In this section we calculate the expectation value \(\langle \xi , \mathcal {H}_{{\mathrm{corr}}} \xi \rangle \), for the trial state \(\xi \) defined in (4.16) and \(\mathcal {H}_{\mathrm{corr}}\) defined in (3.8). We start with some simple estimates for the non-bosonizable terms. Afterwards we linearize the kinetic energy and calculate its contribution to the expectation value, before we eventually turn to the main part of the interaction.
Getting rid of non-bosonizable terms
In the next lemma, we show that the contribution of the terms in (3.6) to the expectation \(\langle \xi , \mathcal {H}_{{\mathrm{corr}}} \xi \rangle \) is negligible for \(N \rightarrow \infty \).
Lemma 5.1
(Non-Bosonizable Interaction Terms). Let \(\mathcal {E}_1 (x,y)\) be defined as in (3.6). Let \(\xi \) be the trial state defined as in (4.16). Then we have
$$\begin{aligned} \Big | \Big \langle \xi , \frac{1}{2N}\int _{\mathbb {T}^3\times \mathbb {T}^3} {{\mathrm{d}}}x{{\mathrm{d}}}y\, V(x-y) \,\mathcal {E}_1(x,y)\, \xi \Big \rangle \Big | \le C N^{-1}. \end{aligned}$$
Proof
We are going to show that for all \(\psi \in \mathcal {F}\) we have
$$\begin{aligned} \Big | \Big \langle \psi , \frac{1}{2N}\int _{\mathbb {T}^3\times \mathbb {T}^3} {{\mathrm{d}}}x{{\mathrm{d}}}y\, V(x-y) \,\mathcal {E}_1(x,y)\, \psi \Big \rangle \Big | \le \frac{2}{N} \sum _{k \in \mathbb {Z}^3} |\hat{V}(k)|\, \langle \psi , (\mathcal {N}+1)^2\psi \rangle . \end{aligned}$$
(5.1)
The final claim then follows using Proposition 4.6. To prove (5.1), let us rewrite the first term on the r. h. s. of (3.6) by using the CAR and \(\langle u_x,u_y\rangle = u(x,y)\), yielding
$$\begin{aligned}&\frac{1}{2N} \int _{\mathbb {T}^3\times \mathbb {T}^3} {{\mathrm{d}}}x{{\mathrm{d}}}y\, V(x-y) a^*(u_x) a^*(u_y) a(u_y) a(u_x)\nonumber \\&\quad = \frac{1}{2N} \int _{\mathbb {T}^3\times \mathbb {T}^3} {{\mathrm{d}}}x{{\mathrm{d}}}y\, V(x-y) \Big ( a^*(u_x) a(u_x) a^*(u_y) a(u_y) - a^*(u_x) \langle u_x,u_y\rangle a(u_y) \Big )\nonumber \\&\quad = \frac{1}{2N} \sum _{k\in \mathbb {Z}^3} \hat{V}(k) \Big ( {{\mathrm{d}}}\Gamma (u e^{ikx}u){{\mathrm{d}}}\Gamma (u e^{-ikx} u) - {{\mathrm{d}}}\Gamma (u e^{ikx} u e^{-ikx} u) \Big )\;. \end{aligned}$$
(5.2)
Recall the two bounds \(||{{\mathrm{d}}}\Gamma (A) \psi || \le ||A||_{{\mathrm{op}}}||\mathcal {N}\psi ||\) and \(|\langle \psi , {{\mathrm{d}}}\Gamma (A) \psi \rangle |\le ||A||_{{\mathrm{op}}}\langle \psi , \mathcal {N}\psi \rangle \) for any bounded one-particle operator A and any \(\psi \in \mathcal {F}\). Thus, using that \(\Vert u \Vert _{\text {op}} \le 1\),
$$\begin{aligned} \Big | \Big \langle \xi , \frac{1}{2N} \sum _{k\in \mathbb {Z}^3} \hat{V}(k) {{\mathrm{d}}}\Gamma (u e^{ikx}u){{\mathrm{d}}}\Gamma (u e^{-ikx} u) \xi \Big \rangle \Big | \le \frac{1}{2N} \sum _{k\in \mathbb {Z}^3} |\hat{V}(k)|||\mathcal {N}\xi ||^2. \end{aligned}$$
The second summand in (5.2) can be estimated in the same way. The same holds true for the other two terms in (3.6). \(\quad \square \)
Let us now consider the error term \(\mathcal {E}_{2}\), defined in (3.7). We prove that this term vanishes in our trial state \(\xi \).
Lemma 5.2
(Interaction Terms of Wrong Parity). Let \(\mathcal {E}_{2}(x,y)\) be defined as in (3.7). Let \(\xi \) be the trial state defined in (4.16). Then we have
$$\begin{aligned} \Big \langle \xi , \frac{1}{2N}\int _{\mathbb {T}^3\times \mathbb {T}^3} {{\mathrm{d}}}x{{\mathrm{d}}}y\, V(x-y) \big ( \mathcal {E}_2(x,y) + {\mathrm{h.c.}}\big ) \xi \Big \rangle =0. \end{aligned}$$
Proof
Since terms in \(\mathcal {E}_2(x,y)\) create exactly two fermions, we have
$$\begin{aligned} i^\mathcal {N}\mathcal {E}_2(x,y) = \mathcal {E}_2(x,y) i^{\mathcal {N}+2} = -\mathcal {E}_2(x,y) i^\mathcal {N}. \end{aligned}$$
Recall that \(\xi = T \Omega \), with \(T= \exp (B)\) and B as in (4.20). We have \([i^\mathcal {N},B] =0\), since B creates or annihilates particles four at a time. This implies \(T i^\mathcal {N}= i^\mathcal {N}T\). Using \(i^\mathcal {N}\Omega = \Omega \), we get
$$\begin{aligned} \langle T\Omega , \mathcal {E}_2 T \Omega \rangle&= \langle T \Omega , \mathcal {E}_2 T i^{\mathcal {N}} \Omega \rangle = -\langle T \Omega , i^\mathcal {N}\mathcal {E}_2 T\Omega \rangle = - \langle (-i)^\mathcal {N}T\Omega , \mathcal {E}_2 T\Omega \rangle \\&= - \langle T (-i)^\mathcal {N}\Omega , \mathcal {E}_2 T\Omega \rangle = - \langle T\Omega , \mathcal {E}_2 T\Omega \rangle \;, \end{aligned}$$
which thus vanishes. \(\quad \square \)
Estimating direct and exchange operators
In this section we estimate the contribution of the direct and exchange terms to \({{\mathrm{d}}}\Gamma (uhu-\overline{v}\overline{h}v)\). Recall that
$$\begin{aligned} h = -\frac{\hbar ^2 \Delta }{2} + (2\pi )^3 \hat{V} (0) + X \end{aligned}$$
where X has the integral kernel \(X(x,y) = -N^{-1} V(x-y) \omega _{{\mathrm{pw}}}(x,y)\). The contribution of the constant direct term \((2\pi )^3 \hat{V} (0)\) is
$$\begin{aligned} (2\pi )^3 \hat{V} (0) \, {{\mathrm{d}}}\Gamma (u^2 - \overline{v} v) = (2\pi )^3 \hat{V} (0) {{\mathrm{d}}}\Gamma (\mathbb {I}-2 \omega _{\mathrm{pw}}) = (2\pi )^3 \hat{V} (0) (\mathcal {N}_{\mathrm{p}} - \mathcal {N}_{\mathrm{h}}) \end{aligned}$$
and therefore it vanishes on \(\xi \) by Lemma 4.3. The next lemma allows us to control the contribution of the exchange term X.
