Instability of pre-existing resonances under a small
constant electric field
I. Herbst∗
J. Rama∗†
October 15, 2013
Abstract
Two simple model operators are considered which have pre-existing resonances. A
potential corresponding to a small electric field, f , is then introduced and the resonances
of the resulting operator are considered as f → 0. It is shown that these resonances
are not continuous in this limit. It is conjectured that a similar behavior will appear
in more complicated models of atoms and molecules. Numerical results are presented.
1
Introduction and Results
It has long been known that atomic bound states below the continuum turn into resonances
when subjected to a weak constant electric field. The resonances move continuously as a
function of the field strength, f , for small |f | and converge to the original bound state as
f → 0; see for example [CReAv, GGr1, GGr2, HSim, He, HeSim1, HeSim2, Re, YaTS]
and references given there. A simple, natural question arises: Is the same true for preexisting atomic resonances? More precisely, suppose an atom has a resonance in the lower
half complex plane, just below and near the continuous spectrum. The resonance might
be for example a shape resonance or due to a broken symmetry. Suppose the resonance
is defined, using for example the dilation analytic framework, as a pole in matrix elements
of the resolvent between dilation analytic vectors (in the sense of Definition 1.2 below).
Suppose then a term is added to the Hamiltonian representing an electric field of strength
f > 0. One can then ask whether there exist new resonance poles near the original atomic
resonance r0 and whether the new poles converge to r0 as f ↓ 0. At first one might expect
that methods entirely similar to those used for the problem of bound states turning into
resonances should be sufficient to treat this problem. But a hint as to the difficulty which
is encountered is the following: Let
H0 (f, θ) := e−2iθ p2 + f xeiθ
∗ University
† Supported
(θ, x ∈ R, f > 0)
of Virginia, Department of Mathematics
by the Deutsche Forschungsgemeinschaft (DFG), research grants RA 2020/1-1 and RA
2020/1-2.
1
2
be the complex dilated Hamiltonian in L2 (R) without potential, which (by [He, Theorem
II.1]) has empty spectrum if |θ| ∈ (0, π3 ). Then if f > 0 and z is in the numerical range
{z ∈ C | Im (e−iθ z) < 0} of H0 (f, θ) one has
k(H0 (f, θ) − z)−1 k ≥ c1 ec2 /f
for some strictly positive constants c1 and c2 ; see [He, Proposition II.6]. (This lower bound
is true in any dimension if x is replaced with a component of x.) Thus from a technical
standpoint there is a loss of control of relevant resolvents for the spectral parameter near
the original resonance.
In this paper we consider two simple models with pre-existing resonances. For these
models we prove (see Theorem 1.7 and Theorem 1.12) that pre-existing resonances, when
perturbed by a constant electric field of strength f > 0 (DC Stark effect), are unstable
in the weak field limit f ↓ 0. Roughly speaking, this is the fact that the analytically
continued resolvent does not converge to the resolvent obtained by first taking f to 0 and
then analytically continuing. In other words, taking the limit f ↓ 0 does not commute with
analytic continuation to the second Riemann sheet. In Section 3 we present some numerical
results illustrating this instability result. Furthermore, for one of our models (Friedrichs
model) we show that if the direct current field is replaced by an alternating current field
(AC Stark effect) pre-existing resonances are stable in the weak field limit (see Theorem
1.10). We conjecture similar instability results in the DC case for more complicated models
of atoms and molecules.
We introduce some notation: Let
Sγ := {z ∈ C | |Im z| < γ}
(γ > 0),
C± := {z ∈ C | Im z ≷ 0}.
(1.1)
∧
By ( · , · )X we denote an inner product in a linear space X. g denote the Fourier transform
∨
√
of a function g, and g its inverse, provided they exist. The symbol · denotes the principal
√
φ
branch of the square root with branch cut (−∞, 0). In particular, z := |z|1/2 e±i 2
(z ∈ C± ) for some φ ∈ (0, π).
Our models use analyticity in the following sense:
Definition 1.1
1. The unitary group of dilations on L2 (R) is defined by
(U (θ)ψ)(x) := eθ/2 ψ(eθ x)
(θ ∈ R, ψ ∈ L2 (R)).
(1.2)
If U (θ)ψ has an L2 -valued analytic extension (in the variable θ) to a strip Sθ0 for some
θ0 > 0, the function ψ is called dilation analytic (in angle θ0 ).
We denote the dense subspace of L2 -functions which are dilation analytic in angle θ0
by Dθ0 .
3
2. Let k0 > 0,
∧
Tk0 := ψ ∈ L2 (R) ψ has an analytic extension to the strip Sk0
∧
and sup | ψ (k)| < ∞ .
(1.3)
k∈Sk0
We remark that for any k0 > 0, Tk0 is a dense subspace of L2 (R). Note that (1.3) is just
∧
a pointwise analyticity condition on ψ and not an L2 -condition (as needed in the classical
Paley-Wiener Theorem for L2 -functions; see, e.g., [RSim, Theorem IX.13]). In particular,
(1.3) does not imply exponential decay of ψ.
Appendix B contains some results on boundedness of dilation analytic vectors in a sector.
1.1
1.1.1
Model I: Friedrichs model
Friedrichs model with DC Stark effect
In the first model the Hilbert space is
H := L2 (R) ⊕ C
(1.4)
with inner product
(u, u)H := (u1 , u1 )L2 + (u2 , u2 )C
u1
(u :=
∈ L2 (R) ⊕ C).
u2
(1.5)
We start out with a Hamiltonian with a simple eigenvalue, 1, embedded in its continuous
spectrum:
!
p2 0
H0 :=
,
(1.6)
0 1
where p denotes the self-adjoint realization of −id/dx in L2 (R). We then add a small rank-2
perturbation which removes the eigenvalue and turns it into a nearby resonance (in the sense
of Definition 1.2 below):
!
p2
Mϕ
(ϕ ∈ Dθ0 ∪ Tk0 for some θ0 , k0 > 0) ,
(1.7)
Hϕ :=
(ϕ , · )L2
1
where θ0 and k0 depend (in a certain sense specified below) on the size of ϕ. Mϕ : C → L2 (R)
denotes the multiplication operator generated by the function ϕ.
We now add the external electric field to the problem to obtain the Hamiltonian
!
p2 + f x Mϕ
Hϕ (f ) :=
(x ∈ R, f ≥ 0, ϕ ∈ Dθ0 ∪ Tk0 ) .
(1.8)
(ϕ , · )L2
1
We are interested in f > 0 small.
4
The operators H0 and Hϕ (f ) (f ≥ 0), Hϕ (0) = Hϕ are essentially self adjoint on C0∞ (R)⊕C.
With the abbreviations
Rf (z) := (p2 + f x − z)−1 ,
Ff,ϕ (z) := 1 − z − (ϕ , Rf (z)ϕ)L2
(1.9)
we compute for later use
(Hϕ (f ) − z)−1 =
Rf (z) 1 + Ff,ϕ (z)−1 (ϕ , Rf (z) · )L2 ϕ
−Ff,ϕ (z)−1 MRf (z)ϕ
−Ff,ϕ (z)−1 (ϕ , Rf (z) · )L2
Ff,ϕ (z)−1
!
,
(1.10)
where Im z 6= 0, ϕ ∈ Dθ0 ∪ Tk0 and f ≥ 0.
There are many ways to see resonances (see, e.g., [Sim]). In this model we use the
following definition:
Definition 1.2
Let κ0 > 0 and 0 < ϑ0 < π/3. Assume (1.4) through (1.8). Fix
ϕ ∈ Dθ0 ∪ Tk0 , where k0 ≥ κ0 and θ0 ≥ ϑ0 . Let

 {z ∈ C | z = |z|eiθ , 0 ≤ |θ| < 2ϑ0 } ∩ C− , ϕ ∈ Dθ0 and f = 0
Sκ0 ∩ C−
, ϕ ∈ Tk0 and f = 0
Ωf :=
.(1.11)

C−
, ϕ ∈ Dθ0 ∪ Tk0 and f > 0
1. Letf ≥ 0. A number ze in C− is defined to be a resonance of Hϕ (f ), if for some Ψ :=
ψ
∈ L2 (R) ⊕ C, with ψ ∈ Dϑ0 if ϕ ∈ Dθ0 and ψ ∈ Tκ0 if ϕ ∈ Tk0 , the meromorphic
u
extension of the resolvent matrix element (Ψ , (Hϕ (f ) − z)−1 Ψ)H (z ∈ C+ ) to the
region Ωf ∪ C+ \(−∞, 0] has a pole at ze.
2. For any function gf,ϕ (·) (depending on given ϕ and f ) which is analytic in C+ the
c,Ω
symbol gf,ϕ f (·) denotes the analytically (or meromorphically) continued function to
the region Ωf ∪ C+ \(−∞, 0].
