal U}} = {\mathcal U}^{\dagger } \mathcal {N}_+ {\mathcal U}$ . We use equation (102) with the bound (95) for the error term $\mathcal {E}_4$ ; we pick $\alpha = - \log \ell / \log N$ so that $N^{\alpha } = 1/\ell $ , for some $\ell> 0$ independent of N, to be specified below. For N large enough, we obtain from Proposition 19

(108) $$ \begin{align} f_{M_0}(\mathcal{N}_+) H_N &f_{M_0}(\mathcal{N}_+) = {\mathcal U} f_{M_0}(\mathcal{N}_{+}^{\, {\mathcal U}}) {\mathcal U}^{\dagger} H_N {\mathcal U} \, f_{M_0}(\mathcal{N}_{+}^{\, {\mathcal U}}) {\mathcal U}^{\dagger} \nonumber \\ &\geq {\mathcal U} f_{M_0}(\mathcal{N}_{+}^{\, {\mathcal U}}) \left( 4\pi \mathfrak{a}_N N + H_1 -C + e^{-{\mathcal{B}}_4}Q_4e^{{\mathcal{B}}_4} + \mathcal E_4 \right)f_{M_0}(\mathcal{N}_{+}^{\, {\mathcal U}}) {\mathcal U}^{\dagger}\nonumber \\ &\geq {\mathcal U} f_{M_0}(\mathcal{N}_{+}^{\, {\mathcal U}}) \big( 4\pi \mathfrak{a}_N N + (1 - C \ell^{1/2} - C\ell^{-5/2} N^{-1}- C \ell^{-7/2}\varepsilon) (H_1+1) \nonumber \\ &\qquad \qquad\qquad\qquad\qquad\qquad\qquad + (1-C \ell^{1/2}) e^{-{\mathcal{B}}_4}Q_4e^{{\mathcal{B}}_4}\big)f_{M_0}(\mathcal{N}_{+}^{\, {\mathcal U}}) {\mathcal U}^{\dagger} \nonumber \\ &\geq f_{M_0}(\mathcal{N}_{+})^2 (4\pi \mathfrak{a}_N N - C + C^{-1} \mathcal{N}_+), \end{align} $$

choosing first $\ell> 0$ small enough and then $\varepsilon> 0$ sufficiently small. Here, we used $(\mathcal {N}_++1)^j \lesssim (\mathcal {N}_+^{\, {\mathcal U}}+1)^j$ (as follows from Lemma 3, Lemma 10 and Lemma 18), to estimate the error terms on the second line of (95). Moreover, we used the bounds $\mathcal {N}_+ , \mathcal {N}^{{\mathcal U}}_+ \lesssim H_1$ .

On the other hand, following an argument from [Reference Boccato, Brennecke, Cenatiempo and Schlein7, Prop. 6.1], we find

(109) $$ \begin{align} g_{M_0}(\mathcal{N}_+) \left(H_N -4\pi \mathfrak{a}_N N\right) &g_{M_0}(\mathcal{N}_+) \geq C^{-1} \mathcal{N}_+ g_{M_0}(\mathcal{N}_+)^2. \end{align} $$

Indeed, otherwise, we could find a normalised sequence $\Psi _N$ , supported on $\{\mathcal {N}_+> \varepsilon N\}$ , satisfying

$$\begin{align*}\Big| \frac{1}{N} \langle \psi_N , H_N \psi_N \rangle_{\Psi_N} - 4\pi \mathfrak{a}_N \Big| \to 0 \end{align*}$$

as $N \to \infty $ , in contradiction with (104).

Finally, we deal with the error term $\mathcal E_{M_0=\varepsilon N}$ . For $\psi _N$ with $\langle \psi _N, H_N \psi _N \rangle \leq C N$ , we immediately find, from the bound (107), that

$$\begin{align*}\langle \psi_N, \mathcal E_{M_0 = \varepsilon N} \psi_N \rangle \lesssim \varepsilon^{-2} N^{-1}. \end{align*}$$

Since the bound (103) holds trivially on states with $\langle \psi _N , H_N \psi _N \rangle \geq CN$ , this, together with the estimates (108) and (109), concludes the proof of Proposition 20.

Proof of Theorem 1.

