Proof of Theorem 2.1.

With equation (2.13) and $H_{\infty }^{\mathrm{Pek}} = H^{\mathrm{Pek}}$ we can rewrite the term of order $\alpha ^{-2}$ as

(2.25) $$ \begin{align} \inf \sigma ({\mathbb H}_{\infty}) -\frac{3}{2}\, = \, \frac{1}{2}\mathrm{Tr}_{L^2} \big( \sqrt {H^{\mathrm{Pek}}} - 1 \big). \end{align} $$

Choosing K proportional to $\alpha $ optimizes the asymptotics of the error in equation (2.24) and thus proves equation (2.1) for $|P|\le \sqrt {2M^{\mathrm{LP}}} \alpha $ by the variational principle. For larger $|P|$ , we use $E_{\alpha }(P) \le E_{\alpha }(0) + \alpha ^{-2}$ as a consequence of equation (1.24) and apply equation (2.1) for $P=0$ .

Proof of Lemma 3.2.

We first observe that

(3.12) $$ \begin{align} e^{- i P_f y } a^{\dagger}(f) e^{i P_f y } & \, =\, a^{\dagger}( e^{- y \nabla } f ) \end{align} $$

which follows from $\frac {d}{ds} e^{- i s P_f y} a^{\dagger }( e^{(s-1)y\nabla } f ) e^{isP_fy} = 0$ . In combination with equation (3.3), this leads to

(3.13) $$ \begin{align} W^{\dagger}(\alpha \varphi_P) e^{ i P_f y } W(\alpha \varphi_P ) & \, =\, e^{iP_f y} W( \alpha ( 1 - e^{- y \nabla}) \varphi_P) \exp\big( i\alpha^2 \operatorname{Im} \langle \varphi_P | e^{-y \nabla } \varphi_P \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} \big). \end{align} $$

Recalling $\varphi _P = \varphi + i \frac {1}{\alpha ^2 M^{\mathrm{LP}}} (P \nabla ) \varphi $ , we compute

(3.14) $$ \begin{align} \alpha^2 \operatorname{Im} \langle \varphi_P | e^{-y \nabla } \varphi_P \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} & = \frac{2}{ M^{\mathrm{LP}}} \langle \varphi | e^{-y \nabla } (P\nabla) \varphi \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} \notag \\ & = - \frac{2}{ M^{\mathrm{LP}}} \langle \varphi | (y\nabla) (P\nabla) \varphi \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} - \frac{2}{M^{\mathrm{LP}}} \int_0^1 \mathrm{d} s\, \langle \varphi | e^{-sy\nabla} (y\nabla)^3 (P\nabla) \varphi \rangle_{\scriptscriptstyle{\text{\(L^2\)}}}, \end{align} $$

where we inserted $e^{-y\nabla } = 1 - (y\nabla ) + \frac {1}{2}( y \nabla )^2 - \int _0^1 \mathrm {d} s\, e^{-s y\nabla } (y\nabla )^3$ and used that, due to rotational invariance of $\varphi $ , $ \langle \varphi | (P\nabla ) \varphi \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} = 0 = \langle \varphi | (y \nabla )^2 (P\nabla ) \varphi \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} $ . Also, because of rotational invariance,

(3.15) $$ \begin{align} \langle \varphi | (y\nabla) (P \nabla) \varphi \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} \, =\, - \frac{(P y )}{3} \vert\hspace{-1pt}\vert \nabla \varphi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}}^2 \, =\, - \frac{ (Py) } {2} M^{\mathrm{LP}} , \end{align} $$

and thus, $ \alpha ^2 \operatorname {Im} \langle \varphi _P | e^{-y \nabla } \varphi _P \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} = P y+ g_{P}(y)$ .

Proof of Lemma 3.1.

Throughout this proof, let $\Xi _i = W ( \alpha \varphi _P ) G^{i}_{K} \in L^2({\mathbb R}^3) \otimes {\mathcal F} $ , $i=0,1$ , with $G_{K}^i$ defined in equation (2.19) and set $\Psi _i = \int \mathrm {d} x \, e^{i (P_f-P)x} \Xi _i(x)$ . First, note that $\Psi _i \in D(H_{\alpha }(P)^{1/2} ) $ for $i,j\in \{ 0,1\}$ which follows from $D(H_{\alpha }(P)^{1/2}) = D( |P_f| + {\mathbb N}^{1/2} ) $ [Reference Lieb and Yamazaki40] together with $| P_f | \Xi _i \in L^1({\mathbb R}^3,{\mathcal F})$ and ${\mathbb N}^{1/2} \Xi _i \in L^1({\mathbb R}^3,{\mathcal F})$ . The integrability of these states is verified using Lemmas 3.16 and 3.14.

Below, we shall employ the identitiesFootnote ¹

(3.16a) $$ \begin{align} \big \langle \Psi_i | \Psi_j \big \rangle_{\scriptscriptstyle{\text{\(\mathcal F\)}}} & = \int\mathrm{d} y \, \big \langle \Xi_i | e^{i (P_f - P) y } T_y \Xi_j \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.16b) $$ \begin{align} \big \langle \Psi_i | H_{\alpha}(P) \Psi_j \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} & = \int\mathrm{d} y\, \big \langle \Xi_i | H_{\alpha} e^{i (P_f - P) y } T_y \Xi_j \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}}, \end{align} $$

where $H_{\alpha }$ is the Fröhlich Hamiltonian given by equation (1.2). To obtain the first identity, write

$$ \begin{align*} \big \langle \Psi_i | \Psi_j \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} = \int \mathrm{d} x \, \big \langle \Xi_i (x) | \int \mathrm{d} y\, e^{ i(P_f-P)(y-x)} \Xi_j (y) \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} = \big \langle \Xi_i | \int \mathrm{d} y\, e^{ i(P_f-P) y } T_y \Xi_j \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align*} $$

and for the second one, use $e^{-i(P_f-P)x} \phi (v_0) = \phi (v_x) e^{-i(P_f-P)x} $ so that

(3.17a) $$ \begin{align} \big \langle \Psi_i | \phi(v_0) \Psi_j \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} & =\int \mathrm{d} x\, \big \langle \Xi_i (x) | \phi(v_x) \int \mathrm{d} y \, e^{ i(P_f-P)( y - x) } \Xi_j (y) \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \end{align} $$

(3.17b) $$ \begin{align} \big \langle \Psi_i | {\mathbb N} \Psi_j \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} & = \int \mathrm{d} x \, \big \langle \Xi_i(x) | {\mathbb N} \int \mathrm{d} y \, e^{ i(P_f-P) (y-x) } \Xi_j (y) \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \end{align} $$

(3.17c) $$ \begin{align} \big \langle \Psi_i | (P_f-P)^2 \Psi_j \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} & = \int \mathrm{d} x \, \big \langle \Xi_i(x) | (P_f-P)^2 \int \mathrm{d} y\, e^{ i(P_f-P) (y-x) } \Xi_j(y) \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \notag \\ & = \int \mathrm{d} x \, \big \langle \Xi_i(x) | (-\Delta_x) \int \mathrm{d} y\, e^{ i(P_f-P) (y-x) } \Xi_j(y) \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}}. \end{align} $$

With equations (3.16a) and (3.16b), the norm and the energy of the trial state are given by

(3.18a) $$ \begin{align} \vert\hspace{-1pt}\vert \Psi_{K,\alpha}(P) \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}}^2 & = \sum_{i\in \{0,1\}} \alpha^{-2i} \int \mathrm{d} y \, \big \langle \Xi_i | e^{i( P_f - P )y} T_y \Xi_i \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag \\ & \hspace{8mm} - 2 \alpha^{-1}\operatorname{Re} \int \mathrm{d} y \, \big \langle \Xi_0 | e^{i( P_f - P )y} T_y \Xi_1 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.18b) $$ \begin{align} \big \langle \Psi_{K,\alpha}(P)| H_{\alpha}(P) \Psi_{K,\alpha}(P)\big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} & = \sum_{i\in \{0,1\}} \alpha^{-2i} \int \mathrm{d} y \, \big \langle \Xi_i | H_{\alpha} e^{i( P_f - P )y} T_y \Xi_i \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag \\ & \hspace{8mm} - 2 \alpha^{-1}\operatorname{Re} \int \mathrm{d} y \, \big \langle \Xi_0 | H_{\alpha} e^{i( P_f - P )y} T_y \Xi_1 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}}. \end{align} $$

Inserting $\Xi _i = W ( \alpha \varphi _P ) G^{i}_{K}$ and applying Lemma 3.2, we find for $i,j\in \{0,1\}$

(3.19a) $$ \begin{align} \big \langle \Xi_i | e^{i( P_f - P )y} T_y \Xi_j \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} & = \big \langle G_K^i | e^{A_{P,y}} W(\alpha w_{P,y}) T_y G_K^j \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.19b) $$ \begin{align} \big \langle \Xi_i | H_{\alpha} e^{i( P_f - P )y} T_y \Xi_j \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} & = \big \langle G_K^i | W ( \alpha \varphi_P )^{\dagger} H_{\alpha} W ( \alpha \varphi_P ) e^{A_{P,y}} W(\alpha w_{P,y}) T_y G_K^j \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}}. \end{align} $$

Using equations (3.10a) and (3.10b) and $2\operatorname {Re} \langle \varphi _P | h_{x} \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} = 2\operatorname {Re} \langle \varphi | h_{x} \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} = V^{\varphi }(x)$ (see equation (1.11)) the Weyl-transformed Hamiltonian becomes

(3.20) $$ \begin{align} W(\alpha \varphi_P)^{\dagger} H_{\alpha} W(\alpha \varphi_P) & = -\Delta_x + V^{\varphi}(x) + \alpha^{-2} {\mathbb N} + \alpha^{-1} \phi(h_x + \varphi_P) + \vert\hspace{-1pt}\vert \varphi_P\vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(L^2\)}}} \notag\\[1mm] & = \widetilde H_{\alpha, P} + e^{\mathrm{Pek}} + \vert\hspace{-1pt}\vert \varphi_P\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}}^2 - \vert\hspace{-1pt}\vert \varphi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}}^2 \end{align} $$

with $\widetilde H_{\alpha ,P}$ defined by equation (3.6). Note that we added and subtracted $e^{\mathrm{Pek}}= \lambda ^{\mathrm{Pek}} + \vert \hspace {-1pt}\vert \varphi \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^2\)}}}^2$ and used equation (3.6). Altogether, this implies

(3.21) $$ \begin{align} \big \langle \Psi_{K,\alpha}(P)| H_{\alpha}(P) \Psi_{K,\alpha}(P)\big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} & \, =\, \Big( e^{\mathrm{Pek}} + \vert\hspace{-1pt}\vert \varphi_P \vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(L^2\)}}} - \vert\hspace{-1pt}\vert \varphi \vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(L^2\)}}} \Big)\, \mathcal N + {\mathcal E} + {\mathcal G} + {\mathcal K}. \end{align} $$

The claimed result now follows from

(3.22) $$ \begin{align} \vert\hspace{-1pt}\vert \varphi_P \vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(L^2\)}}} - \vert\hspace{-1pt}\vert \varphi \vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(L^2\)}}} \, =\, \frac{1}{\alpha^4 ( M^{\mathrm{LP}} )^2} \vert\hspace{-1pt}\vert (P\nabla) \varphi \vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(L^2\)}}} & \, =\, \frac{P^2}{2 \alpha^4 M^{\mathrm{LP}} }, \end{align} $$

where we used $\vert \hspace {-1pt}\vert (P\nabla ) \varphi \vert \hspace {-1pt}\vert ^2_{\scriptscriptstyle{\text{\(L^2\)}}} = \frac {P^2}{3} \vert \hspace {-1pt}\vert \nabla \varphi \vert \hspace {-1pt}\vert ^2_{\scriptscriptstyle{\text{\(L^2\)}}} = \frac {P^2}{2} M^{\mathrm{LP}}$ because of rotational invariance of $\varphi $ .

Proof of Corollary 3.6.

Since

(3.36) $$ \begin{align} \int \mathrm{d} y\, |y|^n e^{ - \eta \lambda \alpha^{2(1-\delta)} y^2} \, =\, ( \eta \lambda \alpha^{2(1-\delta)} )^{- \frac{3+n}{2}} \int \mathrm{d} y\, |y|^n e^{-y^2} \, =\, C_n \alpha^{-(3+n)(1-\delta)}, \end{align} $$

the statement follows immediately from Lemma 3.5.

Proof. It follows from equations (3.18a) and (3.19a) that $\mathcal N = \mathcal N_{0} + \mathcal N_1 + \mathcal N_2 $ with

(3.61a) $$ \begin{align} \mathcal N_0 & \, =\, \int \mathrm{d} y\, \big \langle G_{K }^{0} \big| T_y e^{ A_{P,y}} W(\alpha w_{P,y}) G_{K}^{0} \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.61b) $$ \begin{align} \mathcal N_{1} & \, =\, - \frac{2}{\alpha} \int \mathrm{d} y\, \operatorname{Re} \big \langle G_{K }^{0} \big| T_y e^{ A_{P,y} } W(\alpha w_{P,y}) G_{K}^{1} \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.61c) $$ \begin{align} \mathcal N_{2} & \, =\, \frac{1}{\alpha^2} \int \mathrm{d} y\, \big \langle G_{K}^{1} \big| T_y e^{ A_{P,y} } W(\alpha w_{P,y}) G_{K }^{1} \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}}. \end{align} $$

$\underline {\mathrm {Term}\ {\mathcal N}_0}$ . This part contains the leading order contribution $(\frac {\pi }{\lambda \alpha ^2})^{3/2}$ . With H defined in equation (3.37), let us write

(3.62) $$ \begin{align} \mathcal N_0 & \, = \, \int \mathrm{d} y\, H(y) \big \langle \Upsilon_K \big| W(\alpha w_{P,y}) \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \notag \\ &\quad \, +\, \int \mathrm{d} y\, H(y) \big \langle \Upsilon_K \big| (e^{A_{P,y}}-1) W(\alpha w_{P,y}) \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} = \mathcal N_{01} + \mathcal N_{02}. \end{align} $$

In the first term, we use $\Upsilon _K = \mathbb U_K^{\dagger } \Omega $ and apply equation (3.53d) to transform the Weyl operator with the Bogoliubov transformation. This gives

(3.63) $$ \begin{align} \mathbb U_K W( \alpha w_{P,y} ) \mathbb U_K^{\dagger} & \, = \, W( \alpha \widetilde w_{P,y}) \end{align} $$

with $\widetilde w_{P,y}$ defined in equation (3.31d). From equations (3.2) and (3.33), we thus obtain

(3.64) $$ \begin{align} \mathcal N_{01} \, = \, \int \mathrm{d} y\, H(y) \big \langle \Omega \big| W(\alpha \widetilde w_{P,y}) \Omega \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} & \, = \, \int \mathrm{d} y\, H(y) n_{0,1}(y). \end{align} $$

Since $\vert \hspace {-1pt}\vert H\vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^1\)}}} + \vert \hspace {-1pt}\vert H\vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^{\infty }\)}}} \le C $ , cf. Lemma 3.7, we can apply Lemma 3.5 in order to replace the weight function $n_{0,1}(y)$ by the Gaussian $e^{-\lambda \alpha ^2 y^2}$ . More precisely,

(3.65) $$ \begin{align} \bigg | \int \mathrm{d} y\, H(y) n_{0,1}(y) - \int \mathrm{d} y\, H(y) e^{- \lambda \alpha^2 y^2 } \bigg| \, \le\, C\alpha^{-4} \end{align} $$

for all $|P| /\alpha \le c$ and all $K,\alpha $ large enough. Then we use $| H(y)-1| \le C y^2$ in order to obtain

(3.66) $$ \begin{align} \bigg|\, \mathcal N_{01}- \bigg( \frac{\pi}{\lambda \alpha^2 }\bigg)^{3/2}\bigg|\, \le \, C\alpha^{-4}. \end{align} $$

To treat $\mathcal N_{02}$ , it is convenient to decompose the state $ \Upsilon _K $ into a part with bounded particle number and a remainder. To this end, we choose a small $\delta>0$ and write

(3.67)

Inserting this into $\mathcal N_{02}$ and using unitarity of $e^{A_{P,y}}$ and $\vert \hspace {-1pt}\vert H\vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^1\)}}}\le C$ , we can estimate

(3.68) $$ \begin{align} | \mathcal N_{02}|\, & \le\, \int \mathrm{d} y\, H(y) \big| \big \langle \Upsilon_K^{ <} \big| (e^{A_{P,y}} -1 ) W(\alpha w_{P,y}) \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \big| + C \vert\hspace{-1pt}\vert \Upsilon_K^>\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}}. \end{align} $$

By Corollary 3.14 for $n=0$ , $\vert \hspace {-1pt}\vert \Upsilon _K^>\vert \hspace {-1pt}\vert \le C_{\delta } \, \alpha ^{-10}$ . In the remaining expression, we use equation (3.63),

