Multiresolution analysis and QM∞

Multiresolution analysis and QM _∞

Fabio Bagarello

Montecatini Terme, Assemblea scientifica del GNFM, 2004

Amultiresolution analysis of L²(R) is an increasing sequence of closed subspaces

. . . ⊂ V₋₂ ⊂ V₋₁ ⊂ V₀ ⊂ V₁ ⊂ V₂ ⊂ . . . ⊂ L²(R), (1) with S

j∈ZV_j dense in L²(R) and T

j∈Z V_j = {0}, and such that (1) f (x) ∈ V_j ⇔ f (2x) ∈ V_j+1

(2) There exists a function φ ∈ V₀, called a scaling function, such that {φ(x − k), k ∈ Z} is an o.n. basis of V₀.

(1) + (2) ⇒ {φ_j,k(x) ≡ 2^j/2φ(2^jx − k), k ∈ Z} is an o.n. basis of V_j.

Each V_j can be interpreted as an approximation space: the approximation of f ∈ L²(R) at the resolution 2^j is defined by its projection onto V_j. The additional details needed for increasing the resolution from 2^j to 2^j+1 are given by the projection of f onto the orthogonal complement W_j of V_j in V_j+1:

V_j ⊕ W_j = V_j+1, (2)

and we have:

M

j∈Z

W_j = L²(R). (3)

Now, there exists a function ψ, the mother wavelet, explicitly computable from φ, such that {ψ_j,k(x) ≡ 2^j/2ψ(2^jx − k), j, k ∈ Z} constitutes an orthonormal basis of L²(R).

The construction of ψ proceeds as follows. First, the inclusion V₀ ⊂ V₁ yields the relation

φ(x) = √ 2

X∞ n=−∞

h_nφ(2x − n), h_n = hφ_1,n|φi. (4) Then one uses these coefficients to define the function ψ as

ψ(x) = √ 2

X∞ n=−∞

(−1)ⁿ⁻¹h_−n−1φ(2x − n). (5) ψ(x) can be used to generate the wavelet transform of a given L²-function f (x) as

(T^wavf )(a, b) = |a|^−1/2 Z

dxf (x)ψ

µx − b a

¶ ,

where a ∈ R\{0} and b ∈ R. a and b are the zooming and the translation parameter.

EXAMPLES: (1) Haar MRA:

h(x) =













1, if 0 ≤ x < 1/2

−1, if 1/2 ≤ x < 1 0, otherwise

(6)

This function gives an explicit example of a wavelet o.n. basis in L²(R), obtained via the usual formula

h_mn(x) = 2^−m/2h(2^−mx − n). (7) (2) Littlewood Paley

Ψ(x) = (πx)⁻¹(sin 2πx − sin πx). (8) The behavior of this function is, in a sense, complementary to that of the Haar wavelet: it is very well localized in frequency space (it has a compact support)

Ψ(ω) =b





(2π)^−1/2, if π ≤ |ω| ≤ 2π 0, otherwise,

(3) Journ´e basis

Ψ(ω) =b





(2π)^−1/2 if ^4π₇ ≤ |ω| ≤ π and 4π ≤ |ω| ≤ ^32π₇

0 otherwise

(9) (3) More Examples:

• splines (linear, quadratic, cubic,...),

• Daubechies’ wavelets,

• sinc wavelets,...

Problem: how to find new scaling functions φ and MR?

Definition:– We call relevant any sequence h = {h_n, n ∈ Z}

which satisfies the following properties:

(r1) P

n∈Z h_nh_n+2l = δ_l,0; (r2) h_n = O(_1+|n|¹ ₂), n À 1;

(r3) P

n∈Z h_n = √ 2;

(r4) H(ω) = ^√1₂P

n∈Zh_ne^−iωn 6= 0 ∀ω ∈ [−^π₂, ^π₂].

Relevant sequences ⇒ Scaling functions and Mother Wavelets (Mallat’s algorithm).

Main question: how to produce relevant sequences?

