Pickl’s proof of the quantum mean-field limit and quantum Klimontovich solutions

Abstract

This paper discusses the mean-field limit for the quantum dynamics of N identical bosons in ${\textbf{R}}^3$ interacting via a binary potential with Coulomb-type singularity. Our approach is based on the theory of quantum Klimontovich solutions defined in Golse and Paul (Commun Math Phys 369:1021–1053, 2019) . Our first main result is a definition of the interaction nonlinearity in the equation governing the dynamics of quantum Klimontovich solutions for a class of interaction potentials slightly less general than those considered in Kato (Trans Am Math Soc 70:195–211, 1951). Our second main result is a new operator inequality satisfied by the quantum Klimontovich solution in the case of an interaction potential with Coulomb-type singularity. When evaluated on an initial bosonic pure state, this operator inequality reduces to a Gronwall inequality for a functional introduced in Pickl (Lett Math Phys 97:151-164, 2011), resulting in a convergence rate estimate for the quantum mean-field limit leading to the time-dependent Hartree equation.

1 Introduction and notation

In classical mechanics, the motion equations for a system of N identical point particles of mass m with positions $q_j(t)\in {\textbf{R}}^3$ and momenta $p_j(t)\in {\textbf{R}}^3$ for all $j=1,\ldots ,N$ are:

$$\begin{aligned} \left\{ \begin{aligned} \,\dot{q}_j(t)=&\tfrac{1}{m} p_j(t)={\nabla }_{p_j}H_N(p_1(t),\ldots ,q_N(t))\,, \\ \,\dot{p}_j(t)=&-\sum \nolimits _{\begin{array}{c} k=1\\ k\not =j \end{array}}^N{\nabla }V(q_j(t)-q_k(t))=-{\nabla }_{p_j}H_N(p_1(t),\ldots ,q_N(t))\,, \end{aligned} \right. \end{aligned}$$

(1)

where the N-particle classical Hamiltonian is

$$\begin{aligned} H_N(p_1,\ldots ,q_N):=\sum _{j=1}^N\tfrac{1}{2m}|p_j|^2+\sum _{1\le j<k\le N}V(q_j-q_k). \end{aligned}$$

Assuming that $V\in C^{1,1}({\textbf{R}}^3)$, this differential system has a unique global solution for all initial data. If V is even,¹ the phase-space empirical measure

$$\begin{aligned} \mu _N(t,\hbox {d}x\hbox {d}\xi ):=\tfrac{1}{N}\sum _{j=1}^N{\delta }_{q_j(t/N),p_j(t/N)}(\hbox {d}x\hbox {d}\xi )\,,\qquad Nm=1\,, \end{aligned}$$

(2)

is an exact, weak solution of the Vlasov equation

$$\begin{aligned} {\partial }_t\mu _N+\xi \cdot {\nabla }_x\mu _N-{\nabla }_x(V\star \mu _N(t))\cdot {\nabla }_\xi \mu _N=0 \end{aligned}$$

(3)

with self-consistent, mean-field potential

$$\begin{aligned} V\star _{x,\xi }\mu _N(t,x)=\tfrac{1}{N}\sum _{k=1}^NV(x-q_k(t)). \end{aligned}$$

This remarkable observation is due to Klimontovich, and solutions of the Vlasov equation (3) of the form (2) are referred to as “Klimontovich solutions”. They are discussed in detail in Klimontovich’s own book on the statistical mechanics of plasmas [14]. Thus, if $\mu _N(0)\rightarrow f^{in}\hbox {d}x\hbox {d}\xi$ weakly in the sense of probability measures as $N\rightarrow \infty$, where $f^{in}$ is a probability density on ${\textbf{R}}^3_x\times {\textbf{R}}^3_\xi$, one has

$$\begin{aligned} \mu _N(t,\hbox {d}x\hbox {d}\xi )\rightarrow f(t,x,\xi )\hbox {d}x\hbox {d}\xi \text { weakly } \text {in } \text {the }\text { sense }\text { of } \text { probability } \text { measures } \end{aligned}$$

for all $t\ge 0$ as $N\rightarrow \infty$, where f is the solution of the Vlasov equation

$$\begin{aligned} {\partial }_tf+\xi \cdot {\nabla }_xf-{\nabla }_x(V\star _{x,\xi }f(t,\cdot ,\cdot ))\cdot \nabla _\xi f=0\,,\qquad f{\big |}_{t=0}=f^{in}\,. \end{aligned}$$

(4)

Thus, the mean-field limit in classical mechanics is equivalent to the continuous dependence for the weak topology of probability measures of solutions of the Vlasov equation in terms of their initial data. See [4] for a proof of this result. For instance, the weak convergence of the initial data can be realized by a random choice of $(q_j(0),p_j(0))$, independent and identically distributed with distribution $f^{in}$.

The mean-field limit for bosonic systems in quantum mechanics has been formulated in different settings, by using the so-called BBGKY hierarchy [1, 2, 6, 22], or in the second quantization setting [20]. Interestingly, these techniques allow considering singular potentials such as the Coulomb potential, instead of $C^{1,1}$ potentials as in the classical case. (The mean-field limit with Coulomb potentials in classical mechanics is still an open problem at the time of this writing; see however [21] in the special case of monokinetic particle distributions. See also [9, 10] for potentials less singular than the Coulomb potential).

The quantum mean-field equation analogous to the Vlasov equation (4) is the (time-dependent) Hartree equation

$$\begin{aligned} i{\hbar }{\partial }_t\psi (t,x)=-\tfrac{1}{2}\hbar ^2{\Delta }_x\psi (t,x)+(V\star |\psi (t,\cdot )|^2)(x)\psi (t,x),\qquad \psi {\big |}_{t=0}=\psi ^{in}\,. \end{aligned}$$

(5)

In [15, 18], an original method, close to the second quantization approach in [20], but avoiding the rather heavy formalism of Fock spaces, was proposed and successfully applied to singular potentials including the Coulomb potential.

All these approaches noticeably differ from the classical setting used in [4] for lack of a quantum notion of phase-space empirical measures. However, a quantum analogue of the notion of phase-space empirical measure was recently proposed in [8], along with an equation analogous to (3) governing their evolution. This notion was used in [8] to prove the uniformity of the mean-field limit in the Planck constant $\hbar >0$. However, the discussion in [8] only considers regular potentials (specifically ${\partial }^{\alpha }V\in {\mathcal {F}}L^1({\textbf{R}}^d)$ for $|{\alpha }|\le 3+[d/2]$). Even writing the equation analogous to (3) satisfied by the quantum analogue of the phase-space empirical measure requires $V\in {\mathcal {F}}L^1({\textbf{R}}^d)$ in the setting of [8].

The purpose of the present paper is twofold:

• (a) to extend the formalism of quantum empirical measures considered in [8] to treat the case of singular potentials including the Coulomb potential, which is of particular interest for applications to atomic physics (see Theorem 3.1 in Sect. 3), and

• (b) to explain how the ideas in [15, 18] can be couched in terms of the formalism of quantum empirical measures defined in [8] (see Theorem 4.1 and Corollary 4.2 in Sect. 4).

Specifically, we prove an inequality between operators on the N-particle Hilbert space, of which the key estimates in [15, 18] leading to the quantum mean-field limit are straightforward consequences.

The next section briefly recalls only the essential part of [8] used in the sequel. The main results obtained in the present paper are Theorems 3.1 and 4.1 from Sects. 3 and 4, respectively. The proofs of these results are given in the subsequent sections.

All the discussions in this paper apply to the mean-field limit in the case of bosons only—in particular, the analysis proposed here is not uniform in $\hbar$, exactly as in [15, 18], and at variance with [8]. This precludes using the distinguished limit $\hbar \sim N^{-1/3}$ that is typical of the mean-field limit in the case of fermions.

2 Quantum Klimontovich solutions

Consider the quantum N-body Hamiltonian

$$\begin{aligned} {\mathcal {H}}_N:=\sum _{j=1}^N-\tfrac{1}{2}{\hbar }^2{\Delta }_{x_j}+\tfrac{1}{N}\sum _{1\le j<k\le N}V(x_j-x_k) \end{aligned}$$

(6)

on ${\mathfrak {H}}_N:={\mathfrak {H}}^{\otimes N}\simeq L^2({\textbf{R}}^{3N})$, where ${\mathfrak {H}}:=L^2({\textbf{R}}^3)$. Henceforth, it is assumed that V is a real-valued function such that ${\mathcal {H}}_N$ has a (unique) self-adjoint extension to ${\mathfrak {H}}_N$, still denoted by ${\mathcal {H}}_N$. A well-known sufficient condition for this to be true has been found by Kato (see condition (5) in [12]): there exists $R>0$ such that

$$\begin{aligned} \int _{|z|\le R}V(z)^2\hbox {d}z+{\text {ess sup}}_{|z|>R}|V(z)|<\infty \,. \end{aligned}$$

(7)

In particular, these conditions include the (repulsive) Coulomb potential in ${\textbf{R}}^3$. In fact, ${\mathcal {H}}_N$ has a self-adjoint extension to ${\mathfrak {H}}_N$ under a condition slightly more general than Kato’s original assumption recalled above:

$$\begin{aligned} V\in L^2({\textbf{R}}^3)+L^\infty ({\textbf{R}}^3) \end{aligned}$$

(8)

(see Theorem X.16 and Example 2 in [19], and Theorem V.9 with $m=1$ in [17]).

In the sequel, we adopt the notation in [8]. In particular, we set

$$\begin{aligned} J_kA:=I_{\mathfrak {H}}^{\otimes (k-1)}\otimes A\otimes I_{\mathfrak {H}}^{\otimes (N-k)}\,,\quad 1\le n\le N\,, \end{aligned}$$

(9)

and

$$\begin{aligned} {\mathcal {M}}_N^{in}:=\tfrac{1}{N}\sum _{k=1}^NJ_k\in {\mathcal {L}}({\mathcal {L}}({\mathfrak {H}}),{\mathcal {L}}({\mathfrak {H}}_N))\,. \end{aligned}$$

(10)

The dynamics of the morphism ${\mathcal {M}}_N^{in}$ is defined by conjugation with the N-particle dynamics as follows: for each $A\in {\mathcal {L}}({\mathfrak {H}})$,

$$\begin{aligned} {\mathcal {M}}_N(t)A:={\mathcal {U}}_N(t)^*({\mathcal {M}}_N^{in}A){\mathcal {U}}_N(t)\,,\quad \text { with }{\mathcal {U}}_N(t):=\exp (-it{\mathcal {H}}_N/{\hbar })\,. \end{aligned}$$

(11)

Since ${\mathcal {H}}_N$ is self-adjoint, $t\mapsto {\mathcal {U}}_N(t)$ is a unitary group by Stone’s theorem. The time-dependent morphism $t\mapsto {\mathcal {M}}_N(t)\in {\mathcal {L}}({\mathcal {L}}({\mathfrak {H}}),{\mathcal {L}}({\mathfrak {H}}_N))$ is henceforth referred to as the quantum Klimontovich solution.

Assume henceforth that V is even:

$$\begin{aligned} V(x)=V(-x)\,,\qquad x\in {\textbf{R}}^d\,. \end{aligned}$$

(12)

The first main result in [8] (Theorem 3.3) is that, if ${{\hat{V}}}\in L^1({\textbf{R}}^d)$, the quantum Klimontovich solution ${\mathcal {M}}_N(t)$ satisfies

$$\begin{aligned} i{\hbar }{\partial }_t{\mathcal {M}}_N(t)={\textbf{ad}}^*(K){\mathcal {M}}_N(t)-{\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))\,, \end{aligned}$$

(13)

where $K=-\tfrac{1}{2}{\hbar }^2{\Delta }$ is the quantum kinetic energy, and where

$$\begin{aligned} ({\textbf{ad}}^*(T)\Lambda )A:=-\Lambda ([T,A]) \end{aligned}$$

(14)

for each unbounded self-adjoint operator T on ${\mathfrak {H}}$, each $A\in {\mathcal {L}}({\mathfrak {H}})$ satisfying the condition $[T,A]\in {\mathcal {L}}({\mathfrak {H}})$, and each $\Lambda \in {\mathcal {L}}({\mathcal {L}}({\mathfrak {H}}),{\mathcal {L}}({\mathfrak {H}}_N))$. Moreover,

$$\begin{aligned} {\mathcal {C}}(V,\Lambda _1,\Lambda _2)(A):=\tfrac{1}{(2\pi )^d}\int _{{\textbf{R}}^d}{\hat{V}}({\omega })((\Lambda _1E_{\omega }^*)\Lambda _2(E_{\omega }A)-\Lambda _2(AE_{\omega })(\Lambda _1E_{\omega }^*))\hbox {d}{\omega }\end{aligned}$$

(15)

for each $A\in {\mathcal {L}}({\mathfrak {H}})$ and each $\Lambda _1,\Lambda _2\in {\mathcal {L}}({\mathcal {L}}({\mathfrak {H}}),{\mathcal {L}}({\mathfrak {H}}_N))$, where $E_{\omega }\in {\mathcal {L}}({\mathfrak {H}})$ is the operator defined by

$$\begin{aligned} (E_{\omega }\phi )(x):=e^{i{\omega }\cdot x}\phi (x)\quad \text { for each }\phi \in {\mathfrak {H}}\text { and }{\omega }\in {\textbf{R}}^d\,. \end{aligned}$$

(16)

Since the integrand of the right-hand side of (15) takes its values in the non-separable space ${\mathcal {L}}({\mathfrak {H}}_N)$, it is worth mentioning that this integral is a weak integral for the ultraweak topology in ${\mathcal {L}}({\mathfrak {H}}_N)$: see footnote 3 on p. 1032 in [8]. (We recall that the ultraweak topology on the algebra ${\mathcal {L}}(H)$ of bounded operators on the Hilbert space H is the topology defined by the family of seminorms

$$\begin{aligned} {\mathcal {L}}(H)\ni B\mapsto |{\text {trace}}_H(BT)|\in [0,+\infty ) \end{aligned}$$

as T runs through the set of trace-class operators on H: see for instance §4.6.10 in chapter 4 of the book [16].)

At variance with the classical case recalled in (3), the differential equation (13) satisfied by the quantum Klimontovich solution $t\mapsto {\mathcal {M}}_N(t)$ is not formally identical to the mean-field, time-dependent Hartree equation (5). The relation between (5) and (13) is explained in Theorem 3.5, the second main result in [8], recalled below.

If $\psi$ is a solution of the time-dependent Hartree equation (5) satisfying the normalization condition

$$\begin{aligned} \Vert \psi (t,\cdot )\Vert _{\mathfrak {H}}=1\qquad \text {for all }t\in {\textbf{R}}, \end{aligned}$$

the time-dependent morphism $t\mapsto {\mathcal {R}}(t)\in {\mathcal {L}}({\mathcal {L}}({\mathfrak {H}}),{\mathcal {L}}({\mathfrak {H}}_N))$ defined by the formula²

$$\begin{aligned} {\mathcal {R}}(t)A:=\langle \psi (t,\cdot )|A|\psi (t,\cdot )\rangle I_{{\mathfrak {H}}_N} \end{aligned}$$

is a solution of (13).

3 Extending the definition of ${\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))$ when $V\notin {\mathcal {F}}L^1({\textbf{R}}^3)$

Our first task is to extend the definition (15) of the term ${\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))$ to a more general class of potentials V, including the Coulomb potential in ${\textbf{R}}^3$.

Since

$$\begin{aligned} \begin{aligned}&{\mathcal {M}}_N(t)(E^*_{\omega }){\mathcal {M}}_N(t)(E_{\omega }|\phi \rangle \langle \phi |)-{\mathcal {M}}_N(t)(|\phi \rangle \langle \phi |E_{\omega }){\mathcal {M}}_N(t)(E^*_{\omega }) \\&\quad ={\mathcal {U}}_N(t)^*({\mathcal {M}}_N^{in}(E^*_{\omega }){\mathcal {M}}_N^{in}(E_{\omega }|\phi \rangle \langle \phi |)-{\mathcal {M}}_N^{in}(|\phi \rangle \langle \phi |E_{\omega }){\mathcal {M}}_N^{in}(E^*_{\omega })){\mathcal {U}}_N(t), \end{aligned} \end{aligned}$$

the idea is to define

$$\begin{aligned} \begin{aligned}&\langle \Phi _N^{in}|{\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))(|\phi \rangle \langle \phi |)|\Psi _N^{in}\rangle \\&\quad :=\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })\langle {\mathcal {U}}_N(t)\Phi _N^{in}|S_N[\phi ]({\omega })|{\mathcal {U}}_N(t)\Psi _N^{in}\rangle \hbox {d}{\omega }\end{aligned} \end{aligned}$$

for all $\Phi _N^{in},\Psi _N^{in}\in {\mathfrak {H}}_N$, where

$$\begin{aligned} S_N[\phi ]({\omega }):={\mathcal {M}}_N^{in}(E^*_{\omega }){\mathcal {M}}_N^{in}(E_{\omega }|\phi \rangle \langle \phi |)-{\mathcal {M}}_N^{in}(|\phi \rangle \langle \phi |E_{\omega }){\mathcal {M}}_N^{in}(E^*_{\omega }) \end{aligned}$$

and to take advantage of the decay of $S_N[\phi ]$ in ${\omega }$, assuming that $\phi$ is regular enough. Our argument does not use any regularity on $\Phi _N^{in}$ or $\Psi _N^{in}$. This is quite natural, since anyway Kato’s condition (8) on the interaction potential V does not entail higher than (Sobolev) $H^2$ regularity for ${\mathcal {U}}_N(t)\Phi _N^{in}$ or ${\mathcal {U}}_N(t)\Psi _N^{in}$, as observed in Note V.10 of [17].

Our first main result in this paper is the following result, leading to a definition of ${\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))(|\phi \rangle \langle \phi |)$ in the case of singular, Coulomb-like potentials V, and for bounded wave functions $\phi$. This theorem can be regarded as an extension to the case of singular, Coulomb-like potentials V of the formalism of quantum Klimontovich solutions in [8].

Theorem 3.1

Assume that V is a real-valued measurable function on ${\textbf{R}}^3$ satisfying the parity condition (12), and

$$\begin{aligned} V\in L^2({\textbf{R}}^3)+{\mathcal {F}}L^1({\textbf{R}}^3)\,. \end{aligned}$$

(17)

For each $\phi \in L^2\cap L^\infty ({\textbf{R}}^3)$ and each $\Psi _N\in {\mathfrak {H}}_N$, the function

$$\begin{aligned} {\omega }\mapsto \langle \Psi _N|S_N[\phi ]({\omega })|\Psi _N\rangle \text { belongs to }L^2\cap L^\infty ({\textbf{R}}^3). \end{aligned}$$

The interaction operator ${\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))(|\phi \rangle \langle \phi |)$ is defined by the formula

$$\begin{aligned} {\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))(|\phi \rangle \langle \phi |):=\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega }){\mathcal {U}}_N(t)^*S_N[\phi ]({\omega }){\mathcal {U}}_N(t)\textrm{d}{\omega }\end{aligned}$$

The integral on the right-hand side of the equality above is to be understood as a weak integral and defines

$$\begin{aligned} t\mapsto {\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))(|\phi \rangle \langle \phi |) \end{aligned}$$

as a continuous map from ${\textbf{R}}$ to ${\mathcal {L}}({\mathfrak {H}}_N)$ endowed with the ultraweak topology, which is moreover bounded on ${\textbf{R}}$ for the operator norm on ${\mathcal {L}}({\mathfrak {H}}_N)$.

Obviously, condition (17) is stronger than Kato’s condition (8). However, the repulsive Coulomb potential $z\mapsto 1/|z|$ in ${\textbf{R}}^3$ obviously satisfies (17), since its Fourier transform $\zeta \mapsto {C}/|\zeta |^2$ belongs to $L^1({\textbf{R}}^3)+L^2({\textbf{R}}^3)$. In particular, ${\mathcal {H}}_N$ has a self-adjoint extension to ${\mathfrak {H}}_N$ under condition (17)

Proof

Assuming that $\Psi _N^{in}\in {\mathfrak {H}}_N$, one has

$$\begin{aligned} {\mathcal {U}}_N(t)\Psi _N^{in}\in {\mathfrak {H}}_N\text { with }\Vert {\mathcal {U}}_N(t)\Psi _N^{in}\Vert _{{\mathfrak {H}}_N}=\Vert \Psi _N^{in}\Vert _{{\mathfrak {H}}_N}. \end{aligned}$$

Therefore, we henceforth forget the time dependence in $\Psi _N(t,\cdot )={\mathcal {U}}_N(t)\Psi _N^{in}$, which will be henceforth denoted $\Psi _N\equiv \Psi _N(x_1,\ldots ,x_N)$.

