Zak Transform and non-uniqueness in an extension of Pauli's phase retrieval problem

0.22 MB·English

by  Simon Andreys

Generalized Pauli Problem
Zak Transform and non-uniqueness in an extension of Pauli’s phase retrieval
problem
Simon Andreys & Philippe Jaming1
Institut de Math´ematiques de Bordeaux UMR 5251, Universit´e de Bordeaux, cours de la Lib´eration,
F 33405 Talence cedex, Francea)
(Dated: 19 January 2015)
The aim of this paper is to pursue the investigation of the phase retrieval problem for the fractional Fourier
transform started by the second author. We here extend a method of A.E.J.M Janssen to show that
α
F
there is a countable set such that for everyfinite subset , there exist two functions f,g not multiple
Q A⊂Q
of one an other such that f = g for every α . Equivalently, in quantum mechanics, this result
5 |Fα | |Fα | ∈ A
reformulates as follows: if Q = Qcosα+P sinα (Q,P be the position and momentum observables), then
1 α
0 Qα,α is not informationally complete with respect to pure states.
{ ∈A}
2 This is done by constructing two functions ϕ,ψ such that αϕ and αψ have disjoint support for each
F F
α . To do so, we establish a link between [f], α and the Zak transform Z[f] generalizing the well
n ∈A Fα ∈Q
known marginal properties of Z.
a
J
Keywords: Zak transform; Weyl-Heisenberg transform; Fractional Fourier Transform; Phase Retrieval; Pauli
6
problem
1

]
A
C
I. INTRODUCTION
.
h
t Phase Retrieval Problems arise in many aspects of applied science including optics and quantum mechanics. The
a
m problem consists of reconstructing a function ϕ from phase-less measurements of some transformations of ϕ and
eventualaprioriknowledgeonϕ. Recently,thisfamilyofproblemshasattractedalotofattentioninthemathematical
[
community(seee.g.2,3,14 ortheresearchblog10). Atypicalsuchproblemisthephaseretrievalprobleminopticswhere

1 one wants to reconstruct a compactly supported function ϕ from the modulus of its Fourier transform [ϕ]. An
v other such problem, due to Pauli, is to reconstruct ϕ from its modulus ϕ and that of its Fourier transfo|Frm | [ϕ].
5 | | |F |
The questionshas twosides. Onone hand,one wantstoknowif sucha problemhas aunique solution(upto aglobal
0
phase factor or a more general transformation). On the other hand, one is looking for an algorithm that allows to
9
reconstruct f from phase-less measurements.
3
0 In this paper, we focus on the (non) uniqueness aspects of a generalization of Pauli’s problem in which one has
. phase-lessmeasurementsofFractionalFourierTransform(FrFT)ashasbeenpreviouslystudiedbythesecondauthor7.
1
Recall from11,12 that the FrFT is defined as follows: for α R πZ a parameter that is interpreted as an angle, and
0 ∈ \
5 f L1(R) L2(R),
∈ ∩
1
: f(ξ)=c e−iπ|ξ|2cotα [e−iπ|·|2cotαf](ξ/sinα)
v Fα α F
i
X where c is a normalization constant which ensures that extends to a unitary isomorphism of L2(R). Note that
α α
= . One then defines [ϕ](ξ) = ϕ ( 1)kξ soFthat = . Roughly speaking, the Fractional
r π/2 kπ α β α+β
F F F − F F F
a FourierTransformsareadaptedtothemathematicalexpressionoftheFresneldiffraction,justasthestandardFourier
(cid:0) (cid:1)
transform is adapted to Fraunhofer diffraction. It also occurs in quantum mechanics (see below).
The question we want to address here is the following:
Phase Retrieval Problem for Multiple Fractional Fourier Transforms.
Let (0,π). For f,g L2(R) does [g] = [f] for every α imply that there is a constant c C with
α α
A ⊂ ∈ |F | |F | ∈ A ∈
c =1 such that g =cf.
| |
If this is the case,we will say that Q :α is Informationnaly Complete with respect to Pure States (ICPS).
α
{ ∈A}
The vocabulary comes from quantum mechanics (see e.g.1). For sake of self-containment, let us now recall some
basic notionsfromquantummechanics. The descriptionofa physicalsystemis basedonacomplex separableHilbert
space . For our purpuses, = L2(R). We will denote by = ( ) and = ( ) the bounded and trace class
H H B B H T T H
operators on . The state of the system is then represented by an element ρ of ( ) that is positive ρ 0 and
H T H ≥
a)Electronicmail: [email protected]
Generalized Pauli Problem 2
normalized by tr[ρ]=1. The states form a convex set whose extreme points, the pure states are the one dimensional
projections.
