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arXiv:math-ph/0509027v1 13 Sep 2005
Quadratic Quantum Hamiltonians revisited
Monique Combescure
IPNL, Bˆatiment Paul Dirac
4 rue Enrico Fermi,Universit´e Lyon-1
F.69622 VILLEURBANNE Cedex, France
monique.combescure@ipnl.in2p3.fr
Didier Robert
D´epartement de Math´ematiques
Laboratoire Jean Leray, CNRS-UMR 6629
Universit´e de Nantes, 2 rue de la Houssini`ere,
F-44322 NANTES Cedex 03, France
Didier.Robert@math.univ-nantes.fr
Abstract
Time dependent quadratic Hamiltonians are well known as well in classi-
cal mechanics and in quantum mechanics. In particular for them the corre-
spondance between classical and quantum mechanics is exact. But explicit
formulas are non trivial (like the Mehler formula). Moreover, a good knowlege
of quadratic Hamiltonians is very useful in the study of more general quan-
tum Hamiltonians and associated Schr¨odinger equations in the semiclassical
regime.
Our goal here is to give our own presentation of this important subject. We
put emphasis on computations with Gaussian coherent states. Our main mo-
tivation to do that is application concerning revivals and Loschmidt echo.
1
2
1 Introduction
This paper is a survey concerning exact useful formulas for time dependent Schr¨odinger
equations with quadratic Hamiltonians in the phase space. One of our motivations
is to give a detailed proof for the computation of the Weyl symbol of the propagator.
This formula was used recently by Melhig-Wilkinson [17] to suggest a simpler proof
of the Gutzwiller trace formula [3].
There exist many papers concerning quantum quadratic Hamiltonians and exact
formulas. In 1926, Schr¨odinger [21] has already remarked that quantification of the
harmonic (or Planck) oscillator is exact.
The best known result in this field is certainly the Melher formula for the harmonic
oscillator (see for example [5]).
Quadratic Hamiltonians are very important in partial differential equations on
one side because they give non trivial examples of wave propagation phenomena and
in quantum mechanics and on the otherside the propagation of coherent states by
general classes of Hamiltonians, including −~2△+V, can be approximate modulo
O(~∞) by evolutions of quadratic time dependent Hamiltonians [1, 19, 14].
In his works on pseudodifferential calculus, A. Unterberger in [22, 23] has given
several explicit formulas connecting harmonic oscillators, Gaussian functions and
the symplectic group. This subject was also studied in [8, 13, 12]. More recently de
Gosson [9] has given a different approch for a rigorous proof of the Melhig-Wilkinson
formula for metaplectic operators, using his previous works on symplectic geometry
and the metaplectic group.
Here we shall emphasis on time dependent quadratic Schr¨odinger equation and Gaus-
sian Coherent States. It is well known and clear that this approach is in the heart
of the subject and was more or less present in all papers on quantum quadratic
Hamiltonians. In this survey, we want to give our own presentation of the subject
and cover most of results appearing in particular [13, 7].
Our main motivation to revisit this subject was to prepare useful tools for applica-
tions to revivals and quantum Loschmidt echo [2]. We shall see that in our approach
computations are rather natural direct and explicit.
2 Weyl quantization. Facts and Notations
Let us first recall some well known facts concerning Weyl quantization (for more
details see [11, 18]).
The Planck constant ~>0 is fixed (it is enough to assume ~= 1 in the homogeneous
quadratic case).
The Weyl quantization is a continuous linear map, denoted by ˆ
•or by Opw, de-
fined on the temperate Schwartz space distribution S′(R2n) into Lw(S(Rn),S′(Rn))
where Lw(E, F ) denotes the linear space of continuous linear map from the linear
topological Einto the linear topological Fwith the weak topology.
R2nis a symplectic linear space with the canonical symplectic form σ(X, Y ) = J X·Y
3
where Jis defined by
J=0 1ln
−1ln0
and 1lndenotes the identity n×nmatrix.
Let us introduce the symplectic group Sp(2n): it is the set of linear transformations
of R2npreserving the 2-form σ.
For X∈R2n, we denote X= (q, p)∈Rn×Rn.
The Weyl quantization is uniquely determined by the following conditions:
(W0) A7→ ˆ
Ais continuous.
(W1) (ˆqψ)(q) = qψ(q),(ˆpψ)(q) = Dqψ(q),(Dq=∇q
i)
for every ψ∈ S(Rn).
(W2) exp[i(\
α·q+β·p)] = exp[i(α·ˆq+β·ˆp)],∀α, β ∈Rn
Let us remark that α·ˆq+β·ˆpis self-adjoint on L2(Rn) so, exp[i(α·ˆq+β·ˆp)]
is unitary. In particular, if z= (x, ξ) then
ˆ
T(z) =: exp[ i
~(ξ·ˆq−x·ˆp)
is the quantized translation by zin the phase space (Weyl operators).
From (W1), (W2), using continuity and ~-Fourier transform, defined by
˜
A(Y) = ZR2n
e−i
~X·YA(X)dX
we have
ˆ
A= (2π~)−2nZR2n
˜
A(α, β) exp[ i
~(α·ˆq+β·ˆp)]dαdβ (2.1)
In general, equality (2.1) is only defined in a weak sense i.e through the duality
bracket between S′and S,<ˆ
Aϕ, ψ > for arbitrary ϕ, ψ ∈ S(Rn).
Definition 2.1 Ais the Weyl contravariant symbol of ˆ
Aif they are related through
the formula (2.1).
Using the explicit action of Weyl operators on S(Rn) and Fourier analysis, we
get the following formula (see [11, 18])
(ˆ
Aϕ)(x) = (2π~)−nZZR2n
exp[ i
~(x−y)·ξ]Ax+y
2, ξϕ(y)dydξ. (2.2)
4
In particular, if KAdenotes the Schwartz kernel of ˆ
A, we have
A(q, p) = ZRn
exp[−i
~u·p]KAq+u
2, q −u
2du (2.3)
These formulas are true in the distribution sense in general, and pointwise if ˆ
Ais
smoothing enough ( surely if for example A∈ S(Rn)).
