LECTURE7—10/14/2020WEYL QUANTIZATION:EXAMPLES1.WEYLQUANTIZATIONOFPOLYNOMIAL-TYPEFUNCTIONSToday wefocus onthe Weyl quantization /()asaW(0)(r) =2We shall compute the operator aw for some simple classes of functions. A formulathat we will use several times is the Fourier inversion formulaLeif(y)dydsf(r) = (2h)n Je"ormoreprecisely,itsvariation1eip f(r, y)dyds,f(r,r) =(2元h)nJRwhich can be obtained from the following identity by setting u = r:.Lei--ps f(u, y)dyde.f(u,a) =(2元h）nJRnI Weyl quantization of a(r).We start with the case a = a(r)/ ea()0()dyde =a(a)0(a),aw()(a) =(2元h)nJRm2So, as one can expect (which holds for all t-quantizations)=“multiplication by a(r)".a(r)T Weyl quantization of a($)Next let's consider the casea = a(s)()()()= (Fr-)s→ [a(S)(Fnp)(S) (r)So we get (again the same formula holds (trivially) for all t-quantizations):a()" = Frl oa(E)o Fh
LECTURE 7 — 10/14/2020 WEYL QUANTIZATION: EXAMPLES 1. Weyl quantization of polynomial-type functions Today we focus on the Weyl quantization a❜ W (ϕ)(x) = 1 (2π~) n ❩ Rn ❩ Rn e i (x−y)·ξ ~ a( x + y 2 , ξ)ϕ(y)dydξ. We shall compute the operator a❜W for some simple classes of functions. A formula that we will use several times is the Fourier inversion formula, f(x) = 1 (2π~) n ❩ Rn ❩ Rn e i (x−y)·ξ ~ f(y)dydξ, or more precisely, its variation f(x, x) = 1 (2π~) n ❩ Rn ❩ Rn e i (x−y)·ξ ~ f(x, y)dydξ, which can be obtained from the following identity by setting u = x:. f(u, x) = 1 (2π~) n ❩ Rn ❩ Rn e i (x−y)·ξ ~ f(u, y)dydξ. ¶ Weyl quantization of a(x). We start with the case a = a(x) : a❜ W (ϕ)(x) = 1 (2π~) n ❩ Rn ❩ Rn e i (x−y)·ξ ~ a( x + y 2 )ϕ(y)dydξ = a(x)ϕ(x). So, as one can expect (which holds for all t-quantizations) aÕ(x) W = “multiplication by a(x)”. ¶ Weyl quantization of a(ξ). Next let’s consider the case a = a(ξ) : a❜ W (ϕ)(x) = 1 (2π~) n ❩ Rn ❩ Rn e i (x−y)·ξ ~ a(ξ)ϕ(y)dydξ = (F −1 ~ )ξ→x [a(ξ)(F~ϕ)(ξ)] (x). So we get (again the same formula holds (trivially) for all t-quantizations): aÔ(ξ) W = F −1 ~ ◦ a(ξ) ◦ F~. 1