Proposition 1.3
Let θ0 and k0 be as in Definition 1.2. Let ϕ ∈ Dθ0 ∪ Tk0 , f ≥ 0. Let
Hϕ (f ) and Ff,ϕ be given by (1.8) and (1.9). The resonances of Hϕ (f ) are precisely
the solutions of
c,Ω
Ff,ϕ f (z) = 0
(1.12)
c,Ω
in Ωf . (Ff,ϕ f denotes the analytic extension of Ff,ϕ in the sense of Definition 1.2.)
c,Ω
The real zeros of Ff,ϕ f are eigenvalues of the self-adjoint operator Hϕ (f ).
Proposition 1.3 follows from (1.10).
Proposition 1.4
Let θ0 and k0 be as in Definition 1.2. Let ϕ ∈ Tk0 ∪ Dθ0 . Let Hϕ (f )
and Ff,ϕ be given by (1.8) and (1.9). Then for all f > 0 the function Ff,ϕ has an
extension to an entire function of finite order (cf. Definition 2.1). This entire extension
has infinitely many zeros.
5
The proof of Proposition 1.4 is given in Section 2.1. A consequence of Proposition 1.3 and
Proposition 1.4 is the following corollary:
Corollary 1.5
Suppose the same hypotheses as in Proposition 1.4. Suppose in addition that ϕ is in the domain of hpi(1+ε)/2 for some ε > 0 (as an operator in L2 (R)),
where hpi := (1 + |p|2 )1/2 . Then for all f > 0 the operator Hϕ (f ) has infinitely many
resonances.
Proof: See Appendix A.
In this model the existence of pre-existing resonances (i.e., resonances of Hϕ (0)) is shown
by the following proposition. We further collect some additional useful information about
the spectrum of Hϕ (0):
Proposition 1.6
Let θ0 and k0 be as in Definition 1.2. Let ϕ ∈ Dθ0 ∪ Tk0 and Hϕ (0)
be given by (1.7). Then:
1. Hϕ (0) has an eigenvalue λ < 0 if and only if
Z
∧
| ϕ (k)|2
dk = 1 .
(k 2 + |λ|)(1 + |λ|)
R
2. Hϕ (0) has an eigenvalue λ ≥ 0 if and only if ϕ is in the domain of (p2 − λ)−1 and
(ϕ , (p2 − λ)−1 ϕ)L2 + λ − 1 = 0.
√
∧
3. If ϕ (± λ) 6= 0 for λ ∈ [1 − ε , 1 + ε] with some ε ∈ (0, 1), then for sufficiently
small ϕ the operator Hϕ (0) has exactly one resonance close to 1. The smallness
c,Ω0
| for |z − 1| = %, where %
of ϕ is measured by the size of %−1 |(ϕ , (p2 − z)−1 ϕ)L
2
is some number in (0, ε).
Sketch of proof: 1. and 2. follow from the eigenvalue equation for Hϕ (0) and the fact
that p2 does not have any eigenvalues.
c,Ω0
3. can be seen as follows: By Proposition 1.3, the zeros in C− of 1 − z − (ϕ , (p2 − z)−1 ϕ)L
2
are precisely the resonances of Hϕ (0). Let ∂B% denote the boundary of B% := {z ∈ C | |1 −
z| ≤ %}, % > 0. Let, for some 0 < % < ε,
0
|(ϕ , (p2 − z)−1 ϕ)c,Ω
L2 |
<1
%
(z ∈ ∂B% ) .
Then
0
|(ϕ , (p2 − z)−1 ϕ)c,Ω
L2 | < % = |1 − z|
(z ∈ ∂B% ) .
0
Thus by Rouché’s Theorem (see, e.g., [BN]) the functions 1 − z − (ϕ , (p2 − z)−1 ϕ)c,Ω
and
L2
1 − z have the same number of (nonreal) zeros (namely exactly one) in B% .
The following theorem on the location of resonances of Hϕ (f ) for small f > 0 and their
instability in the weak field limit is our main result in Section 1.1:
6
Theorem 1.7
Let k0 and θ0 be as in Definition 1.2. Let ϕ ∈ Dθ0 ∪ Tk0 . If ϕ is in
Dθ0 \Tk0 , suppose in addition that
dn ∧
π
sup | n ϕ (eiθ k)| k ∈ R\{0}, θ ∈ R, |θ| < min{θ0 , }, n ≤ 2 < ∞ . (1.13)
dk
3
Let Hϕ (f ), f > 0, be given by (1.8). If ϕ ∈ Dθ0 \Tk0 let
M := {z ∈ C− | 0 > arg
√
√
π
z ≥ − min{ , θ0 } + ε1 , Re z ∈ [ε2 , a]}
3
and if ϕ ∈ Tk0 let
M := {z ∈ C− | 0 > arg
√
z≥−
√
√
π
+ ε1 , Re z ∈ [ε2 , a], 0 > Im z ≥ −k0 + ε3 }
3
for any ε1 , ε2 , ε3 > 0 sufficiently small and any a > ε2 . Suppose there exists δ > 0
such that for all z ∈ M
√
∧ √ ∧
| ϕ ( z)ϕ(− z)| ≥ δ .
(1.14)
Then:
∃ c0 > 0 ∃ f0 > 0 sufficiently small ∀ f ∈ (0, f0 ] :
r is a resonance of Hϕ (f ) ⇒ |Im r| ≤ c0 f .
In particular,
Im r → 0
(f ↓ 0) .
Thus r does not converge to any resonance of Hϕ (0) (given by (1.7)) as f ↓ 0.
Proof: See Section 2.2.
Remark 1.8
∧
1. It follows from dilation analyticity that ϕ is analytic in {z ∈ C | z = ±|z|eiθ , θ ∈ R
with |θ| < θ0 , z 6= 0}. But (1.13) is an additional assumption near k = 0.
∧
√
2. If z ∈ M , ± z is in the region of analyticity for ϕ.
∧ √
3. Our proof of Theorem 1.7 uses power series expansions (in ζ) of ϕ ( z + ζ) and
∧
√
ϕ(− z − ζ) up to first order (e.g., in (2.69) and (2.97)). Presumably the results stated
∧ √ ∧
√
in Theorem 1.7 are true without the assumption that ϕ ( z)ϕ(− z) 6= 0. The proof
∧ √
∧
√
would involve going to higher order in the power series for ϕ ( z + ζ) and ϕ(− z − ζ).
7
1.1.2
Friedrichs model with AC Stark effect
If in Hϕ (f ), given by (1.8), the direct current field f x is replaced by an alternating electric
field, then the behavior of pre-existing resonances of Hϕ (0) changes: In contrast to preexisting resonances in the DC Stark effect (which are unstable by Theorem 1.7), pre-existing
resonances in the AC Stark effect are stable (in the sense of Theorem 1.10 below). This
is reminiscent of the situation in classical mechanics that introducing a periodic change in
parameters may cause an unstable equilibrium to become stable. An example is the unstable
equilibrium for a pendulum (see, e.g., [Ar]).
The Hamiltonian in the AC field regime is given by
!
h(t, f ) Mϕ
Hϕ (t, f ) :=
, h(t, f ) := p2 + f x sin(ωt) (f ≥ 0, x, t ∈ R) ,
(1.15)
(ϕ , · )L2
1
where ω > 0 denotes the frequency of an alternating electric field and ϕ is in Dθ0 for some
θ0 ∈ (0, π2 ).
In order to define resonances in this time dependent setting, we adapt the method of
[Y] (which uses Floquet theory) to the Friedrichs model (1.15). (In [Y] Yajima proved
that eigenvalues of a one-body Hamiltonian turn into resonances under the influence of an
alternating electric field.)
We introduce the transformation
!
T (t, f ) 0
2
2
e
e
T (t, f ) : L (R) ⊕ C → L (R) ⊕ C ,
T (t, f ) :=
(1.16)
0
1
for all f ≥ 0 and t ∈ R, where
T (t, f ) := ei2f ω
−2
sin(ωt)p −if ω −1 cos(ωt)x
e
(1.17)
and p := −id/dx. Then, if i∂t Ψ = Hϕ (t, f )Ψ, one has
e ϕ (t, f )Te(t, f )Ψ ,
i∂t (Te(t, f )Ψ) = H
(1.18)
e ϕ (t, f ) := (i∂t Te(t, f ))Te(t, f )−1 + Te(t, f )Hϕ (t, f )Te(t, f )−1
H
(1.19)
where
and
−1
Te(t, f )Hϕ (t, f )Te(t, f )
T (t, f )h(t, f )T (t, f )−1
=
(ϕ , T (t, f )−1 · )L2
!
T (t, f )ϕ
(1.20)
1
for all f ≥ 0, t ∈ R. A calculation shows
f2
f
cos2 (ωt) + 2 cos(ωt)p + p2
2
ω
ω
f
2
=
cos(ωt) + p .
ω
T (t, f )h(t, f )T (t, f )−1 =
(1.21)
8
Combining (1.19) and (1.20) gives
e ϕ (t, f ) =
H
!
T (t, f )ϕ
e
h(t, f )
(T (t, f )ϕ , · )L2
,
(1.22)
1
where
e
h(t, f ) := (i∂t T (t, f ))T (t, f )−1 + T (t, f )h(t, f )T (t, f )−1
= p2 +
f2
f2
cos(2ωt) +
.