Lemma 5.3
(Bound for the Exchange Term). Let \(\xi \) be the trial state defined as in (4.16). Then we have
$$\begin{aligned} |\langle \xi , {{\mathrm{d}}}\Gamma (uXu-\overline{v}\overline{X}v) \xi \rangle |\le C N^{-1}. \end{aligned}$$
Proof
Notice that
$$\begin{aligned} \omega _{\mathrm{pw}}(x,y) = \frac{1}{(2\pi )^3}\sum _{h \in B_{\mathrm{F}}} e^{ih\cdot (x-y)} =: f (x-y). \end{aligned}$$
Thus X is translation invariant, and hence
$$\begin{aligned} ||X||_\text {op} = N^{-1} || \widehat{Vf}||_{L^\infty } \le N^{-1} || \hat{f} ||_{L^\infty } \sum _{k \in \mathbb {Z}^3} |\hat{V}(k)|\, \le C N^{-1}. \end{aligned}$$
Using that \(||u||_{{\mathrm{op}}}= 1= ||v||_{{\mathrm{op}}}\), we get, by Proposition 4.6:
$$\begin{aligned} |\langle \xi , {{\mathrm{d}}}\Gamma (uXu-\overline{v}\overline{X}v) \xi \rangle |\le ||uXu-\overline{v}\overline{X}v||_{{\mathrm{op}}}\langle \xi ,\mathcal {N}\xi \rangle \le C N^{-1}. \end{aligned}$$
\(\square \)
Expectation value of the kinetic energy
In this section we evaluate the contribution of the Laplacian to the expectation value of the correlation Hamiltonian in the trial state \(\xi \) defined as in (4.16). We start by linearizing in Fourier space,
$$\begin{aligned} -\frac{\hbar ^2}{2} \langle \xi , {{\mathrm{d}}}\Gamma \left( u \Delta u - \overline{v} \Delta v\right) \xi \rangle&= \frac{\hbar ^2}{2}\Bigg \langle \xi , \Big [ \sum _{p \in B_{\mathrm{F}}^c} p^2 a^*_p a_p - \sum _{h \in B_{\mathrm{F}}} h^2 a^*_h a_h \Big ] \xi \Bigg \rangle \\&= \frac{\hbar ^2}{2}\Bigg \langle \xi , \sum _{\alpha =1}^M \Big [ \sum _{p \in B_{\mathrm{F}}^c \cap B_\alpha } \left( (p-\omega _\alpha )^2 + 2p\cdot \omega _\alpha - \omega _\alpha ^2\right) a^*_p a_p\\&\qquad \quad \qquad - \sum _{h \in B_{\mathrm{F}} \cap B_\alpha } \left( (h-\omega _\alpha )^2+ 2h \cdot \omega _\alpha -\omega _\alpha ^2\right) a^*_h a_h \Big ] \xi \Bigg \rangle . \end{aligned}$$
Notice that from the first to the second line, momenta p and h that lie in the corridors or are more than a distance R away from the Fermi surface have disappeared from the sums; this is justified since such modes are never occupied in the trial state, i. e., \(a_p \xi = 0\) and \(a_h\xi =0\). Furthermore, thanks to Lemma 4.3, we have
$$\begin{aligned} \left\langle \xi , \sum _{\alpha =1}^M \left[ \sum _{p \in B_{\mathrm{F}}^c \cap B_\alpha } \omega _\alpha ^2 a^*_p a_p- \sum _{h \in B_{\mathrm{F}} \cap B_\alpha } \omega _\alpha ^2 a^*_h a_h\right] \xi \right\rangle = k_{\mathrm{F}}^2 \, \left\langle \xi , \left[ \mathcal {N}_{\mathrm{p}}- \mathcal {N}_{\mathrm{h}}\right] \xi \right\rangle = 0 \end{aligned}$$
where we used that \(|\omega _\alpha |= k_{\mathrm{F}}\) for all \(\alpha \). To estimate \((p-\omega _\alpha )^2\) and \((h-\omega _\alpha )^2\), we recall that the diameter of the patches is bounded by \(C \sqrt{N^{2/3}/M}\) (since the diameter of the patch on the Fermi surface is bounded by \(\sqrt{N^{2/3}/M}\) which is large compared to its thickness of order R). Therefore
$$\begin{aligned} \begin{aligned}&-\frac{\hbar ^2}{2} \langle \xi , {{\mathrm{d}}}\Gamma \left( u \Delta u - \overline{v} \Delta v\right) \xi \rangle = \langle \xi , \mathbb {H}_{\mathrm{kin}} \xi \rangle +\mathfrak {E}_{\mathrm{lin}} \end{aligned} \end{aligned}$$
where we introduced
$$\begin{aligned} \mathbb {H}_{\mathrm{kin}} := {\hbar ^2} \sum _{\alpha =1}^M \Big [ \sum _{p \in B_{\mathrm{F}}^c \cap B_\alpha }p\cdot \omega _\alpha \, a^*_p a_p - \sum _{h \in B_{\mathrm{F}} \cap B_\alpha } h\cdot \omega _\alpha \, a^*_h a_h \Big ] \end{aligned}$$
and where the error \(\mathfrak {E}_{\mathrm{lin}}\) is bounded by
$$\begin{aligned} \begin{aligned} \left|\mathfrak {E}_{\mathrm{lin}} \right|&= \Big | \frac{\hbar ^2}{2} \Bigg \langle \xi , \sum _{\alpha =1}^M \Big [ \sum _{p \in B_{\mathrm{F}}^c \cap B_\alpha } (p - \omega _\alpha )^2 a^*_p a_p- \sum _{h \in B_{\mathrm{F}} \cap B_\alpha } (h - \omega _\alpha )^2 a^*_h a_h \Big ] \xi \Bigg \rangle \Big |\\&\le C \frac{\hbar ^2}{2} \frac{N^{2/3}}{M} \langle \xi , \mathcal {N}\xi \rangle \le \frac{C}{M} \end{aligned} \end{aligned}$$
where in the last step we used Proposition 4.6 to bound the expectation value of the number operator and \(\hbar = N^{-1/3}\).
To compute the expectation of the linearized kinetic energy operator \(\mathbb {H}_{\mathrm{kin}}\), we will make use of the following lemma.
Lemma 5.4
(Kinetic Energy of Particle–Hole Pairs). For all \(k\in \Gamma ^{{\mathrm{nor}}}\) and \(\alpha \in \mathcal {I}_{k}\) we have
$$\begin{aligned}{}[\mathbb {H}_{\mathrm{kin}}, c^*_\alpha (k)] = {\hbar ^2} |k\cdot \omega _\alpha |c^*_\alpha (k). \end{aligned}$$
Proof
We first treat the case \(\alpha \in \mathcal {I}_{k}^{+}\), for which \(k\cdot \omega _\alpha >0\). Using the CAR we calculate
$$\begin{aligned}&[ \mathbb {H}_{\mathrm{kin}}, c^*_{\alpha }(k)]\\&\quad = [ \mathbb {H}_{\mathrm{kin}}, b^*_{\alpha ,k} ] = [\mathbb {H}_{\mathrm{kin}}, \frac{1}{n_{\alpha ,k}} \sum _{\begin{array}{c} p \in B_{\mathrm{F}}^c \cap B_\alpha \\ h \in B_{\mathrm{F}} \cap B_\alpha \end{array}} \delta _{p-h,k} a^*_p a^*_h]\\&\quad = {\hbar ^2} \sum _{\beta =1}^M \frac{1}{n_{\alpha ,k}} \sum _{\begin{array}{c} p \in B_{\mathrm{F}}^c \cap B_\alpha \\ h \in B_{\mathrm{F}} \cap B_\alpha \end{array}} \delta _{p-h,k} \sum _{{\tilde{p}} \in B_{\mathrm{F}}^c \cap B_\beta } {\tilde{p}} \cdot \omega _\beta [a^*_{{\tilde{p}}} a_{{\tilde{p}}}, a^*_p a^*_h] \\&\qquad - {\hbar ^2} \sum _{\beta =1}^M \frac{1}{n_{\alpha ,k}} \sum _{\begin{array}{c} p \in B_{\mathrm{F}}^c \cap B_\alpha \\ h \in B_{\mathrm{F}} \cap B_\alpha \end{array}} \delta _{p-h,k} \sum _{{\tilde{h}} \in B_{\mathrm{F}} \cap B_\beta } {\tilde{h}} \cdot \omega _\beta [a^*_{{\tilde{h}}} a_{{\tilde{h}}}, a^*_p a^*_h] \\&\quad = {\hbar ^2} \sum _{\beta =1}^M \frac{1}{n_{\alpha ,k}} \sum _{\begin{array}{c} p \in B_{\mathrm{F}}^c \cap B_\alpha \\ h \in B_{\mathrm{F}}\cap B_\alpha \end{array}} \delta _{p-h,k} \Big ( \sum _{{\tilde{p}} \in B_{\mathrm{F}}^c \cap B_\beta } {\tilde{p}} \cdot \omega _\beta \delta _{p, {\tilde{p}}} - \sum _{{\tilde{h}} \in B_{\mathrm{F}} \cap B_\beta } {\tilde{h}}\cdot \omega _\beta \delta _{h,{\tilde{h}}}\Big ) a^*_p a^*_h; \end{aligned}$$
notice that the Kronecker deltas \(\delta _{p,{\tilde{p}}}\) and \(\delta _{h,{\tilde{h}}}\) imply \(\beta =\alpha \), so we find
$$\begin{aligned}&= {\hbar ^2} \frac{1}{n_{\alpha ,k}} \sum _{\begin{array}{c} p \in B_{\mathrm{F}}^c \cap B\alpha \\ h \in B_{\mathrm{F}} \cap B_\alpha \end{array}} \delta _{p-h,k} (p-h)\cdot \omega _\alpha a^*_p a^*_h = {\hbar ^2} |k \cdot \omega _\alpha |c^*_\alpha (k). \end{aligned}$$
The absolute value was trivially introduced since the scalar product is anyway non-negative. For \(k\cdot \omega _\alpha < 0\), recall that \(c^*_\alpha (k) = b^*_{\alpha ,-k}\); the calculation then proceeds the same way, but in the second last line we use \((p-h)\cdot \omega _\alpha = (-k)\cdot \omega _\alpha = |k \cdot \omega _\alpha |\). \(\quad \square \)
We are now ready to calculate the kinetic energy of our trial state.
Proposition 5.5
(Kinetic Energy). Let \(\xi \) be defined as in (4.16). Then
$$\begin{aligned} \langle \xi , \mathbb {H}_{\mathrm{kin}} \xi \rangle = {\hbar \kappa }\sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|{\text {tr}}D(k)\sinh ^2(K(k)) + \mathfrak {E}_{\mathrm{kin}}, \end{aligned}$$
where \(D(k)\) is defined in (4.19) and the error term is such that \(|\mathfrak {E}_{\mathrm{kin}}|\le C \hbar / \mathfrak {n}^2\) with \(\mathfrak {n}= N^{1/3-\delta /2} M^{-1/2}\) as in (3.18).
