Mean-Field Limits for Large Particle Systems Lecture 2: From Schrödinger to Hartree
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1 for Large Particle Systems Lecture 2: From Schrödinger to Hartree CMLS, École polytechnique & CNRS, Université Paris-Saclay FRUMAM, Marseilles, March 13-15th 2017
2 A CRASH COURSE ON QUANTUM N-PARTICLE DYNAMICS T. Kato: Perturbation theory for linear operators. Springer
3 N-particle wave function The state at time t of a N-particle system in quantum mechanics is described by its wave function Ψ N Ψ N (t, x 1,..., x N ) C where x j is the position of the jth particle What does the wave function tells us? the probability density of having particle no.1 at the position x 1, particle no.2 at the position x 2,... and particle no. N at the position x N is Ψ N (t, x 1,..., x N ) 2 In particular Ψ N (t, x 1,..., x N ) 2 dx 1... dx N = 1
4 Quantum N-particle Hamiltonian Classical Hamiltonian=function H N (X N, Ξ N ) with (X N, Ξ N ) R 2dN (R 2dN = the N-particle phase space) Quantum Hamiltonian=(unbounded) operator H N on H N := L 2 (R dn ) (R dn = the N-particle configuration space) Quantization rule for H N (X N, Ξ N ) := N j=1 1 2 ξ j N 1 j<k N V (x j x k ) (a) Function f (X N ) multiplication operator by f (X N ) on H N (b) Monomial ξ α ξα N N ( i x 1 ) α 1... ( i xn ) α N H N H N := N xj + 1 N j=1 V (x j x k ) 1 j<k N
5 The (N-particle) Schrödinger equation Planck s constant = J s i t Ψ N = H N Ψ N, Ψ N t=0 = Ψ in N Theorem (Kato Trans. AMS 51) If d = 3, and if for some R > 0, V B(0,R) L 2 (B(0, R)) and V R 3 \B(0,R) L (R 3 \ B(0, R)) then, for each N 1 and each > 0, the quantum Hamiltonian H N := N xk + 1 N k=1 N V (x k x l ) k l=1 has a self-adjoint extension on H N unitary group e ith N on H N := L 2 (R 3N ) and generates a
6 Dirac s notation Dirac s notation Denote Φ N := the vector Φ N H N = L 2 (R dn ) Ψ N := linear functional on H N φ N Ψ N φ N (X N )dx N R dn Hence Ψ N Φ N = Ψ N Φ N (X N )dx N C R dn while Φ N Ψ N = projection on C Φ N along (C Ψ N )
7 Hilbert-Schmidt operators Let H := L 2 (R n ), let L(H) = set of bounded operators on H and let K(H) be the set of compact operators on H (2-sided ideal of L(H)) Hilbert-Schmidt operators on H are integral operators of the form T ψ(x) := T (x, y)ψ(y)dy s.t. T L 2 (R n R n ) R d Set of Hilbert-Schmidt operators on H denoted L 2 (H) K(H); closed 2-sided ideal of L(H) Hilbert-Schmidt inner product and norm (R S) L 2 (H) := R(x, y)s(x, y)dxdy R d R d T L 2 (H) := (T T ) 1/2 L 2 (H) = T L 2 (R n R n )
8 Trace-class operators Set of trace-class operators L 1 (H) := {RS s.t. R and S L 2 (H)} Trace of a trace-class operator: for each T L 1 (H), tr(t ) := T (x, x)dx Trace norm R d T L 1 (H) = tr( T ) = T 1/2 2 L 2 (H), where T = T T Exercise Let T L(H) with integral kernel T. (1) If T L 1 (H), the map z T (x+z, x) belongs to C(R d ; L 1 (R d )) (2) Assume x T (x, x) integrable on R d. Does T L 1 (H)?
