From the N-body problem to the cubic NLS equation

Transcription

1 From the N-body problem to the cubic NLS equation François Golse Université Paris 7 & Laboratoire J.-L. Lions golse@math.jussieu.fr Los Alamos CNLS, January 26th, 2005

2 Formal derivation by N.N. Bogolyubov (1947) see for instance Landau- Lifshitz vol. 9, 25 More recently, there have been rigorous derivations of nonlinear PDEs in the single particle phase-space from the linear, N-body problem. See for instance the derivation of the Boltzmann equation for a hard sphere gas by Lanford (1975) and then Illner-Pulvirenti (1986). Derivation of the Schrödinger-Poisson equation from the quantum N- body problem with Coulomb potential: Bardos-G-Mauser (2000), Erdös- Yau (2001) Work in collaboration with Riccardo Adami et Sandro Teta (preprint June 2005); space dimension 1, global in time. More recent preprint by L. Erdös and H.-T. Yau (preprint August 2005); space dimension 3, global in time.

3 The N-body Schrödinger equation Unknown: the N-particle wave function Ψ N Ψ N (t, X N ) C, X N = (x 1,..., x N ) R N R N Ψ N(t, X N ) 2 dx N = 1 Hamiltonian H N := 1 2 N + U N = N k= x j + 1 k<l N U(x k x l ) where the potential U(z) = U( z) is real-valued, compactly supported (hence short-range), smooth and nonnegative.

4 Hence the wave function Ψ N satisfies i t Ψ N = N k= x j Ψ N + i t Ψ N = H N Ψ N, soit 1 k<l N U(x k x l )Ψ N In the sequel, all particles considered are bosons, meaning that the wave function Ψ N is symmetrical in the variables x k (Bose statistics): Ψ N (t, x 1,..., x N ) = Ψ N (t, x σ(1),..., x σ(n) ) pour tout σ S N. One easily checks the following: if Ψ N t=0 is symmetrical in the x k s and Ψ N solves the N-body Schrödinger equation, then Ψ N (t, ) also is symmetrical in the x k s for each t R.

5 Scaling We shall be using two different scaling assumptions, as follows: a) a collective scaling of mean-field type: U(x k x l ) := 1 N V N(x k x l ) so that the interaction potential per particle is 1 2 k l b) and an ultra-short range scaling 1 N V N(x k x l ) Ψ N (X N ) 2 = O(1) V N (z) := N γ V (N γ z) with 0 < γ < 1 2, and V nonnegative, even and smooth

6 The total energy of the system of particles considered is + 1 k<l N H N Ψ N Ψ N = N k=1 1 2 x k Ψ N 2 L 2 N γ 1 V (N γ (x k x l )) Ψ N (X N ) 2 dx N We shall be using only wave functions for which H N Ψ N Ψ N = O(N) Example: an important example of such wave functions is the case of a tensor product Ψ N (X N ) := N k=1 ψ(x k ) avec ψ H 1 (R)

7 Density matrix, marginals The density matrix is the integral operator on L 2 (R N ) whose kernel is ρ N (t, X N, Y N ) := Ψ N (t, X N )Ψ N (t, Y N ) a standard notation for this operator is ρ N (t) = Ψ N (t, ) Ψ N (t, ) ; it is a rank-one orthogonal projection. For each 1 k < N, defined the k-particle marginal of ρ N to be ρ N:k (t, X k, Y k ) := R N k ρ N(t, X k, Z N k+1, Y k, Z N k+1 )dzn k+1 where Z N k+1 := (z k+1,..., z N ). We denote by ρ N:k (t) the associated integral operator; it is a nonnegative, trace-class operator with trace equal to 1.

8 Theorem. Let 0 V C (R) and γ (0, 2 1 ); assume there exists M > 0 such that t=0 N Ψ N ψ in (x k ), with ( N ) n t=0 ΨN t=0 Ψ N M n N n k=1 for n = 1,..., N. Then, for all t 0, the sequence of single-particle marginals ρ N:1 (t, x, y) ψ(t, x)ψ(t, y) as N in Hilbert-Schmidt norm, where ψ solves i t ψ xψ α ψ 2 ψ = 0, with α := ψ = ψ in t=0 R V (x)dx

9 BBGKY hierarchy We shall be writing a sequence of equations satisfied by the sequence of marginals ρ N:j, where j = 1,..., N. Start from the von Neumann equation satisfied by ρ N : i t ρ N = [H N, ρ N ] In that equation, set x 2 = y 2 = z 2,..., x N = y N = z N, and integrate in z 2,..., z N : = (N 1) i t ρ N: ( 2 x 1 2 y 1 )ρ N:1 [U(x 1 z) U(y 1 z)]ρ N:2 (t, x 1, z, y 1, z)dz We recall that U(z) = N γ 1 V (N γ z) avec 0 < γ < 1 2.

