SYMPLECTIC DIMENSIONAL EXTENSIONS OF A PSEUDO-DIFFERENTIAL OPERATOR - - PowerPoint PPT Presentation

SYMPLECTIC DIMENSIONAL EXTENSIONS OF A PSEUDO-DIFFERENTIAL OPERATOR

M.A. de Gosson (joint work with N.C. Dias and J.N. Prata)

University of Vienna, Faculty of Mathematics, NuHAG

September 2012

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References

• N. C. Dias, M.A. de Gosson, and J.N. Prata. Prata. Dimensional

Extension of Pseudo-Di¤erential Operators: Properties and Spectral

• Results. Preprint 2012 (submitted)

N.C. Dias, M. de Gosson, F. Luef, and J.N. Prata. A pseudo–di¤erential calculus on non–standard symplectic space; spectral and regularity results in modulation spaces. J. Math. Pure

• Appl. 96 (2011) 423–445
• M. de Gosson. Spectral Properties of a Class of Generalized Landau

Operators, Comm. Partial Di¤erential Equations, 33(11) (2008), 2096–2104

• M. de Gosson. Symplectic Methods in Harmonic Analysis and in

Mathematical Physics. Birkhäuser; Springer Basel (2011)

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Introductory example

We begin by making a simple observation: in phase space QM one often replaces the usual "canonical quantization rules" (x, p) ! (b x, b p) with b xψ = xψ, b pψ = i h∂xψ by the "Bopp quantization rules" (x, p) ! (e x, e p) where e x = x + 1

2i

h∂p , e p = p 1

2i

h∂p. (1) This is an example of what we call a "symplectic dimensional extension". Suppose indeed that a function a(x, p) ! e A = a(e x, e p). The operator e A can be viewed as the Weyl quantization e a(x, y; p, q) = a s(x, y; p, q) (2) where s is the symplectic matrix s = I 1

2D

D

1 2I

• , I =

1 1

• , D =

1 1

• .

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Symplectic and metaplectic

The symplectic group

A 2n 2n matrix s is symplectic if sT Js = J where J = I I

• is the

"standard symplectic matrix". More abstract de…nition: s 2 Sp(2n) ( ) s linear automorphism of R2n σ2n(sz, sz0) = σ2n(z, z0) Symplectic automorphisms/matrices form a group: Sp(2n). Note that sT Js = J = ) det s = 1. In fact: s 2 Sp(2n) = ) det s = 1. The importance of symplectic automorphisms in mathematical physics comes from the fact that Sp(2n) is the main group of symmetries in Hamiltonian mechanics. In addition, Hamiltonian ‡ows consist of symplectomorphisms, i.e. of di¤eomorphisms whose Jacobian matrix is, at every point, a symplectic matrix. The symplectic group Sp(2n) is a connected Lie group with dimension n(2n + 1).

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Symplectic and metaplectic

The metaplectic group

We have π1(Sp(2n)) = Z, hence Sp(2n) has covering groups of all orders 2, 3, ..., ∞. One of these plays an essential role, both in theoretical questions and in applications: the twofold covering group Sp2(2n) has a faithful representation by a group of unitary operators acting on L2(Rn): the metaplectic group Mp(2n). It is the group generated by the generalized Fourier transforms SW ,.mψ(x) = 1

2πi

n/2 ∆m(W )

Z

Rn eiW (x,x 0)ψ(x0)dx0

(3) W (x, x0) =

1 2Px x Lx x0 + 1 2Qx0 x0

(4) with P = PT , Q = QT and det L 6= 0, and ∆m(W ) = imp j det Ljwhere m is an integer ("Maslov index") corresponding to a choice of arg det L. The projection of SW ,.m on Sp(2n) is de…ned by (x, p) = s(x0, p0) ( )

• p = ∂xW (x, x0)

p0 = ∂x 0W (x, x0) . (5)

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Symplectic and metaplectic

Covariance

Let b A be the pseudo-di¤erential operator with Weyl symbol a: b A

Weyl

! a. Let S 2 Mp(2n) have projection s 2 Sp(2n). Then Sb AS1 has Weyl symbol a s1: Sb AS1 Weyl ! a s1.