2LECTURE7—10/14/2020 WEYLQUANTIZATION:EXAMPLESSuch operators are known as Fourier multipliers. Note that in particular we getaW = Pa, and thus (lsP/2 + V(r)" =-h2/2+ V.As an application, we calculate the Weyl quantization of the quadratic expo-nential a(s) = esTQs , where Q is a non-singular symmetric n x n matrix. DenotePQ(S) = $TQE.Proposition 1.1. For a(s) = e"Qs we haveawo(z) = [det Q-1/2.-eisgnoLe-hyTQ-yp(r+y)dy(1)(2元h)n/2Proof. In Lecture 4 we showedF(eQr) = (2)n/2eigsgn(Q)e-1TQ-'gIdetQ|?It follows1n etesTosds = Fr (esro(0)(a) - detol1e-eifsgnQe-PQ-1(a)(2元h)n/2(2元h)nJThuset(r-)esTQsp(y)dydeawp(r)=(2元h)nIdet Q|-1/2 L Idet Q-1/2-eisnQ [e-PQ-1()p(r +y)dy.(2元h)n/2口Theformula (1)will beused next timeto computethesymbol ofthecompositionof twoWeyloperators.Weyl quantization of polynomials in both and .By linearity, to compute the Weyl quantization of a polynomial in both r andS, it is enough to compute the Weyl quantization of monomialsa(r,s) = ragL etp("+)aPp(y)dydea(0)(z)= (2元h)n J".inti-KNTTa-EB(0)(r)21a7r(hDa)(ra-(r)2lal
2 LECTURE 7 — 10/14/2020 WEYL QUANTIZATION: EXAMPLES Such operators are known as Fourier multipliers. Note that in particular we get ξ❝α W = P α , and thus (|ξ|Û2/2 + V (x)) W = −~ 2∆/2 + V . As an application, we calculate the Weyl quantization of the quadratic exponential a(ξ) = e i 2~ ξ T Qξ , where Q is a non-singular symmetric n × n matrix. Denote pQ(ξ) = ξ TQξ. Proposition 1.1. For a(ξ) = e i 2~ ξ T Qξ we have (1) a❜ 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. Proof. In Lecture 4 we showed F(e i 2 x T Qx) = (2π) n/2 e i π 4 sgn(Q) | det Q| 1 2 e − i 2 ξ T Q−1 ξ . It follows 1 (2π~) n ❩ Rn e i ~ x·ξ e i 2~ ξ T Qξdξ = F −1 ~ (e i 2~ pQ(ξ) )(x) = | det Q| −1/2 (2π~) n/2 e i π 4 sgnQe − i 2~ pQ−1 (x) . Thus a❜ W ϕ(x) = 1 (2π~) n ❩ Rn ❩ Rn e i ~ (x−y)·ξ e i 2~ ξ T Qξϕ(y)dydξ = | det Q| −1/2 (2π~) n/2 e i π 4 sgnQ ❩ Rn e − i 2~ pQ−1 (x−y)ϕ(y)dy = | det Q| −1/2 (2π~) n/2 e i π 4 sgnQ ❩ Rn e − i 2~ pQ−1 (y)ϕ(x + y)dy. The formula (1) will be used next time to compute the symbol of the composition of two Weyl operators. ¶ Weyl quantization of polynomials in both x and ξ. By linearity, to compute the Weyl quantization of a polynomial in both x and ξ, it is enough to compute the Weyl quantization of monomials a(x, ξ) = x α ξ β : a❜ W (ϕ)(x) = 1 (2π~) n ❩ Rn ❩ Rn e i (x−y)·ξ ~ ( x + y 2 ) α ξ βϕ(y)dydξ = ❳ γ≤α 1 2 |α| ✥ α γ ✦ x γxØα−γξ β anti−KN (ϕ)(x) = ❳ γ≤α 1 2 |α| ✥ α γ ✦ x γ (~Dx) β (x α−γϕ(x)).