10 For j = 2,..., N 1, the analogous equation is i t ρ N:j = (N j) j k=1 + j k=1 ( 2 x k 2 y k )ρ N:j [U(x k z) U(y k z)]ρ N:j+1 (t, X k, z, Y k, z)dz 1 k<l j [U(x k x l ) U(y k y l )]ρ N:j (t, X k, Y k ) For j = N, this equation is nothing but the von Neumann equation for the N-body density matrix ρ N. Conceptually, it is advantageous to deal with infinite hierarchies of equations: in the sequel, we set ρ N:j = 0 whenever j > N.

11 By passing to the limit (at the formal level) in the BBGKY hierarchy as N and for j fixed; remember that U(z) = 1 N V N(z) and that V N αδ z=0 with α = Assuming that ρ N:j ρ j for N, we find that V (x)dx j = α j k=1 i t ρ j k=1 ( 2 x k 2 y k )ρ j [δ(x k z) δ(y k z)]ρ j+1 (t, X k, z, Y k, z)dz Unlike in the case of the BBGKY hierarchy, j 1 is unlimited, so that this new hierarchy has infinitely many equations.

12 Let ψ be a smooth solution of the cubic NLS equation Define then i t ψ xψ = α ψ 2 ψ ρ j (t, X j, Y j ) := j k=1 ψ(t, x k )ψ(t, y k ) We find that i t ρ ( 2 x y 2 )ρ 1 = α(ρ 1 (t, x, x) ρ 1 (t, y, y))ρ 1 More generally, a straightforward computation shows that the sequence ρ j soves the inifinite hierarchy

13 This suggests the following strategy, inspired from the derivation by Lanford of the Boltzmann equation from the classical N-body problem: a) for the N-body Schrödinger equation, pick the initial data Ψ N t=0 := N k=1 ψ (x t=0 k ) ; b) show that the sequence of marginals ρ N:j ρ j as N and for each fixed j in some suitable sense; next show that ρ j solves the infinite hierarchy by passing to the limit in the BBGKY hierarchy; c) prove that the infinite hierarchy has a unique solution which implies that ρ j (t, X j, Y j ) = j k=1 where ψ is the solution of cubic NLS. ψ(t, x k )ψ(t, y k )

14 An abstract uniqueness argument Consider the infinite hierarchy of equations u n + A n u n = L n,n+1 u n+1, u n (0) = 0, n 1 where u n takes its values in a Banach space E n ; here the linear operator L n,n+1 belongs to L(E n+1, E n ) while A n is the generator of a oneparameter group of isometries U n (t) on E n. Defining v n (t) := U n ( t)u n (t), one sees that v n(t) = U n ( t)l n,n+1 U n+1 (t)v n+1 (t), u n (0) = 0.

15 Lemma. Assume there exists C > 0 and R > 0 s.t. L n,n+1 L(En+1,E n ) Cn and u n(t) En R n for each n 1 and each t [0, T ]. Then u n 0 on [0, T ] for each n 1. Proof: Consider the decreasing scale of Banach spaces B r := and set v = (v n) n 0 E n v r = r n v n En < + n 1 n 1 F (v) := (U n ( t)l n,n+1 U n+1 (t)v n+1 ) n 1.

16 A straightforward computation shows that F (v) r1 C nr1 n v n En C n 1 n 1 r n+1 r1 n+1 r r v n 1 En C v r r r 1 We conclude by applying the abstract variant of the Cauchy-Kowalewski theorem proved by Nirenberg and Ovsyanikov. The key idea is to view B r as the analogue of the class of functions with holomorphic extension to a strip of width r. The estimate above is similar to Cauchy s inequality bearing on the derivative of a holomorphic function. Hence F behaves like a differential operator of order 1.

17 Interaction estimate The first difficulty is to find Banach spaces E n such that the interaction term L n,n+1 is bounded by O(n). Set S j := (1 xj ) 1/2 ; define E n := {ρ n L(L 2 (R n )) S 1... S n ρ n S 1... S n is Hilbert-Schmidt} which is a Hilbert space for the norm ρ n En := S 1... S n ρ n S 1... S n L 2 = n j=1 (1 xj ) 1/2 (1 yj ) 1/2 ρ n (X n, Y n ) 2 dx n dy n 1/2

18 Proposition. Let ρ E n+1 and U be a tempered distribution whose Fourier transform is bounded on R. Let σ be the integral operator with kernel σ(x n, Y n ) := U(x 1 z)ρ(x n, z, Y n, z)dz Then σ En C Û L ρ En+1 In the BBGKY hierarchy, the operator L n,n+1 is the sum of 2n terms analogous to the one treated in the proposition above. Hence L n,n+1 L(En+1,E n ) Cn V L 1