2
2ω
2ω 2
(1.23)
Define
V (θ)±1 :=
U (θ)±1
0
!
0
1
(θ ∈ Sθ0 ) ,
(1.24)
where U (·) denotes the group of dilations in L2 (R) (cf. (1.2)). In particular, U (θ)xU (θ)−1 =
eθ x (x ∈ R, θ ∈ Sθ0 ). One gets
!
U (θ)e
h(t, f )U (θ)−1 U (θ)T (t, f )ϕ
−1
e
V (θ)Hϕ (t, f )V (θ) =
,
(1.25)
(U (θ)T (t, f )ϕ , · )L2
1
f2
f2
U (θ)e
h(t, f )U (θ)−1 = e−2θ p2 +
cos(2ωt)
+
2ω 2
2ω 2
(1.26)
for all θ ∈ Sθ0 , f ≥ 0 and t ∈ R. Using Howland’s idea from [Ho] and adapting the outline
in [Y, Introduction] to our matrix model (1.15), we consider
e ϕ (t, f )
K(f ) := −i∂t + H
(f ≥ 0 , t ∈ R)
(1.27)
f := (L2 (R) ⊕ C) ⊗ L2 (Tω ), where
as an operator in H
Tω := R/τ Z ,
τ :=
2π
.
ω
(1.28)
Then
e ϕ (t, f ) ⊗ 1 T
K(f ) = 1 L2 (R)⊕C ⊗ (−i∂t ) + H
ω
(f ≥ 0 , t ∈ R) ,
(1.29)
f, which we also denote by K(f ). The operator
and K(f ) has a self-adjoint realization in H
−isK(f )
K(f ) generates a unitary group {e
}s∈R . By Floquet theory, e−iτ K(f ) is unitarily
e (t + τ, t; f ) ⊗ 1 L2 (T ) (t ∈ R) over the period τ , where
equivalent to the propagator U
ω
e (t, s; f ) = H
e ϕ (t, f )U
e (t, s; f ),
i∂t U
e (t, t; f ) = 1
U
(1.30)
for t, s ∈ R. Similar to [Y], we define
e ϕ (t, f )V (θ)−1
K(f, θ) := −i∂t + V (θ)H
In particular,
K(0, 0) = −i∂t +
p2
(ϕ , · )L2
!
ϕ
.
1
(θ ∈ Sθ0 , f ≥ 0, t ∈ R).
(1.31)
9
Definition 1.9
Let f ≥ 0, θ ∈ Sθ0 with Im θ > 0. The nonreal eigenvalues (in C− )
of K(f, θ) are defined to be the resonances of Hϕ ( · , f ).
Theorem 1.10
Let ω > 0 denote the frequency of an alternating electric field. Let
θ0 ∈ (0, π2 ) and ϕ ∈ Dθ0 . Assume in addition that there exists an f0 > 0 sufficiently
−1
small such that for all β ∈ [−f0 , f0 ] and all θ ∈ Sθ0 the function ef0 |x|ω sin θ0 ϕ(eθ x +
2βω −2 ) (x ∈ R) is analytic in the parameter θ and
kef0 | · |ω
−1
sin θ0
ϕ(eθ · +2βω −2 )k2 ≤ C
(1.32)
for some constant C < ∞, uniformly for (θ, β) ∈ Sθ0 × [−f0 , f0 ]. Let Hϕ (t, f ) (f ≥
0, t ∈ R) be given by (1.15) and K(f, θ) (f ≥ 0, θ ∈ Sθ0 ) by (1.31).
Fix θ ∈ Sθ0 with Im θ > 0. Suppose there exists an eigenvalue r0 (not necessarily close
to 1) of K(0, θ) of multiplicity m. Then for all f > 0 sufficiently small there are exactly
m eigenvalues (counting multiplicities) of K(f, θ) close to r0 and they all converge to
r0 as f ↓ 0. (By Definition 1.9, r0 is a resonance of Hϕ ( · , 0) and the eigenvalues of
K(f, θ) are resonances of Hϕ ( · , f ).)
Proof of Theorem 1.10:
We shall prove that
kK(f, θ) − K(0, θ)k → 0
(f ↓ 0)
(1.33)
for all θ ∈ Sθ0 : If f = 0, then
e−2θ p2
K(0, θ) = −i∂t + Hϕ (0, θ) ,
Hϕ (0, θ) :=
!
U (θ)ϕ
,
(U (θ)ϕ , · )L2
(1.34)
1
which follows from (1.31), (1.25) and (1.26). As a consequence of [Y, Lemma 2.7] the
spectrum of K(0, θ) is
σ(K(0, θ)) = {ωn + z | z ∈ σ (Hϕ (0, θ)) , n ∈ Z} ,
(1.35)
and K(0, θ) has nonreal eigenvalues (which are resonances of Hϕ ( · , 0)) at {ωn + r0 | n ∈ Z}.
It follows from Proposition 1.6 and dilation analyticity (see [ACo]) that
σ (Hϕ (0, θ)) = e−i2Im θ [0, ∞) ∪ {nonreal eigenvalues of Hϕ (0, θ)}
∪ {real eigenvalues of Hϕ (0, θ)}
=e
−i2Im θ
[0, ∞) ∪ {resonances of Hϕ ( · , 0)}
∪ {(real) eigenvalues of Hϕ ( · , 0)} .
(1.36)
We calculate
K(f, θ) − K(0, θ) =
f2
2ω 2 (1
+ cos(2ωt))
(U (θ)(T (t, f ) − 1)ϕ , · )L2
!
U (θ)(T (t, f ) − 1)ϕ
0
(1.37)
10
and get
f2
+ sup(kU (θ)(T (t, f ) − 1)ϕk2 + kU (θ)(T (t, f ) − 1)ϕk2 )
ω 2 t∈R
kK(f, θ) − K(0, θ)k ≤
(1.38)
for all θ ∈ Sθ0 . Furthermore,
U (θ)T (t, f )ϕ (x) = eθ/2 T (t, f )ϕ (eθ x)
(θ ∈ Sθ0 f > 0, t ∈ R)
(1.39)
and
(T (t, f )ϕ)(eθ x)
=
ei2f ω
=
e−if ω
=
e−if ω
(1.17)
e−if
=
2
−2
e
−1
−1
ω
sin(ωt)p −if ω −1 cos(ωt) ·
ϕ( · + 2f ω −2 sin(ωt)) (eθ x)
eθ x+2f ω −2 sin(ωt)
ϕ(eθ x + 2f ω −2 sin(ωt))
cos(ωt)( · +2f ω −2 sin(ωt))
cos(ωt)
−3
ϕ( · ) (eθ x)
sin(2ωt) −if ω −1 cos(ωt)eθ x
e
ϕ(eθ x + 2f ω −2 sin(ωt)) ,
(1.40)
where we have used that {eiαp }α∈R is just the group of translations; in particular
(ei2f ω
−2
sin(ωt)p
g)(y) = g(y + 2f ω −2 sin(ωt))
(g ∈ L2 (R) , y ∈ R) .
Since U (θ) = U (Re θ)U (iIm θ) and kU (Re θ)k = 1 for all θ ∈ Sθ0 , equation (1.40) implies
kU (θ)(T (t, f ) − 1)ϕk2 = kU (iIm θ)(T (t, f ) − 1)ϕk2
Im θ
2 −3
−1
iIm θ
(·)
ϕ(eiIm θ · +2f ω −2 sin(ωt))
= e−i 2 e−if ω sin(ωt) e−if ω cos(ωt)e
− ϕ(eiIm θ · ) 2
= ke−if
2
ω −3 sin(ωt) −if ω −1 cos(ωt)eiIm θ (·)
e
ϕ(eiIm θ · +2f ω −2 sin(ωt)) − ϕ(eiIm θ · )k2 .
(1.41)
for all θ ∈ Sθ0 , f > 0, t ∈ R. Setting
af (x) := f 2 ω −3 sin(ωt) + f ω −1 cos(ωt) cos(Im θ)x ,
bf (x) := f ω
−1
cos(ωt) sin(Im θ)x
(1.42)
(1.43)
gives
e−if
2
ω −3 sin(ωt) −if ω −1 cos(ωt)eiIm θ x
e
= e−if ω
−1
cos(ωt)(cos(Im θ)+i sin(Im θ))x
= e−iaf (x) ebf (x) .
(1.44)
11
Thus, by (1.44), equation (1.41) becomes
kU (θ)(T (t, f ) − 1)ϕk2 = e−iaf ebf ϕ(eiIm θ · +2f ω −2 sin(ωt)) − ϕ(eiIm θ · )
+ ϕ(eiIm θ · ) − ϕ(eiIm θ · )2
≤ e−iaf ebf ϕ(eiIm θ · +2f ω −2 sin(ωt)) − ϕ(eiIm θ · ) 2
+ k(e−iaf ebf − 1)ϕ(eiIm θ · )k2
= ebf ϕ(eiIm θ · +2f ω −2 sin(ωt)) − ϕ(eiIm θ · ) 2
+ k(e−iaf ebf − 1)ϕ(eiIm θ · )k2 .
