MEAN-FIELD DYNAMICS: SINGULAR POTENTIALS AND RATE OF ...

MEAN-FIELD DYNAMICS: SINGULAR POTENTIALS AND RATE OF CONVERGENCE

arXiv:0907.4313v2 [math-ph] 10 Dec 2009

Antti Knowles and Peter Pickl December 10, 2009 We consider the time evolution of a system of N identical bosons whose interaction potential is rescaled by N −1 . We choose the initial wave function to describe a condensate in which all particles are in the same one-particle state. It is well known that in the meanfield limit N → ∞ the quantum N -body dynamics is governed by the nonlinear Hartree equation. Using a nonperturbative method, we extend previous results on the mean-field limit in two directions. First, we allow a large class of singular interaction potentials as well as strong, possibly time-dependent external potentials. Second, we derive bounds on the rate of convergence of the quantum N -body dynamics to the Hartree dynamics.

1. Introduction We consider a system of N identical bosons in d dimensions, described by a wave function ΨN ∈ H(N ) . Here H(N ) := L2+ (RN d , dx1 · · · dxN )

is the subspace of L2 (RN d , dx1 · · · dxN ) consisting of wave functions ΨN (x1 , . . . , xN ) that are symmetric under permutation of their arguments x1 , . . . , xN ∈ Rd . The Hamiltonian is given by HN =

N X i=1

hi +

1 N

X

16i 0. In the present article we adopt yet another approach based on a method of Pickl [10]. We strengthen and generalize many of the results listed above, by treating more singular interaction potentials as well as deriving estimates on the rate of convergence. Moreover, our approach

3

allows for a large class of (possibly time-dependent) external potentials, which might for instance describe a trap confining the particles to a small volume. We also show that if the solution ϕ(·) of the Hartree equation satisfies a scattering condition, all of the error estimates are uniform in time. The outline of the article is as follows. Section 2 is devoted to a short discussion of the (k) (k) indicators of convergence EN and RN , in which we derive estimates relating them to each other. In Section 3 we state and prove our first main result, which concerns the mean-field limit in the case of L2 -type singularities in w; see Theorem 3.1 and Corollary 3.2. In Section 4 we state and prove our second main result, which allows for a larger class of singularities such as the nonrelativistic critical case h = −∆ and w(x) = λ|x|−2 ; see Theorem 4.1. For an outline of the methods underlying our proofs, see the beginnings of Sections 3 and 4. Acknowledgements. We would like to thank J. Fr¨ ohlich and E. Lenzmann for helpful and stimulating discussions. We also gratefully acknowledge discussions with A. Michelangeli which led to Lemma 2.1. Notations. Except in definitions, in statements of results and where confusion is possible, we refrain from indicating the explicit dependence of a quantity aN (t) on the time t and the particle number N . When needed, we use the notations a(t) and a|t interchangeably to denote the value of the quantity a at time t. The symbol C is reserved for a generic positive constant that may depend on some fixed parameters. We abbreviate a 6 Cb with a . b. To simplify notation, we assume that t > 0. We abbreviate Lp (Rd , dx) ≡ Lp and k·kLp ≡ k·kp . We also set k·kL2 (RNd ) = k·k. For s ∈ R

we use H s ≡ H s (Rd ) to denote the Sobolev space with norm kf kH s = (1 + |k|2 )s/2 fˆ 2 , where fˆ is the Fourier transform of f . Integer indices on operators denote particle number: A k-particle operator A (i.e. an operator on H(k) ) acting on the coordinates xi1 , . . . , xik , where i1 < · · · < ik , is denoted by Ai1 ...ik . Also, by a slight abuse of notation, we identify k-particle functions f (x1 , . . . , xk ) with their associated multiplication operators on H(k) . The operator norm of the multiplication operator f is equal to, and will always be denoted by, kf k∞ . We use the symbol Q(·) to denote the form domain of a semibounded operator. We denote the space of bounded linear maps from X1 to X2 by
L(X1 ; X2 ), and abbreviate L(X) = L(X; X). 2 N d We abbreviate the operator norm of L L (R ) by k·k. For two Banach spaces, X1 and X2 , contained in some larger space, we set
kf1 kX1 + kf2 kX2 , kf kX1 +X2 = inf f =f1 +f2

kf kX1 ∩X2 = kf kX1 + kf kX2 ,

and denote by X1 + X2 and X1 ∩ X2 the corresponding Banach spaces.

2. Indicators of convergence This section is devoted to a discussion, which might also be of independent interest, of quanti(k) (k) tative relationships between the indicators EN and RN . Throughout this section we suppress

4

the irrelevant index N . Take a k-particle density matrix γ (k) ∈ L(H(k) ) and a one-particle condensate wave function ϕ ∈ L2 . The following lemma gives the relationship between different elements of the sequence E (1) , E (2) , . . . , where, we recall,

 E (k) = 1 − ϕ⊗k , γ (k) ϕ⊗k .

(2.1)

Lemma 2.1. Let γ (k) ∈ L(H(k) ) satisfy

γ (k) > 0 , Let ϕ ∈ L2 satisfy kϕk = 1. Then (k)
i>1

Proof. Let Φi

ϕ⊗k , γ (k) ϕ⊗k



=

Tr γ (k) = 1 .

E (k) 6 k E (1) .

(2.2) (k)

be an orthonormal basis of H(k) with Φ1 = ϕ⊗k . Then X

i>1

(k−1)

ϕ ⊗ Φi

= hϕ , γ (1) ϕi −

(k−1) 

, γ (k) ϕ ⊗ Φi

X

i>2

ϕ⊗

−

(k−1) Φi , γ (k) ϕ

X

i>2

(k−1)

ϕ ⊗ Φi

(k−1) 

⊗ Φi

(k−1) 

, γ (k) ϕ ⊗ Φi

.

Therefore,

 hϕ , γ (1) ϕi − ϕ⊗k , γ (k) ϕ⊗k X

(k−1)  (k−1) , γ (k) ϕ ⊗ Φi ϕ ⊗ Φi = i>2

6

XX

i>2 j>1

=

XX

i>1 j>1



This yields

(k−1)

, γ (k) Φj ⊗ Φi

(1)

(k−1)

, γ (k) Φj ⊗ Φi

(1)

Φj ⊗ Φi

⊗(k−1)

= 1− ϕ

(1)

(k−1) 

(1)

Φj ⊗ Φi

,γ

(k−1)

⊗(k−1)

ϕ



.

(k−1) 

−

X

j>1

(1)

(1)

Φj ⊗ ϕ⊗(k−1) , γ (k) Φj ⊗ ϕ⊗(k−1)



E (k) 6 E (k−1) + E (1) , and the claim follows. Remark 2.2. The bound in (2.2) is sharp. Indeed, let us suppose that E (k) 6 k f (k) E (1) for some function f . Then f (k) > sup γ (k)

E (k) > kE (1)

1 − (1 − α)k 1 − (1 − α)k > lim = 1, α→0 kα kα 0 0. Remark 3.6. If supt φ(t) < ∞, or in other words if kϕ(t)kq1 and kϕ(t)kq2 are integrable in t over R, then all estimates are uniform in time. This describes a scattering regime where the time evolution is asymptotically free for large times. Such an integrability condition requires large exponents qi , which translates to small exponents pi , i.e. an interaction potential with strong decay. Remark 3.7. The result easily extends to time-dependent one-particle Hamiltonians h ≡ h(t). Replace (A1) and (A2) with (A1’) The Hamiltonian h(t) is self-adjoint and bounded from below. We assume that there is an operatorPh0 >
0 that such that 0 6 h(t) 6 h0 for all t. Define the Hilbert space XN = Q i (h0 )i as in (A1).

