Wave packet decompositions adapted to (non-self-adjoint) operators

Transcription

1 1 / 42 Wave packet decompositions adapted to (non-self-adjoint) operators Joe Viola Laboratoire de Mathématiques Jean Leray Université de Nantes Joseph.Viola@univ-nantes.fr 26 March 2018

2 2 / 42 Outline 1 Translations in phase space 2 The Bargmann transform

3 3 / 42 Wave packet decompositions Our goal is to recall some techniques in wave packet decompositions Wf (ρ) = f (x), ϕ ρ (x) such as the Bargmann transform and to describe how one can adapt the decomposition to the operator investigated, in particular for quadratic non-self-adjoint operators.

4 3 / 42 Wave packet decompositions Our goal is to recall some techniques in wave packet decompositions Wf (ρ) = f (x), ϕ ρ (x) such as the Bargmann transform and to describe how one can adapt the decomposition to the operator investigated, in particular for quadratic non-self-adjoint operators. Example: for the Bargmann transform, the family {ϕ ρ } ρ C is made up of phase-space translations of the Gaussian ϕ 0 (x) = e πx2.

5 4 / 42 Application to non-self-adjoint operators If θ ( π/2, π/2) and ( Q θ = π e iθ x 2 e iθ 1 d 2 ) 4π 2 dx 2, identify the eigenfunctions u k,θ and the L 2 -operator norm of the spectral projections 1 k log Π k,θ ( ) 1 + sin θ 1/2, θ 0. 1 sin θ [Davies, Kuijlaars 2004; Bagarello, 2010; V. 2013]

6 5 / 42 Phase-space translations We know how to translate a function by x 0 in physical space: T (x0,0)f (x) = f (x x 0 ).

7 5 / 42 Phase-space translations We know how to translate a function by x 0 in physical space: T (x0,0)f (x) = f (x x 0 ).

8 5 / 42 Phase-space translations We know how to translate a function by x 0 in physical space: T (x0,0)f (x) = f (x x 0 ).

9 5 / 42 Phase-space translations We know how to translate a function by x 0 in physical space: T (x0,0)f (x) = f (x x 0 ). We know how to translate a function by ξ 0 in momentum: T (0,ξ0 )f (x) = e 2πiξ 0x f (x).

10 5 / 42 Phase-space translations We know how to translate a function by x 0 in physical space: T (x0,0)f (x) = f (x x 0 ). We know how to translate a function by ξ 0 in momentum: T (0,ξ0 )f (x) = e 2πiξ 0x f (x).

11 5 / 42 Phase-space translations We know how to translate a function by x 0 in physical space: T (x0,0)f (x) = f (x x 0 ). We know how to translate a function by ξ 0 in momentum: T (0,ξ0 )f (x) = e 2πiξ 0x f (x).

12 5 / 42 Phase-space translations We know how to translate a function by x 0 in physical space: T (x0,0)f (x) = f (x x 0 ). We know how to translate a function by ξ 0 in momentum: T (0,ξ0 )f (x) = e 2πiξ 0x f (x). To do both simultaneously, there s a natural correction factor. T (x0,ξ 0 )f (x) = e πix 0ξ 0 +2πiξ 0 x f (x x 0 ).

13 6 / 42 Why the correction factor? T (x0,ξ 0 )f (x) = e πix 0ξ 0 +2πiξ 0 x f (x x 0 ).

14 6 / 42 Why the correction factor? T (x0,ξ 0 )f (x) = e πix 0ξ 0 +2πiξ 0 x f (x x 0 ). 1. We d like {T t(x0,ξ 0 )} t R to be a group.

15 6 / 42 Why the correction factor? T (x0,ξ 0 )f (x) = e πix 0ξ 0 +2πiξ 0 x f (x x 0 ). 1. We d like {T t(x0,ξ 0 )} t R to be a group. 2. Note that, if D t = 1 2πi t (as in Folland) Moreover, if D x = 1 2πi x as well, D t e 2πitξ 0x f (x) = ξ 0 xe 2πitξ 0x f (x). D t f (x tx 0 ) = x 0 (D x f )(x tx 0 ).

