LECTURE9—10/21/2020THECOMPOSITIONFORMULA1.THE COMPOSITIONFORMULAFOR WEYLQUANTIZATION The composition.Given two semiclassical pseudo-differential operators aw and w, a natural ques-tion is: what is the composition awobw?is it still a semiclassical pseudo-differentialoperator? If yes, what is its Weyl symbol? Our first task today will be: given a(r,)and b(r, s), find a new symbol function a * b = a * b(r, ), called the Moyal productof aand b,sothata*b"=aw。bwLet's start with a very simple example: we take n = 1, a(r, $) = r and b(r,s) = $.ThenWI(PQ +QP)+aW。bW=QP=hi.Id = re ++Fromthis simple examplewecanobserveeven in this simple case the function a *b is h-dependent:1a*b(c,)=+_hi;.the function a*b(,) we arelooking for is not the product re = a(r, s)b(r,$)(as we can expect, otherwise the composition will be commutative); however, the product re does appear in a b(r, ) as the “leading term"(namelythetermwithlowesthpower).We will show below that given a and b, we can construct a function c (which maydepends on h, with leading term is ab) such that ab= aw 。6w.For simplicity we will assume a,b (Rn × R").[We will extend the results tolarger symbol classes later.J Last time we showed-et(uQ+μ-P)[Fna](u, μ)dudμaw(2元h)2n and similarly1Bwet(-Q+v.P)[Fb](u, v)dudv.(2元h)2r-
LECTURE 9 — 10/21/2020 THE COMPOSITION FORMULA 1. The composition formula for Weyl quantization ¶ The composition. Given two semiclassical pseudo-differential operators a❜W and ❜b W , a natural question is: what is the composition a❜W ◦ ❜b W ? is it still a semiclassical pseudo-differential operator? If yes, what is its Weyl symbol? Our first task today will be: given a(x, ξ) and b(x, ξ), find a new symbol function a ? b = a ? b(x, ξ), called the Moyal product of a and b, so that a ? b ÕW = a❜ W ◦ ❜b W . Let’s start with a very simple example: we take n = 1, a(x, ξ) = x and b(x, ξ) = ξ. Then a❜ W ◦ ❜b W = QP = 1 2 (P Q + QP) + 1 2 ~i · Id = xξÚ+ 1 2 ~i W . From this simple example we can observe • even in this simple case the function a ? b is ~-dependent: a ? b(x, ξ) = xξ + 1 2 ~i; • the function a?b(x, ξ) we are looking for is not the product xξ = a(x, ξ)b(x, ξ) (as we can expect, otherwise the composition will be commutative); • however, the product xξ does appear in a ? b(x, ξ) as the “leading term” (namely the term with lowest ~ power). We will show below that given a and b, we can construct a function c (which may depends on ~, with leading term is ab) such that a ? b ÕW = a❜W ◦ ❜b W . For simplicity we will assume a, b ∈ S (R n × R n ). [We will extend the results to larger symbol classes later.] Last time we showed a❜ W = 1 (2π~) 2n ❩ R2n e i ~ (u·Q+µ·P) [F~a](u, µ)dudµ and similarly ❜b W = 1 (2π~) 2n ❩ R2n e i ~ (v·Q+ν·P) [F~b](v, ν)dvdν. 1