Let us remark that they are consistent and that A7→ ˆ
Ais a bijection from S′(R2n)
into Lw(S(R2n),S′(R2n)). In particular we have the following inversion formula
Proposition 2.2 For every ˆ
A∈ Lw(S(R2n),S′(R2n)), there exists a unique con-
travariant Weyl symbol A∈ S′(R2n)given by the following formula
A(X) = 2nTr[ ˆ
ASym(X)] (2.4)
where Sym(X)is the unitary operator in L2(Rn)defined by
Sym(X)ϕ(q) = (π~)−ne−2iξ
~(x−q)ϕ(2x−q)
for X= (x, ξ).
Sketch of proof:
We first prove the formula for ˆ
A∈ Lw(S′(R2n),S(R2n)).The general case follows
by duality and density.
We start with the following formula, easy to prove if A, B ∈ S(R2n),
Tr[ ˆ
Aˆ
B] = (2π~)−nZR2n
A(X)B(X)dX (2.5)
Assume now that ˆ
Bis a bounded operator in L2(Rn). The Weyl symbol Bof ˆ
B
satisfies:
Tr[ ˆ
Aˆ
B] = (2π~)−n< A, B > 1(2.6)
we have to check that the Weyl symbol of Sym(X) is (π~)nδXwhere δXis the Dirac
mass in X.
To prove that let us consider any ϕ∈ S(Rn) and denote by Wϕthe Weyl symbol
of the projector ψ7→< ψ, ϕ > ϕ.Wϕis called the Wigner function of ϕ. A direct
computation using (2.3) gives
Wϕ(q, p) = ZRn
exp[−i
~u·p]ϕ(q+u
2)ϕ(q−u
2)du
We get the proposition applying formula (2.6) with ˆ
B=Sym(X) and ˆ
A=πϕ.⊓⊔
We shall see later that it may be convenient to introduce the covariant Weyl
symbol for ˆ
Awhich has a nice connection with Weyl translations.
1the bracket <, > denotes the usual bilinear form (integral or distribution pairing). We shall
denote h•|•i the Hermitean sesquilinear form on Hilbert spaces, linear in the second argument.
5
Proposition 2.3 For every ˆ
A∈ Lw(S(R2n),S′(R2n)) there exists a unique temper-
ate distribution A#on R2n, named covariant Weyl symbol of ˆ
A, such that
ˆ
A= (2π~)−nZR2n
A#(X)ˆ
T(X)dX (2.7)
Moreover we have the inverse formula
A#(X) = Tr[ ˆ
Aˆ
T(−X)] (2.8)
( As above, if ˆ
Ais not trace-class, this formula has to be interpreted in a weak
distribution sense).
The covariant and contravariant Weyl symbols are related with the following formula
A#(X) = (2π~)−n˜
A(JX).(2.9)
˜
A◦Jis named the symplectic Fourier transform of A
Proof
These properties are not difficult to prove following for example [20].⊓⊔
We define the (usual) Gaussian coherent states ϕzas follows:
ϕz=ˆ
T(z)ϕ0,∀z∈R2n(2.10)
where
ϕ0(x) := (π~)−n/4e−x2/2~(2.11)
We get the following useful formula for the mean value of observables:
Corollary 2.4 With the above notations, for every ϕ, ψ ∈ S(Rn), we have
hϕ|ˆ
Aψi= (2π~)−nZR2n
A#(X)hϕ|ˆ
T(X)ψidX (2.12)
In particular for Gaussian Coherent States, we have
hϕz|ˆ
Aϕ0i= (2π~)−nZR2n
A#(X) exp −|X−z|2
4~−i
2~σ(X, z)dX (2.13)
Proof:
The first formula is a direct consequence of definition for covariant symbols.
The second formula is a consequence of the first and the following easy to prove
equalities
ˆ
T(z)ˆ
T(z′) = exp iσ
2~(z, z′)ˆ
T(z+z′) (2.14)
hϕz|ϕ0i= e−|z|2
4~(2.15)
⊓⊔ For later use let us recall the following
6
Definition 2.5 Let be ˆ
A, ˆ
B∈ Lw(S(Rn),S′(Rn)) such that the operator compo-
sition ˆ
Aˆ
Bis well defined. Then the Moyal product of Aand Bis the unique
A#B∈ S′(R2n)such that
ˆ
Aˆ
B=\
A#B(2.16)
For details concerning computations rules and properties of Moyal products see
[11, 18].
3 Time evolution of Quadratic Hamiltonians
In this section we consider a quadratic time-dependent Hamiltonian,
Ht(z) = P1≤j,k≤2ncj,k(t)zjzk, with real and continuous coefficients cj,k(t), defined
on the whole real line for simplicity. It is convenient to consider the symplectic
splitting z= (q, p)∈Rn×Rnand to write down Ht(z) as
Ht(q, p) = 1
2(Gtq·q+ 2Ltq·p+Ktp·p)
where Kt, Lt, Gtare real n×nmatrices, Ktand Gtbeing symmetric.
The classical motion in the phase space is given by the linear equation
˙q
˙p=J. GtLT
t
LtKtq
p,(3.17)
where LTis the transposed matrix of L. This equation defines a flow, Ft(linear
symplectic transformations) such that F0= 1l. On the quantum side, c
Htis a family
of self-adjoint operators on the Hilbert space H=L2(Rn) (this will be proved
later). The quantum evolution follows the Schr¨odinger equation, starting with an
initial state ϕ∈ H.
i~∂ψt
∂t =c
Htψt, ψt0=ϕ(3.18)
Suppose that we have proved existence and uniqueness for solution of (3.18), we
write ψt=b
Utϕ. The correspondence between the classical evolution and quantum
evolution is exact. For every A∈ S(R2n), we have
Proposition 3.1
b
Ut.b
A. d
U−1
t=d
A.Ft(3.19)
Sketch of proof
For any quadratic Weyl symbol Bwe have the exact formula
i
~[b
B, b
A] = \
{B, A}(3.20)
where {B, A}=∇B·J∇Adenote the Poisson bracket, and [ ˆ
B, ˆ
A] = ˆ
B. ˆ
A−ˆ
A. ˆ
Bis
the Moyal bracket. So we can prove Proposition (3.1) by taking derivative in time
and using (3.20) (see [18] for more details).
7
Now we want to compute explicitly the quantum propagator Ut0,t in terms of
classical evolution of Ht.
One approach is to compute the time evolution of Gaussian coherent sta(es, b
Ut0,tϕz,
or in other word to solve the Schr¨odinger equation (3.18) with ϕ=ϕz, the Gaus-
sian coherent state in z∈Rn. Let us recall that ϕz=ˆ
T(z)ϕ0and ϕ0(x) =
(π~)−n/4exp −|x|2
2~.