Proof
We write \(T_\lambda = \exp (\lambda B)\), with B as in (4.20), and \(\xi = T \Omega \). Hence
$$\begin{aligned}&\langle \xi , \mathbb {H}_{\mathrm{kin}} \xi \rangle \\&\quad = \int _0^1 {{\mathrm{d}}}\lambda \langle \Omega , T^*_\lambda [\mathbb {H}_{\mathrm{kin}}, B] T_\lambda \Omega \rangle \\&\quad = \int _0^1 {{\mathrm{d}}}\lambda \langle \Omega , T^*_\lambda \Big [ \mathbb {H}_{\mathrm{kin}}, \sum _{k\in \Gamma ^{{\mathrm{nor}}}} \frac{1}{2} \sum _{\alpha ,\beta \in \mathcal {I}_{k}} K(k)_{\alpha ,\beta } c^*_\alpha (k) c^*_\beta (k) - {\mathrm{h.c.}}\Big ] T_\lambda \Omega \rangle \\&\quad = {\text {Re}}\int _0^1 {{\mathrm{d}}}\lambda \sum _{k\in \Gamma ^{{\mathrm{nor}}}} \sum _{\alpha ,\beta \in \mathcal {I}_{k}} K(k)_{\alpha ,\beta } \langle \Omega , T^*_\lambda \left( [\mathbb {H}_{\mathrm{kin}}, c^*_\alpha (k)]c^*_\beta (k) + c^*_\alpha (k) [\mathbb {H}_{\mathrm{kin}},c^*_\beta (k)]\right) T_\lambda \Omega \rangle . \end{aligned}$$
From Lemma 5.4
$$\begin{aligned}&\langle \xi , \mathbb {H}_{\mathrm{kin}} \xi \rangle \\&\quad = {\text {Re}}\int _0^1{{\mathrm{d}}}\lambda \sum _{k\in \Gamma ^{{\mathrm{nor}}}} {\hbar ^2} \sum _{\alpha ,\beta \in \mathcal {I}_{k}} K(k)_{\alpha ,\beta } \left( |k\cdot \omega _\alpha |+ |k\cdot \omega _\beta |\right) \langle \Omega , T^*_\lambda c^*_\alpha (k) c^*_\beta (k) T_\lambda \Omega \rangle . \end{aligned}$$
Recall that \(|k \cdot \omega _\alpha |= |k|\kappa \hbar ^{-1} u_\alpha (k)^2\) with \(u_\alpha (k)\) defined in (4.15). Using Proposition 4.4 then
$$\begin{aligned}&\langle \xi , \mathbb {H}_{\mathrm{kin}} \xi \rangle \\&\quad = {\text {Re}}\int _0^1{{\mathrm{d}}}\lambda \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|{\hbar \kappa } \sum _{\alpha ,\beta \in \mathcal {I}_{k}} K(k)_{\alpha ,\beta } \left( u_\alpha (k)^2 + u_\beta (k)^2 \right) \\&\qquad \times \left\langle \Omega ,\left( \sum _{\delta \in \mathcal {I}_{k}} \cosh (\lambda K(k))_{\alpha ,\delta } c^*_\delta (k) + \sum _{\delta \in \mathcal {I}_{k}} \sinh (\lambda K(k))_{\alpha ,\delta } c_\delta (k) + \mathfrak {E}^*_\alpha (\lambda ,k) \right) \right. \\&\qquad \times \left. \left( \sum _{\gamma \in \mathcal {I}_{k}} \cosh (\lambda K(k))_{\beta ,\gamma } c^*_\gamma (k) + \sum _{\gamma \in \mathcal {I}_{k}} \sinh (\lambda K(k))_{\beta ,\gamma } c_\gamma (k) + \mathfrak {E}^*_\beta (\lambda ,k) \right) \Omega \right\rangle . \end{aligned}$$
Finally, using \(c_\delta (k)\Omega = 0\) and \(\langle \Omega , c_\delta (k) c^*_\gamma (k)\Omega \rangle = \delta _{\delta ,\gamma }\), we get
$$\begin{aligned} \langle \xi , \mathbb {H}_{\mathrm{kin}} \xi \rangle&= {\text {Re}}\int _0^1 {{\mathrm{d}}}\lambda \sum _{k \in \Gamma ^{{\mathrm{nor}}}} |k|\frac{\hbar \kappa }{2} \sum _{\alpha ,\beta \in \mathcal {I}_{k}} K(k)_{\alpha ,\beta } \left( u_\alpha (k)^2 + u_\beta (k)^2 \right) \nonumber \\&\quad \times \sum _{\delta \in \mathcal {I}_{k}} \sinh (\lambda K(k))_{\alpha ,\delta } \sum _{\gamma \in \mathcal {I}_{k}} \cosh (\lambda K(k))_{\beta ,\gamma } \delta _{\delta ,\gamma } + \mathfrak {E}_{\mathrm{kin}} \nonumber \\&= \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|{\hbar \kappa } \sum _{\alpha \in \mathcal {I}_{k}} u_\alpha (k)^2 \int _0^1 {{\mathrm{d}}}\lambda \Big ( \sinh \big (2\lambda K(k)\big ) K(k) \Big )_{\alpha ,\alpha } + \mathfrak {E}_{\mathrm{kin}} \end{aligned}$$
(5.3)
where we defined
$$\begin{aligned} \mathfrak {E}_{\mathrm{kin}}&:= {\text {Re}}\int _0^1 {{\mathrm{d}}}\lambda \sum _{k \in \Gamma ^{{\mathrm{nor}}}} |k|\frac{\hbar \kappa }{2} \sum _{\alpha ,\beta \in \mathcal {I}_{k}} K(k)_{\alpha ,\beta } \left( u_\alpha (k)^2 + u_\beta (k)^2 \right) \\&\quad \times \Big ( \langle \Omega , \mathfrak {E}^*_\alpha (\lambda ,k) \mathfrak {E}^*_\beta (\lambda ,k) \Omega \rangle + \sum _{\delta \in \mathcal {I}_{k}} \sinh (\lambda K(k))_{\alpha ,\delta } \langle \Omega ,c_\delta (k) \mathfrak {E}^*_\beta (\lambda ,k)\Omega \rangle \\&\qquad \qquad + \sum _{\gamma \in \mathcal {I}_{k}} \cosh (\lambda K(k))_{\beta ,\gamma } \langle \Omega , \mathfrak {E}_\alpha ^*(\lambda ,k) c^*_\gamma (k) \Omega \rangle \Big )\\&=: \mathfrak {E}^{(1)}_{\mathrm{kin}} + \mathfrak {E}^{(2)}_{\mathrm{kin}}+ \mathfrak {E}^{(3)}_{\mathrm{kin}}. \end{aligned}$$
We compute the integral in (5.3). We get
$$\begin{aligned} \langle \xi , \mathbb {H}_{\mathrm{kin}} \xi \rangle = {\hbar \kappa }\sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|{\text {tr}}D(k)\sinh ^2(K(k))+ \mathfrak {E}_{\mathrm{kin}}. \end{aligned}$$
Using that \(u_\alpha (k)^2 = |\hat{k}\cdot \hat{\omega }_\alpha |\le 1\), we bound the first error term by
$$\begin{aligned} | \mathfrak {E}^{(1)}_{\mathrm{kin}}|&\le \Big | \int _0^1 {{\mathrm{d}}}\lambda \sum _{k \in \Gamma ^{{\mathrm{nor}}}} |k|{\hbar \kappa } \sum _{\alpha ,\beta \in \mathcal {I}_{k}} K(k)_{\alpha ,\beta } \left( u_\alpha (k)^2 + u_\beta (k)^2 \right) \\&\quad \times \langle \Omega , \mathfrak {E}^*_\alpha (\lambda ,k) \mathfrak {E}^*_\beta (\lambda ,k) \Omega \rangle \Big |\\&\le 2\int _0^1{{\mathrm{d}}}\lambda \sum _{k \in \Gamma ^{{\mathrm{nor}}}} |k|{\hbar \kappa }\sum _{\alpha ,\beta \in \mathcal {I}_{k}} |K(k)_{\alpha ,\beta }|||\mathfrak {E}_\alpha (\lambda ,k)\Omega || ||\mathfrak {E}^*_\beta (\lambda ,k)\Omega || \\&\le 2\int _0^1{{\mathrm{d}}}\lambda \sum _{k \in \Gamma ^{{\mathrm{nor}}}} |k|{\hbar \kappa } \Big [ \sum _{\alpha ,\beta \in \mathcal {I}_{k}} |K(k)_{\alpha ,\beta }|^2 \Big ]^{1/2} \\&\quad \times \Big [ \sum _{\alpha \in \mathcal {I}_{k}} ||\mathfrak {E}_\alpha (\lambda ,k)\Omega ||^2 \sum _{\beta \in \mathcal {I}_{k}} ||\mathfrak {E}^*_\beta (\lambda ,k)\Omega ||^2 \Big ]^{1/2}; \end{aligned}$$
and finally using (4.21)
$$\begin{aligned} | \mathfrak {E}^{(1)}_{\mathrm{kin}}|&\le \frac{C\hbar }{\mathfrak {n}^4} \sum _{k \in \Gamma ^{{\mathrm{nor}}}} |k|||K(k)||_{{\mathrm{HS}}}e^{2||K(k)||_{{\mathrm{HS}}}} \sup _{\lambda \in [0,1]} \langle T_\lambda \Omega , (\mathcal {N}+2)^{3} T_\lambda \Omega \rangle \\&\quad \times \Big ( \sum _{l\in \Gamma ^{{\mathrm{nor}}}} ||K(l)||_{{\mathrm{HS}}}\Big )^{2}. \end{aligned}$$
From Proposition 4.6 and Lemma 4.5, we conclude that \(|\mathfrak {E}^{(1)}_{\mathrm{kin}}| \le C \hbar / \mathfrak {n}^4\). The third error term \(\mathfrak {E}^{(3)}_{\mathrm{kin}}\) can be controlled similarly, using Lemma 4.2:
$$\begin{aligned} |\mathfrak {E}^{(3)}_{\mathrm{kin}}|&\le C\hbar \int _0^1 {{\mathrm{d}}}\lambda ||(\mathcal {N}+1)^{1/2} \Omega || \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|\sum _{\alpha ,\beta \in \mathcal {I}_{k}} |K(k)_{\alpha ,\beta } |||\mathfrak {E}_\alpha (\lambda ,k)\Omega || \\&\quad \times \Big [ \sum _{\gamma \in \mathcal {I}_{k}} |\cosh (\lambda K(k))_{\beta ,\gamma } |^2 \Big ]^{1/2} \\&\le \frac{C\hbar }{\mathfrak {n}^2} ||(\mathcal {N}+1)^{1/2} \Omega || \sup _{\lambda \in [0,1]} ||(\mathcal {N}+2)^{3/2}T_\lambda \Omega || \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|||K(k)||_{{\mathrm{HS}}}\, e^{||K(k)||_{{\mathrm{HS}}}} \\&\quad \times \int _0^1 {{\mathrm{d}}}\lambda \, ||\cosh (\lambda K(k))||_{{\mathrm{HS}}}\sum _{l\in \Gamma ^{{\mathrm{nor}}}} ||K(l)||_{{\mathrm{HS}}}\\&\le \frac{C\hbar }{\mathfrak {n}^2} ||(\mathcal {N}+1)^{1/2} \Omega || \sup _{\lambda \in [0,1]} ||(\mathcal {N}+2)^{3/2}T_\lambda \Omega ||\\&\quad \times \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|||K(k)||_{{\mathrm{HS}}}\, e^{2||K(k)||_{{\mathrm{HS}}}} \sum _{l\in \Gamma ^{{\mathrm{nor}}}} ||K(l)||_{{\mathrm{HS}}}\\&\le \frac{\hbar }{\mathfrak {n}^2} C. \end{aligned}$$
The second error term \(\mathfrak {E}_{\mathrm{kin}}^{(2)}\) can be controlled in the same way. \(\quad \square \)
Expectation value of the interaction energy
We now evaluate the main contribution (3.9) to the interaction energy. This is the content of the next proposition.