(3.69) $$ \begin{align} \big \langle \Upsilon_K^{ <} \big| (e^{A_{P,y}} -1 ) W(\alpha w_{P,y}) \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \, = \, \big \langle \Upsilon_K^< \big| (e^{A_{P,y}}-1) \mathbb U_K^{\dagger} W(\alpha \widetilde w_{P,y}) \Omega \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} , \end{align} $$

and insert the identity

(3.70)

on the left of the Weyl operator (where $b>0$ is the constant from Lemma 3.13). After applying the Cauchy–Schwarz inequality, this leads to

(3.71) $$ \begin{align} & \big| \big \langle \Upsilon_K^{ <} \big| (e^{A_{P,y}} -1 ) W(\alpha w_{P,y}) \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \big| \le \, \vert\hspace{-1pt}\vert e^{\kappa {\mathbb N}} \mathbb U_K (e^{-A_{P,y}}-1) \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \vert\hspace{-1pt}\vert e^{-\kappa {\mathbb N} } W(\alpha \widetilde w_{P,y}) \Omega \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}}. \end{align} $$

In the second factor, we then employ

(3.72) $$ \begin{align} \vert\hspace{-1pt}\vert e^{- \kappa {\mathbb N} } W (\alpha \widetilde w_{P,y}) \Omega \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \, = \, e^{-\frac{\alpha^2}{2} \vert\hspace{-1pt}\vert \widetilde w_{P,y}\vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(L^2\)}}}} \vert\hspace{-1pt}\vert e^{- \kappa {\mathbb N} } e^{ a^{\dagger}(\alpha \widetilde w_{P,y})} e^{\kappa {\mathbb N} } \Omega \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \end{align} $$

and use $e^{-\kappa {\mathbb N}}a^{\dagger }(f)e^{\kappa {\mathbb N}} = a^{\dagger }(e^{-\kappa } f) $ to write

(3.73) $$ \begin{align} e^{- \kappa {\mathbb N} } e^{a^{\dagger}(\alpha \widetilde w_{P,y})} e^{\kappa {\mathbb N} } \Omega \, = \, e^{a^{\dagger}(e^{-\kappa } \alpha \widetilde w_{P,y})} \Omega \, = \, e^{\frac{\alpha^2 e^{-2 \kappa} }{2} \vert\hspace{-1pt}\vert \widetilde w_{P,y}\vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(L^2\)}}}} W (e^{-\kappa } \alpha \widetilde w_{P,y}) \Omega. \end{align} $$

Combining the previous two lines, we obtain

(3.74)

for some $\alpha $ -independent $\eta> 0 $ and $\alpha $ large enough. To estimate the first factor in equation (3.71), we apply Lemma 3.15 (note that )

(3.75) $$ \begin{align} \vert\hspace{-1pt}\vert e^{\kappa {\mathbb N}} \mathbb U_K (e^{ - A_{P,y}}-1) \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \le \sqrt 2 \vert\hspace{-1pt}\vert (e^{ - A_{P,y}}-1) \Upsilon_K\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}}. \end{align} $$

On the right side, we use the functional calculus for self-adjoint operators

(3.76) $$ \begin{align} \vert\hspace{-1pt}\vert (e^{- A_{P,y} }-1) \Upsilon_K\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \le \vert\hspace{-1pt}\vert A_{P,y} \Upsilon_K\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \le \vert\hspace{-1pt}\vert (y P_f) \Upsilon_K \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} + |g_{P}(y)| \le C \big( \sqrt K |y| + \alpha|y|^3 \big), \end{align} $$

where in the last step we applied Lemma 3.16 and used

(3.77) $$ \begin{align} |g_{P}(y)|\, \le \, C \alpha |y|^3, \end{align} $$

which is inferred from equation (3.5) using $\vert \hspace {-1pt}\vert \Delta \varphi \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^2\)}}}< \infty $ . Returning to equation (3.71), we have shown that

(3.78) $$ \begin{align} |\mathcal N_{02} | \, \le\, C \int \mathrm{d} y \, H(y) \big( \sqrt{K} |y| + \alpha |y|^3 \big) n_{\delta , \eta} (y) + C_{\delta}\, \alpha^{-10} , \end{align} $$

and hence we are in a position to apply Corollary 3.6. This implies for all $\alpha $ large

(3.79) $$ \begin{align} |\mathcal N_{02} | \, \le \, C\big( \sqrt K \alpha^{-4 (1-\delta) } + \alpha^{-6(1-\delta)+1} \big) + C_{\delta}\, \alpha^{-10} \le C_{\delta}\, \sqrt K \alpha^{-4(1-\delta) }. \end{align} $$

$\underline {\mathrm {Term}\ {\mathcal N}_1}$ . We start by inserting equation (2.22) for $G_K^1$ in expression (3.61b). Since the Weyl operator commutes with $u_{\alpha }$ , R and $P_{\psi } = |\psi \rangle \langle \psi | $ , we can apply equation (3.10a) to obtain

(3.80) $$ \begin{align} W(\alpha w_{P,y} ) G_K^1\, =\, u_{\alpha} R \big( \phi(h_{K,\cdot}^1) + 2 \alpha \langle h_{K,\cdot} | \operatorname{Re} ( w_{P,y}^1 ) \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} \big) P_{\psi} W(\alpha w_{P,y}) G_K^0, \end{align} $$

where we used that $h_{K,x}$ is real-valued. Note that $\langle h_{K,\cdot } | \operatorname {Re}( w_{P,y}^1 ) \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} $ is a y-dependent multiplication operator in the electron variable. With $( T_y e^{A_{P,y}})^{\dagger } = T_{-y} e^{-A_{P,y}}$ and equation (3.67), we can thus write

(3.81) $$ \begin{align} \mathcal N_1 & \, = \, - \frac{2}{\alpha} \int \mathrm{d} y \, \operatorname{Re} \big \langle R_{1,y} \psi \otimes \big( \Upsilon_K^< + \Upsilon_K^> \big) \big| W(\alpha w_{P,y}) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} = \mathcal N_{1}^< + \mathcal N_{1}^>, \end{align} $$

where we introduced the operator $R_{1,y} = R_{1,y}^1 + R_{1,y}^2$ with

(3.82a) $$ \begin{align} R_{1,y}^1 & \, = \, P_{\psi} \phi(h_{K,\cdot}^1) R u_{\alpha} T_{-y} P_{\psi} e^{-A_{P,y}} , \end{align} $$

(3.82b) $$ \begin{align} R_{1,y}^2 & \, = \, 2\alpha P_{\psi} \big \langle h_{K,\cdot} | \operatorname{Re} (w_{P,y}^1) \big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} R u_{\alpha} T_{-y} P_{\psi} e^{-A_{P,y}}. \end{align} $$

Using Lemma 3.9 in combination with $\vert \hspace {-1pt}\vert \nabla P_{\psi } \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} + \vert \hspace {-1pt}\vert \nabla R^{1/2}\vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} <\infty $ (see Lemmas 3.7 and 3.8), we can bound the first operator, for any $\Psi \in \mathscr H$ , by

(3.83) $$ \begin{align} \vert\hspace{-1pt}\vert R_{1,y}^1 \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} & \le C \vert\hspace{-1pt}\vert ({\mathbb N}+1)^{1/2} u_{\alpha} T_{-y} P_{\psi} e^{-A_{P,y}} \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \le C \vert\hspace{-1pt}\vert u_{\alpha} T_{-y} P_{\psi} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} \vert\hspace{-1pt}\vert ({\mathbb N}+1)^{1/2} \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}}. \end{align} $$

To estimate the second operator, we write out the inner product, use Cauchy–Schwarz twice, apply Corollary 3.11 (with $A=1$ , $X=R$ and $Y=P_{\psi }$ ) and use equation (3.32a),

(3.84) $$ \begin{align} \vert\hspace{-1pt}\vert R_{1,y}^2 \Psi \vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, & = \, 4\alpha^2 \vert\hspace{-1pt}\vert \int \mathrm{d} z\, \operatorname{Re} (w_{P,y}^1(z)) P_{\psi} h_{K,\cdot}(z) R u_{\alpha} T_{-y} P_{\psi} e^{-A_{P,y}} \Psi \vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag\\ & \le \, 4 \alpha^2 \int\mathrm{d} u\, | w_{P,y}^1(u)|^2 \int \mathrm{d} z\, \vert\hspace{-1pt}\vert P_{\psi} h_{K,\cdot}(z) R \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}}^2 \, \vert\hspace{-1pt}\vert u_{\alpha} T_{-y}P_{\psi} e^{-A_{P,y}} \Psi \vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag\\[1mm] & \le \, C \alpha^2 \vert\hspace{-1pt}\vert w_{P,y}^1 \vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(L^2\)}}} \vert\hspace{-1pt}\vert u_{\alpha} T_{-y} P_{\psi} e^{-A_{P,y}} \Psi \vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag \\[2.5mm] & \le \, C \alpha^2 (y^4 + \alpha^{-4}) \vert\hspace{-1pt}\vert u_{\alpha} T_{-y} P_{\psi} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}}^2 \vert\hspace{-1pt}\vert \Psi \vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(\mathscr H\)}}}. \end{align} $$

Combining the above estimates we arrive at

(3.85) $$ \begin{align} \vert\hspace{-1pt}\vert R_{1,y} \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} & \, \le\, C \vert\hspace{-1pt}\vert u_{\alpha} T_{-y} P_{\psi} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} (1+ \alpha y^2) \vert\hspace{-1pt}\vert ({\mathbb N}+1)^{1/2} \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}}. \end{align} $$

Since $\psi (x)$ decays exponentially for large $|x|$ , the function $f_{\alpha }(y) := \vert \hspace {-1pt}\vert u_{\alpha } T_{-y} P_{\psi } \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} $ satisfies

(3.86) $$ \begin{align} \vert\hspace{-1pt}\vert |\cdot |^n f_{\alpha} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^1\)}}} \, \le \, \int \mathrm{d} y\, |y|^n \bigg( \int \mathrm{d} x\, \psi(x+y)^2 u_{\alpha}(x)^2 \bigg)^{1/2} \, \le\, C_n \alpha^{3+n}\ \ \text{for all}\ n \in {\mathbb N}_0. \end{align} $$

With this at hand, we can estimate the part containing the tail. Invoking Corollary 3.14

(3.87) $$ \begin{align} | \mathcal N_{1}^> | \, \le \, \frac{C}{\alpha} \vert\hspace{-1pt}\vert ({\mathbb N}+1)^{1/2} \Upsilon_K^> \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \int \mathrm{d} y\, f_{\alpha} (y) (1+\alpha y^2) \, \le \, C_{\delta}\, \alpha^{-5}. \end{align} $$

To estimate the first term in equation (3.81), we proceed similarly as in the bound for $\mathcal N_{02}$ . We insert the identity (3.70), apply Cauchy–Schwarz and employ equation (3.74). This leads to

(3.88)

In the remaining norm, we use the fact that $R_{1,y}$ changes the number of phonons at most by one, and thus we can apply Lemma 3.15 and equation (3.85), together with equation (3.55a), to get

(3.89)

With Corollary 3.6, equation (3.86) and $\vert \hspace {-1pt}\vert f_{\alpha } \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^{\infty }\)}}} \leq 1 $ , this leads to

(3.90) $$ \begin{align} |\mathcal N_{1}^< | & \, \le \, \frac{C}{\alpha} \int \mathrm{d} y\, f_{\alpha} (y) \big( 1 + \alpha y^2 \big) \, n_{\delta,\eta}(y) \, \le\, C \alpha^{-1-3(1- \delta)}. \end{align} $$

$\underline {\mathrm {Term}\ {\mathcal N}_{2}}$ . The strategy for estimating this term is similar to the one for $\mathcal N_1$ . Proceeding as described before equation (3.81), one obtains

(3.91) $$ \begin{align} \mathcal N_2 & \, = \, \frac{1}{\alpha^2} \int \mathrm{d} y \, \big \langle R_{2,y} \psi \otimes \big( \Upsilon_K^< + \Upsilon_K^> \big) \big| W(\alpha w_{ P ,y} ) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, = \, \mathcal N_{2}^< + \mathcal N_{2}^> \end{align} $$

with $R_{2,y} = R_{2,y}^1 + R_{2,y}^2$ and

(3.92a) $$ \begin{align} R_{2,y}^1 &\, =\, P_{\psi} \phi(h^1_{K,\cdot}) R e^{-A_{P,y}} u_{\alpha} T_{-y} u_{\alpha} R \phi(h^1_{K,\cdot}) P_{\psi} , \end{align} $$

(3.92b) $$ \begin{align} R_{2,y}^2 & \, = \, 2 \alpha P_{\psi} \langle h_{K,\cdot } | \operatorname{Re} (w^1_{P,y})\rangle_{\scriptscriptstyle{\text{\(L^2\)}}} R e^{-A_{P,y}} u_{\alpha} T_{-y} u_{\alpha} R \phi(h^1_{K,\cdot}) P_{\psi}. \end{align} $$

It follows in close analogy as for $R_{1,y}$ in equations (3.82a) and (3.82b) that given any $\Psi \in \mathscr H$ ,

(3.93) $$ \begin{align} \vert\hspace{-1pt}\vert R_{2,y} \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} & \, \le \, C \vert\hspace{-1pt}\vert u_{\alpha} T_{-y} u_{\alpha} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} (1+\alpha y^2) \vert\hspace{-1pt}\vert ({\mathbb N}+1) \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}}, \end{align} $$

and since , we can use Corollary 3.14 to estimate

(3.94)

To bound the first term in equation (3.91), we proceed similarly as for $\mathcal N_{01}$ ,

(3.95)

The last integral is estimated again via Corollary 3.6, and thus $|\mathcal N_{2}^< | \le C \alpha ^{-5+3\delta }$ .

Collecting all relevant estimates and choosing $\delta>0$ small enough completes the proof of the proposition.

Proof. Since $G_K^0 = \psi \otimes \Upsilon _K$ , $ h^{\mathrm{Pek}} \psi = 0$ and ${\mathbb N}\Upsilon _K = {\mathbb N}_1 \Upsilon _K $ , one has

(3.97) $$ \begin{align} {\mathcal E} & \, =\, \int \mathrm{d} y\, \big \langle G_{K}^{0}| \big(\alpha^{-2} {\mathbb N}_1 + \alpha^{-1} \phi(h_{\cdot} + \varphi_P) \big) T_y e^{A_{P,y}} W(\alpha w_{P,y}) | G_{K}^{ 0} \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, =\, {\mathcal E}_1 + {\mathcal E}_2, \end{align} $$

where both terms provide contributions to the energy of order $\alpha ^{-2}$ .

$\underline {\mathrm {Term}\ {\mathcal E}_1}$ . Recall that $H(y) = \langle \psi | T_y \psi \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} $ , and use this to write

(3.98) $$ \begin{align} {\mathcal E}_1 & \, =\, \frac{1}{\alpha^2} \int \mathrm{d} y\, H(y) \big \langle \Upsilon_K | {\mathbb N}_1 W(\alpha w_{P,y}) \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \notag \\ & \quad + \frac{1}{\alpha^{2}} \int \mathrm{d} y\, H(y) \big \langle \Upsilon_K | {\mathbb N}_1 (e^{A_{P,y}} -1 ) W(\alpha w_{P,y}) \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \, =\, {\mathcal E}_{11} + {\mathcal E}_{12}. \end{align} $$

With equations (3.63), (3.3) and (3.33), it follows that

(3.99) $$ \begin{align} W(\alpha w_{P,y}) \Upsilon_K \, =\, \mathbb U_K^{\dagger} W(\alpha \widetilde w_{P,y}) \Omega \, =\, n_{0,1}(y) \mathbb U_K^{\dagger}\, e^{a^{\dagger}( \alpha w_{P,y}^0) } e^{a^{\dagger}( \alpha \widetilde w_{P,y}^1 ) } \Omega , \end{align} $$

and since $e^{a^{\dagger }( \alpha w_{P,y}^0) }$ commutes with $\mathbb U_K {\mathbb N}_1 \mathbb U_K^{\dagger }$ and $e^{a( \alpha w_{P,y}^0) } \Upsilon _K = \Upsilon _K $ (we use $\mathbb U_K a^{\dagger }(f^0) \mathbb U_K^{\dagger } = a^{\dagger } (f^0)$ for $f^0 \in \text {Ran}(\Pi _0)$ ), this leads to

(3.100) $$ \begin{align} {\mathcal E}_{11} & \, =\, \frac{1}{\alpha^2} \int \mathrm{d} y\, H(y) n_{0,1}(y) \big \langle \Omega | \mathbb U_K {\mathbb N}_1 \mathbb U_K^{\dagger} e^{a^{\dagger}( \alpha \widetilde w_{P,y}^1 ) } \Omega \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}}. \end{align} $$

Because $\mathbb U_K {\mathbb N}_1 \mathbb U_K^{\dagger } $ is quadratic in creation and annihilation operators, we can expand the exponential in the inner product and use that only the zeroth- and second-order terms give a nonvanishing contribution,