(A2’) The Hamiltonian HN (t) is self-adjoint and bounded from below. We assume that Q(HN (t)) ⊂ XN for all t. We also assume that the N -body propagator UN (t, s), defined by i∂t UN (t, s) = HN (t)UN (t, s) ,

UN (s, s) = 1 ,

exists and satisfies UN (t, 0)ΨN,0 ∈ Q(HN (t)) for all t. It is then straightforward that Theorem 3.1 holds with the same proof. Remark 3.8. In some cases (see e.g. Section 3.2.1 below) it is convenient to modify the assumptions as follows. Replace (A3) and (A4) with (A3’) The interaction potential w is a real and even function satisfying

2

w ∗ |ϕ|2 6 K kϕk2X 1 ∞

for some constant K > 0. Without loss of generality we assume that K > 1.

(A4’) The solution ϕ(·) of (1.3) satisfies ϕ(·) ∈ C(R; X1 ) ∩ C 1 (R; X1∗ ) .

9

(3.6)

Then Theorem 3.1 and Corollary 3.2 hold with Z t ds kϕ(s)k2X1 . φ(t) = 32K 0

The proof remains virtually unchanged. One replaces (3.24) with (3.6), as well as (3.20) with

w ∗ |ϕ|2 6 2K kϕk2X , 1 ∞ which is an easy consequence of (3.6).

3.2. Examples. We list two examples of systems satisfying the assumptions of Theorem 3.1. 3.2.1. Particles in a trap. Consider nonrelativistic particles in R3 confined by a strong trapping potential. The particles interact by means of the Coulomb potential: w(x) = λ|x|−1 , where λ ∈ R. The one-particle Hamiltonian is of the form h = −∆+v, where v is a measurable function on R3 . Decompose v into its positive and negative parts: v = v+ − v− , where v+ , v− > 0. We assume that v+ ∈ L1loc and that v− is −∆-form bounded with relative bound less than one, i.e. there are constants 0 6 a < 1 and 0 6 b < ∞ such that hϕ , v− ϕi 6 ahϕ , −∆ϕi + bhϕ , ϕi .

(3.7)

Thus h + b1 is positive, and it is not hard to see that h is essentially self-adjoint on Cc∞ (R3 ). This follows by density and a standard argument using Riesz’s representation theorem to show that the equation (h + (b + 1)1)ϕ = f has a unique solution ϕ ∈ {ϕ ∈ L2 : hϕ ∈ L2 } for each f ∈ L2 . It is now easy to see that Assumptions (A1) and (A2) hold with the one-particle Hamiltonian h + c1 for some c > 0. Let us assume without loss of generality that c = 0. Next, we verify Assumptions (A3’) and (A4’) (see Remark 3.8). We find Z

2

λ2 2 2

w ∗ |ϕ|2 = sup dy |ϕ(y)| . hϕ , −∆ϕi . hϕ , hϕi + hϕ , ϕi = kϕkX1 , ∞ |x − y|2 x

where the second step follows from Hardy’s inequality and translation invariance of ∆, and the third step is a simple consequence of (3.7). This proves (A3’). Next, take ϕ0 ∈ X1 . By standard methods (see e.g. the presentation of [7]) one finds that (A4’) holds. Moreover, the mass kϕ(t)k2 and the energy   Z 1 2 2 ϕ dx dy w(x − y)|ϕ(x)| |ϕ(y)| E (t) = hϕ , hϕi + 2 t

are conserved under time evolution. Using the identity |x|−1 6 1{|x|6ε} ε|x|−2 + 1{|x|>ε} ε−1 and Hardy’s inequality one sees that kϕ(t)k2X1 . E ϕ (t) + kϕ(t)k2 , and therefore kϕ(t)kX1 6 C for all t. We conclude: Theorem 3.1 holds with φ(t) = Ct. More generally, the preceding discussion holds for interaction potentials w ∈ L3w + L∞ , where Lpw denotes the weak Lp space (see e.g. [11]). This follows from a short computation using symmetricdecreasing rearrangements; we omit further details. This example generalizes the results of [3], [12] and [4].

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3.2.2. A boson√star. Consider semirelativistic particles in R3 whose one-particle Hamiltonian is given by h = 1 − ∆. The particles interact by means of a Coulomb potential: w(x) = λ|x|−1 . We impose the condition λ > −4/π. This condition is necessary for both the stability of the N -body problem (i.e. Assumption (A2)) and the global well-posedness of the Hartree equation. See [8, 7] for details. It is well known that Assumptions (A1) and (A2) hold in this case. In order to show (A4) we need some regularity of ϕ(·). To this end, let s > 1 and take ϕ0 ∈ H s . Theorem 3 of [7] implies that (1.3) has a unique global solution in H s . Therefore Sobolev’s inequality implies that (A4) holds with 1 s 1 = − . q1 2 3 Thus q1 > 6, and (A3) holds with appropriately chosen values of p1 , p2 . We conclude: Theorem 3.1 holds for some continuous function φ(t). (In fact, as shown in [7], one has the bound φ(t) . eCt .) This example generalizes the result of [1]. 3.3. Proof of Theorem 3.1. 3.3.1. A family of projectors. Define the time-dependent projectors p(t) := |ϕ(t)ihϕ(t)| , Write

q(t) := 1 − p(t) .

1 = (p1 + q1 ) · · · (pN + qN )

(3.8)

and define Pk , for k = 0, . . . , N , as the term obtained by multiplying out (3.8) and selecting all summands containing k factors q. In other words, Pk =

N Y

X

i ai p1−a qi . i

(3.9)

N : i=1 a∈{0,1} P i ai =k

If k 6= {0, . . . , N } we set Pk = 0. It is easy to see that the following properties hold: (i) Pk is an orthogonal projector, (ii) Pk Pl = δkl Pk , P (iii) k Pk = 1 .

Next, for any function f : {0, . . . , N } → C we define the operator X fb := f (k)Pk . k

It follows immediately that

fbgb = fcg ,

and that fb commutes with pi and Pk . We shall often make use of the functions r k k , n(k) := . m(k) := N N 11

(3.10)

We have the relation 1 XX 1 X 1 X qi = qi Pk = kPk = m b. N N N i

i

k

(3.11)

k

Thus, by symmetry of Ψ, we get

α = hΨ , q1 Ψi = hΨ , m b Ψi .

(3.12)

The correspondence q1 ∼ m b of (3.11) yields the following useful bounds.

Lemma 3.9. For any nonnegative function f : {0, . . . , N } → [0, ∞) we have



 Ψ , fbq1 Ψ = Ψ , fbmΨ b ,

  N

Ψ , fbq1 q2 Ψ 6 Ψ , fbm b 2Ψ . N −1

(3.13) (3.14)

Proof. The proof of (3.13) is an immediate consequence of (3.11). In order to prove (3.14) we write, using symmetry of Ψ as well as (3.11),

 Ψ , fbq1 q2 Ψ =

X

 1 Ψ , fbqi qj Ψ N (N − 1) i6=j

X

  N

1 Ψ , fbqi qj Ψ = Ψ , fbm b 2Ψ , N (N − 1) N −1

6

i,j

which is the claim.

Next, we introduce the shift operation τn , n ∈ Z, defined on functions f through (τn f )(k) := f (k + n) .

(3.15)

Its usefulness for our purposes is encapsulated by the following lemma. Lemma 3.10. Let r > 1 and A be an operator on H(r) . Let Qi , i = 1, 2, be two projectors of the form Qi = # 1 · · · # r , where each # stands for either p or q. Then

Q1 A1...r fbQ2 = Q1 τd n f A1...r Q2 ,

where n = n2 − n1 and ni is the number of factors q in Qi . Proof. Define Pkr :=

N Y

X

i ai p1−a qi . i

N−r i=r+1 a∈{0,1} P i ai =k

Then, Qi fb =

X

f (k) Qi Pk =

k

The claim follows from the fact that

X

r = f (k) Qi Pk−n i

k

Pkr

X k

commutes with A1...r .

12

f (k + ni ) Qi Pkr .