Proposition 5.6
(Interaction Energy). Let \(\xi \) be the trial state defined as in (4.16), and let \(Q_{N}^{{\mathrm{B}}}\) be given by (3.9). Then
$$\begin{aligned} \begin{aligned} \langle \xi , Q_N^{{\mathrm{B}}} \xi \rangle&= {\hbar \kappa } \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|{\text {tr}}\left( W(k) \sinh ^2 (K(k)) + \tilde{W}(k) \sinh (K(k)) \cosh (K(k)) \right) \\&\quad + \mathfrak {E}_{\mathrm{int}} + \mathcal {O}(\hbar N^{-\delta /2}) \end{aligned} \end{aligned}$$
where \(W(k)\) and \(\tilde{W}(k)\) are defined in (4.19). The error term is bounded by \(|\mathfrak {E}_{\mathrm{int}}|\le C\hbar / \mathfrak {n}^2\), with \(\mathfrak {n}= N^{1/3-\delta /2} M^{-1/2}\) as in (3.18).
Proof
We start by decomposing the \(b_{k}\)-operators in the interaction Hamiltonian (3.12) by their patch decomposition (3.15),
$$\begin{aligned} \tilde{b}_k = \sum _{\alpha \in \mathcal {I}_{k}^{+}} \tilde{b}_{\alpha ,k} + \mathfrak {r}_k. \end{aligned}$$
(5.4)
We recall that the error terms \(\mathfrak {r}_k\) collect modes in the corridors and close to the equator:
$$\begin{aligned} \mathfrak {r}_k = \tilde{\mathfrak {r}}_k + \sum _{\alpha \not \in \mathcal {I}_{k}} {\tilde{b}}_{\alpha ,k}, \end{aligned}$$
(5.5)
where \(\tilde{\mathfrak {r}}_k\) is a linear combination of products \(a_h a_p\) such that at least one of the two momenta is in the corridors \(B_{\mathrm{corri}}\) (see Fig. 2), and the second term collects the contributions coming from the patches close to the equator. We are going to show that \(\mathfrak {r}_k\) gives a negligible contribution to \(\langle \xi , Q_{N}^{{\mathrm{B}}} \xi \rangle \).
Contribution of corridors. We claim that the error operators \(\tilde{\mathfrak {r}}_k\) do not contribute to \(\langle \xi , Q_{N}^{{\mathrm{B}}} \xi \rangle \). To see this, recall that \(T = e^{B}\), with B not containing any mode \(q\in B_{\mathrm{corri}}\), see (4.20). Since at least one of the two momenta p and h appearing in \(\tilde{\mathfrak {r}}_k\) is in the corridor \(B_{\mathrm{corri}}\), we have \(\tilde{\mathfrak {r}}_k \xi = 0\). Plugging the decomposition (5.4) into \(Q^{{\mathrm{B}}}_{N}\) from (3.12) and taking the expectation value on \(\xi \), we realize that all terms containing at least one error operator \(\tilde{\mathfrak {r}}^\natural _k\) are zero, due to the fact that there is at least one error operator \(\tilde{\mathfrak {r}}_k\) directly acting on \(T \Omega \).
Contribution of patches near the equator. We claim that the contribution to \(\langle \xi , Q^{{\mathrm{B}}}_{N} \xi \rangle \) coming from patches \(\beta \not \in \mathcal {I}_{k}\) is subleading as \(N\rightarrow \infty \). These are the patches \(\beta \) in the collar where \(|\hat{k}\cdot \hat{\omega }_\beta |< N^{-\delta }\). The width of this collar is bounded above by \(C k_{\mathrm{F}} N^{-\delta }\), and its length—approximately equal to the circumference of the equator—is bounded above by \(C k_{\mathrm{F}}\); we conclude that the surface area of the collar is of order \(k_F^2 N^{-\delta }\).
Recall that \(n^2_{\beta ,k}\) is the number of particle–hole pairs with relative momentum k in patch \(\beta \); thus, adding in the corridors for an upper bound, \(\sum _{\beta \not \in \mathcal {I}_{k}} n^2_{\beta ,k}\) is bounded by the number of particle–hole pairs with relative momentum k in the collar. This number is bounded above by the number of hole momenta \(h \in B_{\mathrm{F}}\) that are at most a distance R from the collar (since \(|k|\le R\)). The number of such points of the lattice \(\mathbb {Z}^3\) can be counted by Gauss’ classical argument: assign to each lattice point k the cube \([k_1,k_1+1]\times [k_2,k_2+1]\times [k_3,k_3+1]\). Then the number of cubes belonging to lattice points near the collar is bounded by the Lebesgue measure of the collar “fattened” to a thickness R; i. e.,
$$\begin{aligned} \sum _{\beta \not \in \mathcal {I}_{k}} n^2_{\beta ,k} \le C k_{\mathrm{F}}^2 N^{-\delta } \times R = \mathcal {O}(N^{2/3-\delta }). \end{aligned}$$
(5.6)
We are now ready to estimate the contribution to \(\langle \xi , Q_{N}^{{\mathrm{B}}} \xi \rangle \) coming from the modes close to the equator. Consider, e. g., the term \(\frac{1}{2N} \sum _{k \in \Gamma ^{{\mathrm{nor}}}} \hat{V}(k) 2{\tilde{b}}^*_k {\tilde{b}}_k\) (all the other terms can be dealt with similarly). We get three contributions from (5.5), namely
$$\begin{aligned} \begin{aligned}&\frac{1}{N} \sum _{k \in \Gamma ^{{\mathrm{nor}}}}\hat{V}(k) \sum _{\beta \not \in \mathcal {I}_{k}} \tilde{b}^*_{\beta ,k} \sum _{\alpha \in \mathcal {I}_{k}^{+}} \tilde{b}_{\alpha ,k},\\&\frac{1}{N} \sum _{k \in \Gamma ^{{\mathrm{nor}}}}\hat{V}(k) \sum _{\beta \in \mathcal {I}_{k}^{+}} \tilde{b}^*_{\beta ,k} \sum _{\alpha \not \in \mathcal {I}_{k}} \tilde{b}_{\alpha ,k},\\&\frac{1}{N} \sum _{k \in \Gamma ^{{\mathrm{nor}}}}\hat{V}(k) \sum _{\beta \not \in \mathcal {I}_{k}} \tilde{b}^*_{\beta ,k} \sum _{\alpha \not \in \mathcal {I}_{k}} \tilde{b}_{\alpha ,k}. \end{aligned} \end{aligned}$$
(5.7)
We give the detailed estimate for the first term in the list (the other two terms can be controlled similarly)
$$\begin{aligned}&\frac{1}{N} \sum _{k \in \Gamma ^{{\mathrm{nor}}}}\hat{V}(k) \Big \langle \xi , \sum _{\beta \not \in \mathcal {I}_{k}} \tilde{b}^*_{\beta ,k} \sum _{\alpha \in \mathcal {I}_{k}^{+}} \tilde{b}_{\alpha ,k} \xi \Big \rangle \\&\quad =\frac{1}{N} \sum _{k \in \Gamma ^{{\mathrm{nor}}}}\hat{V}(k) \Big \langle \xi , \sum _{\beta \not \in \mathcal {I}_{k}} n_{\beta ,k} b^*_{\beta ,k} \sum _{\alpha \in \mathcal {I}_{k}^{+}} n_{\alpha ,k} {b}_{\alpha ,k} \xi \Big \rangle \\&\quad \le \frac{1}{N} \sum _{k\in \Gamma ^{{\mathrm{nor}}}}\hat{V}(k) \Big ( \sum _{\beta \not \in \mathcal {I}_{k}} n_{\beta ,k}^2 \Big )^{1/2} \Big ( \sum _{\alpha \in \mathcal {I}_{k}^{+}} n_{\alpha ,k}^2\Big )^{1/2} ||\mathcal {N}^{1/2} \xi ||^2 \\&\quad \le \frac{1}{N} \sum _{k\in \Gamma ^{{\mathrm{nor}}}}\hat{V}(k) \Big ( C N^{2/3-\delta } \Big )^{1/2} \Big ( M \frac{C k_{\mathrm{F}}^2}{M}\Big )^{1/2} \langle \xi ,\mathcal {N}\xi \rangle = \mathcal {O}(\hbar N^{-\delta /2}), \end{aligned}$$
where we used (5.6) to control the sum over \(\beta \not \in \mathcal {I}_{k}\), \(n_{\alpha ,k}^2 \le C k_{\mathrm{F}}^2/M\) due to Proposition 3.1 for the sum over \(\alpha \in \mathcal {I}_{k}^{+}\), and the bound on the number of fermions \(\mathcal {N}\) from Proposition 4.6.
