## 1 Introduction

In this paper, we consider interacting, spin 1/2 fermions, in three dimensions, in the thermodynamic limit. We will focus on the ground state energy of the system, for positive and short-ranged interaction potentials. Let $$\rho _{\sigma }$$ be the density of particles with spin-up, $$\sigma =\, \uparrow$$, or spin-down, $$\sigma =\, \downarrow$$. Let $$e(\rho _{\uparrow }, \rho _{\downarrow })$$ be the ground state energy density of the system, in the thermodynamic limit. We will be interested in the dilute regime, corresponding to $$\rho _{\sigma } \ll 1$$. It is well known that, in units such that $$\hbar = 1$$ and setting the masses of the particles to be equal to 1/2:

\begin{aligned} e(\rho _{\uparrow }, \rho _{\downarrow }) = \frac{3}{5}(6\pi ^{2})^{\frac{2}{3}} (\rho _{\uparrow }^{\frac{5}{3}} + \rho _{\downarrow }^{\frac{5}{3}}) + 8\pi a \rho _{\uparrow } \rho _{\downarrow } + o(\rho ^{2}). \end{aligned}
(1.1)

The first term in the right-hand side of Eq. (1.1) is purely kinetic, and its $$\rho ^{5/3}$$-dependence is a consequence of the fermionic nature of the wave function. It is easy to find a fermionic state that reproduces the correct $$\rho ^{5/3}$$ dependence of the energy; this is the free Fermi gas, i.e., the fermionic state that minimizes the total kinetic energy of the systems, in a way compatible with Pauli principle.

The effect of the interaction is visible at the next order; denoting by V the two-body potential, the constant a in Eq. (1.1) is the scattering length of the potential. For small potentials, it can be computed as a perturbative expansion in V, via the Born series. It is easy to check that taking the free Fermi gas as a trial state for the many-body problem, the $$\rho ^{2}$$-dependence of the ground state energy is off by an order 1 multiplicative constant: instead of $$8\pi a$$ one finds $$\hat{V}(0)$$, which is strictly larger than $$8\pi a$$. To reproduce the correct dependence of the energy in the interaction, one has to understand the effect of correlations in the fermionic ground state, which is not an easy task even from the point of view of an upper bound.

The first proof of (1.1) has been given by Lieb, Seiringer and Solovej in an important work [25]. The proof of [25] covers a large class of positive two-body potentials, including the case of hard spheres. The result of [25] has then been extended by Seiringer to the computation of the thermodynamic pressure for positive temperature Fermi gases [32]. Concerning interacting lattice fermions (Hubbard model), the analogue of Eq. (1.1) follows from the upper bound of Giuliani [18] and from the lower bound of Seiringer and Yin [33].

For bosonic systems, in a seminal paper [26] Lieb and Yngvason proved that the ground state energy density of the dilute Bose gas is, assuming the particles to be spinless:

\begin{aligned} e(\rho ) = 4\pi a \rho ^{2} + o(\rho ^{2}). \end{aligned}
(1.2)

In this expression, the interaction determines the ground state energy at leading order, in contrast to (1.1). This is consistent with the fact that bosons tend to minimize the energy occupying the lowest momentum state, which gives no contribution to the ground state energy. This is of course forbidden for fermions, due to Pauli principle. The result of [26] has been recently improved by Fournais and Solovej in [17]. The work [17] obtained a more refined asymptotics for the ground state energy density, from the point of view of a lower bound. Combined with the upper bound of Yau and Yin in [35], see also [2] for a streamlined proof, the work [17] determined the next-order correction to the ground state energy of dilute bosons and put on rigorous grounds the celebrated Lee–Huang–Yang formula.

Comparing Eq. (1.1) with Eq. (1.2), one is naturally tempted to think the low energy excitations around the free Fermi gas as pairs of fermions, which can be described by emergent bosonic particles, whose ground state energy reconstructs the $$8\pi a \rho _{\uparrow } \rho _{\downarrow }$$ term in (1.1). (The extra factor 2 is due to the spin degrees of freedom.) The main motivation of the present paper is to make this intuition mathematically precise.

For the bosonic problem, a natural trial state that captures the correct dependence of the ground state energy on the scattering length is provided by a suitable unitary transformation, a Bogoliubov rotation, of a coherent state, [16]. Interestingly, for small potentials, the energy of this trial state also reproduces the Lee–Huang–Yang formula for the next-order correction to the energy [16, 27, 28], up to higher-order terms in the interaction. In this paper, we introduce the fermionic analogue of such transformations, roughly by considering pairs of fermions as effective bosons. The main difficulty we have to face is that, in the language of second quantization, quadratic expressions in the bosonic creation and annihilation operators become quartic in terms of the fermionic operators. As a consequence, the nice algebraic properties of Bogoliubov transformations are only approximately true, in the fermionic setting; quantifying the validity of this approximation is a nontrivial task, and it is the main technical challenge faced in the present paper.

The main application of our method is a new proof of (1.1). Our result comes with a substantial improvement of the error estimate. However, it is restricted to more regular interaction potentials with respect to [25]. In particular, the result of [25] includes the case of hard spheres, which we cannot cover at the moment. We believe that a larger class of interactions could be treated by approximation arguments, but we have not tried to extend the result in this direction. Nevertheless, we think that our approach is conceptually simple, and that it gives a new point of view on dilute Fermi gases. That said, our paper is significantly longer than [25]; the length is partially motivated by the fact that our manuscript does not rely on any prior work: the technical tools needed for our analysis are all developed here.

Our method borrows ideas from a series of recent, groundbreaking works of Boccato, Brennecke, Cenatiempo and Schlein [10,11,12,13,14]. There, Bogoliubov theory for interacting Bose gases in the Gross–Pitaevskii regime has been put on rigorous grounds, and it has been used to obtain sharp asymptotics on the ground state energy and on the excitation spectrum. Concerning the energy asymptotics of interacting fermions in the mean-field regime, the first rigorous result about the correlation energy, defined as the difference between the many-body and Hartree–Fock ground state energies, has been obtained in [22]. In [22], the correlation energy has been rigorously computed for small potentials via upper and lower bounds, that agree at second order in the interaction. The proof is based on rigorous second-order perturbation theory, first developed in [19, 20]. The method that we introduce in the present paper is related to the bosonization approach of [5, 6]. The method of [5, 6] allowed to compute the correlation energy of weakly interacting, mean-field fermionic systems, at all orders in the interaction strength. The result of [5, 6] confirmed the prediction of the random phase approximation, see [3] for a review. Both [5, 6, 22] make use of Fock space methods and fermionic Bogoliubov transformations, extending ideas previously introduced in the context of many-body fermionic dynamics [4, 7,8,9, 29].

A key technical ingredient of [5, 6] is the localization of the low energy excitations around the Fermi surface in terms of suitable patches, where the quasi-particle dispersion relation can be approximated by a linear one. This is not needed in the dilute regime considered here, due to the fact that the Fermi momentum is much smaller than the typical momentum exchanged in the two-body scattering. Another difference with respect to [5, 6] is that here we consider interacting systems in the thermodynamic limit; controlling this limit is nontrivial, due to the slow decay of the correlations for the free Fermi gas, which plays the role of reference state in our analysis, and of the solution of the scattering equation. Although the mean-field regime and the dilute regime are somewhat opposite, we find it remarkable that similar bosonization ideas apply in both cases.

As a future perspective, we think that the method presented in this paper might provide a good starting point for the derivation of more refined energy asymptotics, by importing tools that have been developed in the last decade for interacting bosons. An outstanding open problem is to prove the fermionic analogue of the Lee–Huang–Yang formula, due to Huang and Yang in [23], which gives the next-order correction to the ground state energy of dilute fermions, of order $$\rho ^{7/3}$$ (in three dimensions).

The paper is organized as follows. In Sect. 2 we define the model and state our main result, Theorem 2.1. In Sect. 3 we formulate the problem in Fock space, and we introduce fermionic Bogoliubov transformations, which will allow to extract the $$\rho ^{5/3}$$ dependence of the ground state energy, and part of the $$\rho ^{2}$$ dependence. In Sect. 4 we define the fermionic analogue of the bosonic Bogoliubov transformation, called correlation structure, that will allow us to compute the ground state energy at order $$\rho ^{2}$$; see Sect. 4.1 for a heuristic discussion. Section 5 is the main technical section of the paper; here, we discuss the properties of the correlation structure that mimic the algebraic properties of bosonic Bogoliubov transformations at leading order in the density. In Sect. 6 we use the discussion of Sect. 5 to prove a lower bound for the ground state energy that displays the correct dependence of the scattering length at order $$\rho ^{2}$$. Then, in Sect. 7 we conclude the proof of Theorem 2.1 by proving a matching upper bound, by the choice of a suitable trial state. Finally, in Appendix 7 we collect properties of the solution of the scattering equation that we shall use in our proofs; in Appendix 7 we prove some technical estimates for almost-bosonic operators; and in Appendix 7 we collect technical estimates on the infrared and ultraviolet regularizations of various expressions appearing in our proofs.

## 2 Main Result

We consider a system of N interacting, spinning fermions in a cubic box $$\Lambda _{L} = [0,L]^{3}$$, with periodic boundary conditions. The Hamiltonian of the model acts on $$L^{2}(\Lambda _{L}; \mathbb {C}^{2})^{\otimes N}$$, and it is given by:

\begin{aligned} H_{N} = -\sum _{i=1}^{N} \Delta _{x_{i}} + \sum _{i<j = 1}^{N} V(x_{i} - x_{j}), \end{aligned}
(2.1)

with $$\Delta _{x_{i}}$$ the Laplacian acting on the i-th particle, and V the pair interaction potential. We shall suppose that V is the ‘periodization’ on $$\Lambda _{L}$$ of a potential $$V_{\infty }$$ on $$\mathbb {R}^{3}$$, compactly supported and regular enough:

\begin{aligned} V(x-y) = \frac{1}{L^{3}} \sum _{p\in \frac{2\pi }{L}\mathbb {Z}^{3}} e^{ip\cdot (x - y)} \hat{V}_{\infty }(p), \end{aligned}
(2.2)

with $$\hat{V}_{\infty }(p) = \int _{\mathbb {R}^{3}} dx\, e^{-ip\cdot x} V_{\infty }(x)$$. We shall denote by $$N_{\sigma }$$ the number of particles with a given spin $$\sigma \in \{ \uparrow , \downarrow \}$$, and we shall set $$N = N_{\uparrow } + N_{\downarrow }$$. We shall require the wave function of the system, on which the Hamiltonian acts, to be antisymmetric separately in the first $$N_{\uparrow }$$ variables, and in the second $$N_{\downarrow }$$ variables. That is, the space of allowed wave functions is $$\mathfrak {h}(N_{\uparrow }, N_{\downarrow }) := L^{2}_{\text {a}}(\Lambda _{L}^{N_{\uparrow }}) \otimes L^{2}_{\text {a}}(\Lambda _{L}^{N_{\downarrow }})$$, with $$L^{2}_{\text {a}}(\Lambda _{L}^{N_{\sigma }}) = L^{2}(\Lambda _{L})^{\wedge N_{\sigma }}$$ the antisymmetric sector of $$L^{2}(\Lambda _{L})^{\otimes N_{\sigma }}$$. We will focus on the ground state energy of the system:

\begin{aligned} E_{L}(N_{\uparrow }, N_{\downarrow }) = \inf _{\Psi \in \mathfrak {h}(N_{\uparrow }, N_{\downarrow })} \frac{\langle \Psi , H_{N} \Psi \rangle }{\langle \Psi , \Psi \rangle }. \end{aligned}
(2.3)

By translation invariance of the Hamiltonian, the energy is extensive in the system size. Thus, let us define the ground state energy density as:

\begin{aligned} e_{L}(\rho _{\uparrow }, \rho _{\downarrow }) := \frac{E_{L}(N_{\uparrow }, N_{\downarrow })}{L^{3}}. \end{aligned}
(2.4)

Let $$\rho _{\sigma } = N_{\sigma } / L^{3}$$ be the density of particles with spin $$\sigma$$, and let $$\rho = \rho _{\uparrow } + \rho _{\downarrow }$$ be the total density. We shall be interested in the thermodynamic limit, meaning $$N_{\sigma }, L\rightarrow \infty$$, with $$\rho _{\sigma }$$ fixed. The existence of the limit, and the independence of the limit from the choice of the boundary conditions, is well known [30, 31]. We shall focus on the dilute regime, corresponding to $$\rho \ll 1$$. The next theorem is our main result.

### Theorem 2.1

Let $$V\in C^{\infty }(\Lambda _{L})$$, compactly supported, $$V\ge 0$$. There exists $$L_{0}>0$$ large enough such that for $$L\ge L_{0}$$ the following holds:

\begin{aligned} e_{L}(\rho _{\uparrow }, \rho _{\downarrow }) = \frac{3}{5}(6\pi ^{2})^{\frac{2}{3}} (\rho _{\uparrow }^{\frac{5}{3}} + \rho _{\downarrow }^{\frac{5}{3}}) + 8\pi a \rho _{\uparrow } \rho _{\downarrow } + r_{L}(\rho _{\uparrow }, \rho _{\downarrow }), \end{aligned}
(2.5)

where a is the scattering length of the potential V, and for some constant C only dependent on V:

\begin{aligned} -C\rho ^{2 + \xi _{2}} \le r_{L}(\rho _{\uparrow }, \rho _{\downarrow }) \le C\rho ^{2 + \xi _{1}} \end{aligned}
(2.6)

with $$\xi _{1} = \frac{2}{9}$$ and $$\xi _{2} = \frac{1}{9}$$.

### Remark 2.2

1. (i)

As discussed in the introduction, the result is not new. The first proof of (2.5) has been given by Lieb, Seiringer, Solovej in [25]. The extension to positive temperature has been obtained in [32]. The analogue of (2.5) for the Hubbard model follows from the combination of [18, 33].

2. (ii)

With respect to [25], our result is restricted to smooth interaction potentials. However, our result comes with improved error estimates; in [25], $$\xi _{1} = 2/27$$ and $$\xi _{2} = 1/39$$.

3. (iii)

As the inspection of the proof shows, the statement of the main result could actually be easily extended to potentials $$V\in C^{k}(\Lambda _{L})$$ for large enough k. To avoid unnecessary complications, we prefer to state the result in a more restrictive setting.

The rest of the paper is devoted to the proof of Theorem 2.1.

## 3 Second Quantization

### 3.1 Fock Space Representation

In the following, it will be convenient to work in a setting in which the number of particles is not fixed. To this end, we define the fermionic Fock space as:

\begin{aligned} \mathcal {F} = \bigoplus _{n\ge 0} \mathcal {F}^{(n)},\quad \mathcal {F}^{(n)} = L^{2}(\Lambda _{L}; \mathbb {C}^{2})^{\wedge n} \end{aligned}
(3.1)

with the understanding that $$\mathcal {F}^{(0)} = \mathbb {C}$$. Thus, a given element $$\psi \in \mathcal {F}$$ is an infinite sequence of fermionic wave functions, $$\psi = (\psi ^{(0)}, \psi ^{(1)}, \ldots , \psi ^{(n)}, \ldots )$$ with $$\psi ^{(n)} \in \mathcal {F}^{(n)}$$, $$\psi ^{(n)} \equiv \psi ^{(n)}((x_{1}, \sigma _{1}), \ldots , (x_{n}, \sigma _{n}))$$ and $$(x,\sigma ) \in \Lambda _{L} \times \{\uparrow , \downarrow \}$$. An important example of vector in the Fock space is the vacuum vector $$\Omega$$, describing the zero particle state:

\begin{aligned} \Omega = (1, 0,0, \ldots , 0,\ldots ). \end{aligned}
(3.2)

Next, it is convenient to introduce the fermionic creation/annihilation operators, as follows. For $$f\in L^{2}(\Lambda _{L}; \mathbb {C}^{2}) \simeq L^{2}(\Lambda _{L}) \oplus L^{2}(\Lambda _{L})$$, $$f = (f_{\uparrow }, f_{\downarrow })$$, the fermionic annihilation operator $$a(f) : \mathcal {F}^{(n)} \rightarrow \mathcal {F}^{(n-1)}$$ and creation operator $$a^{*}(f): \mathcal {F}^{(n)} \rightarrow \mathcal {F}^{(n+1)}$$ are defined as:

\begin{aligned}&(a(f) \psi )^{(n)}((x_{1}, \sigma _{1}), \ldots , (x_{n}, \sigma _{n})) \nonumber \\&\quad = \sqrt{n+1} \sum _{\sigma = \uparrow \downarrow } \int _{\Lambda _{L}} dx\, \overline{f_{\sigma }(x)} \psi ^{(n+1)}((x, \sigma ), (x_{1}, \sigma _{1}), \ldots , (x_{n}, \sigma _{n})) \nonumber \\&(a^{*}(f) \psi )^{(n)}((x_{1}, \sigma _{1}), \ldots , (x_{n}, \sigma _{n})) \nonumber \\&\quad = \frac{1}{\sqrt{n}} \sum _{j=1}^{n} (-1)^{j+1} f_{\sigma _{j}}(x_{j}) \psi ^{(n-1)}((x_{1}, \sigma _{1}), \ldots , (x_{j-1}, \sigma _{j-1}), (x_{j+1}, \sigma _{j+1}), \ldots , (x_{n}, \sigma _{n})).\nonumber \\ \end{aligned}
(3.3)

The above definitions are complemented by the requirement that the operator a(f) annihilates the Fock space vacuum, $$a(f)\Omega = 0$$. Definition (3.3) implies that $$a(f)^{*} = a^{*}(f)$$, and that:

\begin{aligned} \{ a(f), a(g) \} = \{ a^{*}(f), a^{*}(g) \} = 0,\quad \{ a(f), a^{*}(g) \} = \langle f, g \rangle _{L^{2}(\Lambda _{L}; \mathbb {C}^{2})} \end{aligned}
(3.4)

where $$\langle f, g \rangle _{L^{2}(\Lambda _{L}; \mathbb {C}^{2})} = \sum _{\sigma = \uparrow \downarrow } \int _{\Lambda _{L}} dx\, \overline{f_{\sigma }(x)} g_{\sigma }(x)$$. As a consequence of the canonical anticommutation relations (3.4), we have:

\begin{aligned} \Vert a(f) \Vert \le \Vert f\Vert _{L^{2}(\Lambda _{L}; \mathbb {C}^{2})},\quad \Vert a^{*}(g)\Vert \le \Vert g\Vert _{L^{2}(\Lambda _{L};\mathbb {C}^{2})}. \end{aligned}
(3.5)

It will also be convenient to represent the creation/annihilation operators in terms of the operator-valued distributions $$a^{*}_{x,\sigma }$$, $$a_{x,\sigma }$$,

\begin{aligned} a(f) = \sum _{\sigma = \uparrow \downarrow } \int _{\Lambda _{L}} dx\, a_{x,\sigma } \overline{f_{\sigma }(x)},\quad a^{*}(g) = \sum _{\sigma = \uparrow \downarrow } \int _{\Lambda _{L}} dx\, a^{*}_{x,\sigma } g_{\sigma }(x), \end{aligned}
(3.6)

where, formally, $$a_{x,\sigma } = a(\delta _{x,\sigma })$$. We used the notation $$\delta _{x,\sigma }(y, \sigma ') = \delta _{\sigma , \sigma '} \delta (x-y)$$, with $$\delta _{\sigma , \sigma '}$$ the Kronecker delta, and $$\delta (x-y)$$ the Dirac delta distribution, periodic over $$\Lambda _{L}$$:

\begin{aligned} \delta (x-y) = \frac{1}{L^{3}} \sum _{k\in \frac{2\pi }{L}\mathbb {Z}^{3}} e^{ik\cdot (x-y)}. \end{aligned}
(3.7)

It will also be convenient to introduce momentum-space fermionic creation and annihilation operators. Let $$f_{k}(x) = L^{-3/2} e^{ik\cdot x}$$, for $$k\in (2\pi / L) \mathbb {Z}^{3}$$. Then:

\begin{aligned} \hat{a}_{k,\sigma } \equiv a_{\sigma }(f_{k}) = \frac{1}{L^{\frac{3}{2}}}\int _{\Lambda _{L}} dx\, a_{x,\sigma } e^{-ik\cdot x},\quad \hat{a}^{*}_{k,\sigma } = a_{\sigma }(f_{k})^{*}. \end{aligned}
(3.8)

These relations can be inverted as follows, for all $$x\in \Lambda _{L}$$:

\begin{aligned} a_{x,\sigma } = \frac{1}{L^{\frac{3}{2}}} \sum _{k\in \frac{2\pi }{L}\mathbb {Z}^{3}} e^{ik\cdot x} \hat{a}_{k,\sigma }. \end{aligned}
(3.9)

We then define the number operator $$\mathcal {N}$$ as:

\begin{aligned} \mathcal {N} = \sum _{\sigma = \uparrow \downarrow }\int _{\Lambda _{L}} dx\, a^{*}_{x,\sigma } a_{x,\sigma } \equiv \sum _{\sigma = \uparrow \downarrow } \sum _{k\in \frac{2\pi }{L}\mathbb {Z}^{3}} \hat{a}^{*}_{k,\sigma } \hat{a}_{k,\sigma }. \end{aligned}
(3.10)

The operator $$\mathcal {N}$$ counts the number of particles in a given sector of the fermionic Fock space, $$(\mathcal {N} \psi )^{(n)} = n\psi ^{(n)}$$. We shall also define the number operator associated to particles with a given spin $$\sigma$$ as:

\begin{aligned} \mathcal {N}_{\sigma } = \int _{\Lambda _{L}} dx\, a^{*}_{x,\sigma } a_{x,\sigma }. \end{aligned}
(3.11)

We shall say that $$\psi \in \mathcal {F}$$ is an N-particle state if $$\mathcal {N} \psi = N \psi$$. Also, we shall say that an N-particle state $$\psi$$, with $$N = N_{\uparrow } + N_{\downarrow }$$, has $$N_{\sigma }$$ particles with spin $$\sigma$$ if $$\mathcal {N}_{\sigma } \psi = N_{\sigma }\psi$$. We shall denote by $$\mathcal {F}^{(N_{\uparrow }, N_{\downarrow })} \subset \mathcal {F}$$ the set of such states. Let us now rewrite the ground state energy of the system in the language of second quantization. We define the second-quantized Hamiltonian as:

\begin{aligned} \mathcal {H}= & {} \sum _{\sigma = \uparrow \downarrow } \int _{\Lambda _{L}} dx\, \nabla _{x} a^{*}_{x,\sigma } \nabla _{x} a_{x,\sigma } \nonumber \\&+ \frac{1}{2}\sum _{\sigma , \sigma ' = \uparrow \downarrow }\int _{\Lambda _{L} \times \Lambda _{L}} dxdy\, V(x-y) a^{*}_{x,\sigma } a^{*}_{y,\sigma '} a_{y,\sigma '} a_{x,\sigma }. \end{aligned}
(3.12)

It is not difficult to check that $$(\mathcal {H} \psi )^{(n)} = H_{n} \psi ^{(n)}$$, for $$n\ge 1$$. By the spin independence of the n-particle Hamiltonian $$H_{n}$$, we rewrite the ground state energy (2.3) as:

\begin{aligned} E_{L}(N_{\uparrow }, N_{\downarrow }) = \inf _{\psi \in \mathcal {F}^{(N_{\uparrow }, N_{\downarrow })}} \frac{\langle \psi , \mathcal {H} \psi \rangle }{\langle \psi , \psi \rangle }. \end{aligned}
(3.13)

Equation (3.13) is a convenient starting point for our analysis.

### 3.2 Fermionic Bogoliubov Transformations

#### 3.2.1 The Free Fermi Gas

A simple upper bound for the ground state energy is obtained taking as a trial state the Slater determinant that minimizes the kinetic energy, for given $$N_{\uparrow }$$, $$N_{\downarrow }$$. We shall refer to this state as the free Fermi gas (FFG). Explicitly,

\begin{aligned}&\Psi _{\text {FFG}}\big ( (x_{i}, \uparrow \}_{i=1}^{N_{\uparrow }}, \{ y_{j}, \downarrow \}_{j=1}^{N_{\downarrow }} \big )\nonumber \\&\quad = \frac{1}{\sqrt{N_{\uparrow }!}} \frac{1}{\sqrt{N_{\downarrow }!}} (\det f^{\uparrow }_{k_{i}}(x_{j}))_{1\le i, j \le N_{\uparrow }} (\det f^{\downarrow }_{k_{i}}(y_{j}))_{1 \le i, j\le N_{\downarrow }}, \end{aligned}
(3.14)

where $$f^{\sigma }_{k}(x) \equiv f_{k}(x)$$, with $$f_{k}(x) = L^{-\frac{3}{2}} e^{ik\cdot x}$$ and $$k\in \mathcal {B}_{F}^{\sigma }$$, with $$\mathcal {B}_{F}^{\sigma }$$ the Fermi ball:

\begin{aligned} \mathcal {B}_{F}^{\sigma } = \Big \{ k\in \frac{2\pi }{L} \mathbb {Z}^{3} \, \Big |\, |k| \le k_{F}^{\sigma } \Big \}, \end{aligned}
(3.15)

where the Fermi momentum $$k_{F}^{\sigma }$$ is chosen so that $$N_{\sigma } = |\mathcal {B}_{F}^{\sigma }|$$. Of course, this is not possible for all values of $$N_{\sigma }$$. We shall only consider values of $$N_{\sigma }$$ such that the Fermi ball is completely filled; i.e., for which there exists $$k_{F}^{\sigma }$$ so that $$N_{\sigma } = |\mathcal {B}_{F}^{\sigma }|$$. This is not a loss of generality, by the existence of the thermodynamic limit, and since the densities $$\rho _{\sigma } = N_{\sigma } / L^{3}$$ obtained in this way are dense in $$\mathbb {R}_{+}$$ as $$L\rightarrow \infty$$. Notice that, for fixed density $$\rho _{\sigma }$$, $$k_{F}^{\sigma } = (6\pi ^{2})^{1/3} \rho _{\sigma }^{1/3}+ o(1)$$ as $$L\rightarrow \infty$$.

No repetition occurs in the momenta involved in the definition of each determinant in the right-hand side of (3.14); otherwise, the wave function would be exactly zero, by antisymmetry (Pauli principle). The state (3.14) turns out to be equal to the fermionic ground state of the total kinetic energy operator $$-\sum _{j=1}^{N}\Delta _{x_j}$$, on $$\mathfrak {h}(N_{\uparrow }, N_{\downarrow })$$. The total kinetic energy density of such state is:

\begin{aligned} \frac{\langle \Psi _{\text {FFG}}, -\sum _{j=1}^{N} \Delta _{x_{j}} \Psi _{\text {FFG}} \rangle }{L^{3}}= & {} \frac{1}{L^{3}} \sum _{\sigma } \sum _{k\in \mathcal {B}_{F}^{\sigma }} |k|^{2} \nonumber \\= & {} \frac{3}{5}(6\pi ^{2})^{\frac{2}{3}} (\rho _{\uparrow }^{\frac{5}{3}} + \rho _{\downarrow }^{\frac{5}{3}}) + O(L^{-1}), \end{aligned}
(3.16)

where the error term denotes contribution bounded as $$CL^{-1}$$ for L large enough (it is the error term arising from replacing the Riemann sum in the first line by an integral). The same state allows to obtain a simple upper bound for the ground state energy of the interacting system. It is convenient to introduce the fully antisymmetrized version of the $$\Psi _{\text {FFG}}$$, as:

\begin{aligned} \Phi _{\text {FFG}}(x_{1}, \ldots , x_{N}) = \frac{1}{\sqrt{N!}} \det (f^{\sigma _{i}}_{i}(x_{j}))_{1\le i,j\le N}, \end{aligned}
(3.17)

with the understanding that $$\sigma _{i} = \uparrow$$ for $$i\in [1, N_{\uparrow }]$$ and $$\sigma _{i} = \downarrow$$ for $$i\in [N_{\uparrow } + 1, N]$$. The orbitals satisfy the orthogonality condition $$\langle f^{\sigma }_{k}, f^{\sigma '}_{k'}\rangle = \delta _{\sigma ,\sigma '} \delta _{k,k'}$$. In the Fock space language, the state (3.17) can also be represented as (up to an overall sign):

\begin{aligned} \Phi _{\text {FFG}} = \prod _{\sigma = \uparrow ,\downarrow } \prod _{k\in \mathcal {B}_{F}^{\sigma }} \hat{a}^{*}_{k,\sigma }\Omega . \end{aligned}
(3.18)

Being the Hamiltonian spin independent:

\begin{aligned} \langle \Phi _{\text {FFG}}, H_{N} \Phi _{\text {FFG}}\rangle = \langle \Psi _{\text {FFG}}, H_{N} \Psi _{\text {FFG}}\rangle . \end{aligned}
(3.19)

Eq. (3.17) is an example of quasi-free state, for which all correlation functions can be computed starting from the reduced one-particle density matrix:

\begin{aligned} \omega _{\sigma , \sigma '}(x,x')&:= \langle \Phi _{\text {FFG}}, a^{*}_{x',\sigma '} a_{x,\sigma }\Phi _{\text {FFG}}\rangle \nonumber \\&= \delta _{\sigma , \sigma '} \sum _{k\in \mathcal {B}^{\sigma }_{F}} \frac{1}{L^{3}}\, e^{ik\cdot (x-x')}. \end{aligned}
(3.20)

Eq. (3.20) defines the integral kernel of an operator $$\omega : L^{2}(\Lambda _{L}; \mathbb {C}^{2}) \rightarrow L^{2}(\Lambda _{L}; \mathbb {C}^{2})$$, such that $$\omega = \omega ^{2} = \omega ^{*}$$, $$\mathrm {tr}\, \omega = N$$. In Fourier space, $$\hat{\omega }_{\sigma , \sigma }(k)$$ is the characteristic function of the Fermi ball $$\mathcal {B}_{F}^{\sigma }$$. All higher-order density matrices of the system can be computed starting from $$\omega$$, via the fermionic Wick rule. In particular, the energy of $$\Phi _{\text {FFG}}$$ only depends on $$\omega$$:

\begin{aligned} \langle \Phi _{\text {FFG}}, H_{N} \Phi _{\text {FFG}}\rangle = E_{\text {HF}}(\omega ) \end{aligned}
(3.21)

where $$E_{\text {HF}}(\omega )$$ is the Hartree-Fock energy functional:

\begin{aligned} E_{\text {HF}}(\omega ) = -\mathrm {tr}\, \Delta \omega + \frac{1}{2} \sum _{\sigma , \sigma ' = \uparrow \downarrow }\int _{\Lambda _{L} \times \Lambda _{L}} dxdy\, V(x-y) ( \omega _{\sigma , \sigma }(x;x) \omega _{\sigma ', \sigma '}(y;y) - |\omega _{\sigma , \sigma '}(x;y)|^{2}). \end{aligned}
(3.22)

The first term reproduces the kinetic energy of the free Fermi gas, Eq. (3.16). The second term, called the direct term, only depends on the density of the system, $$\omega _{\sigma , \sigma }(x;x) = \rho _{\sigma }$$:

\begin{aligned} \frac{1}{2} \sum _{\sigma , \sigma ' = \uparrow , \downarrow }\int _{\Lambda _{L} \times \Lambda _{L}} dxdy\, V(x-y) \omega _{\sigma , \sigma }(x;x) \omega _{\sigma ', \sigma '}(y;y) = \frac{L^{3}}{2} \sum _{\sigma , \sigma '} \hat{V}(0) \rho _{\sigma } \rho _{\sigma '}. \end{aligned}
(3.23)

The last term, called the exchange term, can be computed at leading order in the density:

\begin{aligned} - \frac{1}{2}\sum _{\sigma , \sigma ' = \uparrow , \downarrow } \int _{\Lambda _{L} \times \Lambda _{L}} dxdy\, V(x-y) |\omega _{\sigma , \sigma '}(x;y)|^{2}= & {} - \frac{1}{2}\sum _{\sigma , \sigma '} \frac{\delta _{\sigma ,\sigma '}}{L^{3}} \sum _{k,k' \in \mathcal {B}^{\sigma }_{F}} \hat{V} (k - k') \nonumber \\= & {} - \frac{1}{2}\sum _{\sigma } L^{3} \hat{V}(0) \rho ^{2}_{\sigma } + O(\rho ^{7/3}),\nonumber \\ \end{aligned}
(3.24)

where we used that if $$k,k'\in \mathcal {B}_{F}^{\sigma }$$ then $$|k - k'|\le C\rho ^{1/3}$$, which implies $$\hat{V}(k-k') = \hat{V}(0) + O(\rho ^{1/3})$$. By the variational principle, we get, putting (3.16), (3.23), (3.24) together, for L large enough:

\begin{aligned} \frac{E_{L}(N_{\uparrow }, N_{\downarrow })}{L^{3}}\le & {} \frac{E_{\text {HF}}(\omega )}{L^{3}}\nonumber \\= & {} \frac{3}{5}(6\pi ^{2})^{\frac{2}{3}} (\rho _{\uparrow }^{\frac{5}{3}} + \rho _{\downarrow }^{\frac{5}{3}}) + \hat{V}(0) \rho _{\uparrow } \rho _{\downarrow } + O(\rho ^{\frac{7}{3}}). \end{aligned}
(3.25)

In Eq. (3.25), the effect of the interaction is only visible via the average of the potential, $$\hat{V}(0) = \int _{\Lambda _{L}} dx\, V(x)$$. The mismatch between (2.5) and (3.25) will be due to the correlations between the particles in the many-body ground state, which are absent in the free Fermi gas.