3.3.2. A bound on α. ˙ Let us abbreviate W ϕ := w ∗ |ϕ|2 . From (A3) and (A4) we find W ϕ ∈ L∞ (see (3.20) below). Then i∂t ϕ = (h + W ϕ )ϕ, where h + W ϕ ∈ L(X1 ; X1∗ ). Thus, for any ψ ∈ X1 independent of t we have i∂t hψ , p ψi = hψ , [h + W ϕ , p]ψi . On the other hand, it is easy to see from (A3) and (A4) that mΨ b ∈ Q(H). Combining these observations, and noting that Ψ ∈ Q(H) ⊂ X by (A2), we see that α is differentiable in t with derivative

   α˙ = i Ψ , H − H ϕ , m b Ψ , P where H ϕ := i (hi + Wiϕ ). Thus,   X   X ϕ 1 α˙ = i Ψ , Wij − Wi , m b Ψ . N i 0. We also assume that there are constants κ1 , κ2 > 0 such that −∆ 6 κ1 h + κ2 , as an inequality of forms on H(1) .

(B2) The Hamiltonian (3.4) is self-adjoint and bounded from below. We also assume that Q(HN ) ⊂ XN , where XN is defined as in Assumption (A1). (B3) There is a constant κ3 ∈ (0, 1) such that 0 6 (1 − κ3 )(h1 + h2 ) + W12 , as an inequality of forms on H(2) . (B4) The interaction potential w is a real and even function satisfying w ∈ Lp + L∞ , where p0 < p 6 2. (B5) The solution ϕ(·) of (1.3) satisfies ϕ(·) ∈ C(R; X12 ∩ L∞ ) ∩ C 1 (R; L2 ) , where X12 := Q(h2 ) ⊂ L2 is equipped with the norm

kϕkX12 := (1 + h2 )1/2 ϕ .

Next, we define the microscopic energy per particle Ψ EN (t) :=

1 hΨN , HN ΨN i t , N

as well as the Hartree energy   Z 1 2 2 ϕ dx dy w(x − y)|ϕ(x)| |ϕ(y)| . E (t) := hϕ , h ϕi + 2 t

Ψ (t) is independent of t. Also, invoking Assumption (B5) to differentiate By spectral calculus, EN ϕ E (t) with respect to t shows that E ϕ (t) is conserved as well. Summarizing, Ψ Ψ EN (t) = EN (0) ,

E ϕ (t) = E ϕ (0) ,

We may now state the main result of this section.

17

t ∈ R.

Theorem 4.1. Let ΨN,0 ∈ Q(HN ) and assume that Assumptions (B1) – (B5) hold. Then there is a constant K, depending only on d, h, w and p, such that βN (t) 6



βN (0) +

Ψ EN

1 −E + η N ϕ



eKφ(t) ,

where η := and φ(t) :=

Z

0

t

p/p0 − 1 2p/p0 − p/2 − 1

(4.4)

  ds 1 + kϕ(s)k3X 2 ∩L∞ . 1

Ψ = E ϕ and Remark 4.2. We have convergence to the mean-field limit whenever limN EN ⊗N limN βN (0) = 0. For instance if we start in a fully factorized state, ΨN,0 = ϕ0 , then βN (0) = 0 and 1 Ψ EN − Eϕ = hϕ0 ⊗ ϕ0 , W12 ϕ0 ⊗ ϕ0 i , N

so that the Theorem 4.1 yields (1)

EN (t) 6 βN (t) .

1 Kφ(t) e , Nη

and the analogue of Corollary 3.2 holds. Remark 4.3. The following graph shows the dependence of η on p for d = 3, i.e. p0 = 6/5.

0.5 0.4

η

0.3 0.2 0.1 0 1.2

1.4

1.6

1.8

2

p Remark 4.4. Theorem 4.1 remains valid for a large class of time-dependent one-particle Hamiltonians h(t). See Section 4.4 below for a full discussion. Remark 4.5. In three dimensions Assumption (B1) and Sobolev’s inequality imply that kϕk∞ . kϕkX12 , so that Assumption (B5) is equivalent to ϕ ∈ C(R; X12 ) ∩ C 1 (R; L2 ).

18

4.2. Example: nonrelativistic particles with interaction potential of critical type. Consider nonrelativistic particles in R3 with one-particle Hamiltonian h = −∆. The interaction potential is given by w(x) = λ|x|−2 . This corresponds to a critical nonlinearity of the Hartree equation. We require that λ > −1/2, which ensures that the N -body Hamiltonian is stable and the Hartree equation has global solutions. To see this, recall Hardy’s inequality in three dimensions, hϕ , |x|−2 ϕi 6 4hϕ , −∆ϕi .

(4.5)

One easily infers that Assumptions (B1) – (B3) hold. Moreover, Assumption (B4) holds for any p < 3/2. In order to verify Assumption (B5) we refer to [5], where local well-posedness is proven. Global existence follows by standard methods using conservation of the mass kϕk2 , conservation of the energy E ϕ , and Hardy’s inequality (4.5). Together they yield an a-priori bound on kϕkX1 , from which an a-priori bound for kϕkX12 may be inferred; see [5] for details. We conclude: For any η < 1/3 there is a continuous function φ(t) such that Theorem 4.1 holds. 4.3. Proof of Theorem 4.1. 4.3.1. An energy estimate. In the first step of our proof we exploit conservation of energy to derive an estimate on k∇1 q1 Ψk. Lemma 4.6. Assume that Assumptions (B1) – (B5) hold. Then  
1 2 Ψ ϕ 2 . k∇1 q1 Ψk . E − E + 1 + kϕkX 2 ∩L∞ β + √ 1 N Proof. Write

1 E ϕ = hϕ , hϕi + hϕ , W ϕ ϕi , 2

as well as E Ψ = hΨ , h1 Ψi + Inserting

N −1 hΨ , W12 Ψi . 2N

1 = p1 p2 + (1 − p1 p2 )

in front of every Ψ in (4.7) and multiplying everything out yields

Ψ , (1 − p1 p2 )h1 (1 − p1 p2 )Ψ



= E Ψ − hΨ , p1 p2 h1 p1 p2 Ψi N −1 hΨ , p1 p2 W12 p1 p2 Ψi −

2N 

 − Ψ , (1 − p1 p2 )h1 p1 p2 Ψ − Ψ , p1 p2 h1 (1 − p1 p2 )Ψ  N − 1

 N − 1

Ψ , (1 − p1 p2 )W12 p1 p2 Ψ − Ψ , p1 p2 W12 (1 − p1 p2 )Ψ − 2N 2N  N − 1

Ψ , (1 − p1 p2 )W12 (1 − p1 p2 )Ψ . − 2N 19

(4.6)

(4.7)

We want to find an upper bound for the left-hand side. In order to control the last term on the right-hand side for negative interaction potentials, we need to use some of the kinetic energy on the left-hand side. To this end, we split the left-hand side by multiplying it with 1 = κ3 +(1−κ3 ). Thus, using (4.6), we get 

κ3 Ψ , (1 − p1 p2 )h1 (1 − p1 p2 )Ψ = EΨ − Eϕ

− hΨ , p1 p2 h1 p1 p2 Ψi + hϕ , hϕi N −1 1 − hΨ , p1 p2 W12 p1 p2 Ψi + hϕ , W ϕ ϕi

2N  2  − Ψ , (1 − p1 p2 )h1 p1 p2 Ψ − Ψ , p1 p2 h1 (1 − p1 p2 )Ψ  N − 1

 N − 1

Ψ , (1 − p1 p2 )W12 p1 p2 Ψ − Ψ , p1 p2 W12 (1 − p1 p2 )Ψ − 2N 2N 

 N − 1

− Ψ , (1 − p1 p2 )W12 (1 − p1 p2 )Ψ − (1 − κ3 ) Ψ , (1 − p1 p2 )h1 (1 − p1 p2 )Ψ . 2N

(4.8)

The rest of the proof consists in estimating each line on the right-hand side of (4.8) separately. There is nothing to be done with the first line. Line 6. The last line of (4.8) is equal to −

E 1 

N − 1D Ψ , (1 − p1 p2 )W12 (1 − p1 p2 )Ψ − (1 − κ3 ) Ψ , (1 − p1 p2 )(h1 + h2 )(1 − p1 p2 )Ψ 2N 2 E   N − 1D 6 − Ψ , (1 − p1 p2 ) (1 − κ3 )(h1 + h2 ) + W12 (1 − p1 p2 )Ψ 6 0 , 2N

where in the last step we used Assumption (B3).