16 6 / 42 Why the correction factor? T (x0,ξ 0 )f (x) = e πix 0ξ 0 +2πiξ 0 x f (x x 0 ). 1. We d like {T t(x0,ξ 0 )} t R to be a group. 2. Note that, if D t = 1 2πi t (as in Folland) Moreover, if D x = 1 2πi x as well, D t e 2πitξ 0x f (x) = ξ 0 xe 2πitξ 0x f (x). D t f (x tx 0 ) = x 0 (D x f )(x tx 0 ). We d actually like {T t(x0,ξ 0 )} t R to be the group T t(x0,ξ 0 ) = exp(2πit(ξ 0 x x 0 D x )).

17 7 / 42 Composition law One can compute directly that, for v, w R 2n and σ((v x, v ξ ), (w x, w ξ )) = v ξ w x w ξ v x, the shifts obey T v T w = e πiσ(v,w) T v+w.

18 7 / 42 Composition law One can compute directly that, for v, w R 2n and σ((v x, v ξ ), (w x, w ξ )) = v ξ w x w ξ v x, the shifts obey T v T w = e πiσ(v,w) T v+w. Another way of saying this is that {e πis T v : (s, v) R 1+2n } is the Heisenberg group.

19 7 / 42 Composition law One can compute directly that, for v, w R 2n and the shifts obey σ((v x, v ξ ), (w x, w ξ )) = v ξ w x w ξ v x, Another way of saying this is that T v T w = e πiσ(v,w) T v+w. {e πis T v : (s, v) R 1+2n } is the Heisenberg group. Remark: the sympletic product also appears in T v = exp (2πiσ(v, (x, D x )))

20 8 / 42 What the correction factor is good for Let Ff (x) = e πi 4 e 2πixy f (y) dy be the Fourier transform.

21 8 / 42 What the correction factor is good for Let Ff (x) = e πi 4 e 2πixy f (y) dy be the Fourier transform. We know that Ff ( x 0 )(x) = e 2πix 0x Ff (x) and that Fe 2πiξ 0 f ( )(x) = Ff (x ξ 0 ). But the two together...

22 8 / 42 What the correction factor is good for Let Ff (x) = e πi 4 e 2πixy f (y) dy be the Fourier transform. We know that Ff ( x 0 )(x) = e 2πix 0x Ff (x) and that Fe 2πiξ 0 f ( )(x) = Ff (x ξ 0 ). But the two together... We have the general rule with no constants. FT (x0,ξ 0 ) = T (ξ0, x 0 )F,

23 9 / 42 A more complicated example A more involved computation comes from tracking the quantum Schrödinger evolution e itq 0 ϕ 0 (x x 0 ) when ϕ 0 (x) = e πx2, Q 0 = π(d 2 x + x 2 ).

24 10 / 42 Ansatz and ODEs We could guess that e itq 0ϕ 0 (x x 0 ) should take the form for a(0) = 2x 0 and b 0 = x 2 0. e itq 0 ϕ 0 (x x 0 ) = e π(x2 +a(t)x+b(t))

25 10 / 42 Ansatz and ODEs We could guess that e itq 0ϕ 0 (x x 0 ) should take the form e itq 0 ϕ 0 (x x 0 ) = e π(x2 +a(t)x+b(t)) for a(0) = 2x 0 and b 0 = x 2 0. We can obtain ODEs for a and b which give us a(t) = 2e it x 0, b(t) = x 2 0e it cos t + it 2π.

26 11 / 42 Using Egorov Instead, we can use that ϕ 0 (x) is chosen such that ( Q 0 ϕ 0 (x) = π 1 d 2 ) (2π) 2 dx 2 + x2 e πx2 = 1 2 ϕ 0(x) and that e itq 0 follows a rule for shifts like that of F: ( ) t ( e itq 0 T v = T F t ve itq 0, F t 0 1 cos t sin t = = 1 0 sin t cos t ).

27 11 / 42 Using Egorov Instead, we can use that ϕ 0 (x) is chosen such that ( Q 0 ϕ 0 (x) = π 1 d 2 ) (2π) 2 dx 2 + x2 e πx2 = 1 2 ϕ 0(x) and that e itq 0 follows a rule for shifts like that of F: ( ) t ( e itq 0 T v = T F t ve itq 0, F t 0 1 cos t sin t = = 1 0 sin t cos t ). Therefore e itq 0 T (x0,0)ϕ 0 (x) = T x0 (cos t, sin t)e itq 0 ϕ 0 (x) = e it 2 Tx0 (cos t, sin t)ϕ 0 (x).