The observables are normalized Positive Operator Valued Measure E : (R) ( ) where (R) is the Borel set
B →L H B
R, that is, maps satisfying E[X] 0, E[R]=I, and, for every ρ ( ), for every sequence (X ) of pairwise disjoint
j
sets in (R), tr[ρE( X )] = ≥tr[ρE(X )]. It follows that, fo∈r TanyHstate ρ, ρE = tr[ρE] is a probability measure
B j j j j
on R. The number ρE(X) is interpreted as the probability that the measurement of E gives anoutcome from the set
S P
X when the system is initially prepared in the state ρ.
The problem we are concerned here is thus following:
Definition. Let be a collection of observables (R) ( ). Then
A B →L H
– is informationally complete if, for any ρ ,ρ ( ), ρE =ρE for all E implies ρ =ρ ;
A 1 2 ∈T H 1 2 ∈A 1 2
– is informationally complete with respect to pure states if, for any pure states ρ ,ρ , ρE = ρE for all E
A 1 2 1 2 ∈ A
implies ρ =ρ .
1 2
Now let Q,P be the position and momentum observables. For any α [0,2π], define Q (X) = Q =
α −α α
∈ F F
Qcosα+P sinα. For a system in a pure state ρ= ψ ψ , the measurement outcome probabilities are given by
| ih |
ρQα(X)= ψ,Q (X)ψ = ψ(x)2dx.
α α
h i |F |
ZX
This shows that the probability density associatedto a measurementof the observableQ preformedon a pure state
α
ψ ψ is just the intensity ψ(x)2.
α
| ih | |F |
In this setting, Pauli conjectered that 0,π/2 is informationnaly complete with respect to pure states. However,
{ }
thiswassoondisprovedandtherearenowafairlylargenumberofcounterexamples(seee.g.6,4andreferencestherein).
In13,A.VogtmentionedaconjecturebyWrightthatthereexistsathirdobservablessuchthatthisobservabletogether
withpositionandmomentumformsaninformationnalycompletewithrespecttopure statesfamily. Thisamountsto
lookingforaunitaryoperatorU :L2(R) L2(R)suchthat,ifwesetU =I, U = then U f = U g ifandonly
3 1 2 j j
→ F | | | |
if f = cg with c = 1. In7, the second author investigated the above question further. In particular, when α / Qπ
none of the cla|ss|es of counterexamples mentioned in4,6 allowed to show that := Q(θ),θ = 0,α,π/2 woul∈d not
A { }
be ICPS. This lead naturally to conjecture that was ICPS, which was recently disproved by Carmeli, Heinosaari,
Schultz & A. Toigo in1. Actually, those authors eAxhibit an invariance property which shows that the ICPS property
of Q(α ),Q(α ),Q(α ) does notdepend onα ,α ,α providedthey aredifferent. Then they extenda construction
1 2 3 1 2 3
of7{to show that Q ,α} F is not ICPS when F Qπ is finite.
α
{ ∈ } ⊂
The aimofthis paper is to providea somewhatsimilar resultfor an other classof parameterswhich is basedonan
adaptation of an unnoticed counterexample to Pauli’s problem due to A.E.J.M. Janssen9:
Main Theorem.
Let F Q be a finite set of rational numbers. Define by
⊂ A
= α (0,π) : there exists r F with cotα=r .
A { ∈ ∈ }
Then the collection Q ,α is not Informationnaly Complete with respect to Pure States.
α
{ ∈A}
This is based on the fact that a function and its Fourier transform f are recovered from its Zak transform (also
known as Weyl-Heisenberg tranform)5,15
Z[f](x,ξ)= f(x+k)e−2iπkξ.
k∈Z
X
byintegratingalonghorizontalandverticallines. Weadaptthispropertybyshowingthat,whencotα Q,then f
α
∈ F
can also be recoveredfrom Z[f] by integrating in an oblique direction. We believe this result is interesting in itself.
Once this is done, one can construct two functions f and f such that for each α ( defined in the Main
1 2
∈ A A
Theorem), [f ] and [f ] have disjoint support. Therefore, for any c C with c = 1, and any α ,
α 1 α 2
F F ∈ | | ∈ A
[f +cf ] = [f ] + [f ] does not depend on c, but f +cf is not a constant multiple of f +f unless
α 1 2 α 1 α 2 1 2 1 2
|F | |F | |F |
c=1 thus proving the main theorem.
We complete the paper by showing that the approximate phase retrievalproblem for the FrFT has infinitely many
solutions that are far from multiples from one an other:
The Approximate Phase Retrieval Problem for the Fractional Fourier Transform.