2LECTURE9—10/21/2020 THECOMPOSITIONFORMULAIt follows that the composition aw 。6w equals1 et(u-Q+μ-P)et(-Q+vP)[Fra](u, μ)[Frb)](u, v)dodvdudp.(2元h)4nJR2nJROn the other hand,from the Baker-Campbell-Hausdorffformula we can guesse(uQ+μ-P)et(v-Q+vP) =e(μ-uv)et(u+)Q+(μ+v),P)which can be justified rigorously via the formula(ei(aQ+bP)/np)(r)=e%abep(r + tb).It follows by changing of variables (u,v)→ (z= u+ u,(=μ+v) that工awobw(2h)(,)e+dwhere1[[Fra](u, μ)[Frb](z-u, (-μ)e[r(2u)-u(S-μ)]dudμ.c(z,C) =(2元h)2nJR2So we geta*b(r,s) = (Frlc)(c,s)[Fa](u,)[](u,)e()-()le(+)dzddd,(2元h)4n/R2/and by changing variable back from (z,)to (u,v),we arrived at an “ugly"formulaforthe Moyal product:(1)1a*b(,)=A quadratic exponential.To get a better expression for ab, we observe that by Fourier inversion formula.. [a](u, )[b](,)e(u+$μ+y+)dddudμ=a(,)b(y,n),(2元h)4n/R2m/So we need a clever way to relate the product e(u-w-uv)e(r-u+$-μ+-w+$) to thefunction e(a-u+$-μ+y+nv), It is not too hard to do so, at least formally: If we writee(DeD,-DeD,)“="(B素(学(D: D, - De D,)(2)
2 LECTURE 9 — 10/21/2020 THE COMPOSITION FORMULA It follows that the composition a❜W ◦ ❜b W equals 1 (2π~) 4n ❩ R2n ❩ R2n e i ~ (u·Q+µ·P) e i ~ (v·Q+ν·P) [F~a](u, µ)[F~b](v, ν)dvdνdudµ. On the other hand, from the Baker-Campbell-Hausdorff formula we can guess e i ~ (u·Q+µ·P) e i ~ (v·Q+ν·P) = e i 2~ (µ·v−u·ν) e i ~ ((u+v)·Q+(µ+ν)·P) which can be justified rigorously via the formula (e it(a·Q+b·P)/~ϕ)(x) = e it2 2~ a·b e it ~ a·xϕ(x + tb). It follows by changing of variables (v, ν) → (z = u + v, ζ = µ + ν) that a❜ W ◦ ❜b W = 1 (2π~) 2n ❩ R2n c(z, ζ)e i ~ (z·Q+ζ·P) dzdζ, where c(z, ζ) = 1 (2π~) 2n ❩ R2n [F~a](u, µ)[F~b](z − u, ζ − µ)e i 2~ [µ·(z−u)−u·(ζ−µ)]dudµ. So we get a ? b(x, ξ) = (F −1 ~ c)(x, ξ) = 1 (2π~) 4n ❩ R2n ❩ R2n [F~a](u, µ)[F~b](z − u, ζ − µ)e i 2~ [µ·(z−u)−u·(ζ−µ)]e i ~ (x·z+ξ·ζ) dzdζdudµ, and by changing variable back from (z, ζ) to (v, ν), we arrived at an “ugly” formula for the Moyal product: (1) a?b(x, ξ) = 1 (2π~) 4n ❩ R2n ❩ R2n [F~a](u, µ)[F~b](v, ν)e i 2~ (µ·v−u·ν) e i ~ (x·u+ξ·µ+x·v+ξ·ν) dvdνdudµ. ¶ A quadratic exponential. To get a better expression for a?b, we observe that by Fourier inversion formula, 1 (2π~) 4n ❩ R2n ❩ R2n [F~a](u, µ)[F~b](v, ν)e i ~ (x·u+ξ·µ+y·v+η·ν) dvdνdudµ = a(x, ξ)b(y, η). So we need a clever way to relate the product e i 2~ (µ·v−u·ν) e i ~ (x·u+ξ·µ+x·v+ξ·ν) to the function e i ~ (x·u+ξ·µ+y·v+η·ν) . It is not too hard to do so, at least formally: If we write (2) e i~ 2 (Dξ·Dy−Dx·Dη) “ = ” ❳ k≥0 1 k! ❶ i~ 2 (Dξ · Dy − Dx· Dη) ➀k