28 12 / 42 Metaplectic operators There are many such operators K, unitary on L 2 (R n ), associated with linear transformations K such that Generators of this set are KT v = T Kv K. The Fourier transform F 1 in x 1 associated with F 1 (x 1, x, ξ 1, ξ ) = (ξ 1, x, x 1, ξ ). A linear change of variables V G f (x) = (det G) 1/2 f (Gx) is associated with V G (x, ξ) = (G 1 x, G ξ). The multiplication operator W A f (x) = e πix Ax f (x), where A is symmetric, is associated with W A (x, ξ) = (x, ξ + Ax).

29 13 / 42 More on metaplectic operators The (linear) transformation K is canonical (preserves σ) or ( ) ( ) A B K =, K 1 D B =. C D C A

30 13 / 42 More on metaplectic operators The (linear) transformation K is canonical (preserves σ) or ( ) ( ) A B K =, K 1 D B =. C D C A If det B 0, Kf (x) = ±(det( ib)) 1/2 e πi(x B 1 Dx 2x B 1 y+y B 1 Ay) f (y) dy.

31 13 / 42 More on metaplectic operators The (linear) transformation K is canonical (preserves σ) or ( ) ( ) A B K =, K 1 D B =. C D C A If det B 0, Kf (x) = ±(det( ib)) 1/2 e πi(x B 1 Dx 2x B 1 y+y B 1 Ay) f (y) dy. In this way, there are two (and only two) metaplectic operators associated with K.

32 13 / 42 More on metaplectic operators The (linear) transformation K is canonical (preserves σ) or ( ) ( ) A B K =, K 1 D B =. C D C A If det B 0, Kf (x) = ±(det( ib)) 1/2 e πi(x B 1 Dx 2x B 1 y+y B 1 Ay) f (y) dy. In this way, there are two (and only two) metaplectic operators associated with K. These operators are also generated by exp( itq w ) for q(x, ξ) and q w defined on the next slide.

33 A quantization respecting the metaplectic group A quantization is an association function operator, like a Fourier multiplier takes a function of ξ and gives an operator. 14 / 42

34 14 / 42 A quantization respecting the metaplectic group A quantization is an association function operator, like a Fourier multiplier takes a function of ξ and gives an operator. The Weyl quantization takes the Fourier inversion formula a(x, ξ) = e 2πi(x,ξ) (x,ξ )â(x, ξ ) dx dξ and replaces (x, ξ) with (x, D x ). Since e 2πi(x,ξ ) (x,d x) = T ( ξ,x ), we write a w (x, D x ) = â(x, ξ )T ( ξ,x ) dx dξ

35 14 / 42 A quantization respecting the metaplectic group A quantization is an association function operator, like a Fourier multiplier takes a function of ξ and gives an operator. The Weyl quantization takes the Fourier inversion formula a(x, ξ) = e 2πi(x,ξ) (x,ξ )â(x, ξ ) dx dξ and replaces (x, ξ) with (x, D x ). Since e 2πi(x,ξ ) (x,d x) = T ( ξ,x ), we write a w (x, D x ) = â(x, ξ )T ( ξ,x ) dx dξ Egorov theorem: Ka w = â(x, ξ )T K( ξ,x ) dx dξ = (a K 1 ) w K

36 Explicit computations 1 and the integral kernel We can obtain an integral kernel for a w (x, D x )u(x) = e 2πi(x,ξ )(x,ξ ) a(x, ξ )e πix ξ +2πix x u(x + ξ ) dx dξ dx dξ. Upon making the change of variables ξ + x ξ, the exponent becomes 2πix ( x + ξ x ) + 2πi(x ξ )ξ. 2 We traditionally write (y, ξ) instead of (ξ, ξ ); Fourier inversion in x, x gives a w (x, D x )u(x) = e 2πi(x y) ξ a( x + y, ξ)u(y) dy dξ. 2 1 not to be read seriously 15 / 42

37 16 / 42 For polynomials Most concretely, x α D β x can be obtained by expanding ( x+y 2 )α ξ β and using x α 1D β x y α 2 x α 1D β x x α 2: xξ 1 2 (xd x + D x x) and x 3 ξ (x3 D 2 x + 3x 2 D 2 xx + 3xD 2 xx 2 + D 3 xx 2 )

38 17 / 42 Outline 1 Translations in phase space 2 The Bargmann transform

39 18 / 42 As an integral kernel for a metaplectic operator The ( general ) format for a metaplectic operator quantizing a b SL(2, R), if b 0, is c d Mf (x) = 1 ib e πi b (dx2 2xy+ay 2) f (y) dy. (Reminder: F corresponds to a = d = 0 and b = c = 1.)