Approximate Bogoliubov diagonalization of the effective interaction. By the discussion of the previous paragraph
$$\begin{aligned} \begin{aligned}&\langle \xi , Q_N^{{\mathrm{B}}} \xi \rangle \\&\quad = \frac{1}{N} \left\langle \xi , \sum _{k \in \Gamma ^{{\mathrm{nor}}}} \hat{V}(k) \Big ( \sum _{\alpha ,\beta \in \mathcal {I}_{k}^{+}}{\tilde{b}}^*_{\alpha ,k} \tilde{b}_{\beta ,k} + \sum _{\alpha ,\beta \in \mathcal {I}_{k}^{-}}{\tilde{b}}^*_{\alpha ,-k} {\tilde{b}}_{\beta ,-k} \right. \\&~~~~~~~~~~~~~~~~~~~~~~~~~\qquad \qquad + \left. \Big [ \sum _{\alpha \in \mathcal {I}_{k}^{+},\, \beta \in \mathcal {I}_{k}^{-}} {\tilde{b}}^*_{\alpha ,k} {\tilde{b}}^*_{\beta ,-k} + {\mathrm{h.c.}}\Big ]\Big ) \xi \right\rangle + \mathcal {O}(\hbar N^{-\delta /2}). \end{aligned} \end{aligned}$$
Introducing the normalization factors \(n_{\alpha ,k} = k_{\mathrm{F}} \sqrt{|k|} v_\alpha (k)\) and combining the \(b^*_{\alpha ,k}\) and \(b^*_{\alpha ,-k}\) operators to \(c^*_\alpha (k)\) operators as in (4.1), we get
$$\begin{aligned} \langle \xi , Q_N^{{\mathrm{B}}} \xi \rangle = \langle \xi , \mathbb {H}_{\mathrm{int}} \xi \rangle + \mathcal {O}(\hbar N^{-\delta /2}), \qquad \mathbb {H}_{\mathrm{int}} := \mathbb {H}_{\mathrm{int}}^{(1)} + \mathbb {H}_{\mathrm{int}}^{(2)} + \mathbb {H}_{\mathrm{int}}^{(3)}, \end{aligned}$$
(5.8)
where, recalling that \(g(k) = \kappa \hat{V}(k)\),
$$\begin{aligned} \mathbb {H}_{\mathrm{int}}^{(1)}&:= {\hbar \kappa } \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|g(k)\sum _{\alpha ,\beta \in \mathcal {I}_{k}^{+}} v_\alpha (k) v_\beta (k) c^*_\alpha (k) c_\beta (k),\\ \mathbb {H}_{\mathrm{int}}^{(2)}&:= \hbar \kappa \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|g(k)\sum _{\alpha ,\beta \in \mathcal {I}_{k}^{-}} v_\alpha (k) v_\beta (k) c^*_\alpha (k) c_\beta (k),\\ \mathbb {H}_{\mathrm{int}}^{(3)}&:= \hbar \kappa \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|g(k) \sum _{\alpha \in \mathcal {I}_{k}^{+}} \sum _{\beta \in \mathcal {I}_{k}^{-}} v_\alpha (k) v_\beta (k) c^*_\alpha (k) c^*_\beta (k) + {\mathrm{h.c.}}\end{aligned}$$
We shall evaluate \(\langle \xi , \mathbb {H}^{(i)}_{\text {int}} \xi \rangle \), \(i=1,2,3\), with \(\xi = T\Omega \), using the fact that the T operator behaves as an approximate bosonic Bogoliubov transformation, recall Proposition 4.4. Using also \(\langle \Omega , c_\delta (k) c^*_\gamma (k)\Omega \rangle = \delta _{\delta ,\gamma }\), we have
$$\begin{aligned} \langle \xi , \mathbb {H}_{\mathrm{int}}^{(1)} \xi \rangle = \hbar \kappa \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|{\text {tr}}W^{++}(k) \sinh ^2(K(k)) + \mathfrak {E}_{\mathrm{int}}^{(1)}, \end{aligned}$$
(5.9)
where
$$\begin{aligned} W^{++}(k)_{\alpha ,\beta } = \left\{ \begin{array}{cl} g(k) v_\alpha (k) v_\beta (k) &{} \text {for }\alpha ,\beta \in \mathcal {I}_{k}^{+}\\ 0 &{} \text {otherwise} \end{array}\right. \end{aligned}$$
and the error term is
$$\begin{aligned} \mathfrak {E}_{\mathrm{int}}^{(1)}&= \hbar \kappa \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|g(k) \sum _{\alpha ,\beta \in \mathcal {I}_{k}^{+}} v_\alpha (k) v_\beta (k) \Bigg [ \sum _{\gamma \in \mathcal {I}_{k}} \sinh (K(k))_{\alpha ,\gamma } \langle \Omega , c_\gamma (k) \mathfrak {E}_\beta (1,k) \Omega \rangle \\&\quad +\sum _{\gamma \in \mathcal {I}_{k}} \sinh (K(k))_{\beta ,\gamma } \langle \Omega , \mathfrak {E}^*_\alpha (1,k) c^*_\gamma (k) \Omega \rangle +\langle \Omega , \mathfrak {E}^*_\alpha (1,k) \mathfrak {E}_\beta (1,k) \Omega \rangle \Bigg ]. \end{aligned}$$
For the second part of the interaction we find
$$\begin{aligned} \langle \xi , \mathbb {H}_{\mathrm{int}}^{(2)} \xi \rangle = \hbar \kappa \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|{\text {tr}}W^{--}(k) \sinh ^2(K(k)) + \mathfrak {E}_{\mathrm{int}}^{(2)}, \end{aligned}$$
(5.10)
where
$$\begin{aligned} W^{--}(k)_{\alpha ,\beta } = \left\{ \begin{array}{cl} g(k) v_\alpha (k) v_\beta (k) &{}\quad \text {for }\alpha ,\beta \in \mathcal {I}_{k}^{-}\\ 0 &{}\quad \text {otherwise} \end{array}\right. \end{aligned}$$
and
$$\begin{aligned} \mathfrak {E}_{\mathrm{int}}^{(2)} = \hbar \kappa \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|g(k)&\sum _{\alpha ,\beta \in \mathcal {I}_{k}^{-}} v_\alpha (k) v_\beta (k) \Bigg [ \sum _{\gamma \in \mathcal {I}_{k}} \sinh (K(k))_{\alpha ,\gamma } \langle \Omega , c_\gamma (k) \mathfrak {E}_\beta (1,k) \Omega \rangle \\&+\sum _{\gamma \in \mathcal {I}_{k}} \sinh (K(k))_{\beta ,\gamma } \langle \Omega , \mathfrak {E} ^*_\alpha (1,k) c^*_\gamma (k) \Omega \rangle +\langle \Omega , \mathfrak {E}^*_\alpha (1,k) \mathfrak {E}_\beta (1,k) \Omega \rangle \Bigg ]. \end{aligned}$$
Finally, for the third interaction term we find
$$\begin{aligned} \begin{aligned} \langle \xi , \mathbb {H}_{\mathrm{int}}^{(3)} \xi \rangle&= 2 \hbar \kappa {\text {Re}}\sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|{\text {tr}}W^{+-}(k) \sinh (K(k)) \cosh (K(k)) + \mathfrak {E}_{\mathrm{int}}^{(3)} \\&= \hbar \kappa \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|{\text {tr}}\tilde{W}(k) \sinh (K(k)) \cosh (K(k)) + \mathfrak {E}_{\mathrm{int}}^{(3)}, \end{aligned} \end{aligned}$$
(5.11)
where
$$\begin{aligned} W^{+-}(k)_{\alpha ,\beta } = \left\{ \begin{array}{cl} g(k) v_\alpha (k) v_\beta (k) &{} \text {for }\alpha \in \mathcal {I}_{k}^{+}\text { and } \beta \in \mathcal {I}_{k}^{-}\\ 0 &{} \text {otherwise} \end{array}\right. \;; \end{aligned}$$
we used the fact that all terms are real to write the more symmetric expression in terms of \(\tilde{W}(k) = W^{+-}(k) + W^{-+}(k)\) (the latter is defined by exchanging the role of \(\mathcal {I}_{k}^{+}\) and \(\mathcal {I}_{k}^{-}\) in the former). The error term \(\mathfrak {E}_{\mathrm{int}}^{(3)}\) is given by
$$\begin{aligned} \mathfrak {E}_{\mathrm{int}}^{(3)}&= 2\hbar \kappa {\text {Re}}\sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|g(k) \sum _{\alpha \in \mathcal {I}_{k}^{+}} \sum _{\beta \in \mathcal {I}_{k}^{-}} v_\alpha (k) v_\beta (k)\\&\quad \times \Bigg [ \sum _{\gamma \in \mathcal {I}_{k}} \sinh (K(k))_{\gamma ,\alpha } \langle \Omega , c_\gamma (k) \mathfrak {E}^*_\beta (1,k) \Omega \rangle \\&\quad \qquad + \sum _{\gamma \in \mathcal {I}_{k}} \cosh (K(k))_{\gamma ,\beta } \langle \Omega ,\mathfrak {E}^*_\alpha (1,k) c^*_\gamma (k) \Omega \rangle + \langle \Omega , \mathfrak {E}^*_\alpha (1,k) \mathfrak {E}^*_\beta (1,k) \Omega \rangle \Bigg ]. \end{aligned}$$
Combining (5.9), (5.10), (5.11) and (5.8), we conclude that
$$\begin{aligned} \langle \xi , Q_N^{{\mathrm{B}}} \xi \rangle = \hbar \kappa \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|{\text {tr}}\left( W(k) \sinh ^2 (K(k)) + \tilde{W}(k) \sinh (K(k)) \cosh (K(k)) \right) + \mathfrak {E}_{\mathrm{int}} \end{aligned}$$
with \( \mathfrak {E}_{\mathrm{int}} = \mathfrak {E}^{(1)}_{\mathrm{int}} + \mathfrak {E}^{(2)}_{\mathrm{int}} + \mathfrak {E}^{(3)}_{\mathrm{int}}\). To control the error term \(\mathfrak {E}^{(1)}_{\mathrm{int}}\), we decompose it as
