On the Weak Coupling Limit of Quantum Many-body Dynamics and the Quantum Boltzmann Equation
aa r X i v : . [ m a t h . A P ] D ec ON THE WEAK COUPLING LIMIT OF QUANTUM MANY-BODY DYNAMICSAND THE QUANTUM BOLTZMANN EQUATION
XUWEN CHEN AND YAN GUO
Abstract.
The rigorous derivation of the Uehling-Uhlenbeck equation from more fundamental quan-tum many-particle systems is a challenging open problem in mathematics. In this paper, we exam theweak coupling limit of quantum N -particle dynamics. We assume the integral of the microscopic interac-tion is zero and we assume W , per-particle regularity on the coressponding BBGKY sequence so thatwe can rigorously commute limits and integrals. We prove that, if the BBGKY sequence does convergein some weak sense, then this weak-coupling limit must satisfy the infinite quantum Maxwell-Boltzmannhierarchy instead of the expected infinite Uehling-Uhlenbeck hierarchy, regardless of the statistics theparticles obey. Our result indicates that, in order to derive the Uehling-Uhlenbeck equation, one mustwork with per-particle regularity bound below W , . Introduction
The rigorous derivation of the celebrated Uehling-Uhlenbeck equation from more fundamental quan-tum many-particle systems is a challenging open problem in mathematics. This problem has receiveda lot of attentions in recent years. In particular, Erd¨os, Salmhofer and Yau have given, in [4], a for-mal derivation of the spatially homogeneous Uehling-Uhlenbeck equation as the thermodynamic limitfrom the Fock space model. Around the same time, in [1, 2, 3], Benedetto, Castella, Esposito, andPulvirenti initiated a different study of the problem with the ”classical N -particle Bogoliubov–Born–Green–Kirkwood–Yvon (BBGKY) hierarchy” approach. Here, ”classical BBGKY hierarchy” means theusual BBGKY hierarchy in R N +1 . Moreover, Benedetto, Castella, Esposito, and Pulvirenti considerthe N → ∞ limit of N particles in R instead of the thermodynamic limit. In this paper, we follow theclassical BBGKY hierarchy approach in [1, 2, 3]. Let t ∈ R , x k = ( x , ..., x k ) , v k = ( v , ..., v k ) ∈ R k , ε = N − , and φ be an even pair interaction. We consider the following quantum BBGKY hierarchy( ∂ t + v k · ∇ x k ) f ( k ) N = 1 √ ε A ( k ) ε f ( k ) N + N √ ε B ( k +1) ε f ( k +1) N , (1.1)where 1 √ ε A ( k ) ε = X i Key words and phrases. Quantum Many-body Dynamics, Quantum Boltzmann Equation, BBGKY Hierarchy, Uehling-Uhlenbeck Equation, Weak Coupling. Maxwell-Boltzmann statistics are identical, though their initial data are totally different. To be precise,the coefficient of B ( k +1) ε f ( k +1) N is ε − √ ε in [2, (2.18)] while the coefficient of B ( k +1) ε f ( k +1) N in [3, (2.14)] is N − k √ ε . Since N = ε − , the difference − k √ ε B εj,k +1 must tend to zero as long as N √ ε B εj,k +1 = ε − √ ε B εj,k +1 tendsto a definite limit as ε → k . Hence, we have not assumed anything about the statisticsthe particles obey.We shall not go into the details about the rise of hierarchy (1.1). We refer the interested readers to[2] and [3]. The ε → √ ε and the density of particles is 1. Therefore the number of collisionsper unit time is ε − . Since the quantum mechanical cross–section in the Born approximation (justifiedbecause the potential is small) is quadratic in the potential interaction, the accumulated effect is of theorder number of collisions × [potential interaction] = 1 /ε × ε = 1 . We are concerned with the central question of identifying the weak coupling limit ε → N → ∞ ) for such a quantum BBGKY hierarchy (1.1), even at a formal level. The expected ε → ∂ t + v k · ∇ x k ) f ( k ) = k X j =1 Q ,j,k +1 f ( k +1) + k X j =1 Q ,j,k +2 f ( k +2) (1.2)where the two particle term Q ,j,k +1 is given by Q ,j,k +1 f ( k +1) ( x k , v k ) = Z dv ′ j dv k +1 dv ′ k +1 W (cid:0) v j , v k +1 | v ′ j , v ′ k +1 (cid:1) ×{ f ( k +1) ( x k , x j , v , ..., v ′ j , ..., v ′ k +1 ) − f ( k +1) ( x k , x j , v , ..., v k +1 ) } , and the three particle term Q ,j,k +2 is given by Q ,j,k +2 f ( k +2) ( x k , v k ) = 8 π θ Z dv ′ j dv k +1 dv ′ k +1 W (cid:0) v j , v k +1 | v ′ j , v ′ k +1 (cid:1) ×{ f ( k +2) ( x k , x j , x j , v , ..., v ′ j , ..., v ′ k +1 , v j )+ f ( k +2) ( x k , x j , x j , v , ..., v ′ j , ..., v ′ k +1 , v k +1 ) − f ( k +2) ( x k , x j , x j , v , ..., v k +1 , v ′ j ) − f ( k +2) ( x k , x j , x j , v , ..., v k +1 , v ′ k +1 ) } . In the above, W ( v, v ′ | v ∗ , v ′∗ ) = 18 π h ˆ φ ( v ′ − v ) + θ ˆ φ ( v ′ − v ∗ ) i δ ( v + v ∗ − v ′ − v ′∗ ) × δ ( 12 (cid:16) v + v ∗ − ( v ′ ) − ( v ′∗ ) (cid:17) ) . and θ = ± f ( k ) ( t, x k , v k ) = k Y j =1 f ( t, x j , v j )to hierarchy (1.2), provided that f satisfies ∂ t f + v · ∇ x f (1.3)= Z dv ∗ dv ′∗ dv ′ W ( v, v ∗ | v ′ , v ′∗ ) (cid:8) f ′ f ′∗ (cid:0) π θf (cid:1) (cid:0) π θf ∗ (cid:1) − f f ∗ (cid:0) π θf ′ (cid:1) (cid:0) π θf ′∗ (cid:1)(cid:9) . However, to our surprise, we find that, as long as n f ( k +1) N o is of W , per-particle regularity, the infiniteUehling-Uhlenbeck hierarchy (1.2) is not the ε → ε → UST DERIVE THE UEHLING-UHLENBECK EQUATION BELOW W , limit of the BBGKY hierarchy (1.1) is the infinite quantum Maxwell-Boltzmann hierarchy coming from[3], defined as (1.4) in this paper. In fact, if one formally commutes integrals and lim ε → in approriateplaces, one finds that, regardless of the statistics the particles obey, the formal ε → ∂ t + v k · ∇ x k ) f ( k ) = k X j =1 C j,k +1 f ( k +1) (1.4)with C j,k +1 f ( k +1) = 18 π Z R dv j +1 Z S dS ω | ω · ( v j − v k +1 ) | (cid:12)(cid:12)(cid:12) ˆ φ (( ω · ( v j − v k +1 )) ω ) (cid:12)(cid:12)(cid:12) (1.5) × [ f ( k +1) (cid:0) t, x k , x j , v ..., v j − , v ′ j , v j +1 , ..., v ′ k +1 (cid:1) − f ( k +1) ( t, x k , x j , v ..., v j − , v j , v j +1 , ..., v k +1 )] . The C j,k +1 in (1.4) is certainly the Boltzmann collision operator. We use A ε and B ε to denote theinhomogeneous terms in (1.1) because neither of the ε → A ε nor B ε along gives the collisionoperator C. The collision operator C j,k +1 in (1.4) arises as the ε → A ε and B ε . Notice that, the mean-field equation of hierarchy (1.4) is not the Uehling-Uhlenbeck equation (1.3). Ifone assumes Maxwell-Boltzmann statistics as well as ˆ φ (0) = 0 , a term by term convergence from (1.1) to(1.4) was rigorously established in [3]. The mean-field equation in this case is the quantum Boltzmannequation: ( ∂ t + v · ∇ x ) f = Q ( f, f ) (1.6)where the collision operator Q is given by Q ( f, f ) = 18 π Z R dv Z S dS ω | ω · ( v − v ) | (cid:12)(cid:12)(cid:12) ˆ φ (( ω · ( v − v )) ω ) (cid:12)(cid:12)(cid:12) [ f ( t, x, v ′ ) f ( t, x, v ′ ) − f ( t, x, v ) f ( t, x, v )] , and ˆ φ is the Fourier transform of φ , and v ′ = v − ([ v − v ] · ω ) ω, v ′ = v + ([ v − v ] · ω ) ω, because f ( k ) ( t, x k , v k ) = k Y j =1 f ( t, x j , v j )is a solution to hierarchy (1.4) provided that f solves (1.6).In our main theorem, we assume W , per-particle regularity on the BBGKY sequence n f ( k ) N o Nk =1 so that we can rigorously commute limits and integrals in suitable places. We are then able to provethat, if the BBGKY sequence n f ( k ) N o Nk =1 does converge in some weak sense, then the limit sequence n f ( k ) = lim N →∞ f ( k ) N o ∞ k =1 must satisfy the infinite quantum Maxwell-Boltzmann hierarchy (1.4) insteadof the infinite Uehling-Uhlenbeck hierarchy (1.2), regardless of the statistics the particles obey. We workin the space W , k which is W , ( R k × R k ) equiped with the weak topology. We work with the norm (cid:13)(cid:13)(cid:13) f ( k ) (cid:13)(cid:13)(cid:13) W , k = k X j =1 4 X m =0 (cid:16)(cid:13)(cid:13)(cid:13) ∂ mx j f ( k ) (cid:13)(cid:13)(cid:13) L + (cid:13)(cid:13)(cid:13) ∂ mv j f ( k ) (cid:13)(cid:13)(cid:13) L (cid:17) . Our main theorem is the following. Theorem 1 (Main Theorem) . Assume the interaction potential φ is an even Schwarz class function andsatisfies the vanishing condition: ˆ φ vanishes at the origin to at least 11th order. Suppose a subsequenceof (cid:26) Γ N = n f ( k ) N o Nk =1 (cid:27) N converges weakly to some Γ = (cid:8) f ( k ) (cid:9) ∞ k =1 in the following sense:(1) In C (cid:16) [0 , T ] , W , k (cid:17) , we have, f ( k ) N → f ( k ) as N → ∞ , XUWEN CHEN AND YAN GUO (2) There is a C > such that sup k,N,t ∈ [0 ,T ] k (cid:13)(cid:13)(cid:13) f ( k ) N (cid:13)(cid:13)(cid:13) W , k C. Then Γ = (cid:8) f ( k ) (cid:9) ∞ k =1 satisfies the infinite quantum Boltzmann hierarchy (1.4), regardless of the form ofthe initial datum n f ( k ) N (0) o Nk =1 or the statistics (Bose-Einstein / Fermi-Dirac / Maxwell-Boltzmann)it satisfies. In particular, f ( k ) does not satisfy the infinite Uehling-Uhlenbeck hierarchy (1.2). We remark that Theorem 1 certainly does not imply that the Uehling-Uhlenbeck equation (1.3) is notderivable as a mean-field limit. Our result is merely an indication that, in order to derive the Uehling-Uhlenbeck equation, one must work with per-particle regularity bound below W , . It is certainly aninteresting question to lower the regularity requirement of Theorem 1. But we are not able to do socurrently.Before delving into the proof of Theorem 1, we would like to discuss the assumptions of Theorem 1.First of all, not only we are not specifying the statistics n f ( k ) N o satisify, we are not assuming any statisticsor symmetric conditions on the limit f ( k ) either. Moreover, we do not need f ( k ) N ( t ) or f ( k ) ( t ) to take aspecial form, e.g. tensor product form or quasi free form, to make Theorem 1 to hold. Compared withthe work by King [6] and Landford [7] on deriving the classical Boltzmann equation