4 Time evolution of Coherent States
The coherent states system {ϕz}z∈R2nintroduced before is a very convenient tool
to analyze properties of operators in L2(Rn) and their Schwartz distribution kernel.
To understand that let us underline the following consequence of the Plancherel
Formula for the Fourier transform. In all this section we assume ~= 1. For every
u∈L2(Rn) we have
ZRn|u(x)|2dx = (2π)−nZR2n|hu, ϕzi|2dz (4.21)
Let ˆ
Rbe some continuous linear opertor from S(Rn) into S′(Rn) and KRits Schwartz
distribution kernel. By an easy computation, we get the following representation
formula
KR(x, y) = (2π)−nZR2n
(ˆ
Rϕz)(x)ϕz(y)dz. (4.22)
In other words we have the following continuous resolution of the Schwartz distri-
bution kernel of the identity
δ(x−y) = (2π)−nZR2n
ϕz(x)ϕz(y)dz.
This formula explains why the Gaussian coherent system may be an efficient tool
for analysis of operators on the Euclidean space Rn.
Let us consider first the harmonic oscillator
ˆ
H=−1
2
d2
dx2+1
2x2(4.23)
It is well known that for t6=kπ, k ∈Zthe quantum propagator e−it ˆ
Hhas an explicit
Schwartz kernel K(t;x, y) (Mehler formula).
It is easier to compute with the coherent states ϕz.ϕ0is an eigenstate of ˆ
H, so we
have
e−it ˆ
Hϕ0= e−it/2ϕ0(4.24)
Let us compute e−it ˆ
Hϕz,∀z∈R2, with the following ansatz
e−it ˆ
Hϕz= eiδt(z)ˆ
T(zt)e−it/2ϕ0(4.25)
8
where zt= (qt, pt) is the generic point on the classical trajectory (a circle here),
coming from zat time t= 0. Let be ψt,z the state equal to the r.h.s in (4.25), and
let us compute δt(z) such that ψt,z satisfies the equation id
dt ϕ=ˆ
Hϕ ϕ|t=0 =ψ0,z .
We have ˆ
T(zt)u(x) = ei(ptx−qtpt/2)u(x−qt)
and
ψt,z(x) = ei(δt(z)−t/2+ptx−qtpt/2)ϕ0(x−qt) (4.26)
So, after some computations left to the reader, using properties of the classical
trajectories
˙qt=pt˙pt=−qt, p2
t+q2
t=p2+q2,
the equation
id
dtψt,z (x) = 1
2(D2
x+x2)ψt,z(x) (4.27)
is satisfied if and only if
δt(z) = 1
2(ptqt−pq) (4.28)
⊓⊔ Let us now introduce the following general notations for later use.
Ftis the classical flow with initial time t0= 0 and final time t. It is represented as
a 2n×2nmatrix which can be written as four n×nblocks :
Ft=AtBt
CtDt.(4.29)
Let us introduce the following squeezed states. ϕΓis defined as follows.
ϕΓ(x) = aΓexp i
2~Γx·x(4.30)
where Γ ∈Σn, Σnis the Siegel space of complex, symmetric matrices Γ such that
ℑ(Γ) is positive and non degenerate and aΓ∈Cis such that the L2-norm of ϕΓis
one.
We also denote ϕΓ
z=b
T(z)ϕΓ.
For Γ = i1l, we denote ϕ=ϕi1l.
Theorem 4.1 We have the following formulae, for every x∈Rnand z∈R2n,
b
UtϕΓ(x) = ϕΓt(x) (4.31)
b
UtϕΓ
z(x) = b
T(Ftz)ϕΓt(x) (4.32)
where Γt= (Ct+iDtΓ)(At+iBtΓ)−1and aΓt=aΓ(det(At+iBtΓ))−1/2.
9
Beginning of the proof
The first formula can be proven by the ansatz
b
Utϕ0(x) = a(t) exp i
2~Γtx·x
where Γt∈Σnand a(t) is a complex values time dependent function. We get first
a Riccati equation to compute Γtand a linear equation to compute a(t).
The second formula is easy to prove from the first, using the Weyl translation oper-
ators and the following known property
b
Utb
T(z)b
U∗
t=b
T(Ftz).
Let us now give the details of the proof for z= 0.
We begin by computing the action of a quadratic Hamiltonian on a Gaussian (~=
1).
Lemma 4.2
Lx ·Dxei
2Γx·x= (LTx·Γx−i
2TrL)e i
2Γx·x
Proof
This is a straightforward computation, using
Lx ·Dx=1
iX
1≤j,k≤n
Ljk
xjDk+Dkxj
2
and, for ω∈Rn,
(ω.Dx)e i
2Γx·x= (Γx·ω)e i
2Γx·x
⊓⊔
Lemma 4.3
(GDx·Dx)e i
2Γx·x= (GΓx·Γx−iTr(GΓ)) e i
2Γx·x
Proof
As above, we get
b
Hei
2Γx·x=1
2Kx ·x+x·LΓx+1
2GΓx·Γx−i
2Tr(L+GΓ)ei
2Γx·x(4.33)
We are now ready to solve the equation
i∂
∂t ψ=ˆ
Hψ (4.34)
with
ψ|t=0(x) = g(x) := (2π)−n/2e−x2/2.
10
We try the ansatz
ψ(t, x) = a(t)ei
2Γtx·x(4.35)
which gives the equations
˙
Γt=−K−2ΓT
tL−ΓtGΓt(4.36)
˙
f(t) = −1
2(T r(L+GΓt)) f(t) (4.37)
with the initial conditions
Γ0=i1l, a(0) = (2π)−n/2
Remark: ΓTLet LΓ determine the same quadratic forms. So the first equation is
a Ricatti equation and can be written as
˙
Γt=−K−ΓtLT−LΓt−ΓtGΓt,(4.38)
where LTdenotes the transposed matrix for L. We shall now see that equation
(4.38) can be solved using Hamilton equation
˙
Ft=JK L
LTGFt(4.39)
F0= 1l (4.40)
We know that
Ft=AtBt
CtDt
is a symplectic matrix ∀t. So we have det(At+iBt)6= 0 ∀t(see below). Let us
denote
Mt=At+iBt, Nt=Ct+iDt(4.41)
We shall prove that Γt=NtM−1
t. By an easy computation, we get
˙
Mt=LTMt+GNt
˙
Nt=−KMt−LNt(4.42)
Now, compute
d
dt(NtM−1
t) = ˙
NM −1−NM −1˙
MM −1
=−K−LNM−1−N M−1(LTM+GN )M−1
=−K−LNM−1−N M−1LT−N M−1GN M−1(4.43)
which is exactly equation (4.38).