$$\begin{aligned} \mathfrak {E}_{\mathrm{int}}^{(1)}&= \hbar \kappa \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|g(k) \sum _{\alpha ,\beta \in \mathcal {I}_{k}^{+}} v_\alpha (k) v_\beta (k) \Bigg [ \sum _{\gamma \in \mathcal {I}_{k}} \sinh (K(k))_{\alpha ,\gamma } \langle \Omega , c_\gamma (k) \mathfrak {E}_\beta (1,k) \Omega \rangle \\&\quad +\sum _{\gamma \in \mathcal {I}_{k}} \sinh (K(k))_{\beta ,\gamma } \langle \Omega , \mathfrak {E}^*_\alpha (1,k) c^*_\gamma (k) \Omega \rangle +\langle \Omega , \mathfrak {E}^*_\alpha (1,k) \mathfrak {E}_\beta (1,k) \Omega \rangle \Bigg ] \\&=: \mathfrak {E}_{\mathrm{int}}^{(1,1)} + \mathfrak {E}_{\mathrm{int}}^{(1,2)} + \mathfrak {E}_{\mathrm{int}}^{(1,3)}. \end{aligned}$$
Recall that \(u_\alpha (k)^2 = |\hat{k}\cdot \hat{\omega }_\alpha |\le 1\) and hence, by Proposition 3.1, \(v_\alpha (k) \le \sqrt{\frac{C}{M}} u_\alpha (k) \le \sqrt{\frac{C}{M}}\). Thus, using Proposition 4.4 and Cauchy–Schwarz, we find
$$\begin{aligned} | \mathfrak {E}_{\mathrm{int}}^{(1,3)} |&\le \Big | \hbar \kappa \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|g(k) \sum _{\alpha ,\beta \in \mathcal {I}_{k}^{+}} v_\alpha (k) v_\beta (k) \langle \Omega , \mathfrak {E}^*_\alpha (1,k) \mathfrak {E}_\beta (1,k) \Omega \rangle \Big | \\&\le \hbar \kappa \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|g(k) \sum _{\alpha \in \mathcal {I}_{k}^{+}} \sqrt{\frac{C}{M}} ||\mathfrak {E}_\alpha (1,k) \Omega || \sum _{\beta \in \mathcal {I}_{k}^{+}} \sqrt{\frac{C}{M}} ||\mathfrak {E}_\beta (1,k) \Omega ||\\&\le C {\hbar } \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|\hat{V}(k) \sum _{\alpha \in \mathcal {I}_{k}^{+}} ||\mathfrak {E}_\alpha (1,k)\Omega ||^2. \end{aligned}$$
(Recall that \(|\mathcal {I}_{k}^{+}| = I_k \le M/2\).) With (4.21), we get
$$\begin{aligned} | \mathfrak {E}_{\mathrm{int}}^{(1,3)} |&\le \hbar \frac{C}{\mathfrak {n}^4} \sup _{\lambda \in [0,1]} \langle T_\lambda \Omega , (\mathcal {N}+2)^{3} T_\lambda \Omega \rangle \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|\hat{V}(k) e^{2||K(k)||_{{\mathrm{HS}}}} \Big [ \sum _{l\in \Gamma ^{{\mathrm{nor}}}} ||K(l)||_{{\mathrm{HS}}}\Big ]^2. \end{aligned}$$
Lemma 4.5 and Proposition 4.6 imply that \(|\mathfrak {E}_{\mathrm{int}}^{(1,3)}| \le C \hbar / \mathfrak {n}^4\). The term \(\mathfrak {E}_{\mathrm{int}}^{(1,1)}\) can be controlled similarly:
$$\begin{aligned} |\mathfrak {E}_{\mathrm{int}}^{(1,1)}|&\le \Big |\hbar \kappa \sum _{k \in \Gamma ^{{\mathrm{nor}}}} |k|g(k) \sum _{\alpha ,\beta \in \mathcal {I}_{k}^{+}} v_\alpha (k) v_\beta (k) \sum _{\gamma \in \mathcal {I}_{k}} \sinh (K(k))_{\alpha ,\gamma } \langle \Omega , c_\gamma (k) \mathfrak {E}_\beta (k,1) \Omega \rangle \Big |\\&\le \frac{C \hbar }{M} \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|\hat{V}(k) \sum _{\alpha ,\beta \in \mathcal {I}_{k}^{+}} \Big \Vert \sum _{\gamma \in \mathcal {I}_{k}} \sinh (K(k))_{\alpha ,\gamma } c^*_\gamma (k) \Omega \Big \Vert ||\mathfrak {E}_\beta (k,1)\Omega || \\&\le \frac{C \hbar }{M} \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|\hat{V}(k) \sum _{\alpha ,\beta \in \mathcal {I}_{k}^{+}} \Big ( \sum _{\gamma \in \mathcal {I}_{k}} |\sinh (K(k))_{\alpha ,\gamma }|^2 \Big )^{1/2} \\&\quad \times || (\mathcal {N}+1)^{1/2} \Omega || ||\mathfrak {E}_\beta (k,1)\Omega ||\;; \end{aligned}$$
applying Cauchy–Schwarz in \(\alpha \) and in \(\beta \), using \(|\mathcal {I}_{k}^{+}| = I_k \le M/2\) and (4.21) we arrive at
$$\begin{aligned} |\mathfrak {E}_{\mathrm{int}}^{(1,1)}|&\le \frac{C\hbar }{\mathfrak {n}^2} \sup _{\lambda \in [0,\lambda ]}|| (\mathcal {N}+2)^{3/2} T_\lambda \xi || \sum _{k \in \Gamma ^{{\mathrm{nor}}}} |k|\hat{V}(k) e^{2||K(k)||_{{\mathrm{HS}}}} \sum _{l\in \Gamma ^{{\mathrm{nor}}}} ||K(l)||_{{\mathrm{HS}}}. \end{aligned}$$
Again, Lemma 4.5 and Proposition 4.6 show that \(|\mathfrak {E}_{\mathrm{int}}^{(1,1)}| \le C \hbar / \mathfrak {n}^2\). Analogously, we obtain also \(|\mathfrak {E}_{\mathrm{int}}^{(1,2)}| \le C \hbar / \mathfrak {n}^2\). Hence \(|\mathfrak {E}_{\mathrm{int}}^{(1)}| \le C \hbar / \mathfrak {n}^2\).
The error term \(\mathfrak {E}_{\mathrm{int}}^{(2)}\) differs from \(\mathfrak {E}_{\mathrm{int}}^{(1)}\) only in the replacement of the index set \(\mathcal {I}_{k}^{+}\) by \(\mathcal {I}_{k}^{-}\). Therefore, we find \(|\mathfrak {E}_{\mathrm{int}}^{(2)}|\le C \hbar / \mathfrak {n}^2\). As for the error term \(\mathfrak {E}_{\mathrm{int}}^{(3)}\), it also differs from \(\mathfrak {E}_{\mathrm{int}}^{(1)}\) in the index set, some hermitian conjugations, and the appearance of a \(\cosh \) instead of a \(\sinh \). The estimates however remain valid and we also obtain \(|\mathfrak {E}_{\mathrm{int}}^{(3)}| \le C \hbar / \mathfrak {n}^2\). \(\quad \square \)
Proof of the main theorem
Proof of Theorem 2.1
Recall the definition (3.8) of the correlation Hamiltonian and the decomposition (3.9) of the quartic interaction \(Q_N\). Combining the results of Sects. 5.1, 5.2, 5.3, Propositions 5.5 and 5.6, we conclude that
$$\begin{aligned} \begin{aligned}&\langle \xi , \mathcal {H}_{{\mathrm{corr}}} \xi \rangle \\&\quad = \hbar \kappa \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|{\text {tr}}\left( (D(k)+W(k)) \sinh ^2 (K(k)) + \tilde{W}(k) \sinh (K(k)) \cosh (K(k)) \right) + \mathfrak {E} \end{aligned} \end{aligned}$$
for an error \(\mathfrak {E}\) such that
$$\begin{aligned} |\mathfrak {E}|\le C \Big [ \frac{1}{N} + \frac{1}{M} + \frac{\hbar }{\mathfrak {n}^2} + \hbar N^{-\delta /2} \Big ] \end{aligned}$$
with \(\hbar = N^{-1/3}\) and \(\mathfrak {n}= N^{1/3-\delta /2} M^{-1/2}\).
To evaluate the main part of the expectation value explicitly, notice that by definition (4.17) of K(k) we have
$$\begin{aligned} \sinh (K(k)) \,{=}\, \frac{1}{2} \left( |S_1(k)^T|- |S_1(k)^T|^{-1} \right) , \quad \cosh (K(k)) \,{=}\, \frac{1}{2} \left( |S_1(k)^T|+ |S_1(k)^T|^{-1} \right) . \end{aligned}$$
Notice also that \(S_1(k) S_1(k)^T = |S_1(k)^T|^2\) and \( \left( |S_1(k)^T|^{-1} \right) ^2 = S_2(k) S_2(k)^T\), where
$$\begin{aligned}S_2(k) = \left( D(k) + W(k) - \tilde{W}(k)\right) ^{-1/2} E(k)^{1/2}.\end{aligned}$$
Consequently
$$\begin{aligned} \begin{aligned} \sinh (K(k)) \cosh (K(k))&= \frac{1}{4} \left( |S_1(k)^T|- |S_1(k)^T|^{-1} \right) ^T \left( |S_1(k)^T|+ |S_1(k)^T|^{-1} \right) \\&= \frac{1}{4} \left( S_1(k) S_1(k)^T - S_2(k) S_2(k)^T \right) .\end{aligned} \end{aligned}$$
Likewise
$$\begin{aligned} \begin{aligned} \sinh ^2(K(k))&= \frac{1}{4} \left( |S_1(k)^T|- |S_1(k)^T|^{-1} \right) ^T \left( |S_1(k)^T|- |S_1(k)^T|^{-1} \right) \\&= \frac{1}{4} \left( S_1(k) S_1(k)^T + S_2(k) S_2(k)^T - 2 \mathbb {I}\right) . \end{aligned} \end{aligned}$$
Now using the explicit form (4.18) of \(S_1(k)\), E(k), and \(S_2(k)\), this can be simplified to yield
$$\begin{aligned} \langle \xi , \mathcal {H}_{{\mathrm{corr}}} \xi \rangle&= \frac{\hbar \kappa }{4} \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|\Big ( {\text {tr}}\left( D(k)+W(k)+\tilde{W}(k)\right) S_1(k) S_1(k)^T \nonumber \\&\quad + {\text {tr}}\left( D(k)+W(k)-\tilde{W}(k)\right) S_2(k) S_2(k)^T \Big ) \nonumber \\&\quad - \frac{\hbar \kappa }{2} \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|{\text {tr}}\big ( D(k)+W(k) \big ) + \mathfrak {E}\nonumber \\&= \hbar \kappa \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|\left( \frac{1}{2} {\text {tr}}E(k) - \frac{1}{2} {\text {tr}}\big (D(k)+W(k)\big )\right) + \mathfrak {E}. \end{aligned}$$
(5.12)
We are left with evaluating the traces in (5.12).
