LECTURE6—10/12/2020SEMICLASSICALQUANTIZATION1. THE KOHN-NIRENBERG QUANTIZATIONTThesemiclassicalFouriertransformIn semiclassical analysis, instead of the usual Fourier transform, we shall use thesemiclassical Fourier transform (also known as the h-Fourier transform) Fh. Bydefinition it is a dilation of the usual Fourier transform,Fro(s) := (F0)()= /ne-p(r)dr.(1)In PSet1-6 you are supposed to translate known properties of the usual Fouriertransform to the semiclassical Fourier transform.For example, the inverse of Fh ise(E)(2)Fr-(r) =(2元h)n JrRIt followserrv(r)drde(0)=[FhF(0)=(2h) Jawhich can be interpreted as a strange-looking distributional identity1ekrdE=0(2元h)nJeIt is also easy to checkFr(hDr)) =Frp and Fn()= (-1)lal(hDe)FTheKohn-Nirenberg quantization.Wehave mentioned that areasonable way to quantize theposition and momen-tum functions isTj~Qj=multiplicationbyjandha=hDjSwPj=V-10rjWe want to extend the “same" rule to more general functions. Note that the operatorP, and the function S, are related by the semiclassical Fourier transform, namelyPi =hDip = Fr-1(Fn(hD,)0) = F-(S,F)
LECTURE 6 — 10/12/2020 SEMICLASSICAL QUANTIZATION 1. The Kohn-Nirenberg quantization ¶ The semiclassical Fourier transform. In semiclassical analysis, instead of the usual Fourier transform, we shall use the semiclassical Fourier transform (also known as the ~-Fourier transform) F~. By definition it is a dilation of the usual Fourier transform, (1) F~ϕ(ξ) := (Fϕ)( ξ ~ ) = ❩ Rn e − ix·ξ ~ ϕ(x)dx. In PSet1-6 you are supposed to translate known properties of the usual Fourier transform to the semiclassical Fourier transform. For example, the inverse of F~ is (2) F −1 ~ ψ(x) = 1 (2π~) n ❩ Rn e ix·ξ ~ ψ(ξ)dξ. It follows ψ(0) = [F~F −1 ~ ψ](0) = 1 (2π~) n ❩ Rn ❩ Rn e i ~ x·ξψ(x)dxdξ, which can be interpreted as a strange-looking distributional identity 1 (2π~) n ❩ Rn e i ~ x·ξ dξ = δ0. It is also easy to check F~((~Dx) αϕ) = ξ αF~ϕ and F~(x αϕ) = (−1)|α| (~Dξ) αF~ϕ. ¶ The Kohn-Nirenberg quantization. We have mentioned that a reasonable way to quantize the position and momentum functions is xj Qj = multiplication by xj and ξj Pj = ~ √ −1 ∂ ∂xj = ~Dj . We want to extend the “same” rule to more general functions. Note that the operator Pj and the function ξj are related by the semiclassical Fourier transform, namely Pjϕ = ~Djϕ = F −1 ~ (F~(~Dj )ϕ) = F −1 ~ (ξjF~ϕ). 1
2LECTURE 6—10/12/2020 SEMICLASSICAL QUANTIZATIONSimilarly we have1Qi = ro = Fr-1(Fhp).More generally, we may replace Q, by any function V(r) of r to getV(r)p= F-'(V(r)Fhp),that is, the operator “multiplication by V(r)" and the function V(r) are related bythe same formula. Similarly we can relate the polynomialp(s) = pasa[a]<kwith the the constant coefficient semiclassical differential operator p(hD) defined byp(D)= pa(Dr)alal<kvia exactly the same rule:p(hD) = F-(Fh(p(hD))) = Fr-1(p(S)Fh).Bingo! In particular, by using h-Fourier transform we get an explanation ofh2H(c, ) = [EP2+V() H=-△+V(r),22since the Hamiltonian function H and the Schrodinger operator H are related byH = F-(H(r,s)Fhp).We can go further. Still suppose p is a polynomial in s, but now with coefficientsdependingonr,namelyp(r, S) = pa(r)s%.lal<kThen we can do the same computationF-1( pa(r)saF) = Pe(r)F-1Fn(D)) = Pa()(Dr)lal<klal<klal<kand