Now we compute a(t), using the following equality,
Tr LT+G(C+iD)(A+iB)−1= Tr( ˙
M)M−1= Tr (L+GΓt)
11
using TrL= TrLT. Let us recall the Liouville formula
d
dt log(det Mt) = Tr( ˙
MtM−1
t) (4.44)
which give directly
a(t) = (2π)−n/2(det(At+iBt))−1/2(4.45)
To complete the proof, we need to prove the following
Lemma 4.4 Let Sbe a symplectic matrix.
S=A B
C D
Then det(A+iB)6= 0 and ℑ(C+iD)(A+iB)−1is positive definite.
We shall prove a more general result concerning the Siegel space Σn.
Lemma 4.5 If
S=A B
C D
is a symplectic matrix and Z∈Σnthen A+BZ et C+DZ are non singular and
(C+DZ)(A+BZ)−1∈Σn
Proof
Let us denote E:= A+BZ, F := C+DZ.Fis symplectic, so we have FTJF =J.
Using E
F=SI
Z
we get
(ET, F T)JE
F= (I, Z)JI
Z= 0 (4.46)
which gives
ETF=FTE
In the same way, we have
1
2i(ET, F T)J¯
E
¯
F=1
2i(I, Z)FTJF I
¯
Z(4.47)
=1
2i(I, Z)JI
¯
Z=1
2i(¯
Z−Z) = −ℑZ
We get the following equation
FT¯
E−ET¯
F= 2iℑZ(4.48)
12
Because ℑZis non degenerate, from (5.68), we get that Eand Fare injective. If
x∈Cn, Ex = 0, we have ¯
E¯x=xTET= 0
hence
xTℑZ¯x= 0
then x= 0.
So, we can define,
α(S)Z= (C+DZ)(A+BZ)−1(4.49)
Let us prove that α(S)∈Σn. We have:
α(M)Z=F E−1⇒(α(M)Z)T= (E−1)TFT= (E−1)TETF E −1=F E−1=α(M)Z.
We have also:
ETF E−1−¯
F¯
E−1
2i¯
E=FT¯
E−ET¯
F
2i=ℑZ
and this proves that ℑ(α(M)) is positive and non degenerate.
This finishes the proof of the Theorem for z= 0 . ⊓⊔
Remark 4.6 For a different proof of formula (4.31), using the usual approach of
the metaplectic group, see the book [8].
The family {ϕz}z∈R2nspans all of L2(Rn)(see for exemple [19] for properties of
the Fourier-Bargmann transform) so formula (4.31) wholly determines the unitary
group b
Ut. In particular it results that b
Utis a unitary operator and that ˆ
Hthas a
unique self-adjoint extension in L2(Rn). This is left as exercises for the reader.
Remark 4.7 The map S7→ α(S)defines a representation of the symplectic group
Sp(2n)in the Siegel space Σn. It is easy to prove that α(S1S2) = α(S1)α(S2). This
representation is transitive. Many other properties of this representation are studied
in [16].
5 The metaplectic group and Weyl symbols com-
putation
A metaplectic transformation associated with a linear symplectic tranformation F∈
Sp(2n) in R2n, is a unitary operator b
R(F) in L2(Rn) satisfying one of the following
equivalent conditions
b
R(F)∗b
Ab
R(F) = \
A◦F , ∀A∈ S(R2n) (5.50)
b
R(F)∗b
T(X)b
R(F) = b
T[F−1(X)],∀X∈R2n(5.51)
b
R(F)∗b
Ab
R(F) = \
A◦F ,
for A(q, p) = qj,1≤j≤nand A(q, p) = pk,1≤k≤n. (5.52)
13
We shall see below that for every F∈Sp(2n) there exists a metaplectic transforma-
tion b
R(F).
Let us remark that if c
R1(F) and c
R2(F) are two metaplectic operators associ-
ated to the same symplectic map Fthen there exists λ∈C,|λ|= 1, such that
c
R1(F) = λc
R2(F). It is also required that F7→ b
R(F) defines a projective represen-
tation of the real symplectic group Sp(2n) with sign indetermination only. More
precisely, let us denote by Mp(n) the group of metaplectic transformations and
πpthe natural projection: Mp→Sp(2n) then the metaplectic representation is
a group homomorphism F7→ ˆ
R(F), from Sp(2n) onto Mp(n)/{1l,−1l}, such that
πp[b
R(F)] = F,∀F∈Sp(2n) (for more details concerning the metaplectic represen-
tation see [15]). We shall show here that this can be achieved straightforward using
Theorem 4.1.
For every F∈Sp(2n) we can find a C1- smooth curve Ft,t∈[0,1], in Sp(2n),
such that F0= 1l and F1=F. An explicit way to do that is to use the polar
decomposition of F,F=V|F|where Vis a symplectic orthogonal matrix and
|F|=√FTFis positive symplectic matrix. Each of these matrices have a logarithm,
so F= eKeLwith K, L Hamiltonian matrices, and we can choose Ft= etK etL. Any
way, Ftis clearly the linear flow defined by the quadratic Hamiltonian Ht(z) = 1
2Stz·z
where St=−J˙
FtF−1
t. So using above results, we define b
R(F) = c
U1. From
this definition and Theorem 4.1 we can easily recover the usual properties of the
metaplectic representation.
Proposition 5.1 Let us consider two symplectic paths Ftand F′
tjoining 1l (t= 0)
to F(t= 1). Then we have c
U1=±c
U′
1(with obvious notations).
Moreover, if F1, F 2∈Sp(2n)then we have
b
R(F1)ˆ
R(F2) = ±b
R(F1F2).
Proof
Using (4.31) we see that the phase shift between the two paths comes from variation
of argument between 0 and 1 of the complex numbers b(t) = det(At+iBt) and
b′(t) = det(A′
t+iB′
t).