9 AT 2 L 2 (H) = k 1 ATe k 2 H A 2 k 1 ATe k 2 H = A 2 T 2 L 2 (H) With the polar decompositions AT = AT V and T = T U, one has AT L 1 (H) = tr( AT ) = tr(a T UV ) = tr(a T 1/2 T 1/2 UV ) A T 1/2 L 2 (H) T 1/2 UV L 2 (H) A T 1/2 2 L 2 (H) UV = A T L 1 (H)
10 Density operators A density operator on a separable Hilbert space H is R L(H) s.t. R = R 0, and tr(r) = 1. Set of density operators on H denoted D(H) Examples (1) If ψ H satisfies ψ H = 1, then ψ ψ D(H) (pure state) (2) If (ψ k ) k 1 H satisfies ψ k ψ l = δ kl, then λ k ψ k ψ k D(H) iff λ k 0 and λ k = 1 k 1 k 1 (mixed state)
11 The formalism of density operators Key observation Let Ψ N Ψ N (t, X N ) H N = L 2 (R dn ), then i t Ψ N = H N Ψ N i t Ψ N Ψ N = [ H N, Ψ N Ψ N ] Proof compute the time-derivative of e ith N/ ( Ψ N t=0 Ψ N t=0 ) e ith N / von Neumann equation with unknown t R N (t) D(H N ) i t R N (t) = [H N, R N (t)], R N t=0 = R in N D(H N)
12 Indistinguishable particles For σ S N and Ψ N H N = L 2 (R dn ), define U σ Ψ N (X N ) = Ψ N (x σ 1 (1),..., x σ 1 (N)) Observe that U στ = U σ U τ, U σ = U 1 σ Density operator R N of a system of N indistinguishable particles U σ R N U σ = R N for all σ S N Do not mix indistinguishability with U σ Ψ N = Ψ N, U σ Ψ N = ( 1) σ Ψ N, Bose-Einstein statistics Fermi-Dirac statistics
13 Propagation of indistinguishability Lemma Let R in N D(H N), and assume that U σ R in N U σ = R in N. Then, the solution R N (t) of the N-particle von Neumann equation with initial data R in N satisfies U σ R N (t)u σ = R N (t), for all t R Proof Since the N-particle Hamiltonian H N := N xj + 1 N j=1 V (x j x k ) 1 j<k N satisfies [H N, U σ ] = 0, one has U σ e ithn/ Uσ = e ithn/ so that U σ (e ithn/ RN in eithn/ )Uσ = e ithn/ (U σ RN in U σ)e ith N/
14 Hartree equation (wave function formalism) Quantum mean-field theory used mostly for bosons (for fermions, there is a correction exchange term) Mean-field potential in quantum mechanics V ψ (t, x) := V (x z) ψ(t, z) 2 dz = V ψ(t, ) 2 (x) R d Mean-field quantum Hamiltonian H ψ (t) := x + V ψ (t, x) Hartree equation for the 1-particle wave function ψ ψ(t, x) C i t ψ(t, x) = H(t)ψ(t, x), ψ t=0 = ψ in
15 Hartree equation (density operator formalism) Unknown t R(t) D(H) where H = L 2 (R d ) Mean-field potential denoting τ x f (z) := f (z x), assuming (H1) V R (t, x) := tr(τ x VR(t)) = V (x z) R(t, z, z)dz R d Mean-field quantum Hamiltonian H R (t) := x + V R (t, x) Hartree equation i t R(t) = [H R (t), R(t)], R t=0 = R in
16 THE BBGKY FORMALISM C. Bardos, F. Golse, N. Mauser: Methods Appl. Anal. 7, (2000) H. Spohn: Rev. Mod. Phys. 52, (1980) C. Bardos, L. Erdös, F. Golse, N. Mauser, H.T. Yau: C.R. Math. Acad. Sci. Paris 334, (2002)
