《半经典微局部分析 Semiclassical Microlocal Analysis》课程教学资源（英文讲义）Lec08 WEYL QUANTIZATION VIA LINEAR EXPONENTIALS

LECTURE8—10/19/2020WEYL QUANTIZATION VIA LINEAREXPONENTIALS1.QUANTIZING LINEAR EXPONENTIALS↑ From polynomials to the exponential functionLast time, by using symplectic invariance we proved that the Weyl quantizationhas many nice properties on polynomials, e.g.(+)=(Q+P)%InPSet2you will beasked toprovesimilarexpressions like(ar +bs)a=(aQ+bP)and(a-r+b.s)n=(a·Q+b.P)nNote that in the first expression we used abbreviations(ar+bE)=(aia1+bisi)a1..-(ann+bnsn)anwhile in the second expression,(a·r+b.s)n =(airi +..+anan+bisi +..+bnSn)nAs a consequence of the second fact (together with the Taylor expansion of theexponential function),formally we would expectto have=ea-Q+bPea-r+b"Of course we need to justify the meaning of ea-Q+b-P.We can't just formally defineeaQ+b.P=l(a.Q+b.P)kKk≥0because the operators Q and P are unbounded and we will encounter convergenceproblem. However, in what follows we will show that for a,b e Rn, we may definetheoperatoreit(a-Q+b.P)/hwhich is a well-defined (unitary) operator. [This is a special case of Stone's theoremthat we will discuss later.j1

LECTURE 8 — 10/19/2020 WEYL QUANTIZATION VIA LINEAR EXPONENTIALS 1. Quantizing linear exponentials ¶ From polynomials to the exponential function. Last time, by using symplectic invariance we proved that the Weyl quantization has many nice properties on polynomials, e.g. (Ùx + ξ) α W = (Q + P) α . In PSet 2 you will be asked to prove similar expressions like (Ûax + bξ) α W = (aQ + bP) α and (aÛ· x + b · ξ) n W = (a · Q + b · P) n . Note that in the first expression we used abbreviations (ax + bξ) α = (a1x1 + b1ξ1) α1 · · ·(anxn + bnξn) αn while in the second expression, (a · x + b · ξ) n = (a1x1 + · · · + anxn + b1ξ1 + · · · + bnξn) n . As a consequence of the second fact (together with the Taylor expansion of the exponential function), formally we would expect to have e×a·x+b·ξ W = e a·Q+b·P . Of course we need to justify the meaning of e a·Q+b·P . We can’t just formally define e a·Q+b·P = ❳ k≥0 1 k! (a · Q + b · P) k because the operators Q and P are unbounded and we will encounter convergence problem. However, in what follows we will show that for a, b ∈ R n , we may define the operator e it(a·Q+b·P)/~ which is a well-defined (unitary) operator. [This is a special case of Stone’s theorem that we will discuss later.] 1

2LECTURE8-10/19/2020 WEYLQUANTIZATIONVIALINEAR EXPONENTIALSDETOUR:The Baker-Campbell-Hausdorff formula: a special case.To understand the operator et(a.Q+bP)/h, let's start with some general discussionSuppose A is a bounded linear operator defined on a Hilbert space H. Then onecan define the exponential of A to be the operatoreA :=1Akk≥0k!which is also a bounded linear operator on H. It is easy to checkdetA = AetA = e'AA.dtIt follows that for any po E H, the function p(t) := etApo solves the equation[ 显(t) = Ap(t),1 (0) = 0.We say the operator etA is the solution operator to the above equation.Now we assume A and B are bounded linear operators.Then A+ B is againbounded and we have71eA+B=(A +B)*0kHowever, due to the non-commutativity of operator composition, in generaleA+B +eAeBThere is a formula called the Baker-Campbell-Hausdorff formulal which describesthe relation between eA+B and eAeB. Here we prove a special case:Theorem 1.1 (Baker-Campbell-Hausdorff formula, a special case). Suppose A, Barebounded linearoperator onH and[A, [A, B] = 0, [B, [A, B] = 0,theneA+B= e-[A,B]/2eAeB(1)Proof.We will prove: for any t eR.et(A+B) = e-t[A,B]/2etAetB,We calculate the derivative of the right hand side via the Leibnitz rule:de-[4,B/2e+AeB = -t[A, B]e-[A,B/2e'AetB+e-t[4,B/2 Ae'AetB+e-t[4,B/2e'ABetBdtSince [A, B] commutes with A and thus with etA, we have(e'A Be-A) = e'4(AB - BA)e-tA = [A, B)dt1See my Lie group notes for the general Baker-Campbell-Hausdorff formula on Lie groups