MRA ⇔ FQHE

• F. Bagarello, J.P. Antoine, Wavelet-like Orthonormal Basis of the Lowest Landau Level, J. Phys. A, 27, 2471-2481 (1994)

• F. Bagarello, More Wavelet-like Orthonormal Bases for the Lowest Landau Level: Some Considerations, J. Phys. A, 27, 5583-5597 (1994)

• F. Bagarello, Applications of Wavelets to Quantum Mechanics: a Pedagogical Example, J. Phys. A, 29, 565-576 (1996)

• F. Bagarello Multi-Resolution Analysis and Fractional Quantum Hall Effect: an Equivalence Result, J. Math. Phys., 42, 5116-5129, (2001)

• F. Bagarello Multi-Resolution Analysis and Fractional Quantum Hall Effect: More Results, J. Phys. A, 36, 123-138 (2003)

• F. Bagarello, J.P. Antoine, Localization properties and wavelet-like orthonormal bases for the lowest Landau level, in Advances in Gabor Analysis, H.G. Feichtinger, T. Strohmer Eds., Birkh¨auser, Boston, 2003

• F. BagarelloMulti-Resolution Analysis generated by a seed function, J. Math. Phys., 44, 1519-1534 (2003)

The old results: FQHE

Full hamiltonian:

H^(N) = H₀^(N) + λ(H_C^(N) + H_B^(N)), (10) H₀^(N) =

XN i=1

H₀(i). (11)

Here H₀(i) describes the minimal coupling of the i−th electron with the magnetic field:

H₀ = 1 2

¡p + A(r)¢₂

= 1 2

³

p_x − y 2

´₂ + 1

2

³

p_y + x 2

´₂

. (12)

H_C^(N) is the Coulomb interaction between charged particles, and H_B^(N) is the interaction of the charges with the background. Both are small compared with H₀^(N)!

Let us introduce the following operators:

P⁰ = p_x−y/2, Q⁰ = p_y+x/2, P = p_y−x/2, Q = p_x+y/2.

(13) Then we have

H₀ = 1

2(Q⁰² + P⁰²). (14)

and

[Q, P ] = [Q⁰, P⁰] = i, [Q, P⁰] = [Q⁰, P ] = [Q, Q⁰] = [P, P⁰] = 0.

(15) Therefore, defining the magnetic translation operators T (~a_i) for a square lattice with basis ~a₁ = a(1, 0), ~a₂ = a(0, 1), a² = 2π by T₁ := T ( ~a₁) = e^iaQ, T₂ := T ( ~a₂) = e^iaP, (16) we find

[T ( ~a₁), T ( ~a₂)] = [T ( ~a₁), H₀] = [T ( ~a₂), H₀] = 0, (17) and, given a generic function f (x, y) ∈ L²(R²),

f_m,n(x, y) := T₁^mT₂ⁿf (x, y) = (−1)^mne^ia2(my−nx)f (x+ma, y+na).

(18) The wave function in the (x, y)-space is related to its P P⁰- expression by the formula

Ψ(x, y) = e^ixy/2 2π

Z _∞

−∞

Z _∞

−∞

e^{i(xP0+yP +P P0)}Ψ(P, P⁰) dP dP⁰, (19)

which can be easily inverted:

Ψ(P, P⁰) = e^{−iP P0} 2π

Z _∞

−∞

Z _∞

−∞

e^{−i(xP0+yP +xy/2)}Ψ(x, y) dxdy.

(20)

The ground state of H₀ must have the form f₀(P⁰)h(P ), where f₀(P⁰) = π^−1/4e^−P02/2, (21) while the function h(P ) is arbitrary, which manifests the degen- eracy of the LLL, and should be fixed by the interaction. With f₀ as above formula (19) becomes

Ψ(x, y) = e^ixy/2

√2π^3/4

Z _∞

−∞

e^iyPe^{−(x+P )2/2}h(P ) dP, (22)

which produces the lattice as ψ_m,n(x, y) = T₁^mT₂ⁿΨ(x, y).