THE COMPOSITIONFORMULA3LECTURE9—10/21/2020thenCih(au+sμtynv)e(DeDy-DrDn)er(r-u+S-μ+yu+n-v)De·Du-DrDk>0k!er(ru+s-μ+yu+nv)4(u-u.v)2万=e(μw-u)et(ru+s-μ+y-u+n)So we get[e(D-Dy-Dr-Dn)et(r-u+$-μ+yru+m-v)]=e(-)et(r+sμ++s)(3)Let's justify that the formal computation is rigorous.Of course we need tojustify the operator e(De-Dy-Dr-D,): We have a 4n-dimensional Euclidian space R4nwith coordinates r, S, y, n, and we have a densely defined differential operatorDe.Dy-DrDn=(De,Dyi-Dr,Dn,)i=1Now the operator e(DeDy-D-Dn) can be defined via functional calculus, as we ex-plained last time. Equivalently, by Stone's theorem, the (unitary) operator e(De.D-Da-D,)can becharacterized asthe solution operatortothepartial differential equationih(4)a(p(t) =(De - Dy - D-· D,)(t).To justify the validityof(3), weneed an explicit formulafor theoperatore(DeDyDaD).It turns out, as one can expect, that it is the Weyl quantization of a quadrat-ic exponential.To writedown such aquadratic exponential,we have to considerthe8n-dimensional space T*R4n =R4n×R4n, whosecoordinatesaredenoted by,S,y,n,a,β, ,.Now consider thefollowing quadratic exponential:t(r,y,s, n;α, β, , 0) = e(-β-a8),(Wewilldenote=o.)Wefirstprove:Lemma 1.1. ,W = e(DeDy-Da·D)Proof. It is enough to check that otW is the solution operator to the equation (4).This follows from direct computation: since ot is a function that depends only on“cotangent variables", we have (see Lecture 7)otW=FrlootoFh
LECTURE 9 — 10/21/2020 THE COMPOSITION FORMULA 3 then e i~ 2 (Dξ·Dy−Dx·Dη) e i ~ (x·u+ξ·µ+y·v+η·ν) = ❳ k≥0 1 k! ❶ i~ 2 (Dξ · Dy − Dx· Dη) ➀k e i ~ (x·u+ξ·µ+y·v+η·ν) = ❳ k≥0 1 k! ✒ i 2~ (µ · v − u · ν) ✓k e i ~ (x·u+ξ·µ+y·v+η·ν) = e i 2~ (µ·v−u·ν) e i ~ (x·u+ξ·µ+y·v+η·ν) . So we get (3) ❤ e i~ 2 (Dξ·Dy−Dx·Dη) e i ~ (x·u+ξ·µ+y·v+η·ν) ✐ y=x,η=ξ = e i 2~ (µ·v−u·ν) e i ~ (x·u+ξ·µ+x·v+ξ·ν) . Let’s justify that the formal computation is rigorous. Of course we need to justify the operator e i~ 2 (Dξ·Dy−Dx·Dη) : We have a 4n-dimensional Euclidian space R 4n with coordinates x, ξ, y, η, and we have a densely defined differential operator Dξ · Dy − Dx · Dη = ❳n j=1 (DξjDyj − DxjDηj ). Now the operator e i~ 2 (Dξ·Dy−Dx·Dη) can be defined via functional calculus, as we explained last time. Equivalently, by Stone’s theorem, the (unitary) operator e it~ 2 (Dξ·Dy−Dx·Dη) can be characterized as the solution operator to the partial differential equation (4) ∂t(ϕ(t)) = i~ 2 (Dξ · Dy − Dx · Dη)ϕ(t). To justify the validity of (3), we need an explicit formula for the operator e i~ 2 (Dξ·Dy−Dx·Dη) . It turns out, as one can expect, that it is the Weyl quantization of a quadratic exponential. To write down such a quadratic exponential, we have to consider the 8n-dimensional space T ∗R 4n = R 4n × R 4n , whose coordinates are denoted by x, ξ, y, η, α, β, γ, δ. Now consider the following quadratic exponential: σt(x, y, ξ, η; α, β, γ, δ) = e it 2~ (γ·β−α·δ) . (We will denote σ1 = σ.) We first prove: Lemma 1.1. σ❝t W = e it~ 2 (Dξ·Dy−Dx·Dη) . Proof. It is enough to check that σ❝t W is the solution operator to the equation (4). This follows from direct computation: since σt is a function that depends only on “cotangent variables”, we have (see Lecture 7) σ❝t W = F −1 ~ ◦ σt ◦ F~