(3.101) $$ \begin{align} {\mathcal E}_{11} & = \frac{1}{\alpha^2} \int \mathrm{d} y\, H(y) n_{0,1}(y) \big \langle \Upsilon_K | {\mathbb N}_1 \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \notag\\ & \quad + \frac{1}{2\alpha^2} \int \mathrm{d} y\, H(y) n_{0,1}(y) \big \langle \Upsilon_K | {\mathbb N}_1 \mathbb U_K^{\dagger} a^{\dagger}( \alpha \widetilde w_{P,y}^1 ) a^{\dagger}( \alpha \widetilde w_{P,y}^1 ) \Omega \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} = {\mathcal E}_{111} + {\mathcal E}_{112}. \end{align} $$

Next, we add and subtract the Gaussian to separate the leading-order term,

(3.102) $$ \begin{align} {\mathcal E}_{111} & \, =\, \frac{1}{\alpha^2} \int \mathrm{d} y\, H(y) \, e^{- \lambda \alpha^2 y^2} \big \langle \Upsilon_K | {\mathbb N}_1 \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \notag\\ & \quad + \frac{1}{\alpha^2} \int \mathrm{d} y\, H(y) \big( n_{0,1}(y) - e^{- \lambda \alpha^2 y^2} \big) \big \langle \Upsilon_K | {\mathbb N}_1 \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \, =\, {\mathcal E}_{111}^{\mathrm{lo}} + {\mathcal E}_{111}^{\mathrm{err}}. \end{align} $$

In ${\mathcal E}_{111}^{\mathrm{lo}}$ , we use $| H(y) - 1 | \le C y^2$ and Corollary 3.14 to replace $H(y)$ by unity at the cost of an error of order $\alpha ^{-7}$ . In the term where $H(y)$ is replaced by unity, we perform the Gaussian integral and use Proposition 3.17 and again Corollary 3.14. This leads to

(3.103) $$ \begin{align} \Big|\, {\mathcal E}_{111}^{\mathrm{lo}} - \mathcal N \frac{1}{\alpha^2} \big \langle \Upsilon_K | {\mathbb N}_1 \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \, \Big| \, \le \, C_{\varepsilon} \sqrt K \alpha^{-6+\varepsilon}. \end{align} $$

The error in equation (3.102) is bounded with the help of Lemma 3.5,

(3.104) $$ \begin{align} | {\mathcal E}_{111}^{\mathrm{err}} |\, \le\, \frac{C}{\alpha^2} \int \mathrm{d} y\, H(y) | n_{0,1}(y) - e^{-\lambda \alpha^2 y^2 }| \, \le \, C \alpha^{-6}. \end{align} $$

In ${\mathcal E}_{112}$ , we use the Cauchy–Schwarz inequality, Corollary 3.14 and Lemma 3.4, to obtain

(3.105) $$ \begin{align} & \big| \big \langle \Upsilon_K | {\mathbb N}_1 \mathbb U_K^{\dagger} a^{\dagger}( \alpha \widetilde w_{P,y}^1 ) a^{\dagger}( \alpha \widetilde w_{P,y}^1 ) \Omega \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \big| \notag\\[1mm] & \hspace{0.5cm} \, \le \, \vert\hspace{-1pt}\vert {\mathbb N}_1 \Upsilon_K \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \, \vert\hspace{-1pt}\vert a^{\dagger}( \alpha \widetilde w_{P,y}^1 ) a^{\dagger}( \alpha \widetilde w_{P,y}^1 ) \Omega \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \, \le \, 2 \alpha^2 \vert\hspace{-1pt}\vert \widetilde w_{P,y}^1\vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(L^2\)}}} \, \le \, C \alpha^2 (y^4 + \alpha^{-4}). \end{align} $$

With $\vert \hspace {-1pt}\vert |\cdot |^n H \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^1\)}}} \le C_n$ , we can now apply Corollary 3.6 to obtain

(3.106) $$ \begin{align} | {\mathcal E}_{112} | \, \le \, C \int \mathrm{d} y\, H(y) ( y^4 + \alpha^{ - 4}) n_{0,1}(y) \, \le\, C \alpha^{-7}. \end{align} $$

In order to bound ${\mathcal E}_{12}$ in equation (3.98), we decompose $\Upsilon _K = \Upsilon _K^< + \Upsilon _K^>$ for some $\delta>0$ (see equation (3.67)) and then follow similar steps as described below equation (3.69). This way we can estimate

(3.107) $$ \begin{align} | {\mathcal E}_{12} | \le \frac{1}{\alpha^2} \int \mathrm{d} y\, H(y) \vert\hspace{-1pt}\vert e^{\kappa {\mathbb N} } \mathbb U_K (e^{ - A_{P,y}} - 1) {\mathbb N}_1 \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \, n_{\delta,\eta}(y) + \, \frac{2}{\alpha^2} \vert\hspace{-1pt}\vert {\mathbb N}_1 \Upsilon_K^>\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \int \mathrm{d} y\, H(y). \end{align} $$

While the second term is bounded via equation (3.55b) by $C_{\delta }\, \alpha ^{-12}$ , in the first term we apply Lemma 3.15 and use the functional calculus for self-adjoint operators,

(3.108) $$ \begin{align} \vert\hspace{-1pt}\vert e^{\kappa {\mathbb N} } \mathbb U_K (e^{ - A _{P,y} } - 1) {\mathbb N}_1 \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} & \, \le \, \sqrt 2 \vert\hspace{-1pt}\vert ( e^{ - A_{P,y}} - 1 ) {\mathbb N}_1 \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \notag\\[1mm] &\, \le \, \sqrt 2 \vert\hspace{-1pt}\vert (P_fy + g_{P}(y) ) {\mathbb N}_1 \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}}. \end{align} $$

Since $P_f$ changes the number of phonons in ${\mathcal F}_1$ at most by one, we can proceed by

(3.109) $$ \begin{align} \hspace{-1mm}\vert\hspace{-1pt}\vert (P_fy + g_{P}(y) ) {\mathbb N}_1 \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \le (\alpha^{\delta}+1) \vert\hspace{-1pt}\vert (P_f y + g_{P}(y)) \Upsilon_K \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \le C \alpha^{\delta} (\sqrt K |y| + \alpha |y|^3), \end{align} $$

where we used $1\le \alpha ^{\delta }$ , Lemma 3.16 and equation (3.77) in the second step. We conclude via Corollary 3.6 that

(3.110) $$ \begin{align} | {\mathcal E}_{12} | \, \le \, \frac{C}{\alpha^2} \int \mathrm{d} y\, H(y) (\sqrt K |y| + \alpha |y|^3 ) n_{\delta,\eta}(y) + C_{\delta}\, \alpha^{-12} \, \le \, C_{\delta} \, \sqrt K \alpha^{- 6 + 4\delta}. \end{align} $$

$\underline {\mathrm {Term}\ {\mathcal E}_2}$ . Here, we start with

(3.111) $$ \begin{align} {\mathcal E}_2 & \, = \, \alpha^{-1} \int \mathrm{d} y\, \big \langle \Upsilon_K | L_{1,y} W(\alpha w_{P,y}) \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \notag \\ & \quad + \alpha^{-1} \int \mathrm{d} y\, \big \langle \Upsilon_K | L_{1,y} ( e^{A_{P,y}} -1 ) W(\alpha w_{P,y}) \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \, = \, {\mathcal E}_{21} + {\mathcal E}_{22} , \end{align} $$

where

(3.112) $$ \begin{align} L_{1,y} \, =\, \big \langle \psi| \phi(h_{\cdot} + \varphi_P) T_y \psi \big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} = \phi(l_y)+\pi(j_y) \end{align} $$

with

(3.113) $$ \begin{align} l_y \, = \, H(y)\varphi+\big \langle \psi|h_{\cdot}T_y \psi\big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}}, \quad j_ y \, = \, H(y)\xi_P, \end{align} $$

and $\xi _P$ defined in equation (2.18). We record the following properties of $l_y$ and its derivative. The proof of the lemma is postponed until the end of the present section.

Lemma 3.19. For $k=0,1$ and for all $n\in \mathbb {N}_0$ ,

(3.114) $$ \begin{align} \sup_y \|\nabla^k l_y \|_{\scriptscriptstyle{\text{\(L^2\)}}} \, <\, \infty, \quad \int |y |^n\|\nabla^k l_y \|_{\scriptscriptstyle{\text{\(L^2\)}}} \, \mathrm{d} y \, <\, \infty. \end{align} $$

Note that, by Lemma 3.7, $j_y$ clearly has these properties as well. We proceed by writing ${\mathcal E}_{21} = \mathcal {E}_{21}^0 + \mathcal {E}_{21}^P$ with

(3.115a) $$ \begin{align} \mathcal{E}_{21}^0 & \, =\, \alpha^{-1}\int \mathrm{d} y\, \big \langle \Upsilon_K|\phi(l_y)W(\alpha w_{P,y}) \Upsilon_K\big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \end{align} $$

(3.115b) $$ \begin{align} \mathcal{E}^P_{21} & \, =\, \alpha^{-1}\int \mathrm{d} y \, \big \langle \Upsilon_K |\pi(j_y)W(\alpha w_{P,y}) \Upsilon_K\big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} , \end{align} $$

and estimate the two parts separately. Using the canonical commutation relations and (3.53c), we evaluate

(3.116) $$ \begin{align} \mathcal{E}_{21}^0 & \, =\, \int \big \langle \underline{l_y}|\widetilde{w}_{P,y}\big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} n_{0,1}(y) \mathrm{d} y \notag\\ & \, =\, \int \Big( \big \langle l_y^0|w_{P,y}^0\big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} + \big \langle l_y^1 |\mathrm{Re}(w_{P,y}^1)\big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} + i\big \langle l_y^1|\Theta^{-2}_K\mathrm{Im}(w_{P,y}^1)\big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} \Big) n_{0,1}(y) \mathrm{d} y, \end{align} $$

where we used that $l_y$ is real-valued. Note that $l_{-y}(-z)=l_y(z)$ . As discussed in Remark 3.3, $n_{0,1}(y)$ is even, and using the arguments therein one can conclude that $ \Theta ^{-2}_K\mathrm {Im}(w_{P,y}^1)$ and $\mathrm {Im}(w_{P,y}^0)$ are odd functions on $\mathbb {R}^6$ since $(y,z)\mapsto \mathrm {Im}(w_{P,y})(z)$ is odd on this space, and hence

(3.117) $$ \begin{align} \int \big \langle l_y^0 |\mathrm{Im}(w_{P,y}^0) \big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} n_{0,1}(y) \mathrm{d} y \, =\, \int \big \langle l_y^1 |\Theta^{-2}_K\mathrm{Im}(w_{P,y}^1)\big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} n_{0,1}(y)\mathrm{d} y \, =\, 0. \end{align} $$

Thus, with $\mathrm {Re}(w_{P,y})=w_{0,y}$ , and with

(3.118) $$ \begin{align} v(y)\, :=\, \big \langle l_y|w_{0,y}\big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} \end{align} $$

we finally have

(3.119) $$ \begin{align} \mathcal{E}_{21}^0\, =\, \int \big \langle l_y^0 + l_y^1| \operatorname{Re}(w_{P,y}^0)+ \operatorname{Re} (w_{P,y}^1) \big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} n_{0,1}(y)\mathrm{d} y\, =\, \int v(y)n_{0,1}(y)\mathrm{d} y. \end{align} $$

Note that $v\in L^1\cap L^{\infty }$ since $y\mapsto \vert \hspace {-1pt}\vert l_y \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^2\)}}}$ is, while $\vert \hspace {-1pt}\vert w_{0,y}\vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^2\)}}}$ is uniformly bounded in y. Because of $\varphi (z) = -\langle \psi |h_{\cdot }(z)\psi \rangle _{\scriptscriptstyle{\text{\(L^2\)}}}$ and $\nabla _z h(x-z)\, =\, -\nabla _x h(x-z)$ , we have by integration by parts

(3.120) $$ \begin{align} \nabla \varphi\, =\, -2\big \langle \nabla \psi|h_{\cdot}\psi\big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}}. \end{align} $$

Thus,

(3.121) $$ \begin{align} l_y\, =\, -\frac{1}{2}y\nabla\varphi +\varphi(H(y)-1)+\big \langle \psi|h_{\cdot}(T_y \psi -\psi-y\nabla\psi)\big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}}. \end{align} $$

Since $\psi $ is a smooth function with uniformly bounded derivatives, there exists a $C>0$ such that for all y

(3.122) $$ \begin{align} \vert\hspace{-1pt}\vert T_y \psi-\psi-y\nabla \psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^{\infty}\)}}} \, \leq\, Cy^2. \end{align} $$

Moreover, for $k=0,1$ and every $z\in {\mathbb R}^3$ ,

(3.123) $$ \begin{align} x\mapsto (h_{\cdot}(z)\nabla^k\psi)(x)\in L^1(\mathbb{R}^3,\mathrm{d} x)\quad \text{and}\quad z\mapsto \vert\hspace{-1pt}\vert h_{\cdot}(z)\nabla^k \psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^1\)}}} \in L^2({\mathbb R}^3, \mathrm{d} z). \end{align} $$

The first statement follows easily from Lemma 3.7; to show the second one, use

(3.124) $$ \begin{align} \int \mathrm{d} z\, \frac{1}{|u-z|^2|v-z|^2} \, =\, \frac{1}{\pi^3 |u-v|} \end{align} $$

and apply the Hardy–Littlewood–Sobolev inequality. This, together with equation (3.38), shows that there exists a function f in $L^2({\mathbb R}^3, \mathrm {d} z)$ such that

(3.125) $$ \begin{align} |l_y(z)+\frac{1}{2}y\nabla\varphi(z)|\, \leq\, f(z)y^2. \end{align} $$

Now, let

(3.126) $$ \begin{align} b_y(z)\, :=\, w_{0,y}(z)-y\nabla\varphi(z)\, =\, \int_0^1 \mathrm{d} s \int_0^s\mathrm{d} t\, (y\nabla)^2 \varphi(z-ty) \end{align} $$

and note that $\vert \hspace {-1pt}\vert b_y \vert \hspace {-1pt}\vert ^2_{\scriptscriptstyle{\text{\(L^2\)}}} \leq \frac {1}{4} y^4\vert \hspace {-1pt}\vert \Delta \varphi \vert \hspace {-1pt}\vert ^2_{\scriptscriptstyle{\text{\(L^2\)}}}$ which is finite since $\Delta \varphi \in L^2$ . This equation, together with equation (3.125), implies

(3.127) $$ \begin{align} \bigg| v(y)+\frac{1}{2} \vert\hspace{-1pt}\vert y\nabla \varphi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}}^2 \bigg| \, \leq\, C(|y|^3 + |y|^4). \end{align} $$

From this and from $v\in L^1\cap L^{\infty }$ , it is also easy to deduce that $|\cdot |^{-2}v \in L^1\cap L^{\infty }$ . We can thus write

(3.128) $$ \begin{align} \int \mathrm{d} y \, v(y) n_{0,1}(y)\, =\, \int \mathrm{d} y\, v(y) e^{-\alpha^2\lambda y^2}+\int \mathrm{d} y \, |y|^{-2}v(y) y^2 \big( n_{0,1}(y)-e^{-\alpha^2\lambda y^2} \big) \end{align} $$

and use Lemma 3.5 for $g=|\cdot |^{-2}|v|$ to bound

(3.129) $$ \begin{align} \left|\int \mathrm{d} y\, |y|^{-2}v(y) y^2 (n_{0,1}(y)-e^{-\alpha^2\lambda y^2})\right|\, \leq\, C\alpha^{-6}. \end{align} $$

Using equation (3.127), the definition of $\lambda =\frac {1}{6}\vert \hspace {-1pt}\vert \nabla \varphi \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^2\)}}}^2$ as well as $\int y^2 e^{-y^2}\mathrm {d} y= \frac {3}{2}\pi ^{3/2}$ , we further have that

(3.130) $$ \begin{align} \bigg| \int \mathrm{d} y \, v(y) e^{-\alpha^2\lambda y^2} + \frac{3}{2\alpha^2}\left(\frac{\pi}{\lambda \alpha^2}\right)^{3/2} \bigg| \, \le\, C\alpha^{-6} \end{align} $$

which finally gives the estimate

(3.131) $$ \begin{align} \bigg| \mathcal{E}_{21}^0 + \bigg( \frac{3}{2\alpha^2} \bigg) \mathcal N \, \bigg| \, \le\, C_{\varepsilon} \sqrt K \alpha^{- 6 + \varepsilon } \end{align} $$

using Proposition 3.17.