Our main result is that, if the following ONC holds,

< ψ_l1,2l2, ψ_0,0 >=

Z _∞

−∞

dxe^−2il2axh(x − l₁a)h(x) =

=

Z _∞

−∞

dpe^il1apˆh(p − 2l₂a)ˆh(p) = δ_l1,0δ_l2,0, (23) for all l₁, l₂ ∈ Z, where ˆh(p) is the Fourier transform of h(x),

then, defining

h_n = 1

√a

Z _∞

−∞

dpe^−inxah(x), (24)

it follows automatically that X

n∈Z

h_nh_n+2l = δ_l,0. (25) We also have proved the opposite implication: given any rel- evant sequence we can produces an o.n. set in the LLL.

Therefore: At that stage we believed in the existence of a deep relationship between FQHE and MRA. We also analyzed the other requirements of a relevant sequence in that perspective...

New results

...but this is not strictly necessary: the procedure is only related to the unitary map between two pairs of conjugate oper- ators. All the results deduced are model independent !

Consider the operators ((ˆx, ˆp_x), (ˆy, ˆp_y)) and ((ˆx₁, ˆp₁), (ˆx₂, ˆp₂)), satisfying

[ˆx, ˆp_x] = [ˆy, ˆp_y] = i, [ˆx₁, ˆp₁] = [ˆx₂, ˆp₂] = i

Let ξ_x and η_y be (generalized) eigenstates of ˆx and ˆy: ˆxξ_x = xξ_x, ˆyη_y = yη_y, and ξ_x⁰₁ and η_x⁰₂ eigenstates of the new position operators ˆx₁ and ˆx₂. We have

Z dx

Z

dy |ξ_x,y >< ξ_x,y| = Z

dx₁ Z

dx₂|ξ_x⁰_1,x2 >< ξ_x⁰_1,x2| = 11,

where ξ_x,y = ξ_x ⊗ η_y and ξ_s,t⁰ = ξ_s⁰ ⊗ η_t⁰. Any Ψ ∈ H can be written in the (x, y)-coordinates or in the (x₁, x₂)-coordinates:

Ψ(x, y) =< ξ_x,y|Ψ > and Ψ⁰(x₁, x₂) =< ξ_x⁰_1,x2|Ψ >,

which are related as Ψ(x, y) =

Z

dx₁ Z

dx₂ < ξ_x,y|ξ_x⁰_1,x2 > Ψ⁰(x₁, x₂) Ψ⁰(x₁, x₂) =

Z dx

Z

dy < ξ_x⁰_1,x2|ξ_x,y > Ψ(x, y)

Remark:– Ψ(x, y) and Ψ⁰(x₁, x₂) are different representa- tions of the same element of H.

Let us choose Ψ⁰(x₁, x₂) = ϕ(x₁)h(x₂) and let us define three commuting operators: H = H(ˆx₁, ˆp₁) = H^†, T₁ = e^iaˆx2 and T₂ = e^iaˆp2. Here a² = 4π. We still call the unitary operators magnetic translations and H the hamiltonian.

But there is no physical system behind, now.

As for the FQHE we require orthonormality of the functions Ψ_h,~l(x, y) = T₁^l1T₂^l2Ψ_h(x, y), which generate a lattice with cell area = 4π.

Remark:– In the FQHE the function ϕ(x₁) was the ground state of H. We will show that this is unessential.

We find that, independently of the model, that is, of the ex- plicit expressions for H, T₁ and T₂, as well as of the choice of ((ˆx₁, ˆp₁), (ˆx₂, ˆp₂)), (mainly) due to the resolution of the identity above, if ϕ has L²(R)-norm 1, then:

< Ψ_h,(l1,l2), Ψ_h,(0,0) >=

Z

R

ds h(s) h(s + al₂) e^−isal1 =

= Z

R

dp ˆh(p) ˆh(p − al₁) e^−ipal2 =: S_l^(h)

1,l₂,

which does not depend on the choice of ϕ, [B, JMP, 2001], [B, JPA, 1996]. We call ONC the equation

S_l^(h)

1,l₂ = δ_~l,~0.