3LECTURE7—10/14/2020WEYLQUANTIZATION:EXAMPLESIn other words, we get the following McCoy's formula:TERW=PBQ-S021alwhere ≤a means ≤aj for all j, andQ!(a- Weyl quantization of polynomials in .Next let's compute the Weyl quantization of a(r,E) = Zja/≤x aa(r)EUsingthe fact(-hDy)cei-ps= sei-psand the Fourier inversion formula we get+y(eips a()sp(y)dydeaW(p)(r) =(2元h)nJ2lal<kT+y((-hDy)eip)p(y)dydelalk (2元h)n /Rn P(hD,) (aa(“)() dyde0lal/≤x (2h)n /Rn1= Z(D,) [aa(“)0()]2lol<k= 2-hi() [(hD)aa(r)] (D)α-(r),[al<≤Sowe getaw= 2-mi(a) [(hD)aa(α)] (D)a-la<≤aAs a consequenceCorollary 1.2. If a(r,E) = jal<h aa(r)sa is a polynomial of degree k in E, thenaw is a semiclassical differential operator of order k of the formaW = Z aa(r)(hD)°+ "terms of order <k -1".[α]=kNote that the same result holds for the Kohn-Nirenberg quantization and theanti-Kohn-Nirenberg quantization, and in fact for all semiclassical t-quantizations
LECTURE 7 — 10/14/2020 WEYL QUANTIZATION: EXAMPLES 3 In other words, we get the following McCoy’s formula: xÕαξ β W = ❳ γ≤α 1 2 |α| ✥ α γ ✦ Q γP βQ α−γ , where γ ≤ α means γj ≤ αj for all j, and ✥ α γ ✦ := α! γ!(α − γ)! = ✥ α1 γ1 ✦ · · · ✥ αn γn ✦ . ¶ Weyl quantization of polynomials in ξ. Next let’s compute the Weyl quantization of a(x, ξ) = P |α|≤k aα(x)ξ α . Using the fact (−~Dy) α e i (x−y)·ξ ~ = ξ α e i (x−y)·ξ ~ and the Fourier inversion formula we get a❜ W (ϕ)(x) = 1 (2π~) n ❩ Rn ❩ Rn e i (x−y)·ξ ~ ❳ |α|≤k aα( x + y 2 )ξ αϕ(y)dydξ = ❳ |α|≤k 1 (2π~) n ❩ Rn ❩ Rn ⑩ (−~Dy) α e i (x−y)·ξ ~ ❿ aα( x + y 2 )ϕ(y)dydξ = ❳ |α|≤k 1 (2π~) n ❩ Rn ❩ Rn e i (x−y)·ξ ~ (~Dy) α ⑩ aα( x + y 2 )ϕ(y) ❿ dydξ = ❳ |α|≤k (~Dy) α ➉ aα( x + y 2 )ϕ(y) ➌ y=x = ❳ |α|≤k ❳ γ≤α 2 −|γ| ✥ α γ ✦ [(~D) γ aα(x)] · (~D) α−γϕ(x). So we get a❜ W = ❳ |α|≤k ❳ γ≤α 2 −|γ| ✥ α γ ✦ [(~D) γ aα(x)] · (~D) α−γ . As a consequence Corollary 1.2. If a(x, ξ) = P |α|≤k aα(x)ξ α is a polynomial of degree k in ξ, then a❜W is a semiclassical differential operator of order k of the form a❜ W = ❳ |α|=k aα(x)(~D) α + “terms of order ≤ k − 1”. Note that the same result holds for the Kohn-Nirenberg quantization and the anti-Kohn-Nirenberg quantization, and in fact for all semiclassical t-quantizations
4LECTURE7—10/14/2020WEYLQUANTIZATION:EXAMPLES2.SYMPLECTICINVARIANCEAND APPLICATIONSSymplectic invariance of Weyl quantization.According to the computations above.it seems that Wevl guantization ismuchmore complicated than the Kohn-Nirenberg or the anti-Kohn-Nirenberg quantiza-tions. A natural question is: what is the advantage of the Weyl quantization? Wehave seen the first big advantage: Weyl quantization will quantize real-valued func-tions to formally self-adjoint operators. Here we explain the second big advantage:the (unitary) invariance under linear symplectomorphisms '(this conception wil beexplained later).Theorem 2.1 (Symplectic invariance of Weyl quantization). Given any“linearsymplectomorphism" : T*Rn → T*R", there is a metaplectic operator U (whichis an isomorphism on(R")and on(Rn),and is unitary on L?(Rn))such that(2)ao"=U-loaw。