Observe first that

$$\begin{aligned} S_N[\phi ]({\omega })=\tfrac{1}{N^2}\sum _{1\le k\not =l\le N}(J_k(E^*_{\omega })J_l(E_{\omega }|\phi \rangle \langle \phi |)-J_l(|\phi \rangle \langle \phi |E_{\omega })J_k(E^*_{\omega })) \end{aligned}$$

since

$$\begin{aligned} \begin{aligned}&J_k(E^*_{\omega })J_k(E_{\omega }|\phi \rangle \langle \phi |)-J_k(|\phi \rangle \langle \phi |E_{\omega })J_k(E^*_{\omega }) \\&\quad =J_k(E^*_{\omega }E_{\omega }|\phi \rangle \langle \phi |)-J_k(|\phi \rangle \langle \phi |E_{\omega }E^*_{\omega }) \\&\quad =J_k(|\phi \rangle \langle \phi |)-J_k(|\phi \rangle \langle \phi |)=0&. \end{aligned} \end{aligned}$$

Without loss of generality, consider the term

$$\begin{aligned} \begin{aligned}&\langle \Psi _N|J_1(E_{\omega }^*)J_2(E_{\omega }|\phi \rangle \langle \phi |)|\Psi _N\rangle \\&\quad =\int _{{\textbf{R}}^6}\!e^{-i{\omega }\cdot (x_1\!-x_2)}\phi (x_2) \\&\quad \quad \times \left( \int _{{\textbf{R}}^{3N-6}}\!\!\!\overline{\Psi _N(x_1,x_2,Z)}\left( \int _{{\textbf{R}}^3}\!\Psi _N(x_1,y_2,Z)\overline{\phi (y_2)}\hbox {d}y_2\!\right) \hbox {d}Z\!\right) \hbox {d}x_1\hbox {d}x_2 \\&\quad ={\hat{F}}(-{\omega }) \end{aligned} \end{aligned}$$

where

$$\begin{aligned} F(X):=\int _{{\textbf{R}}^3}\!\phi (X\!+\!x_1)f(x_1,x_1+X,x_1)\hbox {d}x_1, \end{aligned}$$

with the notation

$$\begin{aligned} f(x_1,x_2,y_1):=\int _{{\textbf{R}}^{3N-6}}\!\!\overline{\Psi _N(x_1,x_2,Z)}\left( \int _{{\textbf{R}}^3}\!\Psi _N(y_1,y_2,Z)\overline{\phi (y_2)}\hbox {d}y_2\!\right) \hbox {d}Z. \end{aligned}$$

We shall prove that $F\in L^1({\textbf{R}}^3)\cap L^2({\textbf{R}}^3)$, so that ${\hat{F}}\in L^2({\textbf{R}}^3)\cap C_0({\textbf{R}}^3)$.

First

$$\begin{aligned} \begin{aligned} \int _{{\textbf{R}}^3}|F(X)|\hbox {d}X\le&\int _{{\textbf{R}}^6}|\phi (X+x_1)||f(x_1,x_1+X,x_1)|\hbox {d}x_1\hbox {d}X \\ =&\int _{{\textbf{R}}^3}\left( \int _{{\textbf{R}}^3}|\phi (x_2)||f(x_1,x_2,x_1)|\hbox {d}x_2\right) \hbox {d}x_2 \\ \le&\int _{{\textbf{R}}^{3N-3}}\left( \int _{{\textbf{R}}^3}|\phi (x_2)||\Psi _N(x_1,x_2,Z)|\hbox {d}x_2\right) \\&\times \left( \int _{{\textbf{R}}^3}|\Psi _N(x_1,y_2,Z)||\phi (y_2)|\hbox {d}y_2\!\right) \hbox {d}Z\hbox {d}x_1 \\ =&\int _{{\textbf{R}}^{3N-3}}\left( \int _{{\textbf{R}}^3}|\phi (x_2)||\Psi _N(x_1,x_2,Z)|\hbox {d}x_2\right) ^2\hbox {d}Z\hbox {d}x_1 \\ \le&\Vert \phi \Vert _{L^2({\textbf{R}}^3)}^2\int _{{\textbf{R}}^{3N-3}}\int _{{\textbf{R}}^3}|\Psi _N(x_1,x_2,Z)|^2\hbox {d}x_2\hbox {d}Z\hbox {d}x_1 \\ =&\Vert \phi \Vert _{L^2({\textbf{R}}^3)}^2\Vert \Psi _N\Vert ^2_{L^2({\textbf{R}}^{3N})}<\infty , \end{aligned} \end{aligned}$$

where the last inequality is the Cauchy–Schwarz inequality for the inner integral.

On the other hand,

$$\begin{aligned} \int _{{\textbf{R}}^3}|F(X)|^2\hbox {d}X\le \Vert \phi \Vert ^2_{L^\infty ({\textbf{R}}^3)}\int _{{\textbf{R}}^3}\left( \int _{{\textbf{R}}^3}|f(x_1,x_1+X,x_1)|\hbox {d}x_1\right) ^2\hbox {d}X, \end{aligned}$$

and

$$\begin{aligned} \int _{{\textbf{R}}^3}|f(x_1,x_1+X,x_1)|\hbox {d}x_1\le \int _{{\textbf{R}}^{3N-3}}|\Psi _N(x_1,x_1+X,Z)|\Phi _N(x_1,Z)\hbox {d}Z\hbox {d}x_1, \end{aligned}$$

with

$$\begin{aligned} \Phi (x_1,Z):=\int _{{\textbf{R}}^3}|\Psi _N(x_1,y_2,Z)||\phi (y_2)|\hbox {d}y_2, \end{aligned}$$

so that

$$\begin{aligned} \Phi _N(x_1,Z_N)^2\le \Vert \phi \Vert ^2_{L^2({\textbf{R}}^3)}\int _{{\textbf{R}}^3}|\Psi _N(x_1,y_2,Z)|^2\hbox {d}y_2, \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \int _{{\textbf{R}}^{3N-3}}\Phi _N(x_1,Z)^2\hbox {d}Z\hbox {d}x_1\le&\Vert \phi \Vert ^2_{L^2({\textbf{R}}^3)}\int _{{\textbf{R}}^3}|\Psi _N(x_1,y_2,Z)|^2\hbox {d}x_1\hbox {d}y_2\hbox {d}Z \\ =&\Vert \phi \Vert ^2_{L^2({\textbf{R}}^3)}\Vert \Psi _N\Vert ^2_{L^2({\textbf{R}}^{3N})}. \end{aligned} \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned}&\left( \int _{{\textbf{R}}^3}|f(x_1,x_1+X,x_1)|\hbox {d}x_1\right) ^2\\&\quad \le \left( \int _{{\textbf{R}}^{3N-3}}|\Psi _N(x_1,x_1+X,Z)|\Phi _N(x_1,Z)\hbox {d}Z\hbox {d}x_1\right) ^2 \\&\quad \le \int _{{\textbf{R}}^{3N-3}}|\Psi _N(x_1,x_1+X,Z)|^2\hbox {d}Z\hbox {d}x_1\int _{{\textbf{R}}^{3N-3}}\Phi _N(x_1,Z)^2\hbox {d}Z\hbox {d}x_1 \\&\quad \le \int _{{\textbf{R}}^{3N-3}}|\Psi _N(x_1,x_1+X,Z)|^2\hbox {d}Z\hbox {d}x_1\Vert \phi \Vert ^2_{L^2({\textbf{R}}^3)}\Vert \Psi _N\Vert ^2_{L^2({\textbf{R}}^{3N})}, \end{aligned} \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned}&\int _{{\textbf{R}}^3}\left( \int _{{\textbf{R}}^3}|f(x_1,x_1+X,x_1)|\hbox {d}x_1\right) ^2\hbox {d}X \\&\quad \le \Vert \phi \Vert ^2_{L^2({\textbf{R}}^3)}\Vert \Psi _N\Vert ^2_{L^2({\textbf{R}}^{3N})}\int _{{\textbf{R}}^3}\int _{{\textbf{R}}^{3N-3}}|\Psi _N(x_1,x_1+X,Z)|^2\hbox {d}Z\hbox {d}x_1\hbox {d}X \\&\quad =\Vert \phi \Vert ^2_{L^2({\textbf{R}}^3)}\Vert \Psi _N\Vert ^2_{L^2({\textbf{R}}^{3N})}\int _{{\textbf{R}}^3}\int _{{\textbf{R}}^{3N-3}}|\Psi _N(x_1,x_2,Z)|^2\hbox {d}Z\hbox {d}x_1\hbox {d}x_2 \\&\quad =\Vert \phi \Vert ^2_{L^2({\textbf{R}}^3)}\Vert \Psi _N\Vert ^4_{L^2({\textbf{R}}^{3N})}. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{{\textbf{R}}^3}|F(X)|^2\hbox {d}X\le \Vert \phi \Vert ^2_{L^\infty ({\textbf{R}}^3)}\Vert \phi \Vert ^2_{L^2({\textbf{R}}^3)}\Vert \Psi _N\Vert ^4_{L^2({\textbf{R}}^{3N})}<\infty \end{aligned}$$

so that ${\omega }\mapsto {\hat{F}}(-{\omega })$ belongs to $L^2({\textbf{R}}^d)$ by Plancherel’s theorem. Hence, for each $k\not =l\in \{1,\ldots ,N\}$, one has

$$\begin{aligned} {\hat{V}}\in L^1({\textbf{R}}^d)+L^2({\textbf{R}}^d)\implies \left\{ \begin{aligned}{}&\int _{{\textbf{R}}^3}|{\hat{V}}({\omega })||\langle \Psi _N|J_k(E_{\omega }^*)J_l(E_{\omega }|\phi \rangle \langle \phi |)|\Psi _N\rangle | \hbox {d}{\omega }<\infty ,\\&\int _{{\textbf{R}}^3}|{\hat{V}}({\omega })||\langle \Psi _N|J_l(|\phi \rangle \langle \phi |E_{\omega })J_k(E_{\omega }^*)|\Psi _N\rangle | \hbox {d}{\omega }<\infty .\end{aligned}\right. \end{aligned}$$

Hence,

$$\begin{aligned} (t,{\omega })\mapsto {\hat{V}}({\omega })\langle {\mathcal {U}}_N(t)\Psi _N^{in}|S_N[\phi ]({\omega })|{\mathcal {U}}_N(t)\Psi _N^{in}\rangle \text { belongs to }C_b({\textbf{R}}_t,L^1({\textbf{R}}^3_{\omega })). \end{aligned}$$

Since $S_N[\phi ]({\omega })^*=-S_N[\phi ]({\omega })\in {\mathcal {L}}({\mathfrak {H}}_N)$ for each ${\omega }\in {\textbf{R}}^3$ and ${\hat{V}}$ is even because of (12), the formula

$$\begin{aligned} \begin{aligned}&\langle \Psi _N^{in}|{\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))(|\phi \rangle \langle \phi |)|\Psi _N^{in}\rangle \\&\quad :=\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })\langle {\mathcal {U}}_N(t)^*\Psi _N^{in}|S_N[\phi ]({\omega })|{\mathcal {U}}_N(t)^*\Psi _N^{in}\rangle \hbox {d}{\omega }\end{aligned} \end{aligned}$$

defines

$$\begin{aligned} {\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))(|\phi \rangle \langle \phi |)=-{\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))(|\phi \rangle \langle \phi |)^*\in {\mathcal {L}}({\mathfrak {H}}_N) \end{aligned}$$

for each $t\in {\textbf{R}}$ by polarization, and the function

$$\begin{aligned} t\mapsto {\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t)) \end{aligned}$$

is bounded on ${\textbf{R}}$ with values in ${\mathcal {L}}({\mathfrak {H}}_N)$ for the norm topology, and continuous on ${\textbf{R}}$ with values in ${\mathcal {L}}({\mathfrak {H}}_N)$ endowed with the weak operator topology, and therefore, for the ultraweak topology, since the weak operator and the ultraweak topologies coincide on norm bounded subsets of ${\mathcal {L}}({\mathfrak {H}}_N)$. This last point is Proposition 4.6.14 in chapter 4 of [16], where the weak operator and the ultraweak topologies are referred to as the weak and the ${\sigma }$-weak topologies, respectively. $\square$

Remark

In the sequel, we shall also need to consider terms of the form

$$\begin{aligned} \begin{aligned} (I):=&\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })\langle \Psi _N|J_1(E_{\omega }^*|\phi \rangle \langle \phi |)J_2(E_{\omega }|\phi \rangle \langle \phi |)|\Psi _N\rangle \hbox {d}{\omega }\\ (II):=&\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })\langle \Psi _N|J_1(|\phi \rangle \langle \phi |E_{\omega }^*)J_2(E_{\omega }|\phi \rangle \langle \phi |)|\Psi _N\rangle \hbox {d}{\omega }\\ (III):=&\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })\langle \Psi _N|J_1(|\phi \rangle \langle \phi |E_{\omega }^*|\phi \rangle \langle \phi |)J_2(AE_{\omega }B)|\Psi _N\rangle \hbox {d}{\omega }\end{aligned} \end{aligned}$$

where $A,B\in {\mathcal {L}}({\mathfrak {H}})$.

The term (III) is the easiest of all. Indeed,

$$\begin{aligned} \begin{aligned} (III)=&\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })\widehat{|\phi |^2}({\omega })\langle \Psi _N|J_1(|\phi \rangle \langle \phi |)J_2(AE_{\omega }B)|\Psi _N\rangle \hbox {d}{\omega }\\ =&\langle \Psi _N|J_1(|\phi \rangle \langle \phi |)J_2(A(V\star |\phi |^2)B)|\Psi _N\rangle \end{aligned} \end{aligned}$$

and since $V\in L^2({\textbf{R}}^3)+C_b({\textbf{R}}^3)$ while $\phi \in L^1\cap L^\infty ({\textbf{R}}^3)$, one has $V\star |\phi |^2\in C_b({\textbf{R}}^3)$, so that $A(V\star |\phi |^2)B\in {\mathcal {L}}({\mathfrak {H}})$.

The terms (I) and (II) are slightly more delicate, but can be treated by the same method already used in the proof of the theorem above. First,

$$\begin{aligned} \langle \Psi _N|J_1(E_{\omega }^*|\phi \rangle \langle \phi |)J_2(E_{\omega }|\phi \rangle \langle \phi |)|\Psi _N\rangle ={\hat{F}}_1({\omega }), \end{aligned}$$

with

$$\begin{aligned} \begin{aligned} F_1(Y):=&\int _{{\textbf{R}}^{3N-6}}A_1(Y,Z)A_2(Z)\hbox {d}Z, \\ A_1(Y,Z):=&\int _{{\textbf{R}}^3}\phi (X+\tfrac{Y}{2})\phi (X-\tfrac{Y}{2})\overline{\Psi _N(X+\tfrac{Y}{2},X-\tfrac{Y}{2},Z)}\hbox {d}X, \\ A_2(Z):=&\int _{{\textbf{R}}^6}\!\Psi _N(y_1,y_2,Z)\overline{\phi (y_1)\phi (y_2)}\hbox {d}y_1\hbox {d}y_2, \end{aligned} \end{aligned}$$

so that

$$\begin{aligned} (I)=\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega }){\hat{F}}_1({\omega })\hbox {d}{\omega }. \end{aligned}$$

Then,

$$\begin{aligned} \begin{aligned}&\left( \int _{{\textbf{R}}^3}|F_1(Y)|\hbox {d}Y\right) ^2 \\&\quad \le \Vert A_2\Vert ^2_{L^2({\textbf{R}}^{3N-6})}\int _{{\textbf{R}}^{3N-6}}\left( \int _{{\textbf{R}}^3}|A_1(Y,Z)|\hbox {d}Y\right) ^2\hbox {d}Z\le \Vert A_2\Vert ^2_{L^2({\textbf{R}}^{3N-6})} \\&\quad \quad \times \int _{{\textbf{R}}^{3N-6}}\left( \int _{{\textbf{R}}^6}|\phi (X+\tfrac{Y}{2})||\phi (X-\tfrac{Y}{2})||\Psi _N(X+\tfrac{Y}{2},X-\tfrac{Y}{2},Z)|\hbox {d}X\hbox {d}Y\right) ^2\hbox {d}Z \\&\quad \le \Vert A_2\Vert ^2_{L^2({\textbf{R}}^{3N-6})}\int _{{\textbf{R}}^6}|\phi (X+\tfrac{Y}{2})|^2|\phi (X-\tfrac{Y}{2})|^2\hbox {d}X\hbox {d}Y \\&\quad \quad \times \int _{{\textbf{R}}^{3N}}|\Psi _N(X+\tfrac{Y}{2},X-\tfrac{Y}{2},Z)|^2\hbox {d}X\hbox {d}Y\hbox {d}Z \\&\quad =\Vert A_2\Vert ^2_{L^2({\textbf{R}}^{3N-6})}\int _{{\textbf{R}}^6}|\phi (x_1)|^2|\phi (x_2)|^2\hbox {d}x_1\hbox {d}x_2 \\&\quad \quad \times \int _{{\textbf{R}}^{3N}}|\Psi _N(x_1,x_2,Z)|^2\hbox {d}x_1\hbox {d}x_2\hbox {d}Z \\&\quad =\Vert A_2\Vert ^2_{L^2({\textbf{R}}^{3N-6})}\Vert \phi \Vert _{L^2({\textbf{R}}^3)}^4\Vert \Psi _N\Vert _{{\mathfrak {H}}_N}^2. \end{aligned} \end{aligned}$$

Besides

$$\begin{aligned} \begin{aligned} \Vert A_2\Vert ^2_{L^2({\textbf{R}}^{3N-6})}\le&\int _{{\textbf{R}}^6}|\phi (y_1)|^2|\phi (y_2)|^2\hbox {d}y_1\hbox {d}y_2\int _{{\textbf{R}}^{3N-6}}|\Psi _N(y_1,y_2,Z)|^2\hbox {d}y_1\hbox {d}y_2\hbox {d}Z \\ =&\Vert \phi \Vert _{L^2({\textbf{R}}^3)}^4\Vert \Psi _N\Vert _{{\mathfrak {H}}_N}^2, \end{aligned} \end{aligned}$$

so that

$$\begin{aligned} \Vert F_1\Vert _{L^1({\textbf{R}}^3)}\le \Vert \phi \Vert _{L^2({\textbf{R}}^3)}^4\Vert \Psi _N\Vert _{{\mathfrak {H}}_N}^2<\infty . \end{aligned}$$

On the other hand,

$$\begin{aligned} \int _{{\textbf{R}}^3}|F_1(Y)|^2\hbox {d}Y\le \Vert A_2\Vert ^2_{L^2({\textbf{R}}^{3N-6})}\Vert A_1\Vert ^2_{L^2({\textbf{R}}^{3N-3})}, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} \Vert A_1\Vert ^2_{L^2({\textbf{R}}^{3N-3})}\le&\sup _{Y\in {\textbf{R}}^3}\int _{{\textbf{R}}^3}|\phi (X+\tfrac{Y}{2})|^2|\phi (X-\tfrac{Y}{2})|^2\hbox {d}X \\&\times \int _{{\textbf{R}}^{3N}}|\Psi _N(X+\tfrac{Y}{2},X-\tfrac{Y}{2},Z)|^2\hbox {d}X\hbox {d}Y\hbox {d}Z \\ \le&\Vert \phi \Vert _{L^4({\textbf{R}}^3)}^4\Vert \Psi _N\Vert _{{\mathfrak {H}}_N}^2, \end{aligned} \end{aligned}$$

so that

$$\begin{aligned} \int _{{\textbf{R}}^3}|F_1(Y)|^2\hbox {d}Y\le \Vert \phi \Vert _{L^4({\textbf{R}}^3)}^4\Vert \phi \Vert _{L^2({\textbf{R}}^3)}^4\Vert \Psi _N\Vert _{{\mathfrak {H}}_N}^4<\infty . \end{aligned}$$

Thus, we have proved that $F_1\in L^1\cap L^2({\textbf{R}}^d)$, and since ${\hat{V}}\in L^2({\textbf{R}}^d)+L^1({\textbf{R}}^d)$, the product ${\hat{V}}{\hat{F}}\in L^1({\textbf{R}}^d)$, which leads to a definition of (I).