In a similar fashion as for $\mathcal {E}_{21}^0$ , we obtain

(3.132) $$ \begin{align} \mathcal{E}^P_{21} & \, =\, \frac{1}{\alpha^{2}M^{\mathrm{LP}}}\int \big \langle i P\nabla \varphi |w^0_{P,y}\big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} H(y)n_{0,1}(y) \mathrm{d} y. \end{align} $$

Explicit computation, using $\Pi _0 = \frac {3}{\vert \hspace {-1pt}\vert \nabla \varphi \vert \hspace {-1pt}\vert ^2_{\scriptscriptstyle{\text{\(L^2\)}}}}\sum _{i=1}^3 |\partial _i \varphi \rangle \langle \partial _i \varphi |$ and $\langle \varphi | \nabla \varphi \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} = 0$ , gives

(3.133) $$ \begin{align} \frac{1}{3}w^0_{P,y}(z) \, =\, -\frac{(\varphi\ast \nabla \varphi)(y)}{\vert\hspace{-1pt}\vert \nabla \varphi\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}}^2}\nabla \varphi(z)+\frac{iP}{\alpha^2 M^{\mathrm{LP}}}\left(\vert\hspace{-1pt}\vert \nabla \varphi\vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(L^2\)}}} -(\nabla \varphi\ast \nabla \varphi)(y)\right)\frac{\nabla \varphi(z)}{\vert\hspace{-1pt}\vert \nabla \varphi\vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(L^2\)}}}}. \end{align} $$

Note that the real part of the above is odd as a function of y and hence

(3.134) $$ \begin{align} \int \big \langle \nabla \varphi|\mathrm{Re}(w_{P,y}^0)\big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} n_{0,1}(y)H(y)\mathrm{d} y\, =\, 0, \end{align} $$

and, taking rotational invariance of $\varphi $ into account, we arrive at

(3.135) $$ \begin{align} \mathcal{E}_{21}^P\, =\, \frac{P^2}{\alpha^4 (M^{\mathrm{LP}})^2}\int \left(\|\nabla \varphi\|_2^2-\left(\nabla \varphi\ast\nabla \varphi\right)(y)\right)n_{0,1}(y)H(y)\mathrm{d} y. \end{align} $$

Further, note that $|\vert \hspace {-1pt}\vert \nabla \varphi \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^2\)}}}^2 - \left (\nabla \varphi \ast \nabla \varphi \right )(y)|\leq Cy^2$ and thus, by Lemma 3.7 and Corollary 3.6, one obtains

(3.136) $$ \begin{align} |\mathcal{E}_{21}^P|\, \leq\, C\frac{P^2}{\alpha^9} \le \frac{C}{\alpha^7}. \end{align} $$

This completes the analysis of ${\mathcal E}_{21}$ .

In order to estimate the term $\mathcal {E}_{22}$ , we proceed as before by splitting $\Upsilon _K=\Upsilon _K^{<}+\Upsilon _{K}^{>}$ . Using equation (3.44), we can estimate

(3.137) $$ \begin{align} &\bigg| \alpha^{-1}\int \mathrm{d} y\, \big \langle \Upsilon^{>}_K |(\phi(l_y)+\pi(j_y))(e^{A_{P,y}}-1)W(\alpha {w}_{P,y}) \Upsilon_K\big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \bigg| \notag \\ & \quad \quad \, \leq \, C \alpha^{-1}\int \mathrm{d} y \, (\vert\hspace{-1pt}\vert l_y \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}} + \vert\hspace{-1pt}\vert j_y\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}})\|(\mathbb{N}+1)^{1/2}\Upsilon_K^{>}\|_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \, \leq \, C_{\delta} \, \alpha^{-11} \end{align} $$

where we used Corollary 3.14 and Lemmas 3.7 and 3.19. The term involving $\Upsilon _K^{<}$ is split again into two contributions,

(3.138a) $$ \begin{align} \mathcal{E}_{22}^0 & \, =\, \alpha^{-1}\int \mathrm{d} y\, \big \langle \Upsilon^{<}_K | \phi(l_y)(e^{A_{P,y}}-1)W(\alpha {w}_{P,y}) \Upsilon_K\big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \end{align} $$

(3.138b) $$ \begin{align} \mathcal{E}^P_{22} & \, =\, \alpha^{-1}\int \mathrm{d} y\, \big \langle \Upsilon^{<}_K |\pi(j_y) (e^{A_{P,y}}-1)W(\alpha {w}_{P,y})\Upsilon_K\big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}}. \end{align} $$

To bound the first one, we proceed as in equation (3.107), that is, use Lemma 3.15 and the fact that $\phi (l_y)$ changes the number of phonons at most by one. This leads to

(3.139) $$ \begin{align} |{\mathcal E}_{22}^0| & \, \leq\, \alpha^{-1}\int \mathrm{d} y\, \|e^{\kappa \mathbb{N}}\mathbb{U}_{K}(e^{-A_{P,y}}-1)\phi(l_y)\Upsilon_K^{<}\|_{\scriptscriptstyle{\text{\(\mathcal F\)}}}\, n_{\delta,\eta}(y) \notag \\ &\, \leq \, \sqrt 2 \alpha^{-1}\int \mathrm{d} y\, \|(e^{-A_{P,y}}-1)\phi(l_y)\Upsilon_K^{<}\|_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \, n_{\delta,\eta}(y). \end{align} $$

Furthermore, we have

(3.140) $$ \begin{align} \vert\hspace{-1pt}\vert (e^{-A_{P,y}}-1)\phi(l_y)\Upsilon^{<}_K \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} & \, \leq \, \vert\hspace{-1pt}\vert A_{P,y}\phi(l_y)\Upsilon_K^{<}\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \, \leq \, \vert\hspace{-1pt}\vert \phi(l_y)A_{P,y}\Upsilon_K^{<}\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} + \vert\hspace{-1pt}\vert [A_{P,y},\phi(l_y)]\Upsilon_K^{<}\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \notag \\[0.5mm] & \, \leq \, C \alpha^{\delta/2}\left(\vert\hspace{-1pt}\vert l_y \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}} \vert\hspace{-1pt}\vert A_{P,y}\Upsilon_K \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} + \vert\hspace{-1pt}\vert y\nabla l_y \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}}\right), \end{align} $$

where we used $[iP_f y,\phi (f)]=\pi (y\nabla f)$ and . Note that in order to estimate the remaining expression, it is not sufficient to directly apply Corollary 3.6. To obtain a better bound, we first replace $n_{\delta ,\eta }(y)$ by $e^{-\eta \lambda \alpha ^{2(1-\delta )} y^2}$ and then, for the part containing the Gaussian, we use that $\vert \hspace {-1pt}\vert l_y\vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^2\)}}}$ and $\vert \hspace {-1pt}\vert \nabla l_y \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^2\)}}}$ provide additional factors of $|y|$ , as is shown below. More precisely, with Lemma 3.19 and the aid of Lemmas 3.5 and 3.16, we bound

(3.141) $$ \begin{align} &\hspace{-12pc} \alpha^{\frac{\delta}{2}-1}\int \mathrm{d} y\, \vert\hspace{-1pt}\vert l_y\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}} \vert\hspace{-1pt}\vert A_{P,y} \Upsilon_K \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \, n_{\delta,\eta}(y) \end{align} $$

(3.142) $$ \begin{align} & \quad\quad\quad \, \leq \, C\alpha^{\frac{\delta}{2}-1} \int \mathrm{d} y \, \vert\hspace{-1pt}\vert l_y\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}} (\sqrt K |y| + \alpha |y|^3 ) n_{\delta,\eta}(y) \notag \\ & \quad\quad\quad \, \leq\, C\alpha^{\frac{\delta}{2}-1} \int \mathrm{d} y\, \vert\hspace{-1pt}\vert l_y \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}} (\sqrt K |y| + \alpha |y|^3 ) e^{-\eta\lambda \alpha^{2(1-\delta)} y^2}+ C\sqrt{K} \alpha^{-6 + \frac{9\delta}{2}}. \end{align} $$

Next, we use that by equation (3.125) there exists an $L^2$ function f such that

(3.143) $$ \begin{align} |l_y(z)|\, \leq \, \frac{1}{2}|y\nabla\varphi(z)|+f(z) y^2. \end{align} $$

Hence, by integration

(3.144) $$ \begin{align} \alpha^{\delta/2-1}\int \mathrm{d} y \, \vert\hspace{-1pt}\vert l_y \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}} \left(\sqrt K |y| + \alpha |y|^3 \right)e^{-\lambda \eta \alpha^{2(1-\delta)}y^{2}}\, \leq \, C \sqrt{K}\alpha^{-6 + 11/2\delta}. \end{align} $$

Regarding the second term in equation (3.140), we need to bound

(3.145) $$ \begin{align} \alpha^{\delta/2-1}\int \mathrm{d} y\, |y| \vert\hspace{-1pt}\vert \nabla l_y \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}} n_{\delta,\eta}(y), \end{align} $$

where we proceed in a similar way as above, using that

(3.146) $$ \begin{align} \vert\hspace{-1pt}\vert \nabla l_y \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}}\, \leq \, C(|y|+ y ^2). \end{align} $$

In fact, since $\nabla \varphi (z) =-\langle \psi |h_{\cdot }(z) \nabla \psi \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} -\langle \nabla \psi |h_{\cdot }(z) \psi \rangle _{\scriptscriptstyle{\text{\(L^2\)}}}$ , we have the identity

(3.147) $$ \begin{align} \nabla l_y(z) & \, =\, H(y)\nabla \varphi(z) +\big \langle \nabla\psi|h_{\cdot}(z) T_y\psi\big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} + \big \langle \psi|h_{\cdot}(z) \nabla T_y\psi\big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} \\ &\, =\, (H(y)-1)\nabla \varphi (z) + \big \langle \nabla \psi|h_{\cdot}(z)( T_y-1 ) \psi\big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} + \big \langle \psi |h_{\cdot}(z) (T_y -1 ) \nabla \psi \big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} .\notag \end{align} $$

Again, using that $\psi $ has bounded derivatives, we have

(3.148) $$ \begin{align} \vert\hspace{-1pt}\vert (T_y-1)\psi\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^{\infty}\)}}} + \vert\hspace{-1pt}\vert (T_y-1)\nabla \psi\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^{\infty}\)}}} \le C |y|, \end{align} $$

and the desired inequality now follows from $| H(y)-1 | \le Cy^2$ and equation (3.123). Given equation (3.114), we can use Lemma 3.5 to replace $n_{\delta ,\eta }(y)$ in equation (3.145) with $e^{-\lambda \eta \alpha ^{2(1-\delta }) y^2}$ at the energy penalty $C \alpha ^{-6 + 9\delta /2}$ , and then use equation (3.146) to bound the remaining integral involving the Gaussian factor, which yields an error of the same order. Altogether, this gives the estimate

(3.149) $$ \begin{align} |\mathcal{E}_{22}^0|\, \leq\, C \sqrt{K}\alpha^{-6+\frac{11}{2}\delta}. \end{align} $$

For the term $\mathcal {E}^P_{22}$ , we proceed in exactly the same way as in equation (3.139):

(3.150) $$ \begin{align} | \mathcal{E}^P_{22} | &\, \leq\, \sqrt{2} \alpha^{-1} \int \mathrm{d} y\, \|\left(e^{- A_{P,y}}-1\right)\pi (j_y)\Upsilon_K^{<}\|_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \, n_{\delta,\eta}(y) \notag \\ & \, \leq\, C\alpha^{\delta/2-1}\int \mathrm{d} y\, \|j_y\|_{\scriptscriptstyle{\text{\(L^2\)}}} \|A_{P,y}\Upsilon_K\|_{\scriptscriptstyle{\text{\(\mathcal F\)}}}\, n_{\delta,\eta}(y) +C\alpha^{\delta/2-1}\int \mathrm{d} y\, \|y\nabla j_y\|_{\scriptscriptstyle{\text{\(L^2\)}}} n_{\delta,\eta}(y) \notag \\ &\, \leq\, C\alpha^{\delta/2-1} \frac{|P|}{\alpha^2}\int \mathrm{d} y\, H(y)\, (\sqrt{K}|y|+\alpha|y|^3) n_{\delta,\eta}(y) \notag \\ & \quad + C\alpha^{\delta/2-1} \frac{|P|}{\alpha^2} \int \mathrm{d} y\, |y|H(y) \, n_{\delta,\eta}(y) \notag\\ &\, \leq\, C\alpha^{-6 + \frac{9}{2}\delta} \sqrt{K}, \end{align} $$

where the last estimate follows from Corollary 3.6 and the assumption $|P|\le c \alpha $ .

Combining the relevant estimates, that is, equations (3.103), (3.104), (3.106) and (3.110) for ${\mathcal E}_1$ as well as equations (3.131), (3.136), (3.137), (3.149) and (3.150) for ${\mathcal E}_2$ , we arrive at the statement of Proposition 3.18, thus providing an appropriate bound for $\mathcal {E}$ .

Proof of Lemma 3.19.

Since H has the desired properties, we need to show them for

(3.151) $$ \begin{align} l^{(1)}_y \, =\, \big \langle \psi|h_{\cdot}T_y \psi\big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}}. \end{align} $$

To this end, we introduce

(3.152) $$ \begin{align} \mathcal{S}\, =\, \lbrace f\in L^p(\mathbb{R}^3,(1+|y|^n)\mathrm{d} y) \quad \forall 1\leq p\leq \infty, \quad \forall n\geq 0\rbrace \end{align} $$

and start with the following observation: Suppose $f_1,f_2,f_3$ and $f_4$ are functions in $\mathcal {S}$ . Then

(3.153) $$ \begin{align} S(y)\, :=\, \iint \mathrm{d} u \mathrm{d} v \frac{f_1(u)f_2(v)f_3(u+y)f_4(v+y)}{|u-v|} \in \mathcal{S}. \end{align} $$

In fact, $|S(y)| \leq C \|f_4\|_{\scriptscriptstyle{\text{\(L^{\infty }\)}}} \|f_3\|_{\scriptscriptstyle{\text{\(L^{\infty }\)}}} \|f_1\|_{\scriptscriptstyle{\text{\(L^p\)}}} \|f_2\|_{\scriptscriptstyle{\text{\(L^q\)}}}$ for all $1<p<3/2, q=3p/(5p-3)$ by the Hardy–Littlewood–Sobolev inequality. Since $\int \mathrm {d} y |y|^n f_3(u+y)\leq 2^{n-1}\left (|u|^n \| f_3 \|_{\scriptscriptstyle{\text{\(L^1\)}}} +\||\cdot |^nf_3\|_{\scriptscriptstyle{\text{\(L^1\)}}}\right )$ , we have also

(3.154) $$ \begin{align} \int \mathrm{d} y |y|^n S(y)\, \leq\, C \|f_4\|_{\scriptscriptstyle{\text{\(L^{\infty}\)}}} \left(\||\cdot|^nf_1\|_{\scriptscriptstyle{\text{\(L^p\)}}}\|f_2\|_{\scriptscriptstyle{\text{\(L^q\)}}} \|f_3\|_{\scriptscriptstyle{\text{\(L^1\)}}} +\|f_1\|_{\scriptscriptstyle{\text{\(L^p\)}}} \|f_2\|_{\scriptscriptstyle{\text{\(L^q\)}}} \||\cdot|^n f_3\|_{\scriptscriptstyle{\text{\(L^1\)}}}\right) \end{align} $$

from which (3.153) follows. Moreover,

(3.155) $$ \begin{align} f\in \mathcal{S }\implies \sqrt{|f|}\in \mathcal{S}. \end{align} $$

Indeed, we have for all $n\geq 0$ ,

(3.156) $$ \begin{align} \int |y|^n \sqrt{|f|}\mathrm{d} y \, \leq\, \sqrt{\|f\|_{\scriptscriptstyle{\text{\(L^{\infty}\)}}} }\int_{|y|\leq 1}|y|^n\mathrm{d} y+\frac{1}{2}\int |y|^{n+m}|f| \mathrm{d} y+\frac{1}{2}\int_{|y|>1}|y|^{n-m}\mathrm{d} y \, <\, \infty \end{align} $$

since m can be chosen arbitrarily large by assumption. Thus, it suffices to prove the desired statement for the functions $\|\nabla ^k l_y^{(1)}\|_{\scriptscriptstyle{\text{\(L^2\)}}}^2$ . For $k=0$ , we use equation (3.124) to compute

(3.157) $$ \begin{align} \|l^{(1)}_y\|_{\scriptscriptstyle{\text{\(L^2\)}}}^2\, =\, \frac{1}{4\pi}\iint \mathrm{d} u \mathrm{d} v \frac{\psi(u)\psi(v)\psi(y+u)\psi(v+y)}{|u-v|}. \end{align} $$

The statement now follows easily from equation (3.153) and Lemma 3.7. Arguing again via equation (3.155), for $k=1$ it suffices to show the statement for

(3.158) $$ \begin{align} \|\nabla l^{(1)}_y\|_{\scriptscriptstyle{\text{\(L^2\)}}}^2 &\, =\, \|\langle \nabla \psi|h_{\cdot} T_y\psi\rangle_{\scriptscriptstyle{\text{\(L^2\)}}} +\big \langle \psi|h_{\cdot}\nabla T_y\psi\rangle_{\scriptscriptstyle{\text{\(L^2\)}}} \|_{\scriptscriptstyle{\text{\(L^2\)}}}^2 \notag \\ & \, \leq\, 2 \|\langle \nabla \psi|h_{\cdot} T_y\psi\rangle_{\scriptscriptstyle{\text{\(L^2\)}}} \|_{\scriptscriptstyle{\text{\(L^2\)}}}^2 + 2 \|\langle \psi|h_{\cdot}\nabla T_y \psi\rangle_{\scriptscriptstyle{\text{\(L^2\)}}} \|_{\scriptscriptstyle{\text{\(L^2\)}}}^2 \end{align} $$

(the first equality follows from $\nabla _z h_x(z) = - \nabla _x h_x(z)$ and integration by parts). Using (3.124), we find

(3.159a) $$ \begin{align} \|\langle \nabla \psi|h_{\cdot} T_y\psi\rangle_{\scriptscriptstyle{\text{\(L^2\)}}} \|_{\scriptscriptstyle{\text{\(L^2\)}}}^2 & \, \leq\, C \iint \mathrm{d} u \mathrm{d} v \frac{|\nabla\psi(u)||\nabla\psi(v)|\psi(v+y)\psi(u+y)}{|u-v|}, \end{align} $$

(3.159b) $$ \begin{align} \|\langle \psi|h_{\cdot}\nabla T_y\psi\rangle_{\scriptscriptstyle{\text{\(L^2\)}}} \|_{\scriptscriptstyle{\text{\(L^2\)}}}^2 & \, \leq \, C \iint \mathrm{d} u \mathrm{d} v \frac{|\nabla \psi(u+y)||\nabla\psi(v+y)|\psi(v)\psi(u)}{|u-v|}. \end{align} $$

We arrive at the desired conclusion by Lemma 3.7 and equation (3.153).