(1.45)
Next we will prove that the r.h.s. of (1.45) converges to zero as f ↓ 0. We estimate the first
summand by
b e f ϕ(eiIm θ · +2f ω −2 sin(ωt)) − ϕ(eiIm θ · ) 2
f ω−1 | cos(ωt)| | sin(Im θ)| | · | iIm θ
≤ sup
e
ϕ(e
· +β) − ϕ(eiIm θ · ) 2
|β|≤2f /ω 2
≤
sup
|β|≤2f /ω 2
f ω−1 | sin θ | | · | 0
e
ϕ(eiIm θ · +β) − ϕ(eiIm θ · ) 2
(1.46)
for 0 < θ0 < π/2. Obviously,
−1
lim sup ef ω | sin θ0 | | · | ϕ(eiIm θ · +β) − ϕ(eiIm θ · ) 2 = 0 .
(1.47)
f ↓0 |β|≤2f /ω 2
The second summand also converges to zero as f ↓ 0: It follows from (1.42) and (1.43) that
|(e−iaf (x) ebf (x) − 1)ϕ(eiIm θ x)| → 0
(f ↓ 0)
(1.48)
for all x ∈ R. Furthermore,
|(e
−iaf (x) bf (x)
e
− 1)ϕ(e
iIm θ
2
x)|
(1.42)
(1.43)
|(ef ω
−1
| cos(ωt)| | sin(Im θ)| |x|
≤
|(ef ω
−1
| sin θ0 | |x|
=:
g(x)
≤
+ 1)ϕ(eiIm θ x)|2
+ 1)ϕ(eiIm θ x)|2
(1.49)
−1
and g ∈ L1 (R), since by assumption (1.32) the functions ϕ(eθ · ) and eω | sin θ0 | | · | ϕ(eθ · +β)
both are in L2 (R) for all β ∈ [−f0 , f0 ] and θ ∈ Sθ0 . Thus by dominated convergence
k(e−iaf ebf − 1)ϕ(eiIm θ ·)k2 → 0
(f ↓ 0) .
(1.50)
Combining (1.50), (1.47), (1.46) and (1.45) shows
kU (θ)(T (t, f ) − 1)ϕk2 → 0
(f ↓ 0)
(1.51)
for all t ∈ R, θ ∈ Sθ0 . This implies
sup(kU (θ)(T (t, f ) − 1)ϕk2 + kU (θ)(T (t, f ) − 1)ϕk2 ) → 0
t∈R
for all θ ∈ Sθ0 . Using (1.52) in (1.38) proves (1.33).
(f ↓ 0)
(1.52)
12
It follows that K(f, θ) → K(0, θ) as f ↓ 0 in norm resolvent sense for all θ ∈ Sθ0 with
Im θ > 0. In particular, the corresponding spectral projection (Riesz projections) of K(f, θ)
converges in norm to the one of K(0, θ). In this sense the resonances of Hϕ ( · , f ) are stable.
1.2
Model II
In this model we consider the operator
Hε (f ) := p2 + f x + Vε
(ε ≥ 0, f ≥ 0, x ∈ R) ,
(1.53)
in L2 (R), where p denotes the self-adjoint realization of −id/dx in L2 (R) and
Vε := (ϕε , · )ϕε
(1.54)
is a rank one self-adjoint operator in L2 (R). (In this Section 1.2 the inner product in L2 (R)
is denoted by ( · , · ).) We shall construct ϕε in such a way that, for ε = 0, the operator
p2 + V0 will have eigenvalue 1 embedded in the continuous spectrum [0, ∞) of p2 . For ε > 0
this eigenvalue will turn into a resonance of p2 + Vε which we will then perturb by taking
f > 0 in (1.53).
Our second model needs the following observation (similar to Proposition 1.6):
Proposition 1.11
Let H2 (R) denote the Sobolev space of second order. Let p =
−id/dx, ϕ ∈ L2 (R). Clearly, the operators p2 and T := p2 + (ϕ , · )ϕ with domain
D(T ) = D(p2 ) = H2 (R) are self-adjoint in L2 (R). T has an eigenvalue λ > 0 if and
only if
ϕ ∈ Ran (p2 − λ)
and
(ϕ , (p2 − λ)−1 ϕ) = −1 .
(1.55)
Proof: Let λ > 0. If p2 ψ + (ϕ , ψ)ϕ = λψ for some non-zero ψ ∈ H2 (R), then (p2 − λ)ψ =
−(ϕ , ψ)ϕ. Since p2 has no eigenvalues we cannot have (ϕ , ψ) = 0, thus ϕ ∈ Ran (p2 − λ)
and in addition
ψ = −(ϕ , ψ)(p2 − λ)−1 ϕ.
(1.56)
(ϕ , ψ) = −(ϕ , ψ)(ϕ , (p2 − λ)−1 ϕ) ,
(1.57)
Thus
so (ϕ , (p2 − λ)−1 ϕ) = −1. Conversely, assuming (1.55) and defining
ψ := (p2 − λ)−1 ϕ ,
(1.58)
we have
(p2 − λ)ψ + (ϕ , ψ)ϕ = ϕ + (ϕ , (p2 − λ)−1 ϕ)ϕ = 0.
(1.59)
13
We shall now construct Vε and calculate the resonance of Hε (0): Let θ0 > 0, k0 > 0.
Let ϕ0 ∈ Dθ0 ∪ Tk0 (where Dθ0 and Tk0 are given in Definition 1.1) with
ϕ0 = p
(p2 − 1)ψ0
(1.60)
(ψ0 , (1 − p2 )ψ0 )
∧
for some ψ0 ∈ H2 (R) with (ψ0 , (1 − p2 )ψ0 ) < 0. Choose ψ0 so that ψ 0 has no zeros on the
real axis which implies 1 is the only eigenvalue of p2 + (ϕ0 , · )ϕ0 . Define
(p2 − 1 + iε)ψ0
ϕε := p
(ψ0 , (1 − p2 )ψ0 )
(ε ≥ 0)
(1.61)
such that for ε > 0, the self-adjoint operator
Hε (0) = p2 + Vε
(1.62)
has no eigenvalues. Let
Rf (z) := (p2 + f x − z)−1
(f ≥ 0 , Im z 6= 0) .
(1.63)
A short calculation shows that, for f, ε ≥ 0 and z ∈ C+ ,
−1
(Hε (f ) − z)−1 = Rf (z) − Fε,f
(z)(ϕε , Rf (z) · )Rf (z)ϕε ,
(1.64)
Fε,f (z) := 1 + (ϕε , Rf (z)ϕε ) .
(1.65)
where
For f = 0 and ε ≥ 0 one gets
Fε,0 (z) =1 + (ϕε , R0 (z)ϕε )
=
(z − 1)kψ0 k2 + (ε2 + (z − 1)2 )(ψ0 , (p2 − z)−1 ψ0 )
|(ψ0 , (1 − p2 )ψ0 )|
(z ∈ C+ ) .
(1.66)
In this model the resonances of Hε (f ) (for ε, f ≥ 0 sufficiently small) are defined to be the
zeros in C− of Fε,f (·)c,Ωf (i.e., the zeros in C− of the analytically continued Fε,f (·) from
the upper half complex plane into Ωf ∪ C+ \(−∞, 0], where Ωf is given by (1.11) and Fε,f
by (1.65)). (This definition is an analog of Proposition 1.3 for Model I.)
For ε > 0 sufficiently small Hε (0) has exactly one resonance, r0 , near z = 1. (This can
be seen by the same arguments as used in the proof of Proposition 1.6, 3.) This resonance
r0 is a solution (in C− , near 1) of Fε,0 (z)c,C− = 0. One finds
r0 = 1 −
∧
ε2 iπ ∧
2
−1
2
2
P.V.(ψ
,
(p
−
1)
ψ
)
+
(|
ψ
(1)|
+
|
ψ
(−1)|
)
+ O(ε4 )
0
0
0
0
kψ0 k2
2
(1.67)
as ε ↓ 0.
As with Model I (Friedrichs model; cf. Theorem 1.7) there is in general no convergence
of resonances of Hε (f ) to r0 (or to any other resonance of Hε (0)) as f ↓ 0:
14
Theorem 1.12
Let ϕε (ε ≥ 0) be given by (1.61), where ψ0 ∈ Tk0 ∪Dθ0 with k0 , θ0 > 0
as in Definition 1.2. Suppose
Z
∧
| ψ 0 (k + iα)|2 k 4 dk < ∞
(1.68)
|k|>Λ
for large Λ and |α| < k0 . If ψ0 6∈ Tk0 assume in addition that ψ0 satisfies (1.13). Let
Hε (f ), Vε (ε, f ≥ 0) be given by (1.53) and (1.54). Let M be as in Theorem 1.7.