4LECTURE9—10/21/2020 THECOMPOSITIONFORMULAand thus0(otWp) = Fr1 o Otot o Fhp=Frl。((2h(·β-α·)OtoFhy= F-10(2方(-β-Q·)0 Fho F-l oQto Fhpih(De.Dy- Da.D,)oaWo.口As a consequence, we gete(De-Dy-Dr·Dn)et(r-u+s-μ+y-u+n)(et(r-,y-j.s-E,n-)(a,B,0)et(,,g,n)-(u,u,)e(r-B-a0)dadBdyd8didEddi)(2元h)4n/R4n/Re(,n)(a,B,,)e(,,,n)(u-a,μ-B,-8)e[()(μ-)-(u-a)(-) dadBdddzddd(2元h)4n/R4n/R4e(μ-uv)er(r-u+μ+y+nv)So the formula (3) is justified, and thus the"ugly formula"(1) can be simplified to2/n [Fra](u,μ)[Fb](u,)[e(DD-DD)e(cu+sμ+yu+)]a*b(r,) =dvdvdudy.(2元h)4n1= e(De-Dy-Dr-Dn) [02, [Fra](u, )[Fb)](u, )e(ru+$μ+y+) odvdud[(2元h)4n/Rr.n=5[e(De-D-Dr-Dn)(a(r, E)b(y, n)]We concludeTheorem 1.2. Supposea,b E(Rn ×R"), thenaw。bw(a,hD)=ab(5)wheretheMoyal productab isgivenbya * b(r,E) = [e(De-D,-D-D,)(a(r, E)b(y, n)(6)=r,n=5 Asymptotic behavior of a * b.As we explained at the beginning of today's lecture, the Moyal product a*b is afunction that depends on h. Fortunately, the h-dependence is not too bad: at leastformally, by applying (2) we will get1(ih)2(D D, - ,)[(,)(0, )a*b(r,) ~.=
4 LECTURE 9 — 10/21/2020 THE COMPOSITION FORMULA and thus ∂t(σ❝t W ϕ) = F −1 ~ ◦ ∂tσt ◦ F~ϕ = F −1 ~ ◦ ✒ i 2~ (γ · β − α · δ) ✓ σt ◦ F~ϕ = F −1 ~ ◦ ✒ i 2~ (γ · β − α · δ) ✓ ◦ F~ ◦ F −1 ~ ◦ σt ◦ F~ϕ = i~ 2 (Dξ · Dy − Dx · Dη) ◦ σ❝t W ϕ. As a consequence, we get e it~ 2 (Dξ·Dy−Dx·Dη) e i ~ (x·u+ξ·µ+y·v+η·ν) = 1 (2π~) 4n ❩ R4n ❩ R4n e i ~ (x−x,y ˜ −y,ξ ˜ −ξ,η ˜ −η˜)·(α,β,γ,δ) e i ~ (˜x,ξ,˜ y, ˜ η˜)·(u,µ,v,ν) e i 2~ (γ·β−α·δ) dαdβdγdδdxd˜ ˜ξdyd˜ η˜ = 1 (2π~) 4n ❩ R4n ❩ R4n e i ~ (˜x,y, ˜ ξ,˜ η˜)·(˜α,β, ˜ γ, ˜ δ˜) e i ~ (x,y,ξ,η)·(u−α,µ ˜ −β,v ˜ −γ,ν ˜ −δ˜) e i 2~ [(v−γ˜)·(µ−β˜)−(u−α˜)·(ν−δ˜)]dαd˜ βd˜ γd˜ ˜δdxd˜ ˜ξdyd˜ η˜ =e i 2~ (µ·v−u·ν) e i ~ (x·u+ξ·µ+y·v+η·ν) . So the formula (3) is justified, and thus the “ugly formula” (1) can be simplified to a ? b(x, ξ) = 1 (2π~) 4n ❩ R2n ❩ R2n [F~a](u, µ)[F~b](v, ν) ❤ e i~ 2 (Dξ·Dy−Dx·Dη) e i ~ (x·u+ξ·µ+y·v+η·ν) ✐ y=x,η=ξ dvdνdudµ. = e i~ 2 (Dξ·Dy−Dx·Dη) ➊ 1 (2π~) 4n ❩ R2n ❩ R2n [F~a](u, µ)[F~b](v, ν)e i ~ (x·u+ξ·µ+y·v+η·ν) dvdνdudµ➍ y=x,η=ξ = ❤ e i~ 2 (Dξ·Dy−Dx·Dη) (a(x, ξ)b(y, η))✐ y=x,η=ξ . We conclude Theorem 1.2. Suppose a, b ∈ S (R n × R n ), then (5) a❜ W ◦ ❜b W (x, ~D) = a ? b ÕW , where the Moyal product a ? b is given by (6) a ? b(x, ξ) = ❤ e i~ 2 (Dξ ·Dy−Dx·Dη) (a(x, ξ)b(y, η))] ☞ ☞ ☞ y=x,η=ξ . ¶ Asymptotic behavior of a ? b. As we explained at the beginning of today’s lecture, the Moyal product a ? b is a function that depends on ~. Fortunately, the ~-dependence is not too bad: at least formally, by applying (2) we will get a ? b(x, ξ) ∼ ❳∞ k=0 1 k! (i~) k 2 k (Dξ · Dy − Dx · Dη) k [a(x, ξ)b(y, η)] ☞ ☞ ☞ ☞ ☞ y=x,η=ξ ,