Line 2. The second line on the right-hand side of (4.8) is bounded in absolute value by hϕ , hϕi − hΨ , p1 p2 h1 p1 p2 Ψi = hϕ , hϕi hΨ , (1 − p1 p2 )Ψi = hϕ , hϕi hΨ , (q1 p2 + p1 q2 + q1 q2 )Ψi 6 3 α hϕ , hϕi

6 3 β hϕ , hϕi ,

where in the last step we used (4.3). Line 3. The third line on the right-hand side of (4.8) is bounded in absolute value by 1 hϕ , W ϕ ϕi − N − 1 hΨ , p1 p2 W12 p1 p2 Ψi = 1 hϕ , W ϕ ϕi 1 − N − 1 hΨ , p1 p2 Ψi 2 2N 2 N

 1 1 ϕ 6 kW k∞ Ψ , (q1 p2 + p1 q2 + q1 q2 )Ψ + hΨ , p1 p2 Ψi 2 N   3 1 6 kW ϕ k∞ α + 2 N   1 3 ϕ . 6 kW k∞ β + 2 N

20

As in (3.20), one finds that kW ϕ k∞ 6 kwkL1 +L∞ kϕk2L2 ∩L∞ . Line 4. The fourth line on the right-hand side of (4.8) is bounded in absolute value by



  Ψ , (1 − p1 p2 )h1 p1 p2 Ψ = Ψ , (q1 p2 + p1 q2 + q1 q2 )h1 p1 p2 Ψ

 = Ψ , q1 h1 p1 p2 Ψ

 = Ψ , q1 n b−1/2 n b1/2 h1 p1 p2 Ψ

 1/2 = Ψ , q1 n b−1/2 h1 τd p1 p2 Ψ , 1n

where in the last step we used Lemma 3.10. Using Cauchy-Schwarz, we thus get q



   q

1/2 2 1/2 Ψ , (1 − p1 p2 )h1 p1 p2 Ψ 6 Ψ , p1 p2 τd h1 τd p1 p2 Ψ Ψ , q1 n b−1 Ψ 1n 1n q

p p  = hΨ , n b Ψi hϕ , h2 ϕi Ψ , τd 1 n p1 p2 Ψ ,

where in the second step we used Lemma 3.9. Using r 1 k+1 6 n(k) + √ (τ1 n)(k) = N N we find s p p

 1 Ψ , (1 − p1 p2 )h1 p1 p2 Ψ 6 β hϕ , h2 ϕi hΨ , n b Ψi + √ N   p p p 1 = hϕ , h2 ϕi β β + 1/4 N   p 1 6 2 hϕ , h2 ϕi β + √ . N

Line 5. Finally, we turn our attention to the fifth line on the right-hand side of (4.8), which is bounded in absolute value by



  Ψ , p1 p2 W12 (1 − p1 p2 )Ψ = Ψ , p1 p2 W12 (p1 q2 + q1 p2 + q1 q2 Ψ 6 2(a) + (b) ,

where



 (a) := Ψ , p1 p2 W12 q1 p2 Ψ ,



 (b) := Ψ , p1 p2 W12 q1 q2 Ψ .

One finds, using (3.17), Lemma 3.10 and Lemma 3.9,

 (a) = Ψ , p1 p2 W1ϕ q1 Ψ

 = Ψ , p1 p2 W1ϕ n b1/2 n b−1/2 q1 Ψ

 1/2 = Ψ , p1 p2 τd W1ϕ n b−1/2 q1 Ψ 1n q

 q

 Ψ , τd n Ψ Ψ, n b−1 q1 Ψ 6 kW ϕ k∞ 1 s q

  1 6 kW ϕ k∞ Ψ, n bΨ + √ Ψ, n bΨ N   1 6 2kW ϕ k∞ β + √ . N

21

The estimation of (b) requires a little more effort. We start by splitting w = w(p) + w(∞) ,

w(p) ∈ Lp , w(∞) ∈ L∞ .

This yields (b) 6 (b)(p) + (b)(∞) in self-explanatory notation. Let us first concentrate on (b)(∞) :

 (∞) (b)(∞) = Ψ , p1 p2 W12 q1 q2 Ψ

 (∞) = Ψ , p1 p2 W12 n bn b−1 q1 q2 Ψ

 (∞) −1 = Ψ , p1 p2 τd b q1 q2 Ψ 2 n W12 n q

q

 2  (∞) Ψ , τd 6 kW k∞ Ψ, n b−2 q1 q2 Ψ 2n Ψ r 2 √ (∞) α 6 kw k∞ α + N   2 6 2kw(∞) k∞ β + . N Let us now consider (b)(p) . In order to deal with the singularities in w(p) , we write it as the divergence of a vector field ξ, w(p) = ∇ · ξ . (4.9)

This is nothing but a problem of electrostatics, which is solved by x ξ = C d ∗ w(p) , |x|

with some constant C depending on d. By the Hardy-Littlewood-Sobolev inequality, we find

1 1 1 kξkq . w(p) p , = − . (4.10) q p d

Thus if p > p0 then q > 2. Denote by X12 multiplication by ξ(x1 − x2 ). For the following it is convenient to write ∇ · ξ = ∇ρ ξ ρ , where a summation over ρ = 1, . . . , d is implied. Recalling Lemma 3.10, we therefore get

 (p) (b)(p) = Ψ , p1 p2 W12 n bn b−1 q1 q2 Ψ

 (p) −1 = Ψ , p1 p2 τd b q1 q2 Ψ 2 n W12 n

 ρ ρ = Ψ , p1 p2 τd b−1 q1 q2 Ψ . 2 n (∇ X )12 n 1

Integrating by parts yields



 ρ ρ ρ −1 (b)(p) 6 ∇ρ1 τd b−1 q1 q2 Ψ + τd b q1 q2 Ψ . 2 n p1 p2 Ψ , X12 n 2 n p1 p2 Ψ , X12 ∇1 n

(4.11)

Let us begin by estimating the first term. Recalling that p = |ϕihϕ|, we find that the first term on the right-hand side of (4.11) is equal to q

ρ  −1  ρ −1 σ σ

n X p2 (∇ρ p)1 τd (∇ρ p)1 τd d b q 1 q 2 Ψ b q1 q2 Ψ 6 2 n Ψ , p2 X12 X12 p2 (∇ p)1 τ 2n Ψ 2n Ψ , n 12 q

−1

|ϕ|2 ∗ ξ 2 k∇ϕk τd

n

6 n Ψ b q q Ψ 2 1 2 ∞ r 2 √ . kξkq kϕkL2 ∩L∞ kϕkX1 α + α, N

22

where we used Young’s inequality, Assumption (B1), and Lemma 3.9. Recalling that β 6 α, we conclude that the first term on the right-hand side of (4.11) is bounded by   1 2 C kϕkX1 ∩L∞ β + . N Next, we estimate the second term on the right-hand side of (4.11). It is equal to q

ρ   ρ −1 2

∇1 n X p1 p2 τd 6 b−1 q1 q2 Ψ τd d n Ψ , ∇ n b q q Ψ 2 n Ψ , p1 p2 X12 p1 p2 τ 2n Ψ 2 1 2 12 1 q



|ϕ|2 ∗ ξ 2 kd τ2 n Ψk ∇1 n b−1 q1 q2 Ψ 6 ∞ r

2

∇1 n 6 kξkq kϕkL2 ∩L∞ α + b−1 q1 q2 Ψ . N

We estimate ∇1 n b−1 q1 q2 Ψ by introducing 1 = p1 + q1 on the left. The term arising from p1 is bounded by



−1

p1 ∇1 n b−1 q1 q2 Ψ = p1 q2 τd ∇1 q 1 Ψ 1n q

 −2 ∇1 q1 Ψ , q2 τd ∇1 q 1 Ψ 6 1n v u  N u 1 X −2 t = qi τd ∇1 q 1 Ψ ∇1 q 1 Ψ , 1n N −1 i=2 v u  N u 1 X −2 t 6 ∇1 q 1 Ψ , qi τd ∇1 q 1 Ψ 1n N i=1 q

 −2 ∇1 q 1 Ψ , n b2 τd ∇1 q 1 Ψ = 1n 6 k∇1 q1 Ψk .

