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 development (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.