19 Sketch of the proof: Do it for the limiting interaction U = δ 0. Then σ(x n, Y n ) = ρ n+1 (X n, Y n, x 1, x 1 ) If ρ n+1 was the n+1st fold tensor product of functions of a single variable, the inequality that we want to prove reduces to the fact that H 1 (R) is an algebra. The same proof (in Fourier space variables) works for the restriction of functions of arbitrarily many variables to a subspace of arbitrary codimension, provided that cross-derivatives of these functions are bounded in L 2 this is different from the trace problem for functions in H 1 (R n ). This proof extends to the case where Û is an arbitrary function in L

20 An elementary computation shows that Set ˆσ(Ξ n, H n ) = we seek to estimate σ 2 E n = ˆρ n+1 (ξ 1 k, Ξ n 2, k l; H n, l) dkdl 4π 2 Γ n (Ξ n ) := n k=1 1 + ξ 2 k Γ n (Ξ n ) 2 Γ n (H n ) 2 ˆσ(Ξ n, H n ) 2dΞ ndh n (2π) 2n

21 Since it follows that Γ 1 (ξ 1 ) (Γ 1 (ξ 1 k) + Γ 1 (k l) + Γ 1 (l))) 1 3 Γ 1 (ξ 1 ) π 2 2 4π 2 2 ˆρ n+1 (ξ 1 k, Ξ n 2, k l; H n, l) dkdl 4π 2 Γ 1 (ξ 1 k)ˆρ n+1 (ξ 1 k, Ξ n 2, k l; H n, l) dkdl + Γ 1 (k l)ˆρ n+1 (ξ 1 k, Ξ n 2, k l; H n, l) dkdl Γ 1 (l)ˆρ n+1 (ξ 1 k, Ξ n 2, k l; H n, l) dkdl 4π 2

22 By the Cauchy-Schwarz inequality C where Γ 1 (ξ 1 k)ˆρ n+1 (ξ 1 k, Ξ n 2, k l, H n, l) dkdl 4π 2 ˆρ n+1 (ξ 1 k, Ξ n 2, k l; H n, l) 2 Γ 1 (ξ 1 k) 2 Γ 1 (k l) 2 Γ 1 (l) 2 dkdl 4π 2 C := dkdl Γ 1 (k l) 2 Γ 1 (l) 2 <. The two other terms are treated in the same manner. Therefore Γ n (Ξ n ) 2 Γ n (H n ) 2 ˆσ(Ξ n, H n ) 2 dξ ndh n (2π) 2n C Γ n 1 (Ξ n 2 )2 Γ(H n ) 2 ˆρ n+1 (ξ 1 k, Ξ n 2, k l; H n, l) 2 Γ 1 (ξ 1 k) 2 Γ 1 (k l) 2 Γ 1 (l) 2 dkdl 4π 2 with C := 3C. We conclude after changing variables: (ξ 1 k, k l, l) (ξ 1, ξ n+1, η n+1 ) 2 dξ n dh n (2π) 2n

23 Growth estimate for ρ n En Proposition. Let 0 V Cc 2 (R) and γ (0, 1); define H N = 1 2 X N + 1 k<l N N γ 1 V (N γ (x k x l )) Assume that, for each n 1 and each N N 0 (n), HN n Ψin N Ψin N M n N n with Ψ in N (X n) = N k=1 ψ in (x k ). Then, for each M 1 > M, there exists N 1 = N 1 (M 1, n) such that trace(s 1... S n ρ N,n (t)s 1... S n ) M n 1 for each t 0 and each N N 1.

24 Sketch of the proof: This is a variant of an argument by Erdös et Yau for the existence of a solution to the infinite hierarchy in space dimension 3. The only estimate involving derivatives that is propagated by the N-body equation bears on In this quantity, the typical term is H n N Ψ N Ψ N j 1 <...<j n xj1... xj1 Ψ N 2 dx N In all the other terms, either one derivative bears on V, leading to a lesser order term, or there is a multiple derivative in one of the x j s, and there are less many of such terms.