We have arg[b(t)] = ℑRt
0
˙
b(s)
b(s)dsand it is well known (see Lemma (5.4) below and
its proof) that
ℑ Z1
0
˙
b(s)
b(s)ds!=ℑ Z1
0
˙
b′(s)
b′(s)ds!+ 2πN
with N∈Z. So we get
b(1)−1/2= eiNπ b′(1)−1/2
The second part of the proposition is an easy consequence of Theorem 4.1 concern-
ing propagation of squeezed coherent states with little computations. ⊓⊔
14
In a recent paper [17] the authors use a nice explicit formula for the Weyl symbol
of metaplectic operators ˆ
R(F). In what follows we detail a rigorous proof of this
formula including computation of the phase factor. In principle we could use Theo-
rem 4.1 to compute the Weyl symbol of the propagator Ut0,t. But in this approach
it seems difficult to compute phase factors (Maslov- Conley-Zehnder index).
For technical reason, It is easier for us to compute first the contravariant Weyl sym-
bol, Ut, for the propagator b
Utdefined by b
Ht. In any case, Utis a Schwartz temperate
distribution on the phase space R2n.
We follow the approach used in Fedosov [7].
It is enough to assume ~= 1. In a first step we shall solve the following problem
i∂
∂t c
Uε
t=b
Htc
Uε
t
c
Uε
0=b
1lε(5.53)
where 1lεis a smoothing family of operators such that lim
ε→01lε= 1l. It will be conve-
nient to take 1lε(X) = exp(−ε|X|2).
Let us recall that # denotes the Moyal product for Weyl symbols. So for the
contravariant symbol Ut(X) of Utwe have
i∂
∂t Ut(X) = (Ht#Ut)(X) (5.54)
Because Htis a quadratic polynomial we have
(Ht#Ut)(X) = Ht(X)Ut(X) + 1
2i{Ht, Ut}(X)
−1
8(∂ξ∂y−∂x∂η)2Ht(X)Ut(Y)|X=Y(5.55)
where ∂x=∂
∂x,X= (x, ξ), Y= (y, η).
It seems natural to make the following ansatz
Ut(X) = α(t)Et(X),where
Et(X) = exp (iMtX·X).(5.56)
α(t) is a complex time dependent function, Mtis a time dependent 2n×2ncomplex,
symmetric matrix such that ℑMtis positive and non degenerate.
A, B being two classical observables, we have:
{A , B}=∇A . J ∇B
and
(∂x∂η−∂y∂ξ)2A(x, ξ)B(y, η)|X=Y=∂2
x2A ∂2
ξ2B+∂2
x2B ∂2
ξ2A−2∂2
xξA ∂2
xξB
=−Tr(J A′′J B′′ ) (5.57)
15
where A′′ is the Hessian of A(and similarly for B). Applying this with
A(X) = 1
2StX.X, B(X) = exp(iMtX.X)
we get:
Ht#Et(X) = Ht(X)Et(X) + JStX . MtX Et(X) + 1
8Tr(J StJB′′)
However,
∇B= 2iB(X)MtX
(B′′)jk = 2i((Mt)jk + 2i(MtX)j(MtX)k)B(X)
so that:
1
8Tr(J StJB′′) = i
4Tr(J StJMt)−1
2MtX . J StJMtXB(X)
Therefore the Ansatz (5.56) leads to the equation:
i˙α(t)−α(t)˙
MtX . X =1
2(StX . X +MtJStX . X −StJ MtX . X )α(t)
+i
4α(t)Tr(MtSt)−1
2α(t)MtX . StMtX(5.58)
where we have introduced the Hamiltonian matrices
Mt:= JMt,St:= JSt
Then equation (5.58) is equivalent to
˙
Mt=1
2(Mt+ 1)St(Mt−1) (5.59)
˙αt=1
4Tr (MtSt)αt(5.60)
The first equation is a Riccati equation and can be solved with a Cayley transform:
Mt= (1l − Nt)(1l + Nt)−1
which gives the linear equation ˙
Nt=StNt
so we have, recalling that St=˙
FtF−1
t,
Nt=FtN0
Coming back to M, we get
Mt= (1l + M0−Ft(1l − M0)) (1l + M0+Ft(1l − M0))−1(5.61)
16
Let us now compute the phase term. We introduce
χ±
t= 1l + M0±Ft(1l −M0) (5.62)
Using the following properties
χ−
t=χ+
t−2Ft(1l − M0) (5.63)
TrSt= 0 (5.64)
St=˙
FtF−1
t(5.65)
we have
Tr(MtSt) = −2Tr ˙χ+
t(χ+
t)−1)(5.66)
so we get
αt=α0exp −1
2[log detχ+
•]t
0, α0= 1.(5.67)
In formula (5.67) the log is defined by continuity, because we shall see that χ+
tis
always non singular.
Until now we just compute at the formal level. To make the argument rigorous
we state some lemmas.
It is convenient here to introduce the following notations.
sp+(2n, C) is the set of complex, 2n×2nmatrices Msuch that M=JM where M
is symmetric (that means that Mis a complex Hamiltonian matrix) and such that
ℑMis positive non degenerate.
Sp+(2n, C) is the set of complex, symplectic, 2n×2nmatrices Nsuch that the
quadratic form
z7→ ℑ1l −N−1Nz·Jz
is positive and non degenerate on C2n.
Lemma 5.2 If Fis a real symplectic matrix and M ∈ sp+(2n, C), then 1l + M+
F(1l − M)is invertible.
Proof
Write M=J M. It is enough to prove that the adjoint 1l + FT+M∗(1l −FT) is
injective. But M∗=−M J. So if z∈C2nis such that (1l + FT+M∗(1l −FT))z= 0
then we get
((1l + FT)z−MJ (1l −FT)z·J(1l −FT)z) = 0 (5.68)
But, using that Fis symplectic, we have that (1 + F)J(1 −FT) = F J −J F Tis
symmetric so taking the imaginary part in (5.68), we have
ℑMJ (1l −FT)z·J(1l −FT)z= 0.
Then using that ℑMis non degenerate, we get successively (1l −FT)z= 0, (1l +
FT)z= 0 and z= 0. ⊓⊔
17
Lemma 5.3 Assume that -1 is not an eigenvalue of N. Then N ∈ Sp+(2n, C)if
and only if M ∈ sp+(2n, C)where Nand Mare linked by the formula
M= (1l −N)(1l + N)−1.
Proof
Assuming that N+ 1 is invertible, and
M= (1l −N)(1l + N)−1.
then we can easily see that Nis symplectic if and only if Mis Hamiltonian.