40 18 / 42 As an integral kernel for a metaplectic operator The ( general ) format for a metaplectic operator quantizing a b SL(2, R), if b 0, is c d Mf (x) = 1 ib e πi b (dx2 2xy+ay 2) f (y) dy. (Reminder: F corresponds to a = d = 0 and b = c = 1.) The Bargmann transform Bf (x) = 2 1/4 e π 2 x2 +2πxy πy 2 f (y) dy corresponds to B = ( 1 i i/2 1/2 ).

41 19 / 42 Consequences of the Egorov theorem Writing q 0 (x, ξ) = π(x 2 + ξ 2 ), we expect BQ 0 B to be the quantization of And it is true that (q 0 B 1 )(x, ξ) = π ( (x/2 + iξ) 2 + (ix/2 + ξ) 2) = 2πixξ. Q 0 := BQ 0 B = 2πi 2 (xd x + D x x) = x x

42 Formal consequences of Q 0 = x x We obtain the Hermite functions Q 0 f (x) = (k + 1 )f (x) f (x) = Cxk 2 and the Schrödinger evolution e itq 0 (i t Q 0 )F(t, x) = 0 F(t, x) = e it/2 F(0, e it x) 20 / 42

43 Formal consequences of Q 0 = x x We obtain the Hermite functions Q 0 f (x) = (k + 1 )f (x) f (x) = Cxk 2 and the Schrödinger evolution e itq 0 (i t Q 0 )F(t, x) = 0 F(t, x) = e it/2 F(0, e it x) But: The transformation B is complex, F(0, e it x) makes no sense for F(0, ) L 2 (R), x k is not integrable. 20 / 42

44 Formal consequences of Q 0 = x x We obtain the Hermite functions Q 0 f (x) = (k + 1 )f (x) f (x) = Cxk 2 and the Schrödinger evolution e itq 0 (i t Q 0 )F(t, x) = 0 F(t, x) = e it/2 F(0, e it x) But: The transformation B is complex, F(0, e it x) makes no sense for F(0, ) L 2 (R), x k is not integrable. This is solved by saying Bf is holomorphic in a weighted space: f L 2 (R) = e π 2 x 2 Bf (x) L 2 (C). 20 / 42

45 21 / 42 Other points of view We can also view the Bargmann transform as the wave packet decomposition: if x = x 0 + iξ 0, e π 2 x 2 Bf (x) = f, T (x0, ξ 0 )ϕ 0.

46 21 / 42 Other points of view We can also view the Bargmann transform as the wave packet decomposition: if x = x 0 + iξ 0, e π 2 x 2 Bf (x) = f, T (x0, ξ 0 )ϕ 0. We can also formally view the Bargmann transform as the Schrödinger evolution Bf (x) = e π 4 (x2 D 2 x ).

47 Some computations 2 around the ground state If ϕ 0 (x) = 2 1/4 e πx2 is a normalized Gaussian, Bf (x) = 2 1/2 e π 2 x2 +2xy πy 2 e πy2 dy = 2 1/2 e 2π(y x 2 )2 dy = 1. 2 not to be read seriously 22 / 42

48 Some computations 2 around the ground state If ϕ 0 (x) = 2 1/4 e πx2 is a normalized Gaussian, Bf (x) = 2 1/2 e π 2 x2 +2xy πy 2 e πy2 dy = 2 1/2 e 2π(y x 2 )2 dy = 1. Note that (x x )1 = and e π 2 x 2 1 L 2 (C) = e π(x2 1 +x2 2 ) dx 1 dx 2 = 1 = ϕ 0 L 2 (R). 2 not to be read seriously 22 / 42

49 Some computations 3 around the Hermite functions To normalize the Hermite functions, we note that x j e π 2 x 2, x k e π 2 x 2 = x j x k e π x 2 dx 1 dx 2 2π = e i(j k)θ dθ r j+k+1 e πr2 dr 0 = k! δ(j k). πk So the real-side Hermite functions {h k } are π k h k (x) = k! B (x k ) 3 not to be read seriously 23 / 42