Evaluation of the traces. For simplicity, we shall drop the k-dependence in the notation (we will restore it in (5.14)). Recall the block diagonalization (4.25), by which
$$\begin{aligned} \begin{aligned} \frac{1}{2}{\text {tr}}E&= \frac{1}{2}{\text {tr}}\left[ \begin{pmatrix} d &{} 0 \\ 0 &{} d+2b \end{pmatrix}^{1/2} \begin{pmatrix} d+2b &{} 0 \\ 0 &{} d \end{pmatrix} \begin{pmatrix} d &{} 0 \\ 0 &{} d+2b \end{pmatrix}^{1/2}\right] ^{1/2}\\&= \frac{1}{2}{\text {tr}}\left[ d^{1/2} (d+2b) d^{1/2}\right] ^{1/2} + \frac{1}{2}{\text {tr}}\left[ (d+2b)^{1/2} d (d+2b)^{1/2}\right] ^{1/2}\\&= {\text {tr}}\left[ d^{1/2} (d+2b) d^{1/2}\right] ^{1/2}, \end{aligned} \end{aligned}$$
(5.13)
since \(d^{1/2} (d+2b) d^{1/2}\) and \((d+2b)^{1/2} d (d+2b)^{1/2}\) have the same spectrum. To calculate this trace, notice that
$$\begin{aligned} d^{1/2}(d+2b)d^{1/2} = d^2 + 2g |{\tilde{u}}\rangle \langle {\tilde{u}} |\end{aligned}$$
is a rank-one perturbation of a diagonal operator, with diagonal part \(d^2 = {\text {diag}}(u_\alpha ^4: \alpha =1,\ldots I)\) and with \({\tilde{u}} = ( v_1 u_1, \ldots , v_{I} u_{I} ) \in \mathbb {R}^{I}\).
The resolvent of a matrix with rank-one perturbation can easily be calculated: For any invertible matrix \(A \in \mathbb {C}^{n\times n}\), and \(x,y \in \mathbb {C}^n\),
$$\begin{aligned} (A+|x\rangle \langle y|)^{-1} = A^{-1} - \frac{A^{-1}|x\rangle \langle y|A^{-1}}{1+\langle y,A^{-1} x\rangle } \end{aligned}$$
whenever the right-hand side is well-defined. So for \(\lambda \in [0,\infty )\) we find
$$\begin{aligned} \left( d^2 + 2g|{\tilde{u}}\rangle \langle {\tilde{u}}|+\lambda ^2\right) ^{-1} = \left( d^2 + \lambda ^2 \right) ^{-1} - \frac{2g}{1+2g \sum _{\alpha =1}^{I_k} \frac{u_\alpha ^2 v_\alpha ^2}{u_\alpha ^4 + \lambda ^2}} \left|w \right\rangle \left\langle w\right|\;, \end{aligned}$$
with \(w \in \mathbb {R}^{I}\) defined by \(w_\alpha = u_\alpha v_\alpha (u_\alpha ^4 + \lambda ^2)^{-1}\). By functional calculus, for any non-negative operator A we have the identity
$$\begin{aligned} \sqrt{A}= \frac{2}{\pi }\int _0^\infty \left( \mathbb {I}- \frac{\lambda ^2}{A+\lambda ^2} \right) {{\mathrm{d}}}\lambda . \end{aligned}$$
Using the integral identity twice we find
$$\begin{aligned} {\text {tr}}\left[ d^{1/2} (d+2b) d^{1/2}\right] ^{1/2}&= \frac{2}{\pi } \int _0^{\infty }{\text {tr}}\left( \mathbb {I}- \frac{\lambda ^2}{d^2 + \lambda ^2}\right) {{\mathrm{d}}}\lambda \\&\quad + \frac{2}{\pi } \int _0^{\infty } \frac{\lambda ^2\, 2g}{1+2g\sum _{\alpha =1}^{I} \frac{u_\alpha ^2 v_\alpha ^2}{u_\alpha ^4+\lambda ^2}}||w||^2{{\mathrm{d}}}\lambda \\&= {\text {tr}}d + \frac{2}{\pi } \int _0^\infty \frac{\lambda ^2}{1+2g\sum _{\alpha =1}^{I} \frac{u_\alpha ^2 v_\alpha ^2}{u_\alpha ^4+\lambda ^2}} 2g \sum _{\alpha =1}^{I} \frac{u_\alpha ^2 v_\alpha ^2}{(u_\alpha ^4+\lambda ^2)^2}{{\mathrm{d}}}\lambda . \end{aligned}$$
Restoring the k-dependence, let
$$\begin{aligned} f_k (\lambda ) := 1+2g(k)\sum _{\alpha =1}^{I_k} \frac{u_\alpha (k)^2 v_\alpha (k)^2}{u_\alpha (k)^4+\lambda ^2}. \end{aligned}$$
Integrating by parts (noting that the boundary terms vanish since \(\log f_k(\lambda ) \sim 1/\lambda ^2\)), we find
$$\begin{aligned} {\text {tr}}\left[ d^{1/2} (d+2b) d^{1/2}\right] ^{1/2} = \frac{1}{2}{\text {tr}}D - \frac{1}{\pi } \int _0^\infty \lambda \frac{f_k'(\lambda )}{f_k (\lambda )}{{\mathrm{d}}}\lambda&= \frac{1}{2}{\text {tr}}D + \frac{1}{\pi } \int _0^\infty \log f_k (\lambda ){{\mathrm{d}}}\lambda . \end{aligned}$$
Thus, inserting in (5.13) and then in (5.12), we obtain
$$\begin{aligned} \langle \xi ,\mathcal {H}_{{\mathrm{corr}}} \xi \rangle= & {} \hbar \kappa \sum _{k\in \Gamma ^{{\mathrm{nor}}}} |k|\left( \frac{1}{\pi }\int _0^\infty \log f_k (\lambda ){{\mathrm{d}}}\lambda - g(k) \sum _{\alpha =1}^{I_k}v_\alpha (k)^2 \right) + \mathfrak {E} \end{aligned}$$
(5.14)
where we used that according to (4.24) \({\text {tr}}W= 2 {\text {tr}}b =2g\sum _{\alpha =1}^{I} v_\alpha ^2\).
Convergence to the Gell-Mann–Brueckner formula. To conclude the proof of Theorem 2.1, we show that (5.14) reproduces the Gell-Mann–Brueckner formula as stated in the theorem. Let
$$\begin{aligned} \tilde{f}_k (\lambda ) := 1+ 4\pi g(k)\left( 1-\lambda \arctan \left( \frac{1}{{\lambda }} \right) \right) . \end{aligned}$$
We claim that
$$\begin{aligned}&\Big |\Big ( \frac{1}{\pi }\int _0^\infty \log f_k (\lambda ){{\mathrm{d}}}\lambda - g(k) \sum _{\alpha =1}^{I_k}v_\alpha (k)^2 \Big ) - \Big ( \frac{1}{\pi }\int _0^\infty \log \tilde{f}_k (\lambda ){{\mathrm{d}}}\lambda - g(k)\pi \Big ) \Big |\nonumber \\&\quad \le C \left( M^{1/4 }N^{-\frac{1}{6}+\frac{\delta }{2}} + N^{-\frac{\delta }{2}} + M^{-\frac{1}{4}}N^{\frac{\delta }{2}} \right) . \end{aligned}$$
(5.15)
Since \(\log \tilde{f}_k (\lambda ) = g(k) = 0\) for all \(|k|> R\), inserting (5.15) into (5.14) we obtain
$$\begin{aligned} \langle \xi ,\mathcal {H}_{{\mathrm{corr}}} \xi \rangle= & {} \hbar \kappa \!\!\sum _{k\in \Gamma ^{{\mathrm{nor}}}}\!\! |k|\! \left( \frac{1}{\pi }\!\int _0^\infty \! \log \! \left[ 1 + 4\pi g(k) \left( 1- {\lambda } \arctan \left( \frac{1}{{\lambda }}\right) \right) \right] {{\mathrm{d}}}\lambda - g(k)\pi \right) \\&+ \tilde{\mathfrak {E}} \end{aligned}$$
with an error
$$\begin{aligned} \begin{aligned} |\tilde{\mathfrak {E}}|&\le C \Big [ N^{-1} + M^{-1} + N^{-1+\delta } M\Big ] + C\hbar \Big [ M^{1/4 }N^{-\frac{1}{6}+\frac{\delta }{2}} + N^{-\frac{\delta }{2}} + M^{-\frac{1}{4}}N^{\frac{\delta }{2}} \Big ]. \end{aligned} \end{aligned}$$
Recalling that \(M = N^{1/3 + \epsilon }\) and optimizing over \(0< \epsilon < 1/3\), \(0< \delta < 1/6 - \epsilon /2\), we find (with \(\epsilon = 1/27\) and \(\delta = 2/27\)), that \(|\tilde{\mathfrak {E}} |\le C \hbar N^{-1/27}\). Replacing the sum over \(k \in \Gamma ^{{\mathrm{nor}}}\) by 1 / 2 times the sum over \(k \in \mathbb {Z}^3\), and replacing \(\kappa = \kappa _0 + \mathcal {O}(N^{-1/3})\) by \(\kappa _0 = (3/4\pi )^{1/3}\) (using also the Lipschitz continuity of the logarithm), we arrive at (2.1).