#### 3.2.2 Fermionic Bogoliubov Transformation

In this section, we shall introduce a suitable unitary transformation in Fock space that will allow us to efficiently compare the many-body ground state energy with the energy of the free Fermi gas.

Given the reduced one-particle density matrix of the free Fermi gas, $$\omega _{\sigma , \sigma '} = \delta _{\sigma , \sigma '}\sum _{k\in \mathcal {B}^{\sigma }_{F}} |f_{k}\rangle \langle f_{k}|$$, we define the operators $$u: L^{2}(\Lambda _{L}; \mathbb {C}^{2}) \rightarrow L^{2} (\Lambda _{L}; \mathbb {C}^{2})$$ and $$v: L^{2} (\Lambda _{L};\mathbb {C}^{2}) \rightarrow L^{2}(\Lambda _{L}; \mathbb {C}^{2})$$ as:

\begin{aligned} u_{\sigma , \sigma '}(x;y) = \delta _{\sigma , \sigma '}\delta (x-y) - \omega _{\sigma , \sigma '}(x;y),\quad v_{\sigma , \sigma '}(x;y) = \delta _{\sigma , \sigma '}\sum _{k\in \mathcal {B}^{\sigma }_{F}} |\overline{f_{k}}\rangle \langle f_{k}|. \end{aligned}
(3.26)

The symbol $$\delta (x-y)$$ denotes the periodic Dirac delta distribution on $$\Lambda _{L}$$, see Eq. (3.7). Clearly,

\begin{aligned} u \overline{v} = 0,\quad \overline{v} v = \omega . \end{aligned}
(3.27)

By the Shale–Stinespring theorem, see [34] for a pedagogical introduction to the topic, there exists a unitary operator $$R: \mathcal {F} \rightarrow \mathcal {F}$$ such that the following holds.

1. (i)

The vector $$R\Omega$$ is an N-particle state, which reproduces the Slater determinant $$\Phi _{\text {FFG}}$$, Eq. (3.17):

\begin{aligned} (R\Omega )^{(n)} = 0\quad \text {unless } n = N,\text { in which case}\quad (R\Omega )^{(N)} = \Phi _{\text {FFG}}. \end{aligned}
(3.28)
2. (ii)

The map $$R:\mathcal {F} \rightarrow \mathcal {F}$$ implements the following transformation in Fock space:

\begin{aligned} R^{*} a(f) R = a(uf) + a^{*}(\overline{v} \overline{f}),\quad \text {for all } f\in L^{2}(\Lambda _{L}; \mathbb {C}^{2}). \end{aligned}
(3.29)

Equivalently,

\begin{aligned} R^{*} a_{x,\sigma }R = a_{\sigma }(u_{x}) + a^{*}_{\sigma }(\overline{v}_{x}), \end{aligned}
(3.30)

where setting $$u_{\sigma ,\sigma } \equiv u_{\sigma }$$, $$v_{\sigma ,\sigma }\equiv v_{\sigma }$$, and $$u_{x}(y) \equiv u(x;y)$$, $$v_{x}(y) \equiv v(x;y)$$:

\begin{aligned} a_{\sigma }(u_{x}) = \int dy\, a_{y,\sigma } \overline{u_{\sigma }(y;x)},\quad a^{*}_{\sigma }(\overline{v}_{x}) = \int dy\, a^{*}_{y,\sigma } \overline{v_{\sigma }(y;x)}. \end{aligned}
(3.31)

Let $$\hat{a}_{k,\sigma } = a_{\sigma }(f_{k})$$, recall Eq. (3.8). Then, the transformation (3.29) reads:

\begin{aligned} R^{*} \hat{a}_{k,\sigma } R = \left\{ \begin{array}{l@{\quad }l} \hat{a}_{k,\sigma } &{} \text {if } k\notin \mathcal {B}^{\sigma }_{F} \\ \hat{a}^{*}_{k,\sigma } &{} \text {if } k\in \mathcal {B}^{\sigma }_{F}.\end{array}\right. \end{aligned}
(3.32)

Thus, Eq. (3.29) can be seen as implementing a particle-hole transformation. By the unitarity of R, Eq. (3.32) also implies that $$R^{*} \hat{a}_{k,\sigma } R = R \hat{a}_{k,\sigma } R^{*}$$.

The operator R is a Bogoliubov transformation, see [1, 34] for reviews. The property (i) immediately implies:

\begin{aligned} \langle R\Omega , \mathcal {H} R\Omega \rangle = E_{\text {HF}}(\omega ). \end{aligned}
(3.33)

Instead, property (ii) allows to compare the energy of any state in the fermionic Fock space, with the energy of the free Fermi gas. This is the content of the next proposition. From now on, we shall simply write $$\sum _{\sigma }$$ for $$\sum _{\sigma = \uparrow \downarrow }$$ and $$\int dx$$ for $$\int _{\Lambda _{L}}dx$$.

### Proposition 3.1

Let $$\psi \in \mathcal {F}$$ be a normalized state, such that $$\langle \psi , \mathcal {N}_{\sigma } \psi \rangle = N_{\sigma }$$ and $$N = N_{\uparrow } + N_{\downarrow }$$. Then:

1. (i)

The following identity holds true:

\begin{aligned} \langle \psi , \mathcal {H} \psi \rangle = E_{\text {HF}}(\omega ) + \langle R^{*}\psi , \mathbb {H}_{0} R^{*}\psi \rangle + \langle R^{*}\psi , \mathbb {X} R^{*}\psi \rangle + \langle R^{*} \psi , \mathbb {Q} R^{*} \psi \rangle \end{aligned}
(3.34)

where the operators $$\mathbb {H}_{0}$$, $$\mathbb {X}$$ are given by:

\begin{aligned} \mathbb {H}_{0}= & {} \sum _{k,\sigma } | |k|^{2} - \mu _{\sigma } | \hat{a}^{*}_{k,\sigma } \hat{a}_{k,\sigma },\quad \mu _{\sigma } = (k_{F}^{\sigma })^{2},\\ \mathbb {X}= & {} \sum _{\sigma }\int dxdy\, V(x-y) \omega _{\sigma }(x-y) ( a^{*}_{\sigma }(u_{x}) a_{\sigma }(u_{y}) - a^{*}_{\sigma }(\overline{v}_{y}) a_{\sigma }(\overline{v}_{x})). \nonumber \end{aligned}
(3.35)

The operator $$\mathbb {Q}$$ can be written as $$\mathbb {Q} = \sum _{i=1}^{4} \mathbb {Q}_{i}$$ with:

\begin{aligned} \mathbb {Q}_{1}= & {} \frac{1}{2} \sum _{\sigma , \sigma '} \int dxdy\, V(x-y) a_{\sigma }^{*}(u_{x})a_{\sigma '}^{*}(u_{y})a_{\sigma '}(u_{y})a_{\sigma }(u_{x})\nonumber \\ \mathbb {Q}_{2}= & {} \frac{1}{2}\sum _{\sigma , \sigma '}\int dxdy\, V(x-y) \Big [ a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma }(\overline{v}_{x}) a_{\sigma '}(\overline{v}_{y}) a_{\sigma '}(u_{y})\nonumber \\&- 2a_{\sigma }^{*}(u_{x})a_{\sigma '}^{*}(\overline{v}_{y})a_{\sigma '}(\overline{v}_{y})a_{\sigma }(u_{x}) + a_{\sigma '}^{*}(\overline{v}_{y})a_{\sigma }^{*}(\overline{v}_{x})a_{\sigma }(\overline{v}_{x})a_{\sigma '}(\overline{v}_{y})\Big ]\nonumber \\ \mathbb {Q}_{3}= & {} -\sum _{\sigma , \sigma '} \int dxdy\, V(x-y) \Big [ a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma '}(u_{y}) a^{*}_{\sigma }(\overline{v}_{x}) a_{\sigma '}(u_{y})\nonumber \\ {}- & {} a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma '}(\overline{v}_{y}) a^{*}_{\sigma }(\overline{v}_{x}) a_{\sigma '}(\overline{v}_{y})\Big ] + \text {h.c.} \nonumber \\ \mathbb {Q}_{4}= & {} \frac{1}{2} \sum _{\sigma , \sigma '} \int dxdy\, V(x-y) a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma '}(u_{y}) a^{*}_{\sigma '}(\overline{v}_{y}) a^{*}_{\sigma }(\overline{v}_{x}) + \text {h.c.} \end{aligned}
(3.36)
2. (ii)

The following inequality holds true:

\begin{aligned} \langle \psi , \mathcal {H} \psi \rangle \ge E_{\text {HF}}(\omega ) + \langle R^{*}\psi , \mathbb {H}_{0} R^{*}\psi \rangle + \langle R^{*}\psi , \mathbb {X} R^{*}\psi \rangle + \langle R^{*} \psi , \widetilde{\mathbb {Q}} R^{*} \psi \rangle , \end{aligned}
(3.37)

where $$\widetilde{\mathbb {Q}} = \sum _{i=1}^{4} \widetilde{\mathbb {Q}}_{i}$$ and:

\begin{aligned} \widetilde{\mathbb {Q}}_{1}= & {} \frac{1}{2} \sum _{\sigma \ne \sigma '} \int dxdy\, V(x-y) a_{\sigma }^{*}(u_{x})a_{\sigma '}^{*}(u_{y})a_{\sigma '}(u_{y})a_{\sigma }(u_{x})\nonumber \\ \widetilde{\mathbb {Q}}_{2}= & {} \frac{1}{2}\sum _{\sigma \ne \sigma '}\int dxdy\, V(x-y) \Big [ a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma }(\overline{v}_{x}) a_{\sigma '}(\overline{v}_{y}) a_{\sigma '}(u_{y})\nonumber \\&- 2a_{\sigma }^{*}(u_{x})a_{\sigma '}^{*}(\overline{v}_{y})a_{\sigma '}(\overline{v}_{y})a_{\sigma }(u_{x}) + a_{\sigma '}^{*}(\overline{v}_{y})a_{\sigma }^{*}(\overline{v}_{x})a_{\sigma }(\overline{v}_{x})a_{\sigma '}(\overline{v}_{y})\Big ]\nonumber \\ \widetilde{\mathbb {Q}}_{3}= & {} -\sum _{\sigma \ne \sigma '} \int dxdy\, V(x-y) \Big [ a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma '}(u_{y}) a^{*}_{\sigma }(\overline{v}_{x}) a_{\sigma '}(u_{y}) \nonumber \\&\quad - a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma '}(\overline{v}_{y}) a^{*}_{\sigma }(\overline{v}_{x}) a_{\sigma '}(\overline{v}_{y})\Big ] + \text {h.c.} \nonumber \\ \widetilde{\mathbb {Q}}_{4}= & {} \frac{1}{2} \sum _{\sigma \ne \sigma '} \int dxdy\, V(x-y) a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma '}(u_{y}) a^{*}_{\sigma '}(\overline{v}_{y}) a^{*}_{\sigma }(\overline{v}_{x}) + \text {h.c.} \end{aligned}
(3.38)

### Remark 3.2

The operator $$\mathbb {H}_{0}$$ describes the kinetic energy of the quasi-particle excitations around the Fermi energy $$\mu$$. The operator $$\mathbb {X}$$ describes the interaction of quasi-particles inside the Fermi ball or outside the Fermi ball with the filled Fermi sea. The operator $$\mathbb {Q}$$ collects all the possible two-body scattering processes, for particles with momenta in the Fermi ball, outside the Fermi ball, or between them.

### Proof

(i) To prove this identity, we transformed each fermionic operator according to (3.30) and we put the resulting expression into normal order, using the canonical anticommutation relations (3.4) and the properties (3.27). The details of the computation have been given already in a number of places and hence will be omitted; see for instance [4,5,6,7, 22].

(ii) Firstly, let us remark that the only difference between $$\mathbb {Q}$$ and $$\tilde{\mathbb {Q}}$$ is that in the latter the sums over $$\sigma ,\sigma '$$ are restricted to $$\sigma \ne \sigma '$$. We use that:

\begin{aligned}&\langle \psi , \mathcal {H} \psi \rangle \nonumber \\&\quad = \sum _{\sigma } \int dx\, \Vert \nabla a_{x,\sigma } \psi \Vert ^{2} + \frac{1}{2} \sum _{\sigma \ne \sigma '} \int dxdy\, V(x-y) \Vert a_{x,\sigma } a_{y,\sigma '} \psi \Vert ^{2} \nonumber \\&\qquad + \frac{1}{2} \sum _{\sigma } \int dxdy\, V(x-y) \Vert a_{x,\sigma } a_{y,\sigma } \psi \Vert ^{2} \nonumber \\&\quad \ge \sum _{\sigma } \int dx\, \Vert \nabla a_{x,\sigma } \psi \Vert ^{2} + \frac{1}{2} \sum _{\sigma \ne \sigma '} \int dxdy\, V(x-y) \Vert a_{x,\sigma } a_{y,\sigma '} \psi \Vert ^{2},\nonumber \\ \end{aligned}
(3.39)

and then we repeat the proof of (i) for the right-hand side of (3.39). $$\square$$

If $$\psi$$ is energetically close to the ground state, the state $$R^{*}\psi$$ is expected to have ‘few particles’, that is much less that $$\rho L^{3}$$. On these states, the energetic contribution due to $$\mathbb {X}$$ is subleading with respect to $$\rho ^{2} L^{3}$$, as we shall see. This is intuitively due to the fact that the filled Fermi sea has density $$\rho$$, while the density of particles in $$R^{*}\psi$$ is much less than $$\rho$$. Instead, the terms $$\langle R^{*}\psi , \mathbb {H}_{0} R^{*}\psi \rangle$$, $$\langle R^{*} \psi , \mathbb {Q} R^{*} \psi \rangle$$ give a contribution to the ground state energy of order $$L^{3}\rho ^{2}$$, which will allow to reconstruct the scattering length in the final result (2.5). Before proving this, let us establish some useful estimates for the various terms arising in (3.34) that will allow us to identify terms that are subleading with respect to $$L^{3} \rho ^{2}$$.

### Proposition 3.3

Under the assumptions of Theorem 2.1, the following holds.

1. (a)

The operator $$\mathbb {X}$$ satisfies the bound:

\begin{aligned} |\langle \psi , \mathbb {X} \psi \rangle | \le C\rho \langle \psi , \mathcal {N} \psi \rangle . \end{aligned}
(3.40)
2. (b)

The operators $$\mathbb {Q}_{1}$$, $$\widetilde{\mathbb {Q}}_{1}$$ are nonnegative.

3. (c)

The operators $$\mathbb {Q}_{2}$$, $$\widetilde{\mathbb {Q}}_{2}$$ satisfy the bounds:

\begin{aligned} | \langle \psi , \mathbb {Q}_{2}\psi \rangle | \le C\rho \langle \psi , \mathcal {N} \psi \rangle ,\quad | \langle \psi , \mathbb {\widetilde{Q}}_{2}\psi \rangle | \le C\rho \langle \psi , \mathcal {N} \psi \rangle . \end{aligned}
(3.41)
4. (d)

The operators $$\mathbb {Q}_{3}$$, $$\widetilde{\mathbb {Q}}_{3}$$ satisfy the bounds, for any $$\alpha \ge 0$$:

\begin{aligned}&|\langle \psi , \mathbb {Q}_{3} \psi \rangle | \le \rho ^{\alpha }\langle \psi , \mathbb {Q}_{1} \psi \rangle + C\rho ^{1-\alpha } \langle \psi , \mathcal {N} \psi \rangle ,\nonumber \\&\quad |\langle \psi , \mathbb {\widetilde{Q}}_{3} \psi \rangle | \le \rho ^{\alpha }\langle \psi , \mathbb {\widetilde{Q}}_{1} \psi \rangle + C\rho ^{1-\alpha } \langle \psi , \mathcal {N} \psi \rangle . \end{aligned}
(3.42)

Furthermore, suppose that $$\psi$$ is a Fock space vector such that $$\psi ^{(n)} = 0$$ unless $$n = 4k$$ for $$k\in \mathbb {N}$$. Then:

\begin{aligned} \langle \psi , \mathbb {Q}_{3} \psi \rangle =0,\quad \langle \psi , \widetilde{\mathbb {Q}}_{3} \psi \rangle = 0. \end{aligned}
(3.43)
5. (e)

The operators $$\mathbb {Q}_{4}$$, $$\widetilde{\mathbb {Q}}_{4}$$ satisfy the bounds, for any $$\delta > 0$$:

\begin{aligned} | \langle \psi , \mathbb {Q}_{4} \psi \rangle | \le \delta \langle \psi , \mathbb {Q}_{1} \psi \rangle + \frac{C}{\delta } \rho ^{2} L^{3} \Vert \psi \Vert ^{2},\quad | \langle \psi , \mathbb {\widetilde{Q}}_{4} \psi \rangle | \le \delta \langle \psi , \widetilde{\mathbb {Q}}_{1} \psi \rangle + \frac{C}{\delta } \rho ^{2} L^{3} \Vert \psi \Vert ^{2}. \end{aligned}
(3.44)

### Proof

We shall only prove the statements for the $$\mathbb {Q}_{i}$$ operators; the analogous statements for the $$\widetilde{\mathbb {Q}}_{i}$$ operators are proven in exactly the same way.

$$\underline{\hbox {Proof of } a).}$$ We have, using the notation $$\Vert \omega _{\sigma , x} \Vert _{\infty } = \sup _{y} | \omega _{\sigma }(x;y)|$$:

\begin{aligned} | \langle \psi , \mathbb {X} \psi \rangle |\le & {} \sum _{\sigma } \Vert \omega _{\sigma ,x}\Vert _{\infty }\int dxdy\, V(x-y) \big ( \Vert a_{\sigma }(u_{x}) \psi \Vert ^{2} + \Vert a_{\sigma }(\overline{v}_{x}) \psi \Vert ^{2} \big ) \nonumber \\\le & {} \rho \Vert V\Vert _{1} \langle \psi , \mathcal {N} \psi \rangle \end{aligned}
(3.45)

where we used that $$\Vert \omega _{\sigma , x} \Vert _{\infty } \le \rho _{\sigma }$$, and that:

\begin{aligned} \int dx\, \Vert a_{\sigma }(u_{x}) \psi \Vert ^{2} = \sum _{k \notin \mathcal {B}_{F}^{\sigma }} \langle \psi , \hat{a}^{*}_{k,\sigma } \hat{a}_{k,\sigma }\psi \rangle ,\quad \int dx\, \Vert a_{\sigma }(\overline{v}_{x}) \psi \Vert ^{2} = \sum _{k \in \mathcal {B}_{F}^{\sigma }} \langle \psi , \hat{a}^{*}_{k,\sigma } \hat{a}_{k,\sigma }\psi \rangle , \end{aligned}
(3.46)

from which the final bound in (3.45) immediately follows (recall the expression for the number operator, (3.10)).

$$\underline{\hbox {Proof of } b).}$$ We have:

\begin{aligned} \langle \psi , \mathbb {Q}_{1} \psi \rangle = \frac{1}{2} \sum _{\sigma , \sigma '} \int dxdy\, V(x-y) \Vert a_{\sigma }(u_{x}) a_{\sigma '}(u_{y}) \psi \Vert ^{2} \ge 0, \end{aligned}
(3.47)

where we used that $$V(x-y) \ge 0$$.

$$\underline{\hbox {Proof of } c).}$$ We have:

\begin{aligned}&\Big |\sum _{\sigma , \sigma '}\int dxdy\, V(x-y) \langle \psi , a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma }(\overline{v}_{x}) a_{\sigma '}(\overline{v}_{y}) a_{\sigma '}(u_{y}) \psi \rangle \Big | \nonumber \\&\quad \le \sum _{\sigma ,\sigma '} \int dxdy\, V(x-y) \Vert v_{\sigma , x}\Vert _{2} \Vert v_{\sigma ', y}\Vert _{2} \Vert a_{\sigma }(u_{x}) \psi \Vert \Vert a_{\sigma '}(u_{y}) \psi \Vert \nonumber \\&\quad \le C\rho \sum _{\sigma , \sigma '} \int dxdy\, V(x-y) \Vert a_{\sigma }(u_{x}) \psi \Vert ^{2} \nonumber \\&\quad \le C \rho \Vert V\Vert _{1} \langle \psi , \mathcal {N} \psi \rangle . \end{aligned}
(3.48)

The second inequality follows from Cauchy–Schwarz inequality and:

\begin{aligned} \Vert v_{\sigma , x}\Vert _{2}^{2} = \int dy\, |v_{\sigma }(x;y)|^{2} = \omega _{\sigma }(x;x) = \rho _{\sigma }. \end{aligned}
(3.49)

The last inequality follows from:

\begin{aligned} \sum _{\sigma } \int dx\, \Vert a_{\sigma }(u_{x}) \psi \Vert ^{2} \le \langle \psi , \mathcal {N} \psi \rangle . \end{aligned}
(3.50)

The other two terms in the definition of $$\mathbb {Q}_{2}$$ are estimated in exactly the same way.

$$\underline{\hbox {Proof of } d).}$$ Consider the first contribution to $$\mathbb {Q}_{3}$$. We write:

\begin{aligned}&\Big | \sum _{\sigma , \sigma '} \int dxdy\, V(x-y) \langle \psi , a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma '}(u_{y}) a^{*}_{\sigma }(\overline{v}_{x}) a_{\sigma '}(u_{y})\psi \rangle \Big | \nonumber \\&\quad \le \frac{\rho ^{\alpha }}{8}\sum _{\sigma , \sigma '}\int dxdy\, V(x-y) \Vert a_{\sigma }(u_{x}) a_{\sigma '}(u_{y}) \psi \Vert ^{2} \nonumber \\&\qquad + \frac{2}{\rho ^{\alpha }} \sum _{\sigma , \sigma '}\int dxdy\, V(x-y) \Vert a^{*}_{\sigma }(\overline{v}_{x}) a_{\sigma '}(u_{y}) \psi \Vert ^{2} \nonumber \\&\quad \le \frac{\rho ^{\alpha }}{4} \langle \psi , \mathbb {Q}_{1} \psi \rangle + C \Vert V\Vert _{1} \rho ^{1-\alpha } \langle \psi , \mathcal {N} \psi \rangle , \end{aligned}
(3.51)

by Cauchy–Schwarz inequality and $$\Vert v_{\sigma , x}\Vert _{2} \le \rho ^{\frac{1}{2}}$$. Consider now the second contribution to $$\mathbb {Q}_{3}$$. We have:

\begin{aligned}&\Big | \sum _{\sigma , \sigma '} \int dxdy\, V(x-y) \langle \psi , a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma '}(\overline{v}_{y}) a^{*}_{\sigma }(\overline{v}_{x}) a_{\sigma '}(\overline{v}_{y}) \psi \rangle \Big |\nonumber \\&\quad \le \sum _{\sigma , \sigma '} \int dxdy\, V(x-y) \Vert v_{\sigma ', y} \Vert _{2} \Vert v_{\sigma , x} \Vert \Vert a_{\sigma }(u_{x}) \psi \Vert \Vert a_{\sigma '}(\overline{v}_{y}) \psi \Vert \nonumber \\&\quad \le C\rho \Vert V\Vert _{1} \langle \psi , \mathcal {N} \psi \rangle , \end{aligned}
(3.52)

again by Cauchy–Schwarz inequality and $$\Vert v_{\sigma , x}\Vert _{2} \le \rho ^{\frac{1}{2}}$$. The remaining terms in the definition of $$\mathbb {Q}_{3}$$ are estimated in exactly the same way. Let us now prove the identities (3.43). Consider the first; the proof of the second is identical. Let $$\psi$$ be such that $$\psi ^{(n)} = 0$$ unless $$n=4k$$ for $$k\in \mathbb {N}$$. Let $$\varphi = \mathbb {Q}_{3} \psi$$. Then, $$\varphi$$ is a Fock space vector such that $$\varphi ^{(n)} = 0$$ unless $$n = 4k + 2$$ for $$k\in \mathbb {N}$$. Since $$4k + 2$$ is not a multiple of 4, $$\langle \psi , \varphi \rangle = \sum _{n\ge 0} \langle \psi ^{(n)}, \varphi ^{(n)} \rangle = 0$$.

$$\underline{\hbox {Proof of } e).}$$ Consider the first contribution to $$\mathbb {Q}_{4}$$. We write, by Cauchy–Schwarz inequality, for $$\delta > 0$$:

\begin{aligned}&\Big | \sum _{\sigma , \sigma '} \int dxdy\, V(x-y) \langle \psi , a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma '}(u_{y}) a^{*}_{\sigma '}(\overline{v}_{y}) a^{*}_{\sigma }(\overline{v}_{x}) \psi \rangle \Big | \nonumber \\&\quad \le \sum _{\sigma , \sigma '} \int dxdy\, V(x-y) \Big [ \frac{\delta }{2}\Vert a_{\sigma }(u_{x}) a_{\sigma '}(u_{y}) \psi \Vert ^{2} + \frac{1}{2\delta } \Vert a^{*}_{\sigma '}(\overline{v}_{y}) a^{*}_{\sigma }(\overline{v}_{x}) \psi \Vert ^{2} \Big ]\nonumber \\&\quad \le \delta \langle \psi , \mathbb {Q}_{1} \psi \rangle + \frac{C}{\delta } \rho ^{2} L^{3} \Vert \psi \Vert ^{2}. \end{aligned}
(3.53)

The other contribution to $$\mathbb {Q}_{4}$$ is bounded in the same way. This concludes the proof of Proposition 3.3. $$\square$$

In order to make good use of the above estimates, we need a priori information on the size of the expectation of the number operator, on states that are close enough to the ground state of the system. We shall refer to these states as approximate ground states

### Definition 3.4

(Approximate ground state) Let $$\psi \in \mathcal {F}$$ be a normalized state, such that $$\langle \psi , \mathcal {N}_{\sigma } \psi \rangle = N_{\sigma }$$ and $$N = N_{\uparrow } + N_{\downarrow }$$. Suppose that:

\begin{aligned} \Big | \langle \psi , \mathcal {H} \psi \rangle - \sum _{\sigma = \uparrow \downarrow } \sum _{k\in \mathcal {B}_{F}^{\sigma }} |k|^{2} \Big | \le CL^{3} \rho ^{2}. \end{aligned}
(3.54)

Then, we shall say that $$\psi$$ is an approximate ground state of $$\mathcal {H}$$.

We will first get an a priori estimate on the relative kinetic energy operator $$\mathbb {H}_{0}$$. Afterward, we will show how to get information on the number operator from this a priori bound.

### Lemma 3.5

(A priori estimate for $$\mathbb {H}_{0}$$). Under the assumptions of Theorem 2.1, the following holds. Suppose that $$\psi$$ is an approximate ground state. Then:

\begin{aligned} \langle R^{*}\psi , \mathbb {H}_{0} R^{*}\psi \rangle \le C L^{3}\rho ^{2}. \end{aligned}
(3.55)

### Proof

By the positivity of the interaction,

\begin{aligned} \langle \psi , \mathcal {H} \psi \rangle\ge & {} \sum _{\sigma =\uparrow \downarrow }\sum _{k\in \frac{2\pi }{L}\mathbb {Z}^{3}} |k|^{2}\langle \psi , \hat{a}^{*}_{k,\sigma } \hat{a}_{k,\sigma } \psi \rangle \nonumber \\= & {} \sum _{\sigma =\uparrow \downarrow }\sum _{k\in \frac{2\pi }{L}\mathbb {Z}^{3}} |k|^{2}\langle R^{*}\psi , R^{*}\hat{a}^{*}_{k,\sigma } \hat{a}_{k,\sigma } R R^{*} \psi \rangle \nonumber \\= & {} \sum _{\sigma = \uparrow \downarrow } \sum _{k\in \mathcal {B}_{F}^{\sigma }} |k|^{2} + \sum _{\sigma = \uparrow \downarrow } \sum _{k \notin \mathcal {B}_{F}^{\sigma }} |k|^{2} \langle R^{*}\psi , \hat{a}^{*}_{k,\sigma } \hat{a}_{k,\sigma } R^{*} \psi \rangle \nonumber \\&- \sum _{\sigma = \uparrow \downarrow } \sum _{k \in \mathcal {B}_{F}^{\sigma }} |k|^{2} \langle R^{*}\psi , \hat{a}^{*}_{k,\sigma } \hat{a}_{k,\sigma } R^{*}\psi \rangle , \end{aligned}
(3.56)

where the last step follows from (3.32). We then rewrite the last two terms as:

\begin{aligned}&\sum _{\sigma = \uparrow \downarrow } \sum _{k \notin \mathcal {B}_{F}^{\sigma }} (|k|^{2} - \mu _{\sigma }) \langle R^{*}\psi , \hat{a}^{*}_{k,\sigma } \hat{a}_{k,\sigma } R^{*}\psi \rangle - \sum _{\sigma = \uparrow \downarrow } \sum _{k \in \mathcal {B}_{F}^{\sigma }} (|k|^{2} - \mu _{\sigma }) \langle R^{*}\psi , \hat{a}^{*}_{k,\sigma } \hat{a}_{k,\sigma } R^{*}\psi \rangle \nonumber \\&\qquad + \sum _{\sigma = \uparrow \downarrow } \mu _{\sigma } \Bigg [ \sum _{k \notin \mathcal {B}_{F}^{\sigma }} \langle R^{*}\psi , \hat{a}^{*}_{k,\sigma } \hat{a}_{k,\sigma } R^{*}\psi \rangle - \sum _{k \in \mathcal {B}_{F}^{\sigma }} \langle R^{*}\psi , \hat{a}^{*}_{k,\sigma } \hat{a}_{k,\sigma } R^{*}\psi \rangle \Bigg ] \nonumber \\&\qquad \equiv \langle R^{*}\psi , \mathbb {H}_{0} R^{*}\psi \rangle + \sum _{\sigma = \uparrow \downarrow } \mu _{\sigma }\Bigg [ \sum _{k\in \frac{2\pi }{L} \mathbb {Z}^{3}}\langle \psi , \hat{a}^{*}_{k,\sigma } \hat{a}_{k,\sigma } \psi \rangle - N_{\sigma }\Bigg ]. \end{aligned}
(3.57)

To reconstruct the kinetic energy operator $$\mathbb {H}_{0}$$, we used that if $$k\notin \mathcal {B}_{F}^{\sigma }$$ then $$|k|^{2} - \mu _{\sigma } \ge 0$$, while if $$k\in \mathcal {B}_{F}^{\sigma }$$ then $$|k|^{2} - \mu _{\sigma } \le 0$$. To obtain the term in the square brackets, we used again the properties of the Bogoliubov transformation (3.32). The term in the brackets vanishes, by the assumptions on the state. The bound (3.56), combined with the assumption (3.54), implies the final claim. $$\square$$

We are now ready to prove an a priori estimate on the number operator. To do so, the following lemma will play an important role.