Now, using N=N∗,T , we get
1l − N−1N= 2(1 − M)−1(M− M)(1 + M)−1
If z= (1l + M)z′we have, using (J(1l + M))T=−J(1l −M),
(1l −N−1N)z·Jz = 2(M−M)z′·z′.
So the conclusion of the lemma follows easily from the last equality ⊓⊔
The last lemma has the following useful consequence.
Let us start with some M0∈Sp+(2n, C) without the eigenvalue -1. It not difficult
to see that Mtz=−zif and only if M0u=−u, where u= (χ+
t)−1z. In particular
for every time t, -1 is not an eigenvalue for Mt.
Furthermore, using that Nt=FtN0, we have N−1
tNt=N−1
0N0. So we get that the
matrix Mt∈sp+(2n, C) at every time t.
If M0has the eigenvalue −1 it is no more possible to use the Cayley transform but
we see that Mtis still defined by equation (5.61)(from lemma (5.2) ) and solves the
Riccati equation (5.59).
Now we want to discuss in more details the phase factor included in the term
αtand to consider the limiting case M0= 0 to compute the Weyl symbol of the
propagator c
U0
t. So doing we shall recover the Mehlig-Wilkinson formula, including
the phase correction Maslov-Conley-Zehnder index).
Let us denote
δ(Ft,M0) = det 1l + M0+Ft(1l − M0)
2.
Hence we have
αt= exp −1
2Zt
0
˙
δ(Ft,M0)
δ(Ft,M0)!ds.
Lemma 5.4 Let us consider t7→ Fta path in Sp(n, R). Then for every M0∈
sp+(n, C)we have, for the real part:
ℜ"Zt
0
˙
δ(Fs,M0)
δ(Fs,M0)ds#= log |δ(Ft,M0)|
|δ(F0,M0)|,
18
If Ftis τ-periodic, then Zτ
0
˙
δ(Ft,M0)
δ(Ft,M0)ds = 2iπν
with ν∈Z. Furthermore, νis independent on M0∈sp+(2n, C)and depends only
on the homotopy class of the closed path t7→ Ftin Sp(2n).
Proof
For simplicity, let us denote δ(s) = δ(Fs,M0) and
h(t) = Zt
0
˙
δ(s)δ(s)−1ds, g(t) = e−h(t)δ(t)
gis clearly constant in time and g(0) = δ(0) = 1. Then we get ℜ(h(t)) = log |δ(t)|.
In the periodic case we have eh(τ)= 1 so we have
1
2πZτ
0
˙
δ(Ft,M0)
δ(Ft,M0)ds =ν, ν ∈Z.
By a simple continuity argument, we see that νis invariant by continuous deforma-
tion on M0and Ft.⊓⊔
We can now compute the Weyl symbol of b
R(F) when det(1l + F)6= 0. Let us
consider first the case det(1l + F)>0. The case det(1l + F)<0 is a little bit more
complicated because the identity 1l is not in this component.
We start with an arbitrary C1, path t7→ Ftgoing from 1l (t= 0) to F(t= 1).
It is known that Sp+(2n) = {F∈Sp(2n),such that det(1l + F)>0}is an open
connected subset of Sp(2n). So, we can choose a piecewise C1path F′
tin Sp+(2n)
going from 1l to Fand M0=iεJ. We have, using Lemma(5.4),
ℑ Z1
0
˙
δ(Ft, iεJ)
δ(Ft, iεJ)dt!= 2πν +ℑ Z1
0
˙
δ(F′
t, iεJ)
δ(F′
t, iεJ)dt!(5.69)
But det(1l + F′
t) is never 0 on [0,1] and is real; so if ε > 0 is going to zero, the last
term in r.h.s goes to 0 and we get
lim
εց0ℑ Z1
0
˙
δ(Ft, iεJ)
δ(Ft, iεJ)dt!= 2πν
So we have proved for the Weyl symbol R(F, X ) of b
R(F) the following Melhig-
Wilkinson formula
R(F, X ) = eiπν |det(1l + F)|−1/2exp −iJ(1l −F)(1l + F)−1X·X(5.70)
19
Let us now consider F∈Sp−(2n) where Sp−(2n) = {F∈Sp(2n),such that det(1l+
F)<0}. Here we shall replace the identity matrix by
F0
2=−2 0
0−1
2
for n= 1 and F0=F0
2⊗1l2n−2for n≥2 where 1l2n−2is the identity in R2n−2.
Let us consider a path connecting 1l to F0then F0to F1=F. Because Sp(2n) is
open and connected we can find a path in Sp(2n) going from F0to F1=Fand that
part does not contribute to the phase by the same argument as above.
Let us consider the model case n= 1. The following formula gives an explicit path
in Sp(2).
F′
t=cos tπ −sin tπ
sin tπ cos tπ η(t) 0
01
η(t),
where η(t) is analytic on a complex neighborhood of [0,1], η(0) = 1, η(1) = 2,
1≤η(t)≤2 for t∈[0,1]. A simple example is η(t) = 1 + t. Then we can compute:
Lemma 5.5
lim
ε→0ℑ Z1
0
˙
δ(Ft, iεJ)
δ(Ft, iεJ)dt!=π(5.71)
Proof
Let us denote F′
t=R(t)B(t), where R(t) is the rotation matrix of angle tπ and for
ε∈[0,1[,
fε(t) = det[(1l −iεJ) + F′
t(1l −iεJ)]
We have
f0(t) = det(R(−t) + B(t)) = 2 + cos(tπ)1 + t+1
1 + t
f0has exactly one simple zero t1on [0,1], f0(t1) = 0, ˙
f0(t1)6= 0).
This is easy to see by solving the equation cos tπ =h(t), for a suitable h, with a
geometric argument .
Then by a standard complex analysis argument (contour deformation) we get the
equality (5.71). ⊓⊔
So in this case we have the formula (5.70) for the contravariant Weyl symbol of
ˆ
R(F), with index ν∈Z+ 1/2. Summing up the discussion of this section we have
proved:
Theorem 5.6 We can realize the metaplectic representation F7→ ˆ
R(F)of the sym-
plectic group Sp(2n, R)into the unitary group of L2(Rn)by taking for every FaC1-
path γgoing from 1l(t= 0) to F(t= 1) and solving explicitly the corresponding
quadratic Schr¨odinger equation for the Hamiltonian Ht(z) = −1/2J˙
FtF−1
tz·z. So
let us define c
Rγ(F), the propagator at time 1obtained this way.