25 Set n and C ]0, 1[; one shows the existence of N 0 (C, n) st. for each N N 0 and each Ψ D(H N ) (N + H N ) n Ψ Ψ C n N n Ψ S S2 nψ This result is trivial for n = 0, 1 (since V 0 and Ψ N (t, X N ) is symmetrical in the x j s). The general case follows by induction on n: assuming the inequality proved for k = 0,..., n we prove it for n + 2. Write H N + N = N k=1 S 2 k + 1 k<l N N γ 1 V (N γ (x k x l ))

26 Define H n+1,n := H N + N Then N k=n+1 S 2 k 0. Ψ (N + H N )S S2 n(n + H N )Ψ = Ψ Sj 2 1 S S2 n S2 j 2 Ψ + 2R Ψ Sj 2 1 S S2 n H n+1,nψ n<j 1,j 2 N n<j 1 N + Ψ H n+1,n S S2 nh n+1,n Ψ Since H n+1,n S S2 n H n+1,n 0 the last term in the r.h.s. is disposed of.

27 Recall that Ψ N is symmetrical in the space variables, i.e. Ψ N (t, x σ(1),..., x σ(n) ) = Ψ N (t, x 1,..., x N ) for each σ S N Hence, denoting by W kl the multiplication by N γ V (N γ (x k x l )) acting on L 2 (R N ), one has Ψ (N + H N )S S2 n(n + H N )Ψ (N n)(n n 1) Ψ S S2 n S2 n+1 S2 n+2 Ψ +(2n + 1)(N n) Ψ S 4 1 S S2 n+1 Ψ n(n + 1)(N n) + R Ψ W 12 S1 2 N... S2 nsn+1 2 Ψ (n + 1)(N n)(n n 1) + R Ψ W 1,n+2 S1 2 N... S2 nsn+1 2 Ψ

28 By Sobolev embedding, one has the following obvious inequality W (x y) W L 1(1 xx ) Hence all the terms involving V are of a lesser order: and similarly 2R Ψ W 12 S S2 ns 2 n+1 Ψ V L 1 L (N 2γ Ψ S S2 n+1 Ψ +N γ Ψ S 4 1 S S2 n+1 Ψ ) 2R Ψ W 1,n+2 S S2 ns 2 n+1 Ψ V L 1N γ Ψ S S2 ns 2 n+1 S2 n+2 Ψ

29 Growth estimate for initial data We start from an initial data of the form that satisfies Ψ in N (X N) = N j=1 ψ in (x j ) Ψ in N ( N) n Ψ in N M n N n. We prove by induction that, if V Cc (R) and γ (0, 1 2 ), one has for each n 1 and N N (n). ( 1 2 N + U N ) n C n (N N ) n

30 Hence Ψ in N ( 1 2 N + U N ) n Ψ in N 2n 1 C n 1 M n N n for each n 1 and N N (n). To compare powers of the Hamiltonian with powers of the kinetic energy, it suffices to show that U N (N N ) 2n U N (C + C (n)n (4γ 2)n )(N N ) 2n+2 which is done by induction. One has to be careful only with the case n = 0 that sets the constant C uniformly in n. The above computation where the condition γ (0, 2 1 ) comes from.

31 Passing to the limit Let Ψ N be the solution of the N-body Schrödinger equation with factorized initial data; let ρ N be the density matrix and ρ N:n its n-th marginal. The sequence ((ρ N:n ) n 0 ) N 0 is bounded in the product space n 1 L (R, E n ) (each factor being endowed with the weak-* toplogy) On the other hand, if (ρ Nj :n) n 0 converges to (ρ n ) n 0 in that topology, the limit soves the infinite hierarchy in the sense of distributions. Notice in particular that ρ n (t, X n, Y n ) C trace(s 1... S n ρ n (t)s 1... S n ) so that ρ n L t,x n,y n.

32 The recollision term is estimated as follows N γ 1 V (N γ (x 1 x 2 ))ρ N:n (t, X n, Y n )φ(t, X n, Y n )dx n dy n dt CN γ 1 V L ρ N:n L (E n ) φ L 1 0 as N and there are 2n(n 1) such terms in the n-th equation of the infinite hierarchy. As for the interaction term, remember that E n is a Hilbert space, so that the convergence is weak and not only weak-* in the space variables. Since the linear interaction operator L n,n+1 is norm-continuous from E n+1 to E n, it is weakly continuous from E n+1 to E n.

33 NB. The convergence to a solution of the infinite hierarchy follows from a careful analysis involving the conservation of energy. The interaction operator L n,n+1 essentially reduces to taking the restriction of ρ N:n+1 to a (linear) subspace of codimension 2. But ρ N:n+1 is a trace-class operator, which allows taking the restriction to x n+1 = y n+1, with a H 1 estimate that follows from the conservation of energy; this bound allows in turn taking the further restriction x n+1 = x 1 because H 1 functions have H 1/2 L 2 restrictions to hypersurfaces. See Adami-Bardos-G.-Teta, Asympt. Anal. (2004). Analogous result in space dimension 3 (preprint by Erdös-Yau, 2004).