Solutions of the ONC can be easily found using a relevant sequence h = {h_n, n ∈ Z} (P

n∈Zh_nh_n+2l = δ_l0):

h(s) =





√1 a

P

n∈Zh_ne^−isna/2, s ∈ [0, a[,

0 otherwise

(26) Remark 1: any MRA produces an o.n. basis in the LLL (as well as in all the other Landau levels!).

Remark 2: If we have a d-MRA, we still find solutions of the ONC if a² = 2πd.

Remark 3: Since H, T₁, T₂ may have no physical meaning, it is not very relevant to use MRA to produce o.n. sets in the Landau levels. We do the opposite: we use the ONC to produce MRA!!!!

The recipe:

(1) Find a solution h(s) of the ONC.

(2) Put

h_n = 1

√a Z

R

h(s) e^isna/2ds =

r2π a ˆh³

−na 2

´

Then, using the Poisson summation formula, we can easily see that P

n∈Zh_nh_n+2l = δ_l0.

Remark 4: It is much simpler to find solutions of the ONC than, directly, sequences satisfying (r1).

Examples: solutions of the ONC and related sequences

Obvious solutions of the ONC are the ones arising from a MRA. Other solutions are:

(1) the trivial ones:

h(s) =





√1

a, s ∈ [0, a[, 0 otherwise

ˆh(p) =



 q

2

a, p ∈ [0, a/2[, 0 otherwise then h_n = δ_n0. Trivially we have P

n∈Z h_nh_n+2l = δ_l0. (2) the effects of a phase..:

h(s) =





e√^−is

a , s ∈ [0, a[, 0 otherwise then h_n = i(1 − e^−ia)_2πn−a¹ . P

n∈Zh_nh_n+2l = δ_l0 is not trivial!

(3) close to the sinc wavelets:

h(s) =



 q2

a, s ∈ [0, a/2[, 0 otherwise,

h(s) =













√1

2a, s ∈ [0, a/2[∪[2a, 3a[,

−^√1_2a, s ∈ [a/2, a[, 0 otherwise and

h(s) =













√1

2a, s ∈ [0, a/2[∪[a, 2a[,

−^√1_2a, s ∈ [a/2, a[, 0 otherwise

they all satisfy the ONC and produce the following sequence:

h₀ = ^√1₂, h_2n = 0 if n 6= 0, and h_2n+1 = _π(2n+1)^i√2 , which satisfies (r1).

(4) the Haar wavelet:

ˆh(p) =





√1

a, p ∈ [0, a[, 0 otherwise

then h₀ = h₋₁ = ^√1₂, which are the Haar coefficients.

(4) more compactly supported solutions (in p)

ˆh(p) =













√1

2a, p ∈ [0, a/2[∪[2a, 3a[,

−^√1_2a, s ∈ [a/2, a[, 0 otherwise

produces h₀ = h₋₄ = h₋₅ = −h₋₁ = ¹₂, while

ˆh(p) =













√1

2a, s ∈ [0, a/2[∪[a, 2a[,

−^√1_2a, s ∈ [a/2, a[, 0 otherwise produces h₀ = h₋₂ = h₋₃ = −h₋₁ = ¹₂.

Remark: The ONC satisfies an useful symmetry property:

if h(s) is a solution of the ONC, another solution is obtained simply by taking the inverse Fourier transform of a ˆg(p) ≡ h(p).