Ua.Remark.Moreover,it can be shown that such“symplectic invariance"characterizethe Weyl quantization:If there is a“quantization process"Q:'(Rn × Rn)-→C((Rn), (Rn)) which is sequentially continuous, quantizes any bounded func-tion a()to the operator“multiplication by a(r)"and satisfies the symplectic in-variance property above, then it is the Weyl quantization!Here are three special cases of this theorem for which we can easily check (2) bydirect computations:(A) The linear symplectomorphism is the map: Rn ×R"-→R"× R", (C,s)-(S,-r)which“intertwines" r and with a twisting.In this case U=Fh-(So FncanberegardedasthequantizationofJ!)(B) The linear symplectomorphism is the map: R" × R"→Rn ×R", (r,s)-(c,S+Cr),where C is a symmetric n × n matrix. In this case U is the map “multipli-cation by thefunction eiaTCr/2h"(C)The linear symplectomorphism is themapΦ: R" ×Rn →R" × Rn, (c,S) → (Ac, (AT)-1),where A is an invertible matrix. In this case Ua is the map U is given by(Ue)(r) = (Ar)1From the classical-quantum correspondence point of view, a nice quantization should preservesymplecticproperties,twosymplecticallyequivalentclassicalobjectsshouldcorrespondstounitar-ilyequivalent quantumobjects.InLectureI wehavementioned theEgorov theorem,which canbeexplained astheunitaryinvarianceundergeneral symplectomorphisms (which onlyhold inthesemiclassical limit).Here,for linear symplectomorphisms, the invariance is an exact relation
4 LECTURE 7 — 10/14/2020 WEYL QUANTIZATION: EXAMPLES 2. Symplectic invariance and applications ¶ Symplectic invariance of Weyl quantization. According to the computations above, it seems that Weyl quantization is much more complicated than the Kohn-Nirenberg or the anti-Kohn-Nirenberg quantizations. A natural question is: what is the advantage of the Weyl quantization? We have seen the first big advantage: Weyl quantization will quantize real-valued functions to formally self-adjoint operators. Here we explain the second big advantage: the (unitary) invariance under linear symplectomorphisms 1 (this conception will be explained later). Theorem 2.1 (Symplectic invariance of Weyl quantization). Given any “linear symplectomorphism” Φ : T ∗R n → T ∗Rn , there is a metaplectic operator UΦ (which is an isomorphism on S 0 (R n ) and on S (R n ), and is unitary on L 2 (R n )) such that (2) aÖ◦ Φ W = U −1 Φ ◦ a❜ W ◦ UΦ. Remark. Moreover, it can be shown that such “symplectic invariance” characterize the Weyl quantization: If there is a “quantization process” Q : S 0 (R n × R n ) → L(S (R n ), S 0 (R n )) which is sequentially continuous, quantizes any bounded function a(x) to the operator “multiplication by a(x)” and satisfies the symplectic invariance property above, then it is the Weyl quantization! Here are three special cases of this theorem for which we can easily check (2) by direct computations: (A) The linear symplectomorphism is the map Φ : R n × R n → R n × R n , (x, ξ) 7→ (ξ, −x) which “intertwines” x and ξ with a twisting. In this case UΦ = F~. (So F~ can be regarded as the quantization of J!) (B) The linear symplectomorphism is the map Φ : R n × R n → R n × R n , (x, ξ) 7→ (x, ξ + Cx), where C is a symmetric n × n matrix. In this case UΦ is the map “multiplication by the function e ixT Cx/2~ ”. (C) The linear symplectomorphism is the map Φ : R n × R n → R n × R n , (x, ξ) 7→ (Ax,(A T ) −1 ξ), where A is an invertible matrix. In this case UΦ is the map UΦ is given by (UΦϕ)(x) = ϕ(Ax). 1From the classical-quantum correspondence point of view, a nice quantization should preserve symplectic properties, two symplectically equivalent classical objects should corresponds to unitarily equivalent quantum objects. In Lecture 1 we have mentioned the Egorov theorem, which can be explained as the unitary invariance under general symplectomorphisms (which only hold in the semiclassical limit). Here, for linear symplectomorphisms, the invariance is an exact relation