20
If γ′is another path going from 1l (t= 0) to F(t= 1) then there exists an index
N(γ, γ ′)∈Zsuch that
c
Rγ(F) = eiπN (γ,γ ′)c
Rγ′(F)
The metaplectic operator ˆ
R(F)is the two valued unitary operator ±c
Rγ(F).
Moreover if det(1l +F)6= 0,ˆ
R(F)has a smooth contravariant Weyl symbol R(F, X),
given by formula (5.70), where ν∈Zif det(1l + F)>0and ν∈Z+ 1/2if
det(1l + F)<0.
It will be useful to translate the above theorem for the covariant Weyl symbol.
Before that we start to discuss the general case, including det[F±1l] = 0. The Weyl
symbol of ˆ
R(F) may be singular, so it is easier to analyse it using coherent states
(for our application it is exactly what we need).
Let us recall that
R#(F, X ) = (2π~)−nZR2nhϕz+X|ˆ
R(F)ϕzie−i
2σ(X,z)dz (5.72)
Let us denote ˆ
U1=ˆ
R(F) and Uε
1the contravariant Weyl symbol at time 1 con-
structed above, such that Uε
0(X) = e−ε|X|2.
For every ε > 0 we have computed the following formula for the contravariant Weyl
symbol:
Uε(X) = αεexp (iMεX·X),where
Mε(X) = = −J(1l + iεJ −F(1l −iεJ))(1l + iεJ +F(1 −iεJ ))−1(5.73)
αε= det(1l + iεJ +F(1l −iεJ))−1/2(5.74)
At the end we get the result by taking the limit:
lim
ε→+∞hϕz+X|ˆ
Uε
1ϕzi=hϕz+X|ˆ
R(F)ϕzi(5.75)
The computation uses the following formula for the Wigner function, Wz,z+X(Y), of
the pair (ϕz, ϕz+X).
Wz,z+X(Y) = 22nexp −Y−z−X
2
2
−iσ(X, Y −z
2)!(5.76)
So, we have to compute the following Fourier-Gauss integral
hϕz+X|ˆ
Uε
1ϕzi= 22n(2π)−nαεZR2n
dY exp iM εY·Y−Y−z−X
2
2
−iσ(X, Y −z
2)!
(5.77)
So we get
hϕz+X|ˆ
Uε
1ϕzi= 2n(det(1l −iMε))−1/2αε.(5.78)
21
×exp −z+X
2
2
+1
2iσ(X, z) + (1l −iM ε)−1(z+X−iJX
2)·(z+X−iJX
2)!
Now we can compute the limit when εց0. We have
1l −iMε= 1l + iJ(1l + iεJ −F(1l −iεJ))(1l + iεJ −F(1l −iεJ ))−1(5.79)
Then we get, using that (1l + F+iJ(1l −F)) is invertible (see Lemma 5.2),
lim
ε→0(1l −iMε)−1= (1l + F)(1l + F+iJ (1l −F))−1(5.80)
and
lim
ε→0det(1l −iMε)−1/2αε= (det(1l + F+iJ (1l −F))−1/2(5.81)
So, finally, we have proved the following
Proposition 5.7 The matrix elements of b
R(F)on coherent states ϕz, are given by
the following formula:
hϕz+X|ˆ
R(F)ϕzi= 2n(det(1l + F+iJ(1l −F))−1/2.
×exp −z+X
2
2
+1
2iσ(X, z) + KF(z+X−iJX
2)·(z+X−iJX
2)!(5.82)
where
KF= (1l + F)(1l + F+iJ(1l −F))−1(5.83)
Now we can compute the distribution covariant symbol of ˆ
R(F) by plugging formula
(5.82) in formula (5.72).
Let us begin with the regular case det(1l −F)6= 0.
Corollary 5.8 If det(1l −F)6= 0, the covariant Weyl symbol of c
Rγ(F)is computed
by the formula:
R#(F, z) = eiπµ|det(1l −F)|−1/2exp −i
4J(1l + F)(1l −F)−1z·z(5.84)
where µ= ¯ν+n
2,¯ν∈Zis an index computed below in formulas (5.88), (5.89).
Proof
Using Proposition (5.7) and formula (5.72), we have to compute a Gaussian integral
with a complex, quadratic, non degenerate covariance matrix (see [11]).
This covariance matrix is KF−1l and we have clearly
KF−1l = −iJ(1l −F)(1l + F+iJ(1l −F))−1=−(1l −iΛ)−1
22
where Λ = (1l + F)(1l −F)−1Jis a real symmetric matrix. So we have
ℜ(KF−1l) = −(1l + Λ2)−1,ℑ(KF−1l) = −Λ(1l + Λ2)−1(5.85)
So that 1l−KFis in the Siegel space Σ2nand Theorem (7.6.1) of [11] can be applied.
The only serious problem is to compute the index µ.
Let us define a path of 2n×2nsymplectic matrices as follows: Gt= etπJ2nif
det(1l −F)>0 and:
Gt=G2
t⊗etπJ2n−2if det(1l −F)<0, where
G2=η(t) 0
01
η(t)
where ηis a smooth function on [0,1] such that η(0) = 1, η(t)>1 on ]0,1] and
where J2nis the 2n×2nmatrix defining the symplectic matrix on the Euclidean
space R2n.
G1and Fare in the same connected component of Sp⋆(2n) where Sp⋆(2n) = {F∈
Sp(2n),det(1l −F)6= 0}. So we can consider a path s7→ F′
sin Sp⋆(2n) such that
F′
0=G1and F′
1=F.
Let us consider the following “argument of determinant” functions for families of
complex matrices.
θ[Ft] = argc[det(1l + Ft+iJ(1l −Ft)] (5.86)
β[F] = arg+[det(1l −KF)−1] (5.87)
where argcmeans that t7→ θ[Ft] is continuous in tand θ[1l] = 0 (F0= 1l), and
S7→ arg+[det(S)] is the analytic determination defined on the Siegel space Σ2nsuch
that arg+[det(S)] = 0 if Sis real (see [11], vol.1, section (3.4)).
With these notations we have
µ=β[F]−θ[F]
2π.(5.88)
Let us consider first the case det(1l −F)>0.
Using that Jhas the spectrum ±i, we get: det(1l + Gt+iJ(1l −Gt)) = 4nentπi and
1l −KG1= 1l.