More solutions: the orthonormalization trick

More solutions of the ONC, and more relevant sequences as a consequence, can be found using this ONT:

let h(s) be a generic L² function and let us compute S_l^(h)

1,l₂ =<

ψ_l^(h)

1,l₂, ψ_0,0^(h) >. If S_l^(h)

1,l₂ 6= δ_l1,0δ_l2,0 then the ψ_l^(h)

1,l₂(~r) generate a non o.n. lattice (in L²(R²)). However, [BMS, PRB (1993)], we can find an o.n. set in L²(R²), invariant under magnetic translations, simply by considering

ψ_l^(H)

1,l₂(~r) = T₁^l1T₂^l2ψ^(H)_0,0 (~r), where ψ_0,0^(H)(~r) = X

~n∈Z²

f_~nψ_~n^(h)(~r),

and the coefficients f_~n are fixed by requiring that S_l^(H)

1,l₂ =< ψ_l^(H)

1,l₂, ψ^(H)_0,0 >= δ_l1,0δ_l2,0 In this way we deduce that H(s) = P

~n∈Z² f_~ne^isan1h(s + an₂), and the related coefficients are

H_n =

r2π a Hˆ

³

−na 2

´

= . . . = 1

√a Z

R

ds h(s)

pS^(h)(as, 0)e^iasn/2,

where

S^(h)(~p) = X

~n∈Z²

S_~n^(h)e^i~p·~n.

Then P

n∈Z H_nH_n+2l = δ_l,0.

Examples of this construction are contained in [B., JMP, 2003]

Remarks: (1) if < ψ_l^(h)

1,l₂, ψ_0,0^(h) >= δ_l1,0δ_l2,0 then S^(h)(~p) = 1 and we go back to the original result: H_n = h_n and no o.n. trick is needed!

(2) again the procedure is model independent! It only de- pends on the unitary map from ((ˆx, ˆp_x), (ˆy, ˆp_y)) to ((ˆx₁, ˆp₁), (ˆx₂, ˆp₂)) and not on the particular expressions for H, T₁ and T₂.

The other conditions

• Condition (r2), h_n = O(_1+|n|¹ ₂), n À 1

For this to be satisfied we have a very easy criterion: since h_n =

q2π a ˆh¡

−^na₂ a¢

, it is enough to look for solutions of the ONC such that ˆh(p) is compactly supported.

It would be enough to find solutions h(s) of the ONC regular enough: the more regular h(s) is, the more rapidly h_n goes to zero when n → ∞.

If we adopt the o.n. trick, then we can repeat the same considerations with h(s) replaced by ˜h(s) = √ ^h(s)

S^(h)(as,0).

• Condition (r3), P

n∈Z h_n = √ 2.

A necessary condition for (r3) to hold h(s) must satisfies the ONC and one of the following equalities: P

n∈Zh(na) = q

2

a or P

n∈Z ˆh(^na₂ ) = p_a

π. If we adopt the o.n. trick then we can relax the assumption that h(s) satisfies the ONC. It

is enough to have one of the following equalities satisfied:

X

n∈Z

h(an) = vu utX

~l∈Z²

h

µal₁ 2

¶ h

µal₁

2 + al₂

¶

or

X

n∈Z

ˆh³an 2

´

= vu

ut2 X

~l∈Z²

ˆh

µal₂ 2

¶ ˆh

µal₂

2 + al₂1

¶

(The Haar example above satisfies both!)

• Condition (r4), P

n∈Z h_ne^−iωn 6= 0 for all ω ∈ £

−^π₂, ^π₂¤ . For that we find that the following condition must be satis- fied by the seed function h(s):

X

n∈Z

h

³ a

³

n + ω 2π

´´

6= 0, ω ∈ h

−π 2, π

2 i

whether h(s) satisfies the ONC or not.

Future Plains

1. We have to consider still the examples from the point of view of MRA: what kind of mother wavelets do we get? For that we need to find more examples: this is work in progress!

2. How to impose stronger conditions (e.g. on the support of the mother wavelets or its regularity)?

3. More physical applications

4. Explore the underlying symmetry between (ˆx₁, ˆp₁) and (ˆx₂, ˆp₂):

this produces examples of the Tomita-Takesaki procedure for constructing, e.g., a modular operator. A first concrete application has been explored in [Ali, B., JMP, submitted]

where the FQHE has been analyzed.

5. ...