We continue to use the notation $\widetilde {H}_N$ and $E_{\infty }$ introduced in equations (100) and (101). Moreover, we denote by $\lambda _1 (\widetilde {H}_N) \leq \lambda _2 (\widetilde {H}_N) \leq \dots $ and $\lambda _1 (E_{\infty }) \leq \lambda _2 (E_{\infty }) \leq \dots $ the ordered eigenvalues of $\widetilde {H}_N$ and, respectively, $E_{\infty }$ . We now choose $L \in \mathbb {N}$ , with $\lambda _L (\widetilde H_N) \leq \Theta $ for some $1 \leq \Theta \leq N^{1/17}$ . Then we claim that

(110) $$ \begin{align} \lambda_L (\widetilde H_N) = \lambda_L(E_{\infty}) + {\mathcal O} (\Theta N^{-1/17}). \end{align} $$

Since $\lambda _0 (E_{\infty }) = 0$ , the estimate (110) shows that the ground state energy $E_N$ of $H_N$ satisfies the estimate (5). It is then easy to check, using (110), that the excitations of $H_N - E_N$ satisfy the claim in (4).

To prove equation (110), we show first a lower bound and then a matching upper bound. We again use Proposition 19, but this time we choose the exponents $\alpha = 2/17$ .

Lower bound on $\lambda _L (\widetilde {H}_N)$ . We again use the localisation identity given by equation (105), but this time we take $M_0 = N^{1/2 + 1/34}$ . Let Y denote the subspace generated by the first L eigenfunctions of $\widetilde H_N$ , and let us denote $Z = {\mathcal U}^{\dagger } Y$ , which is of dimension L. From the decomposition (105), we have

(111) $$ \begin{align} \lambda_{L}(\widetilde H_N) \geq P_Y \left(f_{M_0}(\mathcal{N}_+) \widetilde H_N f_{M_0}(\mathcal{N}_+) + g_{M_0}(\mathcal{N}_+)\widetilde H_N g_{M_0}(\mathcal{N}_+) + \mathcal E_{M_0}\right) P_Y. \end{align} $$

From Proposition 20, we have

$$ \begin{align*} g_{M_0}(\mathcal{N}_+) \widetilde H_N g_{M_0}(\mathcal{N}_+) \geq C g^2_{M_0}(\mathcal{N}_+) (C^{-1} M_0 - C) \geq 0 \end{align*} $$

for N large enough (recall the choice $M_0 = N^{1/2+1/34}$ ). Here, we used that $g_{M_0}$ is supported on $\mathcal {N}_+> M_0/3$ . Moreover, with the bound (107), we find

$$ \begin{align*} P_Y \mathcal E_{M_0} P_Y &\geq - C M_0^{-2} P_Y (Q_4 + N) P_Y \geq - C M_0^{-2} P_Y (H_N + N) P_Y \geq - C M_0^2 N \geq - C N^{-1/17} \end{align*} $$

because (from the upper bound), we know that $H_N \leq C N$ on Y. From the estimate (111), we obtain

$$ \begin{align*} \lambda_{L}(\widetilde H_N) \geq P_Y f_{M_0}(\mathcal{N}_+) \widetilde H_N f_{M_0}(\mathcal{N}_+) P_Y - CN^{-1/17}. \end{align*} $$

We now use Proposition 19 to estimate

Using that $\mathcal {N}_+ \lesssim H_1 \leq E_{\infty }$ and the choices $M_0 = N^{1/2 + 1/34}$ , $\alpha = 2/17$ , we find

$$ \begin{align*} P_Y f_{M_0}(\mathcal{N}_+) \,&\widetilde H_N \,f_{M_0}(\mathcal{N}_+) P_Y \\ &\geq \left(1- C N^{-1/17}\right) {\mathcal U} P_Z f_{M_0}(\mathcal{N}_+^{{\mathcal U}}) e^{{\mathcal{B}}_4} \, E_{\infty} \, e^{-{\mathcal{B}}_4} f_{M_0}(\mathcal{N}_+^{{\mathcal U}})P_Z {\mathcal U}^{\dagger}. \end{align*} $$