17 Marginals of N-particle quantum densities Let R N D(H N ) satisfy U σ R N U σ = R N for all σ S N Theorem One has L 1 (H N ) = K(H N ) and L(H N ) = L 1 (H N ) Definition For each R N D(H N ) and each n = 1,..., N 1, there exists a unique R N:n D(H n ) s.t. tr Hn (R N:n A) = tr HN (R N (A I HN n )) for all A L(H n ) (Indeed, the r.h.s. defines a linear functional on finite rank operators A which is continuous for the operator norm) For integral operators, denoting Z n+1 N = (z n+1,..., z N ), one has R N:n (X n, Y n ) = RN (X n, Z n+1 N, Y n, Z n+1 n+1 N )dzn R d(n n)
18 BBGKY hierarchy 1 Goal to write an equation for R N:1 Pick A L(H 1 ) s.t. [, A] L(H 1 ); then i t tr H1 (AR N:1 (t)) = tr HN ([H N, A I HN 1 ]R N (t)) Denoting by V kl (= the multiplication by )V (x k x l ), one has [H N, A I HN 1 ] = [ x1, A] I HN N N [V 1k, A I HN k ] k=2 Since U τ R N (t)u τ = R N (t), where τ is the transposition 2 k tr HN ([V 1k, A I HN 1 ]R N (t)) = tr H2 ([V 12, A I H1 ]R N:2 (t))
19 BBGKY hierarchy 1 Thus i t tr H1 (AR N:1 (t)) = tr H1 ( [ x1, A]R N:1 (t) ) + N 1 N for all A L(H 1 ) s.t. [, A] L(H 1 ) tr H 2 ([V 12, A I H1 ]R N:2 (t)) This is the weak formulation of the 1st BBGKY equation: i t R N:1 = [ x1, R N:1 ] + N 1 N [V 12, R N:2 ] :1 This is not a closed equation on R N:1, since it involves R N:2
20 BBGKY hierarchy 2 Seek an equation for R N:2 (t). Pick B L(H 2 ) s.t. [ xj, B] L(H 2 ) for j = 1, 2; then i t tr H2 (BR N:2 (t)) = tr HN ([H N, B I HN 2 ]R N (t)) Then 2 [H N, B I HN 2 ] = [ xj, B] I HN N [V 12, B] I HN 2 j=1 + 1 N 2 N [V jk, B I HN 2 ] j=1 k=3
21 BBGKY hierarchy 2 By indistinguishability so that tr HN ([V jk, A I HN 2 ]R N ) = tr HN (([V j3, A I ] I HN 3 )R N ) i t R N:2 = 2 [ xj, R N:2 ] j=1 + N 2 N = tr HN (([V j3, A I ]R N:3 ) 2 [V j3, R N:3 ] :2 j=1 } {{ } interaction term + 1 N [V 12, R N:2 ] }{{} recollision term
22 BBGKY hierarchy n More generally, for each n = 1,..., N 1 i t R N:n = n [ xj, R N:n ] j=1 + N n N n [V j,n+1, R N:n+1 ] :n j=1 } {{ } interaction term + 1 N j,k=1 n [V jk, R N:n ] } {{ } recollision term For n = N, one has R N:N = R N, and the Nth equation in the BBGKY hierarchy is the N-particle von Neumann equation Hence von Neuman BBGKY von Neumann
23 Hartree hierarchy Passing to the limit formally in the nth BBGKY equation as N, assuming that R N:n (t) R n (t) D(H n ) i t R n = n n [ xj, R n ] + [V j,n+1, R n+1 ] :n, n 1 j=1 j=1 since the recollision term is formally of order O(n 2 /N) Hartree hierarchy formally similar to the BBGKY hierarchy; however, the following differences are fundamental it is an infinite hierarchy there are no recollision terms
24 A key observation Let R R(t) D(H 1 ) be he solution of the Hartree equation i t R(t) = [H R (t), R(t)], R t=0 = R in Then i t R(t) n = n n [ xj, R(t) n ] + [V j,n+1, R(t) (n+1) ] :n j=1 j=1 I.e. the sequence of tensor powers of the solution of the Hartree equation is an exact solution of the infinite Hartree hierarchy Not true in general for the BBGKY hierarchy! (another difference between both hierarchies...)