The case of (II) is essentially similar. Observe that

$$\begin{aligned} \langle \Psi _N|J_1(|\phi \rangle \langle \phi |E_{\omega }^*)J_2(E_{\omega }|\phi \rangle \langle \phi |)|\Psi _N\rangle =\int _{{\textbf{R}}^{3N-6}}{\hat{F}}_2({\omega },Z){\hat{F}}_3({\omega },Z)\hbox {d}Z, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} F_2(y_1,Z):=&\,\overline{\phi (y_1)}\int _{{\textbf{R}}^3}\overline{\phi (y_2)}\Psi _N(y_1,y_2,Z)\hbox {d}y_2, \\ F_3(x_2,Z):=&\,\phi (x_2)\int _{{\textbf{R}}^3}\phi (x_1)\overline{\Psi _N(x_1,x_2,Z)}\hbox {d}x_1. \end{aligned} \end{aligned}$$

And

$$\begin{aligned} (II):=\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })\left( \int _{{\textbf{R}}^{3N-6}}{\hat{F}}_2({\omega },Z){\hat{F}}_3({\omega },Z)\hbox {d}Z\right) \hbox {d}{\omega }. \end{aligned}$$

Observe that

$$\begin{aligned} \begin{aligned} \left( \int |F_2(y_1,Z)|\hbox {d}y_1\right) ^2\le&\left( \int _{{\textbf{R}}^6}|\phi (y_1)||\phi (y_2)||\Psi _N(y_1,y_2,Z)|\hbox {d}y_1\hbox {d}y_2\right) ^2 \\ \le&\Vert \phi \Vert _{L^2({\textbf{R}}^3)}^4\int _{{\textbf{R}}^6}|\Psi _N(y_1,y_2,Z)|^2\hbox {d}y_1\hbox {d}y_2, \end{aligned} \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned}&\sup _{{\omega }\in {\textbf{R}}^3}\left| \int _{{\textbf{R}}^{3N-6}}{\hat{F}}_2({\omega },Z){\hat{F}}_3({\omega },Z)\hbox {d}Z\right| ^2 \\&\quad \le \left( \int _{{\textbf{R}}^{3N-6}}\sup _{{\omega }\in {\textbf{R}}^3}|{\hat{F}}_2({\omega },Z)|\sup _{{\omega }\in {\textbf{R}}^3}|{\hat{F}}_3({\omega },Z)|\hbox {d}Z\right) ^2 \\&\quad \le \int _{{\textbf{R}}^{3N-6}}\sup _{{\omega }\in {\textbf{R}}^3}|{\hat{F}}_2({\omega },Z)|^2\hbox {d}Z\int _{{\textbf{R}}^{3N-6}}\sup _{{\omega }\in {\textbf{R}}^3}|{\hat{F}}_3({\omega },Z)|^2\hbox {d}Z \\&\quad \le \int _{{\textbf{R}}^{3N-6}}\left( \int |F_2(y_1,Z)|\hbox {d}y_1\right) ^2\hbox {d}Z\int _{{\textbf{R}}^{3N-6}}\left( \int |F_3(x_2,Z)|\hbox {d}x_2\right) ^2\hbox {d}Z \\&\quad \le \Vert \phi \Vert _{L^2({\textbf{R}}^3)}^8\Vert \Psi _N\Vert _{{\mathfrak {H}}_N}^4, \end{aligned} \end{aligned}$$

while

$$\begin{aligned} \int _{{\textbf{R}}^{3N-6}}\left( \int |F_2(y_1,Z)|\hbox {d}y_1\right) ^2\hbox {d}Z\le \Vert \phi \Vert _{L^2({\textbf{R}}^3)}^4\Vert \Psi _N\Vert _{{\mathfrak {H}}_N}^2, \end{aligned}$$

with a similar conclusion for $F_3$. On the other hand

$$\begin{aligned} \begin{aligned} \int _{{\textbf{R}}^3}|F_2(y_1,Z)|^2\hbox {d}y_1&\le \int _{{\textbf{R}}^3}|\phi (y_1)|^2\left( \int _{{\textbf{R}}^3}\overline{\phi (y_2)}\Psi _N(y_1,y_2,Z)\hbox {d}y_2\right) ^2\hbox {d}y_1 \\&\le \Vert \phi \Vert _{L^2({\textbf{R}}^3)}^2\int _{{\textbf{R}}^3}|\phi (y_1)|^2\left( \int _{{\textbf{R}}^3}|\Psi _N(y_1,y_2,Z)|^2\hbox {d}y_2\right) \hbox {d}y_1 \\&\le \Vert \phi \Vert ^2_{L^2({\textbf{R}}^3)}\Vert \phi \Vert ^2_{L^\infty ({\textbf{R}}^3)}\int _{{\textbf{R}}^6}|\Psi _N(y_1,y_2,Z)|^2\hbox {d}y_1\hbox {d}y_2, \end{aligned} \end{aligned}$$

so that

$$\begin{aligned}&\int _{{\textbf{R}}^3}\left| \int _{{\textbf{R}}^{3N-6}}{\hat{F}}_2({\omega },Z){\hat{F}}_3({\omega },Z)\hbox {d}Z\right| ^2\hbox {d}{\omega }\\&\quad \le \int _{{\textbf{R}}^3}\int _{{\textbf{R}}^{3N-6}}|{\hat{F}}_2({\omega },Z)|^2\left( \int _{{\textbf{R}}^{3N-6}}|{\hat{F}}_3({\omega },Z)|^2\hbox {d}Z\right) \hbox {d}Z\hbox {d}{\omega }\\&\quad \le \sup _{{\omega }\in {\textbf{R}}^3}\int _{{\textbf{R}}^{3N-6}}|{\hat{F}}_3({\omega },Z)|^2\hbox {d}Z\int _{{\textbf{R}}^{3N-6}}\int _{{\textbf{R}}^3}|{\hat{F}}_2({\omega },Z)|^2\hbox {d}Z\hbox {d}{\omega }\\&\quad \le \int _{{\textbf{R}}^{3N-6}}\sup _{{\omega }\in {\textbf{R}}^3}|{\hat{F}}_3({\omega },Z)|^2\hbox {d}Z\int _{{\textbf{R}}^{3N-6}}(2\pi )^3\left( \int _{{\textbf{R}}^3}|F_2(y_1,Z)|^2\hbox {d}y_1\right) \hbox {d}Z \\&\quad \le (2\pi )^3\int _{{\textbf{R}}^{3N-6}}\left( \int _{{\textbf{R}}^3}|{\hat{F}}_3(x_2,Z)|\hbox {d}x_2\right) ^2\hbox {d}Z\int _{{\textbf{R}}^{3N-6}}\int _{{\textbf{R}}^3}|F_2(y_1,Z)|^2\hbox {d}y_1\hbox {d}Z \\&\quad \le (2\pi )^3\Vert \phi \Vert _{L^2({\textbf{R}}^3)}^6\Vert \phi \Vert _{L^\infty ({\textbf{R}}^3)}^2\Vert \Psi _N\Vert _{{\mathfrak {H}}_N}^4. \end{aligned}$$

Therefore, the map

$$\begin{aligned} {\omega }\mapsto \int _{{\textbf{R}}^{3N-6}}{\hat{F}}_2({\omega },Z){\hat{F}}_3({\omega },Z)\hbox {d}Z \end{aligned}$$

belongs to $L^2\cap L^\infty ({\textbf{R}}^3)$. Since ${\hat{V}}\in L^2({\textbf{R}}^3)+L^1({\textbf{R}}^3)$, this implies that

$$\begin{aligned} {\omega }\mapsto {\hat{V}}\int _{{\textbf{R}}^{3N-6}}{\hat{F}}_2({\omega },Z){\hat{F}}_3({\omega },Z)\hbox {d}Z \end{aligned}$$

belongs to $L^1({\textbf{R}}^3)$, thereby leading to a definition of (II).

4 An operator inequality. Application to the mean-field limit

First consider the Cauchy problem for the time-dependent Hartree equation (5). Assuming that the potential V satisfies (8) and (12), for each $\phi ^{in}\in H^2({\textbf{R}}^3)$, there exists a unique solution $\phi \in C({\textbf{R}},H^2({\textbf{R}}^3))$ of (5) by Theorems 1.4 and 1.3 of [11].

Pickl’s key idea in his proof of the mean-field limit in quantum mechanics is to consider the following functional (see Definition 2.2 and formula (6) in [18], with the choice $n(k):=k/N$, in the notation of [18]):

$$\begin{aligned} {\alpha }_N(\Psi _N,\psi ):=\bigg \langle \Psi _N\Big |\frac{1}{N}\sum _{k=1}^NJ_k(I_{\mathfrak {H}}-|\psi \rangle \langle \psi |)\Big |\Psi _N\bigg \rangle =\langle \Psi _N|{\mathcal {M}}_N^{in}(I_{\mathfrak {H}}-|\psi \rangle \langle \psi |)|\Psi _N\rangle \end{aligned}$$

for all $\Psi _N\in {\mathfrak {H}}_N$ and $\psi \in {\mathfrak {H}}$.

Assuming that $\psi \equiv \psi (t,x)$ is a solution of (5) while $\Psi _N(t,\cdot ):={\mathcal {U}}_N(t)\Psi _N^{in}$, Pickl studies in section 2.1 of [18] the time-dependent function $t\mapsto {\alpha }_N(\Psi _N(t,\cdot ),\psi (t,\cdot ))$, and proves that it satisfies some Gronwall inequality.

Observe first that Pickl’s functional ${\alpha }_N(\Psi _N(t,\cdot ),\psi (t,\cdot ))$ can be recast in terms of the quantum Klimontovich solution ${\mathcal {M}}_N(t)$ as follows

$$\begin{aligned} \begin{aligned} {\alpha }_N({\mathcal {U}}_N(t)\Psi ^{in}_N,\psi (t,\cdot ))=&\langle {\mathcal {U}}_N(t)\Psi ^{in}_N|{\mathcal {M}}_N^{in}(I_{\mathfrak {H}}-|\psi (t,\cdot )\rangle \langle \psi (t,\cdot )|)|{\mathcal {U}}_N(t)\Psi _N^{in}\rangle \\ =&\langle \Psi ^{in}_N|\mathcal U_N(t)^*\left( {\mathcal {M}}_N^{in}(I_{\mathfrak {H}}-|\psi (t,\cdot )\rangle \langle \psi (t,\cdot )|)\right) {\mathcal {U}}_N(t)|\Psi ^{in}_N\rangle \\ =&\langle \Psi _N^{in}|{\mathcal {M}}_N(t)(I_{\mathfrak {H}}-|\psi (t,\cdot )\rangle \langle \psi (t,\cdot )|)|\Psi ^{in}_N\rangle \,. \end{aligned} \end{aligned}$$

(18)

This identity suggests therefore to deduce from (13) and (5) the expression of

$$\begin{aligned} \tfrac{\hbox {d}}{\hbox {d}t}{\mathcal {M}}_N(t)(I_{\mathfrak {H}}-|\psi (t,\cdot )\rangle \langle \psi (t,\cdot )|) \end{aligned}$$

in terms of the interaction operator ${\mathcal {C}}$ defined in (15).

This is done in the first part of the next theorem, which is our second main result in this paper.

Theorem 4.1

Assume that the (real-valued) interaction potential V, viewed as an (unbounded) multiplication operator acting on ${\mathfrak {H}}:=L^2({\textbf{R}}^3)$, satisfies the parity condition (12) and (17).

Let $\psi ^{in}\in H^2({\textbf{R}}^3)$ satisfy $\Vert \psi ^{in}\Vert _{{\mathfrak {H}}}=1$, let $\psi$ be the solution of the Cauchy problem (5) for the time-dependent Hartree equation, and set

$$\begin{aligned} R(t):=|\psi (t,\cdot )\rangle \langle \psi (t,\cdot )|\,,\quad \text { and }\quad P(t):=I_{\mathfrak {H}}-R(t)\,. \end{aligned}$$

(19)

Then,

• (1) the N-body quantum Klimontovich solution $t\mapsto {\mathcal {M}}_N(t)$ satisfies

  $$\begin{aligned} i{\hbar }{\partial }_t({\mathcal {M}}_N(t)(P(t)))={\mathcal {C}}(V,{\mathcal {M}}_N(t)-{\mathcal {R}}(t),{\mathcal {M}}_N(t))(R(t)), \end{aligned}$$

  where

  $$\begin{aligned} {\mathcal {R}}(t)A:=\langle \psi (t,\cdot )|A|\psi (t,\cdot )\rangle I_{{\mathfrak {H}}_N}={\text {trace}}_{{\mathfrak {H}}}(R(t)A)I_{{\mathfrak {H}}_N}; \end{aligned}$$

• (2) the operator ${\mathcal {C}}(V,{\mathcal {M}}_N(t)-{\mathcal {R}}(t),{\mathcal {M}}_N(t))(P(t))$ is skew-adjoint on ${\mathfrak {H}}_N$ and satisfies the operator inequality

  $$\begin{aligned} \pm i{\mathcal {C}}(V,{\mathcal {M}}_N(t)-{\mathcal {R}}(t),{\mathcal {M}}_N(t))(R(t))\le 6L(t)\left( {\mathcal {M}}_N(t)(P(t))+\tfrac{2}{N}I_{{\mathfrak {H}}_N}\right) , \end{aligned}$$

where³

$$\begin{aligned} L(t):=2\max (1,C_S)\Vert V\Vert _{L^2({\textbf{R}}^3)+L^\infty ({\textbf{R}}^3)}\Vert \psi (t,\cdot )\Vert _{H^2({\textbf{R}}^3)}\,, \end{aligned}$$

(20)

and where $C_S$ is the norm of the Sobolev embedding $H^2({\textbf{R}}^3)\subset L^\infty ({\textbf{R}}^3)$.

The operator inequality for quantum Klimontovich solutions in the case of potentials with Coulomb-type singularity obtained in part (2) of Theorem 4.1 can be thought of as the reformulation of Pickl’s argument in terms of the quantum Klimontovich solution ${\mathcal {M}}_N(t)$.

Indeed, we deduce from parts (1) and (2) in Theorem 4.1 the operator inequality

$$\begin{aligned} \tfrac{\hbox {d}}{\hbox {d}t}{\mathcal {M}}_N(t)(P(t))\le \tfrac{6L(t)}{\hbar }\left( {\mathcal {M}}_N(t)(P(t))+\tfrac{2}{N}I_{{\mathfrak {H}}_N}\right) \,. \end{aligned}$$

(21)

Then, evaluating both sides of this inequality on the initial N-particle state $\Psi _N^{in}$ and taking into account the identity (18) lead to the Gronwall inequality

$$\begin{aligned} \tfrac{\hbox {d}}{\hbox {d}t}{\alpha }_N({\mathcal {U}}_N(t)\Psi ^{in}_N,\psi (t,\cdot ))\le \tfrac{6L(t)}{\hbar }\left( {\alpha }_N({\mathcal {U}}_N(t)\Psi ^{in}_N,\psi (t,\cdot ))+\tfrac{2}{N}\right) \end{aligned}$$

satisfied by Pickl’s functional ${\alpha }_N({\mathcal {U}}_N(t)\Psi ^{in}_N,\psi (t,\cdot ))$. This last inequality corresponds to inequality (11) and Lemma 3.2 in [18].

In the sequel, we shall denote by ${\mathcal {L}}^p({\mathfrak {H}})$ for $p\ge 1$ the Schatten two-sided ideal of ${\mathcal {L}}({\mathfrak {H}})$ consisting of operators T such that

$$\begin{aligned} \Vert T\Vert _p:=\left( {\text {trace}}_{\mathfrak {H}}((T^*T)^{p/2})\right) ^{1/p}<\infty . \end{aligned}$$

In particular, ${\mathcal {L}}^1({\mathfrak {H}})$ is the set of trace-class operators on ${\mathfrak {H}}$ and $\Vert \cdot \Vert _1$ the trace norm, while ${\mathcal {L}}^2({\mathfrak {H}})$ is the set of Hilbert–Schmidt operators on ${\mathfrak {H}}$ and $\Vert \cdot \Vert _2$ the Hilbert–Schmidt norm.

Corollary 4.2

Under the same assumptions and with the same notations as in Theorem 4.1, consider the N-body wave function $\Psi _N(t,\cdot ):={\mathcal {U}}_N(t)(\psi ^{in})^{\otimes N}$, and the N-body density operator $F_N(t):=|\Psi _N(t,\cdot )\rangle \langle \Psi _N(t,\cdot )|$. For each $m=1,\ldots ,N$, the m-particle reduced density operator $F_{N:m}(t)$, defined by the identity

$$\begin{aligned} {\text {trace}}_{{\mathfrak {H}}_m}(F_{N:m}(t)A_1\otimes \ldots \otimes A_m)=\langle \Psi _N(t,\cdot )|A_1\otimes \ldots \otimes A_m\otimes I_{{\mathfrak {H}}_{N-m}}|\Psi _N(t,\cdot )\rangle \end{aligned}$$

for all $A_1,\ldots ,A_m\in {\mathcal {L}}({\mathfrak {H}})$, satisfies

$$\begin{aligned} \Vert F_{N:m}(t)-R(t)^{\otimes m}\Vert _1\le 4\sqrt{\frac{m}{N}}\exp \left( \tfrac{3}{{\hbar }}\int _0^tL(s)\hbox {d}s\right) , \end{aligned}$$

with L given by (20).

As already mentioned at the end of the introduction (Sect. 1), the results discussed here apply to the case of bosons, and one reason for this is that the analysis in the present paper is not uniform in $\hbar$ (exactly as in [18]). In particular, one cannot consider the distinguished limit $h\sim N^{-1/3}$ which is typical of the mean-field limit for fermions. Another reason is that the initial condition for $\Psi _N$ is of the form $\Psi _N(0,\cdot )=(\psi ^{in})^{\otimes N}$, which is an example of pure state for bosons.

Let us briefly indicate how one arrives at the operator inequality in part (2) of Theorem 4.1. Let $\Lambda _1,\Lambda _2\in {\mathcal {L}}({\mathcal {L}}({\mathfrak {H}}),{\mathcal {L}}({\mathfrak {H}}_N))$ be such that

$$\begin{aligned} {\omega }\mapsto \langle \Psi _N|\Lambda _1(E_{\omega }^*)\Lambda _2(E_{\omega })|\Psi _N\rangle \text { belongs to }L^1\cap L^2({\textbf{R}}^3) \end{aligned}$$

(22)

for all $\Psi _N\in {\mathfrak {H}}_N$. For all V satisfying (12) and (17), define ${\mathcal {T}}(V,{\Lambda }_1,{\Lambda }_2)\in {\mathcal {L}}({\mathfrak {H}}_N)$ by polarization of the formula

$$\begin{aligned} \langle \Psi _N|{\mathcal {T}}(V,{\Lambda }_1,{\Lambda }_2)|\Psi _N\rangle :=\tfrac{1}{(2\pi )^d}\int _{{\textbf{R}}^d}{\hat{V}}({\omega })\langle \Psi _N|\Lambda _1(E_{\omega }^*)\Lambda _2(E_{\omega })|\Psi _N\rangle \hbox {d}{\omega }. \end{aligned}$$

In other words,

$$\begin{aligned} {\mathcal {T}}(V,\Lambda _1,\Lambda _2):=\tfrac{1}{(2\pi )^d}\int _{{\textbf{R}}^d}{\hat{V}}({\omega })\Lambda _1(E_{\omega }^*)\Lambda _2(E_{\omega })\hbox {d}{\omega }\end{aligned}$$

(23)

where the integral on the right hand is to be understood in the ultraweak sense (see footnote 3 on p. 1032 in [8]).