50 Some computations 3 around the Hermite functions To normalize the Hermite functions, we note that x j e π 2 x 2, x k e π 2 x 2 = x j x k e π x 2 dx 1 dx 2 2π = e i(j k)θ dθ r j+k+1 e πr2 dr 0 = k! δ(j k). πk So the real-side Hermite functions {h k } are π k h k (x) = k! B (x k ) Egorov π k = k! (x id x) k B (1) 3 not to be read seriously 23 / 42

51 Some computations 3 around the Hermite functions To normalize the Hermite functions, we note that x j e π 2 x 2, x k e π 2 x 2 = x j x k e π x 2 dx 1 dx 2 2π = e i(j k)θ dθ r j+k+1 e πr2 dr 0 = k! δ(j k). πk So the real-side Hermite functions {h k } are π k h k (x) = k! B (x k ) Egorov π k = k! (x id x) k B (1) = 2 1/4 π k k! (x id x) k e πx2 3 not to be read seriously 23 / 42

52 Some computations 3 around the Hermite functions To normalize the Hermite functions, we note that x j e π 2 x 2, x k e π 2 x 2 = x j x k e π x 2 dx 1 dx 2 2π = e i(j k)θ dθ r j+k+1 e πr2 dr 0 = k! δ(j k). πk So the real-side Hermite functions {h k } are π k h k (x) = k! B (x k ) Egorov π k = k! (x id x) k B (1) π = 2 1/4 k k! (x id x) k πx2 Egorov π k e = k! ϕ 0(x)(2x id x ) k 1. 3 not to be read seriously 23 / 42

53 24 / 42 Visualizing the Hermite functions The first Hermite function is a Gaussian:

54 24 / 42 Visualizing the Hermite functions We compare the Hermite function to Ie π 2 x 2 Bh k (x) when k = 6:

55 24 / 42 Visualizing the Hermite functions We compare the Hermite function to Ie π 2 x 2 Bh k (x) when k = 15:

56 24 / 42 Visualizing the Hermite functions We compare the Hermite function to Ie π 2 x 2 Bh k (x) when k = 40:

57 24 / 42 Visualizing the Hermite functions We compare the Hermite function to Ie π 2 x 2 Bh k (x) when k = 70:

58 24 / 42 Visualizing the Hermite functions When numerical error makes it difficult to analyze h k, the Bargmann transform is computable and understandable, k = 100:

59 24 / 42 Visualizing the Hermite functions When numerical error makes it difficult to analyze h k, the Bargmann transform is computable and understandable, k = 200:

60 25 / 42 And the Schwartz space In this way, the Hermite decomposition ( = the Taylor series) corresponds to projection onto annuli in phase space.

61 25 / 42 And the Schwartz space In this way, the Hermite decomposition ( = the Taylor series) corresponds to projection onto annuli in phase space. On the Bargmann side, the harmonic oscillator corresponds to multiplying by x 2 : Q 0 u, u = u(x)( x x 1 2 )u(x)e π x 2 dx 1 dx 2 = (π x ) u(x) 2 e π x 2 dx 1 dx 2.

62 25 / 42 And the Schwartz space In this way, the Hermite decomposition ( = the Taylor series) corresponds to projection onto annuli in phase space. On the Bargmann side, the harmonic oscillator corresponds to multiplying by x 2 : Q 0 u, u = u(x)( x x 1 2 )u(x)e π x 2 dx 1 dx 2 = (π x ) u(x) 2 e π x 2 dx 1 dx 2. This allows us to identify S (R) B {f Hol(C) : k N, (1 + x ) k f (x)e π 2 x 2 L 2 (C)}, S (R) B {f Hol(C) : k N, (1 + x ) k f (x)e π 2 x 2 L 2 (C)}.

63 26 / 42 Smoothness estimates Let s suppose that f S (R) such that e π 2 x 2 Bf (x) C. Then, using integration in polar coordinates and Stirling s formula, we can show that Bf (x)e π π k π k 2 x 2 Bf (x), k! xk C x k e π 2 x 2 dx 1 dx 2 k! 2C(2πk) 1/4.