thuswearriveatthesemiclassical differential operatorof theformp(r,D)=pa(r)(hDr)a.We can apply the same construction to many other classes of functions. Forsimplicity, let's first suppose a(r, s) e S(R2n) is a Schwartz function (which is thebest class of functions), we may quantize a to the operator akN given byβ μakN(0) := Fr'(a(r, E)Fn) = (Fr1)→2(a(r, S)(Fh)u(p(y).1Here and in what follows, we can“move" a function depending only on out of F-1becausethe inverse Fourier transform F-1 is an integral with respect to . Warning: although we havebothp=F(rFh)andrp=F-(Fh()),wecan'tconcluderFh=Fh(rp),sincerFhpisa function depending on both and E,whileFr(ro)is afunction only depending on
2 LECTURE 6 — 10/12/2020 SEMICLASSICAL QUANTIZATION Similarly we have 1 Qjϕ = xϕ = F −1 ~ (xF~ϕ). More generally, we may replace Qj by any function V (x) of x to get V (x)ϕ = F −1 ~ (V (x)F~ϕ), that is, the operator “multiplication by V (x)” and the function V (x) are related by the same formula. Similarly we can relate the polynomial p(ξ) = ❳ |α|≤k pαξ α with the the constant coefficient semiclassical differential operator p(~D) defined by p(~D) = ❳ |α|≤k pα(~Dx) α via exactly the same rule: p(~D)ϕ = F −1 ~ (F~(p(~D)ϕ)) = F −1 ~ (p(ξ)F~ϕ). Bingo! In particular, by using ~-Fourier transform we get an explanation of H(x, ξ) = |ξ| 2 2 + V (x) Hˆ = − ~ 2 2 ∆ + V (x), since the Hamiltonian function H and the Schr¨odinger operator Hˆ are related by Hϕˆ = F −1 ~ (H(x, ξ)F~ϕ). We can go further. Still suppose p is a polynomial in ξ, but now with coefficients depending on x, namely p(x, ξ) = ❳ |α|≤k pα(x)ξ α . Then we can do the same computation F −1 ~ ( ❳ |α|≤k pα(x)ξ αF~ϕ) = ❳ |α|≤k pα(x)F −1 ~ F~((~D) αϕ) = ❳ |α|≤k pα(x)(~Dx) α and thus we arrive at the semiclassical differential operator of the form p(x, ~D) = ❳pα(x)(~Dx) α . We can apply the same construction to many other classes of functions. For simplicity, let’s first suppose a(x, ξ) ∈ S(R 2n ) is a Schwartz function (which is the best class of functions), we may quantize a to the operator ˆa KN given by ϕ 7→ aˆ KN (ϕ) := F −1 ~ (a(x, ξ)F~ϕ) = (F −1 ~ )ξ→x(a(x, ξ)(F~)y→ξ(ϕ(y))). 1Here and in what follows, we can “move” a function depending only on x out of F −1 ~ because the inverse Fourier transform F −1 ~ is an integral with respect to ξ. Warning: although we have both xϕ = F −1 ~ (xF~ϕ) and xϕ = F −1 ~ (F~(xϕ)), we can’t conclude xF~ϕ = F~(xϕ), since xF~ϕ is a function depending on both x and ξ, while F~(xϕ) is a function only depending on ξ
3LECTURE6—10/12/2020SEMICLASSICALQUANTIZATIONThequantizationprocessawaknis called the Kohn-Nirenberg quantization (also known as the standard quantization).By using the definition of the semiclassical Fourier and inverse Fourier transformswe can easily write down an explicit formula for aKN:aKN(0) = Fr-(a(r,E)Fnp)= Fr (a(r, ) / e-ip(y)dy)[ er(pa(r,E)p(y)dyde.