Leq 0 α < < α π and T > 0. Let f ,...f L∞(R) with suppf [ T,T] and f 0. Then, for every
≤ 1 ··· n ≤ 2 1 n ∈ j ⊂ − j ≥
Generalized Pauli Problem 3
n
ε>0, there exists ϕ ,...,ϕ L2(R) such that, for c ,...,c C with c = = c =1, ϕ= c ϕ satisfies
1 n 1 n 1 n j j
∈ ∈ | | ··· | |
j=1
X
[ϕ] f ε for k=1,...,n. (I.1)
k|Fαk |− kkL∞([−T,T]) ≤
The remaining of the paper is organized as follows. In the next section, we first make precise definitions of the
FrFT and of the Zak Transform and then show the oblique marginal property of the Zak transform in Theorem II.2.
Section IIB is then devoted to the proof of the main theorem. The last section is devoted to proving our result on
the approximate phase retrieval problem.
II. THE MAIN THEOREM
A. The Zak Transform and the Fractional Fourier Transform
Let α / πZ, the Fractional Fourier Transform of order α of f L1(R) is defined as
∈ ∈
f(ξ)=c e−iπ|ξ|2cotα [e−iπ|·|2cotαf](ξ/sinα)
α α
F F
=cαe−iπcotα ξ2 e−iπ/2cotα x2−2iπsixnξαf(x) dx
R
Z
where c = exp2i(α−sgn(α)π4) is such that c2 = 1 icotα and e c > 0. For α = kπ, k Z, we set g(ξ) =
α √|sinα| α − R α ∈ Fα
g ( 1)kξ .
−
Notice that π = . The FrFT further has the following properties
(cid:0) (cid:1) F2 F
1. For u C let γ (t)=e−uπt2 then [f](ξ)=c γ (ξ) [γ f]( ξ/sinα);
u α α icotα icotα
∈ F F −
2. for every α R, and f L1(R) L2(R), f = f thus f extends into a unitary operator on
∈ ∈ ∩ kFα kL2(R) k kL2(R)
L2(R);
3. = .
α β α+β
F ◦F F
The Zak transform is defined, for f ß(R) (the Schwarz class) as
∈
Zf(x,ξ)= f(x+k)e−2iπkξ.
k∈Z
X
The Zak transform has the following properties
1. Z extends into a unitary operator L2(R) L2(Q) where Q=[0,1] [0,1].
→ ×
2. Zf(x+n,ξ)=e2iπnξZf(x,ξ) and Zf(x,ξ+n)=Zf(x,ξ).
3. Poissonsummation: Zf(x,ξ)=e2iπxξZfˆ(ξ, x)
−
4. Z has the following marginal properties:
1 1
f(x)= Zf(x,ξ) dξ and f(ξ)= e−2iπxξZf(x,ξ) dx.
F
Z0 Z0
WecannowprovethattheZaktransformhasalsomarginalpropertiesinobliquedirectionslinkedtotheFractional
Fourier Transform. Not surprisingly, such a marginal property only occurs when the slope is rational, as otherwise
one would integrate over a dense subset of Q.
In order to establish such a property we will first need to compute the Zak transform of a chirp. This has to be
done in the sense of distributions:
Generalized Pauli Problem 4
Proposition II.1 (Zak transform of a chirp)
Let p Z, q N be relatively prime integers. For n Z, let
∈ ∈ ∈
q−1
cn,p,q = 1 ( 1)kpeiπpqk2e−2iπnqk
q −
k=0
X
and, for x [0,1], n Z, write ξ (x)= px+ 1p+ n and A (x)= n Z 0 ξ (x) 1 .
∈ ∈ n,p,q q 2 q p,q { ∈ | ≤ n,p,q ≤ }
For f ß(R),
∈
1
f(t)e−ipqt2dt= e−iπpqx2 cn,p,q Zf x,ξn,p,q(x) dx. (II.2)
ZR Z0 n∈AXp,q(x) (cid:0) (cid:1)
Remark. This formula is stated in8 without proof in the following form
p 1 n
Zγ (x,ξ)=γ c δ ξ x p (II.3)
−ip/q −ip/q n,p,q − q − 2 − q
n∈Z (cid:18) (cid:19)
X
whereδ isthe Diracdelta function. Wheninterpretedinthe senseoftempereddistributions, this formulais precisely
(II.2). Our first aim here is to provide a rigorous proof of this formula.