Fix ε > 0 sufficiently small. Suppose that there exists δ > 0 such that for all z ∈ M
∧
√ ∧
√
| ϕε ( z)ϕε (− z)| ≥ δ. Then:
∃ c0 > 0 ∃ f0 > 0 sufficiently small ∀ f ∈ (0, f0 ] :
r is a resonance of Hε (f ) ⇒ |Im r| ≤ c0 f .
In particular,
Im r → 0
(f ↓ 0) .
Thus r does not converge to any resonance of Hε (0) as f ↓ 0.
Proof: Theorem 1.12 follows from the estimates in Section 2, used for the proof of Theorem 1.7.
∧
√
√ ∧
For small ε > 0 it is easy to see from (1.61) that | ϕε ( r0 )ϕε (− r0 )| >
∧
√ ∧
√
0 if | ψ 0 ( r0 )ψ 0 (− r0 )| > 0.
Remark 1.13
2
2.1
Proofs
Proof of Proposition 1.4
Our proof of Proposition 1.4 needs the following Theorem 2.2 and Lemmata 2.3 - 2.7 below.
Definition 2.1
Let g be entire. The function g is defined to be of finite order if there
exist an n ∈ R and R > 0 such that for all z ∈ C with |z| ≥ R
n
|g(z)| ≤ e|z| .
Theorem 2.2 [BN, Theorem 16.13]
Let g be an entire function of finite order. Then either g has infinitely many zeros or
g(z) = P (z)eQ(z) (z ∈ C) for some polynomials P and Q.
15
Let p = −id/dx. For all x, f ∈ R, φ ∈ S(R) one has
Lemma 2.3
p3
p3
e−i 3f (p2 + f x)φ = f xe−i 3f φ .
(2.1)
In particular, the unique extension of p2 + f x to an operator in L2 (R) is unitarily
equivalent to the unique extension of f x.
The proof of Lemma 2.3 is by Fourier transform.
Lemma 2.4
∧
Let ϕ∈ L2 (R) be analytic in an open neighborhood of
A := {ζ ∈ C | |Re ζ| ≥ N , |Im ζ| ≤ α0 }
(2.2)
∧
for some α0 > 0, N > 0 and let ϕ be bounded in A. Let
ψf := (e−i
( · )3
3f
∧
ϕ)∨
(f > 0) .
(2.3)
Then ψf , which a priori is only in L2 (R), actually extends to an entire function (which
we also denote by ψf ). (In particular, ψf has a real analytic realization on R.) The
entire function ψf is of finite order ≤ 2. More precisely,
2
|ψf (x)| ≤ a eb|x|
(x ∈ C)
(2.4)
for some finite real constants a and b, possibly depending on f , ϕ and N .
Proof of Lemma 2.4: Lemma 2.4 follows from Lemma 2.6 below and the estimate (2.15)
given in the proof of Lemma 2.6.
Remark 2.5
In distributional sense
Z
ψf (x) = f 1/3 Ai(f 1/3 (y − x))ϕ(y) dy ,
where Ai denotes the Airy function. Ai is an entire function (of order 3/2); see, e.g.,
[Hö, Chapter 7.6]. We conjecture that ψf actually is of order 3/2.
Lemma 2.6
Assume the same conditions and notation as in Lemma 2.4. Let 0 <
α ≤ α0 and f > 0. Let
γ
e−N : [0, α] → C ,
t 7→ −N − i(α − t) ,
γ
e+N : [0, α] → C ,
t 7→ N − it ,
γα := {t − iα | t ∈ (−∞, −N )} ∪ γ
e−N ∪ (−N, N ) ∪ γ
e+N ∪ {t − iα | t ∈ (N, ∞)} .
16
Then the contour integral
Z
3
∧
1
−i k
3f −kx
e
ϕ (k) dk
e
ψα,f (x) := √
2π
(2.5)
γα
exists for all x ∈ C, is independent of α ∈ (0, α0 ] and ψeα,f is an entire function of finite
order ≤ 2. Furthermore, as an element in L2 (R), ψeα,f coincides with ψf (defined in
(2.3)).
∧
Note that γα ⊂ C− lies in the region of analyticity for ϕ. The contour γα is sketched in
Figure 2.1.
Lemma 2.7
Let ϕ be as in Proposition 1.4 or Theorem 1.7. Then ϕ fulfills the
hypotheses of Lemma 2.4.
The proof of Lemma 2.7 is by the definition of Tk0 (see Definition 1.1) and Proposition B.2.
Proof of Lemma 2.6:
We write ζ := κ − iτ for ζ ∈ C− and calculate
(κ − iτ )3
κ3
(−iτ )3
ζ3
=
=
− iτ κ2 − τ 2 κ +
,
3
3
3
3
(2.6)
which implies the estimate
−i
e
(κ−iτ )3
3f
−(κ−iτ )x
(κ−iτ )3
= |e−i 3f | |ei(κ−iτ )x |
τ3
≤ e 3f e−
τ κ2
f
e|κ| |x| eτ |x|
(κ ∈ R , τ ≥ 0 x ∈ C) .
(2.7)
Proof of existence, entireness and finite order:
The following piecewise estimates prove the existence of the integral in (2.5) for all x ∈ C:
Integral from −N to N :
ZN
Φ(x) :=
e
−i
k3
3f
−kx
∧
ϕ (k) dk =
−N
ZN
eikx G(k) dk
(x ∈ R) ,
(2.8)
−N
where
k3
∧
G(k) := e−i 3f ϕ (k)
(k ∈ R) .
(2.9)
∧
Since ϕ∈ L2 (R), the function G is in L2 (R) and in particular in L2 ((−N, N )). Thus (see,
e.g., [Ru, Chapter 19]) Φ(x) (x ∈ R) has an entire extension and
|Φ(x)| ≤ e
N |x|
ZN
IN
(x ∈ C) ,
|G(k)| dk < ∞ .
IN :=
−N
(2.10)
17
Figure 2.1: The contour γα .
For all x ∈ C one gets
Integrals along the paths γ
e±N :
3
Z
∧
−i k
3f −kx
ϕ (k) dk e
Zα
−i
−i e
=
(N −it)3
3f
−(N −it)x ∧
ϕ (N − it) dt
0
γ
e+N
(2.7)
∧
sup | ϕ (ζ)| eN |x|
≤
Zα
e−
tN 2
f
t3
e 3f et|x| dt
ζ∈A
0
∧
α sup | ϕ (ζ)| e
≤
α3
3f
eα|x| eN |x| .
(2.11)
ζ∈A
Similarly,
3
Zα −i (−N −i(α−t))3 −(−N −i(α−t))x ∧
Z
∧
−i k
−kx
3f
3f
ϕ (k) dk = i e
ϕ (−N − i(α − t)) dt
e
0
γ
e−N
(2.7)
Zα
∧
≤ sup | ϕ (ζ)|eN |x|
e
(α−t)3
3f
e−
(α−t)N 2
f
e(α−t)|x| dt
ζ∈A
0
∧
= sup | ϕ (ζ)|e
N |x| α|x| − αN
f
e
2
e
e
α3
3f
Zα
e
tN 2
f
e−t|x| e−
tα2
f
e
t2 α
f
t3
e− 3f dt
ζ∈A
0
∧
≤ α sup | ϕ (ζ)| e
ζ∈A
α3
3f
e
α3
f
e
α|x| N |x|
e
(x ∈ C) .
(2.12)
18
Integrals from −∞ − iα to −N − iα and from N − iα to ∞ − iα:
Z −iα ∞−iα
Z 3
−N
∧
−i k
3f −kx
ϕ (k) dk +
e
−∞−iα
N −iα
≤
−N Z∞
Z
−i (k−iα)3 −(k−iα)x 3f
sup | ϕ (ζ)|
+
e
dk
∧
ζ∈A
(2.7)
≤
−∞
∧
N
α3
Z∞
α3
−∞
Z∞
sup | ϕ (ζ)| e 3f eα|x|
ζ∈A
≤
∧
sup | ϕ (ζ)| e 3f eα|x|
ζ∈A
r
=
2
e−
e−
αk2
f
αk2
f
e|k| |x| dk
(ek |x| + e−k |x| ) dk
−∞
f |x|2
∧
α3
πf
sup | ϕ (ζ)| e 3f eα|x| e 4α
α ζ∈A
(x ∈ C) .
(2.13)
Finally combining (2.13), (2.12), (2.11), (2.10) and (2.8) proves that the integral (2.5)
exists for all x ∈ C and obeys the estimate
3
Z
∧
∧
α3
α3 −i k
3f −kx
ϕ (k) dk ≤ eN |x| IN + α sup | ϕ (ζ)| e 3f eα|x| eN |x| 1 + e f
e
ζ∈A
γα
r
+2
f |x|2
∧
α3
πf
sup | ϕ (ζ)| e 3f eα|x| e 4α
α ζ∈A
(x ∈ C) .
(2.14)
In particular, by (2.5) and (2.14),
2
|ψeα,f (x)| ≤ aeb|x|
(x ∈ C) ,
(2.15)
where a, b < ∞ are some real constants,
depending on α, f, ϕ and N , but not on x.