Let us remark that det(1l −KF)−1= det(1l −F)−1det(1l −F+iJ(1l + F)). Let us
introduce △(E, M) = det(1l −E+M(1l + E)) for E∈Sp(2n) and M ∈ sp+(2n, C).
Let consider the closed path Cin Sp(2n) defined by adding {Gt}0≤t≤1and {F′
s}0≤s≤1.
We denote by 2π¯νthe variation of the argument for △(•,M) along C. Then we get
easily
β(F) = θ[F] + 2π¯ν+nπ, n ∈Z.(5.89)
When det(1l −F)<0, by an explicit computation, we find arg+[det(1l −KG1)] = 0.
So we can conclude as above. ⊓⊔
23
Assume now that Fhas the eigenvalue 1 with some multiplicity 2d. We want to
compute the temperate distribution R♯(F) as a limit of R♯(Fε) where det(1l −Fε)6=
0, ∀ε > 0.
Let us introduce the generalized eigenspace E′=[
j≥1
ker(1l −F)j(dim[E′]=2d). and
E′′ its symplectic orthogonal in R2n. We denote F′the restriction of Fto E′and
F′′ the restriction to E′′. We also denote by J′and J′′ the symplectic applications
defined by the restrictions of the symplectic form σ:σ(u, v) = J′u·v,∀u, v ∈ E′and
the same for J′′. Let us introduce the Hamiltonian maps L′= (1l −F′)(1l + F′)−1
and L′,ε =L′−εJ′.
It is clear that det(L′,ε −1l) 6= 0 for 0 < ε small enough, so we can define
F′,ε = (1l + L′,ε )(1l −L′,ε)−1. Let us remark that
Q′,ε := J′(F′,ε −1l)−1(F′,ε + 1l) = −(L′J′+ε)−1(5.90)
is a symmetric non degenerate matrix in E′defined for every ε > 0.
Lemma 5.9 We have the following properties.
1) F′,ε is symplectic.
2) lim
ε→0F′,ε =F′.
3) For ε6= 0, small enough, det(F′,ε −1l) 6= 0.
Proof.
1) comes from the fact that L′,ε is Hamiltonian.
2) is clear.
For 3), let us assume that F′,εu=u. Then we have L′,εu= 0 hence J′L′u=εu.
Now, choose 0 < ε < dist{0,spec(J′L′)\{0}} then we have u= 0.
Finally we define Fε=F′,ε ⊗F′′,ε. It is clear that Fεsatisfies also properties 1), 2),
3) of the above lemma with Fεin place of F′,ε.⊓⊔
Proposition 5.10 Under the above assumption, the covariant symbol of ˆ
R(F)has
the following form:
R#(F, z1, z2, z′′ ) = eiπµ1|det(1l −F′′)|−1/2δ(z1)
×exp i
4J(1l + F)(F−1l)−1(z2+z′′)·(z2+z′′ )(5.91)
where z:= ((z1, z2), z′′)is the decomposition of the phase-space for which F=
F′⊗F′′ and z′=z1+z2, with z2∈Im(F′−1l),z1∈Im(F′−1l)⊥, the orthogonal
complement in E′for the Euclidean scalar product. δ(z1)denotes the Dirac mass at
point z1= 0.µ1∈Z+ 1/2is given as follows: µ1=µ′′ +sg+Q′
4where µ′′ is the limit
of the µindex for Fε, computed in (5.84), and sg+Q′is the limit for ε→0of the
signature of Q′,ε, defined in (5.90).
24
Proof
We use the same kind of computation as for Corollary (5.8). The new factor comes
from the contribution of E′. So we can forget E′′ . So we assume that E′=R2nand
and we forget the superscripts ’. We have to compute the limit in the distribution
sense of R♯(Fε) which was computed in Corollary (5.8).
Let us consider a test function f∈ S(R2n) and its Fourier transform ˜
f. Using
Plancherel formula, we have
ZR2n
exp −i
4J(1l + Fε)(1l −Fε)−1z·zf(z)dz =
2n(π)2n|det(Qε)|−1/2esg(Qε)iπ
4ZR2n
exp −i(1l + Fε)−1(1l −Fε)Jζ ·ζ˜
f(ζ)dζ (5.92)
So we get the result by taking the limit for ε→0 in (5.92).
Let us remark that we have used that J′(1l + F)(1l −F)−1is an isomorphism from
ker(1l −F′)⊥onto itself. ⊓⊔
In our paper [2] the leading term for the return probability and for fidelity on
coherent states is computed with a Gaussian exponential defined by the quadratic
form which was defined in Proposition 5.7.
γF(X) := 1
4KF(1l −iJ)X·(1l −iJ)X− |X|2(5.93)
In our application we shall have X=zt−zwhere t7→ ztis the classical path starting
from z. So we need to estimate the argument in the exponent of formula (5.82) for
z= 0.
Lemma 5.11 we have, ∀X∈R2n,
ℜ(γF(X)) ≤ − |X|2
2(1 + sF),(5.94)
where sFis the largest eigenvalue of F F T(FTis the transposed matrix of F) .
Proof
Let us begin by assuming that det(1l + F)6= 0. Then we have
KF= (1l + iN)−1,where N=J(1l −F)(1l + F)−1.
So we can compute
ℜ(KF) = (1l + N2)−1=KFK⋆
Fand ℑ(KF) = −N(1l + N2)−1.
So, we get,
γF(X) = 1
4(1l + JN )KFK⋆
F(1l −NJ )X·X−2|X|2(5.95)
25
By definition of KF, we have
(1l + JN )KF= 2 (1l + iJ)F−1+ 1l −iJ−1(5.96)
Le us denote TF= ((1l + iJ )F−1+ 1l −iJ)−1. We have, using that Fis symplectic,
T⋆,−1
FT−1
F= 2(F−1,T F−1+ 1l).
hence we get
TFT⋆
F= (2(F−1,T F−1+ 1l))−1
and the conclusion of the lemma follows for det(1l−F)6= 0 hence for every symplectic
matrix Fby continuity.⊓⊔
Acknowledgement
We begun this work by discussions around the Mehlig-Wilkinson formula and appli-
cation to the Loschmidt echo at Mittag-Leffler Institute, in fall 2002. We thank the
institute and the organizers of the semester on Partial Differential Equations and
Spectral Theory, A. Laptev, V. Guillemin and B. Helffer for their hospitality and
exceptionnal working surroundings.
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