Proof. Define
N′
ZN,N′γ−ip/q(x,ξ)= γ−ip/q(x+k)e−2iπkξ.
k=−N
X
Let f ß(R). Note that the series defining the Zak transform of f is uniformly convergent, thus
∈
1 1 N′
Zf,ZN,N′γ−ip/q L2(Q) = f(x+j)γip/q(x+k)e−2iπjξe2iπkξdξdx
Z0 Z0 j∈Zk=−N
(cid:10) (cid:11) X X
N′ 1 1
= f(x+j)γ (x+k) e2iπ(k−j)ξdξdx
ip/q
k=−NZ0 j∈Z Z0
X X
N′ 1
= γ (x+k)f(x+k)dx
ip/q
k=−NZ0
X
N′+1
= f(t)γ (t)dt.
ip/q
Z−N
It follows that
N,Nli′→m+∞ Zf,ZN,N′γ−ip/q L2(Q) = Rf(t)γip/q(t)dt. (II.4)
Z
(cid:10) (cid:11)
On the other hand,
N′
ZN,N′γ−ip/q(x,ξ)= eiπpq(x+j)2−2iπjξ
j=−N
X
N′
=eiπpqx2 eiπpqj2e2iπj(pqx−ξ).
j=−N
X
We then splitthis sumaccordingto the value ofj modulo 2q. Inordertodo so,note that, ifj =2ql+k, πpj2 = pk2
q q
(mod 2π) so that eiπpqj2 =eiπpqk2.
Generalized Pauli Problem 5
Withoutlossofgenerality,wewillnowassumethatN =2qM andN′ =2q(M+1) 1. Then,everyj N,...,N′
− ∈{− }
decomposes uniquely as j =2qℓ+k with ℓ M,...,M and k 0,...,2q 1 . Therefore
∈{− } ∈{ − }
M 2q−1
ZN,N′γ−ip/q(x,ξ)=eiπpqx2 eiπpqk2e2iπ(2qℓ+k)(pqx−ξ)
ℓ=−M k=0
X X
2q−1 M
=eiπpqx2 eiπ(pqk2+2k(pqx−ξ)) e4iπℓ(px−qξ).
k=0 ℓ=−M
X X
We are thus lead to introduce the Dirichlet kernel
M
D (u)= e2iπu
M
ℓ=−M
X
and
2q−1
P(x,ξ)=e−iπpqx2 e−iπ(pqk2+2k(pqx−ξ)).
k=0
X
so that
ZN,N′γ−ip/q(x,ξ)=P(x,ξ),DM 2(px qξ)
−
We have thus established that, if N =2qM, N′ =2q(M +1) 1 a(cid:0)nd f ß(R(cid:1)), then
− ∈
Zf,ZN,N′γip/q L2(Q) = Zf(x,ξ)ZNγip/q(x,ξ)dxdξ
ZZ[0,1]2
(cid:10) (cid:11)
= Zf(x,ξ)P(x,ξ)D 2(px qξ) dxdξ (II.5)
M
−
ZZ[0,1]2
(cid:0) (cid:1)
Write ϕ(x,ξ)=Zf(x,ξ)P(x,ξ). Note that, as f ß(R), ϕ ∞(R2). We now want to evaluate the limit of
∈ ∈C
ϕ(x,ξ)D (2(px qξ))dxdξ
M
−
ZZQ
when M . Let us first compute the integral with respect to ξ:
→∞
1 1 2q ζ
ϕ(x,ξ)D (2(px qξ))dξ = ϕ x, D (2px ζ) ,dζ
M M
− 2q 2q −
Z0 Z0 (cid:18) (cid:19)
2q−1 1 ℓ+1 ζ
= ϕ x, D (2px ζ)dζ
M
2q 2q −
ℓ=0 Zℓ (cid:18) (cid:19)
X
2q−1 1 1 ζ+ℓ
= ϕ x, D (2px ζ)dζ.
M
2q 2q −
ℓ=0 Z0 (cid:18) (cid:19)
X
As D is 1-periodic, we may remplace 2px by 2px 2px , thus
M
−⌊ ⌋
1 2q−1 1 1 ζ+ℓ
ϕ(x,ξ)D (2(px qξ))dξ = ϕ x, D (2px 2px ζ)dζ
M M
− 2q 2q −⌊ ⌋−
Z0 ℓ=0 Z0 (cid:18) (cid:19)
X
2q−1
1 ζ+ℓ
= ϕ x, D (2px 2px )
ζ M
2q ℓ=0 (cid:18) 2q (cid:19)!∗ −⌊ ⌋
X
where is the (circular) convolution on [0,1] and where we have used the parity of D .