LECTURE9—10/21/2020THECOMPOSITIONFORMULA5in other words, a*b is a nice asymptotic series in h. To justify this formula, we needto applythestationaryphase expansionthatwe studied inLecture5.First according to Proposition 1.1 in Lecture 7, for any non-singular symmetricmatrixQ,W(eseros"0)(n) = det01/2eesnQ e-urQ-1yp(r+y)dy.(7)(2元h)n/2In particular, if we raisento 4n, take11I1Q,sgn(Q)=0Treplace r by r, S, y, n, and replace by α, β, , o, we gete(DeD,-Dr-D)p(r,S, y,n) =e-(g--)p(+,$+Ey+,n+)didedjdn(元h）2n/and thuse-(-g-m)a(r+,E+E)b(y+,n+n)didedjdnab(r,s)(元h)2nr.0=Now we apply the exact stationary phase formula for oscillating integrals with qua-dratic phase (Theorem 1.3 in Lecture5), namely. e+Qra(a)dr ~ (2h)/2 () ([det Q(Po-(D) a(0)2toour setting (with nraised to 4n,Qreplaced by-Q-1 etc)toget[B(岁)a*b(r,s) ~[）(DD-D·D) [a(r+,s+)b(y+,n+)r,n=and thus we concludeTheorem 1.3. Let a,b e. Then as h -→0,ab(a,3)~()(8)(DeDy-DaDn)"[a(r,E)b(y, n)（20An immediate consequence of (8)isCorollary 1.4. If supp(a) n supp(b) = , then a b = O(h).Proof.If supp(a)n supp(b)=, then each term in the expansion (8)vanishes.Remark. If a,b are polynomials in &-variables (namely, if the operators aw and warebothsemiclassicaldifferentialoperators),thentheasymptoticformulaformula(8) will be a polynomial expansion in h and thus be an exact formula
LECTURE 9 — 10/21/2020 THE COMPOSITION FORMULA 5 in other words, a ? b is a nice asymptotic series in ~. To justify this formula, we need to apply the stationary phase expansion that we studied in Lecture 5. First according to Proposition 1.1 in Lecture 7, for any non-singular symmetric matrix Q, (7) (eØi 2~ ξ T Qξ W ϕ)(x) = | det Q| −1/2 (2π~) n/2 e i π 4 sgnQ ❩ Rn e − i 2~ y T Q−1yϕ(x + y)dy. In particular, if we raise n to 4n, take Q = 1 2 ❻ −I I I −I ➅ Q −1 = 2❻ −I I I −I ➅ ,sgn(Q) = 0 replace x by x, ξ, y, η, and replace ξ by α, β, γ, δ, we get e i~ 2 (Dξ·Dy−Dx·Dη)ϕ(x, ξ, y, η) = 1 (π~) 2n ❩ R4n e − 2i ~ (ξ˜·y˜−x˜·η˜)ϕ(x+˜x, ξ+˜ξ, y+˜y, η+˜η)dxd˜ ˜ξdyd˜ η˜ and thus a ? b(x, ξ) = ➊ 1 (π~) 2n ❩ R4n e − 2i ~ (ξ˜·y˜−x˜·η˜) a(x + ˜x, ξ + ˜ξ)b(y + ˜y, η + ˜η)dxd˜ ˜ξdyd˜ η˜ ➍ y=x,η=ξ . Now we apply the exact stationary phase formula for oscillating integrals with quadratic phase (Theorem 1.3 in Lecture 5), namely ❩ Rn e i 2~ x T Qxa(x)dx ∼ (2π~) n/2 e i π 4 sgn(Q) | det Q| 1/2 ❳∞ k=0 1 k! ❶ − i~ 2 pQ−1 (D) ➀k a(0). to our setting (with n raised to 4n, Q replaced by −Q−1 etc) to get a?b(x, ξ) ∼ ✷ ✹ ❳ k≥0 1 k! ❶ i~ 2 ➀k ⑨ Dξ˜ · Dy˜ − Dx˜ · Dη˜ ❾k ☞ ☞ ☞ x˜=0,··· ,η˜=0 [a(x + ˜x, ξ + ˜ξ)b(y + ˜y, η + ˜η)] ✸ ✺ y=x,η=ξ and thus we conclude Theorem 1.3. Let a, b ∈ S . Then as ~ → 0, (8) a ? b(x, ξ) ∼ ❳∞ k=0 1 k! ❶ i~ 2 ➀k (Dξ · Dy − Dx · Dη) k [a(x, ξ)b(y, η)] ☞ ☞ ☞ ☞ ☞ y=x,η=ξ . An immediate consequence of (8) is Corollary 1.4. If supp(a) ∩ supp(b) = ∅, then a ? b = O(~ ∞). Proof. If supp(a) ∩ supp(b) = ∅, then each term in the expansion (8) vanishes. Remark. If a, b are polynomials in ξ-variables (namely, if the operators a❜W and ❜b W are both semiclassical differential operators), then the asymptotic formula formula (8) will be a polynomial expansion in ~ and thus be an exact formula.