64 26 / 42 Smoothness estimates Let s suppose that f S (R) such that e π 2 x 2 Bf (x) C. Then, using integration in polar coordinates and Stirling s formula, we can show that Bf (x)e π π k π k 2 x 2 Bf (x), k! xk C x k e π 2 x 2 dx 1 dx 2 k! 2C(2πk) 1/4. If p > 3/4, we can conclude that Q p f, h k k 1/2, so Q p f L 2 (R).

65 27 / 42 Application to Dirac comb We can apply this to obtain non-optimal results on the Dirac comb u(x) = k Z δ(x k), where IBu(x)e π 2 x 2 is

66 Application to Dirac comb We can apply thispto obtain non-optimal results on the Dirac comb π 2 u(x) = k Z δ(x k), where =Bu(x)e 2 x is 27 / 42

67 Application to Dirac comb We can apply thispto obtain non-optimal results on the Dirac comb π 2 u(x) = k Z δ(x k), where =Bu(x)e 2 x is 27 / 42

68 27 / 42 Application to Dirac comb We can apply this to obtain non-optimal results on the Dirac comb u(x) = k Z δ(x k), where IBu(x)e π 2 x 2 is The lattice of symmetries in phase space become more evident with the absolute value:

69 28 / 42 True values Of course, the reality is much more complicated (NB: these functions particularly like giving numerical errors).

70 28 / 42 True values Of course, the reality is much more complicated (NB: these functions particularly like giving numerical errors).

71 28 / 42 True values Of course, the reality is much more complicated (NB: these functions particularly like giving numerical errors).

72 29 / 42 Gap with interpolation What s more, we showed Q 1/2 0 f L 2 (R) (1 + x 2 )Bf e π 2 x 2 L 2 (C). If this extends to Q p 0, we d have p > 1 2, Bf (x) e π 2 x 2 L = Q p 0 f L2 (R).

73 30 / 42 As a composition of metaplectic operators Can we represent the Bargmann transform using more familiar operators?

74 30 / 42 As a composition of metaplectic operators Can we represent the Bargmann transform using more familiar operators? We have the metaplectic operators F, the Fourier transform; V G, change of variables; and W A, multiplication by a Gaussian. To construct the Bargmann transform, we need a way to pass from R x to C R x R y and to add a holomorphy condition.

75 31 / 42 Doubling variables A natural way to extend a function, preserving the L 2 norm, is to take the tensor product with a normalized Gaussian: E t f (x) = (2t) 1/4 e πty2 f (x).

76 31 / 42 Doubling variables A natural way to extend a function, preserving the L 2 norm, is to take the tensor product with a normalized Gaussian: E t f (x) = (2t) 1/4 e πty2 f (x). But we certainly don t obtain every function in L 2 (R 2 ): we can identify these functions as the kernel of the operator annihilating the Gaussian: E t : L 2 (R) ker(d y ity) L 2 (R 2 (x,y) ).

77 31 / 42 Doubling variables A natural way to extend a function, preserving the L 2 norm, is to take the tensor product with a normalized Gaussian: E t f (x) = (2t) 1/4 e πty2 f (x). But we certainly don t obtain every function in L 2 (R 2 ): we can identify these functions as the kernel of the operator annihilating the Gaussian: E t : L 2 (R) ker(d y ity) L 2 (R 2 (x,y) ). Everything which follows is an attempt to turn D y iy into D z = 1 2 (D x + id y ).

78 32 / 42 Main ideas Without going into details... A change of variables turns D y iy into D y i(y x).

79 32 / 42 Main ideas Without going into details... A change of variables turns D y iy into D y i(y x). The Fourier transform in x gives D y id x iy = 2i(D z Iz).

80 32 / 42 Main ideas Without going into details... A change of variables turns D y iy into D y i(y x). The Fourier transform in x gives D y id x iy = 2i(D z Iz). Since D z ψ = 1 2 Iz when ψ(z) = 1 2i ( z 2 + I(z 2 )), multiplying by e πiψ(z) puts us in ker D z.

81 32 / 42 Main ideas Without going into details... A change of variables turns D y iy into D y i(y x). The Fourier transform in x gives D y id x iy = 2i(D z Iz). Since D z ψ = 1 2 Iz when ψ(z) = 1 2i ( z 2 + I(z 2 )), multiplying by e πiψ(z) puts us in ker D z. But e πiψ(z) = e π 2 x 2, the weight corrects this!