M
∗
2q−1
1 ζ+ℓ
Butnow,foreveryx,defineψ = ϕ x, andnotethatψ isasmooth1-periodicfunction. Moreover,
x x
2q 2q
ℓ=0 (cid:18) (cid:19)
X
ψ D is the partial sum of order M of the Fourier series of ψ . Therefore ψ D ψ uniformly, in particular
x M x x M x
∗ ∗ →
Generalized Pauli Problem 6
at the point 2px 2px . It follows that
−⌊ ⌋
1 2q−1 1 2px 2px +l
ϕ(x,ξ)D (2(px qξ))dξ ϕ x, −⌊ ⌋
M
Z0 − M−→→+∞ l=0 2q (cid:18) 2q (cid:19)
X
2q−1−⌊2px⌋
1 2px+l
= ϕ x, .
2q 2q
l=−X⌊2px⌋ (cid:18) (cid:19)
Replacing ϕ by ZfP and using (II.5) we have thus established that
12q−1−⌊2px⌋ 1 2px+l 2px+l
Zf,Z γ Zf x, P x, dx. (II.6)
(cid:10) N −ip/q(cid:11)L2(Q) M−→→+∞Z0 l=−X⌊2px⌋ 2q (cid:18) 2q (cid:19) (cid:18) 2q (cid:19)
We will now reshape this formula. Let us first simplify P:
2q−1 q−1 2q−1
eiπpqx2P x,2px−ℓ = eiπ(pqk2−kqℓ) = + eiπ(pqk2−kqℓ)
2q
(cid:18) (cid:19) k=0 k=0 k=q
X X X
q−1
= eiπ(pqk2−kqℓ)+eiπ(pq(k+q)2−(k+qq)kℓ)
Xk=0(cid:16) (cid:17)
q−1
= eiπ(pqk2−kqℓ) 1+eiπ(pq−l) .
Xk=0 (cid:16) (cid:17)
As
2 if l =pq (mod 2)
1+eiπ(pq−l) =1+( 1)pq+l = ,
− (0 otherwise
and, writing l pq=2n, we have pk2 kℓ = pk2 2nk kp, we deduce that
− q − q q − q −
q−1
P x,2px−ℓ =e−iπpqx2 2( 1)kpeiπ(pqk2−2nqk) =2qcn,p,qe−iπpqx2
2q −
(cid:18) (cid:19) k=0
X
Finally, the condition 2px 6k 62q 1 2px is equivalentto ξ (x):= px+ l [0,1]. Therefore (II.6)reads
−⌊ ⌋ − −⌊ ⌋ n,p,q q 2 ∈
1
(cid:10)Zf,ZNγ−ip/q(cid:11)L2(Q) M−→→+∞Z0 e−iπpqx2n∈AXp,q(x)cn,p,q Zf(cid:0)x,ξn,p,q(x)(cid:1)dx. (II.7)
It remains to compare this equation with (II.4) to obtain that, for f ß(R),
∈
1
f(t)e−ipqt2dt= e−iπpqx2 cn,p,q Zf(x,ξn,p,q(x)) dx,
ZR Z0 n∈AXp,q(x)
as announced.
We can now prove our main theorem
Theorem II.2
Let p Z 0 , q N be relatively prime integers and let α be defined by cotα = p. For n Z, and x [0,1], let
∈ \{ } ∈ q ∈ ∈
c , ξ and A (x) be defined as in Proposition II.1. Then, for every f L2(R),
n,p,q n,p,q p,q
∈
αf(ξ)=cαe−iπpqξ2 1e−2iπsinξαx−iπpqx2 cn,p,q Zf x, ξ +ξn,p,q(x) dx. (II.8)
F sinα
Z0 n∈AXp,q(x) (cid:18) (cid:19)
The identity has to be taken in the L2(R) sense.
Generalized Pauli Problem 7
Note that sin2α = cot2αcos2α = (1 sin2α)p2 thus sin2α = p2 and as sgnsinα = sgncotα = sgnp,
− q2 p2+q2
sinα= p so that (II.8) can easily be written in terms of p,q only.
√p2+q2
Proof.Letusnowfixω Randf ß(R)anddefinef asf (t)=f(t)e−2iπωt. ThenZf (x,ξ)=Zf(x,ξ+ω)e−2iπωx.
ω ω ω
∈ ∈
Thus, applying (II.2) to f leads to
ω
1
f(t)e−ipqt2e−2iπωtdt= e−2iπωx−iπpqx2 cn,p,q Zf(x,ω+ξn,p,q(x)) dx. (II.9)
ZR Z0 n∈AXp,q(x)
But
αf(ξ)=cαe−iπpqξ2 f(t)e−ipqt2−2iπξt/sinαdt
F R
Z
so that (II.9) reads
αf(ξ)=cαe−iπpqξ2 1e−2iπsinξαx−iπpqx2 cn,p,q Zf x, ξ +ξn,p,q(x) dx. (II.10)
F sinα
Z0 n∈AXp,q(x) (cid:18) (cid:19)
It remains to extend this identity from ß(R) to L2(R). For sake of simplicity, we will assume that p 0.