2 LECTURE 8 — 10/19/2020 WEYL QUANTIZATION VIA LINEAR EXPONENTIALS ¶ DETOUR: The Baker-Campbell-Hausdorff formula: a special case. To understand the operator e it(a·Q+b·P)/~ , let’s start with some general discussion. Suppose A is a bounded linear operator defined on a Hilbert space H. Then one can define the exponential of A to be the operator e A := ❳ k≥0 1 k! A k which is also a bounded linear operator on H. It is easy to check d dte tA = AetA = e tAA. It follows that for any ϕ0 ∈ H, the function ϕ(t) := e tAϕ0 solves the equation ➝ d dtϕ(t) = Aϕ(t), ϕ(0) = ϕ0. We say the operator e tA is the solution operator to the above equation. Now we assume A and B are bounded linear operators. Then A + B is again bounded and we have e A+B = ❳ k≥0 1 k! (A + B) k . However, due to the non-commutativity of operator composition, in general e A+B 6= e A e B . There is a formula called the Baker-Campbell-Hausdorff formula1 which describes the relation between e A+B and e Ae B. Here we prove a special case: Theorem 1.1 (Baker-Campbell-Hausdorff formula, a special case). Suppose A, B are bounded linear operator on H and [A, [A, B]] = 0, [B, [A, B]] = 0, then (1) e A+B = e −[A,B]/2 e A e B . Proof. We will prove: for any t ∈ R, e t(A+B) = e −t 2 [A,B]/2 e tAe tB . We calculate the derivative of the right hand side via the Leibnitz rule: d dt(e −t 2 [A,B]/2 e tAe tB) = −t[A, B]e −t 2 [A,B]/2 e tAe tB+e −t 2 [A,B]/2AetAe tB+e −t 2 [A,B]/2 e tABetB . Since [A, B] commutes with A and thus with e tA, we have d dt(e tABe−tA) = e tA(AB − BA)e −tA = [A, B] 1See my Lie group notes for the general Baker-Campbell-Hausdorff formula on Lie groups

LECTURE8—10/19/2020WEYLQUANTIZATIONVIA LINEAREXPONENTIALS3and thus by integration,etA Be-tA = B + t[A, B].It fllows (using the fact “[A, B], and thus e-t[A,B/2, commutes with everything")d(e-[4,B/2etAeB) = (t[A, B] + A+ B+t[A, B)e-t[4,B/2etAtBdt= (A + B)e-[A,B)/2etAetB,It fllows that for any Po E H, both p(t) = et(A+B)po and p(t) = e-t[A,B/2etAetBposolve the equationd显(t) = (A + B)o(t)with the same initial condition p(O) = Po. So we conclude et(A+B) = e-t?[A,B)/2etAetB口holds for all t and thus the theorem is proved.T The operator eit(a.Q+bP)/h.Suppose a,b e Rn. Before we study the operator eit(a-Q+b-P)/h, let's first look attwo simpler operators eita-Q/h and eitb.P/n.. We can define eita-Q/h to be the solution operator to the equation最(t,a)= (t, ),p(0, r) = po(r).[For simplicity,wealways startwithPo E(Rn),.]It istrivial tocheckeita-Q/h = multiplication by the function eita-a/h(2)Inotherwords,wehaveWeita-Q/heita-r/h(3). Similarly we define eitb.P/h to be the solution operator to the equation(t,r) =(t,),(0, ) = (Po().It is also trivial to check that the solution is(eitb-P/np)(r) = (α + tb)(4)Moreover,bytheFourier inversion formula1Ceipseip(y)dyde(eitbs/h")(r) =(2元h)nJ1-9)-.Tp(y)dyde= p(r+tb)(2元h)nSo we still haveeitb-P/heitb-E/h(5)