### Lemma 3.6

Let $$\alpha \ge \frac{2}{3}$$. The following bound holds true:

\begin{aligned} \langle \psi , \mathcal {N} \psi \rangle \le CL^{3} \rho ^{\frac{1}{3} + \alpha } \Vert \psi \Vert ^{2} + \frac{1}{\rho ^{\alpha }}\langle \psi , \mathbb {H}_{0} \psi \rangle . \end{aligned}
(3.58)

### Proof

We write:

\begin{aligned} \mathcal {N}= & {} \sum _{\sigma } \sum _{k} \hat{a}^{*}_{\sigma , k} \hat{a}_{\sigma , k} = \sum _{\sigma } \sum _{k: |k^{2} - \mu _{\sigma }| \le \rho ^{\alpha }} \hat{a}^{*}_{\sigma , k} \hat{a}_{\sigma , k} + \sum _{\sigma } \sum _{k: |k^{2} - \mu _{\sigma }|> \rho ^{\alpha }} \hat{a}^{*}_{\sigma , k} \hat{a}_{\sigma , k}\nonumber \\\equiv & {} \mathcal {N}^{<} + \mathcal {N}^{>}\;{,} \end{aligned}
(3.59)

For the first term, we use that:

\begin{aligned} \langle \psi , \mathcal {N}^{<} \psi \rangle = \sum _{\sigma } \sum _{k: |k^{2} - \mu _{\sigma }| \le \rho ^{\alpha }} \Vert \hat{a}_{\sigma , k} \psi \Vert ^{2} \le CL^{3} \rho ^{\frac{1}{3} + \alpha } \Vert \psi \Vert ^{2}, \end{aligned}
(3.60)

where the last inequality follows from the boundedness of the fermionic operators and from the estimate (recall that $$\alpha \ge 2/3$$):

\begin{aligned} \frac{1}{L^{3}}\sum _{k: |k^{2} - \mu _{\sigma }| \le \rho ^{\alpha }} 1\le & {} C \int dk\, \chi (|k^{2} - \mu _{\sigma }| \le \rho ^{\alpha }) \nonumber \\\le & {} C\int dk\, \chi \Big ( \rho ^{\frac{1}{3}}( 1 - \rho ^{\alpha - \frac{2}{3}}) \le |k| \le \rho ^{\frac{1}{3}} (1 + \rho ^{\alpha - \frac{2}{3}}) \Big ) \nonumber \\\le & {} C\rho ^{\frac{1}{3} + \alpha }. \end{aligned}
(3.61)

For the second term, we use that:

\begin{aligned} \langle \psi , \mathcal {N}^{>} \psi \rangle= & {} \sum _{\sigma } \sum _{k: |k^{2} - \mu _{\sigma }|> \rho ^{\alpha }} \Vert \hat{a}_{\sigma , k} \psi \Vert ^{2} \nonumber \\\le & {} \frac{1}{\rho ^{\alpha }} \sum _{\begin{array}{c} k,\sigma \\ |k^{2} - \mu _{\sigma }| > \rho ^{\alpha } \end{array}} |k^{2} - \mu _{\sigma }| \Vert \hat{a}_{\sigma , k} \psi \Vert ^{2} \le \frac{1}{\rho ^{\alpha }}\langle \psi , \mathbb {H}_{0} \psi \rangle . \end{aligned}
(3.62)

This concludes the proof. $$\square$$

### Corollary 3.7

(A priori estimate for $$\mathcal {N}$$). Under the assumptions of Lemma 3.5, the following holds:

\begin{aligned} \langle R^{*}\psi , \mathcal {N} R^{*}\psi \rangle \le CL^{3} \rho ^{\frac{7}{6}}. \end{aligned}
(3.63)

### Proof

The bound follows from Eqs. (3.58), (3.55), after optimizing over $$\alpha$$. $$\square$$

### Remark 3.8

(Condensation estimate.) It is well known that the estimate (3.63) can be used to control the difference between the reduced one-particle density matrix of $$\psi$$ and the reduced one-particle density matrix of the free Fermi gas, see, e.g., [7]. Let $$\gamma ^{(1)}_{\psi }$$ be the reduced one-particle density matrix of $$\psi$$,

\begin{aligned} \gamma ^{(1)}_{\sigma , \sigma '}(x;y) = \langle \psi , a^{*}_{y,\sigma '} a_{x,\sigma } \psi \rangle . \end{aligned}
(3.64)

Then, the bound (3.63) implies that, for an approximate ground state $$\psi$$:

\begin{aligned} \mathrm {tr}\, \gamma ^{(1)}_{\psi } (1 - \omega ) \le CL^{3} \rho ^{\frac{7}{6}}. \end{aligned}
(3.65)

This ‘condensation estimate’ is not new: it was an important ingredient of the analysis of [25]. One of the reasons for our improved error estimates in Theorem 2.1 is that we will be able to improve the bound (3.65), see Remark 6.3.

We conclude this section by discussing an a priori estimate for the operator $$\mathbb {Q}_{1}$$, arising after conjugating the Hamiltonian with the fermionic Bogoliubov transformation.

### Lemma 3.9

(A priori estimate for $$\mathbb {Q}_{1}$$). Under the assumptions of Theorem 2.1, the following is true. Suppose that $$\psi$$ an approximate ground state. Then:

\begin{aligned} \langle R^{*}\psi , \mathbb {Q}_{1} R^{*}\psi \rangle \le CL^{3} \rho ^{2}, \quad \langle R^{*}\psi , \widetilde{\mathbb {Q}}_{1} R^{*}\psi \rangle \le CL^{3} \rho ^{2}. \end{aligned}
(3.66)

### Proof

From the estimates for $$\mathbb {X}$$, $$\mathbb {Q}_2$$, $$\mathbb {Q}_3$$, Eqs. (3.40), (3.41) and (3.42), we get:

\begin{aligned} \langle \psi , \mathcal {H}\psi \rangle \ge E_{\text {HF}}(\omega ) + \langle R^{*}\psi , \mathbb {H}_0 R^{*}\psi \rangle +(1-C\rho ^\alpha )\langle R^{*}\psi , \mathbb {Q}_1 R^{*}\psi \rangle + \langle R^{*}\psi , \mathbb {Q}_4 R^{*}\psi \rangle + \mathcal {E}(\psi ), \end{aligned}
(3.67)

with

\begin{aligned} |\mathcal {E}(\psi )|\le C \rho ^{1-\alpha }\langle R^{*} \psi , \mathcal {N} R^{*}\psi \rangle . \end{aligned}
(3.68)

Eq. (3.68) together with the bound $$\pm \mathbb {Q}_{4} \le \delta \mathbb {Q}_{1} + (C/\delta ) L^{3}\rho ^{2}$$, Eq. (3.44), imply, taking, e.g., $$\alpha = 1/12$$:

\begin{aligned} \langle \psi , \mathcal {H} \psi \rangle \ge E_{\text {HF}}(\omega ) + \langle R^{*}\psi , \mathbb {H}_{0} R^{*}\psi \rangle + (1 - C\delta ) \langle R^{*}\psi , \mathbb {Q}_{1} R^{*}\psi \rangle - \frac{C}{\delta }L^{3}\rho ^{2}. \end{aligned}
(3.69)

Taking $$\delta >0$$ small enough, the final claim follows from assumption (3.54), from the explicit expression of $$E_{\text {HF}}(\omega )$$, Eq. (3.25) and from the positivity of $$\mathbb {H}_{0}$$. The inequality for $$\widetilde{\mathbb {Q}}_{1}$$ follows immediately, since $$\mathbb {Q}_{1} \ge \widetilde{\mathbb {Q}}_{1}$$. $$\square$$

## 4 The Correlation Structure

### 4.1 Heuristics

Here, we shall give the intuition behind the method developed in the rest of the paper. Recall the expression for the many-body energy, after conjugating with the Bogoliubov transformation:

\begin{aligned} \langle \psi , \mathcal {H} \psi \rangle = E_{\text {HF}}(\omega ) + \langle R^{*}\psi , (\mathbb {H}_{0} + \mathbb {Q}_{1} + \mathbb {Q}_{4}) R^{*}\psi \rangle + \mathcal {E}_{\text {1}}(\psi ). \end{aligned}
(4.1)

If $$\psi$$ is an approximate ground state, the error term $$\mathcal {E}_{1}(\psi )$$ is subleading with respect to $$\rho ^{2}$$, as a consequence of the estimates proven in the previous section. For the sake of the following heuristic discussion, we shall neglect it. The operator $$\mathbb {Q}_{4}$$ can be rewritten as:

\begin{aligned} \mathbb {Q}_{4} = \frac{1}{2}\sum _{\sigma ,\sigma '} \frac{1}{L^{3}}\sum _{p} \hat{V}(p) \hat{b}_{p,\sigma } \hat{b}_{-p,\sigma '} + \text {h.c.}, \end{aligned}
(4.2)

with:

\begin{aligned} \hat{b}_{p,\sigma } = \int dx\, e^{ip\cdot x} a_{\sigma }(u_{x}) a_{\sigma }(\overline{v}_{x}) = \sum _{\begin{array}{c} k: k+p\notin \mathcal {B}^{\sigma }_{F} \\ k\in \mathcal {B}^{\sigma }_{F} \end{array}} \hat{a}_{k+p,\sigma } \hat{a}_{k,\sigma }.\nonumber \\ \end{aligned}
(4.3)

The b, $$b^{*}$$ operators turn out to behave as ‘bosonic’ operators, if evaluated on states with few particles. To begin, notice that, denoting by $$\delta _{k,k'}$$ the Kronecker delta:

\begin{aligned}&{[} \hat{b}_{p,\sigma }, \hat{b}^{*}_{q,\sigma '} ] = \delta _{p,q} \delta _{\sigma , \sigma '} |\mathcal {B}_{F}^{\sigma }|\nonumber \\&\quad - \delta _{\sigma , \sigma '}\sum _{\begin{array}{c} k,k'\in \mathcal {B}_{F}^{\sigma } \\ k+p, k'+q \notin \mathcal {B}_{F}^{\sigma } \end{array}} ( \hat{a}^{*}_{k'+q,\sigma '} \hat{a}_{k+p,\sigma } \delta _{k,k'} + \hat{a}^{*}_{k',\sigma '} \hat{a}_{k,\sigma } \delta _{k+p, k'+q}) \end{aligned}
(4.4)

and $$[ \hat{b}_{p,\sigma }, \hat{b}_{q,\sigma '} ] = 0$$. In particular, on states $$\psi$$ that contain ‘few’ particles, $$L^{-3}\langle \psi , \mathcal {N}\psi \rangle = o(\rho )$$:

\begin{aligned} L^{-3}\langle \psi , [ \hat{b}_{p,\sigma }, \hat{b}_{q,\sigma '}^{*} ] \psi \rangle = \delta _{p,q}\delta _{\sigma \sigma '} \rho _{\sigma } + o(\rho ),\quad \langle \psi , [ \hat{b}_{p,\sigma }, \hat{b}_{q,\sigma '} ] \psi \rangle = 0. \end{aligned}
(4.5)

Equation (4.5) suggest that, on states with ‘few’ particles, the operators $$b_{p,\sigma }$$, $$b^{*}_{q,\sigma '}$$ satisfy approximate canonical commutation relations. Therefore, $$\mathbb {Q}_{4}$$ is quadratic on these pseudo-bosonic operators, which suggests that one might attempt to evaluate its energetic contribution to the ground state energy via diagonalization. Unfortunately, the $$\mathbb {H}_{0}$$, $$\mathbb {Q}_{1}$$ operators do not have this structure. Nevertheless, if evaluated on a suitable class of states, they behave as quadratic bosonic operators, as we shall see below. For instance, consider $$\mathbb {H}_{0}$$. Let us rescale the b operators so that they satisfy (approximate) canonical commutation relations, by setting $$\hat{c}_{p,\sigma } = \rho _{\sigma }^{-1/2} \hat{b}_{p,\sigma }$$. One has:

\begin{aligned}{}[\mathbb {H}_{0}, \hat{c}^{*}_{q,\sigma }]= & {} \rho _{\sigma }^{-\frac{1}{2}} \sum _{\begin{array}{c} k: k+q\notin \mathcal {B}^{\sigma }_{F} \\ k\in \mathcal {B}^{\sigma }_{F} \end{array}} ( | |k+q|^{2} - \mu _{\sigma } | + ||k|^{2} - \mu _{\sigma }| ) \hat{a}^{*}_{k+q,\sigma } \hat{a}^{*}_{k,\sigma }\nonumber \\= & {} \rho _{\sigma }^{-\frac{1}{2}}\sum _{\begin{array}{c} k: k+q\notin \mathcal {B}^{\sigma }_{F} \\ k\in \mathcal {B}^{\sigma }_{F} \end{array}} ( |k+q|^{2} - |k|^{2}) \hat{a}^{*}_{k+q,\sigma } \hat{a}^{*}_{k,\sigma }. \end{aligned}
(4.6)

Being k inside the Fermi ball, $$|k|^{2}\le C\rho _{\sigma }^{2/3}$$. Thus, for $$|q| \gg \rho _{\sigma }^{1/3}$$, it makes sense to approximate:

\begin{aligned}{}[\mathbb {H}_{0}, c^{*}_{q,\sigma }] \simeq |q|^{2} c^{*}_{q,\sigma }. \end{aligned}
(4.7)

Let:

\begin{aligned} \mathbb {K}_{\text {B}} = \frac{1}{L^{3}}\sum _{p,\sigma } |p|^{2} \hat{c}^{*}_{p,\sigma }\hat{c}_{p,\sigma }. \end{aligned}
(4.8)

Considering the c operators as true bosonic operators, we see that $$\mathbb {K}_{\text {B}}$$ satisfies the same (approximate) commutation relation as $$\mathbb {H}_{0}$$. This suggests that, on states with few bosons, the operators $$\mathbb {H}_{0}$$ and $$\mathbb {K}_{\text {B}}$$ act similarly. For instance, consider a state with one boson, $$\hat{c}^{*}_{q,\sigma }\Omega$$. Then:

\begin{aligned} \mathbb {H}_{0} \hat{c}^{*}_{q,\sigma } \Omega = [\mathbb {H}_{0}, \hat{c}^{*}_{q,\sigma }] \Omega \simeq [ \mathbb {K}_{\text {B}}, \hat{c}^{*}_{q,\sigma } ] \Omega = \mathbb {K}_{\text {B}} \hat{c}^{*}_{q,\sigma }\Omega . \end{aligned}
(4.9)

More generally, it is reasonable to expect that, on states $$R^{*}\psi$$ with few particles:

\begin{aligned} \langle R^{*}\psi , \mathbb {H}_{0} R^{*}\psi \rangle \simeq \langle R^{*}\psi , \mathbb {K}_{\text {B}} R^{*}\psi \rangle . \end{aligned}
(4.10)

Finally, consider now the $$\mathbb {Q}_{1}$$ operator. Again, $$\mathbb {Q}_{1}$$ is not quadratic in the pseudo-bosons. To understand its action in terms of pseudo-bosons, we rewrite it as:

\begin{aligned} \mathbb {Q}_{1}= & {} \frac{1}{2}\sum _{\sigma , \sigma '} \int dxdy\, V(x-y) a_{\sigma }^{*}(u_{x}) a_{\sigma '}^{*}(u_{y})a_{\sigma '}(u_{y})a_{\sigma }(u_{x}) \nonumber \\= & {} \frac{1}{2} \sum _{\sigma , \sigma '}\int dxdy\, V(x-y) a_{\sigma }^{*}(u_{x}) a_{\sigma }(u_{x}) a_{\sigma '}^{*}(u_{y}) a_{\sigma '}(u_{y})\nonumber \\&- \frac{1}{2}\sum _{\sigma , \sigma '} \int dxdy\, V(x-y) u_{\sigma }(y,x) \delta _{\sigma , \sigma '} a_{\sigma }^{*}(u_{x}) a_{\sigma '}(u_{y})\nonumber \\\equiv & {} \frac{1}{2L^{3}} \sum _{p} \hat{V}(p) D_{p} D_{-p} - \frac{1}{2L^{3}} \sum _{p} \hat{V}(p) E_{p}, \end{aligned}
(4.11)

where

\begin{aligned} D_{p} = \sum _{\sigma }\sum _{\begin{array}{c} k: k\notin \mathcal {B}_{F}^{\sigma } \\ k - p \notin \mathcal {B}_{F}^{\sigma } \end{array}} \hat{a}^{*}_{k,\sigma } \hat{a}_{k-p,\sigma }, \quad E_{p} = \sum _{\sigma }\sum _{\begin{array}{c} k: k\notin \mathcal {B}_{F}^{\sigma } \\ k - p \notin \mathcal {B}_{F}^{\sigma } \end{array}} \hat{a}^{*}_{k,\sigma } \hat{a}_{k,\sigma }. \end{aligned}
(4.12)

A simple computation shows that:

\begin{aligned}{}[ D_{p}, \hat{c}^{*}_{q,\sigma }]= & {} \rho _{\sigma }^{-\frac{1}{2}}\sum _{\begin{array}{c} k: k\in \mathcal {B}^{\sigma }_{F} \\ k + q \notin \mathcal {B}^{\sigma }_{F} \\ k+q-p \notin \mathcal {B}^{\sigma }_{F} \end{array}} a^{*}_{k,\sigma } a^{*}_{k+q-p,\sigma } \nonumber \\\simeq & {} \rho _{\sigma }^{-\frac{1}{2}}\sum _{\begin{array}{c} k: k\in \mathcal {B}^{\sigma }_{F} \\ k+q-p \notin \mathcal {B}^{\sigma }_{F} \end{array}} a^{*}_{k,\sigma } a^{*}_{k+q-p,\sigma } \equiv \hat{c}^{*}_{q-p,\sigma }. \end{aligned}
(4.13)

In the last step we neglected the constraint $$k+q\notin \mathcal {B}^{\sigma }_{F}$$: this is reasonable, if $$|q|\gg \rho _{\sigma }^{1/3}$$. Thus, considering the c operators as true bosons, we see that Eq. (4.13) is the same commutation relation satisfied by replacing $$D_{p}$$ with the operator:

\begin{aligned} \mathbb {G}_{p} = \frac{1}{L^{3}}\sum _{\sigma , q} \hat{c}^{*}_{q-p,\sigma } \hat{c}_{q,\sigma }. \end{aligned}
(4.14)

In the same spirit, it is not difficult to see that, for $$|q|\gg \rho _{\sigma }^{1/3}$$:

\begin{aligned}{}[ E_{p}, \hat{c}^{*}_{q,\sigma } ] \simeq \hat{c}^{*}_{q,\sigma }, \end{aligned}
(4.15)

which is the same commutation relation satisfied replacing $$E_{p}$$ by $$\mathbb {G}_{0}$$. All in all, we expect that, on states $$R^{*}\psi$$ with few pseudo-bosonic excitations particles, with momenta $$|q|\gg \rho ^{1/3}$$:

\begin{aligned} \langle \psi , \mathcal {H} \psi \rangle\simeq & {} E_{\text {HF}}(\omega ) + \frac{1}{L^{3}}\sum _{p,\sigma } |p|^{2} \langle R^{*}\psi , \hat{c}^{*}_{p,\sigma } \hat{c}_{p,\sigma } R^{*}\psi \rangle \nonumber \\&+ \frac{1}{2L^{3}}\sum _{p,\sigma ,\sigma '} \rho _{\sigma }^{\frac{1}{2}} \rho _{\sigma '}^{\frac{1}{2}}\hat{V}(p) \langle R^{*}\psi , ( \hat{c}_{p,\sigma }\hat{c}_{-p,\sigma '} + \text {h.c.}) R^{*}\psi \rangle \nonumber \\&+ \frac{1}{2L^{9}}\sum _{\begin{array}{c} p, q, q' \\ \sigma , \sigma ' \end{array}} \hat{V}(p) \langle R^{*}\psi , \hat{c}^{*}_{q-p, \sigma } \hat{c}_{q, \sigma } \hat{c}^{*}_{q'+p,\sigma '} \hat{c}_{q', \sigma '} R^{*}\psi \rangle \nonumber \\&- \frac{1}{2 L^{6}} \sum _{p,q,\sigma } \hat{V}(p) \langle R^{*}\psi , \hat{c}^{*}_{q,\sigma } \hat{c}_{q,\sigma } R^{*}\psi \rangle . \end{aligned}
(4.16)

Now, suppose that $$R^{*}\psi = T\xi$$, with $$\xi$$ ‘close’ to the Fock space vacuum and:

\begin{aligned} T = \exp \left\{ \frac{1}{L^{3}}\sum _{p} \rho _{\uparrow }^{\frac{1}{2}} \rho _{\downarrow }^{\frac{1}{2}} \hat{\varphi }(p) \hat{c}_{p,\uparrow } \hat{c}_{-p,\downarrow } - \text {h.c.} \right\} , \end{aligned}
(4.17)

for some even, real function $$\hat{\varphi }(p)$$, to be chosen in a moment. Treating the c operators as true bosons, the operator T implements a bosonic Bogoliubov transformation. It acts as:

\begin{aligned} T^{*} c_{q,\sigma } T = c_{q,\sigma } - \rho _{\sigma }^{\frac{1}{2}} \rho _{-\sigma }^{\frac{1}{2}} \hat{\varphi }(q) c^{*}_{-q,-\sigma } + o(\rho ). \end{aligned}
(4.18)

The state $$T\Omega$$ is a bosonic quasi-free state, and its energy can be computed via the bosonic Wick rule. We have:

\begin{aligned} \frac{1}{L^{3}}\langle T\Omega , \hat{c}^{*}_{p,\sigma } \hat{c}_{p,\sigma } T\Omega \rangle= & {} \rho _{\sigma } \rho _{-\sigma } \hat{\varphi }(p)^{2} + o(\rho ^{2}) \nonumber \\ \frac{1}{L^{3}}\langle T\Omega , \hat{c}_{p,\sigma } \hat{c}_{-p,\sigma '} T\Omega \rangle= & {} -\delta _{\sigma ,-\sigma '} \rho _{\sigma }^{\frac{1}{2}} \rho _{-\sigma }^{\frac{1}{2}}\varphi (p) + o(\rho )\nonumber \\ \frac{1}{L^{6}} \langle T\Omega , \hat{c}^{*}_{q-p, \sigma } \hat{c}_{q, \sigma } \hat{c}^{*}_{q'+p,\sigma '} \hat{c}_{q', \sigma '} T\Omega \rangle= & {} \delta _{p,0} \delta _{q,q'} \delta _{\sigma , \sigma '} \rho _{\sigma } \rho _{-\sigma }\varphi (q)^{2} \nonumber \\&+ \delta _{q, q'+p} \delta _{\sigma , \sigma '} \rho _{\sigma } \rho _{-\sigma }\varphi (q)^{2} + o(\rho ^{2}).\nonumber \\ \end{aligned}
(4.19)

Supposing that $$R^{*}\psi \simeq T \Omega$$ one has, neglecting all $$o(\rho ^{2})$$ terms:

\begin{aligned} \langle R^{*}\psi , \mathcal {H} R^{*}\psi \rangle\simeq & {} E_{\text {HF}}(\omega ) + \sum _{p,\sigma } \rho _{\sigma }\rho _{-\sigma } |p|^{2} \hat{\varphi }(p)^{2} - \sum _{p, \sigma } \rho _{\sigma } \rho _{-\sigma } \hat{V}(p) \hat{\varphi }(p)\nonumber \\&+ \frac{1}{2L^{3}} \sum _{p,q, \sigma } \rho _{\sigma } \rho _{-\sigma } \hat{V}(p) (\hat{\varphi }(q-p) \hat{\varphi }(-q) + \hat{\varphi }(q-p)^{2}) \nonumber \\&- \frac{1}{2L^{3}} \sum _{p,q,\sigma } \rho _{\sigma }\rho _{-\sigma } \hat{V}(p) \hat{\varphi }(q)^{2} \equiv E_{\text {HF}}(\omega ) + 2L^{3}\rho _{\uparrow }\rho _{\downarrow } e(\varphi )\nonumber \\ \end{aligned}
(4.20)

where

\begin{aligned} e(\varphi ) := \frac{1}{L^{3}}\sum _{p} \Big (|p|^{2} \hat{\varphi }(p)^{2} - \hat{V}(p) \hat{\varphi }(p) + \frac{1}{2} (\hat{V} * \hat{\varphi })(p) \hat{\varphi }(p)\Big ). \end{aligned}
(4.21)

We are interested in the value of $$\varphi$$ that gives the smallest energy. The equation for the minimizer is:

\begin{aligned} 2 |p|^{2} \hat{\varphi }(p) - \hat{V}(p) + (\hat{V}* \hat{\varphi })(p) = 0. \end{aligned}
(4.22)

The infinite volume counterpart of (4.22) is the zero energy scattering equation:

\begin{aligned} -\Delta f + \frac{1}{2} V_{\infty } f = 0,\quad f = 1 - \varphi , \end{aligned}
(4.23)

with boundary condition $$f(x) \rightarrow 1$$ for $$|x| \rightarrow \infty$$. Recall that V on $$\Lambda _{L}$$ and $$V_{\infty }$$ on $$\mathbb {R}^{3}$$ are related by periodization, see Eq. (2.2). With a slight abuse of notation, we temporarily denote by $$\hat{\varphi }(p)$$ the restriction to $$p \in \frac{2\pi }{L}\mathbb {Z}^{3}$$ of the Fourier transform of the function $$\varphi (x)$$ solving (4.23). One can show, see, e.g., Eq. (30) of [16], that this function satisfies Eq. (4.22) up to O(1/L). We shall neglect these error terms for the purpose of the present heuristic discussion. Thus, plugging $$\hat{\varphi }(p)$$ in (4.21) we get, up to subleading error terms:

\begin{aligned} e(\varphi ) = -\frac{1}{2L^{3}}\sum _{p} \hat{V}(p) \hat{\varphi }(p), \end{aligned}
(4.24)

hence:

\begin{aligned} \frac{\langle R^{*}\psi , \mathcal {H} R^{*}\psi \rangle }{L^{3}} = \frac{E_{\text {HF}}(\omega )}{L^{3}} - \rho _{\uparrow } \rho _{\downarrow } \frac{1}{L^{3}}\sum _{p}\hat{V}(p) \hat{\varphi }(p) + o(\rho ^{2}). \end{aligned}
(4.25)

In conclusion, recalling the expression (3.25) for $$E_{\text {HF}}(\omega )$$, for L large enough:

\begin{aligned} \frac{\langle R^{*}\psi , \mathcal {H} R^{*}\psi \rangle }{L^{3}} = \frac{3}{5}(6\pi ^{2})^{\frac{2}{3}} (\rho _{\uparrow }^{\frac{5}{3}} + \rho _{\downarrow }^{\frac{5}{3}}) + \rho _{\uparrow }\rho _{\downarrow } \Big ( \hat{V}(0) - \frac{1}{L^{3}} \sum _{p} \hat{V}(p) \hat{\varphi }(p) \Big ) + o(\rho ^{2}). \end{aligned}
(4.26)

The term in parenthesis can be rewritten as, for $$L\rightarrow \infty$$:

\begin{aligned} \int dx\, V_{\infty }(x) (1 - \varphi (x)) \equiv 8\pi a, \end{aligned}
(4.27)

where a is the scattering length of the potential $$V_{\infty }$$. This reproduces the main term in the final result (2.5). Even at the heuristic level, however, there is a problem: the operators $$\mathbb {H}_{0}$$, $$\mathbb {Q}_{1}$$, $$\mathbb {Q}_{4}$$ can be represented in terms of bosonic operators provided they act on states with few bosonic particles, with momenta $$|q|\gg \rho ^{1/3}$$. The state $$T\Omega$$ is given by a superposition of states by an even number of bosons, with momenta in the support of $$\hat{\varphi }(p)$$. Hence, to enforce the momentum constraint, we would like the function $$\hat{\varphi }(p)$$ to be supported for $$|p|\gg \rho ^{1/3}$$. We can achieve this by regularizing $$\varphi (x)$$, so that it is supported on a ball of radius $$1\ll R\ll \rho ^{-1/3}$$. Let $$\varphi _{\gamma }$$ be the solution of the scattering equation in a ball $$B\equiv B_{\rho ^{-\gamma }}(0) \subset \mathbb {R}^{3}$$ centered at zero and with radius $$\rho ^{-\gamma }$$ with $$0\le \gamma \le 1/3$$, satisfying Neumann boundary conditions:

\begin{aligned} -\Delta (1- \varphi _{\gamma }) + \frac{1}{2}V_{\infty }(1 - \varphi _{\gamma }) = \lambda _{\gamma }(1-\varphi _{\gamma }),\quad \varphi _{\gamma } = \nabla \varphi _{\gamma } = 0\quad \text {on } \partial B,\nonumber \\ \end{aligned}
(4.28)

with $$|\lambda _{\gamma }| \le C\rho ^{3\gamma }$$. See [15, 24] for the study of this equation, and Appendix 7 for a summary of results. For $$x \in B$$ away from the support of $$V_{\infty }$$, the solution of (4.28) behaves as:

\begin{aligned} \varphi _{\gamma }(x) \sim \frac{a_{\gamma }}{|x|},\quad \text {with } 8\pi a_{\gamma } = \int dx\, V(x) (1 - \varphi _{\gamma }(x)). \end{aligned}
(4.29)

The function $$\varphi _{\gamma }$$ is extended to $$\mathbb {R}^{3}$$ by setting $$\varphi _{\gamma }(x) = 0$$ for $$x\notin B$$. To make it compatible with the periodic boundary conditions on $$\Lambda _{L}$$, from now on we shall take $$\varphi$$ as the periodization of $$\varphi _{\gamma }$$,

\begin{aligned} \varphi (x) = \sum _{n\in \mathbb {Z}^{3}} \varphi _{\gamma }(x + n_{1} e_{1} L + n_{2} e_{2} L + n_{3} e_{3} L). \end{aligned}
(4.30)

Equivalently:

\begin{aligned} \varphi (x) = \frac{1}{L^{3}} \sum _{p\in \frac{2\pi }{L}\mathbb {Z}^{3}} e^{ip\cdot x} \hat{\varphi }_{\gamma }(p), \end{aligned}
(4.31)

where $$\hat{\varphi }_{\gamma }(p) = \int _{\mathbb {R}^{3}} dx\, e^{-ip\cdot x} \varphi _{\gamma }(x)$$. Plugging the solution of this equation in $$e(\varphi )$$, we obtain, as $$L\rightarrow \infty$$, using that in this limit $$\varphi (x) \rightarrow \varphi _{\gamma }(x)$$ pointwise:

\begin{aligned} e(\varphi ) = -\frac{1}{2} \int \frac{dp}{(2\pi )^{3}}\, \hat{V}(p) \hat{\varphi }_{\gamma }(p) + \rho ^{3\gamma } \int \frac{dp}{(2\pi )^{3}}\,\hat{\varphi }_{\gamma }(p) (1 - \hat{\varphi }_{\gamma }(p)). \end{aligned}
(4.32)

It is well known that $$|a - a_{\gamma }| \le C\rho ^{\gamma }$$ [15, 24], see Lemma A.1. Thus, the first term combined with the interaction energy of the free Fermi gas reproduces the scattering length, while the last term is bounded by:

\begin{aligned} \Big |\rho ^{3\gamma } \int \frac{dp}{(2\pi )^{3}}\, \hat{\varphi }_{\gamma }(p) (1 - \hat{\varphi }_{\gamma }(p))\Big | \le \rho ^{3\gamma } \varphi _{\gamma }(0) + \rho ^{3\gamma } \Vert \varphi _{\gamma } \Vert _{2}^{2} \le C\rho ^{2\gamma }, \end{aligned}
(4.33)

where we used that $$\varphi _{\gamma }(x) \sim |x|^{-1}$$ for large |x|, and that $$\varphi _{\gamma }(x)$$ is compactly supported. This error term amounts to a small correction to the ground state energy, which does not affect the $$\rho ^{2}$$ term.

### Remark 4.1

1. (i)

As the above heuristics suggests, the choice $$\gamma = 1/3$$ is expected to be the correct one in order to compute the ground state energy density with a $$O(\rho ^{7/3})$$ precision; this is the order of magnitude of the next correction to the ground state energy after $$8\pi a\rho _{\uparrow } \rho _{\downarrow }$$, [23].

2. (ii)

As mentioned in the introduction, similar bosonization ideas have been recently used in order to prove the validity of the random phase approximation, for the ground state energy of interacting fermionic systems in the mean-field regime [5, 6]. There, the emergent bosonic degrees of freedom correspond to particle-hole excitations that are localized around suitable ‘patches’ on the Fermi surface (whose radius grows proportionally to $$N^{1/3}$$), around which the kinetic energy operator $$\mathbb {H}_{0}$$ can be approximated by a linear dispersion for the bosonic modes.

### 4.2 Definition of the Correlation Structure

Here, we shall give a precise definition of the unitary operator T, introduced in the previous section. Before doing this, let us fix some notation that will be used in the rest of the paper.

Notations.

• We shall denote by $$\chi : \mathbb {R}_{+} \rightarrow [0,1]$$ a smooth, nonincreasing cutoff function such that $$\chi (t) = 1$$ for $$t\le 1$$ and $$\chi (t) = 0$$ for $$t\ge 2$$. We shall also use the notation $$\chi ^{c} = 1 - \chi$$.

• We shall denote by C a general constant, possibly dependent on V, whose value might change from line to line.

• We shall denote by $$C_{\beta }$$ a general prefactor of the form $$C \rho ^{-c\beta }$$. We will not keep track of such $$\rho$$-dependence of the bounds. At the end, by taking the interaction potential regular enough, we will be able to consider $$\beta > 0$$ arbitrarily small.

• Unless otherwise stated, we shall use the notation $$\Vert \cdot \Vert _{p}$$ for $$\Vert \cdot \Vert _{L^{p}(\Lambda _{L})}$$.

• We shall use the notations $$\sum _{\sigma }$$ for $$\sum _{\sigma = \uparrow \downarrow }$$, $$\int dx$$ for $$\int _{\Lambda _{L}}dx$$ and $$\sum _{k}$$ for $$\sum _{k\in \frac{2\pi }{L}\mathbb {Z}^{3}}$$.