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Symplectic and metaplectic

Covariance

Let b A be the pseudo-di¤erential operator with Weyl symbol a: b A

Weyl

! a. Let S 2 Mp(2n) have projection s 2 Sp(2n). Then Sb AS1 has Weyl symbol a s1: Sb AS1 Weyl ! a s1. Let W (ψ, φ)(z) = 1

2π

n Z

Rn eipy ψ(x + 1 2y)φ(x 1 2y)dy

(6) be the cross-Wigner transform; then W (Sψ, Sφ)(z) = W (ψ, φ)(s1z). This follows from the fundamental relation between Wigner formalism and Weyl calculus: (b Aψjφ)L2 = ha, W (ψ, φ)i. (7)

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Symplectic dimensional extensions

Let us then make the identi…cations R2n = Rn

x Rn p and

R2(n+k) = R(n+k)

x,y

R(n+k)

p,q

and let the two spaces R2n and R2(n+k) be equipped with the standard symplectic forms. Consider Es : S0(R2n) ! S0(R2(n+k)) (8) a 7 ! e as = Es[a] = (a 12k) s where 12k : R2k ! R; 12k(y, η) = 1 and s 2 Sp(2(n + k), R). We call Es a symplectic dimensional extension map. In the simplest case s = I we write E = EI and the function (or distribution) e a = E[a] is given explicitly by: e a(x, y; p, q) = a(x, p). (9) We note that e a can also be de…ned in terms of the orthogonal projection

• perator Π : R2(n+k)

! R2n: under the identi…cation Π(x, y; p, q) (x, p) we have e as= aΠ s.

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De…nition

Let us write A = Opw (a) and e As = Opw (e as) (Weyl operators with symbols a and e as, respectively). The commutative diagram: a

W

> A j j Es j j Es # # e as

f W

> e As (10) where W and f W are the Weyl correspondences, de…nes, for each Es, a linear embedding map Es Es : L(S(Rn), S0(Rn)) ! L(S(Rn+k), S0(Rn+k)) (11) A 7 ! e As = Es[A] which we call the symplectic dimensional extension map for Weyl

• perators.

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Let denote by e S the elements of Mp(2(n + k), R); they are unitary on L2(Rn+k).

Theorem

Let s 2 Sp(2(n + k), R) and a 2 S0(R2n). Let e A = E[A] and e As = Es[A] be the corresponding dimensional extensions of b

• A. We have

e As = e S1e Ae S (12) where e S 2 Mp(2(n + k), R) is any of the two metaplectic operators with projection s.

Proof.

Formula (12) is equivalent to \ (a 12k) s = e S1 \ a 12k e S (13) which follows from the symplectic covariance of Weyl operators.

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Intertwiners

Let us shortly return to the example in the Introduction, the Bopp

• perators

e A = a(x + 1

2i

h∂p, p 1

2i

h∂p). We have shown that e A is related to the usual Weyl operator b A = a(b x, b p) by the intertwining relations e AWχ = Wχb A , χ 2 S(Rn) , jjχjjL2 = 1 where Wφ is a partial isometry L2(Rn) ! L2(R2n) de…ned by Wχψ = (2π)n/2W (ψ, χ) (14) where W (f , χ)(z) = 1

2π

n Z

Rn eipyf (x + 1 2y)χ(x 1 2y)dy

is the cross-Wigner transform. This allows, among other things, to study the spectral properties of the Bopp operator e A knowing those of b A. We are going to see that similar intertwiners exist for general symplectic dimensional extensions e As.

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Intertwiners: general case

For χ 2 S(Rk), jjχjjL2 = 1, and e S 2 Mp(2(n + k), R) de…ne Te

S,χ : S(Rn)

! S(Rn+k) by Te

S,χψ = e

S1(ψ χ). (15) These operators Te

S,χ are partial isometries L2(Rn)

! L2(Rn+k) and: For a 2 S0(R2n) we have e AsTe

S,χ = Te S,χb

A , T

e S,χe

As = b AT

e S,χ

(16) The operator Te

S,χ extends into a continuous operator

S0(Rn) ! S0(Rn+k) and we have T

e S,χΦ(x) =

Z

Rk χ(y)e

SΦ(x, y)dy. (17) The map T

e S,χ extends into the continuous operator

S0(Rn+k) ! S0(Rn) de…ned for ψ 2 S(Rn) by hT

e S,χΦ, ψi = he

SΦ, χ ψi. (18)

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Proof.

Let us …rst prove that e AsTe

S,χ = Te S,χA. It is clear that Te S,χψ 2 S(Rn+k)

if ψ 2 S(Rn) and χ 2 S(Rk). Let us …rst consider the case e S = e I, and set Tχ = Te

I,χ. We have to show that e

ATχ = TχA. Let ψ 2 S(Rn); we have e A(Tχψ)(x, y) = b A(Tχψ)y (x) = b Aψ(x)χ(y) = Tχ(b Aψ)(x, y). The general cases follows using the symplectic covariance of Weyl calculus e AsTe

S,χ = e

S1e Ae S(e S1Tχ) = e S1e ATχ = e S1Tχb A = Te

S,χb

A. The second formula (16) is deduced from the …rst: since e A

s Te S,χ = Te S,χb

A we have (e A

s Te S,χ) = (Te S,χb

A) that is T

e S,χe

As = b AT

e S,χ.

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Spectral results

A property of the intertwiners

Theorem

Let (φj)j be an orthonormal basis of L2(Rn) and (χl)l an orthonormal basis of L2(Rk). Then (Te

S,χl φj)j,l is an orthonormal basis of L2(Rn+k).