LECTURE7—10/14/2020WEYLQUANTIZATION:EXAMPLES5Remark.Infact,one can prove that anylinear symplectomorphism canbewrittenas a composition of the three classes of linear symplectomorphisms above.As aresult, the general theorem is proven as long as we can check the three cases (A)(B) and (C).Wewill not prove thefull theorem here2.Instead, in what follows wewill prove case (A) and a special case of (B), and give two applications. We willleave the proof of the general cases of (B) and (C) as an exercise. Case (A): Conjugation by Fourier transform.Weprove case (A)by direct computation:Theorem 2.2 (Conjugation by Fourier transform). Let b(r,) = a(E,-r), thenF-loawoFh=6w(3)Proof.We compute[(F-l)m→2(aW)y→n(Fn)z→yl (α)[e-t*p(2)dz]=(F-")n→r(aW)y→n[[a(()e)a(,)p(2)dy(2元h)n(2元h)n1+()a()()(2元h)"（2元h）2+a()d(d(2元h)n(2元h)n1[ek(r+y+2)net(-25-29)-Sa(C,E)dcdedn(y)dy(2元h）m2元h）(2h) / ete(-y)Sa(, ))drdn ) dc(y)dy(2元h）nJB2Using theFourier inversionformula/ etr f(r)drde,f(0) = [FFr"f](0) = 7(2元h）nJ起the expression in (-..) above can be simplified toet(a-)a(c,-) = et(r-)<b(“+,.1口and the conclusion follows.2For a proof, c.f. Folland, Harmonic analysis in phase space, Chapter 4
LECTURE 7 — 10/14/2020 WEYL QUANTIZATION: EXAMPLES 5 Remark. In fact, one can prove that any linear symplectomorphism can be written as a composition of the three classes of linear symplectomorphisms above. As a result, the general theorem is proven as long as we can check the three cases (A), (B) and (C). We will not prove the full theorem here 2 . Instead, in what follows we will prove case (A) and a special case of (B), and give two applications. We will leave the proof of the general cases of (B) and (C) as an exercise. ¶ Case (A): Conjugation by Fourier transform. We prove case (A) by direct computation: Theorem 2.2 (Conjugation by Fourier transform). Let b(x, ξ) = a(ξ, −x), then (3) F −1 ~ ◦ a❜ W ◦ F~ = ❜b W . Proof. We compute ➈ (F −1 ~ )η→x(a❜ W )y→η(F~)z→yϕ ➋ (x) =(F −1 ~ )η→x(a❜ W )y→η ➉❩ Rn e − i ~ z·yϕ(z)dz➌ =(F −1 ~ )η→x ➊ 1 (2π~) n ❩ Rn ❩ Rn ❩ Rn e i ~ (η−y)·ξ a( η + y 2 , ξ)e − i ~ z·yϕ(z)dzdydξ➍ = 1 (2π~) n 1 (2π~) n ❩ Rn ❩ Rn ❩ Rn ❩ Rn e i ~ x·η e i ~ (η−y)·ξ a( η + y 2 , ξ)e − i ~ z·yϕ(z)dzdydξdη = 1 (2π~) n ❩ Rn ➊ 1 (2π~) n ❩ Rn ❩ Rn ❩ Rn e i ~ [x·η+(η−z)·ξ−y·z] a( η + z 2 , ξ)dzdξdη➍ ϕ(y)dy = 1 (2π~) n ❩ Rn ➊ 2 n (2π~) n ❩ Rn ❩ Rn ❩ Rn e i ~ [x·η+2(η−ζ)·ξ−y·(2ζ−η)]a(ζ, ξ)dζdξdη➍ ϕ(y)dy = 1 (2π~) n ❩ Rn ➊ 2 n (2π~) n ❩ Rn ❩ Rn ❩ Rn e i ~ (x+y+2ξ)·η e i ~ (−2ξ−2y)·ζ a(ζ, ξ)dζdξdη➍ ϕ(y)dy = 1 (2π~) n ❩ Rn ➊❩ Rn ❶ 1 (2π~) n ❩ Rn ❩ Rn e i ~ τ·η e i ~ (x−y−τ)·ζ a(ζ, τ − x − y 2 )dτ dη➀ dζ➍ ϕ(y)dy Using the Fourier inversion formula f(0) = [F~F −1 ~ f](0) = 1 (2π~) n ❩ Rn ❩ Rn e i ~ x·ξ f(x)dxdξ, the expression in (· · ·) above can be simplified to e i ~ (x−y)·ζ a(ζ, −x − y 2 ) = e i ~ (x−y)·ζ b( x + y 2 , ζ) and the conclusion follows. 2For a proof, c.f. Folland, Harmonic analysis in phase space, Chapter 4.