The term arising from q1 in the above splitting is dealt with in exactly the same way. Thus we have proven that the second term on the right-hand side of (4.11) is bounded by r 1 CkϕkL2 ∩L∞ β + k∇1 q1 Ψk . N Summarizing, we have (p)

(b)

.

kϕk2X1 ∩L∞



1 β+ N



+ kϕkL2 ∩L∞

r

β+

1 k∇1 q1 Ψk . N

Conclusion of the proof. Putting all the estimates of the right-hand side of (4.8) together, we find

Ψ , (1 − p1 p2 )h1 (1 − p1 p2 )Ψ Ψ

ϕ



. E −E + 1+

kϕk2X 2 ∩L∞ 1






1 β+√ N

23



+ kϕkL2 ∩L∞

r

β+

1 k∇1 q1 Ψk . (4.12) N

Next, from 1 − p1 p2 = p1 q2 + q1 we deduce p p p

p

p

k h1 q1 Ψk = h1 (1 − p1 p2 )Ψ − h1 p1 q2 Ψ 6 h1 (1 − p1 p2 )Ψ + k h1 p1 q2 Ψk . Now, recalling that p = |ϕihϕ|, we find

p p p k h1 p1 q2 Ψk 6 k h1 p1 kkq2 Ψk 6 kϕkX1 β .

Therefore,

p

p

2 k h1 q1 Ψk2 . h1 (1 − p1 p2 )Ψ + kϕk2X1 β .

Plugging in (4.12) yields

r  

p
1

h1 q1 Ψ 2 . E Ψ − E ϕ + 1 + kϕk2 2 ∞ β + √1 k∇1 q1 Ψk . + kϕkL2 ∩L∞ β + X1 ∩L N N

Next, we observe that Assumption (B1) implies

p

p k∇1 q1 Ψk . h1 q1 Ψ + β ,

so that we get

r  

p
1 p

h1 q1 Ψ 2 . E Ψ − E ϕ + 1 + kϕk2 2 ∞ β + √1 k h1 q1 Ψk . β + + kϕk 2 ∩L∞ L X1 ∩L N N

Now we claim that

 

p


h1 q1 Ψ 2 . E Ψ − E ϕ + 1 + kϕk2 2 ∞ β + √1 . X1 ∩L N

(4.13)

This follows from the general estimate

x2 6 C(R + ax)

=⇒

x2 6 2CR + C 2 a2 ,

which itself follows from the elementary inequality 1 1 C(R + ax) 6 CR + C 2 a2 + x2 . 2 2 The claim of the Lemma now follows from (4.13) by using Assumption (B1). ˙ We start exactly as in Section 3. Assumptions (B1) – (B5) imply that β 4.3.2. A bound on β. is differentiable in t with derivative   i  Ψ , (N − 1)W12 − N W1ϕ − N W2ϕ , n b Ψ β˙ = 2 = 2(I) + 2(II) + (III) + complex conjugate ,

24

(4.14)

where    i

Ψ , p1 p2 (N − 1)W12 − N W1ϕ − N W2ϕ , n b q 1 p2 Ψ , 2    i

b q1 q2 Ψ , Ψ , q1 p2 (N − 1)W12 − N W1ϕ − N W2ϕ , n (II) := 2    i

(III) := b q1 q2 Ψ . Ψ , p1 p2 (N − 1)W12 − N W1ϕ − N W2ϕ , n 2 (I) :=

Term (I). Using (3.17) we find

   b q 1 p2 Ψ 2 (I) = Ψ , p1 p2 (N − 1)W12 − N W1ϕ − N W2ϕ , n

   = Ψ , p1 p2 W1ϕ , n b q 1 p2 Ψ


 b − τ[ = Ψ , p1 p2 W1ϕ n −1 n q1 p2 Ψ ,

where we used Lemma 3.10. Define µ(k) := N n(k) − (τ−1 n)(k) Thus,




√ N 6 n−1 (k) , = √ √ k+ k−1

k = 1, . . . , N .

(4.15)



 (I) = 1 Ψ , p1 p2 W ϕ µ 1 b q 1 p2 Ψ N q

 1 Ψ, µ b2 q1 Ψ kW ϕ k∞ 6 N q

 1 6 Ψ, n b−2 q1 Ψ kW ϕ k∞ N 1 . kϕk2L2 ∩L∞ , N by (3.13). Term (II). Using Lemma 3.10 we find

   b q1 q2 Ψ 2|(II)| = Ψ , q1 p2 (N − 1)W12 − N W2ϕ , n     N −1 ϕ W12 − W2 µ b q1 q2 Ψ = Ψ , q 1 p2 N



 6 Ψ , q1 p2 W12 µ b q1 q2 Ψ + Ψ , q1 p2 W2ϕ µ b q1 q2 Ψ . | {z } | {z } =: (a)

=: (b)

One immediately finds

In (a) we split

p b2 q1 q2 Ψi . kϕk2L2 ∩L∞ β . (b) 6 kW ϕ k∞ kq1 Ψk hΨ , µ w = w(p) + w(∞) ,

w(p) ∈ Lp , w(∞) ∈ L∞ ,

with a resulting splitting (a) 6 (a)(p) + (a)(∞) . The easy part is

(a)(∞) 6 kw(∞) k∞ kq1 Ψk2 . β .

25

(4.16) (4.17) (4.18)

In order to deal with (a)(p) we write w(p) = ∇ · ξ as the divergence of a vector field ξ, exactly as in the proof of Lemma 4.6; see (4.9) and the remarks after it. We integrate by parts to find

 (a)(p) = Ψ , q1 p2 (∇ρ1 X ρ )12 µ b q1 q2 Ψ



 ρ ρ 6 ∇ρ1 q1 p2 Ψ , X12 µ b q1 q2 Ψ + q1 p2 Ψ , X12 ∇ρ1 µ b q1 q2 Ψ . (4.19) The first term of (4.19) is equal to

q

ρ  q

  ρ ρ σ p ∇σ q Ψ X p2 ∇ρ q 1 Ψ , µ ∇ q Ψ , p X X Ψ, µ b2 q1 q2 Ψ 6 b q q Ψ 2 12 12 2 1 1 1 2 1 1 12 1 q

p  kξ 2 ∗ |ϕ|2 k∞ k∇1 q1 Ψk Ψ, n b−2 q1 q2 Ψ . r p  N

Ψ, n b2 Ψ 6 kξ 2 ∗ |ϕ|2 k∞ k∇1 q1 Ψk N −1 p . kξkq kϕkL2 ∩L∞ k∇1 q1 Ψk β . k∇1 q1 Ψk2 kϕkL2 ∩L∞ + β kϕkL2 ∩L∞ ,

where in the second step we used (4.15), in the third Lemma 3.9, and in the last (4.3), Young’s inequality, and (4.10). The second term of (4.19) is equal to

 q1 p2 Ψ , X ρ (p1 + q1 )∇ρ µ 1 b q1 q2 Ψ 12



 ρ ρ 6 q1 p2 Ψ , X ρ p1 τd b ∇ρ q1 q2 Ψ , (4.20) 1 µ ∇ q 1 q 2 Ψ + q 1 p2 Ψ , X q 1 µ 12