Proof.

Let us set Φj,l = Te

S,χl φj. We have

(Φj,ljΦj0,l 0) = (e S1(φj χl)je S1(φj0 χl 0)) = (φj χljφj0 χl 0) = (φjjφj0)(χljχl 0) hence the Φj,l form an OS in L2(Rn+k). To prove that this system is complete in L2(Rn+k) it su¢ces to show that if Ψ 2 L2(Rn+k) is such that (ΨjΦj,l) = 0 for all j, l then Ψ = 0.

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Now, (ΨjΦj,l) = (Ψje S1(φj χl)) = (e SΨjφj χl) and (ΨjΦj,l) = 0 for all j, l implies e SΨ = 0 because the tensor product of the two orthonormal bases is an orthonormal basis of L2(Rn+k); it follows that Ψ = 0 as claimed. Note that we have established in the course of the proof the Moyal identity (Te

S,χφjTe S,χ0φ0)L2(Rn+k ) = (φjφ0)L2(Rn)(χjχ0)L2(Rk ).

(19) It is an extension of the well-known eponymous property of the cross-Wigner transform: (W (ψ, φ)jW (ψ0, φ0))L2(R2n) = 1

2π

n (ψjψ0)L2(Rn)(φjφ0)L2(Rn). (20)

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Spectral results

Main result:

Theorem

Let A : S(Rn) 7! S0(Rn) be a linear continuous operator and let e As = Es[A]. Then: (i) The eigenvalues of A and e As are the same. (ii) If ψ is an eigenfunction of A corresponding to the eigenvalue λ then Ψ = Te

S,χψ is an eigenfunction of e

As corresponding to λ, for every χ 2 S(Rk)n f0g, and we have Ψ 2 S(Rn+k). (iii) Conversely, if Ψ is an eigenfunction of e As and ψ = T

e S,χΨ is di¤erent

from zero then ψ is an eigenfunction of A and corresponds to the same eigenvalue. (iv) If (ψj)j is an orthonormal basis of eigenfunctions of A and (χl 2 S(Rk))l is an orthonormal basis of L2(Rk) then (Te

S,χl ψj)j,l is a

complete set of eigenfunctions of e As and forms an orthonormal basis of L2(Rn+k).

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Proof.

That every eigenvalue of A also is an eigenvalue of e As is clear: if Aψ = λψ for some ψ 6= 0 then e As(Te

S,χψ) = Te S,χAψ = λ(Te S,χψ)

and Te

S,χψ 6= 0 because Te S,χ is injective; this shows at the same time that

Ψ = Te

S,χψ is an eigenfunction of e

• As. Since Te

S,χ : S(Rn)

! S(Rn+k) it is clear that Ψ 2 S(Rn+k) which concludes the proof of (ii). Assume conversely that e AsΨ = λΨ and Ψ 6= 0 . For every χ we have, using the second intertwining relation (16), AT

e S,χΨ = T e S,χe

AsΨ = λT

e S,χΨ

hence λ is an eigenvalue of A and T

e S,χΨ will be an eigenfunction of A if it

is di¤erent from zero, hence (iii).

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To conclude the proof of (i) we have to show that if Ψ is a eigenfunction

• f e

As then T

e S,χΨ 6= 0 for some χ 2 S(Rk). Now,

Te

S,χT e S,χ = Pe S,χ

is the orthogonal projection onto the range He

S,χ of Te S,χ. Assume that

T

e S,χΨ = 0 for all χ 2 S(Rk); then Pe S,χΨ = 0 for every χ 2 S(Rk), and

hence Ψ = 0; but this is not possible since Ψ is an eigenfunction. Finally, the implication (ii)= )(iv) is clear.

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Since e As = Es[A] is continuous S(Rn+k) ! S(Rn+k) it can be extended into a continuous operator S0(Rn+k) ! S0(Rn+k). Let (ψ, λ) 2 S0(Rn) C. We will say that ψ is a generalized eigenvector

• f A, corresponding to the generalized eigenvalue λ if ψ 6= 0 and

hψ, Aφi = λhψ, φi for all φ 2 S(Rn). Notation: (ψjAφ) = λ(ψjφ).

Theorem

Assume that A is a continuous linear operator of the form A : S(Rn) ! S(Rn) and let e As = Es[A]. Then: (i) The generalized eigenvalues of the operators A and e As are the same. (ii) Let ψ be a generalized eigenvector of A. Then, for every χ 2 S(Rk)n f0g the vector Ψ = Te

S,χψ is a generalized eigenvector of e

As and corresponds to the same generalized eigenvalue. (iii) Conversely, if Ψ is a generalized eigenvector of e As and ψ = T

e S,χΨ 6= 0 then ψ is a generalized eigenvector of A corresponding to

the same generalized eigenvalue.

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