82 32 / 42 Main ideas Without going into details... A change of variables turns D y iy into D y i(y x). The Fourier transform in x gives D y id x iy = 2i(D z Iz). Since D z ψ = 1 2 Iz when ψ(z) = 1 2i ( z 2 + I(z 2 )), multiplying by e πiψ(z) puts us in ker D z. But e πiψ(z) = e π 2 x 2, the weight corrects this! Note that we have many points of flexibility!

83 33 / 42 Different Bargmann transforms for different operators Johannes Sjöstrand [1974] showed that, for many quadratic forms (q w for q quadratic), one can find B q such that B q q w B q = Gx x tr G for a matrix G in Jordan normal form.

84 33 / 42 Different Bargmann transforms for different operators Johannes Sjöstrand [1974] showed that, for many quadratic forms (q w for q quadratic), one can find B q such that B q q w B q = Gx x tr G for a matrix G in Jordan normal form. Example: For the operator π(e iθ x 2 + e iθ D 2 x), we have the same operator x x + 1 2, but with the weight e π cos θ x 2 R(πie iθ tan θx 2 )

85 34 / 42 Eigenfunctions and evolution for Gx D x + 1 4πi tr G [Sjöstrand, 1974] The generalized eigenfunctions of q w are B K (xα ), α N n.

86 34 / 42 Eigenfunctions and evolution for Gx D x + 1 4πi tr G [Sjöstrand, 1974] The generalized eigenfunctions of q w are As for Schrödinger, B K (xα ), α N n. e 2πit(Gx Dx+tr G/4πi) f (x) = e t 1 2 tr G f (e tg x), so boundedness and compactness depend on Φ(e tg x) Φ(x) [Aleman, V. 2018]

87 35 / 42 Decomposition The projections onto eigenfunctions ( = Taylor series) Π α g(x) = α g(0) x α α! and the low-high energy decomposition ( = truncated Taylor series) Π α N grow (at most) exponentially rapidly in operator norm, Π α, Π α N Ce CN, α N, see [Hitrik, Sjöstrand, V. 2013], [ 3, V. 2013]

88 35 / 42 Decomposition The projections onto eigenfunctions ( = Taylor series) Π α g(x) = α g(0) x α α! and the low-high energy decomposition ( = truncated Taylor series) Π α N grow (at most) exponentially rapidly in operator norm, Π α, Π α N Ce CN, α N, see [Hitrik, Sjöstrand, V. 2013], [ 3, V. 2013] These estimates are elementary from 1 C x 2 Φ(x) C x 2 and the exponent is optimal for the non-self-adjoint harmonic oscillator, [ 3, V. 2013]

89 36 / 42 Resolvents with low-high decomposition The low-high decomposition allows us [Hitrik, Sjöstrand, V. 2013] to obtain (q w z) 1 L(L 2 ) Ce C z if z is not too close to Spec q w.

90 36 / 42 Resolvents with low-high decomposition The low-high decomposition allows us [Hitrik, Sjöstrand, V. 2013] to obtain (q w z) 1 L(L 2 ) Ce C z if z is not too close to Spec q w. From [Dencker, Sjöstrand, Zworski 2005], this type of exponential growth is optimal (and is connected to unsolvability of certain PDEs in the C category).

91 36 / 42 Resolvents with low-high decomposition The low-high decomposition allows us [Hitrik, Sjöstrand, V. 2013] to obtain (q w z) 1 L(L 2 ) Ce C z if z is not too close to Spec q w. From [Dencker, Sjöstrand, Zworski 2005], this type of exponential growth is optimal (and is connected to unsolvability of certain PDEs in the C category). The proof is a low-high energy decomposition (with exponential error), ellipticity on high energies, and straightening to Φ(x) = π 2 x 2 on low energies (with exponential error).

92 37 / 42 Brief explanation of process The process works for quadratic forms such that, for good matrices, q(x, ξ) = B(ξ A x) (ξ A + x), happily Rq positive definite is sufficient.

93 37 / 42 Brief explanation of process The process works for quadratic forms such that, for good matrices, q(x, ξ) = B(ξ A x) (ξ A + x), happily Rq positive definite is sufficient. This is supersymmetric because q w can be written as conjugated derivatives.