LECTURE 8 — 10/19/2020 WEYL QUANTIZATION VIA LINEAR EXPONENTIALS 3 and thus by integration, e tABe−tA = B + t[A, B]. It follows (using the fact “[A, B], and thus e −t 2 [A,B]/2 , commutes with everything”) d dt(e −t 2 [A,B]/2 e tAe tB) = (−t[A, B] + A + B + t[A, B])e −t 2 [A,B]/2 e tAe tB = (A + B)e −t 2 [A,B]/2 e tAe tB . It follows that for any ϕ0 ∈ H, both ϕ(t) = e t(A+B)ϕ0 and ϕ(t) = e −t 2 [A,B]/2 e tAe tBϕ0 solve the equation d dtϕ(t) = (A + B)ϕ(t) with the same initial condition ϕ(0) = ϕ0. So we conclude e t(A+B) = e −t 2 [A,B]/2 e tAe tB holds for all t and thus the theorem is proved. ¶ The operator e it(a·Q+b·P)/~ . Suppose a, b ∈ R n . Before we study the operator e it(a·Q+b·P)/~ , let’s first look at two simpler operators e ita·Q/~ and e itb·P/~ . • We can define e ita·Q/~ to be the solution operator to the equation ➝ ∂ ∂tϕ(t, x) = ia·Q ~ ϕ(t, x), ϕ(0, x) = ϕ0(x). [For simplicity, we always start with ϕ0 ∈ S (R n ).] It is trivial to check (2) e ita·Q/~ = multiplication by the function e ita·x/~ In other words, we have (3) e×ita·x/~ W = e ita·Q/~ . • Similarly we define e itb·P/~ to be the solution operator to the equation ➝ ∂ ∂tϕ(t, x) = ib·P ~ ϕ(t, x), ϕ(0, x) = ϕ0(x). It is also trivial to check that the solution is (4) (e itb·P/~ϕ)(x) = ϕ(x + tb). Moreover, by the Fourier inversion formula, (e×itb·ξ/~ W ϕ)(x) = 1 (2π~) n ❩ Rn ❩ Rn e i (x−y)·ξ ~ e i tb·ξ ~ ϕ(y)dydξ = 1 (2π~) n ❩ Rn ❩ Rn e i (x+tb−y)·ξ ~ ϕ(y)dydξ = ϕ(x + tb) So we still have (5) e×itb·ξ/~ W = e itb·P/~ .

LECTURE8-10/19/2020WEYLQUANTIZATIONVIALINEAREXPONENTIALS1Now we are ready to study the operator eit(a-Q+b-P)/n, which by definition, is thesolution operator to the equation[(t, ) = (aP)(t,2), (0, r) = Po(α).In general, equations of this type could be hard to solve because the non-commutativityof theoperatorsinvolved.However,fortheposition operatorQand themomentumoperator P, thecanonical commutativerelationgivesush[a.Q,b.P] = axbk[Qk, P] =ab Id,which commutes with any operator.It turns out that the Baker-Campbell-Hausdorffformulaalludedtoabovestill holds,namely,it(α-Q+b-P)=e-["a2,h]/2"ebPTo see this,let'sfirst computee-[,P/2eePp(r) = eabeap(r + tb),NowweproveProposition 1.2. We have(e(aQ+bP)/np)() = e%abeap(r + tb)(6)Proof.This follows from a direct computation:[eab(+tb) =a[e(+)]titta.b[eabeap(r+tb) +[eabeaab.Vp(r+t)方ia.r[eabep(a+tb)] +eab(b.)[ea(r+t)]ha.Q+b.P)[eaberarb.Vp(r+tb)] .-方口As a consequence (and as we can expect), we also haveCorollary 1.3.W= eit(a-Q+b-P)/h(7)eit(a-r+bE)/h