For each $A\in {\mathcal {L}}({\mathfrak {H}})$, denote by $\Lambda _j({\bullet }A)$ and $\Lambda _j(A{\bullet })$ the linear maps

$$\begin{aligned} \begin{aligned}&\Lambda _j({\bullet }A):\,{\mathcal {L}}({\mathfrak {H}})\ni B\mapsto \Lambda _j(BA)\in {\mathcal {L}}({\mathfrak {H}}_N) \\&\Lambda _j(A{\bullet }):\,{\mathcal {L}}({\mathfrak {H}})\ni B\mapsto \Lambda _j(AB)\in {\mathcal {L}}({\mathfrak {H}}_N) \end{aligned} \end{aligned}$$

respectively. If $A\in {\mathcal {L}}({\mathfrak {H}})$ is such that $\Lambda _1,\Lambda _2({\bullet }A)$ and $\Lambda _2(A{\bullet }),\Lambda _1$ satisfy (22), then one has

$$\begin{aligned} {\mathcal {C}}(V,\Lambda _1,\Lambda _2)A={\mathcal {T}}(V,\Lambda _1,\Lambda _2({\bullet }A))-{\mathcal {T}}(V,\Lambda _2(A{\bullet }),\Lambda _1)\,. \end{aligned}$$

(24)

Lemma 4.3

Let $\Lambda _1,\Lambda _2\in {\mathcal {L}}({\mathcal {L}}({\mathfrak {H}}),{\mathcal {L}}({\mathfrak {H}}_N))$ be $*$-homomorphisms, in other words

$$\begin{aligned} \Lambda _j(A^*)=\Lambda _j(A)^*,\qquad j=1,2 \end{aligned}$$

for all $A\in {\mathcal {L}}({\mathfrak {H}})$. Assume that $\Lambda _1,\Lambda _2$ satisfy (22). Then

$$\begin{aligned} {\mathcal {T}}(V,\Lambda _2,\Lambda _1)={\mathcal {T}}(V,\Lambda _1,\Lambda _2)^*. \end{aligned}$$

Proof

Indeed

$$\begin{aligned} \begin{aligned} {\mathcal {T}}(V,\Lambda _2,\Lambda _1)=&\tfrac{1}{(2\pi )d}\int _{{\textbf{R}}^d}{\hat{V}}({\omega })\Lambda _2(E_{\omega }^*)\Lambda _1(E_{\omega })\hbox {d}{\omega }\\ =&\tfrac{1}{(2\pi )d}\int _{{\textbf{R}}^d}{\hat{V}}({\omega })\Lambda _2(E_{\omega })^*\Lambda _1(E^*_{\omega })^*\hbox {d}{\omega }={\mathcal {T}}(V,\Lambda _1,\Lambda _2)^* \end{aligned} \end{aligned}$$

where the first equality follows from the fact that $\Lambda _1$ and $\Lambda _2$ are *-homomorphisms, while the second equality uses the fact that ${\hat{V}}$ is real-valued, since V is real-valued and even. $\square$

An easy consequence of (24) and of this lemma is that, for each $A=A^*\in {\mathcal {L}}({\mathfrak {H}})$ such that $\Lambda _1,\Lambda _2\in {\mathcal {L}}({\mathcal {L}}({\mathfrak {H}}),{\mathcal {L}}({\mathfrak {H}}_N))$ are $*$-homomorphisms such that $\Lambda _1,\Lambda _2({\bullet }A)$ satisfy (22), then

$$\begin{aligned} ({\mathcal {C}}(V,\Lambda _1,\Lambda _2)A)^*=-{\mathcal {C}}(V,\Lambda _1,\Lambda _2)A\,. \end{aligned}$$

(25)

The key observations leading to Theorem 4.1 are summarized in the two following lemmas. In the first of these two lemmas, the interaction operator is decomposed into a sum of four terms.

Lemma 4.4

Under the same assumptions and with the same notations as in Theorem 4.1, the interaction operator satisfies the identity

$$\begin{aligned} {\mathcal {C}}(V,{\mathcal {M}}_N(t)-{\mathcal {R}}(t),{\mathcal {M}}_N(t))(R(t))=T_1+T_2+T_3+T_4, \end{aligned}$$

with

$$\begin{aligned} \begin{aligned} T_1:=&{\mathcal {T}}(V,{\mathcal {M}}_N(t)(P(t){\bullet }P(t)),{\mathcal {M}}_N(t)(P(t){\bullet }R(t)) \\&-{\mathcal {T}}(V,{\mathcal {M}}_N(t)(R(t){\bullet }P(t)),{\mathcal {M}}_N(t)(P(t){\bullet }P(t)), \\ T_2:=&{\mathcal {M}}_N(t)(R(t)V_{R(t)}P(t)){\mathcal {M}}_N(t)(P(t)) \\&-{\mathcal {M}}_N(t)(P(t)){\mathcal {M}}_N(t)(P(t)V_{R(t)}R(t)), \\ T_3:=&{\mathcal {T}}(V,{\mathcal {M}}_N(t)(P(t){\bullet }R(t)),{\mathcal {M}}_N(t)(P(t){\bullet }R(t)) \\&-{\mathcal {T}}(V,{\mathcal {M}}_N(t)(R(t){\bullet }P(t)),{\mathcal {M}}_N(t)(R(t){\bullet }P(t)), \\ T_4:=&\tfrac{1}{N}{\mathcal {M}}_N(t)[V_{R(t)},R(t)]. \end{aligned} \end{aligned}$$

All the terms involved in this decomposition can be defined by the same method already used in the proof of Theorem 3.1. Indeed, one can check that all these terms involve only expressions of the type (I), (II) or (III) in the Remark following Theorem 3.1. This easy verification is left to the reader, and we shall henceforth consider this matter as settled by the detailed explanations concerning (I), (II) and (III) given in the previous section.

Each term in this decomposition satisfies an operator inequality involving only the operator norm of the “mean-field squared potential” $(V^2)_{R(t)}$, instead of the “bare” interaction potential V itself.

Lemma 4.5

Under the same assumptions and with the same notations as in Theorem 4.1, set

$$\begin{aligned} \ell (t):=\Vert V^2\star |\psi (t,\cdot )|^2\Vert ^\frac{1}{2}\,. \end{aligned}$$

(26)

Then

$$\begin{aligned} \begin{aligned} \pm iT_1\le&2\ell (t)\left( (1-\tfrac{1}{N}){\mathcal {M}}_N(t)(P(t))+\tfrac{4}{N}I_{{\mathfrak {H}}_N}\right) , \\ \pm iT_2\le&2\ell (t)({\mathcal {M}}_N(t)(P(t))+\tfrac{1}{N}I_{{\mathfrak {H}}_N}), \\ \pm iT_3\le&2\ell (t)((1-\tfrac{1}{N}){\mathcal {M}}_N(t)(P(t))+\tfrac{1}{N}I_{{\mathfrak {H}}_N}), \\ \pm iT_4\le&\tfrac{2}{N}\ell (t)I_{{\mathfrak {H}}_N}. \end{aligned} \end{aligned}$$

Remarks on $\ell (t)$ in (26) and L(t) in (20).

• (1) If V satisfies condition (17) in Theorem 3.1, then $V\!\in L^2({\textbf{R}}^3)\!+\!L^\infty ({\textbf{R}}^3)$, so that $V^2\!\in L^1({\textbf{R}}^d)+L^\infty ({\textbf{R}}^d)$. Thus $(V^2)_{R(t)}$, which is the multiplication operator by the function $V^2\star |\psi (t,\cdot )|^2$, satisfies

  $$\begin{aligned} \begin{aligned} \ell (t)^2:=\Vert V^2\star |\psi (t,\cdot )|^2\Vert _{L^\infty ({\textbf{R}}^3)}\le&\Vert V^2\Vert _{L^1({\textbf{R}}^3)+L^\infty ({\textbf{R}}^3)}\Vert \psi (t,\cdot )\Vert ^2_{L^1\cap L^\infty ({\textbf{R}}^3)} \\ \le&2\Vert V\Vert ^2_{L^1({\textbf{R}}^3)+L^\infty ({\textbf{R}}^3)}\max (1,\Vert \psi (t,\cdot )\Vert _{L^\infty ({\textbf{R}}^3)})^2 \\ \le&2C^2_S\Vert V\Vert ^2_{L^1({\textbf{R}}^3)+L^\infty ({\textbf{R}}^3)}\Vert \psi (t,\cdot )\Vert _{H^2({\textbf{R}}^3)}^2 \end{aligned} \end{aligned}$$

  where we recall that $C_S$ is the norm of the Sobolev embedding $H^2({\textbf{R}}^3)\subset L^\infty ({\textbf{R}}^3)$.

• (2) If V satisfies (8), then $\Vert V(I-{\Delta })^{-1}\Vert \le M$ for some positive constant M (see the discussion in §5.3 of chapter V in [13], so that

  $$\begin{aligned} V^2\le M^2(I-{\Delta })^2. \end{aligned}$$

  In this remark, we shall make a slightly more restrictive assumption, namely that $V^2$ satisfies

  $$\begin{aligned} V^2\le C(I-{\Delta })\,. \end{aligned}$$

  (27)

  In space dimension $d=3$, the Hardy inequality, which can be put in the form⁴

  $$\begin{aligned} \frac{1}{|x|^2}\le 4(-{\Delta }) \end{aligned}$$

  implies that the Coulomb potential satisfies the assumption above on V. If the potential V satisfies the (operator) inequality (27), then

  $$\begin{aligned} \begin{aligned} 0\le (V^2)_{R(t)}(x)&=\int _{{\textbf{R}}^d}V^2(y)|\psi (t,x-y)|^2\hbox {d}y=\langle \psi (t,x-\cdot )|V^2|\psi (t,x-\cdot )\rangle \\&\le C\langle \psi (t,x-\cdot )|(I-{\Delta })|\psi (t,x-\cdot )\rangle =C\Vert \psi (t,x-\cdot )\Vert _{L^2}^2\\&\quad +C\Vert {\nabla }\psi (t,x-\cdot )\Vert _{L^2}^2 \\&=C\Vert \psi (t,\cdot )\Vert _{L^2}^2+C\Vert {\nabla }\psi (t,\cdot )\Vert _{L^2}^2. \end{aligned} \end{aligned}$$

  Thus, if $\psi \in C({\textbf{R}};H^1({\textbf{R}}^d))$ is a solution of the Hartree equation,

  $$\begin{aligned} \ell (t)\le \sqrt{C}\Vert \psi (t,\cdot )\Vert _{H^1({\textbf{R}}^3)}. \end{aligned}$$

• (3) A bound on $\ell (t)$ in terms of $\Vert \psi (t,\cdot )\Vert _{H^1({\textbf{R}}^3)}$ instead of $\Vert \psi (t,\cdot )\Vert _{H^2({\textbf{R}}^3)}$ is advantageous since the former quantity can be controlled rather explicitly by means of the conservation of energy for the Hartree equation (5). This explicit control is useful in particular to assess the dependence in $\hbar$ of the convergence rate for the mean-field limit obtained in Corollary (4.2).

Clearly, the convergence rate for the quantum mean-field limit in Corollary 4.2 is not uniform in the semiclassical regime, in the first place because of the factor $3/{\hbar }$ on the right-hand side of the upper bound for $\Vert F_{N:m}(t)-R(t)^{\otimes m}\Vert _1$, which comes from the $i{\hbar }{\partial }_t$ part of the quantum dynamical equation.

However, one should expect that the function $\ell (t)$, or at least the upper bound for $\ell (t)$ obtained in (2), grows at least as $1/{\hbar }$, since it involves $\Vert {\nabla }_x\psi (t,\cdot )\Vert _{L^2}$, expected to be of order $1/{\hbar }$ for semiclassical wave functions $\psi$ (think for instance of a WKB wave function, or of a Schrödinger coherent state).

We shall discuss this issue by means of the conservation of energy satisfied by the Hartree solution $\psi$ (see formula (5.2) in [3]):

$$\begin{aligned} \begin{aligned}&\tfrac{1}{2}{\hbar }^2\Vert {\nabla }\psi (t,\cdot )\Vert _{L^2}^2+\tfrac{1}{2}\int _{{\textbf{R}}^d}|\psi (t,x)|^2(V\star |\psi (t,x)|^2)\hbox {d}x \\&\quad =\tfrac{1}{2}{\hbar }^2\Vert {\nabla }\psi ^{in}\Vert _{L^2}^2+\tfrac{1}{2}\int _{{\textbf{R}}^d}|\psi ^{in}(x)|^2(V\star |\psi ^{in}(x)|^2)\hbox {d}x. \end{aligned} \end{aligned}$$

Observe that

$$\begin{aligned} |V\star |\psi (t,x)|^2|\le \Vert \psi (t,\cdot )\Vert _{L^2}\Vert (V^2\star |\psi (t,x)|^2)^{1/2} \Vert _{L^\infty }=\ell (t)\,, \end{aligned}$$

(28)

so that

$$\begin{aligned} \begin{aligned}&\tfrac{1}{2}{\hbar }^2\Vert \nabla \psi (t,\cdot )\Vert _{L^2}^2+\tfrac{1}{2}\int _{{\textbf{R}}^d}|\psi (t,x)|^2(V\star |\psi (t,x)|^2)\hbox {d}x \\&\quad \le \tfrac{1}{2}{\hbar }^2\Vert \psi (t,\cdot )\Vert _{H^1}^2+\tfrac{1}{2}\ell (t)\le \tfrac{1}{2}{\hbar }^2\Vert \psi ^{in}\Vert _{H^1}^2+\tfrac{1}{2}\sqrt{C}\Vert \psi ^{in}\Vert _{H^1}. \end{aligned} \end{aligned}$$

Thus, if $V\ge 0$, or if ${\hat{V}}\ge 0$, one has

$$\begin{aligned} \int _{{\textbf{R}}^d}|\psi (t,x)|^2(V\star |\psi (t,x)|^2)\hbox {d}x=\tfrac{1}{(2\pi )^d}\int _{{\textbf{R}}^d}{\hat{V}}({\omega })|{\mathcal {F}}(|\psi (t,\cdot )|^2)|^2({\omega })\hbox {d}{\omega }\ge 0 \end{aligned}$$

(where ${\mathcal {F}}$ designates the Fourier transform on ${\textbf{R}}^d$), so that the conservation of mass and energy for the Hartree solution implies that

$$\begin{aligned} {\hbar }^2\Vert \psi (t,\cdot )\Vert _{H^1}^2\le {\hbar }^2\Vert \psi ^{in}\Vert _{H^1}^2+\sqrt{C}\Vert \psi ^{in}\Vert _{H^1}. \end{aligned}$$

In that case

$$\begin{aligned} \ell (t)\le \tfrac{1}{{\hbar }}\sqrt{C({\hbar }^2\Vert \psi ^{in}\Vert _{H^1}^2+\sqrt{C}\Vert \psi ^{in}\Vert _{H^1})}. \end{aligned}$$

Typical states used in the semiclassical regime (WKB or coherent states, for instance) satisfy ${\hbar }\Vert {\nabla }\psi ^{in}\Vert _{L^2}=O(1)$. Thus, in that case

$$\begin{aligned} \ell (t)\le {\hbar }^{-3/2}\sqrt{C({\hbar }^3\Vert \psi ^{in}\Vert _{H^1}^2+\sqrt{C}{\hbar }\Vert \psi ^{in}\Vert _{H^1})}=O({\hbar }^{-3/2}). \end{aligned}$$

Things become worse if the potential energy is a priori of indefinite sign. With (28), the energy conservation implies that

$$\begin{aligned} \begin{aligned} {\hbar }^2\Vert \psi (t,\cdot )\Vert _{H^1}^2\le&{\hbar }^2\Vert \psi ^{in}\Vert _{H^1}^2+\sqrt{C}\Vert \psi ^{in}\Vert _{H^1}+\sqrt{C}\Vert \psi (t,\cdot )\Vert _{H^1} \\ \le&{\hbar }^2\Vert \psi ^{in}\Vert _{H^1}^2+\sqrt{C}\Vert \psi ^{in}\Vert _{H^1}+\tfrac{C}{2{\hbar }^2}+\tfrac{1}{2}{\hbar }^2\Vert \psi (t,\cdot )\Vert ^2_{H^1}, \end{aligned} \end{aligned}$$

so that

$$\begin{aligned} {\hbar }^2\Vert \psi (t,\cdot )\Vert _{H^1}^2\le 2\left( {\hbar }^2\Vert \psi ^{in}\Vert _{H^1}^2+\sqrt{C}\Vert \psi ^{in}\Vert _{H^1}+\tfrac{C}{2{\hbar }^2}\right) \le 3{\hbar }^2\Vert \psi ^{in}\Vert _{H^1}^2+2\tfrac{C}{{\hbar }^2}, \end{aligned}$$

and thus

$$\begin{aligned} \ell (t)\le {\hbar }^{-2}\sqrt{C(3{\hbar }^4\Vert \psi ^{in}\Vert _{H^1}^2+2C)}=O({\hbar }^{-2}). \end{aligned}$$

Therefore, the exponential amplifying factor in Corollary 4.2 is $\exp (Kt/{\hbar }^{5/2})$ in the first case, and $\exp (Kt/{\hbar }^3)$ in the second. These elementary remarks suggest that Pickl’s clever method for proving the quantum mean-field limit with singular potentials including the Coulomb potential (see [15, 18]) is not expected to give uniform convergence rates (as in [7, 8] in the case of regular interaction potentials) for the mean-field limit in the semiclassical regime.

5 Proof of part (1) in Theorem 4.1

For each ${\sigma }\in {\mathfrak {S}}_N$ and each $\Psi _N\in {\mathfrak {H}}_N$, set

$$\begin{aligned} (U_{\sigma }\Psi _N)(x_1,\ldots ,x_N)=\Psi _N(x_{{\sigma }^{-1}(1)},\ldots ,x_{{\sigma }^{-1}(N)}). \end{aligned}$$

Since $\psi (t,\cdot )\in H^2({\textbf{R}}^3)$, the commutator $[{\Delta },R(t)]$ is a bounded operator on ${\mathfrak {H}}$. According to formula (25) in [8], denoting by $V_{kl}$ the multiplication operator

$$\begin{aligned} (V_{kl}\Psi _N)(x_1,\ldots ,x_N)=V(x_k-x_l)\Psi _N(x_1,\ldots ,x_N)\,, \end{aligned}$$

(29)

one has

$$\begin{aligned} \begin{aligned}&{\text {trace}}_{{\mathfrak {H}}_N}((i{\hbar }{\partial }_t{\mathcal {M}}_N(t)-{\textbf{ad}}^*(-\tfrac{1}{2}{\hbar }^2{\Delta }){\mathcal {M}}_N(t))(P(t))F_N) \\&\quad =-{\text {trace}}_{{\mathfrak {H}}_N}(\tfrac{N-1}{N}([V_{12},J_1P(t)])F_N)={\text {trace}}_{{\mathfrak {H}}_N}(\tfrac{N-1}{N}([V_{12},J_1R(t)])F_N) \end{aligned} \end{aligned}$$

(30)

for all $F_N\in {\mathcal {L}}({\mathfrak {H}}_N)$ such that

$$\begin{aligned} F_N=F_N^*\ge 0\,,\quad {\text {trace}}_{{\mathfrak {H}}_N}(F_N)=1\,,\quad \text { and }\quad U_{\sigma }F_NU_{\sigma }^*=F_N\text { for all }{\sigma }\in {\mathfrak {S}}_N\,. \end{aligned}$$

(31)

The core result in the proof of Theorem 3.1 is that the function

$$\begin{aligned} {\omega }\mapsto \langle \Psi _N|J_k([E_{\omega },R(t)])J_l(E_{\omega }^*)|\Psi _N\rangle \in L^2\cap L^\infty ({\textbf{R}}^3) \end{aligned}$$

for each $k\not =l\in \{1,\ldots ,N\}$. Since ${\hat{V}}\in L^1({\textbf{R}}^3)+L^2({\textbf{R}}^3)$, this has led us to define

$$\begin{aligned} \langle \Psi _N|[V_{kl},J_kR(t)]|\Psi _N\rangle :=\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })\langle \Psi _N|J_k([E_{\omega },R(t)])J_l(E_{\omega }^*)|\Psi _N\rangle \hbox {d}{\omega }, \end{aligned}$$

and more generally, using a spectral decomposition of the trace-class operator $F_N$,

$$\begin{aligned} {\text {trace}}_{{\mathfrak {H}}_N}([V_{kl},J_kR(t)]F_N):=\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega }){\text {trace}}_{{\mathfrak {H}}_N}(J_k([E_{\omega },R(t)])J_l(E_{\omega }^*)F_N)\hbox {d}{\omega }\end{aligned}$$

with

$$\begin{aligned} {\omega }\mapsto {\text {trace}}_{{\mathfrak {H}}_N}(|J_k([E_{\omega },R(t)])J_l(E_{\omega }^*)F_N)\in L^2\cap L^\infty ({\textbf{R}}^3). \end{aligned}$$

Since $U_{\sigma }F_NU_{\sigma }^*=F_N$ for all ${\sigma }\in {\mathfrak {S}}_N$, for each $m\not =n\in \{1,\ldots ,N\}$, one has

$$\begin{aligned} \begin{aligned}&{\text {trace}}_{{\mathfrak {H}}_N}(\tfrac{N-1}{N}([V_{12},J_1R(t)])F_N)\\&\quad ={\text {trace}}_{{\mathfrak {H}}_N}(\tfrac{N-1}{N}([V_{mn},J_mR(t)])F_N) \\&\quad =\tfrac{1}{N^2}\sum _{1\le k\not =l\le N}\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega }){\text {trace}}_{{\mathfrak {H}}_N}(J_k([E_{\omega },R(t)])J_l(E_{\omega }^*)F_N)\hbox {d}{\omega }\\&\quad =\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega }){\text {trace}}_{{\mathfrak {H}}_N}(S_N[\psi (t,\cdot )]({\omega })F_N)\hbox {d}{\omega }. \end{aligned} \end{aligned}$$