-1(2元h)nJRnJDETOUR:Schwartzkernel of anintegral operator.Note that if we denote1[eip-a(r,E)ds,kKN(c,y) :=(2元h）nJgthen the expression of akN can be simplified toakN(0)(r) = L kkN(r, y)p(y)dy.In general, given any reasonable (e.g. smooth or measurable, bounded or integrableetc.)kernel function k(r,y), where r e Rm and y e Rn, we can define a linearoperator K which maps a function on Rn to a function on Rm via the integral(3)K(p)(r) = / k(r,y)p(y)dy.Of course the domain of the integral operator K depends on the function k(r, y):nicer kernel function usually admits larger domain. Here we discuss two extremalcases:. Case 1: The best kernels, namely k E (Rm x Rn),In this case,for the integral (3) to make sense, you may take to be boundedcontinuous functions, or if you want, Lp functions. In fact, you can gofurther:Fact: We may take p to be a tempered distribution!Of course in this case one has to interpret the integral (3) as a pairing betweentempereddistributionsand Schwartzfunctions:foreachrfixed,thefunctionk(a,) is a Schwartz function, and we define(4)K(p)(r) := <g, k(r, 2))A natural question to ask is: what do we get? Is K() a nice function ora bad function? Or maybe only a (very bad) tempered distribution? Theanswer is:Fact:K maps a tempered distribution to a Schwartz function!To see this, we use the following remarkable theorem in distribution theory:
LECTURE 6 — 10/12/2020 SEMICLASSICAL QUANTIZATION 3 The quantization process a aˆ KN is called the Kohn-Nirenberg quantization (also known as the standard quantization). By using the definition of the semiclassical Fourier and inverse Fourier transforms, we can easily write down an explicit formula for ˆa KN : aˆ KN (ϕ) = F −1 ~ (a(x, ξ)F~ϕ) = F −1 ~ ⑩ a(x, ξ) ❩ Rn e −i y·ξ ~ ϕ(y)dy❿ = 1 (2π~) n ❩ Rn ❩ Rn e i (x−y)·ξ ~ a(x, ξ)ϕ(y)dydξ. ¶ DETOUR: Schwartz kernel of an integral operator. Note that if we denote k KN a (x, y) := 1 (2π~) n ❩ Rn e i (x−y)·ξ ~ a(x, ξ)dξ, then the expression of ˆa KN can be simplified to aˆ KN (ϕ)(x) = ❩ Rn k KN a (x, y)ϕ(y)dy. In general, given any reasonable (e.g. smooth or measurable, bounded or integrable etc.)kernel function k(x, y), where x ∈ R m and y ∈ R n , we can define a linear operator K which maps a function on R n to a function on R m via the integral (3) K(ϕ)(x) = ❩ Rn k(x, y)ϕ(y)dy. Of course the domain of the integral operator K depends on the function k(x, y): nicer kernel function usually admits larger domain. Here we discuss two extremal cases: • Case 1: The best kernels, namely k ∈ S (R m × R n ). In this case, for the integral (3) to make sense, you may take ϕ to be bounded continuous functions, or if you want, L p functions. In fact, you can go further: Fact: We may take ϕ to be a tempered distribution! Of course in this case one has to interpret the integral (3) as a pairing between tempered distributions and Schwartz functions: for each x fixed, the function k(x, ·) is a Schwartz function, and we define (4) K(ϕ)(x) := hϕ, k(x, ·)i. A natural question to ask is: what do we get? Is K(ϕ) a nice function or a bad function? Or maybe only a (very bad) tempered distribution? The answer is: Fact: K maps a tempered distribution to a Schwartz function! To see this, we use the following remarkable theorem in distribution theory:
4LECTURE6—10/12/2020SEMICLASSICALQUANTIZATIONTheorem 1.1 (Schwartz representation theorem). For any u E '(R"),there erists a finite collection ua,β : Rn C of bounded continuous functions,with|a|+B|≤k,suchthatu=Z r'Deua,B[α|+/β]≤kNote that both sides of the equation above are understood as distributions:InLecture4wehaveseenhowtorealizeanybounded continuousfunctionasa tempered distribution, and how to realize raDBu as a distribution when uis a distribution. As a consequence, for a tempered distribution E (Rn),the “paring formula"(4) of K() still has an integral representationZy°Depa,b(y), k(r, y)K()(r) =al+IB/<k<pa,p(y),Dyak(r,y))[α]+[B]≤kPa.B(y)DB (y°k(r, y))dy>[0]+1B]≤kand thus by using the fact k E (Rm × Rn), one can check K() e (Rm)Moreover,it can be shown that theoperator K:(Rn)→(Rm)iscontinuous (with respect to the weak-* topology on '(IRn)and the metrictopology on (Rn)). In summary, we haveIf the kernel function k e(Rm x Rn) is a Schwartz functionthen the integral operator K with kernel k can be defined asa continuous linear operator mapping'(Rn)to (Rm).. Case 2: The worst kernels, namely k E '(Rm × Rn).To give an exact meaning of the operator K with a distributional kernel k.we first introduce a notation: Given any p e C(Rn) and b Co(Rm), wedefine E Co(IRn × Rm) to be the function given bypμd(r,y) := p(r)d(y).Obviously if pE(R") and bE(Rm), then E(Rn x Rm)Now suppose k E '(Rm × Rn) is a tempered distribution. We can stilldefine an operator K : (Rn) → '(Rm) via the formulaK(0)() := (k,pμ).and one can show that K is a continuous linear map. In fact, it turns outthat any continuous linear map (Rn)-→ (Rm) arises in this way:2in view of the Schwartz representation theorem, the operator K is really an integral operator
4 LECTURE 6 — 10/12/2020 SEMICLASSICAL QUANTIZATION Theorem 1.1 (Schwartz representation theorem). For any u ∈ S 0 (R n ), there exists a finite collection uα,β : R n → C of bounded continuous functions, with |α| + |β| ≤ k, such that u = ❳ |α|+|β|≤k x αD β x uα,β. Note that both sides of the equation above are understood as distributions: In Lecture 4 we have seen how to realize any bounded continuous function as a tempered distribution, and how to realize x αDβu as a distribution when u is a distribution. As a consequence, for a tempered distribution ϕ ∈ S 0 (R n ), the “paring formula” (4) of K(ϕ) still has an integral representation K(ϕ)(x) = ➦ ❳ |α|+|β|≤k y αD β yϕα,β(y), k(x, y) ➸ = ❳ |α|+|β|≤k ➡ ϕα,β(y), Dβ y y α k(x, y) ➯ = ❳ |α|+|β|≤k ❩ Rn ϕα,β(y)D β y (y α k(x, y)) dy and thus by using the fact k ∈ S (R m × R n ), one can check K(ϕ) ∈ S (R m). Moreover, it can be shown that the operator K : S 0 (R n ) → S (R m) is continuous (with respect to the weak-∗ topology on S 0 (R n ) and the metric topology on S (R m)). In summary, we have If the kernel function k ∈ S (R m ×R n ) is a Schwartz function, then the integral operator K with kernel k can be defined as a continuous linear operator mapping S 0 (R n ) to S (R m). • Case 2: The worst kernels, namely k ∈ S 0 (R m × R n ). To give an exact meaning of the operator K with a distributional kernel k, we first introduce a notation: Given any ϕ ∈ C ∞(R n ) and ψ ∈ C ∞(R m), we define ϕ ψ ∈ C ∞(R n × R m) to be the function given by ϕ ψ(x, y) := ϕ(x)ψ(y). Obviously if ϕ ∈ S (R n ) and ψ ∈ S (R m), then ϕ ψ ∈ S (R n × R m). Now suppose k ∈ S 0 (R m × R n ) is a tempered distribution. We can still define an operator K : S (R n ) → S 0 (R m) via the formula2 K(ϕ)(ψ) := hk, ϕ ψi. and one can show that K is a continuous linear map. In fact, it turns out that any continuous linear map S (R n ) → S 0 (R m) arises in this way: 2 In view of the Schwartz representation theorem, the operator K is really an integral operator.