We still have to show (5.15). To this end, recall from Proposition 3.1 that, in terms of the surface measure \(\sigma \) of the patch \(p_\alpha \) on the unit sphere, we have
$$\begin{aligned} v_\alpha (k)^2 = \sigma (p_\alpha ) u_\alpha (k)^2\Big ( 1 + \mathcal {O}\Big (\sqrt{M}N^{-\frac{1}{3}+\delta }\Big ) \Big ). \end{aligned}$$
Thus
$$\begin{aligned} f_k(\lambda )= & {} 1+2 g(k) \sum _{\alpha =1}^{I_k} \frac{u_\alpha (k)^2 v_\alpha (k)^2}{u_\alpha (k)^4+\lambda ^2} \\= & {} 1+2 g(k) \sum _{\alpha =1}^{I_k}\sigma (p_\alpha )\frac{u_\alpha (k)^4}{u_\alpha (k)^4+\lambda ^2} + \mathcal {O}\left( \sqrt{M}N^{-\frac{1}{3}+\delta }\right) . \end{aligned}$$
We approximate this Riemann sum by the corresponding surface integral over a subset of \(\mathbb {S}^2\). We write \(\cos \theta _\alpha = \hat{k}\cdot \hat{\omega }_\alpha = u_\alpha (k)^2\) and \(\varphi _\alpha \) for the azimuth of \(\omega _\alpha \). We parametrize the surface integrals in the same spherical coordinate systemFootnote 9 (i. e., the inclination \(\theta \) is measured with respect to k, and the azimuth \(\varphi \) in the plane perpendicular to k). We estimate every summand by
$$\begin{aligned}&\left|\int _{p_\alpha } \frac{\cos ^2 \theta }{\cos ^2 \theta + \lambda ^2} {{\mathrm{d}}}\sigma - \sigma (p_\alpha ) \frac{\cos ^2 \theta _\alpha }{\cos ^2 \theta _\alpha +\lambda ^2} \right|\\&\quad \le \int _{p_\alpha } \left|\frac{\cos ^2 \theta }{\cos ^2 \theta + \lambda ^2} - \frac{\cos ^2 \theta _\alpha }{\cos ^2 \theta _\alpha +\lambda ^2} \right|{{\mathrm{d}}}\sigma \\&\quad \le \iint _{\hat{\omega }(\theta ,\varphi ) \in p_\alpha } \left|\frac{\cos ^2 \theta }{\cos ^2 \theta + \lambda ^2} - \frac{\cos ^2 \theta _\alpha }{\cos ^2 \theta _\alpha +\lambda ^2} \right||\sin \theta |{{\mathrm{d}}}\theta {{\mathrm{d}}}\varphi \;. \end{aligned}$$
Bounding the difference using the supremum of the derivative
$$\begin{aligned} \left|\int _{p_\alpha } \frac{\cos ^2 \theta }{\cos ^2 \theta + \lambda ^2} {{\mathrm{d}}}\sigma - \sigma (p_\alpha ) \frac{\cos ^2 \theta _\alpha }{\cos ^2 \theta _\alpha +\lambda ^2} \right|&\le \sup _{\hat{\omega }(\theta ,\varphi ) \in p_\alpha } \left|\frac{{{\mathrm{d}}}}{{{\mathrm{d}}}\theta } \frac{\cos ^2 \theta }{\cos ^2 \theta + \lambda ^2} \right|\frac{C}{\sqrt{M}} \sigma (p_\alpha ), \end{aligned}$$
(5.16)
where we also used that, since the partition is diameter bounded, \(\sup _{(\theta ,\varphi ) \in p_\alpha } |\theta -\theta _\alpha |\le C/\sqrt{M}\). The derivative is bounded by
$$\begin{aligned} \left|\frac{{{\mathrm{d}}}}{{{\mathrm{d}}}\theta } \frac{\cos ^2 \theta }{\cos ^2 \theta + \lambda ^2} \right|\le 2 \frac{\lambda ^2}{\cos ^2 \theta + \lambda ^2 } \frac{|\cos \theta ||\sin \theta |}{\cos ^2 \theta + \lambda ^2 } \le \frac{2}{|\cos \theta |}. \end{aligned}$$
Recall that \(\alpha \in \{1,2,\ldots ,I_k\}\), which by definition of the index set implies \(\cos \theta _\alpha > N^{-\delta }\). The bound \(|\theta - \theta _\alpha |\le CM^{-1/2}\) implies that also \(\cos \theta > CN^{-\delta }\). So (5.16) implies
$$\begin{aligned} \left|\int _{p_\alpha } \frac{\cos ^2 \theta }{\cos ^2 \theta + \lambda ^2} {{\mathrm{d}}}\sigma - \sigma (p_\alpha ) \frac{\cos ^2 \theta _\alpha }{\cos ^2 \theta _\alpha +\lambda ^2} \right|\le C \frac{N^\delta }{M^{3/2}}. \end{aligned}$$
Since the number of patches is at most M we conclude that
$$\begin{aligned} \Big | \sum _{\alpha =1}^{I_k}\sigma (p_\alpha ) \frac{u_\alpha (k)^4}{u_\alpha (k)^4+\lambda ^2} - \int _{\mathbb {S}^2_{\mathrm{reduced}}} \frac{\cos ^2 \theta }{\cos ^2 \theta + \lambda ^2} {{\mathrm{d}}}\sigma \Big | \le C \frac{N^{\delta }}{\sqrt{M}}. \end{aligned}$$
Here we wrote \(\mathbb {S}^2_{\mathrm{reduced}}\) for a unit half-sphere excluding the collar of width \(N^{-\delta }\) and the corridors \(p_{\mathrm{corri}}\). Since \(\cos ^2 \theta /(\cos ^2 \theta + \lambda ^2) \le 1\) we can compare to the integral over the whole unit half-sphere \(\mathbb {S}^2_{\mathrm{half}}\),
$$\begin{aligned} \left|\int _{\mathbb {S}^2_{\mathrm{reduced}}} \frac{\cos ^2 \theta }{\cos ^2 \theta + \lambda ^2} {{\mathrm{d}}}\sigma - \int _{\mathbb {S}^2_{\mathrm{half}}} \frac{\cos ^2 \theta }{\cos ^2 \theta + \lambda ^2} {{\mathrm{d}}}\sigma \right|\le C \left[ N^{-\delta } + M^{1/2} N^{-1/3} \right] . \end{aligned}$$
The surface integral over the unit half-sphere is easy to compute,
$$\begin{aligned} \begin{aligned} \int _{\mathbb {S}^2_{\mathrm{half}}}\!\! \frac{\cos ^2 \theta }{\cos ^2 \theta + \lambda ^2} {{\mathrm{d}}}\sigma&= \int _{0}^{\pi /2}\!\! {{\mathrm{d}}}\theta \sin (\theta ) \frac{\cos (\theta )^2}{\cos (\theta )^2+\lambda ^2} \int _0^{2\pi }\!\! {{\mathrm{d}}}\varphi = 2\pi \Big (1-{\lambda } \arctan \Big ( \frac{1}{{\lambda }} \Big )\Big ). \end{aligned} \end{aligned}$$
(5.17)
Since \(g(k) = \kappa \hat{V}(k)\) is uniformly bounded (by assumption on \(\hat{V}\)), we conclude that
$$\begin{aligned} \left|f(\lambda ) - \tilde{f}(\lambda ) \right|\le C \Big ( \sqrt{M}N^{-\frac{1}{3}+\delta } + N^{-\delta } + \frac{N^\delta }{\sqrt{M}} \Big ). \end{aligned}$$
Since for \(x\ge 0\) the function \(x \mapsto \log (1+x)\) has Lipschitz constant 1 we get
$$\begin{aligned} \left|\log f(\lambda ) - \log \tilde{f}(\lambda ) \right|\le C \Big ( \sqrt{M}N^{-\frac{1}{3}+\delta } + N^{-\delta } + \frac{N^\delta }{\sqrt{M}} \Big ). \end{aligned}$$
It remains to compare the integrals over \(\lambda \). Since \(\log (1+x) \le x\) for all \(x \ge 0\), we have
$$\begin{aligned} \left|\log f(\lambda ) \right|\le 2g(k) \sum _{\alpha =1}^{I_k} \sigma (p_\alpha ) \frac{u_\alpha (k)^4}{u_\alpha (k)^4 + \lambda ^2} \le 2g(k) \sum _{\alpha =1}^{I_k} \frac{C}{M} \frac{1}{\lambda ^2} \le \frac{C}{\lambda ^2}, \end{aligned}$$
where we used the two inequalities \(0 \le u_\alpha (k)^4 \le 1\). Using the integral identity (5.17) it is easy to see that also
$$\begin{aligned} \left|\log \tilde{f}(\lambda ) \right|\le 4\pi g(k) \Big | 1-{\lambda }\arctan \left( \frac{1}{{\lambda }}\right) \Big | \le \frac{C}{\lambda ^2}. \end{aligned}$$
Using the last three estimates, by splitting the integration at some \(\Lambda > 0\) to be optimized in the last step, we obtain
$$\begin{aligned}&\left|\frac{1}{\pi } \int _0^\infty \log f(\lambda ){{\mathrm{d}}}\lambda - \frac{1}{\pi } \int _0^\infty \log {\tilde{f}}(\lambda ){{\mathrm{d}}}\lambda \right|\nonumber \\&\quad \le \frac{1}{\pi } \int _0^{\Lambda } \left|\log f(\lambda ) - \log {\tilde{f}}(\lambda ) \right|{{\mathrm{d}}}\lambda + \frac{1}{\pi } \int _{\Lambda }^\infty \frac{8\pi g(k)}{\lambda ^2}{{\mathrm{d}}}\lambda \nonumber \\&\quad \le C \Lambda \left( \sqrt{M}N^{-\frac{1}{3}+\delta } + N^{-\delta } + \frac{N^\delta }{\sqrt{M}} \right) + C \Lambda ^{-1}\nonumber \\&\quad \le C \left( M^{1/4 }N^{-\frac{1}{6}+\frac{\delta }{2}} + N^{-\frac{\delta }{2}} + M^{-\frac{1}{4}}N^{\frac{\delta }{2}} \right) . \end{aligned}$$
(5.18)
By a similar (simpler) Riemann sum argument we obtain
$$\begin{aligned} -g(k)\sum _{\alpha =1}^{I_k} v_\alpha ^2(k)&= -g(k) \sum _{\alpha =1}^{I_k} \sigma (p_\alpha ) u_\alpha ^2(k) \left( 1 + \mathcal {O}\left( \sqrt{M}N^{-\frac{1}{3}+\delta }\right) \right) \\&= - g(k)\pi + \mathcal {O}\left( \sqrt{M}N^{-\frac{1}{3}+\delta } + N^{-\delta }\right) \end{aligned}$$
where the error is obviously smaller than (5.18). This concludes the proof of (5.15). \(\quad \square \)