With the definition of ${\mathcal {C}}$ in Theorem 3.1, we conclude that the operator

$$\begin{aligned} X_N=(i{\hbar }{\partial }_t{\mathcal {M}}_N(t)-{\textbf{ad}}^*(-\tfrac{1}{2}{\hbar }^2{\Delta }){\mathcal {M}}_N(t))(P(t))-{\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))(R(t)) \end{aligned}$$

satisfies

$$\begin{aligned} {\text {trace}}_{{\mathfrak {H}}_N}(X_N F_N)=0 \end{aligned}$$

for each operator $F_N\in {\mathcal {L}}({\mathfrak {H}}_N)$ satisfying (31). One easily checks that

$$\begin{aligned} U^*_{\sigma }X_NU_{\sigma }=X_N\quad \text { for all }{\sigma }\in {\mathfrak {X}}_N. \end{aligned}$$

Let $D_N\in {\mathcal {L}}({\mathfrak {H}}_N)$ be a density operator on ${\mathfrak {H}}_N$, i.e.

$$\begin{aligned} D_N=D_N^*\ge 0\quad \text { and }\quad {\text {trace}}_{{\mathfrak {H}}_N}(D_N)=1\,. \end{aligned}$$

(32)

Obviously

$$\begin{aligned} F_N:=\tfrac{1}{N!}\sum _{{\sigma }\in {\mathfrak {X}}_N}U_{\sigma }D_NU_{\sigma }^* \end{aligned}$$

satisfies (31), so that

$$\begin{aligned} 0={\text {trace}}_{{\mathfrak {H}}_N}(X_NF_N)=\tfrac{1}{N!}\sum _{{\sigma }\in {\mathfrak {X}}_N}{\text {trace}}_{{\mathfrak {H}}_N}(U_{\sigma }^*X_NU_{\sigma }D_N)={\text {trace}}_{{\mathfrak {H}}_N}(X_ND_N) \end{aligned}$$

for all $D_N\in {\mathcal {L}}({\mathfrak {H}}_N)$ satisfying (32). Since any trace-class operator on ${\mathfrak {H}}_N$ is a linear combination of 4 density operators, we conclude that

$$\begin{aligned} {\text {trace}}_{{\mathfrak {H}}_N}(X_NT_N)=0\qquad \text { for all }T_N\in {\mathcal {L}}^1({\mathfrak {H}}_N), \end{aligned}$$

so that

$$\begin{aligned} X_N=0\,. \end{aligned}$$

(33)

On the other hand,

$$\begin{aligned} \begin{aligned} {\mathcal {M}}_N(t)(i{\hbar }{\partial }_tP(t))&={\mathcal {M}}_N(t)([-\tfrac{1}{2}{\hbar }^2{\Delta }+V\star |\psi (t,\cdot )|^2,P(t)]) \\&={\mathcal {M}}_N(t)([-\tfrac{1}{2}{\hbar }^2{\Delta },P(t)])-{\mathcal {M}}_N(t)([V\star |\psi (t,\cdot )|^2,R(t)]) \end{aligned} \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned} i{\hbar }{\partial }_t({\mathcal {M}}_N(t)(P(t)))=&{\textbf{ad}}^*(-\tfrac{1}{2}{\hbar }^2{\Delta }){\mathcal {M}}_N(t)(P(t))+{\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))(R(t)) \\&+{\mathcal {M}}_N(t)([-\tfrac{1}{2}{\hbar }^2{\Delta },P(t)])-{\mathcal {M}}_N(t)([V\star |\psi (t,\cdot )|^2,R(t)]) \\ {=}&{\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))(R(t)){-}{\mathcal {M}}_N(t)([V\star |\psi (t,\cdot )|^2,R(t)])\,. \end{aligned} \end{aligned}$$

(34)

Finally, by condition (17) on V, one has

$$\begin{aligned} \psi (t,\cdot )\in H^2({\textbf{R}}^3)\subset L^2\cap L^4({\textbf{R}}^3)\implies V\star |\psi (t,\cdot )|^2\in {\mathcal {F}}L^1({\textbf{R}}^3) \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned} V\star |\psi (t,\cdot )|^2=&\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega }){\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })E_{\omega }\hbox {d}{\omega }\\ =&\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })\langle \psi (t,\cdot )|E_{\omega }^*|\psi (t,\cdot )\rangle E_{\omega }\hbox {d}{\omega }\\ =&\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega }){\mathcal {R}}(t)(E_{\omega }^*)E_{\omega }\hbox {d}{\omega }\,. \end{aligned} \end{aligned}$$

(35)

Hence

$$\begin{aligned} \begin{aligned} {\mathcal {M}}_N(t)([V\star |\psi (t,\cdot )|^2,R(t)])=&\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega }){\mathcal {R}}(t)(E_{\omega }^*){\mathcal {M}}_N(t)[E_{\omega },R(t)] \hbox {d}{\omega }\\ =&{\mathcal {C}}(V,{\mathcal {R}}(t),{\mathcal {M}}_N(t))(R(t)) \end{aligned} \end{aligned}$$

(36)

so that, returning to (34), one arrives at the equality

$$\begin{aligned} i{\hbar }{\partial }_t({\mathcal {M}}_N(t)(P(t)))={\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))(R(t))-{\mathcal {C}}(V,{\mathcal {R}}(t),{\mathcal {M}}_N(t))(R(t)), \end{aligned}$$

which proves part (1) in Theorem 4.1.

Remark

In [8], the equality

$$\begin{aligned} i{\hbar }{\partial }_t{\mathcal {M}}_N(t)(A)={\textbf{ad}}^*\left( -\tfrac{1}{2}{\hbar }^2{\Delta }\right) {\mathcal {M}}_N(t)(A)-{\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))(A) \end{aligned}$$

is proved for all $A\in {\mathcal {L}}({\mathfrak {H}})$ such that $[{\Delta },A]\in {\mathcal {L}}({\mathfrak {H}})$, assuming that $V\in {\mathcal {F}}L^1({\textbf{R}}^3)$. This argument cannot be used here since $V\notin {\mathcal {F}}L^1({\textbf{R}}^3)$. Besides, the definition of the operator ${\mathcal {C}}(V,{\mathcal {M}}_N(t),{\mathcal {M}}_N(t))(R(t))$ in Theorem 3.1 makes critical use of the fact that $R(t)=|\psi (t,\cdot )\rangle \langle \psi (t,\cdot )|$ with $\psi (t,\cdot )\in L^2\cap L^\infty ({\textbf{R}}^3)$. This is the reason for the rather lengthy justification of (33) in this section.

6 Proof of Lemma 4.4

In the sequel, we seek to “simplify” the expression of the interaction operator

$$\begin{aligned} {\mathcal {C}}(V,{\mathcal {M}}_N(t)-{\mathcal {R}}(t),{\mathcal {M}}_N(t))(R(t)). \end{aligned}$$

This will lead to rather involved computations which do not seem much of a simplification. However, we shall see that the final result of these computations, reported in Lemma 4.4, although algebraically more cumbersome, has better analytical properties.

6.1 A first simplification

First we decompose $E_{\omega }R(t)$ and $R(t)E_{\omega }$ in the terms ${\mathcal {M}}_N(t)(E_{\omega }R(t))$ and ${\mathcal {M}}_N(t)(R(t)E_{\omega })$ as

$$\begin{aligned} E_{\omega }R(t)=P(t)E_{\omega }R(t)+R(t)E_{\omega }R(t), \end{aligned}$$

and observe that

$$\begin{aligned} \begin{aligned}&{\mathcal {C}}(V,{\mathcal {M}}_N(t)-{\mathcal {R}}(t),{\mathcal {M}}_N(t))(R(t)) \\&\quad =\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })(({\mathcal {M}}_N(t)-{\mathcal {R}}(t))(E_{\omega }^*){\mathcal {M}}_N(t)(P(t)E_{\omega }R(t)) \\&\quad \quad -{\mathcal {M}}_N(t)(R(t)E_{\omega }P(t))({\mathcal {M}}_N(t)-{\mathcal {R}}(t))(E_{\omega }^*))\hbox {d}{\omega }\\&\quad \quad +\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })[({\mathcal {M}}_N(t)-{\mathcal {R}}(t))(E_{\omega }^*),{\mathcal {M}}_N(t)(R(t)E_{\omega }R(t))]\hbox {d}{\omega }. \end{aligned} \end{aligned}$$

All the terms in the right-hand side of the equality above are either similar to the one considered in Theorem 3.1, or of the type denoted (III) in the Remark following Theorem 3.1.

An elementary computation shows that, for all ${\omega }\in {\textbf{R}}^d$,

$$\begin{aligned} \begin{aligned}&{}[({\mathcal {M}}_N(t)-{\mathcal {R}}(t))(E_{\omega }^*),{\mathcal {M}}_N(t)(R(t)E_{\omega }R(t))] \\&\quad =[{\mathcal {M}}_N(t)(E_{\omega }^*),{\mathcal {M}}_N(t)(R(t)E_{\omega }R(t))] \\&\quad =\tfrac{1}{N}{\mathcal {M}}_N(t)[E_{\omega }^*,R(t)E_{\omega }R(t)]. \end{aligned} \end{aligned}$$

Recall indeed that, for each $A,B\in {\mathcal {L}}({\mathfrak {H}})$, one has

$$\begin{aligned} {[}{\mathcal {M}}_N(t)A,{\mathcal {M}}_N(t)B]=\tfrac{1}{N}{\mathcal {M}}_N(t)([A,B]) \end{aligned}$$

— see formula before (41) on p. 1041 in [8]. On the other hand,

$$\begin{aligned} R(t)E_{\omega }R(t)=|\psi (t,\cdot )\rangle \langle \psi (t,\cdot )|E_{\omega }|\psi (t,\cdot )\rangle \langle \psi (t,\cdot )|={\mathcal {F}}(|\psi (t,\cdot )|^2)(-{\omega })R(t)\,, \end{aligned}$$

(37)

so that

$$\begin{aligned} \begin{aligned}&{}[({\mathcal {M}}_N(t)-{\mathcal {R}}(t))(E_{\omega }^*),{\mathcal {M}}_N(t)(R(t)E_{\omega }R(t))] \\&\quad =\tfrac{1}{N}{\mathcal {F}}(|\psi (t,\cdot )|^2)(-{\omega }){\mathcal {M}}_N(t)[E_{\omega }^*,R(t)]. \end{aligned} \end{aligned}$$

Besides

$$\begin{aligned} \begin{aligned} ({\mathcal {M}}_N(t)-{\mathcal {R}}(t))(E_{\omega }^*)=&{\mathcal {M}}_N(t)E_{\omega }^*-\langle \psi (t,\cdot )|E_{\omega }^*\psi (t,\cdot )|\rangle \,I_{{\mathfrak {H}}_N} \\ =&{\mathcal {M}}_N(t)E_{\omega }^*-{\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })I_{{\mathfrak {H}}_N} \\ =&{\mathcal {M}}_N(t)(E_{\omega }^*-{\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })I_{{\mathfrak {H}}}). \end{aligned} \end{aligned}$$

Indeed,

$$\begin{aligned} {\mathcal {M}}_N^{in}I_{\mathfrak {H}}=I_{{\mathfrak {H}}_N}\implies {\mathcal {M}}_N(t)I_{\mathfrak {H}}={\mathcal {U}}_N(t)^*({\mathcal {M}}_N^{in}I_{\mathfrak {H}}){\mathcal {U}}_N(t)=I_{{\mathfrak {H}}_N} \end{aligned}$$

(38)

where we recall that ${\mathcal {U}}_N(t):=e^{-it{\mathcal {H}}_N/{\hbar }}$, while ${\mathcal {H}}_N$ is the N-body Hamiltonian.

Therefore,

$$\begin{aligned} \begin{aligned}&{\mathcal {C}}(V,{\mathcal {M}}_N(t)-{\mathcal {R}}(t),{\mathcal {M}}_N(t))(R(t)) \\&\quad =\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })(({\mathcal {M}}_N(t)(E_{\omega }^*-{\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })I_{\mathfrak {H}}){\mathcal {M}}_N(t)(P(t)E_{\omega }R(t)) \\&\quad \quad -{\mathcal {M}}_N(t)(R(t)E_{\omega }P(t))({\mathcal {M}}_N(t)(E_{\omega }^*-{\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })I_{\mathfrak {H}}))\hbox {d}{\omega }\\&\quad \quad +\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega }){\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })\tfrac{1}{N}{\mathcal {M}}_N(t)[E_{\omega },R(t)]\hbox {d}{\omega }, \end{aligned} \end{aligned}$$

in view of (12). With the formula (36), we conclude that

$$\begin{aligned} \begin{aligned}&{\mathcal {C}}(V,{\mathcal {M}}_N(t)-{\mathcal {R}}(t),{\mathcal {M}}_N(t))R(t) \\&\quad =\tfrac{1}{(2\pi )^3}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })(({\mathcal {M}}_N(t)(E_{\omega }^*-{\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })I_{\mathfrak {H}}){\mathcal {M}}_N(t)(P(t)E_{\omega }R(t)) \\&\quad \quad -{\mathcal {M}}_N(t)(R(t)E_{\omega }P(t))({\mathcal {M}}_N(t)(E_{\omega }^*-{\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })I_{\mathfrak {H}}))\hbox {d}{\omega }\\&\quad \quad +\tfrac{1}{N}{\mathcal {M}}_N(t)[V\star |\psi (t,\cdot )|^2,R(t)]\,. \end{aligned} \end{aligned}$$

(39)

6.2 A second simplification

Next we decompose $E_{\omega }^*$ in ${\mathcal {M}}_N(t)(E_{\omega }^*)$ as

$$\begin{aligned} E_{\omega }^*=P(t)E_{\omega }^*P(t)+P(t)E_{\omega }^*R(t)+R(t)E_{\omega }^*P(t)+R(t)E_{\omega }^*R(t). \end{aligned}$$

The identity (37) shows that

$$\begin{aligned} R(t)E_{\omega }^*R(t)={\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })R(t), \end{aligned}$$

and hence

$$\begin{aligned} R(t)(E_{\omega }^*-{\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })I_{\mathfrak {H}})R(t)=0. \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} {\mathcal {M}}_N(t)(E_{\omega }^*-{\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })I_{\mathfrak {H}})=&{\mathcal {M}}_N(t)(P(t)E_{\omega }^*P(t)-{\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })P(t)) \\&+{\mathcal {M}}_N(t)(P(t)E_{\omega }^*R(t)+R(t)E_{\omega }^*P(t)), \end{aligned} \end{aligned}$$

since $R(t)P(t)=P(t)R(t)=0$. Thus,

$$\begin{aligned} \begin{aligned}&{\mathcal {C}}(V,{\mathcal {M}}_N(t)-{\mathcal {R}}(t),{\mathcal {M}}_N(t))(R(t)) \\&\quad =\int _{{\textbf{R}}^3}{\hat{V}}({\omega })(({\mathcal {M}}_N(t)(P(t)E_{\omega }^*P(t)-{\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })P(t)){\mathcal {M}}_N(t)(P(t)E_{\omega }R(t)) \\&\quad \quad -{\mathcal {M}}_N(t)(R(t)E_{\omega }P(t))({\mathcal {M}}_N(t)(P(t)E_{\omega }^*P(t)-{\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })P(t)))\tfrac{\hbox {d}{\omega }}{(2\pi )^3} \\&\quad \quad +\int _{{\textbf{R}}^3}{\hat{V}}({\omega })({\mathcal {M}}_N(t)(P(t)E_{\omega }^*R(t)+R(t)E_{\omega }^*P(t)){\mathcal {M}}_N(t)(P(t)E_{\omega }R(t)) \\&\quad \quad -{\mathcal {M}}_N(t)(R(t)E_{\omega }P(t)){\mathcal {M}}_N(t)(P(t)E_{\omega }^*R(t)+R(t)E_{\omega }^*P(t)))\tfrac{\hbox {d}{\omega }}{(2\pi )^3} \\&\quad \quad +\tfrac{1}{N}{\mathcal {M}}_N(t)[(V\star |\psi (t,\cdot )|^2),R(t)]. \end{aligned} \end{aligned}$$

Using again (12) implies that

$$\begin{aligned} \begin{aligned}&\int _{{\textbf{R}}^3}{\hat{V}}({\omega })({\mathcal {M}}_N(t)(R(t)E_{\omega }^*P(t)){\mathcal {M}}_N(t)(P(t)E_{\omega }R(t))\hbox {d}{\omega }\\&\quad =\int _{{\textbf{R}}^3}{\hat{V}}({\omega })({\mathcal {M}}_N(t)(R(t)E_{\omega }P(t)){\mathcal {M}}_N(t)(P(t)E_{\omega }^*R(t))\hbox {d}{\omega }, \end{aligned} \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned}&{\mathcal {C}}(V,{\mathcal {M}}_N(t)-{\mathcal {R}}(t),{\mathcal {M}}_N(t))(R(t)) \\&\quad =\int _{{\textbf{R}}^3}{\hat{V}}({\omega })(({\mathcal {M}}_N(t)(P(t)E_{\omega }^*P(t)-{\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })P(t)){\mathcal {M}}_N(t)(P(t)E_{\omega }R(t)) \\&\quad \quad -{\mathcal {M}}_N(t)(R(t)E_{\omega }P(t))({\mathcal {M}}_N(t)(P(t)E_{\omega }^*P(t)-{\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega })P(t)))\tfrac{\hbox {d}{\omega }}{(2\pi )^3} \\&\quad \quad +\int _{{\textbf{R}}^3}{\hat{V}}({\omega })({\mathcal {M}}_N(t)(P(t)E_{\omega }^*R(t)){\mathcal {M}}_N(t)(P(t)E_{\omega }R(t)) \\&\quad \quad -{\mathcal {M}}_N(t)(R(t)E_{\omega }P(t)){\mathcal {M}}_N(t)(R(t)E_{\omega }^*P(t)))\tfrac{\hbox {d}{\omega }}{(2\pi )^3} \\&\quad \quad +\tfrac{1}{N}{\mathcal {M}}_N(t)[(V\star |\psi (t,\cdot )|^2),R(t)]. \end{aligned} \end{aligned}$$

By (35), one can further simplify the term

$$\begin{aligned} \begin{aligned}&\int _{{\textbf{R}}^3}{\hat{V}}({\omega }){\mathcal {F}}(|\psi (t,\cdot )|^2)({\omega }){\mathcal {M}}_N(t)(P(t)){\mathcal {M}}_N(t)(P(t)E_{\omega }R(t)) \\&\quad \quad -{\mathcal {M}}_N(t)(R(t)E_{\omega }P(t))({\mathcal {M}}_N(t)(P(t)))\tfrac{\hbox {d}{\omega }}{(2\pi )^3} \\&\quad ={\mathcal {M}}_N(t)(P(t)){\mathcal {M}}_N(t)(P(t)(V\star |\psi (t,\cdot )|^2)R(t)) \\&\quad \quad -{\mathcal {M}}_N(t)(R(t)(V\star |\psi (t,\cdot )|^2)P(t)){\mathcal {M}}_N(t)(P(t)). \end{aligned} \end{aligned}$$

Finally

$$\begin{aligned} {\mathcal {C}}(V,{\mathcal {M}}_N(t)-{\mathcal {R}}(t),{\mathcal {M}}_N(t))(R(t))=T_1+T_2+T_3+T_4 \end{aligned}$$

with

$$\begin{aligned} \begin{aligned} T_1:=&\int _{{\textbf{R}}^3}{\hat{V}}({\omega })(({\mathcal {M}}_N(t)(P(t)E_{\omega }^*P(t)){\mathcal {M}}_N(t)(P(t)E_{\omega }R(t)) \\&-{\mathcal {M}}_N(t)(R(t)E_{\omega }P(t))({\mathcal {M}}_N(t)(P(t)E_{\omega }^*P(t)))\tfrac{\hbox {d}{\omega }}{(2\pi )^3} \\ T_2:=&{\mathcal {M}}_N(t)(R(t)(V\star |\psi (t,\cdot )|^2)P(t)){\mathcal {M}}_N(t)(P(t)) \\&-{\mathcal {M}}_N(t)(P(t)){\mathcal {M}}_N(t)(P(t)(V\star |\psi (t,\cdot )|^2)R(t)) \\ T_3:=&\int _{{\textbf{R}}^3}{\hat{V}}({\omega })({\mathcal {M}}_N(t)(P(t)E_{\omega }^*R(t)){\mathcal {M}}_N(t)(P(t)E_{\omega }R(t)) \\&-{\mathcal {M}}_N(t)(R(t)E_{\omega }P(t)){\mathcal {M}}_N(t)(R(t)E_{\omega }^*P(t)))\tfrac{\hbox {d}{\omega }}{(2\pi )^3} \\ T_4:=&\tfrac{1}{N}{\mathcal {M}}_N(t)[(V\star |\psi (t,\cdot )|^2),R(t)]. \end{aligned} \end{aligned}$$

Observe again that all the integrals in the right-hand side of the equalities defining $T_1$ and $T_3$ are of the form defined in Theorem 3.1, or of the form (I), (II) or (III), or their adjoint, in the Remark following Theorem 3.1.