• We shall denote by $$|\cdot |$$ the usual Euclidean distance on $$\mathbb {R}^{3}$$, and by $$|\cdot |_{L}$$ the distance on the torus: $$|x - y|_{L} = \min _{n\in \mathbb {Z}^{3}} | x - y - n_{1} e_{1} - n_{2}e_{2} - n_{3} e_{3} |$$, with $$\{e_{i}\}$$ the standard orthonormal basis of $$\mathbb {R}^{3}$$.

• We shall use the notation $$u_{x}$$, $$v_{x}$$ to denote the functions $$y\mapsto u_{x}(y) \equiv u(x;y)$$, $$y\mapsto v_{x}(y)\equiv v(x;y)$$.

• We shall denote by $$\mathfrak {e}_{L}$$ a general finite size correction, subleading with respect to $$L^{3}$$.

Let us introduce regularized versions of the operators u and of v, introduced in Sect. 3.2.2. We define:

\begin{aligned} v^{\text {r}}_{\sigma , \sigma '} = \frac{\delta _{\sigma , \sigma '}}{L^3} \sum _{k} \hat{v}_{\sigma }^{\text {r}}(k) |\overline{f_{k}}\rangle \langle f_{k}|,\quad u^{\text {r}}_{\sigma ,\sigma '} = \frac{\delta _{\sigma ,\sigma '}}{L^{3}} \sum _{k} \hat{u}_{\sigma }^{\text {r}}(k) |f_{k}\rangle \langle f_{k}|, \end{aligned}
(4.34)

where $$\hat{v}_{\sigma }^{\text {r}}(k)$$, $$\hat{u}_{\sigma }^{\text {r}}(k)$$ are smooth, radial functions with the following properties. Let $$\alpha = \frac{1}{3} + \frac{\epsilon }{3}$$, with $$\epsilon >0$$, and let $$\beta >0$$, to be chosen later on:

\begin{aligned} \hat{v}_{\sigma }^{\text {r}}(k) = \left\{ \begin{array}{ll} 1 &{} \quad \text {for } |k| < k_{F}^{\sigma } - \rho _{\sigma }^{\alpha } \\ 0 &{} \quad \text {for } |k|\ge k_{F}^{\sigma } \end{array} \right. {, \qquad }\hat{u}^{\text {r}}_{\sigma }(k) = \left\{ \begin{array}{l@{\quad }l} 0 &{} \text {for } |k|\le k^{\sigma }_{F} \\ 1 &{} \text {for } 2k^{\sigma }_{F}\le |k|\le \frac{3}{2}\rho _{\sigma }^{-\beta } \\ 0 &{} \text {for } |k| \ge 2\rho _{\sigma }^{-\beta } \end{array} \right. {.} \end{aligned}
(4.35)

Concretely, we choose:

\begin{aligned} \hat{v}_{\sigma }^{\text {r}}(k) = \chi \Big ( \frac{|k| - (k_{F}^{\sigma } - 2\rho _{\sigma }^{\alpha })}{\rho _{\sigma }^{\alpha }} \Big ),\quad \hat{u}^{\text {r}}_{\sigma }(k) = \chi (\rho _{\sigma }^{\beta } |k|) \chi ^{c}\Big ( \frac{|k|}{k_{F}^{\sigma }} \Big ). \end{aligned}
(4.36)

Notice that these regularized operators preserve the important orthogonality relation in Eq. (3.27):

\begin{aligned} u^{\text {r}} \overline{v}^{\text {r}} = 0. \end{aligned}
(4.37)

We shall also denote by $$\omega ^{\text {r}}$$ the regularized version of $$\omega$$, $$\omega ^{\text {r}}= \overline{v^{\text {r}}} v^{\text {r}}$$. The reason for the smoothing in momentum space is to ensure fast decay in $$|x-y|_L$$ of kernels $$\omega ^{\text {r}}(x;y)$$ and $$u^{\text {r}}(x;y)$$ (this has to be compared with the nonintegrable decay of $$\omega (x;y)$$). Such improved decay will play an important technical role in the definition of the almost-bosonic Bogoliubov transformation; we postpone the discussion after Definition 4.3. Notice that the definition of the operator $$u^{\text {r}}$$ also includes an ultraviolet regularization, for momenta $$\ge 2\rho ^{-\beta }$$, which implies that $$|u^{\text {r}}(x;y)|$$ is bounded. For regular enough interaction potentials, the effect of this ultraviolet cutoff turns out to be negligible.

The next proposition collects useful bounds for the regularized kernels.

### Proposition 4.2

(Bounds for the regularized kernels). For all $$n\in \mathbb {N}$$ and for L large enough:

\begin{aligned} \Vert u_{x,\sigma }^{\text {r}} \Vert _{2} \le C \rho _{\sigma }^{-\frac{3\beta }{2}},\quad \Vert v_{x,\sigma }^{\text {r}} \Vert _{2} \le \rho _{\sigma }^{\frac{1}{2}},\quad \Vert \omega ^{\text {r}}_{x,\sigma } \Vert _{1} \le C\rho _{\sigma }^{-\frac{\epsilon }{3}},\quad \Vert u^{\text {r}}_{x,\sigma } \Vert _{1} \le C. \end{aligned}
(4.38)

### Proof

The first two estimates of (4.38) easily follow from (4.35). Consider now the last two. Let us prove the estimate for $$\omega ^{\text {r}}(x,y)$$. (We shall omit the spin label for simplicity.) Let $$\omega _{\infty }^{\text {r}}(x,y)$$ be the infinite volume limit of $$\omega ^{\text {r}}(x,y)$$. We have, performing the angular integration:

\begin{aligned} \omega _{\infty }^{\text {r}}(x,y)= & {} \int d^{3}k\, \hat{\omega }^{\text {r}}(k) e^{ik\cdot (x-y)} \nonumber \\= & {} \frac{4\pi }{|x-y|} \int _{0}^{\infty } dt\, t \hat{\omega }^{\text {r}}(t) \sin (t |x-y|), \end{aligned}
(4.39)

where we used that $$\hat{\omega }^{\text {r}}(k)\equiv \hat{\omega }^{\text {r}}(|k|)$$. Recall that $$\hat{\omega }^{\text {r}}(|k|)$$ is a smooth, compactly supported function, given by $$(\hat{v}^{\text {r}}(k))^{2}$$, Eq. (4.36). Using that $$\sin (t |x-y|) = -|x-y|^{-1} \partial _{t} \cos (t|x-y|)$$ and integrating by parts, we get:

\begin{aligned} \omega _{\infty }^{\text {r}}(x,y) = \frac{4\pi }{|x-y|^{2}} \int dt\, ( \hat{\omega }^{\text {r}}(t) + t \partial _{t} \hat{\omega }^{\text {r}}(t) ) \cos (t |x-y|). \end{aligned}
(4.40)

The first term in the integral is easily bounded by $$C \rho ^{\frac{1}{3}}|x-y|^{-2}$$, using that $$\hat{\omega }^{\text {r}}(t)$$ is bounded and supported for $$0\le t\le C \rho ^{\frac{1}{3}}$$. The second term is supported for $$|t - C\rho ^{\frac{1}{3}}| \le K\rho ^{\alpha }$$, and in this domain $$| \partial _{t} \hat{\omega }^{\text {r}}(t) | \le C\rho ^{- \alpha }$$. Thus, the second term in the integral is also bounded by $$C \rho ^{\frac{1}{3}}|x-y|^{-2}$$. All in all:

\begin{aligned} |\omega _{\infty }^{\text {r}}(x,y)| \le \frac{C\rho ^{\frac{1}{3}}}{|x-y|^{2}}. \end{aligned}
(4.41)

Integrating by parts two more times, we get:

\begin{aligned} | \omega ^{\text {r}}_{\infty }(x;y) |\le & {} \frac{C}{|x-y|^{4}} \int dt\, ( | \partial _{t}^{2} \hat{\omega }^{\text {r}}(t)| + t |\partial _{t}^{3} \hat{\omega }^{\text {r}}(t) | ) \nonumber \\\le & {} \frac{C\rho ^{\frac{1}{3} - 2\alpha }}{|x-y|^{4}}\;; \end{aligned}
(4.42)

we used that $$\alpha \ge 1/3$$, to conclude that the last term in the right-hand side of the first line dominates over the first. Therefore, putting together (4.41), (4.42):

\begin{aligned} \Vert \omega ^{\text {r}}_{\infty ,x} \Vert _{1}\le & {} \int _{|y|\le \rho ^{-\alpha }} dy\, \frac{C\rho ^{\frac{1}{3}}}{|y|^{2}} + \int _{|y| > \rho ^{-\alpha }} dy\,\frac{C\rho ^{\frac{1}{3} - 2\alpha }}{|y|^{4}} \nonumber \\\le & {} C\rho ^{\frac{1}{3} - \alpha } \equiv C\rho ^{-\frac{\epsilon }{3}}, \end{aligned}
(4.43)

recall that $$\alpha = \frac{1}{3} + \frac{\epsilon }{3}$$. To prove the estimate for $$\Vert \omega ^{\text {r}}_{x} \Vert _{1}$$, we write, for $$n\ge 0$$ large enough, independent of L, and for L large enough:

\begin{aligned} \Vert \omega ^{\text {r}}_{x} \Vert _{1}\le & {} \Vert \omega ^{\text {r}}_{x} \chi (|\cdot |_{L} \le \rho ^{-n}) \Vert _{1} + \Vert \omega ^{\text {r}}_{x} \chi (|\cdot |_{L}> \rho ^{-n}) \Vert _{1} \nonumber \\\le & {} \Vert \omega ^{\text {r}}_{\infty , x} \chi (|\cdot |_{L} \le \rho ^{-n}) \Vert _{1} + \Vert (\omega ^{\text {r}}_{x} - \omega ^{\text {r}}_{\infty ,x}) \chi (|\cdot |_{L} \le \rho ^{-n}) \Vert _{1} \nonumber \\&+ \Vert \omega ^{\text {r}}_{x} \chi (|\cdot |_{L} > \rho ^{-n}) \Vert _{1} \nonumber \\\le & {} C\rho ^{-\frac{\epsilon }{3}}. \end{aligned}
(4.44)

To prove the last inequality, we used that, for any fixed x, $$|\omega ^{\text {r}}_{\infty }(x) - \omega ^{\text {r}}(x)| \le C/L$$. Also, we used that, by the smoothness of $$\hat{\omega }^{\text {r}}$$, $$\omega ^{\text {r}}(x;y)$$ decays faster than any power in $$|x-y|_{L}$$ (nonuniformly in $$\rho$$), which allows to control the last term in the second line. The proof of the last estimate in Eq. (4.38) is completely analogous, and we shall omit the details; the reason for the uniform bound in $$\rho$$ is that the infrared part of $$\hat{u}^{\text {r}}(k)$$ is smoothened on scale $$\rho ^{\frac{1}{3}}$$ instead of $$\rho ^{\frac{1}{3} + \frac{\epsilon }{3}}$$. $$\square$$

As suggested by the heuristic discussion in Sect. 4.1, in the following an important role will be played by the solution of the scattering equation (4.28). We shall denote by $$\varphi$$ the periodization of $$\varphi _{\gamma }$$ over $$\Lambda _{L}$$, recall Eq. (4.31).

### Definition 4.3

(The correlation structure). Let $$\gamma \ge 0$$, $$\lambda \in [0,1]$$. We define the unitary operator $$T_{\lambda }: \mathcal {F} \rightarrow \mathcal {F}$$ as:

\begin{aligned} T_{\lambda } := e^{\lambda (B - B^{*})},\quad B := \int dzdz'\, \varphi (z-z') a_{\uparrow }(u^{\text {r}}_{z}) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}).\nonumber \\ \end{aligned}
(4.45)

We shall also set $$T \equiv T_{1}$$.

The operator $$T_{\lambda }$$ is a regularized version of the operator introduced in Sect. 4.1. Notice that, despite the presence of two volume integrations, the B operator is bounded proportionally to $$L^{3}$$: to see this, notice that the $$z, z'$$ variables satisfy $$|z - z'|_{L} \le \rho ^{-\gamma }$$, due to the compact support of $$\varphi$$ in $$\Lambda _{L}$$. In particular, thanks to the estimate $$\Vert \varphi _{\gamma }\Vert _{L^{1}(\mathbb {R}^{3})} \le C\rho ^{-2\gamma }$$, see Appendix 7, and recalling (4.30), we have:

\begin{aligned} \Vert \varphi \Vert _{L^{1}(\Lambda _{L})} \le C\Vert \varphi _{\gamma } \Vert _{L^{1}(\mathbb {R}^{3})} \le C\rho ^{-2\gamma }. \end{aligned}
(4.46)

As it will be clear at the end of our analysis, this regularization will not affect the computation of the ground state energy density at order $$\rho ^{2}$$. The boundedness of B follows from the presence of the ultraviolet cutoff in the definition of $$u^{\text {r}}$$. In fact, by (3.5) we have, recalling (4.38):

\begin{aligned} \Vert a(u^{\text {r}}_{x}) \Vert \le \Vert u^{\text {r}}_{x} \Vert _{2} \le C\rho ^{-\frac{3\beta }{2}} \le C_{\beta },\quad \Vert a(\overline{v}^{\text {r}}_{x}) \Vert \le \Vert \overline{v}^{\text {r}}_{x} \Vert _{2} \le C\rho ^{\frac{1}{2}}. \end{aligned}
(4.47)

The operator $$T_{\lambda }$$ plays the role of ‘almost-bosonic’ Bogoliubov transformation. The smoothing of the operators $$u^{\text {r}}$$, $$v^{\text {r}}$$ is needed in order to control the error terms in (4.18). In order to estimate them uniformly in the volume, we will exploit the fast decay of the kernels $$u^{\text {r}}(x;y)$$, $$v^{\text {r}}(x;y)$$: this will allow us to avoid the accumulation of volume factors, paying the price of a small negative power of the density.

From a more conceptual viewpoint, these regularizations allow to make the action of $$T_{\lambda }$$ quasi-local: informally, the unitary conjugation with $$T_{\lambda }$$ of a Fock space operator localized in a region $$\mathcal {R}$$ of $$\Lambda _{L}$$ is an operator that lives in a bigger region $$\widetilde{\mathcal {R}}$$, whose size is bounded uniformly in L (but not in $$\rho$$, in general).

## 5 Bounds on Interpolating States

### 5.1 Introduction

In this section, we shall propagate the a priori bounds on $$\mathcal {N}$$, $$\mathbb {H}_{0}$$, $$\mathbb {Q}_{1}$$ proven in Sect. 3.2.2, over the states

\begin{aligned} \xi _{\lambda } := T^{*}_{\lambda } R^{*}\psi , \end{aligned}
(5.1)

with $$\psi$$ being an approximate ground state, in the sense of Definition 3.4. One of the main results of the section will be that, for $$5/18\le \gamma \le 1/3$$, the following bounds hold true. Let $$\lambda \in [0,1]$$. Then:

\begin{aligned} \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle\le & {} C L^{\frac{3}{2}} \rho ^{\frac{1}{6}} \Vert \mathbb {H}_{0}^{\frac{1}{2}} \xi _{1} \Vert + CL^{3} \rho ^{2 - \gamma },\quad \langle \xi _{\lambda }, \mathbb {H}_{0} \xi _{\lambda } \rangle \le CL^{3} \rho ^{2},\nonumber \\ \langle \xi _{\lambda }, \mathbb {Q}_{1} \xi _{\lambda } \rangle\le & {} CL^{3} \rho ^{2}. \end{aligned}
(5.2)

In particular, these bounds show that the a priori estimates of $$\mathbb {H}_{0}$$, $$\mathbb {Q}_{1}$$ on $$\xi _{0} = R^{*}\psi$$, recall (3.55), (3.66), do not deterioriate over $$\xi _{\lambda } = T^{*}_{\lambda } R^{*} \psi$$, for $$\lambda \in [0,1]$$. Along the way, we shall prove a number of auxiliary results, that will play an important role in the computation of the ground state energy of the system, in Sects. 6, 7. More precisely, we will prove that:

\begin{aligned} \frac{d}{d\lambda } \langle \xi _{\lambda }, \mathbb {H}_{0} \xi _{\lambda } \rangle = -\langle \xi _{\lambda }, \mathbb {T}_{1} \xi _{\lambda }\rangle + \mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda }),\quad \frac{d}{d\lambda } \langle \xi _{\lambda }, \mathbb {Q}_{1} \xi _{\lambda } \rangle = -\langle \xi _{\lambda }, \mathbb {T}_{2} \xi _{\lambda }\rangle + \mathcal {E}_{\mathbb {Q}_{1}}(\xi _{\lambda }) \end{aligned}
(5.3)

where $$\mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda })$$, $$\mathcal {E}_{\mathbb {Q}_{1}}(\xi _{\lambda })$$ are two error terms, subleading with respect to $$L^{3} \rho ^{2}$$, and $$\mathbb {T}_{1}$$, $$\mathbb {T}_{2}$$ are given by:

\begin{aligned} \mathbb {T}_{1}= & {} -\int dxdy\, \theta (|x-y|_{L} < \rho ^{-\gamma })(-2\Delta \varphi )(x-y) a_{\uparrow }(u^{\text {r}}_{x}) a_{\uparrow }(\overline{v}^{\text {r}}_{x}) a_{\downarrow }(u^{\text {r}}_{y}) a_{\downarrow }(\overline{v}^{\text {r}}_{y}) + \text {h.c.}\nonumber \\ \mathbb {T}_{2}= & {} -\int dxdy\, V(x-y) \varphi (x-y) a_{\uparrow }(\overline{v}^{\text {r}}_{x}) a_{\uparrow }(u_{x}) a_{\downarrow }(\overline{v}^{\text {r}}_{y}) a_{\downarrow }(u_{y}) + \text {h.c.}, \end{aligned}
(5.4)

with $$\theta (\cdot )$$ the characteristic function. The important point to notice here is that $$\mathbb {T}_{1}$$, $$\mathbb {T}_{2}$$ have the same ‘bosonic’ structure as $$\widetilde{\mathbb {Q}}_{4}$$. In particular, we will prove the following cancellation, using the fact that the function $$1 - \varphi$$ solves the scattering equation:

\begin{aligned} \langle \xi _{\lambda }, (\mathbb {T}_{1} + \mathbb {T}_{2} + \widetilde{\mathbb {Q}}_{4}) \xi _{\lambda } \rangle = o(L^{3}\rho ^{2}). \end{aligned}
(5.5)

The results (5.2)–(5.5) are the main technical ingredients needed in order to prove our main result, Theorem 2.1, in Sects. 6, 7.

### 5.2 Propagation of the Estimates: Preliminaries

To begin, let us start by propagating the a priori estimate for $$\mathcal {N}$$. The propagation estimate we shall obtain below is not optimal; it will be improved at the end of the section. Nevertheless, Proposition 5.1 will be enough to propagate the a priori bounds for $$\mathbb {H}_{0}$$ and for $$\mathbb {Q}_{1}$$, which is our first task.

### Proposition 5.1

(Propagation estimate for $$\mathcal {N}$$). Let $$\psi \in \mathcal {F}$$ such that $$\Vert \psi \Vert = 1$$. Let $$\gamma \le 1/2$$. Then, the following bound holds, for $$\lambda \in [0,1]$$:

\begin{aligned} |\partial _{\lambda }\langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle | \le C\langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle + C L^3 \rho ^{2- \gamma }. \end{aligned}
(5.6)

### Corollary 5.2

Let $$\psi$$ be an approximate ground state. Then:

\begin{aligned} \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle \le CL^{3} \rho ^{\frac{7}{6}}. \end{aligned}
(5.7)

### Proof

Eq. (5.7) immediately follows from (5.6) and Grönwall lemma, recalling the a priori estimate $$\langle \xi _{0}, \mathcal {N} \xi _{0} \rangle \le CL^{3} \rho ^{\frac{7}{6}}$$, Eq. (3.63). $$\square$$

Let us now discuss the proof of Proposition 5.1. The proof is based on the following lemma.

### Lemma 5.3

Let $$g\in L^{1}(\Lambda _{L}) \cap L^{2}(\Lambda _{L})$$, and let:

\begin{aligned} b_{z,\sigma } = a_{\sigma }(u^{\text {r}}_{z}) a_{\sigma }(\overline{v}^{\text {r}}_{z}),\quad b_{\sigma }(g_{z}) = \int dz'\, \overline{g(z-z')} b_{z',\sigma }. \end{aligned}
(5.8)

Then:

\begin{aligned} \Vert b_{\sigma }(g_{z}) \Vert \le C_{\beta }\rho ^{\frac{1}{2}}\Vert g\Vert _{1},\quad \Vert b^{*}_{\sigma }(g_{z}) \Vert \le C_{\beta }\rho ^{\frac{1}{2}}\Vert g\Vert _{1} \end{aligned}
(5.9)

and:

\begin{aligned} \Vert b^{*}_{\sigma }(g_{z}) \psi \Vert \le \Vert b_{\sigma }(g_{z}) \psi \Vert + C\rho ^{\frac{1}{2}}\Vert g\Vert _{2} \Vert \psi \Vert . \end{aligned}
(5.10)

### Proof

Consider the first of (5.9), the proof of the second is identical. The inequality simply follows from the boundedness of the fermionic operators:

\begin{aligned} \Vert b_{\sigma }(g_{z})\psi \Vert\le & {} \int dz' |g(z-z')| \Vert a_{\sigma }(u^{\text {r}}_{z'}) a_{\sigma }(\overline{v}^{\text {r}}_{z'}) \psi \Vert \nonumber \\\le & {} \int dz' |g(z-z')| \Vert u^{\text {r}}_{z'}\Vert _{2} \Vert \overline{v}^{\text {r}}_{z'} \Vert _{2} \Vert \psi \Vert \nonumber \\\le & {} C_{\beta } \Vert g\Vert _{1} \rho ^{\frac{1}{2}} \Vert \psi \Vert . \end{aligned}
(5.11)

Let us now prove (5.10). It is convenient to rewrite the first norm as:

\begin{aligned} \Vert b^{*}_{\sigma }(g_{z}) \xi _{\lambda }\Vert ^{2} = \Vert b_{\sigma }(g_{z}) \xi _{\lambda }\Vert ^{2} - \langle \xi _{\lambda }, [ b^{*}_{\sigma }(g_{z}), b_{\sigma }(g_{z}) ]\xi _{\lambda } \rangle . \end{aligned}
(5.12)

Let us compute the commutator. We get:

\begin{aligned}&[ b^{*}_{\sigma }(g_{z}), b_{\sigma }(g_{z}) ] = \int dxdy\,g(z-x) \overline{g(z-y)} [ a^{*}_{\sigma }(\overline{v}^{\text {r}}_{x})a^{*}_{\sigma }(u^{\text {r}}_{x}), a_{\sigma }(u^{\text {r}}_{y}) a_{\sigma }(\overline{v}^{\text {r}}_{y}) ] \nonumber \\&\quad = \int dxdy\,g(z-x) \overline{g(z-y)} \Big ( a^{*}_{\sigma }(\overline{v}^{\text {r}}_{x}) [ a^{*}_{\sigma }(u^{\text {r}}_{x}), a_{\sigma }(u^{\text {r}}_{y}) a_{\sigma }(\overline{v}^{\text {r}}_{y}) ] \nonumber \\&\qquad \qquad + [ a^{*}_{\sigma }(\overline{v}^{\text {r}}_{x}), a_{\sigma }(u^{\text {r}}_{y}) a_{\sigma }(\overline{v}^{\text {r}}_{y}) ] a^{*}_{\sigma }(u^{\text {r}}_{x})\Big )\nonumber \\&\quad = \int dxdy\,g(z-x) \overline{g(z-y)} \Big ( a^{*}_{\sigma }(\overline{v}^{\text {r}}_{x}) (u^{\text {r}})^{2}_{\sigma }(x;y) a_{\sigma }(\overline{v}^{\text {r}}_{y}) - a_{\sigma }(u^{\text {r}}_{y}) \omega ^{\text {r}}_{\sigma }(x;y) a^{*}_{\sigma }(u^{\text {r}}_{x}) \Big ).\nonumber \\ \end{aligned}
(5.13)

Putting the last term into normal order, we get:

\begin{aligned}{}[ b^{*}_{\sigma }(g_{z}), b_{\sigma }(g_{z}) ]= & {} -\int dxdy\,g(z-x) \overline{g(z-y)} \omega ^{\text {r}}_{\sigma }(x;y) (u^{\text {r}})^{2}_{\sigma }(y;x) \nonumber \\&+ \int dxdy\,g(z-x) \overline{g(z-y)} \Big ( a^{*}_{\sigma }(\overline{v}^{\text {r}}_{x}) (u^{\text {r}})^{2}_{\sigma }(x;y) a_{\sigma }(\overline{v}^{\text {r}}_{y}) \nonumber \\&+ a^{*}_{\sigma }(u^{\text {r}}_{x}) \omega ^{\text {r}}_{\sigma }(x;y) a_{\sigma }(u^{\text {r}}_{y})\Big ). \end{aligned}
(5.14)

The last two terms are nonnegative. In fact:

\begin{aligned}&\int dxdy\,g(z-x) \overline{g(z-y)} \langle \xi _{\lambda }, a^{*}_{\sigma }(\overline{v}^{\text {r}}_{x}) (u^{\text {r}})^{2}_{\sigma }(x;y) a_{\sigma }(\overline{v}^{\text {r}}_{y}) \xi _{\lambda }\rangle \nonumber \\&\quad = \int dr\, \Big \langle \xi _{\lambda }, \Big ( \int dx\, g(z-x) a^{*}_{\sigma }(\overline{v}^{\text {r}}_{x}) u_{\sigma }^{\text {r}}(x;r) \Big )\Big ( \int dy\, \overline{g(z-y)} a_{\sigma }(\overline{v}^{\text {r}}_{y}) u_{\sigma }^{\text {r}}(r;y) \Big ) \xi _{\lambda } \Big \rangle \nonumber \\&\quad \ge 0, \end{aligned}
(5.15)

and similarly:

\begin{aligned}&\int dxdy\, g(z-x) \overline{g(z-y)} \langle \xi _{\lambda }, a^{*}_{\sigma }(u^{\text {r}}_{x}) \omega ^{\text {r}}_{\sigma }(x;y) a_{\sigma }(u^{\text {r}}_{y}) \xi _{\lambda }\rangle \nonumber \\&\quad = \int dr\, \Big \langle \xi _{\lambda }, \Big ( \int dx\, g(z-x) a^{*}_{\sigma }(u^{\text {r}}_{x}) \overline{v}_{\sigma }^{\text {r}}(x;r) \Big )\Big ( \int dy\, \overline{g(z-y)} a_{\sigma }(u^{\text {r}}_{y}) v_{\sigma }^{\text {r}}(r;y) \Big ) \xi _{\lambda } \Big \rangle \nonumber \\&\quad \ge 0. \end{aligned}
(5.16)

Therefore:

\begin{aligned} \Vert b^{*}_{\uparrow }(g_{z}) \xi _{\lambda }\Vert ^{2} \le \Vert b_{\uparrow }(g_{z}) \xi _{\lambda }\Vert ^{2} + \int dxdy\,g(z-x) \overline{g(z-y)} \omega ^{\text {r}}_{\sigma }(x;y) (u^{\text {r}})^{2}_{\sigma }(y;x). \end{aligned}
(5.17)

The last term can be estimated as, using that $$(u^{\text {r}})^{2}$$ satisfies the same estimates as $$u^{\text {r}}$$, Eqs. (4.38):

\begin{aligned} \Big |\int dxdy\,g(z-x) \overline{g(z-y)} \omega ^{\text {r}}_{\sigma }(x;y) (u^{\text {r}})^{2}_{\sigma }(y;x) \Big |\le & {} \rho \int dxdy\, |g(z-x)|^{2} |(u^{\text {r}})^{2}_{\sigma }(y;x)| \nonumber \\\le & {} C\rho \Vert g\Vert _{2}^{2}. \end{aligned}
(5.18)

In conclusion:

\begin{aligned} \Vert b^{*}_{\uparrow }(g_{z}) \xi _{\lambda }\Vert ^{2} \le \Vert b_{\uparrow }(g_{z}) \xi _{\lambda }\Vert ^{2} + C\rho \Vert g\Vert _{2}^{2}. \end{aligned}
(5.19)

This concludes the proof. $$\square$$

We are now ready to prove Proposition 5.1.

### Proof

(of Proposition 5.1.) We compute:

\begin{aligned} \partial _{\lambda } \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle= & {} -\langle \xi _{\lambda }, [ \mathcal {N}, (B - B^{*}) ] \xi _{\lambda } \rangle \nonumber \\= & {} -4\int dzdz'\, \varphi (z-z') \langle \xi _{\lambda }, a_{\uparrow }(u^{\text {r}}_{z}) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) \xi _{\lambda } \rangle + \text {c.c..}\nonumber \\ \end{aligned}
(5.20)

Using the notation (5.8), we rewrite (5.20) as:

\begin{aligned} \partial _{\lambda } \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle = -4\int dz\, \langle \xi _{\lambda }, b_{\uparrow }(\varphi _{z}) b_{\downarrow ,z} \xi _{\lambda } \rangle + \text {c.c.}. \end{aligned}
(5.21)

We then estimate:

\begin{aligned} |\langle \xi _{\lambda }, b_{\uparrow }(\varphi _{z}) b_{\downarrow ,z} \xi _{\lambda } \rangle |\le \Vert b^{*}_{\uparrow }(\varphi _{z}) \xi _{\lambda }\Vert \Vert b_{\downarrow ,z} \xi _{\lambda } \Vert \;; \end{aligned}
(5.22)

by Lemma 5.3, using that $$\Vert \varphi \Vert _{2} \le C\rho ^{-\frac{\gamma }{2}}$$:

\begin{aligned} |\langle \xi _{\lambda }, b_{\uparrow }(\varphi _{z}) b_{\downarrow ,z} \xi _{\lambda } \rangle |\le \Vert b_{\uparrow }(\varphi _{z}) \xi _{\lambda }\Vert \Vert b_{\downarrow ,z} \xi _{\lambda } \Vert + C\rho ^{\frac{1}{2} - \frac{\gamma }{2}} \Vert b_{\downarrow ,z} \xi _{\lambda } \Vert . \end{aligned}
(5.23)

Consider the first term in (5.23). We have, using that $$\Vert \varphi \Vert _{1} \le C\rho ^{-2\gamma }$$:

\begin{aligned} \int dz\, \Vert b_{\uparrow }(\varphi _{z}) \xi _{\lambda }\Vert \Vert b_{\downarrow ,z} \xi _{\lambda } \Vert\le & {} C\rho \int dzdz'\, \varphi (z-z') (\Vert a_{\uparrow }(u^{\text {r}}_{z}) \xi _{\lambda } \Vert ^{2} + \Vert a_{\downarrow }(u^{\text {r}}_{z'}) \xi _{\lambda }\Vert ^{2})\nonumber \\\le & {} C\rho ^{1-2\gamma } \langle \xi _{\lambda }, \mathcal {N}\xi _{\lambda }\rangle . \end{aligned}
(5.24)

Consider now the second term in (5.23). We have:

\begin{aligned} C\rho ^{\frac{1}{2} - \frac{\gamma }{2}}\int dz\, \Vert b_{\downarrow ,z} \xi _{\lambda } \Vert\le & {} CL^{\frac{3}{2}} \rho ^{1 - \frac{\gamma }{2}} \Vert \mathcal {N}^{\frac{1}{2}}\xi _{\lambda } \Vert \nonumber \\\le & {} CL^{3}\rho ^{2 - \gamma } + \langle \xi _{\lambda }, \mathcal {N}\xi _{\lambda }\rangle . \end{aligned}
(5.25)

All together:

\begin{aligned} \partial _{\lambda } \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle \le C\rho ^{1-2\gamma } \langle \xi _{\lambda }, \mathcal {N}\xi _{\lambda }\rangle + CL^{3} \rho ^{2 - \gamma } + C\langle \xi _{\lambda }, \mathcal {N}\xi _{\lambda }\rangle . \end{aligned}
(5.26)

The final claim follows from $$\gamma \le 1/2$$. This concludes the proof. $$\square$$

The next lemma will allow us to bound recurrent expressions in our computations. We shall use the short-hand notations $$\partial ^{n} \overline{v}^{\text {r}}_{x}$$, $$\partial ^{n} u_{x}^{\text {r}}$$ to denote the functions (in Eq. (5.27) x is fixed, y is the argument of the functions):

\begin{aligned} \frac{\partial ^{n}}{\partial y_{i_{1}} \cdots \partial y_{i_n}} \overline{v}^{\text {r}}(y;x),\quad \frac{\partial ^{n}}{\partial y_{i_{1}} \cdots \partial y_{i_n}} u^{\text {r}}(y;x), \end{aligned}
(5.27)

for some choice of indices $$i_{1}, \ldots , i_{n}$$ (their values will be inessential for the bounds).