94 37 / 42 Brief explanation of process The process works for quadratic forms such that, for good matrices, q(x, ξ) = B(ξ A x) (ξ A + x), happily Rq positive definite is sufficient. This is supersymmetric because q w can be written as conjugated derivatives. To change q(x, ξ) into Mx ξ, it s enough to have K{(x, A + x)} = {(z, 0)}, K{(x, A x)} = {(0, ζ)}.

95 37 / 42 Brief explanation of process The process works for quadratic forms such that, for good matrices, q(x, ξ) = B(ξ A x) (ξ A + x), happily Rq positive definite is sufficient. This is supersymmetric because q w can be written as conjugated derivatives. To change q(x, ξ) into Mx ξ, it s enough to have But the weight e π x 2 will change! K{(x, A + x)} = {(z, 0)}, K{(x, A x)} = {(0, ζ)}.

96 38 / 42 Some purely self-adjoint concerns If q 1 (x, ξ), q 2 (x, ξ) are real-valued positive definite quadratic forms, each is unitarily equivalent to C(x 2 + D 2 x) but not at the same time! We can therefore ask whether e t 1q w 1 e t 2 q w 2 L(L 2 (R n )), t 1, t 2 > 0.

97 38 / 42 Some purely self-adjoint concerns If q 1 (x, ξ), q 2 (x, ξ) are real-valued positive definite quadratic forms, each is unitarily equivalent to C(x 2 + D 2 x) but not at the same time! We can therefore ask whether By bridging the gap with we can show that e t 1q w 1 e t 2 q w 2 L(L 2 (R n )), t 1, t 2 > 0. q 3 (x, ξ) = (creation of q 1 )(annihilation of q 2 ) For t 1, t 2 small it is necessary that t 1 Ct 2 and sufficient that t 1 1 C t 2 and unless the same transformation works for q 1 and q 2, there exists t1 c > 0 such that there exists a t 2 for every 0 < t 1 < t1 c and if t 1 > t1 c then et 1q w 1 e t 2q w 2 is never bounded (because the ground state of q 2 leaves L 2 under e tc 1 qw 1 ).

98 38 / 42 Some purely self-adjoint concerns If q 1 (x, ξ), q 2 (x, ξ) are real-valued positive definite quadratic forms, each is unitarily equivalent to C(x 2 + D 2 x) but not at the same time! We can therefore ask whether By bridging the gap with we can show that e t 1q w 1 e t 2 q w 2 L(L 2 (R n )), t 1, t 2 > 0. q 3 (x, ξ) = (creation of q 1 )(annihilation of q 2 ) For t 1, t 2 small it is necessary that t 1 Ct 2 and sufficient that t 1 1 C t 2 and unless the same transformation works for q 1 and q 2, there exists t1 c > 0 such that there exists a t 2 for every 0 < t 1 < t1 c and if t 1 > t1 c then et 1q w 1 e t 2q w 2 is never bounded (because the ground state of q 2 leaves L 2 under e tc 1 qw 1 ).

99 39 / 42 Summary What I hoped to say is: what are translations in phase space, how their transformation properties are useful, what the Bargmann transform is, how the Bargmann transform simplifies the search for the Hermite functions, how the spectral decomposition of Q 0 corresponds to decomposing into annuli in phase space, how there are different Bargmann transforms which work better for different operators, and how we still have decomposition ( = Taylor series) into low and high energies ( = distance from the origin).

100 40 / 42 Other techniques Other techniques include: Approximating an operator by multiplication ([Cordoba-Fefferman 1978], Martinez, Sjöstrand, and many others.) Adapting a Bargmann space by changing the weight, which can for instance solve a Hamilton-Jacobi equation. ([Herau-Sjöstrand-Stolk 2005], many other works of Sjöstrand, Hitrik, Pravda-Starov.) For quadratic operators, solving differential equations for the phase of the Weyl symbol of the Schrödinger evolution (Howe, Robert, Combescure, Derezinski and Karczmarczyk, Graefe and Schuman...) Studying complex canonical transformations as a holomorphic extension of the metaplectic group ([Hörmander 1983, 1995], also Howe).

101 41 / 42 Some open questions When are simple estimates for spectral projections optimal? Optimal exponential rate of resolvent growth. How rapidly does the resolvent norm decay when restricted to large energies? Geometric understanding of what happens to Gaussians of different shapes under certain operators. Beyond ellipticity : can one make more rigorous claims like B = e π 4 (x2 D 2 x)?

102 42 / 42 Merci! Thanks for listening!