25 The strategy (1) Justify rigorously the formal limit from the BBGKY to the Hartree hierarchy (relatively easy) Idea with R N:n := 0 for n > N, the sequence of sequences (R N:n ) n 1 indexed by N is bounded in L (R + ; L 1 (H n )) n 1 By Banach-Alaoglu+diagonal extraction, one can find an increasing sequence of integers φ(n) such that R φ(n):n R n in L (R + ; L 1 (H n )) weakly as N (2) Prove that the solution of the Hartree hierarchy is uniquely determined by its initial data (quite difficult in general)
26 The mean-field limit theorem Assume that V L (R d ). Let R in D(H 1 ) satisfy [, R in ] L(H 1 ). Let R R(t) be the solution of the Hartree equation i t R(t) = [H R (t), R(t)], R t=0 = R in and let R N (t) be the solution of the N-body von Neumann equation i t R N (t) = [H N, R N (t)], R N t=0 = (R in ) N Then, for each n 1, one has R N:n R n in L (R + ; L 1 (H n )) weak-* as N.
27 Nirenberg-Ovcyannikov s abstract Cauchy-Kovalevska thm Differential equation u(t) = F (t, u(t)) Scale of Banach spaces (B r ) r 0, norms r s.t. r r B r B r with r r Assumptions on F (a) there exists r 0 > 0 s.t. 0 < r < r < r 0 F C(R B r, B r ) and B r z F (t, z) B r (b) there exists C > 0 s.t. is linear for all t R 0 < r < r < r 0 F (t, z) r C z r r r
28 Nirenberg-Ovcyannikov s abstract Cauchy-Kovalevska thm 2 Thm. (Ovcyannikov DAN 71, Nirenberg J Diff Geo 72) For all r (0, r 0 ) and all α > 0, the only solution of u(t) = F (t, u(t)), u(0) = 0 in C 1 (( α, α), B r ) is u = 0. What does this have to do with CK? Consider PDE t u(t, z) = P(t, z, z )u(t, z), u t=0 = 0 where P is a 1st order linear differential operator with coefficients bounded and analytic on strip I(z) < r 0 Assumption (b) is Cauchy estimate for z u on the strip I(z) < r by u on the bigger strip I(z) < r
29 Application to infinite hierarchy Sequence of Banach spaces (E n ) n 1, norm n, with strongly continuous groups of isometries U n (t) in E n Family of bounded linear operators L n,n+1 : E n+1 E n for n 1 Infinite hierarchy: for each n 1 u n (t) = U n (t)l n,n+1 U n+1 ( t)u n+1 (t), u n (0) = 0 Theorem (Bardos-FG-Erdos-Mauser-Yau CRAS02) Assume that L n,n+1 L(En+1,E n) Cn, n 1 Let u n C 1 ([0, t ], E n ) for n 1 satisfy the infinite hierarchy and sup u n (t) n R n for all n 1 0 t t Then u n = 0 on [0, t ] for all n 1