LECTURE6—10/12/2020SEMICLASSICALQUANTIZATION5Theorem 1.2 (Schwartz kernel theorem). There is a one-to-one correspon-dence between continuous linear operators K :(Rn)-→'(Rm) and theirkernels k Eg"(Rm × R").So usually we will call the kernel function k the Schwartz kernel of theintegral operator K.Remark.As we have mentioned, in this course the main objects are pseudodifferen-tial operator sand Fourier integral operators.They are all integral operators definedviaSchwartzkernels.Inwhatfollows,whenwewritedownsuchanintegralexpres-sion in which there could be some convergence issue, we will explain the expressionin the sense of distribution.Remark. Integral operators with Schwartz functions as Schwartz kernels are very nice(we will prove many other nice properties of such operators later), but they are toorestrictive:for example,the Schwartz kernel of semiclassical differential operatorsarepolynomials inEwhich arenotSchwartzfunctions.Onthe other hand,integraloperators with distributional kernel contains all possible integral operators,butusually they don't have nice properties: for example, in general we can't compositetwo such operators.Back to the Kohn-Nirenberg quantization.Now we can say: not only we canquantize Schwartz functions or polynomials, but also we can quantize tempereddistributions a e '(Rn × IR"), in which case the operator akN is the integraloperator whose Schwartz kernel is the tempered distribution1a(r,E)d = (F-1)→(a(r,)).(2元h）nJgWe will introduce many other classes of functions (called symbol classes) so that theresulting classes of integral operators are large enough AND nice enough.2.OTHER SEMICLASSICAL QUANTIZATIONS The anti-Kohn-Nirenberg quantization.Back to semiclassical quantization.By our construction, the Kohn-Nirenbergquantization quantize the function a(r,s) = riSi to the operator akN = QiP.However, in Lecture 1 we have already mentioned that sinceTS = 1 = (S +1)/2insteadofquantizeriSitoQiPonecouldalsoquantizeittoPQior even(QP+PQi)/2. It turns out that there do exist very similar theories, again using thesemiclassical Fouriertransform,which quantizeriSitoPiQto (QP+PQ1)/2.To see how to get the operator PQi out of the function isi, let's do the samecomputation as before:PQ10 =hDr1(1) = Fr-IFh(Dr(19) = F-'(S1Fh(r10)) = Fr-1(Fh(Si10))
LECTURE 6 — 10/12/2020 SEMICLASSICAL QUANTIZATION 5 Theorem 1.2 (Schwartz kernel theorem). There is a one-to-one correspondence between continuous linear operators K : S (R n ) → S 0 (R m) and their kernels k ∈ S 0 (R m × R n ). So usually we will call the kernel function k the Schwartz kernel of the integral operator K. Remark. As we have mentioned, in this course the main objects are pseudodifferential operator sand Fourier integral operators. They are all integral operators defined via Schwartz kernels. In what follows, when we write down such an integral expression in which there could be some convergence issue, we will explain the expression in the sense of distribution. Remark. Integral operators with Schwartz functions as Schwartz kernels are very nice (we will prove many other nice properties of such operators later), but they are too restrictive: for example, the Schwartz kernel of semiclassical differential operators are polynomials in ξ which are not Schwartz functions. On the other hand, integral operators with distributional kernel contains all possible integral operators, but usually they don’t have nice properties: for example, in general we can’t composite two such operators. Back to the Kohn-Nirenberg quantization. Now we can say: not only we can quantize Schwartz functions or polynomials, but also we can quantize tempered distributions a ∈ S 0 (R n × R n ), in which case the operator ˆa KN is the integral operator whose Schwartz kernel is the tempered distribution 1 (2π~) n ❩ Rn e i (x−y)·ξ ~ a(x, ξ)dξ = (F −1 ~ )ξ→x−y(a(x, ξ)). We will introduce many other classes of functions (called symbol classes) so that the resulting classes of integral operators are large enough AND nice enough. 2. Other semiclassical quantizations ¶ The anti-Kohn-Nirenberg quantization. Back to semiclassical quantization. By our construction, the Kohn-Nirenberg quantization quantize the function a(x, ξ) = x1ξ1 to the operator ˆa KN = Q1P1. However, in Lecture 1 we have already mentioned that since x1ξ1 = ξ1x1 = (x1ξ1 + ξ1x1)/2, instead of quantize x1ξ1 to Q1P1 one could also quantize it to P1Q1 or even (Q1P1 + P1Q1)/2. It turns out that there do exist very similar theories, again using the semiclassical Fourier transform, which quantize x1ξ1 to P1Q1 to (Q1P1 + P1Q1)/2. To see how to get the operator P1Q1 out of the function x1ξ1, let’s do the same computation as before: P1Q1ϕ = ~Dx1 (x1ϕ) = F −1 ~ F~(~Dx1 (x1ϕ)) = F −1 ~ (ξ1F~(x1ϕ)) = F −1 ~ (F~(ξ1x1ϕ))