1

12

1

where we used Lemma 3.10. We estimate the first term of (4.20). The second term is dealt with in exactly the same way. We find q



 q

  ρ 2 2 p1 X ρ q1 p2 Ψ , τd Ψ , q1 p2 X12 p2 q1 Ψ ∇1 q1 Ψ , q2 τd 6 1 µ q 2 ∇1 q 1 Ψ 1 µ ∇1 q 1 q 2 Ψ 12 q

p  kξ 2 ∗ |ϕ|2 k∞ kq1 Ψk ∇1 q 1 Ψ , n b−2 q2 ∇1 q1 Ψ 6 v u N  √ u 1 X

. kξkq kϕkL2 ∩L∞ α t ∇1 q 1 Ψ , n b−2 qi ∇1 q1 Ψ N −1 i=2 v u N p u 1 X

 ∇1 q 1 Ψ , n b−2 qi ∇1 q1 Ψ . kϕkL2 ∩L∞ β t N −1 i=1 r p

 N = kϕkL2 ∩L∞ β ∇1 q 1 Ψ , n b−2 n b 2 ∇1 q 1 Ψ N −1 p . kϕkL2 ∩L∞ β k∇1 q1 Ψk 6 β kϕkL2 ∩L∞ + k∇1 q1 Ψk2 kϕkL2 ∩L∞ .

In summary, we have proven that (II) . β kϕkL2 ∩L∞ + k∇1 q1 Ψk2 kϕkL2 ∩L∞ .

Term (III). Using Lemma 3.10 we find




    b − τ[ b q1 q2 Ψ = (N − 1) Ψ , p1 p2 W12 n 2|(III)| = (N − 1) Ψ , p1 p2 W12 , n −2 n q1 q2 Ψ . 26

Defining ν(k) := N n(k) − (τ−2 n)(k) we have As usual we start by splitting




√ N 6 n−1 (k) , = √ √ k+ k−2

k = 2, . . . , N ,

(4.21)



 2 (III) 6 Ψ , p1 p2 W12 νb q1 q2 Ψ

w = w(p) + w(∞) ,

w(p) ∈ Lp , w(∞) ∈ L∞ ,

with the induced splitting (III) = (III)(p) + (III)(∞) . Thus, using Lemma 3.10, we find

 (∞) 1/2 −1/2 2 (III)(∞) = Ψ , p1 p2 W12 n b n b νb q1 q2 Ψ

 (∞) −1/2 1/2 = Ψ , p1 p2 τd W12 n b νb q1 q2 Ψ 2n q

 q

 (∞) 6 kw k∞ Ψ , τd Ψ, n b−1 νb2 q1 q2 Ψ 2n Ψ s r q  2

. β+ Ψ, n b−3 q1 q2 Ψ N s r r 2 N β+ β 6 N N −1 1 .β+√ , N where in the fifth step we used Lemma 3.9. In order to estimate (III)(p) we introduce a splitting of w(p) into “singular” and “regular” parts, (4.22) w(p) = w(p,1) + w(p,2) := w(p) 1{|w(p) |>a} + w(p) 1{|w(p)|6a} , where a is a positive (N -dependent) constant we choose later. For future reference we record the estimates 0 kw(p,1) kp0 6 a1−p/p0 kw(p) kp/p , p

kw(p,2) k2 6 a1−p/2 kw(p) kp/2 p .

(4.23a) (4.23b)

The proof of (4.23) is elementary; for instance (4.23a) follows from kw(p,1) kpp00 =

Z

p p −p dx w(p) w(p) 0 1{|w(p) |>a} Z Z p p 6 ap0 −p dx w(p) 1{|w(p) |>a} 6 ap0 −p dx w(p) .

Let us start with (III)(p,1) . As in (4.9), we use the representation w(p,1) = ∇ · ξ .

27

Then (4.10) and (4.23a) imply that kξk2 . kw(p,1) kp0 . a1−p/p0 . Integrating by parts, we find

 (p,1) 2 (III)(p,1) = Ψ , p1 p2 W12 νb q1 q2 Ψ

 ρ = Ψ , p1 p2 (∇ρ1 X12 ) νb q1 q2 Ψ



 ρ ρ 6 ∇ρ1 p1 p2 Ψ , X12 νb q1 q2 Ψ + p1 p2 Ψ , X12 ∇ρ1 νb q1 q2 Ψ .

(4.24)

(4.25)

Using k∇pk = k∇ϕk and Lemma 3.9 we find that the first term of (4.25) is bounded by q

 q

 √ ρ σ p ∇σ p Ψ ∇ρ1 p1 Ψ , p2 X12 X12 Ψ , νb2 q1 q2 Ψ . k∇pk kϕk∞ kξk2 α 2 1 1 p 6 k∇ϕk kϕk∞ a1−p/p0 β
6 k∇ϕk kϕk∞ β + a2−2p/p0 ,

where in the second step we used the estimate (4.24). Next, using Lemma 3.10, we find that the second term of (4.25) is equal to

 p1 p2 Ψ , X ρ (p1 + q1 )∇ρ νb q1 q2 Ψ 12 1



 ρ ρ ρ p1 τc b ∇ρ1 q1 q2 Ψ . 6 p1 p2 Ψ , X12 1 ν ∇1 q1 q2 Ψ + p1 p2 Ψ , X12 q1 ν

We estimate the first term (the second is dealt with in exactly the same way): q



   q

ρ 2 2 p1 p2 Ψ , X ρ p1 τc 6 ∇1 q1 Ψ , τc Ψ , p1 p2 X12 p1 p2 Ψ 1 ν ∇1 q 1 q 2 Ψ 1 ν q 2 ∇1 q 1 Ψ 12 v u N q u 1 X

 2 6 ∇1 q 1 Ψ , n b−2 qi ∇1 q1 Ψ kp2 X12 p2 kt N −1 i=2 v u N u 1 X

 6 kξk2 kϕk∞ t ∇1 q 1 Ψ , n b−2 qi ∇1 q1 Ψ N −1 i=1 r  N

. a1−p/p0 kϕk∞ ∇1 q 1 Ψ , ∇1 q 1 Ψ N −1
2−2p/p0 + k∇1 q1 Ψk2 . 6 kϕk∞ a Summarizing,

  (III)(p,1) . kϕk∞ βkϕkX + k∇1 q1 Ψk2 + a2−2p/p0 kϕkX . 1 1

Finally, we estimate



  (p,2) (p,2) d d (1) + χ (2) )q q Ψ , (III)(p,2) = Ψ , p1 p2 W12 νb q1 q2 Ψ = Ψ , p1 p2 W12 νb (χ 1 2

where

1 = χ(1) + χ(2) ,

χ(1) , χ(2) ∈ {0, 1}{0,...,N } ,

28

(4.26)

is some partition of the unity to be chosen later. The need for this partitioning will soon become (p,2) clear. In order to bound the term with χ(1) , we note that the operator norm of p1 p2 W12 q1 q2 on the full space L2 (RdN ) is much larger than on its symmetric subspace. Thus, as a first step, (p,2) we symmetrize the operator p1 p2 W12 q1 q2 in coordinate 2. We get the bound   N X 1 (p,2) d (1) p1 pi W1i qi q1 χ νb q1 Ψ Ψ, N −1 i=2 v u N uX

 1 (p,2) d (1) q q W (p−2) p p Ψ . Ψ , p1 pi W1i q1 qi χ 6 νb q1 Ψ t 1 j 1j j 1 N −1



 d (1) q q Ψ = Ψ , p1 p2 W (p,2) νb χ 1 2 12

i,j=2

Using



νb q1 Ψ 6 kb n−1 q1 Ψk 6 1

we find



 d (1) q q Ψ 6 Ψ , p1 p2 W (p,2) νb χ 1 2 12

where

X

A :=

26i6=j6N

B :=

N X

1 √ A+B, N −1

 (p,2) d (1) q W (p,2) p p Ψ , Ψ , p1 pi W1i q1 qi χ j 1j j 1 (p,2)

Ψ , p1 pi W1i

i=2

 d (1) W (p,2) p p Ψ . q1 qi χ i 1 1i

The easy part is B 6

6

N X

i=2 N X i=2

(p,2)
2

Ψ , p1 pi W1i

 pi p1 Ψ

(p,2)
2

w ∗ |ϕ|2 ∞ hΨ , p1 pi Ψi

6 (N − 1)kϕk2∞ kw(p,2) k22 . N a2−p kϕk2∞ .