Proof. Using $h^{\mathrm{Pek}} G_K^0 = 0$ and $\mathbb NG_K^0 = {\mathbb N}_1 G_K^0$ , we can decompose ${\mathcal G}$ into two terms

(3.161) $$ \begin{align} {\mathcal G} & \, =\, - \frac{2}{\alpha} \int \mathrm{d} y\, \operatorname{Re} \big \langle G_{K}^{ 0 } | (\alpha^{-2} {\mathbb N}_1 + \alpha^{-1} \phi (h_{\cdot} + \varphi_P) ) T_y e^{A_{P,y}} W(\alpha w_{P,y}) G_{K}^{ 1} \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag\\ &\, =\, {\mathcal G}_1 + {\mathcal G}_2, \end{align} $$

where the first term will contribute to the error while the second one provides an energy contribution of order $\alpha ^{-2}$ . We proceed for each one separately.

$\underline {\mathrm {Term}\ {\mathcal G}_1}$ . With the aid of equations (3.67) and (3.80) and $(T_y e^{A_{P,y}} )^{\dagger } = T_{-y} e^{-A_{P,y}}$ , one finds

(3.162) $$ \begin{align} {\mathcal G}_1 & \, =\, - \frac{2}{ \alpha^{3} } \int \mathrm{d} y\, \operatorname{Re} \big \langle R_{3,y} \psi \otimes \big( \Upsilon_K^< + \Upsilon_K^>\big) | W(\alpha w_{P,y}) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, =\, {\mathcal G}_{1}^< + {\mathcal G}_{1}^>, \end{align} $$

where we introduced the operator $R_{3,y} = R_{3,y}^1 + R_{3,y}^2$ with

(3.163a) $$ \begin{align} R_{3,y}^1 & \, =\, P_{\psi} \phi(h_{K,\cdot}^1) R u_{\alpha} T_{-y} P_{\psi} e^{-A_{P,y}} {\mathbb N}_1 \end{align} $$

(3.163b) $$ \begin{align} R_{3,y}^2 & \, =\, 2\alpha P_{\psi} \big \langle h_{K,\cdot} | \operatorname{Re} (w_{P,y}^1) \big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} R u_{\alpha} T_{-y} P_{\psi} e^{-A_{P,y}} {\mathbb N}_1. \end{align} $$

Proceeding similarly as for $R_{1,y}^1$ and $R_{2,y}^2$ in equations (3.82a) and (3.82b), one further verifies

(3.164) $$ \begin{align} \vert\hspace{-1pt}\vert R_{3,y} \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, \le \, C \vert\hspace{-1pt}\vert u_{\alpha} T_{-y} P_{\psi}\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} \big( 1 + \alpha y^2\big) \vert\hspace{-1pt}\vert ({\mathbb N}+1)^{3/2} \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}}. \end{align} $$

Recalling the definition $f_{\alpha }(y) = \vert \hspace {-1pt}\vert u_{\alpha } T_{-y} P_{\psi } \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(\mathrm{op}\)}}}$ and equation (3.86), we can use Corollary 3.14 to find

(3.165) $$ \begin{align} |{\mathcal G}_{1}^> | \, \le\, \frac{C}{\alpha^3} \vert\hspace{-1pt}\vert ({\mathbb N}+1)^{3/2} \Upsilon_K^>\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \int \mathrm{d} y \, f_{\alpha}(y) (1+ \alpha y^2) \, \le\, C_{\delta}\, \alpha^{-7 }. \end{align} $$

In the first term, we proceed with equation (3.74) and Lemma 3.15 to obtain

(3.166) $$ \begin{align} |{\mathcal G}_{1}^< | & \, \le \, \frac{2}{ \alpha^{3} } \int \mathrm{d} y\, \vert\hspace{-1pt}\vert e^{\kappa {\mathbb N}} \mathbb U_K ( R_{3,y} \psi \otimes \Upsilon_K^< ) \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, \vert\hspace{-1pt}\vert e^{-\kappa {\mathbb N}} W(\alpha \widetilde w_{P,y} ) \Omega \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \notag\\ & \, \le\, \frac{2\sqrt 2}{ \alpha^{3} } \int \mathrm{d} y\, \vert\hspace{-1pt}\vert R_{3,y} \psi\otimes \Upsilon_K \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, n_{\delta,\eta}(y) \, \le\, \frac{C}{ \alpha^{3} } \int \mathrm{d} y\, f_{\alpha}(y) (1+\alpha y^2) \, n_{\delta,\eta}(y) , \end{align} $$

which brings us again into a position to apply Corollary 3.6. Hence,

(3.167) $$ \begin{align} |{\mathcal G}_{1}^< | \le C \alpha^{- 6 +3 \delta }. \end{align} $$

$\underline {\mathrm {Term}\ {\mathcal G}_2}$ . Here, we have

(3.168) $$ \begin{align} {\mathcal G}_{2} & = - \frac{2}{\alpha^2} \int \mathrm{d} y\, \operatorname{Re} \big \langle G_K^0 | \phi (h_{\cdot} + \varphi_P) T_y W(\alpha w_{P,y}) G_K^1 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag \\ & \quad - \, \frac{2}{\alpha^2} \int \mathrm{d} y \, \operatorname{Re} \big \langle G_K^0 | \phi (h_{\cdot} + \varphi_P ) T_y ( e^{ A_{P,y}} -1 ) W(\alpha w_{P,y}) G_K^1 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} = {\mathcal G}_{21} + {\mathcal G}_{22}. \end{align} $$

To separate the leading order contribution in ${\mathcal G}_{21}$ , we insert $1 = \mathbb U_K^{\dagger } \mathbb U_K$ next to $G_K^0$ and bring $\mathbb U_K^{\dagger }$ to the right side of the inner product. With $\mathbb U_K\Upsilon _K = \Omega $ , equation (3.53c) and equation (3.63) this gives

(3.169) $$ \begin{align} {\mathcal G}_{21} & \, =\, - \frac{2}{\alpha^2} \int \mathrm{d} y \, \operatorname{Re} \big \langle \psi \otimes \Omega | a(\underline { h_{\cdot} + \varphi_P } )T_y W(\alpha \widetilde w_{P,y}) u_{\alpha} R a^{\dagger}(\underline{h_{K,\cdot}^1} ) \psi \otimes \Omega \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}}, \end{align} $$

where $\underline {\, \cdot \, }$ is defined in equation (3.52a). Next, we write $ W(\alpha \widetilde w_{P,y}) = n_{0,1}(y) e^{a^{\dagger }(\alpha \widetilde w_{P,y})} e^{-a(\alpha \widetilde w_{P,y})}$ and move the first exponential to the left side and the second exponential to the right side until they act both on the Fock space vacuum. Using $e^{-a(f)} a^{\dagger }(g) e^{a(f)} = a^{\dagger }(g) - \langle f |g \rangle $ , we find this way

(3.170a) $$ \begin{align} \!\!\!\!\!{\mathcal G}_{21} & = - \frac{2}{\alpha^2} \int \mathrm{d} y \, n_{0,1}(y) \, \operatorname{Re} \big \langle \psi \otimes \Omega | a(\underline { h_{\cdot} + \varphi_P} ) T_y u_{\alpha} R a^{\dagger} ( \underline{h_{K,\cdot}^1} ) \psi \otimes \Omega \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}}\qquad\qquad\qquad \end{align} $$

(3.170b) $$ \begin{align} & \quad + 2 \int \mathrm{d} y \, n_{0,1}(y)\, \operatorname{Re} \big \langle \psi \otimes \Omega | \langle \underline { h_{\cdot} + \varphi_P } | \widetilde w_{P,y} \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} T_y u_{\alpha} R \langle \widetilde w_{P,y} | \underline{h_{K,\cdot}^1} \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} \psi \otimes \Omega \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}}. \end{align} $$

In the first line, we write $h_{\cdot } + \varphi _P = h_{\cdot }^0 + h_{\cdot }^1 + \varphi + i \xi _P$ , with $h_{\cdot }^i = ( \Pi _ih)_{\cdot }$ , and use that

(3.171) $$ \begin{align} \big \langle \psi \otimes \Omega | a(\underline { h_{\cdot}^0 + i \xi_P } ) T_y u_{\alpha} R a^{\dagger} ( \underline{h_{K,\cdot}^1} ) \psi \otimes \Omega \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} = 0 \end{align} $$

since $h_{x}^0 + i \xi _P \in \text {Ran}(\Pi _0)$ whereas $h_{K,x}^1 \in \text {Ran}(\Pi _1)$ . Finally, we can replace a and $a^{\dagger }$ by $\phi $ , and then transform back with $\mathbb U_K$ , using equation (3.53c), in order to obtain

(3.172) $$ \begin{align} (3.170\mathrm{a}) \, =\, - \frac{2}{\alpha^2} \int \mathrm{d} y \, n_{0,1}(y) \, \operatorname{Re} \big \langle \psi \otimes \Upsilon_K | \phi( h_{\cdot}^1 + \varphi ) T_y u_{\alpha} R \phi ( h_{K,\cdot}^1 ) \psi \otimes \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}}. \end{align} $$

To summarize, we have shown that

(3.173) $$ \begin{align} {\mathcal G}_{21} & = - \frac{2}{\alpha^2} \int \mathrm{d} y\, \operatorname{Re} \big \langle G_K^0 | L_{2,y} G_K^0\big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} n_{0,1}(y) + \int \mathrm{d} y \, \ell_2(y) n_{0,1}(y) = {\mathcal G}_{211} + {\mathcal G}_{212} \end{align} $$

with

(3.174a) $$ \begin{align} L_{2,y} & \, =\, P_{\psi} \phi ( h_{\cdot}^1 + \varphi ) T_y u_{\alpha} R \phi( h_{K,\cdot}^1 ) P_{\psi} \end{align} $$

(3.174b) $$ \begin{align} \ell_2(y) & \, =\, 2 \operatorname{Re} \big \langle \psi | \langle \underline{ h_{\cdot} + \varphi_P } | \widetilde w_{P,y} \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} T_y u_{\alpha} R \langle \widetilde w_{P,y}^1 | \underline{ h_{K,\cdot}^1 } \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} \psi \big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}}. \end{align} $$

In the first term, we add and subtract the Gaussian,

(3.175) $$ \begin{align} {\mathcal G}_{211} & = - \frac{2}{\alpha^2} \int \mathrm{d} y \, \operatorname{Re} \big \langle G_K^0 | L_{2,y} G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, e^{-\lambda \alpha^2 y^2 } \notag \\ & \quad - \frac{2}{\alpha^2} \int \mathrm{d} y \, \operatorname{Re} \big \langle G_K^0 | L_{2,y} G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \big( n_{0,1}(y) - e^{-\lambda \alpha^2 y^2 } \big) \, =\, {\mathcal G}_{211}^{\mathrm{lo}} + {\mathcal G}_{211}^{\mathrm{err}}, \end{align} $$

and proceed with ${\mathcal G}_{211}^{\mathrm{lo}}$ by inserting $h_{\cdot }^1 = h_{K,\cdot }^1 +( h_{\cdot }^1 - h_{K,\cdot }^1)$ , $T_y = 1 + ( T_y-1 ) $ and $u_{\alpha } = 1 + ( u_{\alpha } -1 ) $ ,

(3.176) $$ \begin{align} {\mathcal G}_{211}^{\mathrm{lo}} & = - \frac{2}{\alpha^2} \operatorname{Re} \big \langle G_K^0 | \phi( h_{K,\cdot}^1 + \varphi ) R \phi( h_{K,\cdot}^1 ) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \int \mathrm{d} y \, e^{-\lambda \alpha^2 y^2 } \notag \\ & \quad - \frac{2}{\alpha^2} \operatorname{Re} \big \langle G_K^0 | \phi( h_{K,\cdot}^1 + \varphi ) (u_{\alpha} - 1) R \phi( h_{K,\cdot}^1 ) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \int \mathrm{d} y \, e^{-\lambda \alpha^2 y^2 } \notag \\ & \quad - \frac{2}{\alpha^2} \int \mathrm{d} y \, \operatorname{Re} \big \langle G_K^0 | \phi( h_{K,\cdot}^1 +\varphi ) ( T_{y}-1) u_{\alpha} R \phi( h_{K,\cdot}^1 ) G_K^0\big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} e^{-\lambda \alpha^2 y^2 } \notag \\ & \quad - \frac{2}{\alpha^2} \int \mathrm{d} y \, \operatorname{Re} \big \langle G_K^0 | \phi( h_{\cdot}^1 - h_{K, \cdot}^1 ) T_y u_{\alpha} R \phi( h_{K,\cdot}^1 ) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} e^{-\lambda \alpha^2 y^2 } \notag\\ & = \sum_{n=1}^4 {\mathcal G}_{211}^{{\mathrm{lo}},n}. \end{align} $$

Since $P_{\psi } \phi (\varphi ) R = 0$ , we have ${\mathcal G}_{211}^{{\mathrm{lo}},1} = \frac {2}{\alpha ^2} \big \langle \Upsilon _K | ( {\mathbb H}_K -{\mathbb N}_1 ) \Upsilon _K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} (\frac {\pi }{\lambda \alpha ^2})^{3/2}$ , cf. (2.4), and hence we can use Proposition 3.17 to conclude that

(3.177) $$ \begin{align} \bigg|\, {\mathcal G}_{211}^{{\mathrm{lo}},1} - \mathcal N \frac{2}{\alpha^2} \big \langle \Upsilon_K | ({\mathbb H}_K -{\mathbb N}_1) \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \, \bigg| \, \le \, C_{\varepsilon} \sqrt K \alpha^{-6 + \varepsilon}. \end{align} $$

For the other terms, we shall show the combined error estimate

(3.178) $$ \begin{align} |{\mathcal G}_{211}^{{\mathrm{lo}},2} | + |{\mathcal G}_{211}^{{\mathrm{lo}},3} | + |{\mathcal G}_{211}^{{\mathrm{lo}},4} | \, \le\, C \big( \sqrt K \alpha^{-6} + K^{-1/2} \alpha^{-5} \big). \end{align} $$

In the last term, we recall $h_{\cdot }(y)= h_{K=\infty ,\cdot }(y)$ , and apply Lemma 3.9 in combination with $\vert \hspace {-1pt}\vert R^{1/2} u_{\alpha } T_{-y} \nabla \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} \le C$ . This gives

(3.179) $$ \begin{align} |{\mathcal G}_{211}^{\mathrm{lo,4}} | & \, \le\, \frac{2}{\alpha^2} \int \mathrm{d} y \, e^{-\lambda \alpha^2 y^2} \, \vert\hspace{-1pt}\vert R^{1/2} u_{\alpha} T_{-y} \phi( h_{\cdot}^1 - h_{K, \cdot}^1 ) P_{\psi} G_K^0 \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, \vert\hspace{-1pt}\vert R^{1/2} \phi( h_{K,\cdot}^1 ) P_{\psi} G_K^0 \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag\\ & \, \le \, C \alpha^{-5} K^{-1/2}. \end{align} $$

Next, we write $T_y - 1 = \int _0^1 \mathrm {d} s T_{sy} (y \nabla )$ in the third term to obtain an additional $|y|$ ,

(3.180) $$ \begin{align} |{\mathcal G}_{211}^{\mathrm{lo,3}} | & \, \le\, \frac{2}{\alpha^2} \bigg(\int \mathrm{d} y \, |y| e^{-\lambda \alpha^2 y^2} \bigg) \vert\hspace{-1pt}\vert \nabla u_{\alpha} R^{1/2} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} \, \vert\hspace{-1pt}\vert \phi( h_{K,\cdot}^1 +\varphi ) G_K^0 \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \vert\hspace{-1pt}\vert R^{1/2} \phi( h_{K,\cdot}^1 ) G_K^0 \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag\\ & \, \le \, C \alpha^{-6} \sqrt K, \end{align} $$

where the factor $\sqrt K$ comes from the $L^2$ norm of $h_{K,0}^1$ in the bound on the first field operator (since $\Delta R^{1/2}$ is unbounded, we cannot apply the commutator method to this part). In the second term, we use $\psi (x)\le C e^{-|x|/C}$ for some $C>0$ , and thus $\vert \hspace {-1pt}\vert (u_{\alpha } -1) \psi \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^2\)}}} \le C e^{-\alpha /C}$ , to estimate

(3.181) $$ \begin{align} |{\mathcal G}_{211}^{\mathrm{lo,2}} | \, \le\, \frac{C}{\alpha^5} \vert\hspace{-1pt}\vert (u_{\alpha}-1) \psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}} \, \vert\hspace{-1pt}\vert \phi( h_{K,\cdot}^1 + \varphi ) R \phi( h_{K,\cdot}^1 ) G_K^0 \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, \le \, C \sqrt K e^{-\alpha/C}. \end{align} $$

This proves equation (3.178).