30 Application to infinite hierarchies Scale of Banach spaces B r := v = (v n) n 1 E n s.t. v r r n v n n < n 1 n 1 Function F : F (t, v) := (U n (t)l n,n+1 U n+1 ( t)v n+1 ) n 1 so that F (t, v) r C nr n vn n C r n+1 r n+1 v n n r r 1 n 1 n 1 C r r r n+1 v n+1 C v r r r n 1
31 PICKL S METHOD P. Pickl: Lett. Math. Phys. 97, (2011)
32 Pickl s theorem Assume that V is even, that V L 2r (R 3 ) for some r 1, and that V 1 z >R L (R 3 ) for some R > 0. Set = 1, and let ψ in L 2 (R 3 ) satisfy ψ in L 2 = 1. Assume that the Cauchy problem for Hartree s equation has a unique solution ψ C(R; L 2r (R 3 )), with r = r r 1. Let Ψ N = e ith N (ψ in ) N, and let R N = Ψ N Ψ N. Then ( 1 3 min 1, (exp N t 0 R N:1 (t) ψ(t, ) ψ(t, ) ) ) 1/2 10 V L 2r ψ(s, ) L 2r ds 1
33 Pickl s functional 1 Let = 1 and R N (t) := Ψ N (t, ) Ψ N (t, ), where Ψ N solution of the N-particle Schrödinger equation i t Ψ N = H N Ψ N, Ψ t=0 N = Ψ in N is the Assume that, for all σ S N U σ Ψ in N = Ψin N so that U σ Ψ(t, ) = Ψ(t, ) for all t R Let ψ ψ(t, x) C be the wave function solution of i t ψ = H ψ (t)ψ, ψ t=0 = ψ in Set ρ(t) := ψ(t, ) ψ(t, ) and define E N (t) := tr H1 (R N:1 (t)(i ρ(t)))
34 E N controls the operator norm Lemma R N:1 ρ 2E 1/2 N Proof Observe that R N:1 =(p+ρ)r N:1 (p+ρ), + E N ρr N:1 ρ= ψ R N:1 ψ ρ=tr(r N:1 ρ)ρ so that R N:1 ρ = R N:1 tr(r N:1 ρ)ρ E N ρ = pr N:1 + ρr N:1 p E N ρ Since E N = tr H1 (R N:1 p) = tr HN (R N (p I HN 1 )) = (p I HN 1 )Ψ N 2 H N one has pr N:1 = (p I HN 1 )Ψ N Ψ N E 1/2 N ρr N:1 p R N:1 p = (R N:1 p) = pr N:1
35 Evolution of E N (t) Lemma Denote p(t) = I ρ(t); then E N (t) = F N (t) = 2I tr(r N:2 (t)(ρ(t) I )W (t)(p(t) I )) with W (t) := N 1 N V 12 (V ψ(t, ) 2 ) I The core result in Pickl s argument is the following proposition Proposition Let r 1, and set r = r r 1 ; then F N (t) 10 V L 2r ψ(t, ) L 2r ( EN (t) + 1 N )
36 The term F N (t) Expand F N (t) as F N (t) =2I tr(r N:2 (t)(ρ(t) I )W (t)(p(t) I )) =2I tr(r N:2 (t)((ρ(t) ρ(t))w (t)(p(t) ρ(t))) + 2I tr(r N:2 (t)((ρ(t) ρ(t))w (t)(p(t) p(t))) + 2I tr(r N:2 (t)((ρ(t) p(t))w (t)(p(t) p(t))) =2I( (1) + (2) + (3) ) since, by indistinguishability tr(r N:2 ((ρ p)w (p ρ)) = tr((p ρ)w (ρ p)r N:2 ) R Goal to control F N by a linearly growing function of E N and apply Gronwall s inequality