4 LECTURE 8 — 10/19/2020 WEYL QUANTIZATION VIA LINEAR EXPONENTIALS Now we are ready to study the operator e it(a·Q+b·P)/~ , which by definition, is the solution operator to the equation ➝ ∂ ∂tϕ(t, x) = i(a·Q+b·P) ~ ϕ(t, x), ϕ(0, x) = ϕ0(x). In general, equations of this type could be hard to solve because the non-commutativity of the operators involved. However, for the position operator Q and the momentum operator P, the canonical commutative relation gives us [a · Q, b · P] = ❳ k akbk[Qk, Pk] = − ~ i a · b Id, which commutes with any operator. It turns out that the Baker-Campbell-Hausdorff formula alluded to above still holds, namely, e it(a·Q+b·P ) ~ = e −[ ita·Q ~ , itb·P ~ ]/2 e ita·Q ~ e itb·P ~ . To see this, let’s first compute e −[ ita·Q ~ , itb·P ~ ]/2 e ita·Q ~ e itb·P ~ ϕ(x) = e it2 2~ a·b e it ~ a·xϕ(x + tb). Now we prove Proposition 1.2. We have (6) (e it(a·Q+b·P)/~ϕ)(x) = e it2 2~ a·b e it ~ a·xϕ(x + tb). Proof. This follows from a direct computation: ∂ ∂t ➉ e it2 2~ a·b e it ~ a·xϕ(x + tb) ➌ = i ~ a · x ➉ e it2 2~ a·b e it ~ a·xϕ(x + tb) ➌ + it ~ a · b ➉ e it2 2~ a·b e it ~ a·xϕ(x + tb) ➌ + ➉ e it2 2~ a·b e it ~ a·x b · ∇ϕ(x + tb) ➌ = i ~ a · x ➉ e it2 2~ a·b e it ~ a·xϕ(x + tb) ➌ + e it2 2~ a·b (b · ∇) ❤ e it ~ a·xϕ(x + tb) ✐ =( i ~ a · Q + i ~ b · P) ➉ e it2 2~ a·b e it ~ a·x b · ∇ϕ(x + tb) ➌ . As a consequence (and as we can expect), we also have Corollary 1.3. (7) eÛit(a·x+b·ξ)/~ W = e it(a·Q+b·P)/~ .

LECTURE 8—10/19/2020WEYLQUANTIZATION VIA LINEAREXPONENTIALS5Proof.Toseethis,wecalculateet(a-)se(a+bs)p(y)dydeeit(a-r+b-s)/hn"(r) (2元h)n著aret(r-y+tb)ea-yp(y)dyde(2元h)near / e(r-y)sea-(y+t)p(y+tb)dyde(2元h)n=eabeap(r + tb),口 Weyl quantization: Weyl's definition.Now we are ready to give a different way to define the Weyl quantization awwhich is in fact Weyl's original definition!Given a symbol a(r, ) (which could be a tempered distribution), consider thesemiclassical Fourier transform (Fh)(r,s)(y,),e-t(ry+-m)a(r,E)drde.(8)[(Fh)(ar,)(y,n)al(y,n) =The Fourier inversion formula gives12e(+[(F)()(yn)al(,)dya(r,)=(2元h)2nJTo quantize a function f, Weyl wrote?:A quantity f is consequently carried over from classical to quantummechanics in accordance with the rule: replace p and q in Fourier de-velopment (14.8)of f bytheHermitian operators representing themin quantum mechanics.In other words, Weyl quantize the function a(r,)tothe operator e (y-Q+n-P)[(Fh)(s,)(y,n)al(y, n)dydn,aw(9)(2元h)2nwhich, when acting on a Schwartz function ,yields1 [et(yQ+-P)0] (r)[(Fh)(s)-(y,n)a](y, n)dydn.(awp)(r) =(2元h)2nJRWe have to show that Weyl's original definition coincides with the one we gaveearlier:2c.f.H.Weyl,The Theory of Groups and Quantum Mechanics,page275,Dover,1950.Thefirst German edition was published in 1928