As is continuous from L2(R) L2(R), it is enough to check that the right hand side of (II.10) is≥continuous as
α
F →
well.
Tostart,multiplicationbyboundedfunctionsanddilationsarecontinuousfromL2(R) L2(R)sothatitisenough
→
to check that the functional F defined by
1
F[f](ξ)= e−2iπξx−iπpqx2 cn,p,q Zf(x,ξ+ξn,p,q(x)) dx
Z0 n∈AXp,q(x)
is continuous from L2(R) L2(R).
→
Let us first re-write this functional. To do so, note that n A (x) if and only if pq/2 p p(x+q/2) n
p,q
∈ − − ≤− ≤ ≤
p(x+q/2)+q pq/2+q. Let =Z [pq/2 p,pq/2+q] then, for each n there is a subset B (n) of
p,q p,q p,q
− ≤− F ∩ − ∈F
[0,1] such that n A (x) if and only if x B (n). Then F[f]= c F [f] where
p,q p,q n,p,q n
∈ ∈
n∈XFp,q
Fn[f](ω)= e−2iπωxe−iπpqx2Zf(x,ω+ξn,p,q(x)) dx.
ZBp,q(n)
As is finite, it is enoughto showthat eachF is continuous. Further, if ω =η+k with k Z and η [0,1)then
p,q n
F ∈ ∈
Fn[f](η+k)= e−2iπηxe−iπpqx2Zf(x,η+ξn,p,q(x))e−ikxdx
ZBp,q(n)
1
= (x,η)e−ikxdx
Z
Z0
with
Z(x,η)=1Bp,q(n)(x)e−iπpqx2Zf(x,η+ξn,p,q(x)).
Thus
1
F 2 = F [f](η+k)2dη
k nkL2(R) | n |
Z0 k∈Z
X
1 1
= (x,η)2dxdη
|Z |
Z0 Z0
with Plancherel.
It remains to notice that, for any fixed compact set A R2, f Z[f] is bounded L2(R) L2(A) thus f
Zf(x,η+ξ (x)) is bounded L2(R) L2(Q) and so is f⊂ . → → →
n,p,q
→ →Z
Generalized Pauli Problem 8
✟ ✟
✟ ✟
✟ ✟
✟ ✟
✟ ✟
✟ ✟
✟ ✟
J ✟✟✟✟✟✟✟✟✟⑦✟✟✟✟✟✟✟✟✟✟✟✟✟⑦✟✟✟✟✟✟✟✟✟⑦✟✟✟✟✟✟⑦✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟②
J ✟✟ ✟✟
✟ ✟
✟ ✟
✟ ✟
✟ ✟
FIG. 1. In this figure p=2,q =1. The support of Z[f] is the union of the black discs. Then the set J is such that, for each
η J,the(periodic)linestartingat(0,η)withslopeq/p=1/2(cotanα=p/q)intersectsatleastoncethesupportofZf (one
∈
suchlineisdrawndotted). HereJ =[0,1/4] [1/2,3/4]Finally,thesupportof α[f]isthenincludedinthesetofallξsuchthat
ξ/sinα+p/2∈J+Z=Sk∈Z[k/2,(k+1/2∪)/2]. As p=2,q=1, sinα=2/√5F, thussuppFα[f]⊂Sk∈Z[k/√5,(k+1/2)/√5].
B. Application to the generalized Pauli problem
Fix p Z, q N relatively prime and α given by cotα= p.
∈ ∈ q
p
Let T =R/Z. We will identify T with [0,1). Consider the line Γ (ξ) of slope through the point 0,p + ξ
p,q q 2 sinα
in T2. Note that this line has finite length p2+q2 since its slope is rational. This wellknownfactcan(cid:16)be proveda(cid:17)s
follows: first the length does not depend on the starting point so we may assume that p + ξ =0. Then the line is
p 2 sinα
the periodization of the segment starting at (0,0) of slope p/q reaching the first point with integer coordinates, that
is (p,q).
Let π : T2 T be the projection on the second coordinate π(x,v) = v where we identify T = [0,1). For
→
(x,ω) Γ (ξ), let n(x,ω) Z be the number defined as follows: ω = x+ p + ξ +n(x,ω). Then we may write
∈ p,q ∈ 2 sinα
(II.8) as
αf(ξ)= qcα e−iπ(cid:16)pq+psinq2α(cid:17)ξ2 e−iπpq(π(u)−p2)2cn(u),p,q Zf(u)du.