To bound the remaining contributions in ${\mathcal G}_{211}^{\mathrm{err}}$ and ${\mathcal G}_{212}$ , we shall use

(3.182a) $$ \begin{align} \big| \big \langle G_K^0 | L_{2,y} G_K^0 \big \rangle \big| & \, \le\, C f_{2,\alpha}(y) \end{align} $$

(3.182b) $$ \begin{align} |\ell_2 (y) | & \, \le\, C f_{2,\alpha}(y) ( y^2 + \alpha^{-2} ) ( | y | + |y|^3 + \alpha^{-2} ), \end{align} $$

where

(3.183) $$ \begin{align} f_{2,\alpha}(y) = \vert\hspace{-1pt}\vert u_{\alpha} T_{-y} P_{\psi} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} + \vert\hspace{-1pt}\vert \nabla u_{\alpha} T_{-y} P_{\psi} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}}. \end{align} $$

Using the exponential decay of $\psi $ and , for $k=0,1$ , it is easy to show that

(3.184) $$ \begin{align} \vert\hspace{-1pt}\vert f_{2,\alpha} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^{\infty}\)}}} \le C \quad \text{and} \quad \vert\hspace{-1pt}\vert |\cdot |^n f_{2,\alpha} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^1\)}}} \le C_n \alpha^{3+n} \quad \text{for all}\ n\in {\mathbb N}_0. \end{align} $$

To verify equations (3.182a) and (3.182b), use $u_{\alpha } T_{-y}\phi (h_{\cdot })= \phi (h_{\cdot - y } ) u_{\alpha } T_{-y}$ and Cauchy–Schwarz to bound

(3.185) $$ \begin{align} \big| \big \langle G_K^0 | L_{2,y} G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \big| & \, \le \, \vert\hspace{-1pt}\vert R^{1/2} \phi(h_{\cdot-y}^1 + \varphi) u_{\alpha} T_{-y} P_{\psi} G_K^0 \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, \vert\hspace{-1pt}\vert R^{1/2} \phi(h_{K,\cdot}^1) P_{\psi} G_K^0 \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}}. \end{align} $$

Now, we can use equation (3.44) and Lemma 3.9 to obtain equation (3.182a). To estimate $\ell _2(y)$ , defined in equation (3.174b), we proceed with

(3.186a) $$ \begin{align} |\ell_2(y)| & \, \le\, 2 \big| \big \langle \psi | T_y u_{\alpha} \langle \underline{h_{\cdot-y}} | \widetilde w_{P,y} \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} R \langle \widetilde w_{P,y}^1 | \underline{h_{K,\cdot}^1} \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} \psi \big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} \big| \end{align} $$

(3.186b) $$ \begin{align} &\quad \, +\, 2 \big| \big \langle \psi | T_y u_{\alpha} \langle \underline{\varphi_P} | \widetilde w_{P,y} \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} R \langle \widetilde w_{P,y}^1 | \underline{h_{K,\cdot}^1} \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} \psi \big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} \big| , \end{align} $$

and considering the first line, we use Cauchy–Schwarz, write out the two inner products (in the phonon variable) and then use Cauchy–Schwarz again,

(3.187) $$ \begin{align} | ((3.186\mathrm{a}) | \, & \le \, 2 \int \mathrm{d} u\, |\widetilde w_{P,y}(u)| \, \vert\hspace{-1pt}\vert P_{\psi} T_y u_{\alpha} \underline{h_{\cdot-y}}(u) R^{1/2} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} \int \mathrm{d} z \, |\widetilde w_{P,y}^1 (z) |\, \vert\hspace{-1pt}\vert R^{1/2} \underline{ h_{K,\cdot}^1}(z) \psi \vert\hspace{-1pt}\vert \notag\\ & \le \, 2 \vert\hspace{-1pt}\vert \widetilde w_{P,y}\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}} \vert\hspace{-1pt}\vert \widetilde w_{P,y}^1\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}} \,\bigg( \int \mathrm{d} u \vert\hspace{-1pt}\vert P_{\psi} T_y u_{\alpha} \underline{h_{\cdot-y}}(u) R^{1/2} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}}^2 \, \int \mathrm{d} z \vert\hspace{-1pt}\vert R^{1/2} \underline{h_{K,\cdot}^1}(z) \psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}}^2 \bigg)^{1/2} \notag\\ & \le \, C f_{2,\alpha}(y) ( |y| + y^3 + \alpha^{-2}) ( y^2 + \alpha^{-2}) , \end{align} $$

where the last step follows from Lemma 3.4 and Corollary 3.11 together with $\underline { h_{K,\cdot } } = h_{K,\cdot }^0 + \Theta _K^{-1} h_{K,\cdot }^1$ . Since the second line is estimated similarly, we arrive at equation (3.182b). With equation (3.182a) at hand, we can apply Lemma 3.5 and equation (3.184) to get

(3.188) $$ \begin{align} | {\mathcal G}_{211}^{\mathrm{err}} | \, \le \, \frac{2}{\alpha^2} \int \mathrm{d} y \, \big| \big \langle G_K^0 | L_{2,y} G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \big| \, \big| n_{0,1}(y) - e^{-\lambda \alpha^2 y^2 } \big| \, \le \, C \alpha^{-6}, \end{align} $$

and further, using equation (3.182b) and Corollary 3.6, we obtain

(3.189) $$ \begin{align} |{\mathcal G}_{212} | \, \le \, C \int \mathrm{d} y\, |\ell_2 (y) |\, n_{0,1}(y) \, \le \, C \alpha^{-6}. \end{align} $$

This completes the analysis of ${\mathcal G}_{21}$ .

Next, we introduce $ R_{4,y} = R_{4,y}^1 + R_{4,y}^2 $ with

(3.190a) $$ \begin{align} R_{4,y}^1 & \, =\, P_{\psi} \phi( h_{K,\cdot}^1 ) R^{\frac{1}{2}} ( e^{ - A_{P,y} } -1 ) R^{\frac{1}{2}} \phi ( h_{\cdot-y} + \varphi_P ) u_{\alpha} T_{-y} P_{\psi} \end{align} $$

(3.190b) $$ \begin{align} R_{4,y}^2 & \, =\, 2 \alpha P_{\psi} \big \langle h_{K,\cdot} | \operatorname{Re} ( w_{P,y}^1 ) \big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} R^{\frac{1}{2}} ( e^{ - A_{P,y} } -1 ) R^{\frac{1}{2}} \phi ( h_{\cdot-y} + \varphi_P )u_{\alpha} T_{-y} P_{\psi}. \end{align} $$

Inserting equations (3.67) and (3.80) into equation (3.168) it follows that

(3.191) $$ \begin{align} {\mathcal G}_{22}& \, =\, - \frac{2}{\alpha^2} \int \mathrm{d} y\, \operatorname{Re} \big \langle R_{4,y} \psi \otimes \big( \Upsilon_K^< + \Upsilon_K^> \big) | W(\alpha w_{P,y}) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, = \, {\mathcal G}_{22}^< + {\mathcal G}_{22}^>. \end{align} $$

With the aid of Lemma 3.9, we obtain

(3.192) $$ \begin{align} \vert\hspace{-1pt}\vert R_{4,y}^1 \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} & \, \le\, C \vert\hspace{-1pt}\vert ( e^{ - A_{P,y} } -1 ) ({\mathbb N}+1)^{1/2} R^{1/2} \phi ( h_{\cdot-y} + \varphi_P ) u_{\alpha} T_{-y} P_{\psi} \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} , \end{align} $$

and proceeding similarly as in equation (3.84), we find

(3.193) $$ \begin{align} \vert\hspace{-1pt}\vert R_{4,y}^2 \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} &\, \le\, C \alpha (y^2 + \alpha^{-2} ) \vert\hspace{-1pt}\vert ( e^{ - A_{P,y} } -1 ) R^{1/2} \phi ( h_{\cdot-y} + \varphi_P ) u_{\alpha} T_{-y} P_{\psi} \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}}. \end{align} $$

For $\Psi = \psi \otimes \Upsilon _K^>$ , a second application of Lemma 3.9 (after using unitarity of $e^{-A_{P,y}}$ ) together with $\vert \hspace {-1pt}\vert \varphi _P \vert \hspace {-1pt}\vert ^2_{\scriptscriptstyle{\text{\(L^2\)}}} \le C$ for $|P|/ \alpha \le c$ and Corollary 3.14 is sufficient to find

(3.194) $$ \begin{align} \vert\hspace{-1pt}\vert R_{4,y} \psi \otimes \Upsilon_K^> \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} & \, \le\, C \big( \vert\hspace{-1pt}\vert u_{\alpha} T_{- y} P_{\psi} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} + \vert\hspace{-1pt}\vert \nabla u_{\alpha} T_{ - y} P_{\psi} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} \big) (1+ \alpha y^2) \vert\hspace{-1pt}\vert ({\mathbb N}+1) \Upsilon_K^>\vert\hspace{-1pt}\vert_{{\mathcal F}} \notag\\[1mm] & \, \le\, C_{\delta} \, \alpha^{-10} f_{2,\alpha}(y)(1+\alpha y^2) \end{align} $$

with $f_{2,\alpha }$ defined in equation (3.183). Using this bound in $G_{22}^>$ and recalling Corollary 3.14 and equation (3.184) we thus obtain

(3.195) $$ \begin{align} |{\mathcal G}_{22}^>| \le C_{\delta} \, \alpha^{- 6}. \end{align} $$

In ${\mathcal G}_{22}^<$ , we proceed by inserting equation (3.70) and use equation (3.74) and Lemma 3.15. This gives

(3.196) $$ \begin{align} | {\mathcal G}_{22}^< | \, \le\, \frac{2\sqrt 2}{\alpha^2} \int \mathrm{d} y\, \vert\hspace{-1pt}\vert R_{4,y} \psi \otimes \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}}\, n_{\delta,\eta}(y). \end{align} $$

The derivation of a suitable bound for the norm in the integrand is more cumbersome, so we go through it step by step. To shorten the notation let $G_K^{0<} = \psi \otimes \Upsilon _K^<$ . We start from equations (3.192) and (3.193) where we insert $h_{\cdot } = h_{K,\cdot } + ( h_{\cdot } - h_{K, \cdot } ) $ and use the triangle inequality,

(3.197a) $$ \begin{align} \vert\hspace{-1pt}\vert R_{4,y}^1 G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} & \, \le\, C \vert\hspace{-1pt}\vert ( e^{ - A_{P,y} } -1 ) ({\mathbb N}+1)^{1/2} R^{\frac{1}{2}} \phi ( h_{K,\cdot-y} + \varphi_P ) u_{\alpha} T_{-y} G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.197b) $$ \begin{align} & \quad + \, C \vert\hspace{-1pt}\vert ( e^{ - A_{P,y} } -1 ) ({\mathbb N}+1)^{1/2} R^{\frac{1}{2}} \phi ( h_{\cdot-y} - h_{K, \cdot -y } ) u_{\alpha} T_{-y} G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} , \end{align} $$

(3.197c) $$ \begin{align} \vert\hspace{-1pt}\vert R_{4,y}^2 G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} & \, \le \, C \alpha ( y^2 + \alpha^{-2} ) \vert\hspace{-1pt}\vert ( e^{ - A_{P,y} } -1 ) R^{\frac{1}{2}} \phi ( h_{K,\cdot-y} + \varphi_P )u_{\alpha} T_{-y} G_K^{0<}\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.197d) $$ \begin{align} & \hspace{-0.3cm} + \, C \alpha ( y^2 + \alpha^{-2} ) \vert\hspace{-1pt}\vert ( e^{ - A_{P,y} } -1 ) R^{\frac{1}{2}} u_{\alpha} \phi ( h_{\cdot-y} - h_{K,\cdot-y} ) u_{\alpha} T_{-y} G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}}. \end{align} $$

For the second and fourth line, we apply Lemma 3.9 a second time (after bringing $({\mathbb N}+1)^{1/2}$ to the right of a and $a^{\dagger }$ ) to find

(3.198) $$ \begin{align} (3.197\mathrm{b}) + (3.197\mathrm{d}) & \, \le\, C K^{-1/2} (1 + \alpha y^2) ( \vert\hspace{-1pt}\vert u_{\alpha} T_{-y} P_{\psi} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} + \vert\hspace{-1pt}\vert \nabla u_{\alpha} T_{-y} P_{\psi} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} ) \vert\hspace{-1pt}\vert ({\mathbb N}+1) \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \notag \\[0.5mm] & \, \le\, C K^{-1/2} (1 + \alpha y^2) f_{2,\alpha}(y). \end{align} $$

In the first and third line, we use the functional calculus and write out $A_{P,y}= i P_f y + i g_{P}(y)$ ,

(3.199a) $$ \begin{align} \!\!\!(3.197\mathrm{a}) + (3.197\mathrm{c}) & \le C \vert\hspace{-1pt}\vert (P_f y) ({\mathbb N}+1)^{1/2} R^{\frac{1}{2}} \phi ( h_{K,\cdot-y} + \varphi_P ) u_{\alpha} T_{-y} G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.199b) $$ \begin{align} & \quad + C \alpha (y^2 + \alpha^{-2} ) \vert\hspace{-1pt}\vert (P_f y) R^{\frac{1}{2}} \phi ( h_{K,\cdot-y} + \varphi_P ) u_{\alpha} T_{-y} G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.199c) $$ \begin{align} & \quad + C |g_{P}(y)| \vert\hspace{-1pt}\vert ({\mathbb N}+1)^{1/2} R^{1/2} \phi ( h_{K,\cdot-y} + \varphi_P ) u_{\alpha} T_{-y} G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.199d) $$ \begin{align} & \quad + C \alpha (y^2 + \alpha^{-2} ) |g_{P}(y)| \vert\hspace{-1pt}\vert R^{\frac{1}{2}} \phi ( h_{K,\cdot-y} + \varphi_P ) u_{\alpha} T_{-y} G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}}. \end{align} $$

Now, we use $[iP_f y,\phi (f)]=\pi (y\nabla f)$ such that we can estimate the first line by

(3.200) $$ \begin{align} (3.199\mathrm{a}) &\, \le \, C \big( \vert\hspace{-1pt}\vert ({\mathbb N}+1)^{1/2} R^{1/2} \phi ( h_{K,\cdot-y} + \varphi_P ) (P_f y ) u_{\alpha} T_{-y} G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \nonumber \\[1.5mm] & \quad \quad \quad + \vert\hspace{-1pt}\vert ({\mathbb N}+1)^{1/2} R^{1/2} \pi ( y \nabla h_{K,\cdot-y} + y\nabla \varphi_P )u_{\alpha} T_{-y} G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \big). \end{align} $$

To bound the first line, we use again Lemma 3.9, while in the second line we use $(\nabla h_K)_{\cdot } = - \nabla (h_{K,\cdot }) = - [\nabla , h_{K,\cdot }]$ and (3.44) together with $\vert \hspace {-1pt}\vert \nabla \varphi _P \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^2\)}}} \le C $ for $|P|/ \alpha \le c$ . Together, we obtain

(3.201) $$ \begin{align} (3.199\mathrm{a}) & \, \le\, C |y| \big( \vert\hspace{-1pt}\vert u_{\alpha} T_{-y} P_{\psi} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} + \vert\hspace{-1pt}\vert \nabla u_{\alpha} T_{-y} P_{\psi} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} \big) \big( \vert\hspace{-1pt}\vert ({\mathbb N}+1) P_f \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} + \sqrt K \vert\hspace{-1pt}\vert ({\mathbb N}+1) \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \big)\notag \\[1.5mm] & \, \le\, C \alpha^{\delta}|y| f_{2,\alpha}(y) \big( \vert\hspace{-1pt}\vert P_f \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} + \sqrt K \big) \notag\\[2mm] & \, \le\, C \alpha^{\delta} \sqrt K |y| f_{2,\alpha}(y), \end{align} $$

where the factor $\sqrt K$ in the first step comes from the $L^2$ -norm of $h_{K,0}$ , and the last step follows from Lemma 3.16. In a similar fashion, one shows

(3.202) $$ \begin{align} (3.199\mathrm{b}) \, \le\, C \alpha^{\delta} \sqrt K |y| (1 + \alpha y^2) f_{2,\alpha}(y), \end{align} $$

and, with equation (3.77), one also verifies

(3.203) $$ \begin{align} (3.199\mathrm{c}) + (3.199\mathrm{d}) \, \le \, C \alpha^{\delta} (\alpha^2 |y|^5 + \alpha |y|^3 ) f_{2,\alpha}(y). \end{align} $$

Collecting the estimates (3.198), (3.201), (3.202) and (3.203), we arrive at

(3.204) $$ \begin{align} \vert\hspace{-1pt}\vert R_{4,y} \psi \otimes \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, \le\, C f_{2,\alpha}(y)\alpha^{\delta} \Big( K^{-\frac{1}{2}} ( 1 + \alpha y^2) + \alpha^2 |y|^5 + \sqrt K ( |y| + \alpha |y|^3) \Big). \end{align} $$

Now, we can apply Corollary 3.6 together with equation (3.184) to bound the right side of equation (3.196). The result is

(3.205) $$ \begin{align} | {\mathcal G}_{22}^< | \, \le \, C \alpha^{-2+\delta } \big( K^{-1/2} \alpha^{-3} + \sqrt K \alpha^{-4+4\delta}\big). \end{align} $$

In view of the estimates (3.165), (3.167), (3.177), (3.178), (3.188), (3.189), (3.195) and (3.205), the proof of Proposition 3.20 is now complete.