from models withhard spheres collision and singular potentials, the interparticle interaction φ we are considering here, issmooth, and hence the regularity assumption in Theorem 1 is not impossible. The proof of Theorem1 suggests that the assumption R φ = 0 or ˆ φ (0) = 0 might actually be a necessary condition such thatthe quantum BBGKY hierarchy (1.1) has a N → ∞ limit. See § φ (0) = 0.1.1. Acknowledgement. The first author would like to thank P. Germain, E. Lieb, B. Schlein, C.Sulem, and J. Yngvason for discussions related to this work.2. Proof of the Main Theorem For notational simplicity, it suffices to prove the main theorem for k = 1 and with the assumptionthat the whole sequence (cid:26) Γ N = n f ( k ) N o Nk =1 (cid:27) N has only one limit point. Our goal is to prove the absenceof cubic Uehling-Uhlenbeck terms in the limit. Let S ( k ) ( t ) be the solution operator to the equation( ∂ t + v k · ∇ x k ) f ( k ) = 0 . We will prove that every limit point Γ = (cid:8) f ( k ) (cid:9) ∞ k =1 of (cid:26) Γ N = n f ( k ) N o Nk =1 (cid:27) N in the sense of Theorem 1satisfies Z J ( x , v ) f (1) ( t , x , v ) dx dv (2.1)= Z J ( x , v ) S (1) f (1) (0 , x , v ) dx dv + Z J ( x , v ) (cid:18)Z t S (1) ( t − t ) C , f (2) ( t , x , v ) dt (cid:19) dx dv , for all real test function J ( x , v ).To this end, we use the BBGKY hierarchy (1 . . Write hierarchy (1 . 1) in integral form, we have f ( k ) N ( t k ) = S ( k ) ( t k ) f ( k ) N (0) + Z t k S ( k ) ( t k − t k +1 ) 1 √ ε A ( k ) ε f ( k ) N ( t k +1 ) dt k +1 (2.2)+ Z t k S ( k ) ( t k − t k +1 ) N √ ε B ( k +1) ε f ( k +1) N ( t k +1 ) dt k +1 . Iterate hierarchy (2.2) once and get to f (1) N ( t ) = I + II + III + IV + V, (2.3) See also [5]. UST DERIVE THE UEHLING-UHLENBECK EQUATION BELOW W , where I = S (1) ( t ) f (1) N (0) ,II = 1 √ ε Z t S (1) ( t − t ) A (1) ε f (1) N ( t ) dt ,III = N √ ε Z t S (1) ( t − t ) B (2) ε S (2) ( t ) f (2) N (0) dt ,IV = Nε Z t S (1) ( t − t ) B (2) ε Z t S (2) ( t − t ) A (2) ε f (2) N ( t ) dt dt ,V = N ε Z t S (1) ( t − t ) B (2) ε Z t S (2) ( t − t ) B (3) ε f (3) N ( t ) dt dt . On the one hand, iterating hierarchy (2.2) once gives the terms which are quadratic in φ and henceare the central part of the quantum Boltzmann hierarchy (1.4) and the Uehling-Uhlenbeck hierarchy(1.2). On the other hand, we remark that one will not obtain the infinite Uehling-Uhlenbeck hierarchy(1.2) corresponding to the Uehling-Uhlenbeck equation (1.3) even one iterates (2.2) more than once andthen considers its limit as ε → 0. The easiest way to see this is to notice that the new terms will not bequadratic in φ. If one believes the mean-field limit f ( k ) N ( t, x j , v j ) ∼ k Y j =1 f ( t, x j , v j )where f satisfies some mean-field equation, then in the ε → IV in (2.3) will generate a nonlinearitywhich is quadratic in f and φ in the mean-field equation, and V in (2.3) will produce a term which iscubic in f and quadratic in φ . With the above discussion in mind, alert reader can immediately tellthat the main part of the proof of Theorem 1 is proving that the Boltzmann collision operator C j,k +1 defined in (1.5) arises as the ε → IV, and the ε → V is zero and thus there is noUehling-Uhlenbeck term in the limit.Since f ( k ) N → f ( k ) in the sense stated in the main theorem (Theorem 1), we know by definition thatlim N →∞ Z J ( x , v ) f (1) N ( t , x , v ) dx dv = Z J ( x , v ) f (1) ( t , x , v ) dx dv , lim N →∞ Z J ( x , v ) S (1) ( t ) f (1) N (0 , x , v ) dx dv = Z J ( x , v ) S (1) ( t ) f (1) (0 , x , v ) dx dv . Moreover, it has been shown in [1, 3] that the terms II and III tend to zero as ε → 0. We are left toprove the emergence of the quardratic collision kernel C j,k from IV and the possible cubic term V is infact zero as ε → . Emergence of the Quardratic Collision Kernel. IV is the most important term since it con-tributes (1.5) in the limit. Recall IVIV = Nε Z t S (1) ( t − t ) B ε , Z t S (2) ( t − t ) A ε , f (2) N ( t ) dt dt . We write C ε , f (2) N = Nε B ε , Z t S (2) ( t − t ) A ε , f (2) N ( t ) dt . We would like to provelim N →∞ Z J ( x , v ) (cid:18)Z t S (1) ( t − t ) C ε , f (2) N ( t , x , v ) dt (cid:19) dx dv (2.4)= Z J ( x , v ) (cid:18)Z t S (1) ( t − t ) C , f (2) ( t , x , v ) dt (cid:19) dx dv , and hence obtain the quardratic collision kernel which is the rightmost term in (2.1). Notice that S (1) ( t − t ) J ( x , v ) is simply another test function for all t and t . Hence, to establish (2.4), it sufficesto prove the following proposition. Proposition 1. Under the assumptions in Theorem 1, we have lim ε → Z J ( x , v ) C ε , f (2) N ( t , x , v ) dx dv = Z J ( x , v ) C , f (2) ( t , x , v ) dx dv . XUWEN CHEN AND YAN GUO Proof. We prove the propopsed limit with a direct computation. We start by writing out C ε , f (2) N stepby step. First, Z t S (2) ( t − t ) A ε , f (2) N ( t ) dt = ( − i )(2 π ) X σ = ± σ Z t dt Z R dh S (2) ( t − t ) e ih · ( x − x ε ˆ φ ( h ) f (2) N (cid:18) t , x , x , v − σ h , v + σ h (cid:19) = ( − i )(2 π ) X σ = ± σ Z t dt Z R dh e ih · ( x − ( t − t v − x t − t v ε ˆ φ ( h ) f (2) N (cid:18) t , x − ( t − t ) v , x − ( t − t ) v , v − σ h , v + σ h (cid:19) then B ε , Z t S (2) ( t − t ) A ε , f (2) ( t ) dt = ( − i ) (2 π ) X σ ,σ = ± σ σ Z R dx Z R dv Z R dh e ih · ( x − x ε ˆ φ ( h ) Z t dt Z R dh e ih · [ x − ( t − t ( v − σ h ) − x t − t ( v σ h )] ε ˆ φ ( h ) f (2) N ( t , x − ( t − t ) (cid:18) v − σ h (cid:19) , x − ( t − t ) (cid:18) v + σ h (cid:19) ,v − σ h − σ h , v + σ h σ h . Rearrange, we have= ( − i ) (2 π ) X σ ,σ = ± σ σ Z R dx Z R dv Z t dt Z R dh Z R dh e ih · ( x − x ε e ih · [ x − x − ( t − t v − v − σ h ε ˆ φ ( h )ˆ φ ( h ) f (2) N ( t , x − ( t − t ) (cid:18) v − σ h (cid:19) , x − ( t − t ) (cid:18) v + σ h (cid:19) ,v − σ h − σ h , v + σ h σ h . So Z J ( x , v ) C ε , f (2) N ( t , x , v ) dx dv = Nε ( − i ) (2 π ) X σ ,σ = ± σ σ Z d x Z d v Z t dt Z R dh Z R dh J ( x , v ) e ih · ( x − x ε e ih · [ x − x − ( t − t v − v − σ h ε ˆ φ ( h )ˆ φ ( h ) f (2) N ( t , x − ( t − t ) (cid:18) v − σ h (cid:19) , x − ( t − t ) (cid:18) v + σ h (cid:19) ,v − σ h − σ h , v + σ h σ h . The h and h integrals are highly oscillatory. We change variables to move the h ′ s away from f (2) N :first the x part, x ,new = x ,old − ( t − t ) (cid:18) v − σ h (cid:19) ,x ,new = x ,old − ( t − t ) (cid:18) v + σ h (cid:19) , UST DERIVE THE UEHLING-UHLENBECK EQUATION BELOW W , which gives Z J ( x , v ) C ε , f (2) N ( t , x , v ) dx dv = Nε ( − i ) (2 π ) X σ ,σ = ± σ σ Z d x Z d v Z t dt Z R dh Z R dh J ( x + ( t − t ) (cid:18) v − σ h (cid:19) , v ) e ih · ( x − x t − t v − v − σ h ε e ih · ( x − x ε ˆ φ ( h )ˆ φ ( h ) f (2) N (cid:18) t , x , x , v − σ h − σ h , v + σ h σ h (cid:19) . Then the v part v ,new = v ,old − σ h − σ h ,v ,new = v ,old + σ h σ h , which yields Z J ( x , v ) C ε , f (2) N ( t , x , v ) dx dv = Nε ( − i ) (2 π ) X σ ,σ = ± σ σ Z d x Z d v Z t dt Z R dh Z R dh J ( x + ( t − t ) (cid:18) v + σ h (cid:19) , v + σ h σ h e ih · ( x − x t − t v − v σ h ε e ih · ( x − x ε ˆ φ ( h )ˆ φ ( h ) f (2) N ( t , x , x , v , v ) . To evaluate the above integral, we substitute like [1, (2.15)] t = t − εs , h = εξ − h , and have Z J ( x , v ) C ε , f (2) N ( t , x , v ) dx dv = ε ε ( − i ) (2 π ) X σ ,σ = ± σ σ Z d x Z d v Z t ε ds Z R dξ Z R dh J ( x + εs (cid:18) v + σ h (cid:19) , v + σ h σ ( εξ − h )2 ) e i ( εξ − h · ( x − x εs v − v σ h ε e ih · ( x − x ε ˆ φ ( εξ − h )ˆ φ ( h ) f (2) N ( t − εs , x , x , v , v )which simplifies to= ( − i ) (2 π ) X σ ,σ = ± σ σ Z d x Z d v Z t ε ds Z R dξ Z R dh (2.5) J ( x + εs (cid:18) v + σ h (cid:19) , v + σ h σ ( εξ − h )2 )ˆ φ ( εξ − h )ˆ φ ( h ) e iξ · ( x − x ) e i ( εξ − h ) · s ( v − v + σ h ) f (2) N ( t − εs , x , x , v , v ) . XUWEN CHEN AND YAN GUO Taking the ε → § ε → Z J ( x , v ) C ε , f (2) N ( t , x , v ) dx dv (2.6)= ( − i ) (2 π ) X σ ,σ = ± σ σ Z d x Z d v Z + ∞ ds Z R dξ Z R dh J ( x , v + σ h − σ h φ ( − h )ˆ φ ( h ) e iξ · ( x − x ) e − ih · s ( v − v + σ h ) f (2) ( t , x , x , v , v )Using the fact that Z R e iξ · ( x − x ) dξ = (2 π ) δ ( x − x ) , (2.6) becomes= ( − i ) (2 π ) X σ ,σ = ± σ σ Z d x Z d v Z + ∞ ds Z R dh J ( x , v + σ h − σ h φ ( − h )ˆ φ ( h ) δ ( x − x ) e − ih · s ( v − v + σ h ) f (2) ( t , x , x , v , v )= ( − i ) (2 π ) X σ ,σ = ± σ σ Z dx Z d v Z + ∞ ds Z R dh J ( x , v + σ h − σ h φ ( − h )ˆ φ ( h ) e − ih · s ( v − v + σ h ) f (2) ( t , x , x , v , v )Put in spherical coordinates for the dh integration: we let h = rω , where r ∈ R + and ω ∈ S , to get( − i ) (2 π ) X σ ,σ = ± σ σ Z dx Z d v Z + ∞ ds Z ∞ r dr Z S dS ω J ( x , v + σ rω − σ rω (cid:12)(cid:12)(cid:12) ˆ φ ( rω ) (cid:12)(cid:12)(cid:12) e − irω · s ( v − v + σ rω ) f (2) ( t , x , x , v , v ) . Substitute u = rs , = ( − i ) (2 π ) X σ ,σ = ± σ σ Z dx Z d v Z + ∞ du (2.7) Z ∞ rdr Z S dS ω J ( x , v + σ rω − σ rω (cid:12)(cid:12)(cid:12) ˆ φ ( rω ) (cid:12)(cid:12)(cid:12) e − iu [( v − v ) · ω + σ r ] f (2) ( t , x , x , v , v ) . Write J ( x , v + σ rω − σ rω ) = g (( σ − σ ) rω ) for short at the moment. Notice that in the middleof (2.7), we have X σ ,σ = ± σ σ (cid:18)Z + ∞ du Z S dS ω g (( σ − σ ) rω (cid:12)(cid:12)(cid:12) ˆ φ ( rω ) (cid:12)(cid:12)(cid:12) e − iu [( v − v ) · ω + σ r ] (cid:19) (2.8)= (cid:18)Z + ∞ du Z S dS ω g (0) (cid:12)(cid:12)(cid:12) ˆ φ ( rω ) (cid:12)(cid:12)(cid:12) e − iu [( v − v ) · ω + r ] (cid:19) − (cid:18)Z + ∞ du Z S dS ω g ( rω ) (cid:12)(cid:12)(cid:12) ˆ φ ( rω ) (cid:12)(cid:12)(cid:12) e − iu [( v − v ) · ω + r ] (cid:19) + (cid:18)Z + ∞ du Z S dS ω g (0) (cid:12)(cid:12)(cid:12) ˆ φ ( rω ) (cid:12)(cid:12)(cid:12) e − iu [( v − v ) · ω − r ] (cid:19) − (cid:18)Z + ∞ du Z S dS ω g ( − rω ) (cid:12)(cid:12)(cid:12) ˆ φ ( rω ) (cid:12)(cid:12)(cid:12) e − iu [( v − v ) · ω − r ] (cid:19) = A − B + C − D. UST DERIVE THE UEHLING-UHLENBECK EQUATION BELOW W , Do the substitution, ω new = − ω old in terms C and D , we then find that C = ¯ A and D = ¯ B. So (2.8) isactually = Z S dS ω (cid:18) g (0) (cid:12)(cid:12)(cid:12) ˆ φ ( rω ) (cid:12)(cid:12)(cid:12) Z + ∞ due − iu [( v − v ) · ω + r ] (cid:19) − Z S dS ω (cid:18) g ( rω ) (cid:12)(cid:12)(cid:12) ˆ φ ( rω ) (cid:12)(cid:12)(cid:12) Z + ∞ due − iu [( v − v ) · ω + r ] (cid:19) , where Re Z + ∞ due − iu [( v − v ) · ω + r ] = Re Z + ∞−∞ due − iu [( v − v ) · ω + r ] H ( u )= Re ˆ H ([( v − v ) · ω + r ]) = πδ (( v − v ) · ω + r ) . if we denote the Heaviside function by H .Putting the above computation of (2.8) into (2.7), we havelim ε → Z J ( x , v ) C ε , f (2) N ( t , x , v ) dx dv = − π Z dx Z d v Z ∞ rdr Z S dS ω [ J ( x , v ) − J ( x , v + rω )] δ (( v − v ) · ω + r ) (cid:12)(cid:12)(cid:12) ˆ φ ( rω ) (cid:12)(cid:12)(cid:12) f (2) ( t , x , x , v , v )Insert a Heaviside function H ( r ) to do the dr integral,= 28 π Z dx Z d v Z ∞−∞ H ( r ) rdr Z S dS ω ( J ( x , v + rω ) − J ( x , v )) πδ (( v − v ) · ω + r ) (cid:12)(cid:12)(cid:12) ˆ φ ( rω ) (cid:12)(cid:12)(cid:12) f (2) ( t , x , x , v , v )= 28 π Z dx Z d v Z S dS ω ( J ( x , v − [( v − v ) · ω ] ω ) − J ( x , v )) H ( − ( v − v ) · ω ) ( − ( v − v ) · ω ) (cid:12)(cid:12)(cid:12) ˆ φ ([( v − v ) · ω ] ω ) (cid:12)(cid:12)(cid:12) f (2) ( t , x , x , v , v )That is = 18 π Z dx Z d v Z S dS ω ( J ( x , v − [( v − v ) · ω ] ω ) − J ( x , v )) | ( v − v ) · ω | (cid:12)(cid:12)(cid:12) ˆ φ ([( v − v ) · ω ] ω ) (cid:12)(cid:12)(cid:12) f (2) ( t , x , x , v , v ) , which is exactly Z J ( x , v ) C , f (2) ( t , x , v ) dx dv . Whence we conclude the proof of Proposition 1. (cid:3) The Cubic Term is Zero. Here, we investigate the limit of V = N ε Z t S (1) ( t − t ) B (2) ε Z t S (2) ( t − t ) B (3) ε f (3) N ( t ) dt dt . We write Q ε , f (3) N = N ε B ε , Z t S (2) ( t − t ) (cid:0) B ε , + B ε , (cid:1) f (3) N ( t ) dt = Q ε, , f (3) N + Q ε, , f (3) N . If the ε → Q ε , f (3) N is nonzero, it will correspond to a cubic nonlinearity in the mean-fieldequation. On the one hand, as we remarked earlier in the paper, for the Uehling-Uhlenbeck equation (1.3)to rise as the mean-field equation, lim ε → Q ε , f (3) N must not be zero. On the other hand, lim ε → Q ε , f (3) N has to be zero for Theorem 1 to hold. Hence, we compute lim ε → Q ε, , f (3) N and lim ε → Q ε, , f (3) N in completedetail. Treatment of Q ε, , f (3) N . We prove that the limitlim ε → Z J ( x , v ) Q ε, , f (3) N ( t ) dx dv = 0by direct computation. Since the proposed limit is zero, we drop the prefactor ( − i )(2 π ) in B ε so that we donot need to keep track of it. We write Z t S (2) ( t − t ) B ε , f (3) N ( t ) dt = X σ = ± σ Z t dt Z R dx Z R dv Z R dh S (2) ( t − t ) e ih · ( x − x ε ˆ φ ( h ) f (3) N (cid:18) t , x , x , x , v − σ h , v , v + σ h (cid:19) . Different from the quadratic term treated in § S (2) has no effect on ( x , v ), so it becomes= X σ = ± σ Z t dt Z R dx Z R dv Z R dh e ih · [( x − ( t − t v − x ε ˆ φ ( h ) × f (3) N (cid:18) t , x − ( t − t ) v , x − ( t − t ) v , x , v − σ h , v , v + σ h (cid:19) . Then N ε B ε , Z t S (2) ( t − t ) B ε , f (3) N ( t ) dt = N ε X σ ,σ = ± σ σ Z t dt Z R dx Z R dx Z R dv Z R dv Z R dh Z R dh e ih · ( x − x ε e ih · [( x − ( t − t ( v − σ h )) − x ] ε ˆ φ ( h )ˆ φ ( h ) × f (3) N ( t , x − ( t − t ) (cid:18) v − σ h (cid:19) , x − ( t − t ) (cid:18) v + σ h (cid:19) , x ,v − σ h − σ h , v + σ h , v + σ h , thus Z J ( x , v ) Q ε, , f (3) N dx dv = 1 ε X σ ,σ = ± σ σ Z d x Z d v Z t dt Z R dh Z R dh e ih · ( x − x ε e ih · [( x − ( t − t ( v − σ h )) − x ] ε ˆ φ ( h )ˆ φ ( h ) J ( x , v ) × f (3) N ( t , x − ( t − t ) (cid:18) v − σ h (cid:19) , x − ( t − t ) (cid:18) v + σ h (cid:19) , x ,v − σ h − σ h , v + σ h , v + σ h . We move all h ’s away from f (3) N . First, we substitute the x -part with x ,new = x ,old − ( t − t ) (cid:18) v − σ h (cid:19) x ,new = x ,old − ( t − t ) (cid:18) v + σ h (cid:19) UST DERIVE THE UEHLING-UHLENBECK EQUATION BELOW W , which gives Z J ( x , v ) Q ε, , f (3) N dx dv = 1 ε X σ ,σ = ± σ σ Z d x Z d v Z t dt Z R dh Z R dh e ih · ( x t − t ( v − σ h ) − x − ( t − t ( v σ h ) ) ε e ih · ( x − x ε ˆ φ ( h )ˆ φ ( h ) J ( x + ( t − t ) (cid:18) v − σ h (cid:19) , v ) f (3) N ( t , x , x , x , v − σ h − σ h , v + σ h , v + σ h . Then the v -substitution: v ,new = v ,old − σ h − σ h , v ,new = v ,old + σ h , v ,new = v ,old + σ h Z J ( x , v ) Q ε, , f (3) N dx dv = 1 ε X σ ,σ = ± σ σ Z d x Z d v Z t dt Z R dh Z R dh e ih · ( x t − t ( v σ h ) − x − ( t − t v ε e ih · ( x − x ε ˆ φ ( h )ˆ φ ( h ) J ( x + ( t − t ) (cid:18) v + σ h (cid:19) , v + σ h σ h f (3) N ( t , x , x , x , v , v , v ) . Redo the change of variable: t = t − εs , h = εξ − h , we then have Z J ( x , v ) Q ε, , f (3) N dx dv = ε ε X σ ,σ = ± σ σ Z d x Z d v Z t ε ds Z R dξ Z R dh e i ( εξ − h · ( x εs ( v σ h ) − x − εs v ε e ih · ( x − x ε ˆ φ ( εξ − h )ˆ φ ( h ) J ( x + εs (cid:18) v + σ h (cid:19) , v + σ h − σ h σ εξ f (3) N ( t − εs , x , x , x , v , v , v ) . Write out the phase,= 1 ε X σ ,σ = ± σ σ Z d x Z d v Z t ε ds Z R dξ Z R dh e iεξ · ( x εs ( v σ h ) − x − εs v ε e − ih · ( εs ( v σ h ) − εs v ε e − ih · ( x − x ε e ih · ( x − x ε ˆ φ ( εξ − h )ˆ φ ( h ) J ( x + εs (cid:18) v + σ h (cid:19) , v + σ h − σ h σ εξ f (3) N ( t − εs , x , x , x , v , v , v ) . Rearrange the phase,= 1 ε X σ ,σ = ± σ σ Z d x Z d v Z t ε ds Z R dξ Z R dh e iξ · ( x + εs ( v + σ h ) − x − εs v ) e − ih · ( s ( v + σ h ) − s v ) e ih · ( x − x ε ˆ φ ( εξ − h )ˆ φ ( h ) J ( x + εs (cid:18) v + σ h (cid:19) , v + σ h − σ h σ εξ f (3) N ( t − εs , x , x , x , v , v , v )Now we need to perform one more change of variable to take care of [ ih · ( x − x )] /ε . We do x ,old = x − εx ,new , (2.9)and arrive at Z J ( x , v ) Q ε, , f (3) N dx dv = ε ε X σ ,σ = ± σ σ Z d x Z d v Z t ε ds Z R dξ Z R dh e iξ · ( x + εs ( v + σ h ) − x − εs v ) e − ih · ( s ( v + σ h ) − s v ) e ih · x ˆ φ ( εξ − h )ˆ φ ( h ) J ( x + εs (cid:18) v + σ h (cid:19) , v + σ h f (3) N ( t − εs , x , x , x − εx , v , v , v )which simplifies to= X σ ,σ = ± σ σ Z d x Z d v Z t ε ds Z R dξ Z R dh (2.10) e iξ · ( x + εs ( v + σ h ) − x − εs v ) e − ih · ( s ( v + σ h ) − s v ) e ih · x ˆ φ ( εξ − h )ˆ φ ( h ) J ( x + εs (cid:18) v + σ h (cid:19) , v + σ h − σ h σ εξ f (3) N ( t − εs , x , x , x − εx , v , v , v ) . Taking the ε → § ε → Z J ( x , v ) Q ε, , f (3) N dx dv (2.11)= X σ ,σ = ± σ σ Z d x Z d v Z ∞ ds Z R dξ Z R dh e iξ · ( x − x ) e − ih · ( s ( v + σ h ) − s v ) e ih · x ˆ φ ( − h )ˆ φ ( h ) J ( x , v + σ h − σ h f (3) ( t , x , x , x , v , v , v ) . Do the dx (not d x ) integration,= X σ ,σ = ± σ σ Z d x Z d v Z ∞ ds Z R dξ Z R dh e iξ · ( x − x ) e − ih · ( s ( v + σ h ) − s v ) δ ( h )ˆ φ ( − h )ˆ φ ( h ) J ( x , v + σ h − σ h f (3) ( t , x , x , x , v , v , v )Do the dx dξ dh integration,= X σ ,σ = ± σ σ Z dx Z d v Z ∞ ds (cid:12)(cid:12)(cid:12) ˆ φ (0) (cid:12)(cid:12)(cid:12) J ( x , v ) f (3) ( t , x , x , x , v , v , v )Since ˆ φ (0) = 0, the above is zero and hencelim ε → Z J ( x , v ) Q ε, , f (3) N dx dv = 0 . UST DERIVE THE UEHLING-UHLENBECK EQUATION BELOW W , Notice that if ˆ φ (0) = 0, the ds integral yields an infinity. We formally see that it is necessary for φ tohave zero integration in order to have the quantum Boltzmann hierarchy (1.4) and hence the quantumBoltzmann equation (1.6). We will go back to (2.10) in § Q ε, , f (3) N will carry a negative sign and hence cancel out Q ε, , f (3) N . Such a guess is not true.The term lim ε → Z J ( x , v ) Q ε, , f (3) N dx dv actually equals to (2.11) with no sign difference. (See (2.13).) In below we treat Q ε, , f (3) N . Treatment of Q ε, , f (3) N . We write Z t S (2) ( t − t ) B ε , f (3) N ( t ) dt = X σ = ± σ Z t S (2) ( t − t ) dt Z R dx Z R dv Z R dh e ih · ( x − x ε ˆ φ ( h ) f (3) N (cid:18) t , x , x , x , v , v − σ h , v + σ h (cid:19) = X σ = ± σ Z t dt Z R dx Z R dv Z R dh e ih · [( x − ( t − t v − x ε ˆ φ ( h ) f (3) N (cid:18) t , x − ( t − t ) v , x − ( t − t ) v , x , v , v − σ h , v + σ h (cid:19) then N ε B ε , Z t S (2) ( t − t ) B ε , f (3) N ( t ) dt = N ε X σ ,σ = ± σ σ Z t dt Z R dx Z R dx Z R dv Z R dv Z R dh Z R dh e ih · ( x − x ε e ih · [( x − ( t − t ( v σ h )) − x ] ε ˆ φ ( h )ˆ φ ( h ) f (3) N ( t , x − ( t − t ) (cid:18) v − σ h (cid:19) , x − ( t − t ) (cid:18) v + σ h (cid:19) , x ,v − σ h , v + σ h − σ h , v + σ h Z J ( x , v ) Q ε, , f (3) N dx dv = 1 ε X σ ,σ = ± σ σ Z d x Z d v Z t dt Z R dh Z R dh e ih · ( x − x ε e ih · [( x − ( t − t ( v σ h )) − x ] ε ˆ φ ( h )ˆ φ ( h ) J ( x , v ) f (3) N ( t , x − ( t − t ) (cid:18) v − σ h (cid:19) , x − ( t − t ) (cid:18) v + σ h (cid:19) , x ,v − σ h , v + σ h − σ h , v + σ h . Again, we change variables to move all h ’s away from f (3) N . We use the new x -variables: x ,new = x ,old − ( t − t ) (cid:18) v − σ h (cid:19) x ,new = x ,old − ( t − t ) (cid:18) v + σ h (cid:19) which gives Z J ( x , v ) Q ε, , f (3) N dx dv = 1 ε X σ ,σ = ± σ σ Z d x Z d v Z t dt Z R dh Z R dh e ih · ( x t − t ( v − σ h ) − x − ( t − t ( v σ h ) ) ε e ih · ( x − x ε ˆ φ ( h )ˆ φ ( h ) J ( x + ( t − t ) (cid:18) v − σ h (cid:19) , v ) f (3) N ( t , x , x , x , v − σ h , v + σ h − σ h , v + σ h . Then the new velocity variables: v ,new = v ,old − σ h , v ,new = v ,old + σ h − σ h , v ,new = v ,old + σ h Z J ( x , v ) Q