### Lemma 5.4

Under the assumptions of Theorem 2.1, the following holds. Let $$n_{2} \in \mathbb {N}$$, and let $$\psi$$ be an approximate ground state. Let $$\gamma \le 7/18$$. Then:

\begin{aligned}&\Big | \int dxdy\, \varphi (x-y) \langle \xi _{\lambda }, a_{\uparrow }(u^{\text {r}}_{x}) a_{\uparrow }(\partial ^{n_{2}}\overline{v}^{\text {r}}_{x}) a_{\downarrow }(u^{\text {r}}_{y}) a_{\downarrow }(\overline{v}^{\text {r}}_{y}) \xi _{\lambda } \rangle \Big | \le C L^{\frac{3}{2}} \rho ^{1 - \frac{\gamma }{2} + \frac{n_{2}}{3}} \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda } \Vert \\&\Big | \int dxdy\, \varphi (x-y) \langle \xi _{\lambda }, a_{\uparrow }(\partial u^{\text {r}}_{x}) a_{\uparrow }(\partial ^{n_{2}}\overline{v}^{\text {r}}_{x}) a_{\downarrow }(u^{\text {r}}_{y}) a_{\downarrow }(\overline{v}^{\text {r}}_{y}) \xi _{\lambda } \rangle \Big | \\&\quad \le CL^{\frac{3}{2}} \rho ^{1 - \frac{\gamma }{2} + \frac{n_{2}}{3}} ( \Vert \mathbb {H}_{0}^{\frac{1}{2}}\xi _{\lambda }\Vert + \rho ^{\frac{1}{3}} \Vert \mathcal {N}^{\frac{1}{2}}\xi _{\lambda } \Vert ). \end{aligned}

We refer the reader to Appendix 7 for the proof. We shall use Lemma 5.4 to propagate the a priori estimates on $$\xi _{0} = R^{*}\psi$$ for $$\mathbb {H}_{0}$$, $$\mathbb {Q}_{1}$$ and $$\widetilde{\mathbb {Q}}_{1}$$, Eqs. (3.55), (3.66), to the interpolating states $$\xi _{\lambda } = T^{*}_{\lambda } R^{*}\psi$$, via a Grönwall-type argument. To do so, the next proposition will play an important role.

### Proposition 5.5

(Propagation estimate for $$\mathbb {H}_{0}$$, $$\mathbb {Q}_{1}$$ - Part 1) Let $$\gamma \le 7/18$$. Under the assumptions of Theorem 2.1 the following is true:

\begin{aligned} \frac{d}{d\lambda }\langle \xi _{\lambda }, \mathbb {H}_{0} \xi _{\lambda } \rangle= & {} -\langle \xi _{\lambda }, \mathbb {T}_{1} \xi _{\lambda }\rangle + \mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda }),\quad \frac{d}{d\lambda } \langle \xi _{\lambda }, \mathbb {Q}_{1} \xi _{\lambda } \rangle = -\langle \xi _{\lambda }, \mathbb {T}_{2} \xi _{\lambda } \rangle + \mathcal {E}_{\mathbb {Q}_{1}}(\xi _{\lambda }),\nonumber \\ \frac{d}{d\lambda } \langle \xi _{\lambda }, \widetilde{\mathbb {Q}}_{1} \xi _{\lambda } \rangle= & {} -\langle \xi _{\lambda }, \mathbb {T}_{2} \xi _{\lambda } \rangle + \mathcal {E}_{\widetilde{\mathbb {Q}}_{1}}(\xi _{\lambda }) \end{aligned}
(5.28)

where

\begin{aligned} \mathbb {T}_{1}= & {} -\int dxdy\, \theta (|x-y|_{L} < \rho ^{-\gamma })(-2\Delta \varphi )(x-y) a_{\uparrow }(u^{\text {r}}_{x}) a_{\uparrow }(\overline{v}^{\text {r}}_{x}) a_{\downarrow }(u^{\text {r}}_{y}) a_{\downarrow }(\overline{v}^{\text {r}}_{y}) + \text {h.c.}\nonumber \\ \mathbb {T}_{2}= & {} -\int dxdy\, V(x-y) \varphi (x-y) a_{\uparrow }(\overline{v}^{\text {r}}_{x}) a_{\uparrow }(u_{x}) a_{\downarrow }(\overline{v}^{\text {r}}_{y}) a_{\downarrow }(u_{y}) + \text {h.c.,} \end{aligned}
(5.29)

and the error terms are bounded as, for $$0\le \eta < \min \{\gamma , \frac{1}{3}\}$$:

\begin{aligned} | \mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda }) |\le & {} CL^{\frac{3}{2}} \rho ^{\frac{4}{3} - \frac{\gamma }{2}} ( \Vert \mathbb {H}_{0}^{\frac{1}{2}}\xi _{\lambda }\Vert + \rho ^{\frac{1}{3}} \Vert \mathcal {N}^{\frac{1}{2}}\xi _{\lambda } \Vert )\nonumber \\ |\mathcal {E}_{\widetilde{\mathbb {Q}}_{1}}(\xi _{\lambda })|\le & {} C_{\beta }\rho ^{1 - 2\gamma - \eta } \Vert \widetilde{\mathbb {Q}}^{\frac{1}{2}}_{1} \xi _{\lambda } \Vert ( \Vert \mathbb {H}_{0}^{\frac{1}{2}} \xi _{\lambda } \Vert + \rho ^{\frac{5\eta }{2}} \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda } \Vert ) + CL^{\frac{3}{2}} \rho ^{2 - 2\gamma } \Vert \widetilde{\mathbb {Q}}^{\frac{1}{2}}_{1} \xi _{\lambda } \Vert \nonumber \\ |\mathcal {E}_{\mathbb {Q}_{1}}(\xi _{\lambda })|\le & {} C_{\beta }\rho ^{1 - 2\gamma - \eta } \Vert \widetilde{\mathbb {Q}}^{\frac{1}{2}}_{1} \xi _{\lambda } \Vert ( \Vert \mathbb {H}_{0}^{\frac{1}{2}} \xi _{\lambda } \Vert + \rho ^{\frac{5\eta }{2}} \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda } \Vert ) + CL^{\frac{3}{2}} \rho ^{2 - 2\gamma } \Vert \mathbb {Q}^{\frac{1}{2}}_{1} \xi _{\lambda } \Vert . \nonumber \\ \end{aligned}
(5.30)

### Proof

$$\underline{\hbox {Derivative of } \langle \xi _{\lambda }, \mathbb {H}_{0} \xi _{\lambda } \rangle }$$. We compute:

\begin{aligned} \frac{d}{d\lambda } \langle \xi _{\lambda }, \mathbb {H}_{0} \xi _{\lambda } \rangle = -\langle \xi _{\lambda }, [ \mathbb {H}_{0}, B ] \xi _{\lambda } \rangle + \text {c.c.} \end{aligned}
(5.31)

with

\begin{aligned}{}[ \mathbb {H}_{0}, B ] = \int dzdz'\, \varphi (z-z')[ \mathbb {H}_{0}, a_{\uparrow }(u^{\text {r}}_{z}) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) ]. \end{aligned}
(5.32)

We write the commutator as:

\begin{aligned}{}[ \mathbb {H}_{0}, a_{\uparrow }(u^{\text {r}}_{z}) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) ]= & {} [ \mathbb {H}_{0}, a_{\uparrow }(u_{z}^{\text {r}}) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) ] a_{\downarrow }(u^{\text {r}}_{z'}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) \nonumber \\&+ a_{\uparrow }(u_{z}^{\text {r}}) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) [ \mathbb {H}_{0}, a_{\downarrow }(u^{\text {r}}_{z'}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) ] ,\nonumber \\ \end{aligned}
(5.33)

where recalling that $$\mathbb {H}_{0} = \sum _{\sigma } | |k|^{2} - \mu _{\sigma } | \hat{a}^{*}_{k,\sigma } \hat{a}_{k,\sigma }$$, with $$\mu _{\sigma } = (k_{F}^{\sigma })^{2}$$:

\begin{aligned}{}[ \mathbb {H}_{0}, a_{\sigma }(u_{z}^{\text {r}}) a_{\sigma }(\overline{v}^{\text {r}}_{z}) ]= & {} -a_{\sigma }((-\Delta - \mu _{\sigma })u_{z}^{\text {r}}) a_{\sigma }(\overline{v}^{\text {r}}_{z}) - a_{\sigma }(u_{z}^{\text {r}}) a_{\sigma }((\Delta + \mu _{\sigma })\overline{v}^{\text {r}}_{z}) \nonumber \\= & {} \Delta _{z} a_{\sigma }(u^{\text {r}}_{z}) a_{\sigma }(\overline{v}^{\text {r}}_{z}) - 2a_{\sigma }(u_{z}^{\text {r}}) a_{\sigma }(\Delta \overline{v}^{\text {r}}_{z}) - 2a_{\sigma }(\nabla u^{\text {r}}_{z}) a_{\sigma }(\nabla \overline{v}^{\text {r}}_{z}).\nonumber \\ \end{aligned}
(5.34)

To write this identity, we used that $$u^{\text {r}}(z';z) \equiv u^{\text {r}}(z-z')$$ and $$v^{\text {r}}(z';z) \equiv v^{\text {r}}(z' + z)$$; recall Definition (4.34). Hence, $$\Delta _{z'} u^{\text {r}}(z';z) = \Delta _{z} u^{\text {r}}(z';z)$$ and $$\Delta _{z'} v^{\text {r}}(z';z) = \Delta _{z} v^{\text {r}}(z';z)$$. We define:

\begin{aligned} \text {I}= & {} \int dzdz'\, \varphi (z-z') \Delta _{z} \langle \xi _{\lambda }, a_{\uparrow }(u^{\text {r}}_{z}) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) \xi _{\lambda }\rangle + (\uparrow , z) \leftrightarrow (\downarrow , z') \nonumber \\ \text {II}= & {} -2\int dzdz'\, \varphi (z-z') \langle \xi _{\lambda }, a_{\uparrow }(u_{z}^{\text {r}}) a_{\uparrow }(\Delta \overline{v}^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'})\xi _{\lambda }\rangle + (\uparrow , z) \leftrightarrow (\downarrow , z') \nonumber \\ \text {III}= & {} -2\int dzdz'\, \varphi (z-z') \langle \xi _{\lambda }, a_{\uparrow }(\nabla u^{\text {r}}_{z}) a_{\uparrow }(\nabla \overline{v}^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) \xi _{\lambda } \rangle + (\uparrow , z) \leftrightarrow (\downarrow , z').\nonumber \\ \end{aligned}
(5.35)

To estimate $$\text {II}$$ and $$\text {III}$$, we shall use Lemma 5.4. We have:

\begin{aligned} |\text {II}|\le & {} C L^{\frac{3}{2}} \rho ^{\frac{5}{3} - \frac{\gamma }{2}} \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda } \Vert \nonumber \\ |\text {III}|\le & {} CL^{\frac{3}{2}} \rho ^{\frac{4}{3} - \frac{\gamma }{2}} ( \Vert \mathbb {H}_{0}^{\frac{1}{2}}\xi _{\lambda }\Vert + \rho ^{\frac{1}{3}} \Vert \mathcal {N}^{\frac{1}{2}}\xi _{\lambda } \Vert ). \end{aligned}
(5.36)

Consider now $$\text {I}$$. We rewrite it as:

\begin{aligned} \text {I} = \int _{|z-z'|_{L} \le \rho ^{-\gamma }} dzdz'\, \varphi (z-z') (\Delta _{z} + \Delta _{z'}) \langle \xi _{\lambda }, a_{\uparrow }(u^{\text {r}}_{z}) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) \xi _{\lambda }\rangle . \end{aligned}
(5.37)

The condition on the integration domain follows from the fact that $$\varphi$$ is the periodization of $$\varphi _{\gamma }$$, Eq. (4.30), which is compactly supported in the ball B of radius $$\rho ^{-\gamma }$$, Eq. (4.28). Therefore, using that $$\varphi _{\gamma } = \nabla \varphi _{\gamma } = 0$$ on $$\partial B$$, we can integrate by parts in Eq. (5.37), without producing boundary terms. This is the point of our analysis where we take advantage of the Neumann boundary conditions of the scattering equation. We have:

\begin{aligned} \text {I} = \int dzdz'\, \theta (|z-z'|_{L} < \rho ^{-\gamma })(2\Delta \varphi )(z-z') \langle \xi _{\lambda }, a_{\uparrow }(u^{\text {r}}_{z}) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) \xi _{\lambda } \rangle , \end{aligned}
(5.38)

with $$\theta (\cdot )$$ the characteristic function. All in all:

\begin{aligned}&\frac{d}{d\lambda }\langle \xi _{\lambda }, \mathbb {H}_{0} \xi _{\lambda } \rangle = -\langle \xi _{\lambda }, \mathbb {T}_{1} \xi _{\lambda } \rangle + \mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda }), \end{aligned}
(5.39)

where

\begin{aligned} \mathbb {T}_{1} = \int dxdy\, \theta (|z-z'|_{L} < \rho ^{-\gamma }) (2\Delta \varphi )(x-y) a_{\uparrow }(u^{\text {r}}_{x}) a_{\uparrow }(\overline{v}^{\text {r}}_{x}) a_{\downarrow }(u^{\text {r}}_{y}) a_{\downarrow }(\overline{v}^{\text {r}}_{y}) + \text {h.c.} \end{aligned}

and $$\mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda })$$ collects the error terms, produced by the contributions $$\text {II}$$ and $$\text {III}$$. Thanks to the estimates in (5.36):

\begin{aligned} |\mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda })| \le CL^{\frac{3}{2}} \rho ^{\frac{4}{3} - \frac{\gamma }{2}} ( \Vert \mathbb {H}_{0}^{\frac{1}{2}}\xi _{\lambda }\Vert + \rho ^{\frac{1}{3}} \Vert \mathcal {N}^{\frac{1}{2}}\xi _{\lambda } \Vert ). \end{aligned}
(5.40)

This concludes the proof of the first of Eq. (5.28).

$$\underline{\hbox {Derivative of } \langle \xi _{\lambda }, \widetilde{\mathbb {Q}}_{1} \xi _{\lambda } \rangle }$$. We compute:

\begin{aligned} \frac{d}{d\lambda } \langle \xi _{\lambda }, \widetilde{\mathbb {Q}}_{1} \xi _{\lambda } \rangle = -\langle \xi _{\lambda }, [ \widetilde{\mathbb {Q}}_{1}, B] \xi _{\lambda }\rangle + \text {c.c.,} \end{aligned}
(5.41)

with

\begin{aligned}{}[ \widetilde{\mathbb {Q}}_{1}, B ]= & {} \frac{1}{2} \sum _{\sigma \ne \sigma '} \int dxdydzdz'\, V(x-y) \varphi (z-z') \nonumber \\&[a^{*}_{\sigma }(u_{x})a^{*}_{\sigma '}(u_{y}) a_{\sigma '}(u_{y})a_{\sigma }(u_{x}), a_{\uparrow }(u^{\text {r}}_{z}) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'})].\qquad \end{aligned}
(5.42)

We rewrite the commutator as:

\begin{aligned}&[ a^{*}_{\sigma }(u_{x})a^{*}_{\sigma '}(u_{y}) a_{\sigma '}(u_{y})a_{\sigma }(u_{x}), a_{\uparrow }(u^{\text {r}}_{z}) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'})] \nonumber \\&\qquad = -[ a^{*}_{\sigma }(u_{x})a^{*}_{\sigma '}(u_{y}), a_{\uparrow }(u^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) ] a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) a_{\sigma '}(u_{y})a_{\sigma }(u_{x}).\nonumber \\ \end{aligned}
(5.43)

where

\begin{aligned}&\big [ a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma '}(u_{y}) , a_{\uparrow }(u^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) \big ]\nonumber \\&\quad = a^{*}_{\sigma }(u_{x}) \big ( \delta _{\sigma '\uparrow } u^{\text {r}}_{\sigma '}(z;y) a_{\downarrow }(u^{\text {r}}_{z'}) \nonumber \\&\qquad - \delta _{\sigma '\downarrow } u^{\text {r}}_{\sigma '}(z';y) a_{\uparrow }(u^{\text {r}}_{z}) \big )\nonumber \\&\qquad - a^{*}_{\sigma '}(u_{y})\big ( \delta _{\sigma \uparrow } u^{\text {r}}_{\sigma }(z;x) a_{\downarrow }(u^{\text {r}}_{z'}) - \delta _{\sigma \downarrow } u^{\text {r}}_{\sigma }(z';x) a_{\uparrow }(u^{\text {r}}_{z}) \big )\nonumber \\&\qquad + \delta _{\sigma \uparrow } \delta _{\sigma '\downarrow } u^{\text {r}}_{\sigma }(z;x) u^{\text {r}}_{\sigma '}(z';y) - \delta _{\sigma \downarrow }\delta _{\sigma '\uparrow } u^{\text {r}}_{\sigma }(z';x) u^{\text {r}}_{\sigma '}(z;y). \end{aligned}
(5.44)

Consider the first term in the first line of Eq. (5.44). All the other terms in the first and second line of (5.44) give rise to contributions to (5.42) that can be estimated in exactly the same way. We have, recall the definition (5.8) of $$b_{\sigma }(\varphi _{z})$$:

\begin{aligned}&\int dxdydzdz'\, V(x-y) \varphi (z-z') u^{\text {r}}_{\uparrow }(z;y) \langle \xi _{\lambda }, a^{*}_{\downarrow }(u_{x}) a_{\downarrow }(u^{\text {r}}_{z'}) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) a_{\uparrow }(u_{y})a_{\downarrow }(u_{x})\xi _{\lambda }\rangle \nonumber \\&\quad \equiv -\int dxdydz\, V(x-y) u^{\text {r}}_{\uparrow }(z;y) \langle \xi _{\lambda }, a^{*}_{\downarrow }(u_{x}) b_{\downarrow }(\varphi _{z}) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\uparrow }(u_{y})a_{\downarrow }(u_{x}) \xi _{\lambda }\rangle =: \text {A}.\nonumber \\ \end{aligned}
(5.45)

We then estimate, by Lemma 5.3:

\begin{aligned} |\text {A}|\le & {} \int dxdydz\, V(x-y) |u^{\text {r}}_{\uparrow }(z;y)| \Vert b^{*}_{\downarrow }(\varphi _{z}) a_{\downarrow }(u_{x}) \xi _{\lambda }\Vert \Vert a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\uparrow }(u_{y})a_{\downarrow }(u_{x}) \xi _{\lambda }\Vert \nonumber \\\le & {} C\rho ^{\frac{1}{2}} \int dxdydz\, V(x-y) |u^{\text {r}}_{\uparrow }(z;y)| \Vert b_{\downarrow }(\varphi _{z}) a_{\downarrow }(u_{x}) \xi _{\lambda }\Vert \Vert a_{\uparrow }(u_{y})a_{\downarrow }(u_{x}) \xi _{\lambda }\Vert \nonumber \\&+ C\rho ^{1 - \frac{\gamma }{2}} \int dxdydz\, V(x-y) |u^{\text {r}}_{\uparrow }(z;y)| \Vert a_{\downarrow }(u_{x}) \xi _{\lambda }\Vert \Vert a_{\uparrow }(u_{y})a_{\downarrow }(u_{x}) \xi _{\lambda }\Vert \nonumber \\\equiv & {} \text {A}_{1} + \text {A}_{2}. \end{aligned}
(5.46)

To get the second inequality, we used the estimate (5.10), together with $$\Vert a_{\uparrow }(\overline{v}^{\text {r}}_{z}) \Vert \le C\rho ^{\frac{1}{2}}$$. Consider the term $$\text {A}_{2}$$. By Cauchy–Schwarz inequality, using that $$\Vert u^{\text {r}}_{y} \Vert _{1} \le C$$:

\begin{aligned} |\text {A}_{2}| \le C\rho ^{1 - \frac{\gamma }{2}} \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda } \Vert \Vert \widetilde{\mathbb {Q}}^{\frac{1}{2}}_{1} \xi _{\lambda } \Vert . \end{aligned}
(5.47)

Consider now the term $$\text {A}_{1}$$. Here, we split $$u_{x}$$ as $$u^{<}_{x} + u^{>}_{x}$$, with $$\hat{u}^{<}(k)$$ supported for $$|k| \le \rho ^{\eta }$$, with $$\eta < \min \{\gamma , \frac{1}{3}\}$$ to be chosen later. Correspondingly:

\begin{aligned} \text {A}_{1}\le & {} C\rho ^{\frac{1}{2}} \int dxdydz\, V(x-y) |u^{\text {r}}_{\uparrow }(z;y)| \Vert b_{\downarrow }(\varphi _{z}) a_{\downarrow }(u^{<}_{x}) \xi _{\lambda }\Vert \Vert a_{\uparrow }(u_{y})a_{\downarrow }(u_{x}) \xi _{\lambda }\Vert \nonumber \\&+ C\rho ^{\frac{1}{2}} \int dxdydz\, V(x-y) |u^{\text {r}}_{\uparrow }(z;y)| \Vert b_{\downarrow }(\varphi _{z}) a_{\downarrow }(u^{>}_{x}) \xi _{\lambda }\Vert \Vert a_{\uparrow }(u_{y})a_{\downarrow }(u_{x}) \xi _{\lambda }\Vert \nonumber \\\equiv & {} \text {A}_{1;1} + \text {A}_{1;2}. \end{aligned}
(5.48)

Consider $$\text {A}_{1;1}$$. Using that $$\Vert u^{<}_{x} \Vert _{2} \le C\rho ^{\frac{3\eta }{2}}$$, we get:

\begin{aligned} |\text {A}_{1;1}|\le & {} C\rho ^{\frac{1}{2} + \frac{3\eta }{2}} \int dxdydz\, V(x-y) |u^{\text {r}}_{\uparrow }(z;y)| \Vert b_{\downarrow }(\varphi _{z}) \xi _{\lambda }\Vert \Vert a_{\uparrow }(u_{y})a_{\downarrow }(u_{x}) \xi _{\lambda }\Vert \nonumber \\\le & {} C\rho ^{1 + \frac{3\eta }{2}} \int dxdydzdz'\, V(x-y) |u^{\text {r}}_{\uparrow }(z;y)| \varphi (z-z')\nonumber \\&\cdot \Vert a_{\downarrow }(u^{\text {r}}_{z}) \xi _{\lambda } \Vert \Vert a_{\uparrow }(u_{y})a_{\downarrow }(u_{x}) \xi _{\lambda }\Vert \nonumber \\\le & {} C \rho ^{1 + \frac{3\eta }{2} - 2\gamma } \Vert \mathcal {N}^{\frac{1}{2}}\xi _{\lambda } \Vert \Vert \widetilde{\mathbb {Q}}_{1}^{\frac{1}{2}} \xi _{\lambda } \Vert , \end{aligned}
(5.49)

where the last step follows from Cauchy–Schwarz inequality. Finally, consider $$\text {A}_{1;2}$$. We have:

\begin{aligned} |\text {A}_{1;2}|\le & {} C_{\beta }\rho ^{1 - 2\gamma } \int dxdydz\, V(x-y) |u^{\text {r}}_{\uparrow }(z;y)| \Vert a_{\downarrow }(u^{>}_{x}) \xi _{\lambda }\Vert \Vert a_{\uparrow }(u_{y})a_{\downarrow }(u_{x}) \xi _{\lambda }\Vert \nonumber \\\le & {} C_{\beta }\rho ^{1-2\gamma - \eta } \Vert \mathbb {H}_{0}^{\frac{1}{2}} \xi _{\lambda } \Vert \Vert \widetilde{\mathbb {Q}}^{\frac{1}{2}}_{1} \xi _{\lambda } \Vert , \end{aligned}
(5.50)

where the last step follows from Cauchy–Schwarz inequality, combined with:

\begin{aligned} \int dx\, \Vert a_{\downarrow }(u^{>}_{x}) \xi _{\lambda }\Vert ^{2} \le C\rho ^{-2\eta } \langle \xi _{\lambda }, \mathbb {H}_{0} \xi _{\lambda } \rangle . \end{aligned}
(5.51)

This inequality can be proven in a way completely analogous to (3.62). In fact:

\begin{aligned} \int dx\, \Vert a_{\sigma }(u^{>}_{x}) \xi _{\lambda }\Vert ^{2}= & {} \sum _{k} (\hat{u}_{\sigma }^{>}(k))^{2} \langle \xi _{\lambda }, \hat{a}^{*}_{k,\sigma } a_{k,\sigma } \xi _{\lambda }\rangle \nonumber \\\le & {} \rho ^{-2\eta } \sum _{k} |k|^{2} (\hat{u}_{\sigma }^{>}(k))^{2} \langle \xi _{\lambda }, \hat{a}^{*}_{k,\sigma } a_{k,\sigma } \xi _{\lambda }\rangle \nonumber \\\le & {} C\rho ^{-2\eta } \sum _{k} | |k|^{2} - \mu _{\sigma } | \langle \xi _{\lambda }, \hat{a}^{*}_{k,\sigma } a_{k,\sigma } \xi _{\lambda }\rangle \nonumber \\\le & {} C\rho ^{-2\eta } \langle \xi _{\lambda }, \mathbb {H}_{0} \xi _{\lambda }\rangle , \end{aligned}
(5.52)

where we used that $$\hat{u}_{\sigma }^{>}(k)$$ is supported for $$|k| \ge \rho ^{\eta }$$, that $$0\le \hat{u}_{\sigma }^{>}(k) \le 1$$, and that $$\mu _{\sigma } \le C\rho ^{\frac{2}{3}} \ll C\rho ^{2\eta }$$. Hence:

\begin{aligned} |\text {A}| \le C_{\beta }\rho ^{1 - 2\gamma - \eta } \Vert \widetilde{\mathbb {Q}}^{\frac{1}{2}}_{1} \xi _{\lambda } \Vert ( \Vert \mathbb {H}_{0}^{\frac{1}{2}} \xi _{\lambda } \Vert + \rho ^{\frac{5\eta }{2}} \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda } \Vert ). \end{aligned}
(5.53)

The other three terms arising from the first two lines of the right-hand side of (5.44) can be estimated in exactly the same way.

Next, let us plug the last two terms in the right-hand side of Eq. (5.44) in Eq. (5.42). We get:

\begin{aligned} \text {I}_{\text {main}} = -\int dxdydzdz'\, V(x-y) \varphi (z-z') u^{\text {r}}_{\uparrow }(z;x) u^{\text {r}}_{\downarrow }(z';y) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\uparrow }(u_{x}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) a_{\downarrow }(u_{y}). \end{aligned}
(5.54)

We shall use that $$u^{\text {r}}_{\uparrow }(z;x)$$ behaves as a delta function, to leading order in $$\rho$$. More precisely, we write:

\begin{aligned} u_{\sigma }^{\text {r}}(z;x) = \delta _{\sigma }^{\text {r}}(z;x) - \nu _{\sigma }(z;x), \end{aligned}
(5.55)

where $$\nu _{\sigma }(z;x)\equiv \nu _{\sigma }(z-x)$$ with Fourier transform given by:

\begin{aligned} \hat{\nu }_{\sigma }(k) = \left\{ \begin{array}{l@{\quad }l} 1 &{} \text {for } 0\le |k| \le k^{\sigma }_{F} \\ 0 &{} \text {for } |k| > 2k_{F}^{\sigma }, \end{array} \right. \end{aligned}
(5.56)

and it smoothly interpolates between 1 and 0 for $$k_{F}^{\sigma } \le |k| \le 2k_{F}^{\sigma }$$. The function $$\nu _{\sigma , x}(y)$$ satisfies the bounds:

\begin{aligned} \Vert \nu _{\sigma , x} \Vert _{1} \le C,\quad \Vert \nu _{\sigma , x} \Vert _{\infty } \le C\rho . \end{aligned}
(5.57)

Instead, $$\delta _{\sigma }^{\text {r}}(z;x) \equiv \delta ^{\text {r}}(z-x)$$, with:

\begin{aligned} \hat{\delta }_{\sigma }^{\text {r}}(k) = \left\{ \begin{array}{l@{\quad }l} 1 &{} \text {for } 0\le |k| \le \frac{3}{2}\rho ^{-\beta } \\ 0 &{} \text {for } |k| \ge 2\rho ^{-\beta } \\ \end{array} \right. \end{aligned}
(5.58)

and it smoothly interpolates between 1 and 0 in the region $$(3/2)\rho ^{-\beta } \le |k| \le 2\rho ^{-\beta }$$. The function $$\delta ^{\text {r}}_{\sigma ,x}(y)$$ is an approximate Dirac delta function at x, such that

\begin{aligned} \Vert \delta ^{\text {r}}_{\sigma ,x} \Vert _{1} \le C,\quad \Vert \delta ^{\text {r}}_{\sigma , x}\Vert _{\infty } \le C\rho ^{-3\beta }. \end{aligned}
(5.59)

Let us perform the replacement (5.55) in Eq. (5.54). Consider the terms with one $$\nu$$. We have:

\begin{aligned}&\int dxdydzdz'\, V(x-y) \varphi (z-z') |\delta _{\sigma }^{\text {r}}(z;x)| |\nu _{\sigma '}(z';y)|\nonumber \\&\qquad \cdot |\langle \xi _{\lambda }, a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\uparrow }(u_{x}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) a_{\downarrow }(u_{y}) \xi _{\lambda }\rangle |\nonumber \\&\quad \le C\int dxdy\, V(x-y) \rho ^{2-2\gamma } \Vert a_{\uparrow }(u_{x}) a_{\downarrow }(u_{y}) \xi _{\lambda }\Vert \nonumber \\&\quad \le CL^{\frac{3}{2}} \rho ^{2 - 2\gamma } \Vert \widetilde{\mathbb {Q}}^{\frac{1}{2}}_{1} \xi _{\lambda } \Vert . \end{aligned}
(5.60)

Next, consider the terms with two $$\nu$$. Proceeding as above we have:

\begin{aligned}&\int dxdydzdz'\, V(x-y) \varphi (z-z') |\nu _{\sigma }(z;x)| |\nu _{\sigma '}(z';y)|\nonumber \\&\qquad \cdot |\langle \xi _{\lambda }, a(\overline{v}^{\text {r}}_{z}) a_{\uparrow }(u_{x}) a(\overline{v}^{\text {r}}_{z'}) a_{\downarrow }(u_{y}) \xi _{\lambda }\rangle | \nonumber \\&\quad \le C\int dxdy\, V(x-y) \rho ^{2 - 2\gamma } \Vert a_{\uparrow }(u_{x}) a_{\downarrow }(u_{y}) \xi _{\lambda }\Vert \nonumber \\&\quad \le CL^{\frac{3}{2}} \rho ^{2 - 2\gamma } \Vert \widetilde{\mathbb {Q}}^{\frac{1}{2}}_{1} \xi _{\lambda } \Vert . \end{aligned}
(5.61)

Therefore, the main contribution to Eq. (5.54) is:

\begin{aligned} -\int dxdydzdz'\, V(x-y) \varphi (z-z') \delta ^{\text {r}}_{\uparrow }(z;x) \delta ^{\text {r}}_{\downarrow }(z';y) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\uparrow }(u_{x}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) a_{\downarrow }(u_{y}). \end{aligned}
(5.62)

The next lemma will allow us to replace the approximate $$\delta$$ functions with true $$\delta$$ functions, up to a small error. This is one of the points where we use the regularity of the potential.

### Lemma 5.6

Under the assumptions of Proposition 5.5, the following is true:

\begin{aligned}&-\int dxdydzdz'\, V(x-y) \varphi (z-z') \delta ^{\text {r}}_{\uparrow }(z;x) \delta ^{\text {r}}_{\downarrow }(z';y) \langle \xi _{\lambda }, a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\uparrow }(u_{x}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) a_{\downarrow }(u_{y}) \xi _{\lambda } \rangle \nonumber \\&\quad = -\int dxdy\, V(x-y) \varphi (x-y) \langle \xi _{\lambda }, a_{\uparrow }(\overline{v}^{\text {r}}_{x}) a_{\uparrow }(u_{x}) a_{\downarrow }(\overline{v}^{\text {r}}_{y}) a_{\downarrow }(u_{y}) \xi _{\lambda } \rangle + \widehat{\mathcal {E}}_{\widetilde{\mathbb {Q}}_{1}}(\xi _{\lambda }),\nonumber \\ \end{aligned}
(5.63)

where, for all $$n\ge 4$$, taking $$V\in C^{k}$$ with k large enough:

\begin{aligned} |\widehat{\mathcal {E}}_{\widetilde{\mathbb {Q}}_{1}}(\xi _{\lambda })| \le C_{n} \rho ^{\beta (n - 3)} ( CL^{3} \rho ^{2} + \langle \xi _{\lambda }, \widetilde{\mathbb {Q}}_{1} \xi _{\lambda }\rangle ). \end{aligned}
(5.64)

The proof of Lemma 5.6 is deferred to Appendix 7. Putting together (5.53), (5.60), (5.61), (5.63) and recalling the definition of $$\mathbb {T}_{2}$$, Eq. (5.29), the claim follows. The proof of the statement about $$\langle \xi _{\lambda },\mathbb {Q}_{1} \xi _{\lambda }\rangle$$ is exactly the same, and we shall omit the details. $$\square$$

### 5.3 Scattering Equation Cancellation

The next result will imply an important cancellation that will be used to propagate the a priori bounds for $$\mathbb {H}_{0}$$, $$\mathbb {Q}_{1}$$, and to compute the ground state energy at order $$\rho ^{2}$$.

### Proposition 5.7

(Scattering equation cancellation). Let:

\begin{aligned} \widetilde{\mathbb {Q}}^{\text {r}}_{4} = \int dxdy\, V(x-y) a^{*}_{\uparrow }(u^{\text {r}}_{x}) a^{*}_{\downarrow }(u^{\text {r}}_{y}) a^{*}_{\downarrow }(\overline{v}^{\text {r}}_{y}) a^{*}_{\uparrow }(\overline{v}^{\text {r}}_{x}) + \text {h.c..} \end{aligned}
(5.65)

Let $$\gamma \le 7/18$$. Under the assumptions of Theorem 2.1, the following holds:

\begin{aligned}&\langle \xi _{\lambda }, ( \mathbb {T}_{1} + \mathbb {T}_{2} + \widetilde{\mathbb {Q}}^{\text {r}}_{4}) \xi _{\lambda } \rangle \nonumber \\&\quad = \lambda _{\gamma } \int dxdy\,\theta (|x - y|_{L}< \rho ^{-\gamma }) (1-\varphi (x-y)) \nonumber \\&\qquad \cdot \langle \xi _{\lambda }, (a_{\uparrow }(\overline{v}^{\text {r}}_{x}) a_{\uparrow }(u^{\text {r}}_{x}) a_{\downarrow }(\overline{v}^{\text {r}}_{y}) a_{\downarrow }(u^{\text {r}}_{y}) + \text {h.c.}) \xi _{\lambda } \rangle + \mathcal {E}_{\mathbb {T}_{2}}(\xi _{\lambda }) \end{aligned}
(5.66)

with $$|\lambda _{\gamma }| \le C\rho ^{3\gamma }$$ and, for $$\delta > 0$$ and L large enough:

\begin{aligned} | \mathcal {E}_{\mathbb {T}_{2}}(\xi _{\lambda }) | \le CL^{\frac{3}{2}}\rho ^{\frac{3}{2}} \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda } \Vert . \end{aligned}
(5.67)

Moreover:

\begin{aligned}&\lambda _{\gamma } \int dxdy\,\theta (|x - y|_{L} < \rho ^{-\gamma }) (1-\varphi (x-y))\nonumber \\&\qquad \cdot \Big |\langle \xi _{\lambda }, (a_{\uparrow }(\overline{v}^{\text {r}}_{x}) a_{\uparrow }(u^{\text {r}}_{x}) a_{\downarrow }(\overline{v}^{\text {r}}_{y}) a_{\downarrow }(u^{\text {r}}_{y}) + \text {h.c.}) \xi _{\lambda } \rangle \Big | \nonumber \\&\quad \le C L^{\frac{3}{2}} \rho ^{1 + \frac{3\gamma }{2}} \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda } \Vert . \end{aligned}
(5.68)

### Proof

Let $$\mathbb {T}_{2}^{\text {r}}$$ be the operator obtained from $$\mathbb {T}_{2}$$ after replacing all u with $$u^{\text {r}}$$. The error term $$\mathcal {E}_{\mathbb {T}_{2}}(\xi _{\lambda })$$ takes into account the difference $$\langle \xi _{\lambda }, (\mathbb {T}_{2} - \mathbb {T}_{2}^{\text {r}}) \xi _{\lambda }\rangle$$. We postpone its estimate (5.67) to Appendix 7.

Recall the notation $$b_{x,\sigma } = a_{\sigma }(u^{\text {r}}_{x}) a_{\sigma }(\overline{v}^{\text {r}}_{x})$$, Eq. (5.8). We then have, using that by the compact support of the potential $$V(x) \equiv V(x) \theta (|x|_{L} < \rho ^{-\gamma })$$:

\begin{aligned}&\langle \xi _{\lambda }, ( \mathbb {T}_{1} + \mathbb {T}^{\text {r}}_{2} + \widetilde{\mathbb {Q}}^{\text {r}}_{4}) \xi _{\lambda } \rangle \nonumber \\&\quad =\int dxdy\, \theta (|x-y|_{L}<\rho ^{-\gamma })\big (2\Delta \varphi (x-y) + V(x-y)(1 - \varphi (x-y))\big ) \nonumber \\&\qquad \cdot \langle \xi _{\lambda }, b_{x,\uparrow } b_{y,\downarrow } \xi _{\lambda } \rangle + \text {c.c.}\nonumber \\&\quad \equiv 2\int dxdy\, \theta (|x-y|_{L}<\rho ^{-\gamma }) \Big (-\Delta f(x-y) + \frac{1}{2}V(x-y) f(x-y)\Big ) \nonumber \\&\qquad \cdot \langle \xi _{\lambda }, b_{x,\uparrow } b_{y,\downarrow } \xi _{\lambda } \rangle + \text {c.c.,} \end{aligned}
(5.69)

with $$f = 1 - \varphi$$. Recall that $$\varphi$$ is the periodization of $$\varphi _{\gamma }$$, the solution of the Neumann problem (4.28), over $$\Lambda _{L}$$: $$\varphi (x) = \sum _{n \in \mathbb {Z}^{3}} \varphi _{\gamma }(x + n_{1} e_{1} L + n_{2} e_{2} L + n_{3} e_{3} L)$$. The function $$\varphi _{\gamma }(x)$$ is compactly supported in $$\mathbb {R}^{3}$$, with support in $$B = \{ x\in \mathbb {R}^{3} \mid |x| \le \rho ^{-\gamma } \}$$. Thus, up to a boundary term we can replace f with $$f_{\gamma } = 1 - \varphi _{\gamma }$$:

\begin{aligned}&\int _{\Lambda _{L} \times \Lambda _{L}} dxdy\, \theta (|x-y|_{L}<\rho ^{-\gamma }) \Big (-\Delta f(x-y) + \frac{1}{2}V(x-y) f(x-y)\Big ) \nonumber \\&\qquad .\langle \xi _{\lambda }, b_{x,\uparrow } b_{y,\downarrow } \xi _{\lambda } \rangle \nonumber \\&\quad = \int _{\tilde{\Lambda }_{L} \times \tilde{\Lambda }_{L}} dxdy\, \theta (|x-y|_{L}<\rho ^{-\gamma }) \Big (-\Delta f_{\gamma }(x-y) + \frac{1}{2}V_{\infty }(x-y) f_{\gamma }(x-y)\Big ) \nonumber \\&\qquad .\langle \xi _{\lambda }, b_{x,\uparrow } b_{y,\downarrow } \xi _{\lambda } \rangle + \mathfrak {e}_{L} \end{aligned}
(5.70)

where $$\tilde{\Lambda }_{L} = [\rho ^{-\gamma }, L - \rho ^{-\gamma }]^{3}$$, and $$\mathfrak {e}_{L}$$ is a boundary term:

\begin{aligned} | \mathfrak {e}_{L} | \le CL^{2} \rho ^{-4\gamma } ( \Vert \theta (|\cdot |_{L} < \rho ^{-\gamma })\Delta \varphi \Vert _{\infty } + \Vert \varphi \Vert _{\infty } ) \rho ^{-3\beta + 1}. \end{aligned}
(5.71)

In Eq. (5.70), recall that V is the periodization of $$V_{\infty }$$, Eq. (2.2). Therefore, using that $$f_{\gamma }$$ solves the scattering equation in a ball of radius $$\rho ^{-\gamma }$$, Eq. (4.28), we easily get:

\begin{aligned}&\langle \xi _{\lambda }, ( \mathbb {T}_{1} + \mathbb {T}^{\text {r}}_{2} + \widetilde{\mathbb {Q}}^{\text {r}}_{4}) \xi _{\lambda } \rangle \nonumber \\&\quad = 2\lambda _{\gamma } \int _{\tilde{\Lambda }_{L} \times \tilde{\Lambda }_{L}} dxdy\, \theta (|x - y|_{L}< \rho ^{-\gamma })f_{\gamma }(x-y) \langle \xi _{\lambda }, (b_{x,\uparrow } b_{y,\downarrow } + \text {h.c.}) \xi _{\lambda } \rangle + \mathfrak {e}_{L} \nonumber \\&\quad = 2\lambda _{\gamma } \int _{\Lambda _{L} \times \Lambda _{L}} dxdy\, \theta (|x - y|_{L} < \rho ^{-\gamma })f(x-y) \langle \xi _{\lambda }, (b_{x,\uparrow } b_{y,\downarrow } + \text {h.c.})\xi _{\lambda } \rangle + \mathfrak {e}_{L},\nonumber \\ \end{aligned}
(5.72)

up to a redefinition of the boundary term (still satisfying (5.71)). To conclude, we estimate the integral using Lemma 5.3, setting $$g(x) = \lambda _{\gamma } \theta (|x|_{L} < \rho ^{-\gamma })f(x)$$. We shall use that:

\begin{aligned} \Vert g \Vert _{1} \le C,\quad \Vert g \Vert _{2} \le C\rho ^{\frac{3\gamma }{2}}. \end{aligned}
(5.73)

We get, thanks to Lemma 5.3:

\begin{aligned}&\Big |\int dxdy\, g(x-y) \langle \xi _{\lambda }, a_{\uparrow }(\overline{v}^{\text {r}}_{x}) a_{\uparrow }(u^{\text {r}}_{x}) a_{\downarrow }(\overline{v}^{\text {r}}_{y}) a_{\downarrow }(u^{\text {r}}_{y}) \xi _{\lambda } \rangle \Big | \nonumber \\&\quad \le \int dy\, \Vert b^{*}_{\uparrow }(g_{y}) \xi _{\lambda } \Vert \Vert b_{\downarrow ,y} \xi _{\lambda } \Vert \nonumber \\&\quad \le \int dy\, \Vert b_{\uparrow }(g_{y}) \xi _{\lambda } \Vert \Vert b_{\downarrow ,y} \xi _{\lambda } \Vert + C\rho \Vert g\Vert _{2} \int dy\, \Vert a_{\downarrow }(u_{y}) \xi _{\lambda }\Vert \nonumber \\&\quad \le C\rho \int dxdy\, |g(x-y)| \Vert a_{\uparrow }(u_{x}) \xi _{\lambda } \Vert \Vert a_{\downarrow }(u_{y}) \xi _{\lambda } \Vert + CL^{\frac{3}{2}} \rho ^{1 + \frac{3\gamma }{2}} \Vert \mathcal {N}^{\frac{1}{2}}\xi _{\lambda } \Vert .\nonumber \\ \end{aligned}
(5.74)

Thus, by Cauchy–Schwarz inequality, using that $$\Vert g \Vert _{1} \le C$$:

\begin{aligned}&\Big |\int dxdy\, g(x-y) \langle \xi _{\lambda }, a_{\uparrow }(\overline{v}^{\text {r}}_{x}) a_{\uparrow }(u^{\text {r}}_{x}) a_{\downarrow }(\overline{v}^{\text {r}}_{y}) a_{\downarrow }(u^{\text {r}}_{y}) \xi _{\lambda } \rangle \Big | \nonumber \\&\quad \le C\rho \langle \xi _{\lambda }, \mathcal {N}\xi _{\lambda } \rangle + CL^{\frac{3}{2}} \rho ^{1 + \frac{3\gamma }{2}} \Vert \mathcal {N}^{\frac{1}{2}}\xi _{\lambda } \Vert \nonumber \\&\quad \le CL^{\frac{3}{2}} \rho ^{1 + \frac{3\gamma }{2}} \Vert \mathcal {N}^{\frac{1}{2}}\xi _{\lambda } \Vert ,\nonumber \\ \end{aligned}
(5.75)

where in the last step we used the propagation of the a priori estimate for the number operator, and the assumption $$\gamma \le 7/18$$. This concludes the proof. $$\square$$

### 5.4 Propagation of the Estimates

We now have all the ingredients needed in order to propagate the a priori estimates for $$\mathbb {H}_{0}$$ and for $$\mathbb {Q}_{1}$$.

### Proposition 5.8

(Propagation estimate for $$\mathbb {H}_{0}$$, $$\mathbb {Q}_{1}$$ - Part 2.). Let $$5/18\le \gamma \le 1/3$$. Under the assumptions of Theorem 2.1 the following is true. For all $$\lambda \in [0,1]$$:

\begin{aligned} \langle \xi _{\lambda }, \mathbb {H}_{0}\xi _{\lambda }\rangle \le CL^{3} \rho ^{2},\quad \langle \xi _{\lambda }, \mathbb {Q}_{1} \xi _{\lambda } \rangle \le CL^{3} \rho ^{2},\quad \langle \xi _{\lambda }, \widetilde{\mathbb {Q}}_{1} \xi _{\lambda } \rangle \le CL^{3} \rho ^{2}. \end{aligned}
(5.76)

### Proof

The last bound immediately follows from the second, using that $$\widetilde{\mathbb {Q}}_{1} \le \mathbb {Q}_{1}$$. Let us prove the first two bounds. Using Eqs. (5.28), we get:

\begin{aligned} \frac{d}{d\lambda } \langle \xi _{\lambda }, (\mathbb {H}_{0} + \mathbb {Q}_{1}) \xi _{\lambda } \rangle= & {} -\langle \xi _{\lambda }, (\mathbb {T}_{1} + \mathbb {T}_{2}) \xi _{\lambda }\rangle + \mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda }) + \mathcal {E}_{\mathbb {Q}_{1}}(\xi _{\lambda })\nonumber \\= & {} -\langle \xi _{\lambda }, (\mathbb {T}_{1} + \mathbb {T}_{2} + \widetilde{\mathbb {Q}}^{\text {r}}_{4}) \xi _{\lambda }\rangle + \langle \xi _{\lambda }, \widetilde{\mathbb {Q}}^{\text {r}}_{4} \xi _{\lambda }\rangle + \mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda }) + \mathcal {E}_{\mathbb {Q}_{1}}(\xi _{\lambda }).\nonumber \\ \end{aligned}
(5.77)

As proven in Appendix 7:

\begin{aligned} | \langle \xi _{\lambda }, (\widetilde{\mathbb {Q}}_{4} - \widetilde{\mathbb {Q}}_{4}^{\text {r}}) \xi _{\lambda }\rangle | \le \frac{C}{\delta }\rho ^{2 + \frac{\epsilon }{3}} + \delta \langle \xi _{\lambda }, \widetilde{\mathbb {Q}}_{1} \xi _{\lambda }\rangle . \end{aligned}
(5.78)

Also, thanks to (3.44):

\begin{aligned} \pm \widetilde{\mathbb {Q}}_{4} \le \delta \widetilde{\mathbb {Q}}_{1} + (C/\delta ) L^{3} \rho ^{2}. \end{aligned}
(5.79)

Therefore, using that $$\widetilde{\mathbb {Q}}_{1} \le \mathbb {Q}_{1}$$:

\begin{aligned} \frac{d}{d\lambda } \langle \xi _{\lambda }, ( \mathbb {H}_{0} + \mathbb {Q}_{1} ) \xi _{\lambda }\rangle\le & {} -\langle \xi _{\lambda }, (\mathbb {T}_{1} + \mathbb {T}_{2} + \widetilde{\mathbb {Q}}^{\text {r}}_{4}) \xi _{\lambda }\rangle + C\langle \xi _{\lambda }, \mathbb {Q}_{1} \xi _{\lambda } \rangle \nonumber \\&+ CL^{3} \rho ^{2} + \mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda }) + \mathcal {E}_{\mathbb {Q}_{1}}(\xi _{\lambda }). \end{aligned}
(5.80)

To estimate the various terms, we shall use the bounds of Propositions 5.5, 5.7. We have, for $$5/18\le \gamma \le 1/3$$, from Eqs. (5.67), (5.68):

\begin{aligned} |\langle \xi _{\lambda }, (\mathbb {T}_{1} + \mathbb {T}_{2} + \widetilde{\mathbb {Q}}^{\text {r}}_{4}) \xi _{\lambda }\rangle |\le & {} CL^{\frac{3}{2}}\rho ^{1 + \frac{3\gamma }{2}} \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda } \Vert \nonumber \\\le & {} CL^{3}\rho ^{2}, \end{aligned}
(5.81)

where we used the propagation of the a priori estimate for number operator, Eq. (5.6). The bound (5.81) is the only point where we need the lower bound on $$\gamma$$.

Consider now $$\mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda })$$. We have, from the first bound in (5.30):

\begin{aligned} \mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda })\le & {} CL^{\frac{3}{2}} \rho ^{\frac{7}{6}} \Vert \mathbb {H}_{0}^{\frac{1}{2}} \xi _{\lambda } \Vert + CL^{\frac{3}{2}} \rho ^{\frac{3}{2}} \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda }\Vert \nonumber \\\le & {} C\langle \xi _{\lambda }, \mathbb {H}_{0} \xi _{\lambda } \rangle + CL^{3} \rho ^{2}, \end{aligned}
(5.82)

where we used again the propagation of the apriori estimate for number operator (5.7). Also, from the third of (5.30), it is not difficult to see that, for $$\frac{4}{3}\gamma - \frac{7}{18}< \eta < \gamma$$ (which is a nonempty set for $$\eta$$, since $$\gamma \le 1/3$$), and for $$\beta$$ small enough:

\begin{aligned} \mathcal {E}_{\mathbb {Q}_{1}}(\xi _{\lambda }) \le C \langle \xi _{\lambda }, (\mathbb {H}_{0} + \mathbb {Q}_{1}) \xi _{\lambda }\rangle + CL^{3} \rho ^{2}. \end{aligned}
(5.83)

The final claim follows from Eqs. (5.80)–(5.83), together with Grönwall lemma. $$\square$$

Let us rewrite the bounds of Propositions 5.5, 5.7, using the propagation of the a priori estimates (5.76). We have, for $$5/18\le \gamma \le 1/3$$ and $$\eta < \gamma$$:

\begin{aligned} | \mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda }) |\le & {} CL^{3} \rho ^{\frac{7}{3} - \frac{\gamma }{2}} + CL^{\frac{3}{2}} \rho ^{\frac{5}{3} - \frac{\gamma }{2}} \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda }\Vert \nonumber \\ |\mathcal {E}_{\widetilde{\mathbb {Q}}_{1}}(\xi _{\lambda })|\le & {} C_{\beta } L^{3} \rho ^{3 - 2\gamma - \eta } + C_{\beta } L^{\frac{3}{2}} \rho ^{2 + \frac{3\eta }{2} - 2\gamma } \Vert \mathcal {N}^{\frac{1}{2}}\xi _{\lambda } \Vert \nonumber \\ |\mathcal {E}_{\mathbb {Q}_{1}}(\xi _{\lambda })|\le & {} C_{\beta } L^{3} \rho ^{3 - 2\gamma - \eta } + C_{\beta } L^{\frac{3}{2}} \rho ^{2 + \frac{3\eta }{2} - 2\gamma } \Vert \mathcal {N}^{\frac{1}{2}}\xi _{\lambda } \Vert , \end{aligned}
(5.84)

and:

\begin{aligned} |\langle \xi _{\lambda }, ( \mathbb {T}_{1} + \mathbb {T}_{2} + \widetilde{\mathbb {Q}}^{\text {r}}_{4}) \xi _{\lambda } \rangle | \le C L^{\frac{3}{2}} \rho ^{1 + \frac{3\gamma }{2}} \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda } \Vert . \end{aligned}
(5.85)

The reason why we kept the dependence on the number operator is that the estimate on $$\Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda }\Vert$$ obtained propagating the a priori bound for $$\mathcal {N}$$ on $$\xi _{0}$$ is not optimal, in contrast to the estimates for $$\mathbb {H}_{0}$$ and $$\mathbb {Q}_{1}$$. We shall conclude the section by proving an improved version of the bound for the number operator.

### Proposition 5.9

(Improved a priori estimate for the number operator). Let $$\gamma \le 1/2$$. We have:

\begin{aligned} \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle \le C L^{\frac{3}{2}} \rho ^{\frac{1}{6}} \Vert \mathbb {H}_{0}^{\frac{1}{2}} \xi _{1} \Vert + CL^{3} \rho ^{2 - \gamma }. \end{aligned}
(5.86)

### Proof

Lemma 5.1, together with Gronwall lemma, immediately implies:

\begin{aligned} \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle \le \langle \xi _{1}, \mathcal {N} \xi _{1} \rangle + CL^{3} \rho ^{2 - \gamma }. \end{aligned}
(5.87)

To estimate $$\langle \xi _{1}, \mathcal {N} \xi _{1} \rangle$$ in terms of the kinetic energy, we use Lemma 3.6:

\begin{aligned} \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle \le CL^{3} \rho ^{\frac{1}{3} + \alpha } + \frac{1}{\rho ^{\alpha }}\langle \xi _{1}, \mathbb {H}_{0} \xi _{1} \rangle + CL^{3} \rho ^{2 - \gamma }. \end{aligned}
(5.88)

We choose $$\alpha$$ such that:

\begin{aligned} \rho ^{\frac{1}{3} + \alpha } = \max \Big \{ \frac{1}{L^{3} \rho ^{\alpha }} \langle \xi _{1}, \mathbb {H}_{0}\xi _{1} \rangle , \rho ^{2 - \gamma } \Big \}. \end{aligned}
(5.89)

By doing so, we get:

\begin{aligned} \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle \le C L^{\frac{3}{2}} \rho ^{\frac{1}{6}} \Vert \mathbb {H}_{0}^{\frac{1}{2}} \xi _{1} \Vert + CL^{3} \rho ^{2 - \gamma }. \end{aligned}
(5.90)

$$\square$$

Improved bounds for the error terms. To conclude the section, let us reexpress the bounds (5.84), (5.85) in view of the improved estimate (5.86). Take $$5/18\le \gamma \le 1/3$$, in order to be able to use the propagation of the a priori estimates (5.8). The bound (5.86) implies:

\begin{aligned} \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda } \Vert \le C L^{\frac{3}{4}} \rho ^{\frac{1}{12}} \Vert \mathbb {H}_{0}^{\frac{1}{2}} \xi _{1} \Vert ^{\frac{1}{2}} + C L^{\frac{3}{2}} \rho ^{1 - \frac{\gamma }{2}}. \end{aligned}
(5.91)

Plugging this bound in the first of (5.84), we get, for $$0<\delta < 1$$:

\begin{aligned} | \mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda }) |\le & {} CL^{3} \rho ^{\frac{7}{3} - \frac{\gamma }{2}} + CL^{\frac{9}{4}} \rho ^{\frac{7}{4} - \frac{\gamma }{2}} \Vert \mathbb {H}_{0}^{\frac{1}{2}} \xi _{1} \Vert ^{\frac{1}{2}}\nonumber \\\le & {} C_{\delta }L^{3} \rho ^{\frac{7}{3} - \frac{2\gamma }{3}} + \delta \langle \xi _{1}, \mathbb {H}_{0} \xi _{1}\rangle , \end{aligned}
(5.92)

where in the last step we used Young’s inequality $$|ab| \le C_{pq} (|a|^{p} + |b|^{q})$$ with $$1/p + 1/q = 1$$, with $$p=4$$ and $$q=4/3$$. Similarly, using the bound (5.91) in the second of (5.84):

\begin{aligned} |\mathcal {E}_{\widetilde{\mathbb {Q}}_{1}}(\xi _{\lambda })|\le & {} C_{\beta } L^{3} \rho ^{3 - 2\gamma - \eta } + C_{\beta } L^{3} \rho ^{3 + \frac{3\eta }{2} - \frac{5\gamma }{2}} + C_{\beta } L^{\frac{9}{4}} \rho ^{\frac{25}{12} + \frac{3\eta }{2} - 2\gamma } \Vert \mathbb {H}_{0}^{\frac{1}{2}} \xi _{1}\Vert ^{\frac{1}{2}}\nonumber \\\le & {} C_{\beta } L^{3} \rho ^{3 - 2\gamma - \eta } + C_{\beta } L^{3} \rho ^{3 + \frac{3\eta }{2} - \frac{5\gamma }{2}} + C_{\beta ,\delta } L^{3} \rho ^{\frac{25}{9} + 2\eta - \frac{8}{3}\gamma } + \delta \langle \xi _{1}, \mathbb {H}_{0} \xi _{1}\rangle .\nonumber \\ \end{aligned}
(5.93)

For $$\eta \le \frac{4}{9} + \frac{\gamma }{3}$$, which is implied by $$\eta < \gamma$$ and $$\gamma \le 1/3$$:

\begin{aligned} |\mathcal {E}_{\widetilde{\mathbb {Q}}_{1}}(\xi _{\lambda })|\le & {} C_{\beta } L^{3} \rho ^{3 - 2\gamma - \eta } + C_{\beta ,\delta } L^{3} \rho ^{\frac{25}{9} + 2\eta - \frac{8}{3}\gamma } + \delta \langle \xi _{1}, \mathbb {H}_{0} \xi _{1}\rangle \nonumber \\\le & {} C_{\beta } L^{3} \rho ^{\frac{79}{27} - \frac{20}{9}\gamma } + \delta \langle \xi _{1}, \mathbb {H}_{0} \xi _{1}\rangle , \end{aligned}
(5.94)

where in the last step we optimized over $$\eta$$, $$\eta = \frac{2}{27} + \frac{2}{9}\gamma$$, which is strictly less than $$\frac{4}{9} + \frac{\gamma }{3}$$. The same bound holds for $$|\mathcal {E}_{\mathbb {Q}_{1}}(\xi _{\lambda })|$$. Finally, for $$\gamma \le 1/3$$, plugging the bound (5.91) in (5.85):

\begin{aligned} |\langle \xi _{\lambda }, ( \mathbb {T}_{1} + \mathbb {T}_{2} + \widetilde{\mathbb {Q}}^{\text {r}}_{4}) \xi _{\lambda } \rangle |\le & {} CL^{3} \rho ^{2 + \gamma } + CL^{\frac{9}{4}} \rho ^{\frac{13}{12} + \frac{3\gamma }{2}} \Vert \mathbb {H}_{0}^{\frac{1}{2}} \xi _{1}\Vert ^{\frac{1}{2}} \nonumber \\\le & {} C_{\delta }L^{3} \rho ^{\frac{13}{9} + 2\gamma } + \delta \langle \xi _{1}, \mathbb {H}_{0} \xi _{1}\rangle . \end{aligned}
(5.95)

These bounds will play an important role in the proof of the lower bound for the ground state energy, discussed in the next section.

## 6 Lower Bound on the Ground State Energy

In this section, we shall prove the lower bound for the ground state energy of the dilute Fermi gas. In what follows, $$\psi$$ will be approximate ground state, in the sense of Definition 3.4. In the following we shall always assume that $$5/18\le \gamma \le 1/3$$, which is the range of values of $$\gamma$$ for which the estimates (5.91)–(5.95) hold.

The starting point is, recall Proposition 3.1 and the bounds of Proposition 3.3:

\begin{aligned} \langle \psi , \mathcal {H} \psi \rangle \ge E_{\text {HF}}(\omega ) + \langle \xi _{0}, (\mathbb {H}_{0} + \widetilde{\mathbb {Q}}_{1} + \widetilde{\mathbb {Q}}_{4}) \xi _{0} \rangle + \mathcal {E}_{\text {1}}(\psi ), \end{aligned}
(6.1)

where the error term $$\mathcal {E}_{1}(\psi )$$ is bounded as:

\begin{aligned} | \mathcal {E}_{1}(\psi ) | \le C\rho ^{\alpha } \langle \xi _{0}, \widetilde{\mathbb {Q}}_{1} \xi _{0}\rangle + C\rho ^{1 - \alpha } \langle \xi _{0}, \mathcal {N} \xi _{0} \rangle . \end{aligned}
(6.2)

From the estimate (5.86) for the number operator, together with the a priori estimate (3.66) for $$\widetilde{\mathbb {Q}}_{1}$$, we get, optimizing over $$\alpha$$, $$\alpha = 1/9$$:

\begin{aligned} | \mathcal {E}_{1}(\psi ) | \le C_{\delta }L^{3} \rho ^{2 + \frac{1}{9}} + \delta \langle \xi _{1}, \mathbb {H}_{0} \xi _{1} \rangle . \end{aligned}
(6.3)

To extract the correlation energy at order $$\rho ^{2}$$, we shall use an interpolation argument. We write:

\begin{aligned} \langle \xi _{0}, (\mathbb {H}_{0} + \widetilde{\mathbb {Q}}_{1} + \widetilde{\mathbb {Q}}_{4}) \xi _{0} \rangle= & {} \langle \xi _{1}, (\mathbb {H}_{0} + \widetilde{\mathbb {Q}}_{1}) \xi _{1} \rangle - \int _{0}^{1} d\lambda \, \frac{d}{d\lambda } \langle \xi _{\lambda }, (\mathbb {H}_{0} + \widetilde{\mathbb {Q}}_{1}) \xi _{\lambda }\rangle \nonumber \\&+ \langle \xi _{0}, \widetilde{\mathbb {Q}}_{4} \xi _{0} \rangle \nonumber \\= & {} \langle \xi _{1}, (\mathbb {H}_{0} + \widetilde{\mathbb {Q}}_{1}) \xi _{1} \rangle + \int _{0}^{1} d\lambda \, \langle \xi _{\lambda }, (\mathbb {T}_{1} + \mathbb {T}_{2}) \xi _{\lambda } \rangle \nonumber \\&+ \langle \xi _{0}, \widetilde{\mathbb {Q}}_{4} \xi _{0} \rangle + \mathcal {E}_{2}(\psi ), \end{aligned}
(6.4)

where the $$\mathbb {T}_{1}, \mathbb {T}_{2}$$ operators are defined in (5.29) and, thanks to the bounds (5.92), (5.94):

\begin{aligned} |\mathcal {E}_{2}(\psi )|\le & {} \max _{\lambda \in [0;1]} |\mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda })| + \max _{\lambda \in [0;1]} |\mathcal {E}_{\widetilde{\mathbb {Q}}_{1}}(\xi _{\lambda })| \nonumber \\\le & {} C_{\delta }\rho ^{\frac{7}{3} - \frac{2\gamma }{3}} + \delta \langle \xi _{1}, \mathbb {H}_{0}\xi _{1}\rangle , \end{aligned}
(6.5)

where we used the condition $$\gamma \le 1/3$$. Next, in order to make use of the cancellation due to the scattering equation, we rewrite:

\begin{aligned} \langle \xi _{0}, (\mathbb {H}_{0} + \widetilde{\mathbb {Q}}_{1} + \widetilde{\mathbb {Q}}_{4}) \xi _{0} \rangle= & {} \langle \xi _{1}, (\mathbb {H}_{0} + \widetilde{\mathbb {Q}}_{1}) \xi _{1} \rangle + \int _{0}^{1} d\lambda \, \langle \xi _{\lambda }, (\mathbb {T}_{1} + \mathbb {T}_{2} + \widetilde{\mathbb {Q}}_{4}^{\text {r}}) \xi _{\lambda } \rangle \nonumber \\&- \int _{0}^{1} d\lambda \, \langle \xi _{\lambda }, \widetilde{\mathbb {Q}}_{4}^{\text {r}} \xi _{\lambda } \rangle + \langle \xi _{0}, \widetilde{\mathbb {Q}}_{4} \xi _{0} \rangle + \mathcal {E}_{2}(\psi ),\nonumber \\\equiv & {} \langle \xi _{1}, (\mathbb {H}_{0} + \widetilde{\mathbb {Q}}_{1}) \xi _{1} \rangle - \int _{0}^{1} d\lambda \, \langle \xi _{\lambda }, \widetilde{\mathbb {Q}}_{4}^{\text {r}} \xi _{\lambda } \rangle + \langle \xi _{0}, \widetilde{\mathbb {Q}}_{4} \xi _{0} \rangle \nonumber \\&+ \mathcal {E}_{3}(\psi ) + \mathcal {E}_{2}(\psi ), \end{aligned}
(6.6)

where, using (5.95):

\begin{aligned} |\mathcal {E}_{3}(\psi )|\le & {} \max _{\lambda \in [0;1]} |\langle \xi _{\lambda }, (\mathbb {T}_{1} + \mathbb {T}_{2} + \widetilde{\mathbb {Q}}_{4}^{\text {r}}) \xi _{\lambda } \rangle |\nonumber \\\le & {} C_{\delta }L^{3} \rho ^{\frac{13}{9} + 2\gamma } + \delta \langle \xi _{1}, \mathbb {H}_{0} \xi _{1}\rangle . \end{aligned}
(6.7)

Let us now consider the second and third term in Eq. (6.6). We rewrite it as:

\begin{aligned} - \int _{0}^{1} d\lambda \, \langle \xi _{\lambda }, \widetilde{\mathbb {Q}}_{4}^{\text {r}} \xi _{\lambda } \rangle + \langle \xi _{0}, \widetilde{\mathbb {Q}}_{4} \xi _{0} \rangle= & {} \langle \xi _{1},(\widetilde{\mathbb {Q}}_{4} - \widetilde{\mathbb {Q}}^{\text {r}}_{4}) \xi _{1} \rangle - \int _{0}^{1}d\lambda \, \frac{d}{d\lambda } \langle \xi _{\lambda } \widetilde{\mathbb {Q}}_{4} \xi _{\lambda }\rangle \nonumber \\&+ \int _{0}^{1} d\lambda \int _{\lambda }^{1} d\lambda ' \frac{d}{d\lambda '} \langle \xi _{\lambda '}, \widetilde{\mathbb {Q}}_{4}^{\text {r}} \xi _{\lambda '} \rangle \nonumber \\\equiv & {} \mathcal {E}_{4}(\psi ) - \int _{0}^{1}d\lambda \, \frac{d}{d\lambda } \langle \xi _{\lambda } \widetilde{\mathbb {Q}}_{4} \xi _{\lambda }\rangle \nonumber \\&+ \int _{0}^{1} d\lambda \int _{\lambda }^{1} d\lambda ' \frac{d}{d\lambda '} \langle \xi _{\lambda '}, \widetilde{\mathbb {Q}}_{4}^{\text {r}} \xi _{\lambda '} \rangle , \end{aligned}
(6.8)

where, as proven in Appendix 7:

\begin{aligned} | \mathcal {E}_{4}(\psi ) |= & {} |\langle \xi _{1},(\widetilde{\mathbb {Q}}_{4} - \widetilde{\mathbb {Q}}^{\text {r}}_{4}) \xi _{1} \rangle | \nonumber \\\le & {} \frac{C}{\delta } L^{3} \rho ^{2 + \frac{\epsilon }{3}} +\delta \langle \xi _{1}, \widetilde{\mathbb {Q}}_{1} \xi _{1}\rangle +CL^{\frac{3}{2}}\rho ^{\frac{3}{2}}\Vert \mathcal {N}^{\frac{1}{2}}\xi _1\Vert \nonumber \\\le & {} \frac{C}{\delta } L^{3} \rho ^{2 + \frac{\epsilon }{3}} +\delta \langle \xi _{1}, \widetilde{\mathbb {Q}}_{1} \xi _{1}\rangle +C_\delta L^{{3}}\rho ^{{2}+\frac{1}{9}}+\delta \langle \xi _1,\mathbb {H}_0\xi _1\rangle , \end{aligned}
(6.9)

where in the last step we proceeded as for (6.7) with $$\gamma =1/3,$$ recall (5.85), (5.95). The correlation energy at order $$\rho ^{2}$$ arises from the last two terms in Eq. (6.8). We will prove that:

\begin{aligned} \frac{d}{d\lambda }\langle \xi _{\lambda }, \widetilde{\mathbb {Q}}_{4} \xi _{\lambda } \rangle= & {} 2 L^{3}\rho _{\uparrow } \rho _{\downarrow } \int dx\, V(x) \varphi (x) + \text {subleading terms.} \nonumber \\ \frac{d}{d\lambda }\langle \xi _{\lambda }, \widetilde{\mathbb {Q}}^{\text {r}}_{4} \xi _{\lambda } \rangle= & {} 2 L^{3}\rho _{\uparrow } \rho _{\downarrow } \int dx\, V(x) \varphi (x) + \text {subleading terms.} \end{aligned}
(6.10)

The explicit terms are precisely what we need to compute the ground state energy at order $$\rho ^{2}$$. The quantitative version of the statement (6.10) is the content of the next proposition.