We abbreviate u(k, x) := e
Z
ψeα,f ( · ) = u(k, · ) dk
−i
k3
3f
−kx
∧
ϕ (k). Then
γα
is holomorphic in all of C, since u(k, · ) is complex differentiable for all k ∈ γα , and
1
e
2 (∂Re x − i∂Im x )u(k, x) is continuous on γα × C. This proves the entireness of ψα,f . Then,
by (2.15), the entire function ψeα,f is of order ≤ 2.
Proof of independence of α: Since for the path Γ±R : t 7→ ±R − it (0 < α1 ≤ t ≤ α0 ) we
have
Zα0 3
3
Z
(2.7)
∧
∧
tR2
t
−kx
−i k
3f
ϕ (k) dk ≤
e
sup | ϕ (±R − it)| e 3f e− f e|R| |x| e|t| |x| dt
t∈[αa ,α0 ]
Γ±R
=
o(1)
α1
(R → ∞, x ∈ C) ,
19
∧
it follows from the analyticity properties of ϕ that ψeα,f (x) (x ∈ C) is independent of
α ∈ (0, α0 ].
Thus it remains to show that ψeα,f and ψf coincide. By an easy density argument, to
identify ψeα,f with ψf it suffices to show
lim(g , ψeα,f )L2 = (g , ψf )L2
α↓0
(g ∈ C0∞ (R)) ,
(2.16)
since (g , ψeα,f )L2 is actually independent of α. One finds
Z
Z
3
√
∧
(2.5)
−i k
3f −kx
ϕ (k) dk dx
2π(g , ψeα,f )L2
=
g(x) e
γα
R
−N Z∞
−i (k−iα)3 −(k−iα)x ∧
Z
3f
ϕ (k − iα) dk dx
+
e
g(x)
Z
=
−∞
R
Z
N
Z
+
g(x)e
3
∧
−i k
3f −kx
ϕ (k) dk dx + o(1)
(α ↓ 0) .
Rx k∈[−N,N ]
(2.17)
Setting
Fα (k, x) := g(x)e
(k−iα)3
−i
−(k−iα)x ∧
3f
ϕ (k − iα) ,
(2.18)
∧
by (2.7) and the uniform bound for ϕ in A, one gets
∧
α3
|Fα (k, x)| ≤ sup | ϕ (ζ)| e 3f |g(x)|e−
αk2
f
e|k| |x| eα|x|
((k, x) ∈ Ω) ,
(2.19)
ζ∈A
where Ω = (R\[−N, N ]) × R. Thus Fα ∈ L1 (Ω, dk dx). Then Fubini’s Theorem gives
Z
−N Z∞
Z
−i (k−iα)3 −(k−iα)x ∧
3f
ϕ (k − iα) dk dx
+
e
g(x)
−∞
R
=
−N
Z Z
Rx
N
+
−∞
Z∞ Fα (k, x) dk dx .
(2.20)
N
∧
By Cauchy estimates, the uniform boundedness of ϕ in an open neighborhood of A
∧0
implies that its derivative ϕ is uniformly bounded in a slightly smaller domain, and one has
∧
∧0
∧
| ϕ (k − iα)− ϕ (k)| ≤ α sup | ϕ (ζ)| ,
ζ∈γ
(2.21)
20
where γ denotes the path k − iα(1 − t) (t ∈ [0, 1]). We claim that
Z
Z
(k−iα)3
∧
Jα :=
g(x)e−i 3f eikx (eαx − 1) ϕ (k − iα) dk dx → 0
(α ↓ 0) . (2.22)
Rx Rk \[−N,N ]
∧
Indeed, by (2.6) and the uniform bound on ϕ in A,
Z
α3
Z
|g(x)| e−
|Jα | ≤ e 3f
αk2
f
∧
|eαx − 1| | ϕ (k − iα)| dk dx
Rx Rk \[−N,N ]
≤e
∧
α3
3f
sup | ϕ (ζ)|
ζ∈A
Z
Z
αk2
|g(x)| |eαx − 1| dx
e− f dk ,
supp g
(2.23)
R
where
Z
e−
αk2
f
√
πf
dk = √ ,
α
|eαx − 1| ≤ c α|x|
(x ∈ supp g)
(2.24)
R
for some constant c depending on (the support of) g. Thus from (2.24) and (2.23) it follows
that
|Jα | ≤ C
where C
α3
e 3f
∧
g,ϕ,f
√
α → 0
(α ↓ 0) ,
(2.25)
∧
∧
g,ϕ,f
is some constant (depending on g, ϕ and f ).
Now combining (2.24), (2.21) and (2.7) shows
Z
Z
g(x)e−i
(k−iα)3
3f
∧
∧
eikx ϕ (k − iα)− ϕ (k) dk dx
Rx Rk \[−N,N ]
∧0
α3
≤ e 3f sup | ϕ (ζ)|
ζ∈A
Z
Z
|g(x)| dx
supp g
√
= O( α)
αe−
αk2
f
dk
R
(α ↓ 0) .
(2.26)
By (2.18), (2.22) and (2.26) it follows that
lim
−N
Z Z
α↓0
Rx
−∞
+
Z∞ Fα (k, x) dk dx
N
= lim
−N
Z Z
α↓0
Rx
−∞
+
Z∞ N
g(x)eikx e−i
(k−iα)3
3f
∧
ϕ (k) dk dx .
(2.27)
21
Note that g( · ) and e−i
(2.20) and (2.17) imply
( · −iα)3
3f
∧
ϕ ( · ) (for α > 0) both are in L1 (R) ∩ L2 (R). Then (2.27),
Z Z
(k−iα)3 ∧
1
lim(g , ψeα,f )L2 = lim √
g(x)eikx e−i 3f ϕ (k) dk dx
α↓0
α↓0 2π
R R
Z
( · −iα)3 ∧ ∨
= lim g(x) e−i 3f ϕ (x) dx ,
(2.28)
α↓0
R
where
∨
: L (R) ∩ L (R) → L∞ (R) ∩ L2 (R). By Plancherel’s Theorem
1
2
lim (g , ψeα,f )L2
∧
(2.28)
= lim (g , e−i
α↓0
( · −iα)3
3f
α↓0
∧
ϕ)L2
(2.29)
and
∧
∧
(2.3) ∧
(g , ψf )L2 = (g , ψf )L2 = (g , e−i
( · )3
3f
∧
ϕ)L2 .
(2.30)
Then
lim (g , ψeα,f )L2 − (g , ψf )L2 (2.29)
(2.30)
=
∧
( · −iα)3
( · )3
∧ lim g , (e−i 3f − e−i 3f ) ϕ L2 α↓0
∧
αk2
iα2 k
α3 ∧
lim(g , e− f − f + 3f ϕ)L2 ≤
k g k2 lim ke−
=
α↓0
(2.6)
α↓0
∧
α( · )2
f
α↓0
α3
∧
e 3f ϕ k2 ,
and applying Lebesgue’s Theorem on dominated convergence to lim ke−
α( · )2
f
α↓0
α3
∧
e 3f ϕ k2 implies
lim (g , ψeα,f )L2 − (g , ψf )L2 = 0 ,
α↓0
which is (2.16).
Proof of Proposition 1.4:
Let f > 0. We rewrite (1.9) as
Ff,ϕ (z) = 1 − z − rf,ϕ (z) ,
Proof of entireness:
gives
rf,ϕ (z) := (ϕ , (p2 + f x − z)−1 ϕ)L2
(z ∈ C+ ). (2.31)
Obviously, the functions rf,ϕ and Ff,ϕ are analytic in C+ . Lemma 2.3
p3
p3
rf,ϕ (z) = (e−i 3f ϕ , (f x − z)−1 e−i 3f ϕ)L2 = (ψf , (f x − z)−1 ψf )L2
Z
ψ f (x)ψf (x)
1
=
dx (z ∈ C+ ) ,
f
x − fz
(2.32)
R
where ψf is given by (2.3). By Lemma 2.4 and Lemma 2.7 the function ψf has an entire
extension. In (2.32) we can deform the path R into a contour γ in C− such that the
22
singularity x = z/f for Im z < 0 is enclosed by this contour γ and the real line. Then using
the Residue Theorem one obtains
z z
2πi
ψf
+ (ϕ , (p2 + f x − z)−1 ϕ)L2
rf,ϕ (z)c,Ωf =
ψf
(z ∈ C− ) .
(2.33)
f
f
f
For z ∈ R close to a fixed x0 ∈ R (more precisely, |z − x0 | <
extension is given by
δ
2
· f for some fixed δ > 0), this
x0
c,Ωf
rf,ϕ (z)
f −δ
Z
z z 1 Z
2πi
+
ψf
ψf
+
=
f
f
f
f
−∞
Kδ (
+∞
Z
+
x0
f
)
x0
f
+δ
ψ f (x)ψf (x)
dx, (2.34)
x − fz
Kδ ( xf0 ),
denotes the clockwise oriented semicircle in the closed upper half complex
where
plane with radius δ and center x0 /f . Since ψf is entire by Lemma 2.4 and the integral in
(2.34) exists for all z ∈ R, it follows from (2.32), (2.33) and (2.34) that the extension of rf,ϕ
c,Ω
is entire. Thus, by (2.31), Ff,ϕ f is entire.