Now it turns out that for N large enough, $\dim f_{M_0}(\mathcal {N}_+^{{\mathcal U}})P_Z = L$ because

$$ \begin{align*} \max_{\xi \in P_Z} \frac{\bigg\|\sqrt{1-f_{M_0}(\mathcal{N}_+^{{\mathcal U}})^2}\xi\bigg\|^2}{\|\xi\|^2} \leq C M_0^{-1} \max_{\xi \in P_Y} \frac{\|\mathcal{N}_+^{1/2}\xi\|^2}{\|\xi\|^2} \leq C \lambda_{L}(\widetilde H_N) M_0^{-1} \underset{N \to \infty}{\longrightarrow} 0; \end{align*} $$

see, for instance, [Reference Lewin, Nam, Serfaty and Solovej17, Prop. 6.1 ii)]. Thus

$$ \begin{align*} \lambda_{L}(\widetilde H_N) &\geq \max_{\xi \in Y} \frac{\langle e^{-{\mathcal{B}}_4} f_{M_0}(\mathcal{N}_+^{{\mathcal U}}) P_Z \mathcal{U}^{\dagger} \xi , E_{\infty} e^{-{\mathcal{B}}_4} f_{M_0} (\mathcal{N}_+^{{\mathcal U}}) P_Z {\mathcal U}^{\dagger}\xi \rangle}{\|\xi\|^2} - C \Theta N^{-1/17} \\ &\geq \max_{\xi \in e^{-{\mathcal{B}}_4} f_{M_0}(\mathcal{N}_+^{{\mathcal U}}) Z} \frac{\langle \xi, E_{\infty} \xi \rangle}{\|\xi\|^2}(1- C \lambda_{L}(\widetilde H_N) M_0^{-1}) - C \Theta N^{-1/17} \\ &\geq \min_{\dim X = L} \max_{\xi \in X} \frac{\langle \xi , E_{\infty} \xi \rangle}{\|\xi\|^2}(1- C \lambda_{L}(\widetilde H_N) M_0^{-1})- C\Theta N^{-1/17} \\ &\geq \lambda_L (E_{\infty}) (1 -C\Theta N^{-1/2}) -C \Theta N^{-1/17}, \end{align*} $$

which implies $\lambda _L(\widetilde H_N) \geq \lambda _{L}(E_{\infty }) + {\mathcal O}(\Theta N^{-1/17})$ .

Upper bound on $\lambda _L (\widetilde {H}_N)$ . Let Z denote the subspace generated by the first L eigenfunctions of $E_{\infty }$ and $P_Z$ be the orthogonal projection onto Z. The normalised eigenfunctions of $E_{\infty }$ have the form

(112) $$ \begin{align} \xi = \prod_{j=1}^k \frac{a^{\dagger}(p_j)^{n_j}}{\sqrt{n_j!}} \Omega \end{align} $$

for some $k\geq 1$ , $p_j \in \Lambda _+^*$ , $n_j \geq 1$ and where $\Omega $ is the vacuum. Note that

$$ \begin{align*} P_Z \mathcal{N}_+ P_Z \lesssim P_Z H_1 P_Z \leq P_Z E_{\infty} P_Z \leq \lambda_{L}(E_{\infty}) \leq C \Theta \leq C N^{1/17}, \end{align*} $$

where we used the lower bound that we proved above. Since $[P_Z, \mathcal {N}_+]=0$ , this bound can also be applied to powers of $\mathcal {N}_+$ . Note also that $e^{{\mathcal {B}}_4} E_{\infty } e^{-{\mathcal {B}}_4}$ almost commutes with $\mathcal {N}_+$ in the sense that

Hence, we have

Together with Proposition 19, we find

(113)

Again because

we have

for N large enough. With Lemma 18, we obtain

To estimate $P_Z Q_4 P_Z$ , we use an argument from [Reference Boccato, Brennecke, Cenatiempo and Schlein6, Lemma 6.1]: from $P_Z E_{\infty } P_Z \leq \Theta $ , we must have $a_p \xi = 0$ for all $|p|> \Theta ^{1/2}$ and $\xi \in Z$ . This implies that

$$ \begin{align*} \langle \xi, Q_4 \xi \rangle &\leq \sum_{p,q,r} \hat V_N(r) \chi_{|r|\leq \Theta^{1/2}} \|a_{p+r} a_q \xi \| \|a_{p} a_{q+r} \xi \| \\ & \leq C \Theta^{3/2} N^{-1} \|(\mathcal{N}_++1)\xi\|^2 \leq C \Theta^{7/2} N^{-1} \|\xi\|^2 \leq C N^{-1/17} \|\xi\|^2, \end{align*} $$

for all $\xi \in Z$ . Applying the min-max principle, we conclude from the estimate (113) that