6LECTURE7—10/14/2020 WEYLQUANTIZATION:EXAMPLES Application: A second formula for the Weyl quantization of polynomi-als in .As an application, to compute the Weyl quantization of a(r,s) = Zjal<k aasa,we can intertwine r and first, so that instead of handling the function a("), weonly need to handle the polynomial (“) by using the binomial theorem:Proposition 2.3. The Weyl quantization of a(r, E) =jal<x aas isaw = Z 2-lal((4)(D)0 a(r) (hD)a-Jal<k<aProof.Wewill apply the previous theorem to“intertwine and s".For this purposewe let b(r,) = aa(-)r, then a(r,) = b(s, -r) and thus we haveaw=F-lo6WoFh.Note that by definition and the binomial theorem,1L et(e-)s a(-5)(“) p()dyds(6W 0)(a) = (2元h)n JR Jelal<kr= 2-la (α)L et(r-)saa(-E)ya-p(y)dyde.(2元h）nJR10kaSo toprove (4), it remains to check/ et(a-w)-saa(-)ga-p(y)dyds.[Fn 0 (hD) 0a(r) (hD)a-1 0 Fr'0] (a) = (2元h)nJRnThis follows from direct computations:The left hand side of the above expressionshould be interpreted as(Fh)→r 0 (hD)2 0aa(s) 0 (hD)-0 (Frl)y-→P(y),which, by using the property (Fh)s→ o (hD)2 = r(Fn)→, equalsr (Fh)s- 0 aa(s) 0 (Fr1)y-e(ya-p(y),which equalst?et(u-r)Eaa(E)ga-p(y)dyd(2元h)nJrL et(r-)-Eaa(-E)ya-p(g)dyds.(2元h)nJRn口This completes the proof.Remark. You may have noticed that if we apply (4) to monomial ragp, we will get=≥2-I(PrQ°pB-rae≤B
6 LECTURE 7 — 10/14/2020 WEYL QUANTIZATION: EXAMPLES ¶ Application: A second formula for the Weyl quantization of polynomials in ξ. As an application, to compute the Weyl quantization of a(x, ξ) = P |α|≤k aαξ α , we can intertwine x and ξ first, so that instead of handling the function a( x+y 2 ), we only need to handle the polynomial ( x+y 2 ) α by using the binomial theorem: Proposition 2.3. The Weyl quantization of a(x, ξ) = P |α|≤k aαξ α is (4) a❜ W = ❳ |α|≤k ❳ γ≤α 2 −|α| ✥ α γ ✦ (~D) γ ◦ aα(x) ◦ (~D) α−γ Proof. We will apply the previous theorem to “intertwine x and ξ”. For this purpose we let b(x, ξ) = P aα(−ξ)x α , then a(x, ξ) = b(ξ, −x) and thus we have a❜ W = F −1 ~ ◦ ❜b W ◦ F~. Note that by definition and the binomial theorem, ( ❜b W ϕ)(x) = 1 (2π~) n ❩ Rn ❩ Rn e i ~ (x−y)·ξ ❳ |α|≤k aα(−ξ) ⑩ x + y 2 ❿α ϕ(y)dydξ = ❳ |α|≤k ❳ γ≤α 2 −|α| ✥ α γ ✦ x γ (2π~) n ❩ Rn ❩ Rn e i ~ (x−y)·ξ aα(−ξ)y α−γϕ(y)dydξ. So to prove (4), it remains to check ➈ F~ ◦ (~D) γ ◦ aα(x) ◦ (~D) α−γ ◦ F −1 ~ ϕ ➋ (x) = x γ (2π~) n ❩ Rn ❩ Rn e i ~ (x−y)·ξ aα(−ξ)y α−γϕ(y)dydξ. This follows from direct computations: The left hand side of the above expression should be interpreted as (F~)ξ→x ◦ (~D) γ ξ ◦ aα(ξ) ◦ (~D) α−γ ξ ◦ (F −1 ~ )y→ξϕ(y), which, by using the property (F~)ξ→x ◦ (~D) γ ξ = x γ (F~)ξ→x, equals x γ ◦ (F~)ξ→x ◦ aα(ξ) ◦ (F −1 ~ )y→ξ(y α−γϕ(y)), which equals x γ (2π~) n ❩ Rn ❩ Rn e i ~ (y−x)·ξ aα(ξ)y α−γϕ(y)dydξ = x γ (2π~) n ❩ Rn ❩ Rn e i ~ (x−y)·ξ aα(−ξ)y α−γϕ(y)dydξ. This completes the proof. Remark. You may have noticed that if we apply (4) to monomial x α ξ β , we will get xÕαξ β W = ❳ γ≤β 2 −|β| ✥ β γ ✦ P γQ αP β−γ