F p2+q2 ZΓp,q(ξ)
In particular, this identity allopws to relate the support of f to the support of Z[f]:
α
F
supp f ξ :Γ (ξ) suppZf = . (II.11)
α p,q
F ⊂{ ∩ 6 ∅}
We can now extend the idea of Janssen9 to provide counterexamples for the generalized Pauli problem.
Corollary II.3 Let r ,...,r Q be m different rational numbers and N N. For j = 1,...,m let α [0,π) be
1 m j
defined via cotα =r . Then th∈ere exists f ,...,f L2(R) such that, for j∈=1,...,m and k =ℓ=1,...∈,N,
j j 1 N
∈ 6
supp f supp f = . (II.12)
Fαj k∩ Fαj ℓ ∅
In particular, for every j =1,...,m and for every c ,...,c C with c =1,
1 N k
∈ | |
m m
c f = f (II.13)
(cid:12)Fαj" k k#(cid:12) (cid:12)Fαj" k#(cid:12)
(cid:12) kX=1 (cid:12) (cid:12) Xk=1 (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
Generalized Pauli Problem 9
so that Q ,j =1,...,m is not informationnaly complete with respect to pure states.
{ αj }
Proof. Of course (II.13) follows directly from (II.12) since then
m
c [f ](ξ) if ξ supp [f ]
c f (ξ)= kFαj k ∈ Fαj k .
Fαj" k k# (0 otherwise
k=1
X
Let us now prove (II.12). Write each r = p /q as an irreducible fraction with q > 0. Let L = p2+q2 and
j j j j j j j
L= Lj. q
The proof is based on the following observation: Let I [0,1) be an open interval, r = p/q Q and Γ (I) =
P ⊂ ∈ pj,qj
Γ (ξ). NotethatΓ (I)hasareaL I . Notethat,ifJ [0,1)iaanintervalsuchthat( 0 J) Γ (I)=
ξ∈I pj,qj pj,qj j| | ⊂ { }× ∩ pj,qj
then Γ (I) Γ (J)= . Moreover,the existence of such a J is guarantied when Γ (I)=T2.
S∅ pj,qj ∩ pj,qj ∅ pj,qj 6
Further Γ (I) has area L I = LI . If LI < 1 then Γ (I) = T2. Therefore there exists
j pj,qj ≤ j j| | | | | | j pj,qj 6
J (0,1) such that J Γ (I)= and therefore Γ (I) Γ (J)= . Write I =I and I =J. If further
⊂ S ∩ j pj,qj P ∅ pj,qj ∩ pj,qj S ∅ 1 2
L(I + I )<1then Γ (I ) Γ (I )=T2. onecanthusfindI [0,1)suchthatΓ (I ) Γ (I )=
| 1| | 2| jSpj,qj 1 ∪ j pj,qj 2 6 3 ⊂ pj,qj k ∩ pj,qj ℓ
if k =ℓ.
∅ 6 S S
In general, we are thus able to find open intervals I ,...,I (0,1) such that
1 m
⊂
Γ (I ) Γ (I )= if k =ℓ. (II.14)
pj,qj k ∩ pj,qj ℓ ∅ 6
Next, note that K := Γ (I ) has non-empty interior as it contains a neighborhood of 0 I . Let Z be any
k pj,qj k { }× k k
function in L2(Q) and let f L2(R) be given by Z[f ]=Z . Then (II.11) and (II.14) imply (II.12).
k k k
T ∈
III. APPROXIMATE PAULI PROBLEM
Theorem III.1 Leq 0 α < < α π and T > 0. Let f ,...f L∞(R) with suppf [ T,T] and f 0.
≤ 1 ··· n ≤ 2 1 n ∈ j ⊂ − j ≥
Then, for every ε>0, there exists a function ϕ L2(R) such that, for every k =1,...,n,
∈
[ϕ] f ε.
k|Fαk |− kkL∞([−T,T]) ≤
Proof. It is enough to show that, given ε > 0, for every j = k 1 n , there exists ϕ L2(R) such that
k
6 ∈ { ··· } ∈
[ϕ ]=f on [ T,T] and [ϕ ] 6ε/(n 1). Taking ϕ= ϕ then gives the result.
Fαk k k − Fαj k L∞([−T,T]) − k
Without loss of generality, it is enough to construct ϕ .