37 Bounding (3) Writing ρ p = (ρ I )(I p) and p p = (p I )(I p), one has (3) (ρ I )W (p I ) tr H2 ((I p)r N:2 (I p)) (ρ I )W tr H1 (R N:1 p) = (ρ I )W E N Observe that ψ τ x2 V 2 ψ = (V 2 ) ψ (x 2 ) so that (ρ I )V 2 12(ρ I ) = ρ (V 2 ) ψ and therefore (ρ I )W 2 = (ρ I )W 2 (ρ I ) (2 V 2 L r + 2 (V ψ(t, ) 2 ) 2 L r ) ψ(t, ) 2 L r 4 V 2 L ψ(t, ) 2 2r L 2r
38 Bounding (1) Since R N:2 = tr(r N:2 ) = 1 and p 1 (1) (ρ ρ)w (t)(p ρ) (ρ I )(I ρ)w (I ρ) Observe that (I ρ)w (I ρ) = 1 N (V ψ(t, ) 2 ) ρ Hence (ρ I )(I ρ)w (I ρ) 2 (ρ I )((I ρ)w (I ρ)) 2 (ρ I ) = 1 N 2 ψ V ψ 2 1 ψ ρ ρ N 2 V ψ (t, x) 2 ψ(t, x) 2 dx 1 N 2 V ψ(t, ) 2 L 2r ψ(t, ) 2 L 2r 1 N 2 V 2 L 2r ψ(t, ) 2 L 2r
39 Key lemma Notation (1) For T L(H 1 ), set T j,n = I Hj 1 T I HN j L(H N ) (2) For a {0, 1} {1,...,N} and π = π = π 2 L(H 1 ), set P N [a, π] := N j=1 π 1 a(j) N,j (I π N,j ) a(j) Lemma The spectral decomposition of M N := 1 N p j,n is M N = N j=1 where the spectral projections are Π k,n [ρ] = a(1)+...+a(n)=k Π k,n [ρ]π l,n [ρ] = δ kl Π k,n [ρ]. N k=1 k N Π k,n[ρ] P N [a, ρ] = Π k,n [ρ]
40 Bounding (2) Corollary Let M 1/2 N be the pseudo-inverse of M 1/2 N extended by 0 on ker(m N ); then M 1/2 N M 1/2 N = I Π 0,N [ρ] = I ρ N M 1/2 N M 1/2 N p j,n = p j,n Therefore (2) = tr H2 (R N:2 ((ρ ρ)w (p p)) = tr HN (R N ρ 1,N ρ 2,N Wp 1,N p 2,N ) = tr HN (R N ρ 1,N ρ 2,N WM 1/2 N M 1/2 N p 1,N p 2,N ) M 1/2 N p 1,N p 2,N Ψ N HN M 1/2 }{{} N W ρ 1,Nρ 2,N ψ N HN }{{} (2a) (2b)
41 Bounding (2a) By the bosonic symmetry (symmetry of the wave function Ψ N ) Hence N(N 1)(2a) 2 =N(N 1) Ψ N M 1 N p 1,Np 2,N Ψ N =2 Ψ N M 1 N p j,np k,n Ψ N 1 j<k N Ψ N M 1 N (NM N) 2 Ψ N = N 2 Ψ N M N Ψ N N(N 1)(2a) 2 tr HN (R N p 1,N ) = N 2 tr H1 (R N:1 p) = N 2 E N
42 Bounding (2b) Using again the bosonic symmetry M 1/2 N W ρ 1,Nρ 2,N ψ N 2 H N = Ψ N ρ 1,N ρ 2,N WM N W ρ 1,N ρ 2,N Ψ N = N 2 N Ψ N ρ 1,N ρ 2,N Wp 3,N W ρ 1,N ρ 2,N Ψ N + 2 N Ψ N ρ 1,N ρ 2,N Wp 1,N W ρ 1,N ρ 2,N Ψ N Since p 3,N commutes with ρ 1,N, ρ 2,N and W, one has = (2b1) + (2b2) (2b1) N 2 N p 3,NΨ N 2 H N ρ 1,N W 2 ρ 1,N ρ 1,N W 2 ρ 1,N E N 4 V 2 L ψ(t, ) 2 2r L E 2r N Since ρ 2,N commutes with ρ 1,N and p 1,N is a projection (2b2) 2 N ρ 1,Nρ 2,N Wp 1,N W ρ 1,N ρ 2,N 2 N ρ 1,Nρ 2,N Wp 1,N 2 2 N ρ 1,NW 2 2 N ρ 1,NW 2 ρ 1,N 8 N V 2 L 2r ψ(t, ) 2 L 2r
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