LECTURE7—10/14/2020WEYL QUANTIZATION:EXAMPLES7which is different from the McCoy's formula on page 3.In particular, for example,we will get two different formula for re2(QP?+2PQP+ P2Q)(QP2 + P2Q)E2rE2andThere is no mistake: inview of the canonical commutativerelation[Q, P] = i- Id,we haveQP2+P2Q=PQP+P+PQP-P=2PQPand thus the two formulae for r?coincide.Wecanalsowritedownanexpressionforre2which looks evenmoresymmetric:CE2W(QP2+ PQP+P2Q),In what follows we will prove that such symmetricformula holds for the Weyl quan-tization of any monomial. Case (B) with C = Id: Adding r to &.To prove the special case of (B) where C= Id, we need to check:Theorem 2.4. We havee)Proof. Let's compute1We-eip(y)dyds(2元h)）nJ起1-y)-(E-sap(y)dyde(2元h)n1a+yy2p(y)dyde(2元h)n=(+$)a6口 Application: Symmetry in Weyl quantization.Since weknowa=Pa, Theorem 2.4 allows us to compute the Weyl quantiza-tion of (r +)directly, and the result is neat:Corollary 2.5. We have(r+E)a"= (Q + P)°
LECTURE 7 — 10/14/2020 WEYL QUANTIZATION: EXAMPLES 7 which is different from the McCoy’s formula on page 3. In particular, for example, we will get two different formula for xξÔ2 W : xξÔ2 W = 1 2 (QP2 + P 2Q) and xξÔ2 W = 1 4 (QP2 + 2P QP + P 2Q). There is no mistake: in view of the canonical commutative relation [Q, P] = i~ · Id, we have QP2 + P 2Q = P QP + i~P + P QP − i~P = 2P QP, and thus the two formulae for xξÔ2 W coincide. We can also write down an expression for xξÔ2 W which looks even more symmetric: xξÔ2 W = 1 3 (QP2 + P QP + P 2Q). In what follows we will prove that such symmetric formula holds for the Weyl quantization of any monomial. ¶ Case (B) with C = Id: Adding x to ξ. To prove the special case of (B) where C = Id, we need to check: Theorem 2.4. We have e −i |x| 2 2~ ξ❝α W e i |x| 2 2~ = (Ùx + ξ) α W . Proof. Let’s compute e −i |x| 2 2~ ξ❝α W e i |x| 2 2~ ϕ = 1 (2π~) n ❩ Rn ❩ Rn e −i |x| 2 2~ e i (x−y)·ξ ~ ξ α e i |y| 2 2~ ϕ(y)dydξ = 1 (2π~) n ❩ Rn ❩ Rn e i (x−y)·(ξ− x+y 2 ) ~ ξ αϕ(y)dydξ = 1 (2π~) n ❩ Rn ❩ Rn e i (x−y)·ξ ~ ⑩ ξ + x + y 2 ❿α ϕ(y)dydξ = (Ùx + ξ) α W ϕ. ¶ Application: Symmetry in Weyl quantization. Since we know ξ❝α = P α , Theorem 2.4 allows us to compute the Weyl quantization of (x + ξ) α directly, and the result is neat: Corollary 2.5. We have (Ùx + ξ) α W = (Q + P) α .