### Proposition 6.1

(Extracting the correlation energy). Under the assumptions of Theorem 2.1 the following holds. Let $$\psi$$ be an approximate ground state. Take $$\frac{5}{18} \le \gamma \le \frac{1}{3}$$, $$\epsilon \ge 0$$ and $$\frac{1}{6} + \frac{\epsilon }{12} < \gamma$$. Then:

\begin{aligned} \frac{d}{d\lambda }\langle \xi _{\lambda }, \widetilde{\mathbb {Q}}^{\text {r}}_{4} \xi _{\lambda } \rangle= & {} 2\rho ^{\text {r}}_{\uparrow }\rho ^{\text {r}}_{\downarrow } \int dxdy\, V(x-y) \varphi (x-y) + \mathcal {E}_{\widetilde{\mathbb {Q}}^{\text {r}}_{4}}(\xi _{\lambda }) \nonumber \\ \frac{d}{d\lambda }\langle \xi _{\lambda }, \widetilde{\mathbb {Q}}_{4} \xi _{\lambda } \rangle= & {} 2\rho ^{\text {r}}_{\uparrow }\rho ^{\text {r}}_{\downarrow } \int dxdy\, V(x-y) \varphi (x-y) + \mathcal {E}_{\widetilde{\mathbb {Q}}_{4}}(\xi _{\lambda }), \end{aligned}
(6.11)

where, for any $$0<\delta < 1$$:

\begin{aligned} |\mathcal {E}_{\widetilde{\mathbb {Q}}^{\text {r}}_{4}}(\psi )|\le & {} CL^{3} \rho ^{\frac{7}{3}} + C_{\beta ,\alpha } L^{3}\rho ^{\frac{26}{9} - \frac{4}{3}\gamma - \frac{7}{18}\epsilon } + \delta \langle \xi _{1}, \mathbb {H}_{0} \xi _{1}\rangle \nonumber \\ |\mathcal {E}_{\widetilde{\mathbb {Q}}_{4}}(\psi )|\le & {} CL^{3} \rho ^{\frac{7}{3}} + C_{\beta ,\alpha } L^{3}\rho ^{\frac{26}{9} - \frac{4}{3}\gamma - \frac{7}{18}\epsilon } + \delta \langle \xi _{1}, \mathbb {H}_{0} \xi _{1}\rangle . \end{aligned}
(6.12)

### Remark 6.2

(Notations.). Unless otherwise stated, with a slight abuse of notation in the following we will use the notation $$u^{\text {r}}(x;y)$$ to denote the function $$(u^{\text {r}})^{2}(x;y)$$. The two functions satisfy the same estimates, recall Proposition 4.2.

### Proof

We shall only discuss the proof of the statement concerning $$\widetilde{\mathbb {Q}}^{\text {r}}_{4}$$, the one for $$\widetilde{\mathbb {Q}}_{4}$$ being completely analogous (it is actually simpler). We write:

\begin{aligned} \frac{d}{d\lambda } \langle \xi _{\lambda }, \widetilde{\mathbb {Q}}_{4}^{\text {r}} \xi _{\lambda }\rangle = \langle \xi _{\lambda }, [ \widetilde{\mathbb {Q}}^{\text {r}}_{4}, B] \xi _{\lambda }\rangle + \text {c.c.}, \end{aligned}
(6.13)

where

\begin{aligned}&[ \widetilde{\mathbb {Q}}^{\text {r}}_{4}, B] = \frac{1}{2} \sum _{\sigma \ne \sigma '} \int dxdydzdz'\, V(x-y)\varphi (z-z') \\&\quad [ a^{*}_{\sigma }(u^{\text {r}}_{x}) a^{*}_{\sigma '}(u^{\text {r}}_{y}) a^{*}_{\sigma '}(\overline{v}^{\text {r}}_{y}) a^{*}_{\sigma }(\overline{v}^{\text {r}}_{x}), a_{\uparrow }(u^{\text {r}}_{z}) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'})]. \end{aligned}

We rewrite the commutator as:

\begin{aligned}&\big [ a^{*}_{\sigma }(u^{\text {r}}_{x}) a^{*}_{\sigma '}(u^{\text {r}}_{y}) a^{*}_{\sigma '}(\overline{v}^{\text {r}}_{y}) a^{*}_{\sigma }(\overline{v}^{\text {r}}_{x}), a_{\uparrow }(u^{\text {r}}_{z}) a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) \big ] \nonumber \\&\quad = -a^{*}_{\sigma }(u^{\text {r}}_{x}) a^{*}_{\sigma '}(u^{\text {r}}_{y}) \big [ a^{*}_{\sigma '}(\overline{v}^{\text {r}}_{y}) a^{*}_{\sigma }(\overline{v}^{\text {r}}_{x}), a_{\uparrow }(\overline{v}^{\text {r}}_{z}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'})\big ] a_{\uparrow }(u^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'})\nonumber \\&\qquad - a_{\uparrow }(\overline{v}^{\text {r}}_{z})a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) \big [ a^{*}_{\sigma }(u^{\text {r}}_{x}) a^{*}_{\sigma '}(u^{\text {r}}_{y}) , a_{\uparrow }(u^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) \big ] a^{*}_{\sigma '}(\overline{v}^{\text {r}}_{y}) a^{*}_{\sigma }(\overline{v}^{\text {r}}_{x}).\nonumber \\ \end{aligned}
(6.14)

As it will be clear, the only terms contributing at the order $$\rho ^{2}$$ will be those antinormal ordered, which arise from the last line. The first contribution in the right-hand side of Eq. (6.14) is partially normal ordered; as such, it will give rise to an error term. The commutator produces contractions between the fermionic operators; we have, omitting the spin label for simplicity:

\begin{aligned} \big [ a^{*}(\overline{v}^{\text {r}}_{y}) a^{*}(\overline{v}^{\text {r}}_{x}), a(\overline{v}^{\text {r}}_{z}) a(\overline{v}^{\text {r}}_{z'})\big ]= & {} a^{*}(\overline{v}^{\text {r}}_{y}) \omega ^{\text {r}}(x;z) a(\overline{v}^{\text {r}}_{z'}) - a^{*}(\overline{v}^{\text {r}}_{y}) \omega ^{\text {r}}(x;z') a(\overline{v}^{\text {r}}_{z}) \nonumber \\&+ a(\overline{v}^{\text {r}}_{z'}) \omega ^{\text {r}}(y;z) a^{*}(\overline{v}^{\text {r}}_{x}) - a(\overline{v}^{\text {r}}_{z}) \omega ^{\text {r}}(y;z') a^{*}(\overline{v}^{\text {r}}_{x}).\nonumber \\ \end{aligned}
(6.15)

These four terms give rise to contributions to (6.14) that will be estimated in exactly the same way (the lack of normal ordering in the last two terms in Eq. (6.15) will not matter). For instance, consider:

\begin{aligned}&\text {I} := \sum _{\sigma \ne \sigma '}\int dxdydzdz'\, V(x-y) \varphi (z-z') \tilde{\omega }^{\text {r}}(z;y) \\&\quad \langle \xi _{\lambda }, a^{*}_{\sigma }(u^{\text {r}}_{x}) a^{*}_{\sigma '}(u^{\text {r}}_{y}) a^{*}_{\sigma }(\overline{v}^{\text {r}}_{x}) a_{\downarrow }(\overline{v}^{\text {r}}_{z'}) a_{\uparrow }(u^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) \xi _{\lambda }\rangle . \end{aligned}

To begin, we write:

\begin{aligned} \text {I} = \text {I}_{1} + \text {I}_{2}, \end{aligned}
(6.16)

where $$\text {I}_{1}$$ is obtained from $$\text {I}$$ replacing $$u^{\text {r}}_{x}$$, $$u^{\text {r}}_{y}$$ with $$u_{x}$$, $$u_{y}$$, and $$\text {I}_{2}$$ is an error term. Let us first consider the term $$\text {I}_{1}$$. To improve the estimate for this term, we shall study separately different contributions in momentum space. Let $$0\le \eta < \gamma \le 1/3$$, $$\delta > 1$$, to be chosen later. We write $$u^{\text {r}}_{z} = u^{<}_{z} + u^{0}_{z} + u^{>}_{z}$$, with:

\begin{aligned} \hat{u}^{\sharp }(k) = \hat{u}^{\text {r}}(k) \chi _{\sharp }(k),\quad \sharp =\, <\,, 0\,, >, \end{aligned}
(6.17)

where

\begin{aligned} \chi _{<}(k) = \chi \Big (\frac{|k|}{\rho ^{\eta }}\Big ),\quad \chi _{0}(k) = \chi \Big (\frac{|k|}{\rho ^{\frac{\eta }{\delta }}}\Big ) - \chi \Big (\frac{|k|}{\rho ^{\eta }}\Big ),\quad \chi _{>}(k) = 1 - \chi \Big (\frac{|k|}{\rho ^{\frac{\eta }{\delta }}}\Big ). \end{aligned}
(6.18)

Correspondingly, we write $$\text {I}_{1} = \text {I}_{1}^{<} + \text {I}_{1}^{0} + \text {I}_{1}^{>}$$. The key observation is that in each term $$\text {I}^{\sharp }_{1}$$ we can replace $$\varphi$$ by $$\varphi _{\sharp }$$, with $$\hat{\varphi }_{\sharp }$$ with similar support properties as $$u_{\sharp }$$. In fact, recalling that:

\begin{aligned} a_{\sigma }(u^{\text {r}}_{z}) = \sum _{k} \overline{u_{\sigma }^{\text {r}}(k)} \overline{f_{k}(z)} \hat{a}_{k,\sigma }, \end{aligned}
(6.19)

with $$\hat{a}_{k,\sigma } = a_{\sigma }(f_{k})$$ and $$f_{k}(z) = L^{-3/2} e^{-ik\cdot z}$$, we get:

\begin{aligned} \int dz\, \varphi (z-z') \tilde{\omega }^{\text {r}}(z;y) a_{\uparrow }(u^{\text {r}}_{z}) = \frac{1}{L^{\frac{3}{2}}}\sum _{k} \overline{\hat{u}_{\uparrow }^{\text {r}}(k)} \hat{a}_{k,\uparrow } \int dz\, e^{ik\cdot z} \varphi (z-z') \tilde{\omega }^{\text {r}}(z;y), \end{aligned}
(6.20)

where

\begin{aligned} \int dz\, e^{ik\cdot z} \varphi (z-z') \tilde{\omega }^{\text {r}}(z;y) = \frac{1}{L^{3}} \sum _{q} \hat{\varphi }(k+q) \hat{\omega }^{\text {r}}(q) e^{-iz'\cdot (k+q)} e^{-iy\cdot q}. \end{aligned}
(6.21)

Suppose that $$k\in \text {supp}\, \hat{u}^{\text {r}}_{\sharp }$$. Due to the fact that $$\hat{\omega }^{\text {r}}(q)$$ is supported for $$|q| \le C\rho ^{\frac{1}{3}}$$ and that $$\rho ^{\eta } \gg \rho ^{\frac{1}{3}}$$, we can freely replace $$\hat{\varphi }(k+q)$$ in Eq. (6.21) with:

\begin{aligned} \hat{\varphi }_{\sharp }(k+q) = \hat{\varphi }(k+q) \tilde{\chi }_{\sharp }(k+q), \end{aligned}
(6.22)

with $$\tilde{\chi }_{\sharp }$$ defined as:

\begin{aligned} \tilde{\chi }_{<}(p) = \chi \Big (\frac{|k|}{2\rho ^{\eta }}\Big ),\quad \tilde{\chi }_{0}(p) = \chi \Big (\frac{|k|}{2\rho ^{\frac{\eta }{\delta }}}\Big ) - \chi \Big (\frac{2|k|}{\rho ^{\eta }}\Big ),\quad \tilde{\chi }_{>}(p) = \chi (4\rho ^{\beta }|k|) - \chi \Big (\frac{2|k|}{\rho ^{\frac{\eta }{\delta }}}\Big ). \end{aligned}
(6.23)

As discussed in Appendix 7:

\begin{aligned} \Vert \varphi _{<} \Vert _{1} \le C\rho ^{-2\gamma } |\log \rho |,\quad \Vert \varphi _{0} \Vert _{1} \le C\rho ^{-2\eta } |\log \rho |,\quad \Vert \varphi _{>} \Vert _{1} \le C\rho ^{-\frac{2\eta }{\delta }} |\log \rho |. \end{aligned}
(6.24)

The last two estimates improve on $$\Vert \varphi \Vert _{1} \le C\rho ^{-2\gamma }$$ since $$\eta < \gamma$$. To estimate $$\text {I}_{1}^{\sharp }$$, we shall also use that:

\begin{aligned} \Vert u^{<}_{z} \Vert _{2} \le C\rho ^{\frac{3\eta }{2}},\quad \Vert u^{0}_{z} \Vert _{2} \le C\rho ^{\frac{3\eta }{2\delta }},\quad \Vert u^{>}_{z} \Vert _{2} \le C_{\beta }. \end{aligned}
(6.25)

The first two estimates are better than $$\Vert u^{\text {r}}_{z} \Vert _{2} \le C_{\beta }$$. Moreover, we shall use that:

\begin{aligned} \int dz\, \Vert a(u^{<}_{z}) \xi _{\lambda } \Vert ^{2}\le & {} C\langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle ,\quad \int dz\, \Vert a(u^{0}_{z}) \xi _{\lambda } \Vert ^{2} \le C\rho ^{-2\eta }\langle \xi _{\lambda }, \mathbb {H}_{0} \xi _{\lambda } \rangle , \nonumber \\ \int dz\, \Vert a(u^{>}_{z}) \xi _{\lambda } \Vert ^{2}\le & {} C\rho ^{-\frac{2\eta }{\delta }}\langle \xi _{\lambda }, \mathbb {H}_{0} \xi _{\lambda } \rangle . \end{aligned}
(6.26)

$$\underline{Estimate\,for\,\text {I}_{1}^{<}.}$$ Here, we replace $$\varphi$$ with $$\varphi _{<}$$. Also, using the fact that $$\hat{v}^{\text {r}}(k)$$ is supported for momenta $$|k| \le C\rho ^{\frac{1}{3}}$$, repeating an argument similar to the one of (6.20)–(6.23) but this time for the $$z'$$ integration, we can freely replace $$u^{\text {r}}_{z'}$$ with $$\tilde{u}^{<}_{z'}$$; in the following, we shall denote by $$\tilde{u}^{\sharp }_{z'}$$ function whose Fourier transform has similar support properties as $$\hat{\varphi }_{\sharp }$$, possibly replacing the factors 2 and 1/2 in Eq. (6.23) by 4 and 1/4. We then have:

\begin{aligned} |\text {I}_{1}^{<}|\le & {} \sum _{\sigma \ne \sigma '} \int dxdydzdz'\, V(x-y) | \varphi _{<}(z-z')| |\tilde{\omega }^{\text {r}}(z;y)| \Vert \overline{v}^{\text {r}}_{x} \Vert _{2} \Vert \overline{v}^{\text {r}}_{z'} \Vert _{2} \Vert u^{<}_{z} \Vert _{2} \nonumber \\&\Vert a_{\sigma }(u_{x}) a_{\sigma '}(u_{y}) \xi _{\lambda } \Vert \Vert a_{\downarrow }(\tilde{u}^{<}_{z'}) \xi _{\lambda } \Vert \nonumber \\\le & {} C\rho ^{1 + \frac{3\eta }{2}} \sum _{\sigma \ne \sigma '} \int dxdydzdz'\, V(x-y) | \varphi _{<}(z-z')| |\tilde{\omega }^{\text {r}}(z;y)| \nonumber \\&\cdot \Vert a_{\sigma }(u_{x}) a_{\sigma '}(u_{y}) \xi _{\lambda } \Vert \Vert a_{\downarrow }(\tilde{u}^{<}_{z'}) \xi _{\lambda } \Vert \nonumber \\\le & {} C|\log \rho |\rho ^{1 + \frac{3\eta }{2} - 2\gamma - \frac{\epsilon }{3}} \Vert \widetilde{\mathbb {Q}}^{\frac{1}{2}}_{1} \xi _{\lambda } \Vert \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda } \Vert , \end{aligned}
(6.27)

the last step following from Cauchy–Schwarz inequality, combined with $$\Vert \varphi _{<} \Vert _{1} \le C\rho ^{-2\gamma }$$, $$\Vert \omega ^{\text {r}} \Vert _{1} \le C\rho ^{-\frac{\epsilon }{3}}$$.

$$\underline{Estimate\,for\,\text {I}_{1}^{0}.}$$ Here, we replace $$\varphi$$ with $$\varphi _{0}$$ and $$u^{\text {r}}_{z'}$$ with $$\tilde{u}^{0}_{z'}$$. We have:

\begin{aligned}&|\text {I}_{1}^{0}| \le \sum _{\sigma \ne \sigma '} \int dxdydzdz'\, V(x-y) | \varphi _{0}(z-z')| |\tilde{\omega }^{\text {r}}(z;y)| \Vert \overline{v}^{\text {r}}_{x} \Vert _{2} \Vert \overline{v}^{\text {r}}_{z'} \Vert _{2} \Vert u^{0}_{z} \Vert _{2} \nonumber \\&\qquad \cdot \Vert a_{\sigma }(u_{x}) a_{\sigma '}(u_{y}) \xi _{\lambda } \Vert \Vert a_{\downarrow }(\tilde{u}^{0}_{z'}) \xi _{\lambda } \Vert \nonumber \\&\quad \le C\rho ^{1 + \frac{3\eta }{2\delta }} \sum _{\sigma \ne \sigma '} \int dxdydzdz'\, V(x-y) | \varphi _{0}(z-z')| |\tilde{\omega }^{\text {r}}(z;y)| \nonumber \\&\qquad \cdot \Vert a_{\sigma }(u_{x}) a_{\sigma '}(u_{y}) \xi _{\lambda } \Vert \Vert a_{\downarrow }(\tilde{u}^{0}_{z'}) \xi _{\lambda } \Vert \nonumber \\&\quad \le C|\log \rho |\rho ^{1 + \frac{3\eta }{2\delta } - 3\eta - \frac{\epsilon }{3}} \Vert \widetilde{\mathbb {Q}}^{\frac{1}{2}} \xi _{\lambda } \Vert \Vert \mathbb {H}_{0}^{\frac{1}{2}}\xi _{\lambda } \Vert , \end{aligned}
(6.28)

by Cauchy–Schwarz inequality, this time using that $$\Vert \varphi _{0}\Vert \le C|\log \rho |\rho ^{-2\eta }$$ and the second of (6.26).

$$\underline{Estimate\,for\,\text {I}^{>}_{1}.}$$ Here, we replace $$\varphi$$ with $$\varphi _{>}$$ and $$u^{\text {r}}_{z'}$$ with $$\tilde{u}^{>}_{z'}$$. We then have:

\begin{aligned}&|\text {I}_{1}^{>}| \le \sum _{\sigma \ne \sigma '} \int dxdydzdz'\, V(x-y) | \varphi _{>}(z-z')| |\tilde{\omega }^{\text {r}}(z;y)| \Vert \overline{v}^{\text {r}}_{x} \Vert _{2} \Vert \overline{v}^{\text {r}}_{z'} \Vert _{2} \Vert u^{>}_{z} \Vert _{2} \nonumber \\&\qquad \cdot \Vert a_{\sigma }(u_{x}) a_{\sigma '}(u_{y}) \xi _{\lambda } \Vert \Vert a_{\downarrow }(\tilde{u}^{>}_{z'}) \xi _{\lambda } \Vert \nonumber \\&\quad \le C_{\beta }\rho \sum _{\sigma \ne \sigma '} \int dxdydzdz'\, V(x-y) | \varphi _{>}(z-z')| |\tilde{\omega }^{\text {r}}(z;y)| \nonumber \\&\qquad \cdot \Vert a_{\sigma }(u_{x}) a_{\sigma '}(u_{y}) \xi _{\lambda } \Vert \Vert a_{\downarrow }(\tilde{u}^{>}_{z'}) \xi _{\lambda } \Vert \nonumber \\&\quad \le C_{\beta } \rho ^{1 - \frac{3\eta }{\delta } - \frac{\epsilon }{3}} \Vert \widetilde{\mathbb {Q}}^{\frac{1}{2}} \xi _{\lambda } \Vert \Vert \mathbb {H}_{0}^{\frac{1}{2}}\xi _{\lambda } \Vert , \end{aligned}
(6.29)

again by CS inequality, using that $$\Vert \varphi _{>}\Vert \le C|\log \rho |\rho ^{-2\eta /\delta }$$ and the last of (6.26).

$$\underline{Putting\,it\,together:\,estimate\,for\,\text {I}_{1}.}$$ From (6.27), (6.28), (6.29), we get:

\begin{aligned} | \text {I}_{1} |\le & {} C|\log \rho |\rho ^{1 + \frac{3\eta }{2} - 2\gamma - \frac{\epsilon }{3}} \Vert \widetilde{\mathbb {Q}}^{\frac{1}{2}}_{1} \xi _{\lambda } \Vert \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda } \Vert \nonumber \\&+ C|\log \rho |\rho ^{1 + \frac{3\eta }{2\delta } - 3\eta - \frac{\epsilon }{3}} \Vert \widetilde{\mathbb {Q}}^{\frac{1}{2}} \xi _{\lambda } \Vert \Vert \mathbb {H}_{0}^{\frac{1}{2}}\xi _{\lambda } \Vert + C_{\beta } \rho ^{1 - \frac{3\eta }{\delta } - \frac{\epsilon }{3}} \Vert \widetilde{\mathbb {Q}}^{\frac{1}{2}} \xi _{\lambda } \Vert \Vert \mathbb {H}_{0}^{\frac{1}{2}}\xi _{\lambda } \Vert .\nonumber \\ \end{aligned}
(6.30)

The optimal value of $$\delta$$ is $$\delta = 3/2$$. For this value, using also the propagation of the a priori estimates for $$\widetilde{\mathbb {Q}}_{1}$$, $$\mathbb {H}_{0}$$, Eqs. (5.76), we get, for $$0<\alpha < 1$$:

\begin{aligned} |\text {I}_{1}|\le & {} C|\log \rho |L^{\frac{3}{2}}\rho ^{2 + \frac{3\eta }{2} - 2\gamma - \frac{\epsilon }{3}} \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda } \Vert + C_{\beta } L^{3} \rho ^{3 - 2\eta - \frac{\epsilon }{3}} \nonumber \\\le & {} C|\log \rho |L^{3} \rho ^{3 + \frac{3\eta }{2} - \frac{5}{2}\gamma - \frac{\epsilon }{3}} + C_{\beta } L^{3} \rho ^{3 - 2\eta - \frac{\epsilon }{3}} \nonumber \\&+ C_{\alpha }L^{3} |\log \rho |^{\frac{4}{3}} \rho ^{\frac{25}{9} + 2\eta - \frac{8}{3}\gamma - \frac{4\epsilon }{9}} + \alpha \langle \xi _{1}, \mathbb {H}_{0} \xi _{1}\rangle . \end{aligned}
(6.31)

The last estimate follows after using the bound (5.86) for the number operator, and using Young’s inequality. For $$\eta \le 1/3$$, the third term is bigger than the first. Hence:

\begin{aligned} |\text {I}_{1}|\le & {} C_{\beta } L^{3} \rho ^{3 - 2\eta - \frac{\epsilon }{3}} + C_{\alpha }L^{3} |\log \rho |^{\frac{4}{3}} \rho ^{\frac{25}{9} + 2\eta - \frac{8}{3}\gamma - \frac{4\epsilon }{9}} + \alpha \langle \xi _{1}, \mathbb {H}_{0} \xi _{1}\rangle \nonumber \\\le & {} C_{\beta ,\alpha } L^{3}\rho ^{\frac{26}{9} - \frac{4}{3}\gamma - \frac{7}{18}\epsilon } + \alpha \langle \xi _{1}, \mathbb {H}_{0} \xi _{1}\rangle , \end{aligned}
(6.32)

where we optimized over $$\eta$$, $$\eta = \frac{1}{18} + \frac{2\gamma }{3} + \frac{\epsilon }{36}$$, which is less $$\gamma$$ for $$\frac{1}{6} + \frac{\epsilon }{12} < \gamma$$. All the other contributions arising from the first line of (6.14) are estimated in this way.

$$\underline{Estimate\,for\,\text {I}_{2}.}$$ Consider now the term $$\text {I}_{2}$$ in Eq. (6.16). As discussed in Appendix 7, this term satisfies the same estimate as $$\text {I}_{1}$$:

\begin{aligned} |\text {I}_{2}| \le C_{\beta ,\alpha } L^{3}\rho ^{\frac{26}{9} - \frac{4}{3}\gamma - \frac{7}{18}\epsilon } + \alpha \langle \xi _{1}, \mathbb {H}_{0} \xi _{1}\rangle . \end{aligned}
(6.33)

$$\underline{Other\,contributions\,to\,}$$(6.14). Consider now the second line of (6.14). We rewrite it as:

\begin{aligned}&-a_\uparrow (\overline{v}^{\text {r}}_z)a_\downarrow (\overline{v}^{\text {r}}_{z'})[a_{\sigma }^{*}(u^{\text {r}}_{x})a_{\sigma '}^{*}(u^{\text {r}}_{y}), a_{\uparrow }(u^{\text {r}}_z)a_{\downarrow }(u^{\text {r}}_{z'})]a^{*}_{\sigma '}(\overline{v}_y)a^{*}_\sigma (\overline{v}_x) \nonumber \\&\quad = -a^{*}_{\sigma '}(\overline{v}_y)a^{*}_\sigma (\overline{v}_x)[a^{*}_\sigma (u^{\text {r}}_x)a^{*}_{\sigma '}(u^{\text {r}}_y), a_\uparrow (u^{\text {r}}_z)a_\downarrow (u^{\text {r}}_{z'})]a_{\uparrow }(\overline{v}^{\text {r}}_z)a_\downarrow (\overline{v}^{\text {r}}_{z'})\nonumber \\&\qquad - [a^{*}_\sigma (u_x^{\text {r}})a^{*}_{\sigma '}(u^{\text {r}}_y),a_\uparrow (u^{\text {r}}_z)a_\downarrow (u^{\text {r}}_{z'}) ][a_\uparrow (\overline{v}^{\text {r}}_z)a_\downarrow (\overline{v}^{\text {r}}_{z'}), a^{*}_{\sigma '}(\overline{v}_y^{\text {r}})a_{\sigma }(\overline{v}_{x}^{\text {r}})].\nonumber \\ \end{aligned}
(6.34)

Let us consider the first term in the right-hand side. We rewrite the commutator as, omitting the spin for simplicity:

\begin{aligned}{}[a^{*}(u^{\text {r}}_x)a^{*}(u^{\text {r}}_y), a(u^{\text {r}}_z)a(u^{\text {r}}_{z'})]= & {} a^{*}(u^{\text {r}}_{y}) u^{\text {r}}(x;z) a(u^{\text {r}}_{z'}) - a^{*}(u^{\text {r}}_{y}) u^{\text {r}}(x;z') a(u^{\text {r}}_{z}) \nonumber \\&+ a(u^{\text {r}}_{z'}) u^{\text {r}}(y;z) a^{*}(u^{\text {r}}_{x}) - a(u^{\text {r}}_{z}) u^{\text {r}}(y;z') a^{*}(u^{\text {r}}_{x}).\nonumber \\ \end{aligned}
(6.35)

The first two terms give rise to error terms of the form:

\begin{aligned} \text {II}_{a}= & {} \int dxdy dzdz'\, V(x-y) \varphi (z-z') u^{\text {r}}(y;z)\nonumber \\&\cdot \langle \xi _{\lambda }, a^{*}(u^{\text {r}}_{x}) a^{*}(\overline{v}_y)a^{*}(\overline{v}_x) a(\overline{v}^{\text {r}}_z)a(\overline{v}^{\text {r}}_{z'}) a(u^{\text {r}}_{z'})\xi _{\lambda } \rangle . \end{aligned}
(6.36)

We then get, using that $$\Vert u^{\text {r}}_{y} \Vert _{1} \le C$$:

\begin{aligned} |\text {II}_{a}|\le & {} C \rho ^{2} \int dxdydzdz'\, V(x-y) \varphi (z-z') |u^{\text {r}}(y;z) | (\Vert a(u^{\text {r}}_{x}) \xi _{\lambda } \Vert ^{2} + \Vert a(u^{\text {r}}_{z'}) \xi _{\lambda } \Vert ^{2} ) \nonumber \\\le & {} C\rho ^{2 - 2\gamma } \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle . \end{aligned}
(6.37)

To bound the error terms produced by the last two terms in (6.35), we normal order them. The normal ordered contribution can be estimated as $$\text {II}_{a}$$. The new contraction produces an error term of the form:

\begin{aligned} \text {II}_{b} =\int dxdy dzdz'\, V(x-y) \varphi (z-z') u^{\text {r}}(y;z) u^{\text {r}}(x;z') \langle \xi _{\lambda }, a^{*}(\overline{v}_y)a^{*}(\overline{v}_x) a(\overline{v}^{\text {r}}_z)a(\overline{v}^{\text {r}}_{z'}) \xi _{\lambda } \rangle . \end{aligned}
(6.38)