That

$$\begin{aligned} \begin{aligned} T_1=&{\mathcal {T}}(V,{\mathcal {M}}_N(t)(P(t){\bullet }P(t)),{\mathcal {M}}_N(t)(P(t){\bullet }R(t))) \\&-{\mathcal {T}}(V,{\mathcal {M}}_N(t)(R(t){\bullet }P(t)),{\mathcal {M}}_N(t)(P(t){\bullet }P(t))) \\ T_3=&{\mathcal {T}}(V,{\mathcal {M}}_N(t)(P(t){\bullet }R(t)),{\mathcal {M}}_N(t)(P(t){\bullet }R(t))) \\&-{\mathcal {T}}(V,{\mathcal {M}}_N(t)(R(t){\bullet }P(t)),{\mathcal {M}}_N(t)(R(t){\bullet }P(t))) \end{aligned} \end{aligned}$$

follows from (12) and the definition (23). This concludes the proof of Lemma 4.4.

7 Proof of Lemma 4.5

In the sequel, we shall estimate these four terms in increasing order of technical difficulty.

7.1 Bound for $T_4$

The easiest term to treat is obviously $T_4$. We first recall that

$$\begin{aligned} \Vert {\mathcal {M}}_N(t)(A)\Vert \le \Vert A\Vert \qquad \text { for each }A\in {\mathcal {L}}({\mathfrak {H}}) \end{aligned}$$

(40)

— see the formula following (41) on p. 1041 in [8]. Thus

$$\begin{aligned} \begin{aligned} \Vert T_4\Vert \le&\tfrac{1}{N}\Vert [V\star |\psi (t,\cdot )|^2,R(t)]\Vert \le \tfrac{1}{N}(\Vert R(t)V\star |\psi (t,\cdot )|^2\Vert +\Vert (V\star |\psi (t,\cdot )|^2)R(t)\Vert ) \\ =&\tfrac{2}{N}\Vert (V\star |\psi (t,\cdot )|^2)R(t)\Vert , \end{aligned} \end{aligned}$$

where the equality follows from the fact that $R(t)=R(t)^*$, which implies that

$$\begin{aligned} ((V\star |\psi (t,\cdot )|^2)R(t))^*=R(t)(V\star |\psi (t,\cdot )|^2)\,. \end{aligned}$$

(41)

On the other hand, by Jensen’s inequality

$$\begin{aligned} (|V|\star |\psi (t,\cdot )|^2)^2\le V^2\star |\psi (t,\cdot )|^2, \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned} \Vert (V\star |\psi (t,\cdot )|^2)R(t)\Vert ^2\le&\Vert V\star |\psi (t,\cdot )|^2\Vert ^2_{L^\infty } \\ \le&\Vert \,|V|\star |\psi (t,\cdot )|^2\,\Vert ^2_{L^\infty }\le \Vert (V^2)\star |\psi (t,\cdot )|^2\Vert _{L^\infty }=\ell (t)^2\,, \end{aligned} \end{aligned}$$

(42)

and therefore

$$\begin{aligned} \Vert T_4\Vert \le \tfrac{2}{N}\ell (t)\,. \end{aligned}$$

(43)

Finally, we recall that

$$\begin{aligned} ({\mathcal {M}}_N^{in}A)^*=\tfrac{1}{N}\sum _{k=1}^N(J_kA)^*=\tfrac{1}{N}\sum _{k=1}^NJ_k(A^*)={\mathcal {M}}_N^{in}(A^*) \end{aligned}$$

for each $A\in {\mathcal {L}}({\mathfrak {H}})$, so that

$$\begin{aligned} \begin{aligned} ({\mathcal {M}}_N(t)A)^*=&({\mathcal {U}}_N(t)^*({\mathcal {M}}_N^{in}A){\mathcal {U}}_N(t))^*={\mathcal {U}}_N(t)^*({\mathcal {M}}_N^{in}A)^*{\mathcal {U}}_N(t) \\ =&{\mathcal {U}}_N(t)^*{\mathcal {M}}_N^{in}(A^*){\mathcal {U}}_N(t)={\mathcal {M}}_N(t)(A^*)\,. \end{aligned} \end{aligned}$$

(44)

Then, (41) and (44) imply that

$$\begin{aligned} \begin{aligned} ({\mathcal {M}}_N(t)[V\star |\psi (t,\cdot )|^2,R(t)])^*=&{\mathcal {M}}_N(t)([V\star |\psi (t,\cdot )|^2,R(t)]^*) \\ =&{\mathcal {M}}_N(t)(-[V\star |\psi (t,\cdot )|^2,R(t)]) \\ =&-{\mathcal {M}}_N(t)([V\star |\psi (t,\cdot )|^2,R(t)]) \end{aligned} \end{aligned}$$

so that $T_4^*=-T_4$. Hence, $\pm iT_4$ are self-adjoint operators on ${\mathfrak {H}}_N$, so that

$$\begin{aligned} \Vert T_4\Vert \le \tfrac{2}{N}\ell (t)\implies \pm iT_4\le \tfrac{2}{N}\ell (t)I_{{\mathfrak {H}}_N}\,. \end{aligned}$$

(45)

7.2 Bound for $T_2$

Set

$$\begin{aligned} S_2:={\mathcal {M}}_N(t)(P(t)){\mathcal {M}}_N(t)(P(t)(V\star |\psi (t,\cdot )|^2)R(t))\,. \end{aligned}$$

(46)

One has

$$\begin{aligned} \begin{aligned} S_2=&{\mathcal {U}}_N(t)^*{\mathcal {M}}_N^{in}(P(t)){\mathcal {M}}_N^{in}(P(t)(V\star |\psi (t,\cdot )|^2)R(t)){\mathcal {U}}_N(t) \\ =&\tfrac{1}{N}\sum _{k=1}^N{\mathcal {U}}_N(t)^*(J_kP(t)){\mathcal {M}}_N^{in}(P(t)(V\star |\psi (t,\cdot )|^2)R(t))(J_kP(t)){\mathcal {U}}_N(t) \\&+\tfrac{1}{N}\sum _{k=1}^N{\mathcal {U}}_N(t)^*(J_kP(t))[J_kP(t),{\mathcal {M}}_N^{in}(P(t)(V\star |\psi (t,\cdot )|^2)R(t))]{\mathcal {U}}_N(t). \end{aligned} \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned}&{}[J_kP(t),{\mathcal {M}}_N^{in}(P(t)(V\star |\psi (t,\cdot )|^2)R(t))] \\&\quad =\tfrac{1}{N}[J_kP(t),J_k(P(t)(V\star |\psi (t,\cdot )|^2)R(t))] \\&\quad =\tfrac{1}{N}J_k(P(t)(V\star |\psi (t,\cdot )|^2)R(t)), \end{aligned} \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned} S_2=&\tfrac{1}{N}\sum _{k=1}^N{\mathcal {U}}_N(t)^*(J_kP(t)){\mathcal {M}}_N^{in}(P(t)(V\star |\psi (t,\cdot )|^2)R(t))(J_kP(t)){\mathcal {U}}_N(t) \\&+\tfrac{1}{N^2}\sum _{k=1}^N{\mathcal {U}}_N(t)^*J_k(P(t)(V\star |\psi (t,\cdot )|^2)R(t)){\mathcal {U}}_N(t). \end{aligned} \end{aligned}$$

By cyclicity of the trace, for each $F_N^{in}$ satisfying (31), denoting

$$\begin{aligned} F_N(t):={\mathcal {U}}_N(t)F_N^{in}{\mathcal {U}}_N(t)^*, \end{aligned}$$

one has

$$\begin{aligned} \begin{aligned}&{\text {trace}}_{{\mathfrak {H}}_N}(S_2F_N^{in}) \\&\quad =\tfrac{1}{N}\sum _{k=1}^N{\text {trace}}_{{\mathfrak {H}}_N}({\mathcal {M}}_N^{in}(P(t)(V\star |\psi (t,\cdot )|^2)R(t))(J_kP(t))F_N(t)(J_kP(t))) \\&\quad \quad +\tfrac{1}{N^2}\sum _{k=1}^N{\text {trace}}_{{\mathfrak {H}}_N}(J_k(P(t)(V\star |\psi (t,\cdot )|^2)R(t))F_N(t)), \end{aligned} \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned}&|{\text {trace}}_{{\mathfrak {H}}_N}(S_2F_N^{in})| \\&\quad \le \tfrac{1}{N}\sum _{k=1}^N\Vert {\mathcal {M}}_N^{in}(P(t)(V\star |\psi (t,\cdot )|^2)R(t))\Vert \Vert (J_kP(t))F_N(t)(J_kP(t)))\Vert _1 \\&\quad \quad +\tfrac{1}{N^2}\sum _{k=1}^N\Vert J_k(P(t)(V\star |\psi (t,\cdot )|^2)R(t))\Vert \Vert F_N(t)\Vert _1 \\&\quad \le \Vert (V\star |\psi (t,\cdot )|^2)R(t)\Vert \tfrac{1}{N}\sum _{k=1}^N{\text {trace}}_{{\mathfrak {H}}_N}((J_kP(t))F_N(t)(J_kP(t)))) \\&\quad \quad +\Vert (V\star |\psi (t,\cdot )|^2)R(t)\Vert \tfrac{1}{N^2}\sum _{k=1}^N\Vert F_N(t)\Vert _1 \\&\quad =\Vert (V\star |\psi (t,\cdot )|^2)R(t)\Vert ({\text {trace}}_{{\mathfrak {H}}_N}({\mathcal {M}}_N(t)(P(t))F_N^{in})+\tfrac{1}{N}\Vert F_N^{in}\Vert _1)\,. \end{aligned} \end{aligned}$$

(47)

By (44),

$$\begin{aligned} S_2^*={\mathcal {M}}_N(t)(R(t)(V\star |\psi (t,\cdot )|^2)P(t)){\mathcal {M}}_N(t)(P(t)), \end{aligned}$$

so that

$$\begin{aligned} T_2=S_2^*-S_2=-T_2^*. \end{aligned}$$

Thus,

$$\begin{aligned} \begin{aligned} |{\text {trace}}_{{\mathfrak {H}}_N}(T_2F_N^{in})|\le&|{\text {trace}}_{{\mathfrak {H}}_N}(S_2^*F_N^{in})|+|{\text {trace}}_{{\mathfrak {H}}_N}(S_2F_N^{in})| \\ =&|{\text {trace}}_{{\mathfrak {H}}_N}(F_N^{in}S_2)|+|{\text {trace}}_{{\mathfrak {H}}_N}(S_2F_N^{in})|=2|{\text {trace}}_{{\mathfrak {H}}_N}(S_2F_N^{in})|, \end{aligned} \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned}&|{\text {trace}}_{{\mathfrak {H}}_N}(T_2F_N^{in})| \\&\quad \le 2\Vert (V\star |\psi (t,\cdot )|^2)R(t)\Vert ({\text {trace}}_{{\mathfrak {H}}_N}({\mathcal {M}}_N(t)(P(t))F_N^{in})+\tfrac{1}{N}\Vert F_N^{in}\Vert _1) \\&\quad \le 2\ell (t)({\text {trace}}_{{\mathfrak {H}}_N}({\mathcal {M}}_N(t)(P(t))F_N^{in})+\tfrac{1}{N}{\text {trace}}_{{\mathfrak {H}}_N}(F_N^{in})) \end{aligned} \end{aligned}$$

(48)

by (42).

Next we use the following elementary observation.

Lemma 7.1

Let $T=T^*\in {\mathcal {L}}({\mathfrak {H}}_N)$ satisfy

$$\begin{aligned} U_{\sigma }TU_{\sigma }^*=T\text { for all }{\sigma }\in {\mathfrak {S}}_N,\quad \text { and }{\text {trace}}_{{\mathfrak {H}}_N}(TF)\ge 0 \end{aligned}$$

for each $F\in {\mathcal {L}}({\mathfrak {H}}_N)$ satisfying (31). Then $T\ge 0$.

Proof

Indeed, we seek to prove that

$$\begin{aligned} \langle \Psi |T|\Psi \rangle \ge 0\qquad \text { for each }\Psi \in {\mathfrak {H}}_N. \end{aligned}$$

For each $\Psi \in {\mathfrak {H}}_N$ such that $\Vert \Psi _N\Vert _{{\mathfrak {H}}_N}=1$, set

$$\begin{aligned} F=\tfrac{1}{N!}\sum _{{\sigma }\in {\mathfrak {S}}_N}|U_{\sigma }\Psi \rangle \langle U_{\sigma }\Psi |. \end{aligned}$$

Then F satisfies (31), so that

$$\begin{aligned} 0\le {\text {trace}}(TF)=\tfrac{1}{N!}\sum _{{\sigma }\in {\mathfrak {S}}_N}\langle U_{\sigma }\Psi |T|U_{\sigma }\Psi \rangle =\tfrac{1}{N!}\sum _{{\sigma }\in {\mathfrak {S}}_N}\langle \Psi |U_{\sigma }^*TU_{\sigma }|\Psi \rangle =\langle \Psi |T|\Psi \rangle \end{aligned}$$

since $U_{\sigma }^*TU_{\sigma }=T$ for each ${\sigma }\in {\mathfrak {S}}_N$. Thus, $\langle \Psi |T|\Psi \rangle \ge 0$ for each $\Psi \in {\mathfrak {H}}_N$ such that $\Vert \Psi _N\Vert _{{\mathfrak {H}}_N}=1$, and thus for each $\Psi \in {\mathfrak {H}}_N\setminus \{0\}$ by normalization. $\square$

The inequality (48) implies that

$$\begin{aligned} \begin{aligned} 2\ell (t){\text {trace}}_{{\mathfrak {H}}_N}(({\mathcal {M}}_N(t)(P(t))+\tfrac{1}{N}I_{{\mathfrak {H}}_N})F_N^{in})\ge&|{\text {trace}}_{{\mathfrak {H}}_N}(T_2F_N^{in})| \\ \ge&{\text {trace}}_{{\mathfrak {H}}_N}(\pm iT_2F_N^{in}), \end{aligned} \end{aligned}$$

and we conclude from Lemma 7.1 that

$$\begin{aligned} \pm iT_2\le 2\ell (t)({\mathcal {M}}_N(t)(P(t))+\tfrac{1}{N}I_{{\mathfrak {H}}_N})\,. \end{aligned}$$

(49)

7.3 Bound for $T_1$

Next we estimate

$$\begin{aligned} \begin{aligned} S_1:=&{\mathcal {T}}(V,{\mathcal {M}}_N(t)(P(t){\bullet }P(t)),{\mathcal {M}}_N(t)(P(t){\bullet }R(t))) \\ =&{\mathcal {U}}_N(t)^*{\mathcal {T}}(V,{\mathcal {M}}_N^{in}(P(t){\bullet }P(t)),{\mathcal {M}}_N^{in}(P(t){\bullet }R(t))){\mathcal {U}}_N(t) \\ =&\tfrac{1}{N}\sum _{k=1}^N{\mathcal {U}}_N(t)^*{\mathcal {T}}(V,J_k(P(t){\bullet }P(t)),{\mathcal {M}}_N^{in}(P(t){\bullet }R(t))){\mathcal {U}}_N(t). \end{aligned} \end{aligned}$$

Observe that

$$\begin{aligned} \begin{aligned}&{\mathcal {T}}(V,J_k(P(t){\bullet }P(t)),{\mathcal {M}}_N^{in}(P(t){\bullet }R(t))) \\&\quad =\int _{{\textbf{R}}^3}{\hat{V}}({\omega })(J_kP(t))J_k(P(t)E_{\omega }^*P(t)){\mathcal {M}}_N^{in}(P(t)E_{\omega }R(t))(J_kP(t))\tfrac{\hbox {d}{\omega }}{(2\pi )^3} \\&\quad \quad +\tfrac{1}{N}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })(J_kP(t))J_k(P(t)E_{\omega }^*P(t))[J_kP(t),J_k(P(t)E_{\omega }R(t))]\tfrac{\hbox {d}{\omega }}{(2\pi )^3}, \end{aligned} \end{aligned}$$

since $P(t)=P(t)^2$, so that $J_kP(t)=(J_kP(t))^2$. Then

$$\begin{aligned} {[}J_kP(t),J_k(P(t)E_{\omega }R(t))]=J_k(P(t)E_{\omega }R(t)), \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned}&J_k(P(t)E_{\omega }^*P(t))[J_kP(t),J_k(P(t)E_{\omega }R(t))]\\&\quad =J_k(P(t)E_{\omega }^*P(t)E_{\omega }R(t)) \\&\quad =J_k(P(t)E_{\omega }^*(I-R(t))E_{\omega }R(t))=-{\mathcal {F}}(|\psi (t,\cdot )|^2)(-{\omega })J_k(P(t)E_{\omega }^*R(t)). \end{aligned} \end{aligned}$$

Hence, (12) implies that

$$\begin{aligned} \begin{aligned}&\tfrac{1}{N}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })(J_kP(t))J_k(P(t)E_{\omega }^*P(t))[J_kP(t),J_k(P(t)E_{\omega }R(t))]\tfrac{\hbox {d}{\omega }}{(2\pi )^3} \\&\quad =-\tfrac{1}{N}(J_kP(t))J_k(P(t)(V\star |\psi (t,\cdot )|^2)R(t)). \end{aligned} \end{aligned}$$

On the other hand,

$$\begin{aligned} \begin{aligned}&\int _{{\textbf{R}}^3}{\hat{V}}({\omega })(J_kP(t))J_k(P(t)E_{\omega }^*P(t)){\mathcal {M}}_N^{in}(P(t)E_{\omega }R(t))(J_kP(t))\tfrac{\hbox {d}{\omega }}{(2\pi )^3} \\&\quad =\tfrac{1}{N}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })(J_kP(t))J_k(P(t)E_{\omega }^*P(t))\sum _{\begin{array}{c} l=1\\ l\not =k \end{array}}^NJ_l(P(t)E_{\omega }R(t))J_kP(t)\tfrac{\hbox {d}{\omega }}{(2\pi )^3} \\&\quad =\tfrac{1}{N}\int _{{\textbf{R}}^3}{\hat{V}}({\omega })(J_kP(t))J_k(E_{\omega }^*)\sum _{\begin{array}{c} l=1\\ l\not =k \end{array}}^NJ_l(P(t)E_{\omega }R(t))J_kP(t)\tfrac{\hbox {d}{\omega }}{(2\pi )^3} \\&\quad =(J_kP(t))\left( \tfrac{1}{N}\sum _{\begin{array}{c} l=1\\ l\not =k \end{array}}^N(J_kP(t))(J_lP(t))V_{kl}(J_lR(t))(J_kP(t))\right) (J_kP(t)), \end{aligned} \end{aligned}$$

since $J_k(P(t)E_{\omega }R(t))J_k(P(t))=0$, with $V_{kl}$ defined as in (29).