8LECTURE7—10/14/2020WEYLQUANTIZATION:EXAMPLESProof. We have seen that faw = Pa. So we only need to checke-=(Q+P),which follows easily from the facte- pe=(Q+P)and thefacteee口Remark.Note that other t-quantizations like the Kohn-Nirenberg quantization doesnot satisfy this property. For example,=Q?+2QP+P+(Q+P)2(c+)2More generally, one by proving case (B) for general C and then taking C to bediagonal matrices, one can easily prove: for any a,b e Rn,(ar+bE)a= (aQ+bP),where we used the abbreviation(ar+bE)=(aiti+bisi)a1...(anan+bnSn)anand(aQ+bP)°= (aiQ1+biP)a1 0...0 (anQn +bnPn)anThe proof will beleft as a simple exercise.This has a further interesting application,namely the Weyl quantization is the most symmetric way to quantize monomials:Corollary 2.6.Q!B!7TaEYiY2... Yjal+IBl,a +β al+1where Yi, Y2, .. Yal+8l range over all tuples which contains a1 copies of Qi, Q2copies of Q2,.., and Bn copies of Pn.Proof.Comparing the coeficients of aobp of both sides of(a + bE)a+β= (aQ +bP)a+β,we getQ!B!MraEYY2. .. Yal+IBl(@+ B)! Yi...where Yi, Y2, .. Yai+βi range over all tuples which contains Q1 copies of Qi, βi..Ya2+B,range over all tuples which contains Q2 copies ofcopiesof Pi,Ya1+B+1,Q2, β2 copies of P2 and so on. The conclusion follows from the fact “"Pi, Q; commutes口with Pj,Q; for j "and an elementary combinatorics argument
8 LECTURE 7 — 10/14/2020 WEYL QUANTIZATION: EXAMPLES Proof. We have seen that ξ❝α W = P α . So we only need to check e −i |x| 2 2~ P α e i |x| 2 2~ ϕ = (Q + P) αϕ, which follows easily from the fact e −i |x| 2 2~ P ei |x| 2 2~ ϕ = (Q + P)ϕ and the fact e −i |x| 2 2~ PiPje i |x| 2 2~ ϕ = e −i |x| 2 2~ Pie i |x| 2 2~ e −i |x| 2 2~ Pje i |x| 2 2~ ϕ. Remark. Note that other t-quantizations like the Kohn-Nirenberg quantization does not satisfy this property. For example, (Ùx + ξ) 2 KN = Q 2 + 2QP + P 2 6= (Q + P) 2 . More generally, one by proving case (B) for general C and then taking C to be diagonal matrices, one can easily prove: for any a, b ∈ R n Û , (ax + bξ) α W = (aQ + bP) α , where we used the abbreviation (ax + bξ) α = (a1x1 + b1ξ1) α1 · · ·(anxn + bnξn) αn and (aQ + bP) α = (a1Q1 + b1P1) α1 ◦ · · · ◦ (anQn + bnPn) αn . The proof will be left as a simple exercise. This has a further interesting application, namely the Weyl quantization is the most symmetric way to quantize monomials: Corollary 2.6. xÕαξ β W = α!β! |α + β|! ❳ Y1,··· ,Y|α|+|β| Y1Y2 · · · Y|α|+|β| , where Y1, Y2, · · · Y|α|+|β| range over all tuples which contains α1 copies of Q1, α2 copies of Q2, · · · , and βn copies of Pn. Proof. Comparing the coefficients of a α b β Û of both sides of (ax + bξ) α+β W = (aQ + bP) α+β , we get xÕαξ β W = α!β! (α + β)! ❳ Y1,··· ,Y|α|+|β| Y1Y2 · · · Y|α|+|β| , where Y1, Y2, · · · Yα1+β1 range over all tuples which contains α1 copies of Q1, β1 copies of P1, Yα1+β1+1, · · · Yα2+β2 range over all tuples which contains α2 copies of Q2, β2 copies of P2 and so on. The conclusion follows from the fact “Pi , Qi commutes with Pj , Qj for j 6= i” and an elementary combinatorics argument.