Fix f > 0. For |Im fz | > 1 one gets, using the identity in (2.32),
Z
|ψ f (x)ψf (x)|
z
1
1
dx
|(ϕ , (p2 + f x − z)−1 ϕ)L2 | = |(ψf , (x − )−1 ψf )L2 | ≤
f
f
f
|x − fz |
Proof of finite order:
R
≤
1
f
Z
R
|ψ f (x)ψf (x)|
1
dx <
|Im fz |
f
Z
|ψf (x)|2 dx
R
1
= kψf k22 .
f
(2.35)
For |Im fz | ≤ 1 boundedness of the resolvent matrix element is shown by contour deformation
and analytically continuing from {z ∈ C | Imf z > 1} into the strip σf := {z ∈ C | | Imf z | ≤ 1}:
Denoting by K2 ( Ref z ) the counterclockwise oriented semicircle in the closed lower half complex plane with radius 2 and center Re z/f , we use the contour (−∞ , Ref z − 2) ∪ K2 ( Ref z ) ∪
( Ref z + 2 , +∞) to get (using the analyticity of rf,ϕ established above)
Re z
f −2
2
−1
(ϕ , (p + f x − z)
c,σ
ϕ)L2 f
1
=
f
Z
Z
+
−∞
for all z ∈ C+ ∪ σf . For z ∈ σf , one finds
Z
1
ψ f (x)ψf (x) 1
dx ≤
z
f
x− f
f
z
Re z
R\( Re
f −2, f +2)
z
K2 ( Re
f )
+∞
Z
ψ (x)ψf (x)
f
+
dx
x − fz
Z
z
Re z
R\( Re
f −2, f +2)
1
≤
2f
Z
R
(2.36)
Re z
f +2
|ψf (x)|2
dx
|x − fz |
|ψf (x)|2 dx =
1
kψf k22 .
2f
(2.37)
23
Since |x − z/f | ≥ 1 for x ∈ K2 ( Ref z ) and z ∈ σf , one has
ψ f (x)ψf (x) 1
dx ≤
x − fz
f
Z
1
f
z
K2 ( Re
f )
Z
|ψ f (x)||ψf (x)| |dx| .
(2.38)
z
K2 ( Re
f )
By (2.4),
Z
1
f
Z
1
|ψ f (x)||ψf (x)| |dx| ≤
f
z
K2 ( Re
f )
2
2
aeb|x| aeb|x| |dx|
z
K2 ( Re
f )
≤
2π
f
≤
2
e
a eb |z|
e f2
f
2
(a2 e2b|x| )
max
z
x∈K2 ( Re
f )
(
|Im z|
≤ 1)
f
(2.39)
for some finite real constants e
a and eb. Finally, for Im z/f < −1, we use (2.33) and Lemma
2.4 (more precisely, (2.4)) to get
B
c,Ω
|rf,ϕ f (z)| ≤ A e
|z|2
f2
(2.40)
for some finite constants A and B (possibly depending on f and ϕ). Combining the estic,Ω
mates (2.35), (2.39) together with (2.37), and (2.40) proves that rf,ϕ f is of order ≤ 2. Thus,
c,Ω
by (2.31), Ff,ϕ f is of order ≤ 2.
c,Ω
Proof of existence of infinitely many zeros: In order to prove that Ff,ϕ f has infinitely many
zeros it suffices to show (by Theorem 2.2) that there exist no polynomials (in one variable)
c,Ω
P and Q such that Ff,ϕ f (z) = P (z)eQ(z) for all z ∈ C:
Since p2 + f x is self-adjoint, one has |rf,ϕ (z)| ≤ const. |Im z|−1 , Im z > 0. In particular,
lim
Im z→∞
rf,ϕ (z) = 0 .
(2.41)
Now assume that
c,Ω
Ff,ϕ f (z) = P (z)eQ(z)
(z ∈ C)
(2.42)
for some polynomials P and Q. Then
0=
lim
Im z→∞
c,Ω
Ff,ϕ f (z) − P (z)eQ(z) =
lim
Im z→∞
1 − z − P (z)eQ(z) .
(2.43)
Equation (2.43) implies that Q(z) is at most of degree one since otherwise the exponential
would approach infinity rapidly along certain rays in the upper half plane. In addition if
Q(z) = az+b with a 6= 0 the same reasoning implies a = ie
a with e
a > 0 which also contradicts
(2.43) for z ∈ C+ . Thus
c,Ω
c,Ω
Ff,ϕ f (z) = 1 − z − rf,ϕ f (z) = c P (z)
(z ∈ C)
(2.44)
24
for some constant c. But, for Im z > 0,
(2.31)
c,Ω
lim (−i Im z)rf,ϕ f (i Im z) =
Im z→∞
Z
lim
Im z→∞
|ϕ(x)|2 (−i Im z)
dx = kϕk22 .
p2 − f x − i Im z
(2.45)
R
Thus there exists no polynomial P fulfilling (2.44) (except if ϕ = 0). This finishes the proof
of Proposition 1.4.
2.2
Proof of Theorem 1.7
Our proof of Theorem 1.7 uses the expansions and estimates given in Proposition 2.8, Corollary 2.9, Proposition 2.10, Proposition 2.12 and Corollary 2.13 below.
Proposition 2.8
Let ϕ and M be as in Theorem 1.7. Let α ∈ (0, min{tan θ0 , k0 }).
Define the paths
t − iα , |t| > 1
C± : R± → C , t 7→ k(t) :=
(2.46)
(sgn (t) − iα)|t| , −1 ≤ t ≤ 1
∧
(which lie in the region of analyticity for ϕ). Then, for f > 0 sufficiently small,
Z
3
r πf
2 3/2
∧ √
− fi k3 −kz ∧
−iπ/4 ϕ
ϕ (k) dk =
e
(
(2.47)
z)
+
O(f
)
ei 3f z ,
e
z 1/2
C+
Z
e
− fi
k3
3
−kz
∧
ϕ (k) dk =
C−
r πf
√
∧
2 3/2
iπ/4 ϕ
e
(−
z)
+
O(f
)
e−i 3f z + O(f )
z 1/2
r
=
√
2 3/2
πf iπ/4 ∧
ϕ (− z)e−i 3f z + O(f )
e
1/2
z
(2.48)
(2.49)
as f ↓ 0, uniformly for z ∈ M .
∧
Note that by analyticity of ϕ, Lemma 2.4, Lemma 2.6 and Lemma 2.7 and contour deformation, one has
Z
Z
3
3
√
∧
∧
(2.5)
−i k
−i k
3f −kx
3f −kx
ϕ (k) dk =
ϕ (k) dk (x ∈ C) . (2.50)
2π ψf (x) =
e
e
γα
C− ∪C+
Then as a corollary of Proposition 2.8 and (2.50), one gets
Corollary 2.9
Let ϕ and M be as in Theorem 1.7. Let ψf be given by (2.3). Then
z r πf
2 3/2
√
∧ √
2πψf
=
e−iπ/4 ϕ ( z) + O(f ) ei 3f z
1/2
f
z
r πf
√
∧
2 3/2
iπ/4 ϕ
+
e
(−
z)
+
O(f
)
e−i 3f z + O(f )
(2.51)
1/2
z
25
Figure 2.2: The contour C− ∪ C+ defined in (2.46).
and
r πf
2 3/2
√
∧
−iπ/4 ϕ
=
2π ψ f
e
(−
z)
+
O(f
)
ei 3f z
f
z 1/2
r πf
∧ √
2 3/2
iπ/4 ϕ
(
+
e
z)
+
O(f
)
e−i 3f z + O(f )
1/2
z
as f ↓ 0, uniformly for z ∈ M .
√
z
(2.52)
Proof of Proposition 2.8: We break up each of the integrals in (2.47) and (2.48) into
three parts after deforming the contour C± as shown in Figure 2.3. There
√ is one contribution handled by the method of
steepest
descents
(see,
e.g.,
[E])
near
±
z. There is one
√
√
contribution integrating from − z to 0 (from 0 to z, respectively).
And
finally there is
√
an√error term where we integrate from near −∞ in R to near − z and similarly from near
+ z to near +∞
√ in R.
The bounds on z in the definition of M in Theorem 1.7 make sure that for z ∈ M the
∧
deformed integration contours used below lie in the region of analyticity for ϕ.
First we prove (2.47). For this we write
Z
3
− i k −kz ∧
ϕ (k) dk = I1 (z, f ) + I2 (z, f ) + I3 (z, f )
e f 3
(z ∈ M , f > 0) ,
(2.53)
C+
where Ij (z, f ) (j ∈ {1, 2, 3}) are defined in (2.55), (2.61) and (2.74) below.