Hence,

$$\begin{aligned} \begin{aligned} S_1=&\tfrac{1}{N^2}\sum _{1\le k\not =l\le N}{\mathcal {U}}_N(t)^*(J_kP(t))^2(J_lP(t))V_{kl}(J_lR(t))(J_kP(t))^2{\mathcal {U}}_N(t) \\&-\tfrac{1}{N^2}\sum _{k=1}^N{\mathcal {U}}_N(t)^*(J_kP(t))J_k(P(t)(V\star |\psi (t,\cdot )|^2)R(t)){\mathcal {U}}_N(t). \end{aligned} \end{aligned}$$

Therefore, by cyclicity of the trace, for each $F_N^{in}\in {\mathcal {L}}({\mathfrak {H}}_N)$ satisfying (31), denoting $F_N(t):={\mathcal {U}}_N(t)F_N^{in}{\mathcal {U}}_N(t)^*$, one has

$$\begin{aligned} \begin{aligned}&{\text {trace}}_{{\mathfrak {H}}_N}(S_1F_N^{in}) \\&\quad =\tfrac{1}{N^2}\sum _{1\le k\not =l\le N}{\text {trace}}_{{\mathfrak {H}}_N}((J_kP(t))(J_lP(t))V_{kl}(J_lR(t))(J_kP(t))^2F_N(t)(J_kP(t))) \\&\quad \quad -\tfrac{1}{N^2}\sum _{k=1}^N{\text {trace}}_{{\mathfrak {H}}_N}(J_k(P(t)(V\star |\psi (t,\cdot )|^2)R(t))F_N(t)(J_kP(t))), \end{aligned} \end{aligned}$$

so that

$$\begin{aligned}&|{\text {trace}}_{{\mathfrak {H}}_N}(S_1F_N^{in})| \nonumber \\&\quad \le \!\tfrac{1}{N^2}\!\sum _{1\le k\not =l\le N}\!\Vert \!(J_kP(t))\!(J_lP(t))\!V_{kl}\!(J_lR(t))\!(J_kP(t))\!\Vert \Vert \!(J_kP(t))\!F_N(t)\!(J_kP(t)))\!\Vert _1 \nonumber \\&\quad \quad +\tfrac{1}{N^2}\sum _{k=1}^N\Vert J_k(P(t)(V\star |\psi (t,\cdot )|^2)R(t))\Vert \Vert F_N(t)\Vert _1\Vert J_kP(t))\Vert \nonumber \\&\quad \le (1-\tfrac{1}{N})\Vert V_{12}J_2R(t)\Vert {\text {trace}}_{{\mathfrak {H}}_N}(F_N^{in}{\mathcal {M}}_N(t)(P(t))) \nonumber \\&\quad \quad +\tfrac{2}{N}\Vert (V\star |\psi (t,\cdot )|^2)R(t)\Vert \Vert F_N(t)\Vert _1. \end{aligned}$$

(50)

Finally

$$\begin{aligned} \begin{aligned} T_1=&{\mathcal {T}}(V,{\mathcal {M}}_N(t)(P(t){\bullet }P(t)),{\mathcal {M}}_N(t)(P(t){\bullet }R(t))) \\&-{\mathcal {T}}(V,{\mathcal {M}}_N(t)(R(t){\bullet }P(t)),{\mathcal {M}}_N(t)(P(t){\bullet }P(t)))=S_1-S_1^*=-T_1^* \end{aligned} \end{aligned}$$

because of Lemma 4.3, so that

$$\begin{aligned} \begin{aligned} |{\text {trace}}_{{\mathfrak {H}}_N}(T_1F_N^{in})|\le&2(1-\tfrac{1}{N})\Vert V_{12}J_2R(t)\Vert {\text {trace}}_{{\mathfrak {H}}_N}(F_N^{in}{\mathcal {M}}_N(t)(P(t))) \\&+\tfrac{4}{N}\Vert (V\star |\psi (t,\cdot )|^2)R(t)\Vert \Vert F_N(t)\Vert _1. \end{aligned} \end{aligned}$$

Since R(s) is a rank-one orthogonal projection

$$\begin{aligned} \begin{aligned} \Vert V_{12}(J_2R(t))\Vert ^2=&\Vert (J_2R(t))V^2_{12}(J_2R(t))\Vert \\ =&\Vert (V^2\star |\psi (t,\cdot )|^2)\otimes R(s)\Vert \le \Vert (V^2\star |\psi (t,\cdot )|^2\Vert _{L^\infty }=\ell (t)^2\,. \end{aligned} \end{aligned}$$

(51)

Thus,

$$\begin{aligned} \begin{aligned} |{\text {trace}}_{{\mathfrak {H}}_N}(T_1F_N^{in})|\le&2\ell (t)\left( (1\!-\!\tfrac{1}{N}){\text {trace}}_{{\mathfrak {H}}_N}(F_N^{in}{\mathcal {M}}_N(t)(P(t)))\!+\!\tfrac{2}{N}\Vert F_N^{in}\Vert _1\right) \\ =&2\ell (t)\left( (1\!-\!\tfrac{1}{N}){\text {trace}}_{{\mathfrak {H}}_N}(F_N^{in}{\mathcal {M}}_N(t)(P(t)))\!+\!\tfrac{2}{N}{\text {trace}}_{{\mathfrak {H}}_N}(F_N^{in})\right) . \end{aligned} \end{aligned}$$

(52)

In particular,

$$\begin{aligned} 2\ell (t){\text {trace}}_{{\mathfrak {H}}_N}\left( F_N^{in}\left( (1\!-\!\tfrac{1}{N}){\mathcal {M}}_N(t)(P(t)))+\tfrac{2}{N}I_{{\mathfrak {H}}_N}\right) \right) \ge {\text {trace}}_{{\mathfrak {H}}_N}(\pm iT_1F_N^{in}) \end{aligned}$$

and since this inequality holds for each $F_N^{in}\in {\mathcal {L}}_s({\mathfrak {H}}_N)$ such that $F_N^{in}=(F_N^{in})^*\ge 0$, we conclude from Lemma 7.1 that

$$\begin{aligned} \pm iT_1\le 2\ell (t)\left( (1\!-\!\tfrac{1}{N}){\mathcal {M}}_N(t)(P(t)))+\tfrac{2}{N}I_{{\mathfrak {H}}_N}\right) . \end{aligned}$$

(53)

7.4 The operator $\Pi _N$

In order to treat the last term $T_3$, we need the following auxiliary lemma—see the formula preceding (13) in [18].

Lemma 7.2

Let $R=R^*$ be a rank-one projection on ${\mathfrak {H}}$ and let $P:=I-R$. Set $\Pi _N:={\mathcal {M}}_N^{in}P$. For each $N>1$,

$$\begin{aligned} \Pi ^*_N=\Pi _N,\quad \Pi _N^2\ge \tfrac{1}{N}\Pi _N,\qquad \text { and }\quad {\text {Ker}}\Pi _N={\text {Ker}}(I-R^{\otimes N}), \end{aligned}$$

so that

$$\begin{aligned} \Pi _N\ge \tfrac{1}{N}(1-R^{\otimes N}). \end{aligned}$$

In particular, there exists a pseudo-inverse $\Pi _N^{-1}:\,({\text {Ker}}\Pi _N)^\perp \rightarrow ({\text {Ker}}\Pi _N)^\perp$, with extension by 0 on ${\text {Ker}}\Pi _N$ also (abusively) denoted $\Pi _N$, such that

$$\begin{aligned} \Pi _N^{-1}\Pi _N=\Pi _N\Pi _N^{-1}=I-R^{\otimes N}\,. \end{aligned}$$

(54)

In [18], the definition of the pseudo-inverse of $\Pi _N$ immediately follows from formula (6), which can be viewed as the spectral decomposition of $\Pi _N$. The proof below is quite straightforward and avoids using the clever argument leading to formula (6) in [18], which is not entirely obvious unless one already knows the result.

Proof

That $\Pi _N$ is self-adjoint is obvious by definition of ${\mathcal {M}}_N^{in}$. Then,

$$\begin{aligned} \Pi _N^2=\frac{1}{N^2}\left( \sum _{k=1}^NJ_kP+2\sum _{1\le k<l\le N}J_kPJ_lP\right) \ge \frac{1}{N^2}\sum _{k=1}^NJ_kP=\frac{1}{N}\Pi _N. \end{aligned}$$

If $X\in {\text {Ker}}\Pi _N$, one has, for each $k=1,\ldots ,N$,

$$\begin{aligned} 0=\sum _{k=1}^N\langle X|J_kP|X\rangle \implies \langle X|J_kP|X\rangle =0\implies J_kPX=0. \end{aligned}$$

Hence,

$$\begin{aligned} X=J_NRX=J_{N-1}RX=\ldots =J_2RX=J_1RX \end{aligned}$$

so that

$$\begin{aligned} X=J_NRX=J_NRJ_{N-1}RX=\ldots =J_NRJ_{N-1}R\ldots J_2RJ_1RX=R^{\otimes N}X. \end{aligned}$$

Thus, ${\text {Ker}}\Pi _N={\text {Ker}}(I-R^{\otimes N})$. Finally,

$$\begin{aligned} \Pi _N^3=\Pi _N^{1/2}\Pi _N^2\Pi _N^{1/2}\ge \frac{1}{N}\Pi _N^{1/2}\Pi _N\Pi _N^{1/2}=\frac{1}{N}\Pi _N^2. \end{aligned}$$

Therefore, for each $X\in {\mathfrak {H}}_N$, one has

$$\begin{aligned} \langle \Pi _NX|\Pi _N|\Pi _NX\rangle \ge \frac{1}{N}\Vert \Pi _NX\Vert ^2. \end{aligned}$$

Since $\Pi _N=\Pi _N^*$, one has

$$\begin{aligned} \overline{{\text {Im}}\Pi _N}=({\text {Ker}}\Pi _N)^\perp \end{aligned}$$

(see for instance Corollary 2.18 (iv) in chapter 2 of [5]). Since

$$\begin{aligned} \langle Y|\Pi _N|Y\rangle \ge \frac{1}{N}\Vert Y\Vert ^2,\qquad Y\in {\text {Im}}\Pi _N, \end{aligned}$$

and since one has obviously $\Vert \Pi _N\Vert \le 1$, a straightforward density argument shows that

$$\begin{aligned} \langle Y|\Pi _N|Y\rangle \ge \frac{1}{N}\Vert Y\Vert ^2,\qquad Y\in ({\text {Ker}}\Pi _N)^\perp . \end{aligned}$$

Hence

$$\begin{aligned} \Pi _N\ge \tfrac{1}{N}(1-R^{\otimes N}). \end{aligned}$$

The existence of the pseudo-inverse $\Pi _N^{-1}$ follows from this inequality. $\square$

7.5 Bound for $T_3$

Finally, we treat the term $T_3$. Set

$$\begin{aligned} \begin{aligned} S_3=&{\mathcal {T}}(V,{\mathcal {M}}_N(t)(P(t){\bullet }R(t)),{\mathcal {M}}_N(t)(P(t){\bullet }R(t))) \\ =&{\mathcal {U}}_N(t)^*{\mathcal {T}}(V,{\mathcal {M}}_N^{in}(P(t){\bullet }R(t)),{\mathcal {M}}_N^{in}(P(t){\bullet }R(t))){\mathcal {U}}_N(t). \end{aligned} \end{aligned}$$

One easily checks that

$$\begin{aligned} \begin{aligned}&{\mathcal {T}}(V,{\mathcal {M}}_N^{in}(P(t){\bullet }R(t)),{\mathcal {M}}_N^{in}(P(t){\bullet }R(t))) \\&\quad =\int _{{\textbf{R}}^3}{\hat{V}}({\omega }){\mathcal {M}}_N^{in}(P(t)E_{\omega }^*R(t)){\mathcal {M}}_N^{in}(P(t)E_{\omega }R(t))\tfrac{\hbox {d}{\omega }}{(2\pi )^3} \\&\quad =\tfrac{1}{N^2}\sum _{1\le k\not =l\le N}(J_lP(t))(J_kP(t))V_{kl}(J_kR(t))(J_lR(t)). \end{aligned} \end{aligned}$$

At this point, we set $\Pi _N(t):={\mathcal {M}}_N^{in}P(t)$ and use Lemma 7.2 to define the pseudo-inverse $\Pi _N(t)^{-1}$. One has $\Pi _N(t)=\Pi _N(t)^*\ge 0$, thus $\Pi _N(t)^{-1}=(\Pi _N(t)^{-1})^*\ge 0$ on ${\text {Ker}}(I-R(t)^{\otimes N})$. Abusing the notation $\Pi _N(t)^{-1/2}$ to designate the linear map $(\Pi _N(t)^{-1})^{1/2}$, we deduce from (54) that

$$\begin{aligned} \Pi _N(t)^{1/2}\Pi _N(t)^{-1/2}=I-R(t)^{\otimes N}, \end{aligned}$$

so that

$$\begin{aligned} (J_kP(t))\Pi _N(t)^{1/2}\Pi _N(t)^{-1/2}=\Pi _N(t)^{1/2}\Pi _N(t)^{-1/2}(J_kP(t))=J_lP(t). \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned}&{\mathcal {T}}(V,{\mathcal {M}}_N^{in}(P(t){\bullet }R(t)),{\mathcal {M}}_N^{in}(P(t){\bullet }R(t))) \\&\quad =\tfrac{1}{N^2}\!\sum _{1\le k\not =l\le N}\!(J_lP(t))(J_kP(t))\Pi _N(t)^{-\frac{1}{2}}\Pi _N(t)^{\frac{1}{2}}V_{kl}(J_kR(t))(J_lR(t)), \end{aligned} \end{aligned}$$

and we study the quantity

$$\begin{aligned} \begin{aligned} {\text {trace}}_{{\mathfrak {H}}_N}(S_3F_N^{in})&={\text {trace}}_{{\mathfrak {H}}_N}((F_N^{in})^\frac{1}{2}S_2(F_N^{in})^\frac{1}{2}) \\&{=}{\text {trace}}_{{\mathfrak {H}}_N}(F_N(t)^\frac{1}{2}{\mathcal {T}}(V,{\mathcal {M}}_N^{in}(P(t){\bullet }R(t)),{\mathcal {M}}_N^{in}(P(t){\bullet }R(t)))F_N(t)^\frac{1}{2}), \end{aligned} \end{aligned}$$

where $F_N(t)={\mathcal {U}}_N(t)F_N^{in}{\mathcal {U}}_N(t)^*$, for each $F_N^{in}\in {\mathcal {L}}({\mathfrak {H}}_N)$ satisfying (31). Observe that

$$\begin{aligned} \begin{aligned}&|\!{\text {trace}}_{{\mathfrak {H}}_N}\!(F_N(t)^\frac{1}{2}\!(J_lP(t))(J_kP(t))\Pi _N(t)^{-\frac{1}{2}}\!\Pi _N(t)^{\frac{1}{2}}\!V_{kl}(J_kR(t))(J_lR(t))F_N(t)^\frac{1}{2}\!)| \\&\quad \le \Vert \Pi _N(t)^{-\frac{1}{2}}(J_kP(t))(J_lP(t))F_N(t)^\frac{1}{2}\Vert _2\Vert \Pi _N(t)^{\frac{1}{2}}V_{kl}(J_kR(t))(J_lR(t))F_N(t)^\frac{1}{2}\Vert _2, \end{aligned} \end{aligned}$$

so that, by the Cauchy–Schwarz inequality,

$$\begin{aligned} \begin{aligned} |{\text {trace}}_{{\mathfrak {H}}_N}(S_3F_N^{in})|\le&\tfrac{1}{N^2}\left( \sum _{1\le k\not =l\le N}\Vert \Pi _N(t)^{-\frac{1}{2}}(J_kP(t))(J_lP(t))F_N(t)^\frac{1}{2}\Vert _2^2\right) ^{1/2} \\&\times \left( \sum _{1\le k\not =l\le N}\Vert \Pi _N(t)^{\frac{1}{2}}V_{kl}(J_kR(t))(J_lR(t))F_N(t)^\frac{1}{2}\Vert _2^2\right) ^{1/2}. \end{aligned} \end{aligned}$$

First, one has

$$\begin{aligned} \begin{aligned}&\Vert \Pi _N(t)^{-\frac{1}{2}}(J_kP(t))(J_lP(t))F_N(t)^\frac{1}{2}\Vert _2^2 \\&\quad ={\text {trace}}_{{\mathfrak {H}}_N}(F_N(t)^\frac{1}{2}(J_lP(t))(J_kP(t))\Pi _N(t)^{-1}(J_kP(t))(J_lP(t))F_N(t)^\frac{1}{2}) \\&\quad ={\text {trace}}_{{\mathfrak {H}}_N}(F_N(t)^\frac{1}{2}\Pi _N(t)^{-1}(J_kP(t))(J_lP(t))F_N(t)^\frac{1}{2}) \\&\quad ={\text {trace}}_{{\mathfrak {H}}_N}(\Pi _N(t)^{-1}(J_kP(t))(J_lP(t))F_N(t)), \end{aligned} \end{aligned}$$

(the second equality follows from the fact that $J_k(P(t))$ commutes with $\Pi _N(t)$ and $\Pi _N(t)^{-1}$), so that

$$\begin{aligned} \begin{aligned}&\sum _{1\le k\not =l\le N}\Vert \Pi _N(t)^{-\frac{1}{2}}(J_kP(t))(J_lP(t))F_N(t)^\frac{1}{2}\Vert _2^2 \\&\quad \le {\text {trace}}_{{\mathfrak {H}}_N}\left( \Pi _N(t)^{-1}\sum _{1\le k,l\le N}(J_kP(t))(J_lP(t))F_N(t)\right) \\&\quad =N^2{\text {trace}}_{{\mathfrak {H}}_N}(\Pi _N(t)^{-1}\Pi _N(t)^2F_N(t))=N^2{\text {trace}}_{{\mathfrak {H}}_N}(\Pi _N(t)F_N(t)). \end{aligned} \end{aligned}$$

The inequality above follows from the fact that

$$\begin{aligned} \begin{aligned}&{\text {trace}}_{{\mathfrak {H}}_N}(\Pi _N(t)^{-1}(J_kP(t))^2F_N(t)) \\&\quad ={\text {trace}}_{{\mathfrak {H}}_N}(F_N(t)^\frac{1}{2}(J_kP(t))\Pi _N(t)^{-1}(J_kP(t))F_N(t)^\frac{1}{2})\ge 0. \end{aligned} \end{aligned}$$

On the other hand,

$$\begin{aligned}{} & {} \sum _{1\le k\not =l\le N}\Vert \Pi _N(t)^{\frac{1}{2}}V_{kl}(J_kR(t))(J_lR(t))F_N(t)^\frac{1}{2}\Vert _2^2 \\{} & {} \quad =\sum _{1\le k\not =l\le N}{\text {trace}}(F_N(t)^\frac{1}{2}(J_lR(t))(J_kR(t))V_{kl}\Pi _N(t)V_{kl}(J_kR(t))(J_lR(t))F_N(t)^\frac{1}{2}) \\{} & {} \quad =\tfrac{1}{N}\sum _{1\le k\not =l\le N}\Vert J_k(P(t))V_{kl}(J_kR(t))(J_lR(t))F_N(t)^\frac{1}{2}\Vert _2^2 \\{} & {} \quad \quad +\tfrac{1}{N}\sum _{1\le k\not =l\le N}\Vert J_l(P(t))V_{kl}(J_kR(t))(J_lR(t))F_N(t)^\frac{1}{2}\Vert _2^2 \\{} & {} \quad \quad +\tfrac{1}{N}\sum _{1\le m\not =k\not =l\le N}\Vert (J_mP(t))V_{kl}(J_kR(t))(J_lR(t))F_N(t)^\frac{1}{2}\Vert _2^2 \\{} & {} \quad \le \tfrac{2}{N}\sum _{1\le k\not =l\le N}\Vert V_{kl}(J_kR(t))(J_lR(t))F_N(t)^\frac{1}{2}\Vert _2^2 \\{} & {} \quad \quad +\tfrac{1}{N}\sum _{1\le m\not =k\not =l\le N}\Vert (J_mP(t))V_{kl}(J_kR(t))(J_lR(t))F_N(t)^\frac{1}{2}\Vert _2^2. \end{aligned}$$

Now, $m\notin \{k,l\}$ implies that $J_mP(t)$ commutes with $V_{kl}$, $J_kR(t)$ and $J_lR(t)$, so that

$$\begin{aligned}{} & {} \Vert (J_mP(t))V_{kl}(J_kR(t))(J_lR(t))F_N(t)^\frac{1}{2}\Vert _2^2 \\{} & {} \quad =\Vert V_{kl}(J_kR(t))(J_lR(t))(J_mP(t))F_N(t)^\frac{1}{2}\Vert _2^2 \\{} & {} \quad \le \Vert V_{kl}(J_kR(t))(J_lR(t))\Vert ^2\Vert (J_mP(t))F_N(t)^\frac{1}{2}\Vert _2^2 \\{} & {} \quad =\Vert V_{12}R(t)\otimes R(t)\Vert ^2{\text {trace}}_{{\mathfrak {H}}_N}((J_mP(t))F_N(t)(J_mP(t))) \\{} & {} \quad =\Vert V_{12}R(t)\otimes R(t)\Vert ^2{\text {trace}}_{{\mathfrak {H}}_N}(\Pi _N(t)F_N(t)). \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned}&|{\text {trace}}_{{\mathfrak {H}}_N}(S_3F_N^{in})| \\&\quad \le \tfrac{1}{N^2}\cdot N{\text {trace}}_{{\mathfrak {H}}_N}(\Pi _N(t)F_N(t))^\frac{1}{2}\left( \tfrac{2N(N-1)}{N}\Vert V_{12}R(t)\!\otimes \!R(t)\Vert ^2\Vert F_N(t)\Vert _1\right. \\&\quad \quad \left. \!+\!\tfrac{N(N-1)(N-2)}{N}\Vert V_{12}R(t)\!\otimes \!R(t)\Vert ^2{\text {trace}}_{{\mathfrak {H}}_N}(\Pi _N(t)F_N(t))\right) ^\frac{1}{2} \\&\quad \le \!\tfrac{1}{\sqrt{N}}\Vert V_{12}R(t)\!\otimes \!R(t)\Vert {\text {trace}}_{{\mathfrak {H}}_N}({\mathcal {M}}_N(t)(P(t))F_N^{in})^\frac{1}{2} \\&\quad \quad \times \left( 2\Vert F_N(t)\Vert _1+\!(N\!-\!2)\!{\text {trace}}_{{\mathfrak {H}}_N}({\mathcal {M}}_N(t)(P(t))F_N^{in})\right) ^\frac{1}{2}. \end{aligned} \end{aligned}$$