LECTURE 8 — 10/19/2020 WEYL QUANTIZATION VIA LINEAR EXPONENTIALS 5 Proof.ÛTo see this, we calculate e it(a·x+b·ξ)/~ W ϕ(x) = 1 (2π~) n ❩ Rn ❩ Rn e i ~ (x−y)·ξ e it ~ (a· x+y 2 +b·ξ)ϕ(y)dydξ = 1 (2π~) n e it 2~ a·x ❩ Rn ❩ Rn e i ~ (x−y+tb)·ξ e it 2~ a·yϕ(y)dydξ = 1 (2π~) n e it 2~ a·x ❩ Rn ❩ Rn e i ~ (x−y)·ξ e it 2~ a·(y+tb)ϕ(y + tb)dydξ = e it2 2~ a·b e it ~ a·xϕ(x + tb). ¶ Weyl quantization: Weyl’s definition. Now we are ready to give a different way to define the Weyl quantization a❜W , which is in fact Weyl’s original definition! Given a symbol a(x, ξ) (which could be a tempered distribution), consider the semiclassical Fourier transform (F~)(x,ξ)→(y,η) , (8) [(F~)(x,ξ)→(y,η)a](y, η) = ❩ R2n e − i ~ (x·y+ξ·η) a(x, ξ)dxdξ. The Fourier inversion formula gives a(x, ξ) = 1 (2π~) 2n ❩ R2n e i ~ (y·x+η·ξ) [(F~)(x,ξ)→(y,η)a](y, η)dydη. To quantize a function f, Weyl wrote2 : A quantity f is consequently carried over from classical to quantum mechanics in accordance with the rule: replace p and q in Fourier de￾velopment (14.8) of f by the Hermitian operators representing them in quantum mechanics. In other words, Weyl quantize the function a(x, ξ) to the operator (9) a❜ W = 1 (2π~) 2n ❩ R2n e i ~ (y·Q+η·P) [(F~)(s,ξ)→(y,η)a](y, η)dydη, which, when acting on a Schwartz function ϕ, yields (a❜ W ϕ)(x) = 1 (2π~) 2n ❩ R2n ❤ e i ~ (y·Q+η·P)ϕ ✐ (x)[(F~)(s,ξ)→(y,η)a](y, η)dydη. We have to show that Weyl’s original definition coincides with the one we gave earlier: 2 c.f. H. Weyl, The Theory of Groups and Quantum Mechanics, page 275, Dover, 1950. The first German edition was published in 1928.

6LECTURE8-10/19/2020 WEYLQUANTIZATIONVIALINEAREXPONENTIALSTheorem 1.4. The two definitions of Weyl quantization coincides: et(Q+P[(Fh)(ss)-(u)al(y, )dydn.aw(2元h）2nJProof of Theorem 1.4.Applying theright hand sideof (9)to ,and using Proposi-tion 1.2, we get1[et(Q+P)] ()[(F)(s.s)(y,m)al(y,n)dydn(2元h）2n/1et+ynp(r+n)e(s-y+s-m)a(s,)dsdedydn(2元h)2n[. ety($+-s)ets:(r-T)p()a(s, E)dsdedydT(n→T=r+n)(2元h)2nJR2m1(S→S=C+T ewsets(a-)p(T)a(“-,)ddydrdS(2元h)2nJR2n221+ets-(r-T)(T)a(,$)drde,(2元h)n2口where in the last step we used the Fourier inversion formula.2. DETOUR: SOME FUNCTIONAL ANALYSISIn this section we list some definitions and theorems related to self-adjoint op-erators from functional analysis. For details, c.f. Reed-Simon, Methods of modernmathematical physics Vol I, or B. Hall, Quantum Theory for Mathematicians.T Self-adjointness.Definition 2.1. Let H be a Hilbert space and D(A) C H a subspace.(1) A linear operator A : D(A) → H is called closed if its graphgraph(A) = ((r, Ar) I r E D(A))is a closed subspace of H × H.(2) We say B : D(B)-→ H an ertension of A if D(A) C D(B) and Ar = Br forall & E D(A). In this case we will denote A C B(3) For any linear operator A : D(A) → H, the smallest closed extension of A(if exists) is called the closure of A and is denoted by A.(4) The adjoint of a densely defined linear operator A : D(A) → H is an linearoperator A*: D(A*) → H so that(Ar,y)= (r,A*y),where D(A*) := (y E H /FC > 0 s.t. (Ar,y) ≤Crl for Vr E D(A))(5) A densely defined linear operator A : D(A) → H is symmetric if A C A*(6) A densely defined linear operator A : D(A) → H is self-adjoint if A = A*