We have, using that $$|\varphi (z-z')| \le C$$:

\begin{aligned} |\text {II}_{b}|\le & {} C\rho \int dxdy dzdz'\, V(x-y) |u^{\text {r}}(y;z)| |u^{\text {r}}(x;z')| (\Vert a(\overline{v}_y) \xi _{\lambda } \Vert ^{2} + \Vert a(\overline{v}^{\text {r}}_{z'}) \xi _{\lambda } \Vert ^{2}) \nonumber \\\le & {} C\rho \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle . \end{aligned}
(6.39)

This concludes the analysis of the error terms produced by the first term in the right-hand side of (6.34). We are left with the second term in (6.34), involving two commutators. We compute:

\begin{aligned}&[a^{*}_\sigma (u_x^{\text {r}})a^{*}_{\sigma '}(u^{\text {r}}_y),a_\uparrow (u^{\text {r}}_z)a_\downarrow (u^{\text {r}}_{z'}) ] \nonumber \\&\quad = \delta _{\sigma ',\uparrow } u^{\text {r}}_{\sigma '}(z;y)a^{*}_\sigma (u_x^{\text {r}})a_{\downarrow }(u^{\text {r}}_{z'})-\delta _{\sigma ',\downarrow } u^{\text {r}}_{\sigma '}(z';y)a^{*}_\sigma (u_x^{\text {r}})a_\uparrow (u^{\text {r}}_z)\nonumber \\&\qquad -\delta _{\sigma , \uparrow } u^{\text {r}}_\sigma (z;x)a^{*}_{\sigma '}(u^{\text {r}}_y)a_\downarrow (u^{\text {r}}_{z'})-\delta _{\sigma ,\downarrow }u^{\text {r}}_\sigma (z';x)a^{*}_{\sigma '}(u^{\text {r}}_y)a_\uparrow (u^{\text {r}}_z),\nonumber \\&\qquad +\delta _{\sigma ,\uparrow }\delta _{\sigma ',\downarrow }u^{\text {r}}_\sigma (z;x) u^{\text {r}}_{\sigma '}(z';y) - \delta _{\sigma ,\downarrow }\delta _{\sigma ', \uparrow }u^{\text {r}}_{\sigma }(z';x) u^{\text {r}}_{\sigma '}(z;y) \end{aligned}
(6.40)

and

\begin{aligned}&[a_\uparrow (\overline{v}^{\text {r}}_z)a_{\downarrow }(\overline{v}^{\text {r}}_{z'}), a^{*}_{\sigma '}(\overline{v}^{\text {r}}_y)a^{*}_{\sigma }(\overline{v}_x)]\nonumber \\&\quad = \delta _{\sigma ', \uparrow }\omega ^{\text {r}}_{\sigma '}(z;y)a^{*}_\sigma (\overline{v}_x)a_\downarrow (\overline{v}^{\text {r}}_{z'}) -\delta _{\sigma ',\downarrow }\omega ^{\text {r}}_{\sigma '}(z';y)a^{*}_\sigma (\overline{v}^{\text {r}}_x)a_\uparrow (\overline{v}^{\text {r}}_z)\nonumber \\&\qquad +\delta _{\sigma ,\uparrow }\omega ^{\text {r}}_\sigma (z;x)a^{*}_{\sigma '}(\overline{v}_y^{\text {r}})a_\downarrow (\overline{v}^{\text {r}}_{z'})-\delta _{\sigma ,\downarrow }\omega ^{\text {r}}_\sigma (z';x)a^{*}_{\sigma '}(\overline{v}^{\text {r}}_y)a_{\downarrow }(\overline{v}^{\text {r}}_{z'})\nonumber \\&\qquad +\delta _{\sigma ,\uparrow }\delta _{\sigma ',\downarrow }\omega ^{\text {r}}_\sigma (z;x)\omega _{\sigma '}^{\text {r}}(z';y) -\delta _{\sigma ,\downarrow }\delta _{\sigma ', \uparrow }\omega ^{\text {r}}_{\sigma }(z';x)\omega ^{\text {r}}_{\sigma '}(z;y). \end{aligned}
(6.41)

The last two terms in (6.40) times the last two terms in (6.41) produce the explicit $$O(\rho ^{2})$$ term in the final claim. Summing also the complex conjugate, and using that $$\varphi (x) = \varphi (-x)$$, we have:

\begin{aligned} \text {I}_{\text {main}}= & {} \sum _{\sigma \ne \sigma '}\int dxdydzdz'\, V(x-y) \varphi (z-z') (\delta _{\sigma ,\uparrow }\delta _{\sigma ',\downarrow }u^{\text {r}}_\sigma (z;x) u^{\text {r}}_{\sigma '}(z';y) \nonumber \\&\qquad \qquad - \delta _{\sigma ,\downarrow }\delta _{\sigma ', \uparrow }u^{\text {r}}_{\sigma }(z';x) u^{\text {r}}_{\sigma '}(z;y))\nonumber \\&\quad \cdot (\delta _{\sigma ,\uparrow }\delta _{\sigma ',\downarrow }\omega ^{\text {r}}_\sigma (z;x)\omega _{\sigma '}^{\text {r}}(z';y) -\delta _{\sigma ,\downarrow }\delta _{\sigma ', \uparrow }\omega ^{\text {r}}_{\sigma }(z';x)\omega ^{\text {r}}_{\sigma '}(z;y)) \nonumber \\= & {} 2\int dxdydzdz'\, V(x-y) \varphi (z-z') u^{\text {r}}_{\uparrow } (z;x) u^{\text {r}}_{\downarrow }(z';y) \omega ^{\text {r}}_{\uparrow }(z;x)\omega _{\downarrow }^{\text {r}}(z';y),\nonumber \\ \end{aligned}
(6.42)

where in the last step we used that $$\varphi (x) = \varphi (-x)$$. As proven in Appendix , thanks to the regularity of the potential, the function $$u^{\text {r}}_{\sigma }(z;x)$$ can be replaced by the Dirac delta $$\delta (z-x)$$, up to higher-order terms in the density:

\begin{aligned} \text {I}_{\text {main}} = 2\rho ^{\text {r}}_{\uparrow } \rho ^{\text {r}}_{\downarrow } \int dxdy\, V(x-y) \varphi (x-y) + \mathcal {E}_{\text {main}},\quad |\mathcal {E}_{\text {main}}| \le CL^{3}\rho ^{3 - 2\gamma }. \end{aligned}
(6.43)

All the other terms arising in the product of (6.40) and of (6.41) give rise to subleading contributions. For instance, a typical term is:

\begin{aligned} \text {III}_{a}= & {} \int dxdydzdz'\, V(x-y) \varphi (z-z') u^{\text {r}}(z;y)\omega ^{\text {r}}(z;y)\nonumber \\&\cdot \langle \xi _{\lambda }, a^{*}(u_x^{\text {r}})a(u^{\text {r}}_{z'})a^{*}(\overline{v}^{\text {r}}_x)a(\overline{v}^{\text {r}}_{z'}) \xi _{\lambda }\rangle \nonumber \\\equiv & {} \int dxdydzdz'\, V(x-y) \varphi (z-z') u^{\text {r}}(z;y)\omega ^{\text {r}}(z;y)\nonumber \\&\cdot \langle \xi _{\lambda }, a^{*}(u_x^{\text {r}})a^{*}(\overline{v}^{\text {r}}_x)a(\overline{v}^{\text {r}}_{z'}) a(u^{\text {r}}_{z'})\xi _{\lambda }\rangle ,\nonumber \\ \end{aligned}
(6.44)

where we used the orthogonality between $$u^{\text {r}}$$ and $$\overline{v}^{\text {r}}$$. Proceeding as for $$\text {II}_{a}$$, using that $$\Vert u^{\text {r}}_{y} \Vert _{1} \le C$$ and that $$\Vert \omega ^{\text {r}}_{y} \Vert _{\infty } \le C\rho$$, we get:

\begin{aligned} |\text {III}_{a}| \le C\rho ^{2 - 2\gamma } \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle . \end{aligned}
(6.45)

Another typical term is:

\begin{aligned} \text {III}_{b} = \int dxdydzdz'\, V(x-y)\varphi (z-z')\omega ^{\text {r}}(z;x)\omega ^{\text {r}}(z';y)u^{\text {r}}(z;y) \langle \xi _{\lambda }, a^{*}(u^{\text {r}}_x)a(u^{\text {r}}_{z'}) \xi _{\lambda }\rangle , \end{aligned}
(6.46)

which we bound as:

\begin{aligned} |\text {III}_{b}| \le C \rho ^{2 - 2\gamma } \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda } \rangle . \end{aligned}
(6.47)

The last type of error term arising in the product of (6.40) and (6.41) is:

\begin{aligned} \text {III}_{c} = \int dxdydzdz'\, V(x-y)\varphi (z-z')u^{\text {r}}(z;x)u^{\text {r}}(z';y)\omega ^{\text {r}}(z;y) \langle \xi _{\lambda }, a^{*}(\overline{v}_x^{\text {r}})a(\overline{v}^{\text {r}}_{z'}) \xi _{\lambda }\rangle , \end{aligned}
(6.48)

which we estimate as:

\begin{aligned} |\text {III}_{c}| \le C \rho \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda }\rangle . \end{aligned}
(6.49)

Conclusion. Putting together (6.32), (6.33), (6.37), (6.39), (6.43), (6.45)–(6.48), we have, using that $$\rho ^{2 - 2\gamma } \le \rho$$ for $$\gamma \le 1/2$$:

\begin{aligned} |\mathcal {E}_{\widetilde{\mathbb {Q}}_{4}}(\xi _{\lambda })|\le & {} C \rho \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda }\rangle + C_{\beta ,\alpha } L^{3}\rho ^{\frac{26}{9} - \frac{4}{3}\gamma - \frac{7}{18}\epsilon } + \alpha \langle \xi _{1}, \mathbb {H}_{0} \xi _{1}\rangle \nonumber \\\le & {} CL^{3} \rho ^{\frac{7}{3}} + C_{\beta ,\alpha } L^{3}\rho ^{\frac{26}{9} - \frac{4}{3}\gamma - \frac{7}{18}\epsilon } + \alpha \langle \xi _{1}, \mathbb {H}_{0} \xi _{1}\rangle , \end{aligned}
(6.50)

where in the last step we used the bound (5.86) for the number operator. This concludes the proof. $$\square$$

Conclusion: proof of the lower bound. We are now ready to prove a lower bound for the ground state energy. We shall collect all the error terms, starting from Eq. (6.1). We have, for $$0<\alpha <1$$:

\begin{aligned}&\langle \psi , \mathcal {H} \psi \rangle \ge E_{\text {HF}}(\omega )\nonumber \\&\quad - \rho _{\uparrow }\rho _{\downarrow } \int dxdy\, V(x-y) \varphi (x-y) + \langle \xi _{1}, (\mathbb {H}_{0} + \widetilde{\mathbb {Q}}_{1}) \xi _{1} \rangle (1 - \alpha ) \nonumber \\&\quad - C_{\alpha } L^{3} \rho ^{2 + \frac{1}{9}} - C_{\alpha } L^{3} \rho ^{\frac{7}{3} - \frac{2\gamma }{3}} - C_{\alpha } L^{3} \rho ^{\frac{13}{9} + 2\gamma } - C_{\alpha } L^{3} \rho ^{2 + \frac{\epsilon }{3}} \nonumber \\&\quad - CL^{3} \rho ^{\frac{7}{3}} - C_{\beta ,\alpha } L^{3}\rho ^{\frac{26}{9} - \frac{4}{3}\gamma - \frac{7}{18}\epsilon }, \end{aligned}
(6.51)

where we also used that $$|\rho _{\sigma } - \rho _{\sigma }^{\text {r}}| \le \rho ^{1 + \frac{\epsilon }{3}}$$. The integral in the right-hand side can be written as, up to a boundary term:

\begin{aligned} \int dxdy\, V(x-y) \varphi (x-y) = L^{3}\int _{\mathbb {R}^{3}} dx\, V_{\infty }(x) \varphi _{\gamma }(x) + \mathfrak {e}_{L}, \end{aligned}
(6.52)

with $$\mathfrak {e}_{L} = O(L^{2})$$. Recall that, Eq. (4.29):

\begin{aligned} 8\pi a_{\gamma } = \int _{\mathbb {R}^{3}} dx\, V_{\infty }(x) (1 - \varphi _{\gamma }(x)),\quad |a - a_{\gamma }| \le C\rho ^{\gamma }. \end{aligned}
(6.53)

The optimal choice of parameters is:

\begin{aligned} \epsilon = \frac{1}{3},\quad \gamma = \frac{1}{3}, \end{aligned}
(6.54)

which fulfills the assumptions of Proposition 6.1. Taking $$\frac{1}{2} \le \alpha < 1$$, we finally have, for L large enough:

\begin{aligned} \frac{\langle \psi , \mathcal {H} \psi \rangle }{L^{3}} \ge \frac{3}{5}(6\pi ^{2})^{\frac{2}{3}} (\rho _{\uparrow }^{\frac{5}{3}} + \rho _{\downarrow }^{\frac{5}{3}}) + 8\pi a \rho _{\uparrow } \rho _{\downarrow } - CL^{3}\rho ^{2 + \frac{1}{9}} + \langle \xi _{1}, (\mathbb {H}_{0} + \widetilde{\mathbb {Q}}_{1}) \xi _{1} \rangle (1 - \alpha ). \end{aligned}
(6.55)

This concludes the proof of the lower bound.

### Remark 6.3

(Improved condensation estimate.). The inequality (6.55) can be used to prove an improved estimate for $$\langle \xi _{1}, \mathbb {H}_{0} \xi _{1}\rangle$$, for states $$\psi$$ that are energetically close enough to the ground state. Let $$\psi$$ be a fermionic state such that:

\begin{aligned} \frac{\langle \psi , \mathcal {H} \psi \rangle }{L^{3}} - \frac{3}{5}(6\pi ^{2})^{\frac{2}{3}} (\rho _{\uparrow }^{\frac{5}{3}} + \rho _{\downarrow }^{\frac{5}{3}}) + 8\pi a \rho _{\uparrow } \rho _{\downarrow } \le C\rho ^{2 + \frac{1}{9}}. \end{aligned}
(6.56)

As we will see in the next section, such states exists; in particular, the ground state satisfies the inequality (6.56). Eqs. (6.55), (6.56) imply:

\begin{aligned} \langle \xi _{1}, \mathbb {H}_{0}\xi _{1}\rangle \le CL^{2} \rho ^{2 + \frac{1}{9}}\;; \end{aligned}
(6.57)

plugging this bound in (5.86), we get, for $$\gamma = 1/3$$:

\begin{aligned} \langle R^{*}\psi , \mathcal {N} R^{*}\psi \rangle \le CL^{3} \rho ^{\frac{11}{9}}. \end{aligned}
(6.58)

This inequality can be used to prove, see [7]:

\begin{aligned} \mathrm {tr}\, \gamma ^{(1)}_{\psi } (1 - \omega ) \le CL^{3} \rho ^{\frac{11}{9}}. \end{aligned}
(6.59)

This bound improves on the condensation estimate (3.65). The optimal condensation estimate is expected to be of order $$\rho ^{\frac{4}{3}}$$: this is consistent with the fact that the next-order correction to the ground state energy is of order $$\rho ^{\frac{7}{3}}$$, [23].

## 7 Upper Bound on the Ground State Energy

In this section, we shall conclude the proof of Theorem 2.1, by proving an upper bound on the ground state energy that matches the lower bound we obtained in Sect. 6, up to $$o(\rho ^{2})$$. This will be done taking the natural trial state $$\psi = R T \Omega$$, with T the correlation structure defined in Sect. 4, for a suitable value of the parameter $$\gamma$$, to be optimized.

To begin, notice that $$\psi$$ is an N-particle state, with $$N_{\uparrow }$$ particles with spin $$\uparrow$$ and $$N_{\downarrow }$$ particles with spin $$\downarrow$$. In fact, we can rewrite $$\psi$$ as:

\begin{aligned}&\psi = \widetilde{T} R\Omega ,\quad \widetilde{T} := e^{\widetilde{B} - \widetilde{B}^{*}},\nonumber \\&\widetilde{B} := \int dzdz'\, \varphi (z-z') a_{\uparrow }(u^{\text {r}}_{z}) a^{*}_{\uparrow }(\overline{v} \overline{v}^{\text {r}}_{z}) a_{\downarrow }(u^{\text {r}}_{z'}) a^{*}_{\downarrow }(\overline{v} \overline{v}^{\text {r}}_{z'}). \end{aligned}
(7.1)

and $$[\widetilde{B}, \mathcal {N}_{\sigma }] = 0$$, with $$\mathcal {N}_{\sigma } = \int dx\, a^{*}_{x,\sigma } a_{x,\sigma }$$. Also, by the defining properties of fermionic Bogoliubov transformations (3.28), we know that $$R\Omega$$ is an N-particle state, with $$N_{\uparrow }$$ particles with spin $$\uparrow$$ and $$N_{\downarrow }$$ particles with spin $$\downarrow$$. Therefore:

\begin{aligned} \mathcal {N}_{\sigma } \psi = \mathcal {N}_{\sigma } \widetilde{T} R\Omega = \widetilde{T} \mathcal {N}_{\sigma } R\Omega = N_{\sigma } \psi . \end{aligned}
(7.2)

Moreover, being R and T unitary operators, $$\Vert \psi \Vert = \Vert \Omega \Vert = 1$$.

To compute the energy of $$\psi$$, we shall rely on the estimates we have already proved for the lower bound. An important role in the upper bound is played by the following bound for the number operator, for $$\gamma \le 1/2$$:

\begin{aligned} \langle \xi _{\lambda }, \mathcal {N} \xi _{\lambda }\rangle \le CL^{3} \rho ^{2-\gamma },\quad \xi _{\lambda } := T_{1-\lambda } \Omega . \end{aligned}
(7.3)

This bound follows from (5.86), using that now $$\xi _{1} = \Omega$$. Thanks to (7.3), it is not difficult to see that to prove the propagation of the estimates in Proposition 5.8 it is enough to assume $$\gamma \le 1/3$$. The only point where we required a lower bound for $$\gamma$$ is the estimate (5.81), which now holds for all $$\gamma$$, as it is clear from the bound (7.3).

By Propositions 3.1, 3.3, we have:

\begin{aligned} E_{L}(N_{\uparrow }, N_{\downarrow })\le & {} E_{\text {HF}}(\omega ) + \langle T\Omega , \mathbb {H}_{0} T\Omega \rangle + \langle T\Omega , \mathbb {X} T\Omega \rangle + \langle T\Omega , \mathbb {Q} T\Omega \rangle \nonumber \\\le & {} E_{\text {HF}}(\omega ) + \langle T\Omega , (\mathbb {H}_{0} + \mathbb {Q}_{1} + \mathbb {Q}_{4}) T\Omega \rangle + \mathcal {E}_{1}(\psi ), \end{aligned}
(7.4)

with:

\begin{aligned} |\mathcal {E}_{1}(\psi )| \le C\rho \langle T\Omega , \mathcal {N} T\Omega \rangle \ \le CL^{3} \rho ^{3 - \gamma }. \end{aligned}
(7.5)

Here, we crucially used that the state $$\xi _{0} = T\Omega$$ is such that $$\xi _{0}^{(n)} = 0$$ unless $$n = 4k$$ for $$k\in \mathbb {N}$$, and hence that:

\begin{aligned} \langle T\Omega , \mathbb {Q}_{3} T\Omega \rangle = 0, \end{aligned}
(7.6)

recall Eq. (3.43). Consider now the $$\langle T\Omega , \mathbb {Q}_{4} T\Omega \rangle$$ term. We rewrite it as:

\begin{aligned} \langle T\Omega , \mathbb {Q}_{4} T\Omega \rangle = \langle T\Omega , \widetilde{\mathbb {Q}}_{4} T\Omega \rangle + \langle T\Omega , \widehat{\mathbb {Q}}_{4} T\Omega \rangle , \end{aligned}
(7.7)

with $$\widehat{\mathbb {Q}}_{4}$$ the contribution to $$\mathbb {Q}_{4}$$ with aligned spins,

\begin{aligned} \widehat{\mathbb {Q}}_{4} = \frac{1}{2} \sum _{\sigma } \int dxdy\, V(x-y) a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma }(u_{y}) a^{*}_{\sigma }(\overline{v}_{y}) a^{*}_{\sigma }(\overline{v}_{x}) + \text {h.c..} \end{aligned}
(7.8)

We claim that $$\langle T\Omega , \widehat{\mathbb {Q}}_{4} T\Omega \rangle = 0$$. To prove this, we shall use that $$T\Omega$$ and $$\widehat{\mathbb {Q}}_{4} T\Omega$$ belong to different spin sectors, and hence, they are orthogonal vectors in the Fock space. Let $$\mathcal {S}$$ be the spin operator,

\begin{aligned} \mathcal {S} = \sum _{\sigma } \sigma \mathcal {N}_{\sigma },\quad \mathcal {N}_{\sigma } = \int dx\, a^{*}_{x,\sigma } a_{x,\sigma }, \end{aligned}
(7.9)

where we identify $$\uparrow \, \equiv +$$ and $$\downarrow \, \equiv -$$. Clearly, $$\mathcal {S} \Omega = 0$$. Also, since $$[ B, \mathcal {S} ] = 0$$, we have $$[T, \mathcal {S}] = 0$$. Therefore, $$\mathcal {S} T \Omega = 0$$. At the same time,

\begin{aligned}&\mathcal {S} \int dxdy\, V(x-y) a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma }(u_{y}) a^{*}_{\sigma }(\overline{v}_{y}) a^{*}_{\sigma }(\overline{v}_{x}) T\Omega \nonumber \\&\quad = \int dxdy\, V(x-y) a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma }(u_{y}) a^{*}_{\sigma }(\overline{v}_{y}) a^{*}_{\sigma }(\overline{v}_{x}) (\mathcal {S} + \sigma 4) T\Omega \nonumber \\&\quad = \sigma 4 \int dxdy\, V(x-y) a^{*}_{\sigma }(u_{x}) a^{*}_{\sigma }(u_{y}) a^{*}_{\sigma }(\overline{v}_{y}) a^{*}_{\sigma }(\overline{v}_{x}) T\Omega \ne 0. \end{aligned}
(7.10)

Therefore, $$\langle T\Omega , \widehat{\mathbb {Q}}_{4} T\Omega \rangle = 0$$ by orthogonality between different spin sectors. We are then left with:

\begin{aligned} E_{L}(N_{\uparrow }, N_{\downarrow }) \le E_{\text {HF}}(\omega ) + \langle T\Omega , (\mathbb {H}_{0} + \mathbb {Q}_{1} + \widetilde{\mathbb {Q}}_{4}) T\Omega \rangle + \mathcal {E}_{1}(\psi ), \end{aligned}
(7.11)

with $$\mathcal {E}_{1}(\psi )$$ bounded as in (7.5). Proceeding exactly as in Sect. 6, Eq. (6.4), we get:

\begin{aligned} \langle T\Omega , (\mathbb {H}_{0} + \mathbb {Q}_{1} + \widetilde{\mathbb {Q}}_{4}) T\Omega \rangle = \int _{0}^{1} d\lambda \, \langle \xi _{\lambda }, (\mathbb {T}_{1} + \mathbb {T}_{2}) \xi _{\lambda } \rangle + \langle \xi _{0}, \widetilde{\mathbb {Q}}_{4} \xi _{0} \rangle + \mathcal {E}_{2}(\psi ),\nonumber \\ \end{aligned}
(7.12)

where $$\mathcal {E}_{2}(\psi )$$ can be bounded as, for $$0\le \gamma \le 1/3$$, thanks to the estimates (5.84) and the bound (7.3) for the number operator:

\begin{aligned} |\mathcal {E}_{2}(\psi )|\le & {} \max _{\lambda \in [0;1]} | \mathcal {E}_{\mathbb {H}_{0}}(\xi _{\lambda }) | + \max _{\lambda \in [0;1]} | \mathcal {E}_{\mathbb {Q}_{1}}(\xi _{\lambda }) | \nonumber \\\le & {} CL^{3} \rho ^{\frac{7}{3} - \frac{\gamma }{2}} + C_{\beta }L^{3} \rho ^{3 - \frac{11}{5}\gamma }\nonumber \\\le & {} CL^{3} \rho ^{\frac{7}{3} - \frac{\gamma }{2}}. \end{aligned}
(7.13)

To get the second inequality, we optimized over the parameter $$\eta$$ appearing in the bound for $$\mathcal {E}_{\mathbb {Q}_{1}}(\xi _{\lambda })$$, $$\eta = \frac{\gamma }{5}$$, and to get the third we used that $$\gamma \le 1/3$$. Then we write, proceeding as in Eq. (6.6):

\begin{aligned} \langle T\Omega , (\mathbb {H}_{0} + \mathbb {Q}_{1} + \widetilde{\mathbb {Q}}_{4}) T\Omega \rangle = - \int _{0}^{1}d\lambda \langle \xi _{\lambda }, \widetilde{\mathbb {Q}}_{4}^{\text {r}} \xi _{\lambda } \rangle + \langle \xi _{0}, \widetilde{\mathbb {Q}}_{4} \xi _{0} \rangle + \mathcal {E}_{3}(\psi ) + \mathcal {E}_{2}(\psi ),\nonumber \\ \end{aligned}
(7.14)

with:

\begin{aligned} |\mathcal {E}_{3}(\psi )|\le & {} \max _{\lambda \in [0;1]} |\langle \xi _{\lambda }, (\mathbb {T}_{1} + \mathbb {T}_{2} + \widetilde{\mathbb {Q}}_{4}^{\text {r}}) \xi _{\lambda } \rangle |\nonumber \\\le & {} C L^{3} \rho ^{2 + \gamma }, \end{aligned}
(7.15)

where we used the estimate (5.85), and the bound for the number operator (7.3). Next, proceeding as in Eq. (6.8) we have:

\begin{aligned} - \int _{0}^{1} d\lambda \, \langle \xi _{\lambda }, \widetilde{\mathbb {Q}}_{4}^{\text {r}} \xi _{\lambda } \rangle + \langle \xi _{0}, \widetilde{\mathbb {Q}}_{4} \xi _{0} \rangle= & {} - \int _{0}^{1}d\lambda \, \frac{d}{d\lambda } \langle \xi _{\lambda } \widetilde{\mathbb {Q}}_{4} \xi _{\lambda }\rangle \\&+ \int _{0}^{1} d\lambda \int _{\lambda }^{1} d\lambda ' \frac{d}{d\lambda '} \langle \xi _{\lambda '}, \widetilde{\mathbb {Q}}_{4}^{\text {r}} \xi _{\lambda '} \rangle .\nonumber \end{aligned}
(7.16)

We compute the derivatives using Proposition 6.1. We obtain:

\begin{aligned} \frac{d}{d\lambda }\langle \xi _{\lambda }, \widetilde{\mathbb {Q}}^{\text {r}}_{4} \xi _{\lambda } \rangle= & {} 2\rho ^{\text {r}}_{\uparrow }\rho ^{\text {r}}_{\downarrow } \int dxdy\, V(x-y) \varphi (x-y) + \mathcal {E}_{\widetilde{\mathbb {Q}}^{\text {r}}_{4}}(\xi _{\lambda }) \nonumber \\ \frac{d}{d\lambda }\langle \xi _{\lambda }, \widetilde{\mathbb {Q}}_{4} \xi _{\lambda } \rangle= & {} 2\rho ^{\text {r}}_{\uparrow }\rho ^{\text {r}}_{\downarrow } \int dxdy\, V(x-y) \varphi (x-y) + \mathcal {E}_{\widetilde{\mathbb {Q}}_{4}}(\xi _{\lambda })\;; \end{aligned}
(7.17)

the bound for the error terms can be improved with respect to (6.11), making use of the estimate for the number operator (7.3). Inspection of the proof of Proposition 6.1 shows that the estimate for the error terms $$\mathcal {E}_{\widetilde{\mathbb {Q}}^{\text {r}}_{4}}(\xi _{\lambda })$$, $$\mathcal {E}_{\widetilde{\mathbb {Q}}_{4}}(\xi _{\lambda })$$, are determined by the bound for the term $$\text {I}_{1}$$ in the first line of Eq. (6.31):

\begin{aligned} |\text {I}_{1}|\le & {} CL^{\frac{3}{2}}\rho ^{2 + \frac{3\eta }{2} - 2\gamma - \frac{\epsilon }{3}} \Vert \mathcal {N}^{\frac{1}{2}} \xi _{\lambda } \Vert + C_{\beta } L^{3} \rho ^{3 - 2\eta - \frac{\epsilon }{3}}\nonumber \\\le & {} C_{\beta } L^{3} \rho ^{3 - \frac{10}{7}\gamma - \frac{\epsilon }{3}}, \end{aligned}
(7.18)

where in the last step we used the bound (7.3) and we optimized over $$\eta$$, $$\eta = \frac{5}{7}\gamma$$. Notice that, with respect to the original proof of Proposition 6.1, the optimal value of $$\eta$$ is now independent of $$\epsilon$$. We find:

\begin{aligned} |\mathcal {E}_{\widetilde{\mathbb {Q}}^{\text {r}}_{4}}(\psi )| \le C_{\beta }L^{3} \rho ^{3 - \frac{10}{7}\gamma - \frac{\epsilon }{3}},\quad |\mathcal {E}_{\widetilde{\mathbb {Q}}_{4}}(\psi )| \le C_{\beta }L^{3} \rho ^{3 - \frac{10}{7}\gamma - \frac{\epsilon }{3}}. \end{aligned}
(7.19)

With respect to the original proof of Proposition 6.1, the bound (7.19) holds for all $$\epsilon \ge 0$$, as a consequence of the fact that the optimal value of $$\eta$$ does not depend on $$\epsilon$$.

Conclusion: proof of the upper bound. Putting together (7.11)–(7.19), we find, for $$0 \le \gamma \le 1/3$$:

\begin{aligned} E_{L}(N_{\uparrow }, N_{\downarrow })\le & {} E_{\text {HF}}(\omega ) - \rho _{\uparrow }\rho _{\downarrow } \int dxdy\, V(x-y) \varphi (x-y)\nonumber \\&+ CL^{3} \rho ^{2 + \frac{\epsilon }{3}} + C L^{3} \rho ^{2 + \gamma } + C L^{3} \rho ^{\frac{7}{3} - \frac{\gamma }{2}} + C_{\beta }L^{3} \rho ^{3 - \frac{10}{7}\gamma - \frac{\epsilon }{3}},\nonumber \\ \end{aligned}
(7.20)

where we replaced $$\rho _{\sigma }^{\text {r}}$$ with $$\rho _{\sigma }$$, thus giving rise to an error term $$O(\rho ^{2 + \frac{\epsilon }{3}})$$. Using Eqs. (6.52), (6.53), we get, for L large enough:

\begin{aligned} \frac{E_{L}(N_{\uparrow }, N_{\downarrow })}{L^{3}} \le \frac{3}{5}(6\pi ^{2})^{\frac{2}{3}} (\rho _{\uparrow }^{\frac{5}{3}} + \rho _{\downarrow }^{\frac{5}{3}}) + 8\pi a \rho _{\uparrow } \rho _{\downarrow } + C\rho ^{2 + \frac{\epsilon }{3}} + C \rho ^{2 + \gamma } + C\rho ^{\frac{7}{3} - \frac{\gamma }{2}} + C_{\beta } \rho ^{3 - \frac{10}{7}\gamma - \frac{\epsilon }{3}}. \end{aligned}

The optimal value of $$\epsilon$$ is $$\epsilon = \frac{3}{2} - \frac{15}{7}\gamma$$ (recall that we are assuming $$\gamma \le 1/3$$, so that $$\epsilon \ge 0$$). For $$\gamma \le 7/9$$, and for this choice of $$\epsilon$$, $$\rho ^{2 + \frac{\epsilon }{3}}$$ is smaller than $$\rho ^{\frac{7}{3} - \frac{\gamma }{2}}$$. Therefore:

\begin{aligned} \frac{E_{L}(N_{\uparrow }, N_{\downarrow })}{L^{3}} \le \frac{3}{5}(6\pi ^{2})^{\frac{2}{3}} (\rho _{\uparrow }^{\frac{5}{3}} + \rho _{\downarrow }^{\frac{5}{3}}) + 8\pi a \rho _{\uparrow } \rho _{\downarrow } + C \rho ^{2 + \gamma } + C \rho ^{\frac{7}{3} - \frac{\gamma }{2}}. \end{aligned}
(7.21)

Optimizing over $$\gamma$$, $$\gamma = 2/9$$, we finally get:

\begin{aligned} \frac{E_{L}(N_{\uparrow }, N_{\downarrow })}{L^{3}} \le \frac{3}{5}(6\pi ^{2})^{\frac{2}{3}} (\rho _{\uparrow }^{\frac{5}{3}} + \rho _{\downarrow }^{\frac{5}{3}}) + 8\pi a \rho _{\uparrow } \rho _{\downarrow } + C \rho ^{2 + \frac{2}{9}}. \end{aligned}
(7.22)

This concludes the proof of the upper bound, and of Theorem 2.1. $$\square$$