Proof. We split this contribution into three terms

(3.207) $$ \begin{align} {\mathcal K} & \, = \, \frac{1}{\alpha^2} \int \mathrm{d} y \, \big \langle G_K^1 | \big( h^{\mathrm{Pek}} + \alpha^{-2} {\mathbb N} + \alpha^{-1} \phi(h_{\cdot} + \varphi_P) \big) T_y e^{A_{P,y}} W(\alpha w_{P,y}) G_K^1 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag\\ & \, = \, {\mathcal K}_1 + {\mathcal K}_2 + {\mathcal K}_3, \end{align} $$

and note that ${\mathcal K}_1$ provides the energy contribution of order $\alpha ^{-2}$ .

$\underline {\mathrm {Term}\ {\mathcal K}_1}$ . We start again by writing

(3.208) $$ \begin{align} {\mathcal K}_1 &\, =\, \frac{1}{\alpha^2} \int \mathrm{d} y\, \big \langle G_K^1 | h^{\mathrm{Pek}} T_y W(\alpha w_{P,y}) G_K^1 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag\\ & \quad + \frac{1}{\alpha^2} \int \mathrm{d} y \, \big \langle G_K^1 |h^{\mathrm{Pek}} T_y (e^{A_{P,y}}-1) W(\alpha w_{P,y}) G_K^1 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, =\, {\mathcal K}_{11} + {\mathcal K}_{12}, \end{align} $$

and proceed for the first term similarly as in the computation of ${\mathcal G}_2$ ; see equation (3.168). This leads to

(3.209) $$ \begin{align} {\mathcal K}_{11} & \, =\, \frac{1}{\alpha^2} \int \mathrm{d} y \, \big \langle G_K^0 | \phi( h_{K,\cdot}^1 ) R u_{\alpha} h^{\mathrm{Pek}} T_y W(\alpha w_{P,y}) u_{\alpha} R \phi( h_{K,\cdot}^1 ) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag\\ & \, = \, \frac{1}{\alpha^2} \int \mathrm{d} y \, \big \langle \psi \otimes \Omega| a (\underline{h_{K,\cdot}^1}) R u_{\alpha} h^{\mathrm{Pek}} T_y W(\alpha \widetilde w_{P,y}) u_{\alpha} R a^{\dagger}(\underline {h_{K,\cdot}^1 } ) \psi \otimes \Omega \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag\\ & \, =\, \frac{1}{\alpha^2} \int \mathrm{d} y\, \big \langle G_K^0 | L_{3,y} G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} n_{0,1}(y) - \int \mathrm{d} y \, \ell_3(y) n_{0,1}(y) \, =\, {\mathcal K}_{111} + {\mathcal K}_{112}, \end{align} $$

where

(3.210a) $$ \begin{align} L_{3,y} &\, =\, P_{\psi} \phi({h_{K,\cdot}^1}) R u_{\alpha} h^{\mathrm{Pek}} T_y u_{\alpha} R \phi({h_{K,\cdot}^1 } ) P_{\psi} \end{align} $$

(3.210b) $$ \begin{align} \ell_3(y) & \, = \, \big \langle \psi| \langle \underline{h_{K,\cdot}^1}| \widetilde w_{P,y}^1 \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} R u_{\alpha} h^{\mathrm{Pek}} T_y u_{\alpha} R \langle \widetilde w_{ P,y }^1 | \underline{h_{K,\cdot}^1} \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} \psi \big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}}. \end{align} $$

We go on with

(3.211) $$ \begin{align} {\mathcal K}_{111} & \, =\, \frac{1}{\alpha^2} \int \mathrm{d} y\, \big \langle G_K^0 | L_{3,y} G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, e^{-\lambda \alpha^2 y^2 } \notag\\ & \quad + \frac{1}{\alpha^2} \int \mathrm{d} y\, \big \langle G_K^0 | L_{3,y} G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \big( n_{0,1}(y) - e^{-\lambda \alpha^2 y^2 } \big) = {\mathcal K}_{111}^{\mathrm{lo}} + {\mathcal K}_{111}^{\mathrm{err}}, \end{align} $$

and in the leading-order term, we insert $T_y= 1 + (T_y-1)$ and $u_{\alpha } = 1 + (u_{\alpha } -1 )$ ,

(3.212) $$ \begin{align} {\mathcal K}_{111}^{\mathrm{lo}} & \, =\, \frac{1}{\alpha^2} \big \langle G_K^0 | \phi( h_{K,\cdot}^1 ) R h^{\mathrm{Pek}} R \phi ( h_{K,\cdot}^1 ) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \int \mathrm{d} y\, e^{-\lambda \alpha^2 y^2 } \notag\\ & \quad + \frac{1}{\alpha^2} \big \langle G_K^0 | \phi( h_{K,\cdot}^1 ) R (u_{\alpha}-1) h^{\mathrm{Pek}} R \phi ( h_{K,\cdot}^1 ) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \int \mathrm{d} y\, e^{-\lambda \alpha^2 y^2 } \notag\\ & \quad + \frac{1}{\alpha^2} \big \langle G_K^0 | \phi( h_{K,\cdot}^1 ) R u_{\alpha} h^{\mathrm{Pek}} (u_{\alpha}-1) R \phi ( h_{K,\cdot}^1 ) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \int \mathrm{d} y \, e^{-\lambda \alpha^2 y^2 } \notag\\ & \quad + \frac{1}{\alpha^2} \int \mathrm{d} y \, \big \langle G_K^0 | \phi( h_{K,\cdot}^1 ) R u_{\alpha} h^{\mathrm{Pek}} (T_{y}-1) u_{\alpha} R \phi ( h_{K,\cdot}^1 ) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} e^{-\lambda \alpha^2 y^2 } \notag\\ & \, =\, \sum_{n=1}^4 {\mathcal K}_{111}^{\mathrm{lo,n}}. \end{align} $$

Since $R h^{\mathrm{Pek}} R = R$ , one finds $ {\mathcal K}_{111}^{\mathrm{lo,1}} = - \frac {1}{\alpha ^2} \big \langle \Upsilon _K | ( {\mathbb H}_K - {\mathbb N}_1 ) \Upsilon _K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} (\frac {\pi }{\lambda \alpha ^2})^3$ , cf. equation (2.4), and with the aid of Proposition 3.17, this gives the leading-order contribution

(3.213) $$ \begin{align} \bigg| {\mathcal K}_{111}^{\mathrm{lo,1}} + \mathcal N \frac{1}{\alpha^2} \big \langle \Upsilon_K | ( {\mathbb H}_K - {\mathbb N}_1 ) \Upsilon_K \big \rangle _{\scriptscriptstyle{\text{\(\mathcal F\)}}} \bigg| \, \le\, C_{\varepsilon} \sqrt K \alpha^{ - 6 + \varepsilon}. \end{align} $$

For the other terms, we shall show that

(3.214) $$ \begin{align} | {\mathcal K}_{111}^{\mathrm{lo,2}} | + | {\mathcal K}_{111}^{\mathrm{lo,3}} | + | {\mathcal K}_{111}^{\mathrm{lo,4}} | \le C \sqrt K \alpha^{-6}. \end{align} $$

In the second term, we use $h^{\mathrm{Pek}}R = Q_{\psi } = 1- P_{\psi }$ to write

(3.215) $$ \begin{align} K_{111}^{\mathrm{lo,2}} & \, =\, \alpha^{-2} \big \langle G_K^0 | \phi( h_{K,\cdot}^1 ) R (u_{\alpha}-1) (1 - P_{\psi}) \phi ( h_{K,\cdot}^1 ) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \bigg( \frac{\pi}{\lambda \alpha^2}\bigg)^{3/2} \end{align} $$

which is exponentially small in $\alpha $ , since $\vert \hspace {-1pt}\vert (u_{\alpha }-1) \psi \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^2\)}}} \le C e^{-\alpha / C}$ , and thus with Lemma 3.9 one obtains $|{\mathcal K}_{111}^{\mathrm{lo,2}}| \le C \sqrt K e^{-\alpha / C }$ . In the next term, we use $ [h^{\mathrm{Pek}}, u_{\alpha }-1] = - [\Delta , u_{\alpha }] $ and again $h^{\mathrm{Pek}} R = 1-P_{\psi }$ to get

(3.216) $$ \begin{align} {\mathcal K}_{111}^{\mathrm{lo,3}} &\, =\, \alpha^{-2} \big \langle G_K^0 | \phi( h_{K,\cdot}^1 ) R u_{\alpha} (u_{\alpha}-1) (1 - P_{\psi} )\phi ( h_{K,\cdot}^1 ) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \bigg( \frac{\pi}{\lambda \alpha^2}\bigg)^{3/2} \notag\\ & \quad - \alpha^{-2} \big \langle G_K^0 | \phi( h_{K,\cdot}^1 ) R [\Delta , u_{\alpha}] R \phi ( h_{K,\cdot}^1 ) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \bigg( \frac{\pi}{\lambda \alpha^2}\bigg)^{3/2}. \end{align} $$

Here, the first line is bounded again exponentially in $\alpha $ , whereas in the second line we use $[\Delta ,u_{\alpha }] = 2 (\nabla u_{\alpha }) \nabla + (\Delta u_{\alpha })$ and $\vert \hspace {-1pt}\vert \nabla u_{\alpha } \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^{\infty }\)}}} + \vert \hspace {-1pt}\vert \Delta u_{\alpha } \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(L^{\infty }\)}}} \le C\alpha ^{-1}$ , see (2.20). Together with Lemmas 3.8 and 3.9, this implies $ |{\mathcal K}_{111}^{\mathrm{lo,3}}| \le C \alpha ^{- 6 }$ . In the last term, we employ $T_y-1 = \int _0^1 \mathrm {d} s T_{sy} (y\nabla )$ , $[h^{\mathrm{Pek}}, u_{\alpha }] = - [\Delta , u_{\alpha }]$ and $h^{\mathrm{Pek}}R = Q_{\psi }$ to find

(3.217) $$ \begin{align} {\mathcal K}_{111}^{\mathrm{lo,4}} & = \alpha^{-2} \int \mathrm{d} y\, \int_0^1 \mathrm{d} s\, \big \langle G_K^0| \phi( h_{K,\cdot}^1 ) Q_{\psi} u_{\alpha} T_{sy} (y\nabla) u_{\alpha} R \phi ( h_{K,\cdot}^1 ) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} e^{-\lambda \alpha^2 y^2 } \notag\\ & \quad + \, \alpha^{-2} \int \mathrm{d} y\, \int_0^1 \mathrm{d} s\, \big \langle G_K^0 | \phi( h_{K,\cdot}^1 ) R [\Delta, u_{\alpha}] T_{sy} (y\nabla) u_{\alpha} R \phi ( h_{K,\cdot}^1 ) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} e^{-\lambda \alpha^2 y^2 }. \end{align} $$

In both lines, there is an additional factor y, and together with equation (2.20), we thus obtain

(3.218) $$ \begin{align} |{\mathcal K}_{111}^{\mathrm{lo,4}}| & \le C\alpha^{-6} \vert\hspace{-1pt}\vert \phi( h_{K,\cdot}^1 ) G_K^0 \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, \vert\hspace{-1pt}\vert \nabla u_{\alpha} R^{1/2}\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} \vert\hspace{-1pt}\vert R^{1/2} \phi( h_{K,\cdot}^1 ) G_K^0 \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag \\[2mm] & \quad + C\alpha^{-6} \vert\hspace{-1pt}\vert R^{1/2} \phi( h_{K,\cdot}^1 ) G_K^0 \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \vert\hspace{-1pt}\vert R^{1/2} [\Delta,u_{\alpha}]\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} \vert\hspace{-1pt}\vert \nabla u_{\alpha} R^{1/2}\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} \vert\hspace{-1pt}\vert R \phi( h_{K,\cdot}^1 ) G_K^0 \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag \\[2mm] & \le C ( \alpha^{-6} \sqrt K + \alpha^{-7}). \end{align} $$

This proves (3.214).

To estimate ${\mathcal K}_{112}$ and ${\mathcal K}_{111}^{\mathrm{err}}$ , we make use of

(3.219a) $$ \begin{align} \big| \big \langle G_K^0 | L_{3,y} G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \big| & \, \le\, C f_{3,\alpha}(y) \end{align} $$

(3.219b) $$ \begin{align} |\ell_3(y) | &\, \le \, C f_{3,\alpha}(y) (y^4 + \alpha^{-4}), \end{align} $$

where

(3.220) $$ \begin{align} f_{3,\alpha}(y) & = \vert\hspace{-1pt}\vert u_{\alpha} T_{y} u_{\alpha} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} + \vert\hspace{-1pt}\vert (\nabla u_{\alpha} ) T_{y} u_{\alpha} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} + \vert\hspace{-1pt}\vert u_{\alpha} T_{y} (\nabla u_{\alpha}) \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} + \vert\hspace{-1pt}\vert (\nabla u_{\alpha}) T_{y} (\nabla u_{\alpha}) \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}}. \end{align} $$

Recalling that by definition for $k=0,1$ , it follows that and thus

(3.221) $$ \begin{align} \vert\hspace{-1pt}\vert f_{3,\alpha}\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^{\infty}\)}}}\le 4 \quad \text{and} \quad \vert\hspace{-1pt}\vert |\cdot|^n f_{3,\alpha} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^1\)}}} \le C_n \alpha^{3+n}\quad \text{for all}\ n\in {\mathbb N}_0. \end{align} $$

In order to verify equation (3.219a), use $h^{\mathrm{Pek}} = -\Delta + V^{\varphi } - \lambda ^{\mathrm{Pek}}$ to write

(3.222) $$ \begin{align} R^{\frac{1}{2}} u_{\alpha} T_{y} h^{\mathrm{Pek}} u_{\alpha} R^{\frac{1}{2}} \, & =\, R^{\frac{1}{2}} u_{\alpha} \big( (-i\nabla) T_y (-i\nabla) + T_y ( V^{\varphi} - \lambda^{\mathrm{Pek}})\big) u_{\alpha} R^{\frac{1}{2}} \notag \\[1mm] & = - R^{\frac{1}{2}} (-i\nabla u_{\alpha}) T_y (-i \nabla u_{\alpha}) R^{\frac{1}{2}} + R^{\frac{1}{2}} (-i\nabla) u_{\alpha} T_y u_{\alpha} (-i\nabla) R^{\frac{1}{2}} \notag \\[1mm] & \quad + R^{\frac{1}{2}} (-i\nabla) u_{\alpha} T_y (-i \nabla u_{\alpha}) R^{\frac{1}{2}} - R^{\frac{1}{2}} (-i \nabla u_{\alpha}) T_y u_{\alpha} (-i\nabla) R^{\frac{1}{2}} \notag \\[1mm] & \quad + R^{\frac{1}{2}} u_{\alpha} T_y u_{\alpha} (V^{\varphi} - \lambda^{\mathrm{Pek}}) R^{\frac{1}{2}}. \end{align} $$

Since $\vert \hspace {-1pt}\vert V^{\varphi } R^{1/2} \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} \le C (\vert \hspace {-1pt}\vert R\vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} + \vert \hspace {-1pt}\vert \nabla R^{1/2} \vert \hspace {-1pt}\vert _{\scriptscriptstyle{\text{\(\mathrm{op}\)}}}) \le C$ (see Lemma 3.8), it thus follows that

(3.223) $$ \begin{align} \vert\hspace{-1pt}\vert R^{\frac{1}{2}} u_{\alpha} T_{y} h^{\mathrm{Pek}} u_{\alpha} R^{\frac{1}{2}} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} \, \le \, C f_{3,\alpha}(y), \end{align} $$

With this at hand, one applies Lemma 3.9 to conclude the bound stated in equation (3.219a). For $\ell _3(y)$ , we proceed similarly as in equation (3.187), that is,