(55)

Now

$$\begin{aligned} \begin{aligned} T_3=&{\mathcal {T}}(V,{\mathcal {M}}_N(t)(P(t){\bullet }R(t)),{\mathcal {M}}_N(t)(P(t){\bullet }R(t)))\\&-{\mathcal {T}}(V,{\mathcal {M}}_N(t)(R(t){\bullet }P(t)),{\mathcal {M}}_N(t)(R(t){\bullet }P(t)))=S_3-S_3^*=-T_3^*, \end{aligned} \end{aligned}$$

according to Lemma 4.3. Thus, (55) implies that

$$\begin{aligned} \begin{aligned} |{\text {trace}}_{{\mathfrak {H}}_N}(T_3F_N^{in})|\le&2\Vert V_{12}R(t)\!\otimes \!R(t)\Vert {\text {trace}}_{{\mathfrak {H}}_N}({\mathcal {M}}_N(t)(P(t))F_N^{in})^\frac{1}{2} \\&\times \left( \tfrac{2}{N}\Vert F_N(t)\Vert _1+\tfrac{N-2}{N}{\text {trace}}_{{\mathfrak {H}}_N}({\mathcal {M}}_N(t)(P(t))F_N^{in})\right) ^\frac{1}{2}. \end{aligned} \end{aligned}$$

According to (51)

$$\begin{aligned} \Vert V_{12}R(t)\otimes R(t)\Vert =\Vert V_{12}(J_1R(t))(J_2R(t))\Vert \le \Vert V_{12}J_2R(t)\Vert \le \ell (t), \end{aligned}$$

so that

$$\begin{aligned} |{\text {trace}}_{{\mathfrak {H}}_N}(T_3F_N^{in})|\le 2\ell (t)\left( (1-\tfrac{1}{N}){\text {trace}}_{{\mathfrak {H}}_N}({\mathcal {M}}_N(t)(P(t))F_N^{in})+\tfrac{2}{N}\Vert F_N^{in}\Vert _1\right) \,. \end{aligned}$$

(56)

In particular,

$$\begin{aligned} {\text {trace}}_{{\mathfrak {H}}_N}(\pm iT_3F_N^{in})|\le 2\ell (t){\text {trace}}_{{\mathfrak {H}}_N}\left( F_N^{in}\left( (1-\tfrac{1}{N}){\mathcal {M}}_N(t)(P(t))+\tfrac{1}{N}I_{{\mathfrak {H}}_N}\right) \right) . \end{aligned}$$

Since this last inequality holds for each $F_N^{in}\in {\mathcal {L}}({\mathfrak {H}}_N)$ satisfying (31), we deduce from Lemma 7.1 that

$$\begin{aligned} \pm iT_3\le 2\ell (t)\left( (1-\tfrac{1}{N}){\mathcal {M}}_N(t)(P(t))+\tfrac{1}{N}I_{{\mathfrak {H}}_N}\right) \,. \end{aligned}$$

(57)

8 Proofs of part (2) in Theorem 4.1 and Corollary 4.2

8.1 Proof of part (2) in Theorem 4.1

Applying Lemma 4.4 shows that

$$\begin{aligned} \pm i{\mathcal {C}}(V,{\mathcal {M}}_N(t)-{\mathcal {R}}(t),{\mathcal {M}}_N(t))(R(t))=\pm i(T_1+T_2+T_3+T_4). \end{aligned}$$

With Lemma 4.5, this shows that

$$\begin{aligned} (\pm i{\mathcal {C}}(V,{\mathcal {M}}_N(t)-{\mathcal {R}}(t),{\mathcal {M}}_N(t))(R(t)))^*=\pm i{\mathcal {C}}(V,{\mathcal {M}}_N(t)-{\mathcal {R}}(t),{\mathcal {M}}_N(t))(R(t)) \end{aligned}$$

and that

$$\begin{aligned} \pm i{\mathcal {C}}(V,{\mathcal {M}}_N(t)-{\mathcal {R}}(t),{\mathcal {M}}_N(t))(R(t))\le 6\ell (t)\left( {\mathcal {M}}_N(t)(P(t))+\tfrac{2}{N}I_{{\mathfrak {H}}_N}\right) . \end{aligned}$$

It remains to bound the function

$$\begin{aligned} \ell (t):=\Vert V^2\star |\psi (t,\cdot )|^2\Vert _{L^\infty ({\textbf{R}}^3)}^{1/2}. \end{aligned}$$

Since

$$\begin{aligned} V=V_1+V_2\text { with }V_1\in {\mathcal {F}}L^1({\textbf{R}}^3)\subset L^\infty ({\textbf{R}}^3)\text { and }V_2\in L^2({\textbf{R}}^3) \end{aligned}$$

one has

$$\begin{aligned} \begin{aligned} 0\le&V^2\star |\psi (t,\cdot )|^2\le 2V_1^2\star |\psi (t,\cdot )|^2+2V_2^2\star |\psi (t,\cdot )|^2 \\ \le&2\Vert V_1\Vert _{L^\infty ({\textbf{R}}^3)}^2\Vert \psi (t,\cdot )\Vert _{L^2({\textbf{R}}^3)}^2+2\Vert V_2\Vert _{L^2({\textbf{R}}^3)}^2\Vert \psi (t,\cdot )\Vert _{L^\infty ({\textbf{R}}^3)}^2. \end{aligned} \end{aligned}$$

Minimizing $\Vert V_1\Vert _{L^\infty ({\textbf{R}}^3)}+\Vert V_2\Vert _{L^2({\textbf{R}}^3)}$ over all possible decompositions of $V=V_1+V_2$ as above, one has

$$\begin{aligned} \begin{aligned} 0\le V^2\star |\psi (t,\cdot )|^2\le&4\Vert V\Vert ^2_{L^2({\textbf{R}}^3)+L^\infty ({\textbf{R}}^3)}\max (\Vert \psi (t,\cdot )\Vert _{L^2({\textbf{R}}^3)}^2,\Vert \psi (t,\cdot )\Vert _{L^\infty ({\textbf{R}}^3)}^2) \\ \le&4\Vert V\Vert ^2_{L^2({\textbf{R}}^3)+L^\infty ({\textbf{R}}^3)}\max (\Vert \psi (t,\cdot )\Vert _{L^2({\textbf{R}}^3)}^2,C_S^2\Vert \psi (t,\cdot )\Vert _{H^2({\textbf{R}}^3)}^2) \\ \le&4\max (1,C_S)^2\Vert V\Vert ^2_{L^2({\textbf{R}}^3)+L^\infty ({\textbf{R}}^3)}\Vert \psi (t,\cdot )\Vert _{H^2({\textbf{R}}^3)}^2=:L(t)^2. \end{aligned} \end{aligned}$$

8.2 Proof of Corollary 4.2

Pickl’s functional defined in [18] and recalled in formula (18) can be recast as

$$\begin{aligned} {\alpha }_N(t):={\text {trace}}_{{\mathfrak {H}}}(F_{N:1}(t)P(t)) \end{aligned}$$

(58)

(see Definition 2.2 and formula (6) in [18]), where $F_{N:1}(t)$ is the single-body reduced density operator deduced from

$$\begin{aligned} F_N(t):={\mathcal {U}}_N(t)F_N^{in}{\mathcal {U}}_N(t)^*, \end{aligned}$$

where $F_N^{in}\in {\mathcal {L}}({\mathfrak {H}}_N)$ satisfies (31). Specifically, $F_{N:1}(t)$ is defined by the formula

$$\begin{aligned} {\text {trace}}_{\mathfrak {H}}(F_{N:1}(t)A)={\text {trace}}_{{\mathfrak {H}}_N}(F_N(t)J_1A),\qquad \text { for all }A\in {\mathcal {L}}({\mathfrak {H}}). \end{aligned}$$

Since $F_N(t)$ satisfies (31), it holds

$$\begin{aligned} \begin{aligned} {\text {trace}}_{\mathfrak {H}}(F_{N:1}(t)A)=&{\text {trace}}_{{\mathfrak {H}}_N}(F_N(t){\mathcal {M}}_N^{in}A) \\ =&{\text {trace}}_{{\mathfrak {H}}_N}(F_N^{in}{\mathcal {M}}_N(t)A)\,,\qquad \text { for all }A\in {\mathcal {L}}({\mathfrak {H}})\,. \end{aligned} \end{aligned}$$

(59)

This is Lemma 2.3 in [8], and the raison d’être of ${\mathcal {M}}_N(t)$. Thus, formula (18) and (58) are indeed equivalent.

8.2.1 The Gronwall inequality for Pickl’s functional

One deduces from part (2) in Theorem 4.1 that

$$\begin{aligned} \begin{aligned} {\mathcal {M}}_N(t)(P(t))=&{\mathcal {M}}_N^{in}(P(0))+\tfrac{1}{{\hbar }}\int _0^t-i{\mathcal {C}}(V,{\mathcal {M}}_N(s)-{\mathcal {R}}(s),{\mathcal {M}}_N(s))(R(s))\hbox {d}s \\ \le&{\mathcal {M}}_N^{in}(P(0))+\tfrac{6}{{\hbar }}\int _0^tL(s)\left( {\mathcal {M}}_N(s)(P(s))+\tfrac{2}{N}I_{{\mathfrak {H}}_N}\right) \hbox {d}s. \end{aligned} \end{aligned}$$

This inequality implies that

$$\begin{aligned} \begin{aligned}&{\text {trace}}_{{\mathfrak {H}}_N}((F_N^{in})^\frac{1}{2}{\mathcal {M}}_N(t)(P(t))(F_N^{in})^\frac{1}{2})\\&\quad \le {\text {trace}}_{{\mathfrak {H}}_N}((F_N^{in})^\frac{1}{2}{\mathcal {M}}_N^{in}(P(0))(F_N^{in})^\frac{1}{2}) \\&\quad \quad +\tfrac{6}{{\hbar }}\int _0^tL(s)\left( {\text {trace}}_{{\mathfrak {H}}_N}((F_N^{in})^\frac{1}{2}{\mathcal {M}}_N(s)(P(s))(F_N^{in})^\frac{1}{2})+\tfrac{2}{N}{\text {trace}}_{{\mathfrak {H}}_N}(F_N^{in})\right) \hbox {d}s. \end{aligned} \end{aligned}$$

Now, by cyclicity of the trace and (59),

$$\begin{aligned} \begin{aligned} {\text {trace}}_{{\mathfrak {H}}_N}((F_N^{in})^\frac{1}{2}{\mathcal {M}}_N(t)(P(t))(F_N^{in})^\frac{1}{2})=&{\text {trace}}_{{\mathfrak {H}}_N}(F_N^{in}{\mathcal {M}}_N(t)(P(t))) \\ =&{\text {trace}}_{\mathfrak {H}}(F_{N:1}(t)P(t))={\alpha }_N(t), \end{aligned} \end{aligned}$$

so that, by Gronwall’s inequality,

$$\begin{aligned} {\alpha }_N(t)\le {\alpha }_N(0)\exp \left( \tfrac{6}{{\hbar }}\int _0^tL(s)\hbox {d}s\right) +\tfrac{2}{N}\left( \exp \left( \tfrac{6}{{\hbar }}\int _0^tL(s)\hbox {d}s\right) -1\right) . \end{aligned}$$

For instance, if $F_N^{in}=|\psi ^{in}\rangle \langle \psi ^{in}|^{\otimes N}$ with $\psi ^{in}\in {\mathfrak {H}}$ and $\Vert \psi ^{in}\Vert _{\mathfrak {H}}=1$, one has

$$\begin{aligned} {\alpha }_N(0)={\text {trace}}_{{\mathfrak {H}}_N}(R(0)^{\otimes N}{\mathcal {M}}_N^{in}(P(0)))={\text {trace}}_{\mathfrak {H}}(R(0)P(0))=0, \end{aligned}$$

so that

$$\begin{aligned} {\alpha }_N(t)\le \frac{2}{N}\left( \exp \left( \tfrac{6}{{\hbar }}\int _0^tL(s)\hbox {d}s\right) -1\right) =O\left( \frac{1}{N}\right) . \end{aligned}$$

8.2.2 Pickl’s functional and the trace norm

How the inequality above implies the mean-field limit is explained by the following lemma, which recaps the results stated as Lemmas 2.1 and 2.2 in [15], and whose proof is given below for the sake of keeping the present paper self-contained.

If $F_N^{in}\in {\mathcal {L}}({\mathfrak {H}}_N)$ satisfies (31), for each $m=1,\ldots ,N$, we denote by $F_{N:m}(t)$ the m-particle reduced density operator deduced from $F_N(t)={\mathcal {U}}_N(t)F_N^{in}{\mathcal {U}}_N(t)^*$, i.e.

$$\begin{aligned} {\text {trace}}_{{\mathfrak {H}}_m}(F_{N:m}(t)A_1\otimes \ldots \otimes A_m)={\text {trace}}_{{\mathfrak {H}}_N}(F_N(t)(J_1A_1)\ldots (J_mA_m)) \end{aligned}$$

for all $A_1,\ldots ,A_m\in {\mathcal {L}}({\mathfrak {H}})$.

Lemma 8.1

The Pickl functional satisfies the inequality

$$\begin{aligned} \Vert F_{N:m}(t)-R(t)^{\otimes m}\Vert _1\le 2\sqrt{2m{\text {trace}}_{{\mathfrak {H}}}(F_{N:1}(t)P(t))},\qquad m=1,\ldots ,N. \end{aligned}$$

Proof

Call ${\mathcal {P}}_-$ the spectral projection on the direct sums of eigenspaces of the trace-class operator $F_{N:m}(t)-R(t)^{\otimes m}$ corresponding to negative eigenvalues. Then, the self-adjoint operator

$$\begin{aligned} {\mathcal {P}}_-F_{N:m}(t){\mathcal {P}}_--{\mathcal {P}}_-R(t)^{\otimes m}{\mathcal {P}}_-={\mathcal {P}}_-F_{N:m}(t){\mathcal {P}}_--|{\mathcal {P}}_-\psi (t,\cdot )^{\otimes m}\rangle \langle {\mathcal {P}}_-\psi (t,\cdot )^{\otimes m}| \end{aligned}$$

must have only negative eigenvalues by definition of ${\mathcal {P}}_-$ and is obviously nonnegative on the orthogonal complement of ${\mathcal {P}}_-\psi (t,\cdot )^{\otimes m}$ in the range of ${\mathcal {P}}_-$. By definition of ${\mathcal {P}}_-$, this orthogonal complement must be $\{0\}$. Hence, ${\mathcal {P}}_-$ is a rank-one projection, so that $F_{N:m}(t)-R(t)^{\otimes m}$ has only one negative eigenvalue $\lambda _0$, with all its other eigenvalues ${\lambda }_1,{\lambda }_2,\ldots$ being nonnegative. Since

$$\begin{aligned} {\text {trace}}_{{\mathfrak {H}}_m}(F_{N:m}(t)-R(t)^{\otimes m})=\sum _{j\ge 1}\lambda _j+\lambda _0=0, \end{aligned}$$

one has⁵

$$\begin{aligned} \begin{aligned} \Vert F_{N:m}(t)-R(t)^{\otimes m}\Vert _1=\sum _{j\ge 1}\lambda _j+|\lambda _0|=2|\lambda _0|=&2\Vert F_{N:m}(t)-R(t)^{\otimes m}\Vert \\ \le&2\Vert F_{N:m}(t)-R(t)^{\otimes m}\Vert _2. \end{aligned} \end{aligned}$$

Now $F_{N:m}(t)$ is self-adjoint, and therefore

$$\begin{aligned} \begin{aligned} \Vert F_{N:m}(t)-R(t)^{\otimes m}\Vert _2^2=&{\text {trace}}_{{\mathfrak {H}}_m}((F_{N:m}(t)-R(t)^{\otimes m})^2) \\ =&{\text {trace}}_{{\mathfrak {H}}_m}(F_{N:m}(t)^2+R(t)^{\otimes m}) \\&-{\text {trace}}_{{\mathfrak {H}}_m}(F_{N:m}(t)R(t)^{\otimes m}+R(t)^{\otimes m}F_{N:m}(t)) \\ \le&2-2{\text {trace}}_{{\mathfrak {H}}_m}(R(t)^{\otimes m}F_{N:m}(t)R(t)^{\otimes m}) \\ =&2{\text {trace}}_{{\mathfrak {H}}_m}(F_{N:m}(t)(I_{\mathfrak {H}}^{\otimes m}-R(t)^{\otimes m})). \end{aligned} \end{aligned}$$

Hence,

$$\begin{aligned} \Vert F_{N:m}(t)-R(t)^{\otimes m}\Vert _1\le 2\sqrt{2{\text {trace}}_{{\mathfrak {H}}_m}(F_{N:m}(t)(I_{\mathfrak {H}}^{\otimes m}-R(t)^{\otimes m}))}. \end{aligned}$$

Since $R(t)=|\psi (t,\cdot )\rangle \langle \psi (t,\cdot )|$ is a self-adjoint projection

$$\begin{aligned} \begin{aligned}&R(t)\otimes I_{\mathfrak {H}}^{\otimes (m-1)}-R(t)^{\otimes m} \\&\quad = (I_{\mathfrak {H}}^{\otimes m}-I_{\mathfrak {H}}\otimes R(t)^{\otimes (m-1)})R(t)\otimes I_{\mathfrak {H}}^{\otimes (m-1)}(I_{\mathfrak {H}}^{\otimes m}-I_{\mathfrak {H}}\otimes R(t)^{\otimes (m-1)}) \\&\quad \le (I_{\mathfrak {H}}^{\otimes m}-I_{\mathfrak {H}}\otimes R(t)^{\otimes (m-1)})^2=(I_{\mathfrak {H}}^{\otimes m}-I_{\mathfrak {H}}\otimes R(t)^{\otimes (m-1)}), \end{aligned} \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned}&{\text {trace}}_{{\mathfrak {H}}}(F_{N:1}(t)R(t))-{\text {trace}}_{{\mathfrak {H}}_m}(F_{N:m}(t)R(t)^{\otimes m}) \\&\quad ={\text {trace}}_{{\mathfrak {H}}_m}(F_{N:m}(t)(R(t)\otimes I_{\mathfrak {H}}^{\otimes (m-1)}-R(t)^{\otimes m})) \\&\quad \le {\text {trace}}_{{\mathfrak {H}}_m}(F_{N:m}(t)(I_{\mathfrak {H}}^{\otimes m}-I_{\mathfrak {H}}\otimes R(t)^{\otimes (m-1)})) \\&\quad =1-{\text {trace}}_{{\mathfrak {H}}_{m-1}}(F_{N:m-1}(t)R(t)^{\otimes (m-1)}). \end{aligned} \end{aligned}$$

Since $F_{N}^{in}$ satisfies (31), the reduced m-particle operator $F_{N:m}(t)\in {\mathcal {L}}({\mathfrak {H}}_m)$ also satisfies (31) (with N replaced by m), and hence,

$$\begin{aligned} \begin{aligned} {\text {trace}}_{{\mathfrak {H}}_m}(F_{N:m}(t)(I_{\mathfrak {H}}^{\otimes m}-R(t)^{\otimes m}))\le&1-{\text {trace}}_{{\mathfrak {H}}}(F_{N:1}(t)R(t)) \\&+1-{\text {trace}}_{{\mathfrak {H}}_{m-1}}(F_{N:m-1}(t)R(t)^{\otimes (m-1)}) \\ \le&m(1-{\text {trace}}_{{\mathfrak {H}}}(F_{N:1}(t)R(t))) \\ =&m{\text {trace}}_{\mathfrak {H}}(F_{N:1}(t)P(t)), \end{aligned} \end{aligned}$$

by induction, which implies the inequality in the lemma. $\square$

With this lemma, the consequence of the Gronwall inequality above implies that, under the assumptions of Corollary 4.2,

$$\begin{aligned} \Vert F_{N:m}(t)-R(t)^{\otimes m}\Vert _1\le \sqrt{8m{\alpha }_N(t)}\le 4\sqrt{\frac{m}{N}}\exp \left( \tfrac{3}{{\hbar }}\int _0^tL(s)\hbox {d}s\right) . \end{aligned}$$

This completes the proof of Corollary 4.2.