ε, , f (3) N dx dv = 1 ε X σ ,σ = ± σ σ Z d x Z d v Z t dt Z R dh Z R dh e ih · ( x t − t v − x − ( t − t ( v σ h ) ) ε e ih · ( x − x ε ˆ φ ( h )ˆ φ ( h ) J ( x + ( t − t ) v , v + σ h f (3) N ( t , x , x , x , v , v , v ) . With the change of variables t = t − εs , h = εξ − h , we then have Z J ( x , v ) Q ε, , f (3) N dx dv = ε ε X σ ,σ = ± σ σ Z d x Z d v Z t ε ds Z R dξ Z R dh e i ( εξ − h · ( x εs v − x − εs ( v σ h ) ) ε e ih · ( x − x ε ˆ φ ( εξ − h )ˆ φ ( h ) J ( x + εs v , v + σ εξ − h f (3) N ( t − εs , x , x , x , v , v , v ) . Write out the phase,= 1 ε X σ ,σ = ± σ σ Z d x Z d v Z t ε ds Z R dξ Z R dh e iεξ · ( x εs v − x − εs ( v σ h ) ) ε e − ih · ( εs v − εs ( v σ h ) ) ε e − ih · ( x − x ε e ih · ( x − x ε ˆ φ ( εξ − h )ˆ φ ( h ) J ( x + εs v , v + σ εξ − h f (3) N ( t − εs , x , x , x , v , v , v ) , that is, = 1 ε X σ ,σ = ± σ σ Z d x Z d v Z t ε ds Z R dξ Z R dh e iξ · ( x + εs v − x − εs ( v + σ h ) ) e − ih · ( s v − s ( v + σ h ) ) e ih · (2 x − x − x ε ˆ φ ( εξ − h )ˆ φ ( h ) J ( x + εs v , v + σ εξ − h f (3) N ( t − εs , x , x , x , v , v , v ) . UST DERIVE THE UEHLING-UHLENBECK EQUATION BELOW W , Another change of variable x ,old = 2 x − x − εx ,new , takes us to Z J ( x , v ) Q ε, , f (3) N dx dv (2.12)= ε ε X σ ,σ = ± σ σ Z d x Z d v Z t ε ds Z R dξ Z R dh e iξ · ( x + εs v − x − εs ( v + σ h ) ) e − ih · ( s v − s ( v + σ h ) ) e ih · x ˆ φ ( εξ − h )ˆ φ ( h ) J ( x + εs v , v + σ εξ − h f (3) N ( t − εs , x , x , x − x − εx , v , v , v ) . Putting the ε → § ε → Z J ( x , v ) Q ε, , f (3) N dx dv (2.13)= X σ ,σ = ± σ σ Z d x Z d v Z ∞ ds Z R dξ Z R dh e iξ · ( x − x ) e − ih · ( s v − s ( v + σ h ) ) e ih · x ˆ φ ( − h )ˆ φ ( h ) J ( x , v − σ h f (3) ( t , x , x , x − x , v , v , v )= X σ ,σ = ± σ σ Z d x Z d v Z R dξ Z R dh e iξ · ( x − x ) e − ih · ( s v − s ( v + σ h ) ) δ ( h )ˆ φ ( − h )ˆ φ ( h ) J ( x , v − σ h f (3) ( t , x , x , x − x , v , v , v )which is zero under the same reasoning as in the treatment of Q ε, , f (3) N . At this point, we have proven that the possible cubic term V = N ε Z t S (1) ( t − t ) B (2) ε Z t S (2) ( t − t ) B (3) ε f (3) N ( t ) dt dt is zero in the ε → f ( k ) and henceestablished Theorem 1. The rest of this section is to prove that we can take the limits inside theintegrals under the assumptions of Theorem 1.2.3. Justifying lim ε → R = R lim ε → . We interchanged ”lim ε → ” and ” R ” in going from going from(2.5) to (2.6), from (2.10) to (2.11), and from (2.12) to (2.13). We justify (2.12) to (2.13). The proof ofthe other two is similar.We claim that, if sup N X j =1 4 X m =0 (cid:16)(cid:13)(cid:13)(cid:13) ∂ mx j f (3) N ( t, · ) (cid:13)(cid:13)(cid:13) L + (cid:13)(cid:13)(cid:13) ∂ mv j f (3) N ( t, · ) (cid:13)(cid:13)(cid:13) L (cid:17) < + ∞ , then let ε → 0, we havelim ε → Z J ( x , v ) Q ε, , f (3) N dx dv = X σ ,σ = ± σ σ Z d x Z d v Z ∞ ds Z R dξ Z R dh e iξ · ( x − x ) e − ih · ( s v − s ( v + σ h ) ) e ih · x ˆ φ ( − h )ˆ φ ( h ) J ( x , v − σ h f (3) ( t , x , x , x − x , v , v , v )if R J ( x , v ) Q ε, , f (3) N dx dv is given by (2.10). In fact, rewrite Z J ( x , v ) Q ε, , f (3) N dx dv = Z t ε ds Z R dξ A ε ( s , ξ )where A ε ( s , ξ ) ≡ Z d x Z d v Z R dh e iξ · ( x + εs v − x − εs ( v + σ h ) ) e − ih · s [ v − ( v + σ h ) ] e ih · x ˆ φ ( εξ − h )ˆ φ ( h ) J ( x + εs v , v + σ εξ − h f (3) N ( t − εs , x , x , x − x − εx , v , v , v ) . Let x − x = y , x + x = y as well as v − v − σ h w , v + v − σ h w , h = h , which makes v = w + w − σ h , v = w − w , h = h , we can then transform A ε ( s , ξ ) into A ε ( s , ξ )= Z e iξ ·{ y + εs [ w w − σ h ] − εs ( w − w + σ h ) } e − ih · s w e ih · x ˆ φ ( εξ − h )ˆ φ ( h ) J ( y + y εs w + w − σ h , w + w − σ h σ εξ − h f (3) N ( t − εs , y + y , y − y , y − y − εx , w + w − σ h , w − w , v ) . ≡ Z e iξ · y e − ih · s w e ih · x B, where B ( εs , y , y , εx , w , w , εξ , h , v ) ≡ e iξ · εs { [ w w − σ h ] − ( w − w + σ h ) } ˆ φ ( εξ − h )ˆ φ ( h ) J ( y + y εs w + w − σ h , w + w − σ h σ εξ − h f (3) N ( t − εs , y + y , y − y , y − y − εx , w + w − σ h , w − w , v )Clearly, for bounded ξ , x and s , such a integral is finite. We only need to control large ξ , x and s to pass to the limit.Upon using standard smooth cutoff functions, we only need to concentrate the most singular regionof | ξ | > , | x | > , | s | > . (2.15)we may further assume that in such a region, (cid:12)(cid:12) ξ (cid:12)(cid:12) & | ξ | , (cid:12)(cid:12) x (cid:12)(cid:12) & | x | , | h | & | h | (2.16)All the other cases are simpler and can be controlled similarly.We first integrate by part in y , w repeatedly, (since εs is bounded), to obtain Z dy dy dx dw dw dv dh { ξ } m s m { h } m e iξ · y e − ih · s w e ih · x ∂ my ∂ mw B ( εs , y , y , x , w , w , ξ , h , v )= Z dy dy dx dw dw dv dh { ξ } m s m { h } m e iξ · y e − ih · s w x d { e ih x } dh ∂ my ∂ mw B ( εs , y , y , x , w , w , ξ , h , v ) . UST DERIVE THE UEHLING-UHLENBECK EQUATION BELOW W , we then take integration by part in h four times as above to get the worse term, in terms of vanishingorder of ˆ φ ( εξ − h )ˆ φ ( h ) in B as ∽ X j Z dy dy dx dw dw dv dh s j { w } j { ξ } m s m { h } m { x } j e iξ · y e − ih · s w e ih · x ∂ − jh ∂ mh ∂ my ∂ mw B ( εs , y , y , εx , w , w , εξ , h , v ) ∽ m X j =0 Z dy dy dx dw dw dv dh s j { w } j { ξ } m s m { h } m { x } j e iξ · y e − ih · s w e ih · x ∂ − j + mh [ˆ φ ( h )ˆ φ ( εξ − h )] × B j ( εs , y , y , εx , w , w , εξ , h , v ) . Here B j is some nice function with decay in w so that the growth in w is under control. Hence, thisis uniformly integrable for large ξ , s , x , h if m > h for | h | < . We now use the vanishing condition of ˆ φ : ˆ φ ( h ) = h n for | h | ≤ , so that for 0 ≤ j ≤ , | ∂ m +4 − jh [ˆ φ ( h )ˆ φ ( εξ − h )] | ≤ h n − m − . Hence by (2.16) near h = 0 , the integral has a singularity of h − n +2 m . If 4 − n + 2 m < , or n ≥ m > , then we know that R | h |≤ h − n + m dh < + ∞ , and it is uniformly bounded integrable, by(2.16). Hence, we can interchange ”lim ε → ” and ” R ” in going from (2.12) to (2.13) as claimed.2.4. It is Necessary to Have R φ = 0 . Recall (2.10), Z J ( x , v ) Q ε, , f (3) N dx dv (2.17)= X σ ,σ = ± σ σ Z d x Z d v Z t ε ds Z R dξ Z R dh e iξ · ( x + εs ( v + σ h ) − x − εs v ) e − ih · ( s ( v + σ h ) − s v ) e ih · x ˆ φ ( εξ − h )ˆ φ ( h ) J ( x + εs (cid:18) v + σ h (cid:19) , v + σ h − σ h σ εξ f (3) N ( t − εs , x , x , x − εx , v , v , v ) . To see that it is necessary to have R φ = 0 in order to have a ε → J ( ... ) f (3) N ( ... ) is a test function g, because the main point here is the phase e − ih · ( s ( v + σ h ) − s v ) e ih · x . Do the dx integrals in (2.17), we have= X σ ,σ = ± σ σ Z d x Z d v Z t ε ds Z R dξ Z R dh e iξ · ( x + εs ( v + σ h ) − x − εs v ) e − ih · ( s ( v + σ h ) − s v ) ˆ φ ( εξ − h )ˆ φ ( h ) 1 ε ˆ g ( t − εs , x , x , h ε , v , v , v ) . Here ˆ g means the Fourier transform in x . We then find that, for every t , x , v , we effectively have a δ ( h ), so that the dh integral is restricted to have size | h | . ε. Now, say t . 1, we know | s h | . e − ih · ( s ( v + σ h ) − s v ) ∼ ds integral to blow up as ε → φ (0) = 0 . Appendix A. The Cubic Term of (1.3) when ˆ φ (0) = 0Theorem 1 is unexpected. It rules out the possibility to have the Uehling-Uhlenbeck equation (1.3)as the mean-field equation from the BBGKY hierarchy (1.1) in the sense that if f solves (1.3), then f ( k ) ( t, x k , v k ) = k Y j =1 f ( t, x j , v j )is not a solution to the N → ∞ limit of the BBGKY hierarchy (1.1). It is then natural to wonder if theassumption: ˆ φ (0) = 0 , implies that the cubic term in the Uehling-Uhlenbeck equation (1.3) is zero. Suchan statement is unlikely to be true. We include a discussion here for completeness. On the one hand, ifˆ φ (0) = 0, the ε → § M ( f ) = 8 π θ Z dv ∗ Z dv ′∗ Z dv ′ W ( v, v ′ | v ∗ , v ′∗ ) [( f ′ f ′∗ f + f ′ f ′∗ f ∗ ) − ( f f ∗ f ′∗ + f f ∗ f ′ )] , and W ( v, v ′ | v ∗ , v ′∗ ) = 18 π h ˆ φ ( v ′ − v ) + θ ˆ φ ( v ′ − v ∗ ) i δ ( v + v ∗ − v ′ − v ′∗ ) × δ ( 12 (cid:16) v + v ∗ − ( v ′ ) − ( v ′∗ ) (cid:17) ) . Let us suppress the ( t, x ) dependence in f, f ′ , f ∗ , f ′∗ and write f = f ( v ) , f ′ = f ( v ′ ) , f ∗ = f ( v ∗ ) , f ′∗ = f ( v ′∗ ) . since the integral we are considering has nothing to do with that. With the usual parametrization: v ′ = v + [( v − v ∗ ) · ω ] ω, v ′∗ = v ∗ − [( v − v ∗ ) · ω ] ω, we reach M ( f ) = πθ Z R dv ∗ Z S dS ω | v − v ∗ |× h ˆ φ ([( v − v ∗ ) · ω ] ω ) + θ ˆ φ (( v − v ∗ ) + [( v − v ∗ ) · ω ] ω ) i × [ f ′ f ′∗ ( f + f ∗ ) − f f ∗ ( f ′ + f ′∗ )] . Let θ = 1 . Assume that ˆ φ does not change sign and ˆ φ ( ξ ) = 0 only at ξ = 0 , then | v − v ∗ | h ˆ φ ([( v − v ∗ ) · ω ] ω ) + θ ˆ φ (( v − v ∗ ) + [( v − v ∗ ) · ω ] ω ) i = 0only when v ∗ = v which is a measure zero set. It is then hard to believe that if ˆ φ ( ξ ) = 0 only at ξ = 0will make M ( f ) = 0 for every f . References [1] D. Benedetto, F. Castella, R. Esposito, M. Pulvirenti, Some considerations on the derivation of the nonlinear quantumboltzmann equation , J. Stat. Phys. (2004), 381–410.[2] D. Benedetto, F. Castella, R. Esposito, M. Pulvirenti, On the weak-coupling limit for bosons and fermions , Math. Mod.Meth. Appl. Sci. (2005), 1811–1843.[3] D. Benedetto, F. Castella, R. Esposito, M. 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Department of Mathematics, Brown University, 151 Thayer Street, Providence, RI 02912 E-mail address : [email protected] URL : Division of Applied Mathematics, Brown University, 182 George Street, Providence, RI 02912 E-mail address : yan [email protected] URL : Such a potential φφ