(cid:13) (cid:13) 1 P
Let ω R be a parameter t(cid:13)hat we w(cid:13)ill fix later. As f L1(R), we may define h = f (t)e2iπωt . Note that
∈ 1 ∈ ω F−α1 1
[h ] =f on [ T,T]. Moreover,for k =2,...,n, [h ]= f (t)e2iπωt .
|Fα1 ω | 1 − Fαk ω Fαk−α1 1 (cid:2) (cid:3)
Let α˜ =α α . Then, for ξ R,
k k 1
− ∈ (cid:2) (cid:3)
Fα˜k f1(t)eiωt (ξ) =(cid:12)cαZRf1(x)e−iπcotαx2e2iπx(ω−sinξα)dx(cid:12)=(cid:12)ZRuk(x)e2iπx(ω−sinξα)dx(cid:12)=(cid:12)uk(cid:18)sinξα˜k −ω(cid:19)(cid:12)
(cid:12) (cid:2) (cid:3) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
wh(cid:12)ere u (x) = f (x)(cid:12)e−iπ(cid:12)cotαx2 is an L1(R) function. Accord(cid:12)ing(cid:12)to the Riemann-Lebesgu(cid:12)e Le(cid:12)cmma, u (η) 0(cid:12) as
k 1 (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) k (cid:12)
ε →
η . Thus, there exists A such that, if η > A, u (η) < . But then, if ω A+T/sinα˜ and ξ < T,
k k
→ ±∞ | | n 1 ≥ | |
ε − c
ω ξ >A thus [h ](ξ) < . It remains to chose ω =A+T/sinα˜ >A+T/sinα˜ and ϕ =h .
− sinα˜k |Fαk ω | n 1 c 2 k 1 ω
−
Remark. The proof actually shows a bit more, namely that there are infinitely many solutions to the problem that
are not constant multiples of one an other. Indeed, one could as well take any ϕ= c ϕ with c =1.
k k k
| |
However,aninspectionoftheproofofthetheoremshowsthatthefunctionwebuildhasahugesupport(ifα =0).
1
P
Thus, if one imposes additional support constraint (as would be imposed in real life applications), the theorem may
no longer be valid.
Appendix A: The modulus of the coefficients cn,p,q
The factor c appearing in Proposition II.1 is similar to a Gauss sum. As for Gauss sums, it is possible to
n,p,q
compute their modulus.
Generalized Pauli Problem 10
Lemma A.1 Let p Z, q N be relatively prime integers. Fon n Z, let
∈ ∈ ∈
q−1
cn,p,q = 1 ( 1)kpeiπpqk2e−2iπnqk.
q −
k=0
X
1
Then c = .
n,p,q
| | √q
Proof. Indeed, setting ω =exp(2iπ/q), we get
q2 cn,p,q 2 = ( 1)p(k−l)wn(l−k)+p2(k2−l2) = ( 1)psωns ωp2(k2−(k−s)2)
| | − −
06l,k<q −q<s<q 0≤k,l<q:k−l=s
X X X
= ( 1)psωns−p2s2 ωskp.
−
−q<s<q k−l=s,06k,l<q
X X
Now note that, for s 0 the second sum is over k =s ...,q 1 while for s<0, this sum is over k=0,...,q 1+s.
≥ − −
Finally, for s=0 this sum is just =q. The sum thus splits into three parts
s=−1 q−1+s q q−1
q2 cn,p,q 2 = ( 1)psωns−p2s2 ωskp+q+ ( 1)psωns−p2s2 ωskp
| | − −
s=−q k=0 s=1 k=s
X X X X
q s−1 q q−1
= ( 1)ps−pqωns−nq−p2s2+pqs−p2q2 ω(s−q)kp+q+ ( 1)psωns−p2s2 ωskp
− −
s=1 k=0 s=1 k=s
X X X X
changing s s q in the first sum. But ωq =1 and ωp2q2 =( 1)pq thus
→ − −
q s−1 q q−1
q2 cn,p,q 2 = ( 1)psωns−p2s2−p2q2 ωskp+q+ ( 1)psωns−p2s2 ωskp
| | − −
s=1 k=0 s=1 k=s
X X X X
q q−1
=q+ ( 1)psωns−p2s2 ωskp.
−
s=1 k=0
X X
q−1
Finally,itremainstonoticethat ωskp =0unlessq dividessp. Asqandparerelativelyprimeands 1, ,q 1,
∈ ··· −
k=0
X
this can not happen, thus q2 c 2 =q.
n,p,q
| |
ACKNOWLEDGEMENTS
The first author kindly acknowledge financial support from the French ANR programs ANR 2011 BS01 007 01
(GeMeCod), ANR-12-BS01-0001 (Aventures). This study has been carried out with financial support from the
FrenchState,managedbytheFrenchNationalResearchAgency(ANR)intheframeoftheInvestmentsforthefuture
Programme IdEx Bordeaux - CPU (ANR-10-IDEX-03-02).
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