6 LECTURE 8 — 10/19/2020 WEYL QUANTIZATION VIA LINEAR EXPONENTIALS Theorem 1.4. The two definitions of Weyl quantization coincides: a❜ W = 1 (2π~) 2n ❩ R2n e i ~ (y·Q+η·P) [(F~)(s,ξ)→(y,η)a](y, η)dydη. Proof of Theorem 1.4. Applying the right hand side of (9) to ϕ, and using Proposi￾tion 1.2, we get 1 (2π~) 2n ❩ R2n ❤ e i ~ (y·Q+η·P)ϕ ✐ (x)[(F~)(s,ξ)→(y,η)a](y, η)dydη = 1 (2π~) 2n ❩ R2n ❩ R2n e i ~ y·x+ i 2~ y·ηϕ(x + η)e − i ~ (s·y+ξ·η) a(s, ξ)dsdξdydη = 1 (2π~) 2n ❩ R2n ❩ R2n e i ~ y·( x 2 + τ 2 −s) e i ~ ξ·(x−τ)ϕ(τ )a(s, ξ)dsdξdydτ (η → τ = x + η) = 1 (2π~) 2n ❩ R2n ❩ R2n e i ~ y·ζ e i ~ ξ·(x−τ)ϕ(τ )a( x + τ 2 − ζ, ξ)dζdydτ dξ (s → ζ = x + τ 2 − s) = 1 (2π~) n ❩ R2n e i ~ ξ·(x−τ)ϕ(τ )a( x + τ 2 , ξ)dτ dξ, where in the last step we used the Fourier inversion formula. 2. DETOUR: Some Functional Analysis In this section we list some definitions and theorems related to self-adjoint op￾erators from functional analysis. For details, c.f. Reed-Simon, Methods of modern mathematical physics Vol I, or B. Hall, Quantum Theory for Mathematicians. ¶ Self-adjointness. Definition 2.1. Let H be a Hilbert space and D(A) ⊂ H a subspace. (1) A linear operator A : D(A) → H is called closed if its graph graph(A) = {(x, Ax) | x ∈ D(A)} is a closed subspace of H × H. (2) We say B : D(B) → H an extension of A if D(A) ⊂ D(B) and Ax = Bx for all x ∈ D(A). In this case we will denote A ⊂ B. (3) For any linear operator A : D(A) → H, the smallest closed extension of A (if exists) is called the closure of A and is denoted by A. (4) The adjoint of a densely defined linear operator A : D(A) → H is an linear operator A∗ : D(A∗ ) → H so that hAx, yi = hx, A∗ yi, where D(A∗ ) := {y ∈ H |∃C > 0 s.t. hAx, yi ≤ Ckxk for ∀x ∈ D(A)}. (5) A densely defined linear operator A : D(A) → H is symmetric if A ⊂ A∗ . (6) A densely defined linear operator A : D(A) → H is self-adjoint if A = A∗ .