Let us therefore concentrate on A =

X

26i6=j6N

=

X

26i6=j6N

 (p,2) d d (1) χ (1) q W (p,2) p p Ψ Ψ , p1 pi W1i q1 qi χ j 1j j 1

 (1) W (p,2) q W (p,2) τ\ (1) q p p Ψ Ψ , p1 pi qj τ\ 2χ 1 1j 2χ i j 1 1i

= A1 + A2 ,

29

(4.27)

with A = A1 + A2 arising from the splitting q1 = 1 − p1 . We start with |A1 | 6 =

X

26i6=j6N

X

26i6=j6N

6

X

26i6=j6N



 (1) W (p,2) W (p,2) τ\ (1) q p p Ψ Ψ , p1 pi qj τ\ 2χ 2χ i j 1 1i

1j

q q q q

 (p,2) (p,2) (p,2) (p,2) \ (1) Ψ , p1 pi qj τ\ W W W W τ2 χ(1) qi pj p1 Ψ 2χ 1i



1j

1i

1j

 (1) Ψ , (1) q p p W (p,2) W (p,2) p p q τ\ Ψ , τ\ 1 i j 2χ 2χ j 1 i 1j 1i

√ by Cauchy-Schwarz and symmetry of Ψ. Here · is any complex square root. In order to estimate this we claim that, for i 6= j,



2 (p,2) (p,2)

p1 pi W1i W1j p1 pi 6 w(p,2) ∗ |ϕ|2 ∞ .

(4.28)

Indeed, by (3.17), we have

(p,2) (p,2) (p,2) (p,2)
(p,2) p1 pi W1i W1j p1 pi = p1 pi W1i pi W1j p1 = p1 pi w(p,2) ∗ |ϕ|2 1 W1j p1 .


(p,2) The operator p1 w(p,2) ∗ |ϕ|2 1 W1j p1 is equal to fj p1 , where f (xj ) =

Thus,

Z


dx1 ϕ(x1 ) w(p,2) ∗ |ϕ|2 (x1 ) w(p,2) (x1 − xj ) ϕ(x1 ) .

2

kf k∞ 6 w(p,2) ∗ |ϕ|2 ∞ ,

from which (4.28) follows immediately. Using (4.28), we get |A1 | 6

Now let us choose

X

26i6=j6N

2

(p,2)



w ∗ |ϕ|2 2 τ\ χ(1) q1 Ψ ∞ 2

 (1) q Ψ 6 N 2 kw(p) k2p kϕk4L2 ∩L∞ Ψ , τ\ 2χ 1

 (1) n . N 2 kϕk4L2 ∩L∞ Ψ , τ\ b2 Ψ . 2χ χ(1) (k) := 1{k6N 1−δ }

for some δ ∈ (0, 1). Then

(τ2 χ(1) ) n2 6 N −δ

implies |A1 | . kϕk4L2 ∩L∞ N 2−δ .

30

(4.29)

Similarly, we find |A2 | 6

X

26i6=j6N

X

6

26i6=j6N



 (1) p p W (p,2) p W (p,2) p p τ\ (1) q Ψ Ψ , qj τ\ 2χ i 1 1 1 j 2χ i 1i

1j

(p,2)

2 (1) q Ψi

w ∗ |ϕ|2 ∞ hΨ , τ\ 2χ 1

. N 2 kϕk4L2 ∩L∞ N −δ

= kϕk4L2 ∩L∞ N 2−δ . Thus we have proven

|A| . kϕk4L2 ∩L∞ N 2−δ .

Going back to (4.27), we see that

 −δ/2 d (1) q q Ψ . kϕk2 Ψ , p1 p2 W (p,2) νb χ + kϕk∞ N −1/2 a1−p/2 . 1 2 12 L2 ∩L∞ N What remains is to estimate is the term of (III)(p,2) containing χ(2) ,

  N X 1 (p,2) 1/2 1/2 d (2) ν p p W q q χ b ν b q Ψ Ψ , 1 i 1i i 1 1 N −1 i=2 v u N

1/2

u X

 (p,2) (p−2) d (2) ν

νb q1 Ψ t Ψ , p1 pi W1i q1 qi χ b q1 qj W1j pj p1 Ψ .



 d (2) q q Ψ = Ψ , p1 p2 W (p,2) νb χ 1 2 12 6

1 N −1

Using

i,j=2

p p

1/2

νb q1 Ψ 6 hΨ , n b−1 n β b2 Ψi =

we find



 d (2) q q Ψ 6 Ψ , p1 p2 W (p,2) νb χ 1 2 12

where

A :=

X

26i6=j6N

B :=



(p,2)

Ψ , p1 pi W1i

√

β √ A+B, N −1

(4.30)

 (p,2) d (2) ν q1 qi χ b qj W1j pj p1 Ψ ,

N X

 (p,2) (p,2) d (2) ν Ψ , p1 pi W1i q1 qi χ b W1i pi p1 Ψ . i=2

Since

χ(2) (k) = 1{k>N 1−δ }

we find χ(2) ν 6 χ(2) n−1 6 N δ/2 . d (2) ν Thus, kq1 qi χ bk 6 N δ/2 and we get B 6 N

δ/2

N X

i=2

(p,2)
2

Ψ , p1 pi W1i

pi p1 Ψ




2 6 N 1+δ/2 w(p,2) ∗ |ϕ|2 ∞

6 N 1+δ/2 kw(p,2) k22 kϕk2∞ . N 1+δ/2 a2−p kϕk2∞ ,

31

by (4.23b). Next, using Lemma 3.10, we find X

 (p,2) (p,2) d d (2) ν (2) ν A = b1/2 q1 χ b1/2 W1j qi pj p1 Ψ Ψ , p1 pi qj W1i χ 26i6=j6N

X

=

26i6=j6N

 (2) τcν 1/2 W (p,2) q W (p,2) τ\ (2) τcν 1/2 q p p Ψ Ψ , p1 pi qj τ\ 2χ 2 1 1j 2χ 2 i j 1 1i

= A1 + A2 ,

where, as above, the splitting A = A1 + A2 arises from writing q1 = 1 − p1 . Thus, X

 (2) τcν 1/2 W (p,2) W (p,2) τ\ (2) τcν 1/2 q p p Ψ Ψ , p1 pi qj τ\ |A1 | 6 2χ 2 2χ 2 i j 1 1i 1j 26i6=j6N

X

=

26i6=j6N

X

6

26i6=j6N

q q q q

 (p,2) (p,2) (p,2) (p,2) \ 1/2 1/2 (2) Ψ , p1 pi qj τ\ W W W W τc τ2 χ(2) τc q i pj p1 Ψ 2χ 2ν 2ν 1i



1j

1i

1j

 (2) τcν 1/2 p p W (p,2) W (p,2) p p τ\ (2) τcν 1/2 q Ψ , Ψ , qj τ\ 2χ 2 1 i i 1 2χ 2 j 1i 1j

by Cauchy-Schwarz and symmetry of Ψ. Using (4.28) we get



2

 |A1 | 6 N 2 w(p,2) ∗ |ϕ|2 ∞ Ψ , τc 2 ν q1 Ψ b Ψi 6 N 2 kw(p,2) k2p kϕk4L2 ∩L∞ hΨ , n . N 2 kϕk4L2 ∩L∞ β .