(3.224) $$ \begin{align} | \ell_3(y) | & \, \le \, \vert\hspace{-1pt}\vert R^{1/2} u_{\alpha} h^{\mathrm{Pek}} T_y u_{\alpha} R^{1/2} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} \vert\hspace{-1pt}\vert R^{1/2} \langle \widetilde w_{P,y}^1 | \underline{h_{K,\cdot}^1} \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} \psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}}^2 \notag\\[1mm] &\, \le \, f_{3,\alpha}(y) \, \vert\hspace{-1pt}\vert \widetilde w_{P,y}^1 \vert\hspace{-1pt}\vert^2_{\scriptscriptstyle{\text{\(L^2\)}}} \, \int \mathrm{d} z \vert\hspace{-1pt}\vert P_{\psi} \underline{h_{K,\cdot}^1}(z) R^{1/2} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}}^2 \, \le \, C f_{3,\alpha}(y) (y^4 + \alpha^{-4}). \end{align} $$

Now, we can apply Lemma 3.5 and (3.221) to estimate

(3.225) $$ \begin{align} | {\mathcal K}_{111}^{\mathrm{err}} |\, \le\, \frac{C}{\alpha^2} \int \mathrm{d} y\, f_{3,\alpha}(y)\, | n_{0,1}(y) - e^{-\lambda \alpha^2 y^2 } | \, \le \, C \alpha^{-6}, \end{align} $$

and further invoke Corollary 3.6 to obtain

(3.226) $$ \begin{align} | {\mathcal K}_{112} | \, \le\, \int \mathrm{d} y\, f_{3,\alpha}(y) (|y|^4 + \alpha^{-4}) n_{0,1}(y) \le C \alpha^{-7}. \end{align} $$

Next, we come to ${\mathcal K}_{12}$ which we rewrite with the aid of equations (3.67) and (3.80) as

(3.227) $$ \begin{align} {\mathcal K}_{12} \, =\, \frac{1}{\alpha^2} \int \mathrm{d} y\, \big \langle R_{5,y} \psi \otimes \big( \Upsilon_K^< + \Upsilon_K^> \big) | W(\alpha w_{P,y}) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} = {\mathcal K}_{12}^< + {\mathcal K}_{12}^> \end{align} $$

with the operator $ R_{5,y} = R_{5,y}^1 + R_{5,y}^2$ and

(3.228a) $$ \begin{align} R_{5,y}^1 & \, =\, P_{\psi} \phi( h_{K,\cdot}^1 ) R u_{\alpha} ( e^{ - A_{P,y} } -1 ) T_{-y} h^{\mathrm{Pek}} u_{\alpha} R \phi ( h_{K,\cdot}^1 ) P_{\psi} \end{align} $$

(3.228b) $$ \begin{align} R_{5,y}^2 & \, =\, 2 \alpha P_{\psi} \big \langle h_{K,\cdot} |\operatorname{Re} ( w_{P,y}^1) \big \rangle _{\scriptscriptstyle{\text{\(L^2\)}}} R u_{\alpha} ( e^{ - A_{P,y} } -1 ) T_{-y} h^{\mathrm{Pek}} u_{\alpha} R \phi ( h_{K,\cdot}^1 ) P_{\psi}. \end{align} $$

Utilizing Lemma 3.9 and (3.32a), we have

(3.229) $$ \begin{align} \vert\hspace{-1pt}\vert R_{5,y}^1 \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} & \, \le\, C \vert\hspace{-1pt}\vert ( e^{ - A_{P,y} } -1 ) ({\mathbb N}+1)^{1/2} R^{\frac{1}{2}} u_{\alpha} T_{-y} h^{\mathrm{Pek}} u_{\alpha} R \phi ( h_{K,\cdot}^1 ) P_{\psi} \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}}, \end{align} $$

and following the same steps as in equation (3.84),

(3.230) $$ \begin{align} \vert\hspace{-1pt}\vert R_{5,y}^2 \Psi \vert\hspace{-1pt}\vert & \, \le\, C \alpha ( y^2 + \alpha^{-2}) \vert\hspace{-1pt}\vert ( e^{ - A_{P,y} } -1 ) R^{\frac{1}{2}} u_{\alpha} T_{-y} h^{\mathrm{Pek}} u_{\alpha} R \phi ( h_{K,\cdot}^1 ) P_{\psi} \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}}. \end{align} $$

After using unitarity of $e^{-A_{P,y}}$ and equation (3.221), we can apply Lemma 3.9 another time to obtain

(3.231) $$ \begin{align} \vert\hspace{-1pt}\vert R_{5,y} \psi\otimes \Upsilon_K^> \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} & \, \le\, C f_{3,\alpha}(-y) (1+\alpha y^2) \vert\hspace{-1pt}\vert ({\mathbb N}+1) \Upsilon_K^> \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}}. \end{align} $$

Thus, we can estimate the tail with the aid of Corollary 3.14 and equation (3.221),

(3.232) $$ \begin{align} | {\mathcal K}_{12}^>| \, \le\, \frac{C}{\alpha^2} \vert\hspace{-1pt}\vert ({\mathbb N}+1) \Upsilon_K^> \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \int \mathrm{d} y \, f_{3,\alpha}(-y) (1+\alpha y^2)\, \le\, C_{\delta}\, \alpha^{- 6 }. \end{align} $$

Then we use equation (3.63), equaiton (3.74) and apply Lemma 3.15 to get

(3.233) $$ \begin{align} | {\mathcal K}_{12}^< |& \, \le \, \frac{1}{\alpha^2} \int \mathrm{d} y\, \vert\hspace{-1pt}\vert \mathbb U_K e^{\kappa {\mathbb N}} R_{5,y} \psi \otimes \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \vert\hspace{-1pt}\vert e^{-\kappa {\mathbb N}} W(\alpha \widetilde w_{P,y}) \Omega \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \notag \\ & \, \le\, \frac{\sqrt 2}{\alpha^2} \int \mathrm{d} y\, \vert\hspace{-1pt}\vert R_{5,y} \psi \otimes \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, n_{\delta,\eta}(y). \end{align} $$

To bound the norm in the integral, we proceed in close analogy to the steps following equation (3.196). We abbreviate again $G_K^{0<} = \psi \otimes \Upsilon _K^<$ and start from equations (3.229) and (3.230). With equation (3.221), the functional calculus and $A_{P,y} = iP_fy + i g_{P}(y)$ , one finds

(3.234a) $$ \begin{align} \vert\hspace{-1pt}\vert R_{5,y} G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} &\, \le\, C \big( f_{3,\alpha}(-y) \vert\hspace{-1pt}\vert ( e^{ - A_{P,y} } -1 ) ({\mathbb N}+1)^{1/2} R^{\frac{1}{2}} \phi ( h_{K,\cdot}^1 ) G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \notag \\[1mm] & \quad + \alpha ( y^2 + \alpha^{-2})f_{3,\alpha}(-y)\vert\hspace{-1pt}\vert ( e^{ - A_{P,y} } -1 ) R^{\frac{1}{2}} \phi ( h_{K,\cdot}^1 ) G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \big) \notag \\[1mm] & \, \le \, C \big( f_{3,\alpha}(-y) \big( \vert\hspace{-1pt}\vert (yP_f) ({\mathbb N}+1)^{1/2} R^{\frac{1}{2}} \phi ( h_{K,\cdot}^1 ) G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.234b) $$ \begin{align} & \quad + f_{3,\alpha}(-y) |g_{P}(y)| ({\mathbb N}+1)^{1/2} R^{\frac{1}{2}} \phi ( h_{K,\cdot}^1 ) G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.234c) $$ \begin{align} & \quad + f_{3,\alpha}(-y) ( \alpha y^2 + \alpha^{-1}) \vert\hspace{-1pt}\vert (P_f y) R^{\frac{1}{2}} \phi ( h_{K,\cdot}^1 ) G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.234d) $$ \begin{align} & \quad + f_{3,\alpha}(-y) ( \alpha y^2 + \alpha^{-1}) |g_{P}(y)| \vert\hspace{-1pt}\vert R^{\frac{1}{2}} \phi ( h_{K,\cdot}^1 ) G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \big). \end{align} $$

In the second and fourth line, we use $|g_{P}(y)| \le C\alpha |y|^3$ and Lemma 3.9,

(3.235) $$ \begin{align} (3.234\mathrm{b}) + (3.234\mathrm{d}) & \, \le\, C (\alpha^2 |y|^5 + \alpha |y|^3) f_{3,\alpha}(-y) \vert\hspace{-1pt}\vert ({\mathbb N}+1) \Upsilon_K^<\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \notag\\ & \, \le\, C (\alpha^2 |y|^5 + \alpha |y|^3) f_{3,\alpha}(-y). \end{align} $$

In the first and third line, we employ the commutator $[iP_f y,\phi (f)]=\pi (y\nabla f)$ to get

(3.236a) $$ \begin{align} (3.234\mathrm{a}) + (3.234\mathrm{c}) &\, \le\, C \big( f_{3,\alpha}(-y) \vert\hspace{-1pt}\vert ({\mathbb N}+1)^{1/2} R^{\frac{1}{2}} \phi ( h_{K,\cdot}^1 ) (yP_f) G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.236b) $$ \begin{align} & \quad + f_{3,\alpha}(-y) \vert\hspace{-1pt}\vert ({\mathbb N}+1)^{1/2} R^{\frac{1}{2}} \pi (y \nabla h_{K,\cdot}^1 ) G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.236c) $$ \begin{align} & \quad + f_{3,\alpha}(-y) ( \alpha y^2 + \alpha^{-1}) \vert\hspace{-1pt}\vert R^{\frac{1}{2}} \phi ( h_{K,\cdot}^1 ) (yP_f) G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

(3.236d) $$ \begin{align} & \quad + f_{3,\alpha}(-y) ( \alpha y^2 + \alpha^{-1}) \vert\hspace{-1pt}\vert R^{\frac{1}{2}} \pi (y \nabla h_{K,\cdot}^1 ) G_K^{0<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \big). \end{align} $$

After another application of Lemma 3.9, we can use equation (3.67) and then Lemma 3.16 for the terms involving $P_f$ ,

(3.237) $$ \begin{align} (3.236\mathrm{a}) + (3.236\mathrm{c}) \, & \le\, C f_{3,\alpha}(-y) (\alpha y^2 + 1)\, |y|\, \vert\hspace{-1pt}\vert ({\mathbb N}+1) P_f \Upsilon_K^<\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \notag \\[1mm] & \le \, C f_{3,\alpha}(-y) (\alpha |y|^3 + |y|) \alpha^{\delta} \sqrt K ,, \end{align} $$

while in the other two lines, we use $(\nabla h_{K})_{\cdot } = - [\nabla ,h_{K,\cdot }]$ , to obtain

(3.238) $$ \begin{align} (3.236\mathrm{b}) + (3.236\mathrm{d}) & \, \le \, C f_{3,\alpha}(-y)\, |y|\, (\alpha y^2 + 1) \vert\hspace{-1pt}\vert h_{K,0}\vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(L^2\)}}} \vert\hspace{-1pt}\vert ({\mathbb N}+1) \Upsilon_K^{<} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \notag \\[1mm] & \, \le \, C f_{3,\alpha}(-y)(\alpha |y|^3 + |y|)\sqrt K. \end{align} $$

Collecting all estimates we have thus shown that

(3.239) $$ \begin{align} \vert\hspace{-1pt}\vert R_{5,y} \psi \otimes \Upsilon_K^< \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, \le\, C f_{3,\alpha}(-y) \alpha^{\delta} \Big( \alpha^2 |y|^5 + \sqrt K ( \alpha |y|^3 + |y| ) \Big). \end{align} $$

Using this bound in equation (3.233), we can invoke Corollary 3.6 together with equation (3.221) in order to obtain

(3.240) $$ \begin{align} |{\mathcal K}_{12}^<| \le C \sqrt K \alpha^{-6+5\delta}. \end{align} $$

$\underline {\mathrm {Term}\ {\mathcal K}_2}$ . Using equations (3.67) and (3.80), one finds

(3.241) $$ \begin{align} {\mathcal K}_2 & \, =\, \frac{1}{\alpha^{4}} \int \mathrm{d} y \, \big \langle R_{6,y} \psi \otimes \big( \Upsilon_K^< + \Upsilon_K^> \big) | W(\alpha w_{P,y}) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, =\, {\mathcal K}_{2}^< + {\mathcal K}_{2}^> \end{align} $$

with the operator $R_{6,y} = R_{6,y}^1 + R_{6,y}^2$ and

(3.242a) $$ \begin{align} R_{6,y}^1 & \, =\, P_{\psi} \phi(h_{K,\cdot}^1) R u_{\alpha} {\mathbb N} T_{-y} e^{-A_{P,y}} u_{\alpha} R \phi(h_{K,\cdot}^1) P_{\psi} \end{align} $$

(3.242b) $$ \begin{align} R_{6,y}^2 & \, =\, 2 \alpha P_{\psi} \phi(h_{K,\cdot}^1) R u_{\alpha} {\mathbb N} T_{-y} e^{-A_{P,y}} u_{\alpha} R \langle \operatorname{Re} (w_{P,y}^1) | h_{K,\cdot} \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} P_{\psi}. \end{align} $$

With Lemma 3.9 and equation (3.32a) it is not difficult to verify

(3.243) $$ \begin{align} \vert\hspace{-1pt}\vert R_{6,y} \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} & \, \le\, C \vert\hspace{-1pt}\vert u_{\alpha} T_{-y} u_{\alpha} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathrm{op}\)}}} (1+ \alpha y^2) \vert\hspace{-1pt}\vert ({\mathbb N}+1)^2 \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} , \end{align} $$

and since , we can use Corollary 3.14 to estimate the part with the tail by

(3.244)

To treat ${\mathcal K}_{2}^<$ we proceed as in (3.233), that is

(3.245)

It now follows from Corollary 3.6 that

(3.246) $$ \begin{align} | {\mathcal K}_{2}^<| \le C \alpha^{-7}. \end{align} $$

$\underline {\mathrm {Term}\ {\mathcal K}_3}$ . This term is similarly estimated as the previous one. With the aid of equations (3.67) and (3.80), we have

(3.247) $$ \begin{align} {\mathcal K}_3 & \, =\, \frac{1}{\alpha^3} \int \mathrm{d} y \, \big \langle R_{7,y} \psi \otimes \big( \Upsilon_K^< + \Upsilon_K^> \big) | W(\alpha w_{P,y}) G_K^0 \big \rangle _{\scriptscriptstyle{\text{\(\mathscr H\)}}} \, =\, {\mathcal K}_{3}^< + {\mathcal K}_{3}^> \end{align} $$

with the operator $R_{7,y} = R_{7,y}^1 + R_{7,y}^2$ and

(3.248a) $$ \begin{align} R_{7,y}^1 & \, =\, P_{\psi} \phi(h_{K,\cdot}^1) R u_{\alpha} e^{-A_{P,y}} T_{-y} \phi(h_{\cdot} + \varphi_P) u_{\alpha} R \phi(h_{K,\cdot}^1) P_{\psi} \end{align} $$

(3.248b) $$ \begin{align} R_{7,y}^2 & \, =\, 2 \alpha P_{\psi} \langle \operatorname{Re} ( w_{P,y}^1 ) | h_{K,\cdot} \rangle_{\scriptscriptstyle{\text{\(L^2\)}}} R u_{\alpha} e^{-A_{P,y}} T_{-y} \phi(h_{\cdot} + \varphi_P) u_{\alpha} R \phi(h_{K,\cdot}^1) P_{\psi}. \end{align} $$

Utilizing again Lemma 3.9 and equation (3.32a), one shows that

(3.249) $$ \begin{align} \vert\hspace{-1pt}\vert R_{7,y} \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} & \, \le\, C f_{3,\alpha}(-y) (1 + \alpha y^2) \vert\hspace{-1pt}\vert ({\mathbb N}+1)^{3/2} \Psi \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}} \end{align} $$

with $f_{3,\alpha }$ defined in (3.220). Invoking Corollary 3.14 and equation (3.221), we thus find

(3.250) $$ \begin{align} | {\mathcal K}_{3}^>| \, \le\, \frac{C}{\alpha^3} \vert\hspace{-1pt}\vert ({\mathbb N}+1)^{3/2} \Upsilon_K^{>} \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathcal F\)}}} \int \mathrm{d} y\, f_{3,\alpha}(-y) (1 + \alpha y^2) \, \le\, C_{\delta}\, \alpha^{- 7 }. \end{align} $$

Similarly, as in equation (3.233), we also obtain

(3.251) $$ \begin{align} | {\mathcal K}_{3}^<| \, \le\, \frac{\sqrt 2}{\alpha^{3}} \int \mathrm{d} y \, \vert\hspace{-1pt}\vert R_{7,y} \psi \otimes \Upsilon_K^{ < } \vert\hspace{-1pt}\vert_{\scriptscriptstyle{\text{\(\mathscr H\)}}}\, n_{\delta,\eta}(y)\, \le\, \frac{C}{\alpha^3} \int \mathrm{d} y\, f_{3,\alpha}(-y)(1 + \alpha y^2) n_{\delta,\eta}(y). \end{align} $$

By Corollary 3.6 and equation (3.220) it follows that

(3.252) $$ \begin{align} | {\mathcal K}_{3}^<| \le C \alpha^{-6 + 3 \delta}. \end{align} $$

This completes the analysis of ${\mathcal K}$ . The proof of Proposition 3.21 follows from combining equations (3.213), (3.214), (3.225), (3.226), (3.232), (3.240), (3.244), (3.246), (3.250) and (3.252).