LECTURE810/19/2020WEYLQUANTIZATIONVIALINEAREXPONENTIALS(7) A densely defined linear operator A: D(A)→H is essentially self-adjoint ifA=A*A useful criteria for a symmetric operator to be self-adjoint isTheorem 2.2. Let A be a symmetric operator on H.Then(1) A is essentially self-adjoint if and only if the images of the operators A±iare dense in H.(2) A is self-adjoint if and only if the images of the operators A±i are H.One of the most important theorems for self-adjoint operators acting on a Hilbertspace HisTheorem 2.3 (The spectral theorem (multiplication form)). Suppose A is a self-adjoint operator on H. Then there is measurable space (X,μ), a measurable real-valued function h on X, and a unitary map V : H -→ L?(X,p) such that A is unitaryequivalent via V to the operator "multiplication by h(r)" on L?(X,μ), namely(10)VoAoV*(b)(r)=h(c)b(c).As a consequence, for any measurable function f on R (or on o(A), we maydefinean operatorf(A)viaf(A) = V*o Mf(h(r) o V.In general this might be complicated and might have a very small domain if f isunbounded.However,if fisaboundedfunction,thenf(A)willbeabounded linearoperator on H. In particular, if we take f(r) = eitr, then the operatoreitA - V*Meith(a) o V(11)is defined on the whole of H and is unitary.As a consequence of the spectral theorem (multiplication form), together withPSet 1-8 we immediately get the following theorem which will be used later in theproof of Weyl law:Corollary 2.4 (Helffer-Sjostrand).Let P be a self-adjoint operator on H, f be aSchwartz function and f be an almost analytic ertension of f. Thenf(P)== /,.j(2)(P- 2)-1dm.(12) Stone's theorem.We need to justify that the operators eitA defined by the spectral theorem co-incides with the ones defined as the solution operator to suitable partial differentialequations.Thisisknown astheStone'stheorem.Definition 2.5. Let H be a Hilbert space

LECTURE 8 — 10/19/2020 WEYL QUANTIZATION VIA LINEAR EXPONENTIALS 7 (7) A densely defined linear operator A : D(A) → H is essentially self-adjoint if A = A∗ . A useful criteria for a symmetric operator to be self-adjoint is Theorem 2.2. Let A be a symmetric operator on H. Then (1) A is essentially self-adjoint if and only if the images of the operators A ± i are dense in H. (2) A is self-adjoint if and only if the images of the operators A ± i are H. One of the most important theorems for self-adjoint operators acting on a Hilbert space H is Theorem 2.3 (The spectral theorem (multiplication form)). Suppose A is a self￾adjoint operator on H. Then there is measurable space (X, µ), a measurable real￾valued function h on X, and a unitary map V : H → L 2 (X, µ) such that A is unitary equivalent via V to the operator “multiplication by h(x)” on L 2 (X, µ), namely (10) V ◦ A ◦ V ∗ (ψ)(x) = h(x)ψ(x). As a consequence, for any measurable function f on R (or on σ(A)), we may define an operator f(A) via f(A) = V ∗ ◦ Mf(h(x)) ◦ V. In general this might be complicated and might have a very small domain if f is unbounded. However, if f is a bounded function, then f(A) will be a bounded linear operator on H. In particular, if we take f(x) = e itx, then the operator (11) e itA = V ∗ ◦ Me ith(x) ◦ V is defined on the whole of H and is unitary. As a consequence of the spectral theorem (multiplication form), together with PSet 1-8 we immediately get the following theorem which will be used later in the proof of Weyl law: Corollary 2.4 (Helffer-Sj¨ostrand). Let P be a self-adjoint operator on H, f be a Schwartz function and ˜f be an almost analytic extension of f. Then (12) f(P) = 1 πi ❩ C ¯∂z ˜f(z)(P − z) −1 dm. ¶ Stone’s theorem. We need to justify that the operators e itA defined by the spectral theorem co￾incides with the ones defined as the solution operator to suitable partial differential equations. This is known as the Stone’s theorem. Definition 2.5. Let H be a Hilbert space