Similarly, |A2 | 6 6

X

26i6=j6N

X

26i6=j6N



 (2) τcν 1/2 p W (p,2) p W (p,2) p τ\ (2) τcν 1/2 q p Ψ Ψ , pi qj τ\ 2χ 2 1 1 1 2χ 2 i j 1i

1j

(p,2)

2



w ∗ |ϕ|2 ∞ Ψ , τc 2 ν q1 Ψ

6 N 2 kw(p) k2p kϕk4L2 ∩L∞ hΨ , n b Ψi

. N 2 kϕk4L2 ∩L∞ β .

Plugging all this back into (4.30), we find that


d 2−p δ/2−1 (2) q q Ψ . β kϕk2 Ψ , p1 p2 W (p,2) νb χ N 1 2 12 L2 ∩L∞ + kϕk∞ + kϕk∞ a Summarizing: (III)(p,2) .

1 + kϕk2L2 ∩L∞

from which we deduce (III)(p) . kϕk∞ k∇1 q1 Ψk2

+ 1 + kϕkX1 ∩L∞


 β + a2−p N δ/2−1 + N −δ/2 + N −1/2 a1−p/2 ,


 β + a2−p N δ/2−1 + N −δ/2 + N −1/2 a1−p/2 + a2−2p/p0 . 32

Let us set a ≡ aN = N ζ and optimize in δ and ζ. This yields the relations   p δ , ζ(2 − p) + δ = 1 , − = 2ζ 1 − 2 p0 which imply δ p/p0 − 1 = , 2 2p/p0 − p/2 − 1 with δ 6 1. Thus, 
 (III)(p) . kϕk∞ k∇1 q1 Ψk2 + 1 + kϕkX ∩L∞ β + N −η , 1

where η = δ/2 satisfies (4.4).

Conclusion of the proof. We have shown that β˙ . kϕkL2 ∩L∞ k∇1 q1 Ψk2 + 1 + kϕkX1 ∩L∞ Using Lemma 4.6 we find





β + N −η .

   1 3 Ψ ϕ β˙ . 1 + kϕkX 2 ∩L∞ β + E − E + η . 1 N

(4.31)

The claim then follows from the Gr¨ onwall estimate (3.3). 4.4. A remark on time-dependent external potentials. Theorem 4.1 can be extended to timedependent external potentials h(t) without too much sweat. The only complication is that energy is no longer conserved. We overcome this problem by observing that, while the energies E Ψ (t) and E ϕ (t) exhibit large variations in t, their difference remains small. In the following we estimate the quantity E Ψ (t) − E ϕ (t) by controlling its time derivative. We need the following assumptions, which replace Assumptions (B1) – (B3). (B1’) The Hamiltonian h(t) is self-adjoint and bounded from below. We assume that there is an operatorPh0 > 0
that such that 0 6 h(t) 6 h0 for all t. We define the Hilbert space 2 2 XN = Q i (h0 )i as in (A1), and the space X1 = Q(h0 ) as in (B5) using h0 . We also assume that there are time-independent constants κ1 , κ2 > 0 such that −∆ 6 κ1 h(t) + κ2 for all t. We make the following assumptions on the differentiability of h(t). The map t 7→ hψ , h(t)ψi ˙ is continuously differentiable for all ψ ∈ X1 , with derivative hψ , h(t)ψi for some self-adjoint ˙ operator h(t). Moreover, we assume that the quantities

˙ (1 + h(t))−1/2 ˙ 2 ϕ(t)i ,

(1 + h(t))−1/2 h(t) hϕ(t) , h(t) are continuous and finite for all t.

33

(B2’) The Hamiltonian HN (t) is self-adjoint and bounded from below. We assume that Q(HN (t)) ⊂ XN for all t. We also assume that the N -body propagator UN (t, s), defined by i∂t UN (t, s) = HN (t)UN (t, s) ,

UN (s, s) = 1 ,

exists and satisfies UN (t, 0)ΨN,0 ∈ Q(HN (t)) for all t. (B3’) There is a time-independent constant κ3 ∈ (0, 1) such that 0 6 (1 − κ3 )(h1 (t) + h2 (t)) + W12 for all t. Theorem 4.7. Assume that Assumptions (B1’) – (B3’), (B4), and (B5) hold. Then there is a continuous nonnegative function φ, independent of N and ΨN,0 , such that   1 Ψ ϕ βN (t) 6 φ(t) βN (0) + EN (0) − E (0) + η , N with η defined in (4.4). Proof. We start by deriving an upper bound on the energy difference E(t) := E Ψ (t) − E ϕ (t). Assumptions (B1’) and (B2’) and the fundamental theorem of calculus imply E(t) = E(0) +

Z

0

t

  ˙ ds hΨ(s) , h˙ 1 (s)Ψ(s)i − hϕ(s) , h(s)ϕ(s)i . | {z } =: G(s)

By inserting 1 = p1 (s) + q1 (s) on both sides of h˙ 1 (s) we get (omitting the time argument s) ˙ + 2 RehΨ , p1 h˙ 1 q1 Ψi + hΨ , q1 h˙ 1 q1 Ψi . G = hΨ , p1 h˙ 1 p1 Ψi − hϕ , hϕi The first two terms of (4.32) are equal to
˙ ˙ ˙ hΨ , p1 Ψi − 1 hϕ , hϕi = αhϕ , hϕi 6 β|hϕ , hϕi| . The third term of (4.32) is bounded, using Lemmas 3.9 and 3.10, by





  1/2 2 Ψ , p1 h˙ 1 n b1/2 n b−1/2 q1 Ψ = 2 h˙ 1 p1 τd Ψ, n b−1/2 q1 Ψ 1n q

1/2 1/2  −1/2 τd Ψ , p1 h˙ 21 p1 τd Ψ n b q 1 Ψ 6 1n 1n q q

 q

 6 Ψ , τd n Ψ Ψ, n b−1 q1 Ψ |hϕ , h˙ 2 ϕi| 1 s q 1 p 6 |hϕ , h˙ 2 ϕi| β + √ β, N   q 1 2 ˙ . |hϕ , h ϕi| β + √ . N

34

(4.32)

The last term of (4.32) is equal to

 Ψ , q1 (1 + h1 )1/2 (1 + h)−1/2 h˙ 1 (1 + h1 )−1/2 (1 + h)1/2 q1 Ψ



˙ 1 + h)−1/2 (1 + h1 )1/2 q1 Ψ 2 . 6 (1 + h)−1/2 h(

Thus, using Assumption (B1’) we conclude that

 

2 1 1/2

G(t) 6 C(t) β(t) + √ + h1 (t) q1 (t)Ψ(t) N

(4.33)

for all t. Here, and in the following, C(t) denotes some continuous nonnegative function that does not depend on N . Next, we observe that, under Assumptions (B1’) – (B3’), the proof of Lemma 4.6 remains valid for time-dependent one-particle Hamiltonians. Thus, (4.13) implies  




h1 (t)1/2 q1 (t)Ψ(t) 2 . E(t) + 1 + kϕ(t)k2 2 ∞ β(t) + √1 . X1 ∩L N

Plugging this into (4.33) yields

  1 G(t) 6 C(t) β(t) + √ + E(t) . N Therefore, E(t) 6 E(0) +

Z

0

t

  1 ds C(s) β(s) + E(s) + √ , N

(4.34)

Next, we observe that, under Assumptions (B1’) – (B3’), the derivation of the estimate (4.31) in the proof of Theorem 4.1 remains valid for time-dependent one-particle Hamiltonians. Therefore,   Z t 1 ds C(s) β(s) + E(s) + η . β(t) 6 β(0) + (4.35) N 0 Applying Gr¨ onwall’s lemma to the sum of (4.34) and (4.35) yields β(t) + E(t) 6




Rt

β(0) + E(0) e

0

C

1 + η N

Z

t

0

Plugging this back into (4.35) yields   1 β(t) 6 C(t) β(0) + E(0) + η , N which is the claim.

35

Rt

ds C(s) e

0

C

.

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