LECTURE8-10/19/2020WEYLQUANTIZATIONVIALINEAREXPONENTIALS8(1) A one-parameter unitary group U(t), t e R, is a family of unitary operatorsonHsuchthatU(o)=IdandU(s+t) =U(s)U(t), Vs,teR.(2)A one-parameter unitary group U(t)is said to be (strongly continuous iflim U(t)r = U(to)xfor each to E R and each r E H.Forexample, one can check that the threefamilieseita-Q/h,eitb.P/heit(a-Q+b.P)/hform three strongly continuous one-parameterunitary groups.It turns out that self-adjoint operators on H and strongly continuous one-parameter unitary groups on H are closely related to each other:Theorem 2.6 (Stone's Theorem). Suppose A:D(A)-→H is self-adjoint. ThenU(t) = eitAis a strongly continuous one-parameter unitary group on H, and satisfies1U(t+h)-U(t)± = AU(t)=U(t)Ab(13)limh-0ihfor all b EDom(A) and all t ER.Conversely, if U(t) is a strongly continuous one-parameter unitary group on H,then the operator A (called the infinitesimal generator of U(t)) defined via1U(h)-Aw = lim0his densely defined and self-adjoint, and U(t) = eita.This justifies our definitions of eita-Q/h, eitb-P/h and eit(a-Q+b.P)/h as the solutionoperators to suitable partial differential equations A functional calculus.We just mentioned how to define f(A) via the spectral theorem. There is how-ever another way to define f(A) via the unitary group U(t) = eitA, which is verysimilar to the way Weyl define his quantization:start with the Fourier inversionformula[ eitr f(t)dtf(r) =2元J斤and define二[etAf(t)dt(14)f(A) := 2元 JkOf course inthis definitiononeneed toassume fto bea functionwhose Fouriertransform f = Ff eLl

8 LECTURE 8 — 10/19/2020 WEYL QUANTIZATION VIA LINEAR EXPONENTIALS (1) A one-parameter unitary group U(t), t ∈ R, is a family of unitary operators on H such that U(0) = Id and U(s + t) = U(s)U(t), ∀s, t ∈ R. (2) A one-parameter unitary group U(t) is said to be (strongly continuous if lim t→t0 U(t)x = U(t0)x for each t0 ∈ R and each x ∈ H. For example, one can check that the three families e ita·Q/~ , eitb·P/~ , eit(a·Q+b·P)/~ form three strongly continuous one-parameter unitary groups. It turns out that self-adjoint operators on H and strongly continuous one￾parameter unitary groups on H are closely related to each other: Theorem 2.6 (Stone’s Theorem). Suppose A : D(A) → H is self-adjoint. Then U(t) = e itA is a strongly continuous one-parameter unitary group on H, and satisfies (13) limh→0 1 i U(t + ~)ψ − U(t)ψ h = AU(t)ψ = U(t)Aψ for all ψ ∈ Dom(A) and all t ∈ R. Conversely, if U(t) is a strongly continuous one-parameter unitary group on H, then the operator A (called the infinitesimal generator of U(t)) defined via Aψ = lim h→0 1 i U(h)ψ − ψ h is densely defined and self-adjoint, and U(t) = e itA. This justifies our definitions of e ita·Q/~ , eitb·P/~ and e it(a·Q+b·P)/~ as the solution operators to suitable partial differential equations. ¶ A functional calculus. We just mentioned how to define f(A) via the spectral theorem. There is how￾ever another way to define f(A) via the unitary group U(t) = e itA, which is very similar to the way Weyl define his quantization: start with the Fourier inversion formula f(x) = 1 2π ❩ R e itx ˆf(t)dt. and define (14) f(A) := 1 2π ❩ R e itA ˆf(t)dt. Of course in this definition one need to assume f to be a function whose Fourier transform ˆf = Ff ∈ L 1