√
Integration from 0 to z: Defining
x :=
z 3/2
,
f
s
u := s(1 − )1/2
3
(2.54)
26
Figure 2.3: Deformed C− ∪ C+ as used in the method of steepest descents.
gives, for z ∈ M and f > 0,
√
Zz
Z1
3
√ 2ix/3
√
∧
2
− fi k3 −kz ∧
ϕ
e
(k) dk = z e
e−ixs (1−s/3) ϕ ((1 − s) z) ds
I1 (z, f ) :=
0
=
=
√
√
0
z
z
√
Z2/3
√ ds(u)
2 ∧
2ix/3
du
e
e−ixu ϕ ((1 − s(u)) z)
du
0
√
Z2/3
∧ √
2
ϕ ( z)e2ix/3
e−ixu du + E1 (z, f )e2ix/3 ,
(2.55)
(2.56)
0
where
√
∧
√ ds(u) ∧ √
√ Z2/3
d −ixu2 ϕ ((1 − s(u)) z) du − ϕ ( z)
i z
e
du .
E1 (z, f ) :=
x
du
2u
(2.57)
0
After integrating by parts one finds
∧ (n) √ |E1 (z, f )| ≤ cf sup | ϕ (t z)| t ∈ (0, 1] , n ≤ 2
(2.58)
27
for some c > 0. We calculate
√
r
Z2/3
Z∞
1 π
−ixu2
−ixu2
−1
e
du = e
du + O(|x| ) =
+ O(|x|−1 )
2 ix
0
0
r
πf −iπ/4
1
(2.54)
√
e
=
+ O(f ) (f ↓ 0) .
2 z
z 1/2
Then using (2.59) and (2.58) in (2.56) gives
2
1 r πf
∧ √
−iπ/4 ϕ
e
(
z)
+
O(f
)
ei 3f
I1 (z, f ) =
2 z 1/2
z 3/2
(|x| → ∞)
(f ↓ 0) ,
(2.59)
(2.60)
uniformly for z ∈ M .
√
Integration near + z: Define the integral
Z
3
− fi k3 −kz ∧
ϕ (k) dk (z ∈ M, f > 0)
e
I2 (z, f ) :=
(2.61)
e+
Γ
e + near k =
along a particular path Γ
gives
−
√
e + is given as follows: We write k = ζ + √z which
z. Γ
k3
ζ3 √ 2iz 3/2
−i
− kz = −i
+ zζ 2 .
3
3
3
(2.62)
Using the method of steepest descents we want ζ to be on a curve, Γ+ , satisfying
Re
ζ3
3
+
√
zζ 2 = 0.
(2.63)
We want 0 ≤ Re ζ ≤ δ1 for some sufficiently small δ1 > 0 and Im ζ ≤ 0. With the
abbreviations
√
√
(2.64)
γ := Re z , −ν := Im z < 0 , x := Re ζ , y := Im ζ
we obtain for ζ = x + iy on Γ+
y = −γ −1 (|z|1/2 − ν)x +
∞
X
An xn
(x ∈ [0, δ1 ])
(2.65)
n=2
for some An ∈ R. On this curve we have
−i
ζ3
3
+
√
zζ
2
2 2
= −b x +
∞
X
Bn xn ,
(2.66)
n=3
where
b2 := 2γ −2 |z|(|z|1/2 − ν)
(2.67)
28
and Bn is real. The power series in (2.65) and (2.66) converge for |x| < γ. (Note that for
z ∈ M , |z|1/2 − ν > 0.) We have
Z
Z
3
3 √
√
2 3/2
(2.62)
− fi ζ3 + zζ 2 ∧
− fi k3 −kz ∧
−i 3f
z
ϕ
ϕ (ζ + z) dζ ,
e
(k) dk =
e
(2.68)
e
Γ+
e+
Γ
where Γ+ is given by (2.65). Substituting (2.65) and (2.66) into (2.68) and expanding
∧
√
ϕ (ζ + z) in a power series we obtain
r
Z
3 √
√
1
πf −iπ/4 ∧ √
− fi ζ3 + zζ 2 ∧
ϕ ( z) + O(f ) (f ↓ 0) , (2.69)
ϕ (ζ + z) dζ =
e
e
2 z 1/2
Γ+
and thus
I2 (z, f ) = e
2 3/2
i 3f
z
1 r πf
∧ √
−iπ/4 ϕ
e
(
z)
+
O(f
)
2 z 1/2
(f ↓ 0) ,
(2.70)
uniformly for z ∈ M .
e + to infinity: We can give an exact expression for the path Γ+ (defined by
Connecting Γ
(2.65)) as follows:


s
2
γ + x3
ν
ν
,
y = x
−
(2.71)
+
γ+x
γ+x
γ+x
from which it follows that
s
y
γ + x3
≤
x
γ+x
(2.72)
for all ζ = x + iy on Γ+ . We continue the path Γ+ by starting at ζ(δ1 ) =: t0 − iΘ0 (some
t0 , Θ0 > 0). From (2.72) with x = t0 and y = −Θ0 we get t0 > Θ0 > 0. Thus, for
ζ(t) = t − iΘ0 (t ≥ t0 ), we have
3
t
ζ (t) √ 2
Θ0 2
2
Re −i
+ zζ (t)
= Re −i(t − Θ0 − 2itΘ0 )
+ γ − i(ν +
)
3
3
3
t
Θ0 = −(t2 − Θ20 ) ν +
− 2tΘ0
+γ
3
3
=: −G(t) < 0 (t ≥ t0 ) .
(2.73)
The integral
2
I3 (z, f ) := ei 3f
z 3/2
∞−iΘ
Z 0
e
t0 −iΘ0
− fi
ζ3
3
√
+ zζ 2 ∧
ϕ (ζ +
√
z) dζ
(z ∈ M , f > 0)
(2.74)
29
obeys the estimate
∞−iΘ
Z 0
e
− fi
√
+ zζ 2 ∧
ζ3
3
ϕ (ζ +
√
Z∞
√
∧
z) dζ ≤ e−G(t)/f | ϕ (t + z − iΘ0 )| dt
t0
t0 −iΘ0
∧
≤ e−c/f sup | ϕ (t +
√
z − iΘ0 )|
(2.75)
t≥t0
for some c > 0 independent of f , uniformly for z ∈ M . Note that Θ0 is of the order of
δ1 so that c can be taken independent of ν. Note also that if δ1 is taken sufficiently small
∧
relatively to γ, this path is in a region of analyticity of ϕ.
Finally combining (2.75), (2.74), (2.70), (2.60) and (2.53) proves (2.47).
Next we prove (2.48). For this we write
Z
3
− fi k3 −kz ∧
ϕ (k) dk = I4 (z, f ) + I5 (z, f ) − I6 (z, f )
e
(z ∈ M , f > 0) ,
(2.76)
C−
where Ij (z, f ) (j ∈ {4, 5, 6}) are defined in (2.77), (2.90) and (2.109) below.
√
Integration from − z to 0:
Z0
I4 (z, f ) :=
e
− fi
Using (2.54) we obtain, for z ∈ M and f > 0,
k3
3
−kz
∧
ϕ (k) dk
(2.77)
√
− z
=
√
−2ix/3
Z1
ze
eixs
2
∧
(1−s/3) ϕ
√
(−(1 − s) z) ds
0
√
Z2/3
√ −2ix/3
√ ds(u)
2 ∧
= ze
du
eixu ϕ (−(1 − s(u)) z)
du
0
√
Z2/3
√ ∧
√
2
−2ix/3
= z ϕ (− z) e
eixu du + E2 (z, f )e−2ix/3 ,
(2.78)
0
where
√
−i z
E2 (z, f ) :=
x
√
Z2/3
0
∧
√
√ ds(u) ∧
d ixu2 ϕ (−(1 − s(u)) z) du − ϕ (− z)
e
du .
du
2u
(2.79)
30
Note that |e−2ix/3 | ≤ 1, possibly much smaller than 1. On the other hand if |Im z| ≤ cf for
some c > 0, the term |e−2ix/3 | may not be small. An integration by parts shows
∧ (n)
√ |E2 (z, f )| ≤ c|x|−1 sup | ϕ (−t z)| t ∈ (0, 1], n ≤ 2 |e2ix/3 |,
where by (2.54) |x|−1 = O(f ) as f ↓ 0, uniformly for z ∈ M .√The integral
(2.78) is easily handled by Cauchy’s Theorem: Using v := u x we obtain
√
√
Z2/3
Z2x/3
2
1
ixu2
e
du = √
eiv dv
x
0
0
√
Re 2x/3
Z
2
1
1
=√
eiv dv + √
x
x
0
√
Z2x/3
Re
√
(2.80)
R√
0
2
eiv dv .
2/3 ixu2
e
du in
(2.81)
2x/3
We set
p
2x/3 > 0 ,
−b := Im
eiv dv = O(a−1 )
(a → ∞)
a := Re
p
2x/3 < 0 .
(2.82)
Since
Z∞
2
(2.83)
a
and
Z∞
2
eiv dv =
1 √ iπ/4
πe
2
(2.84)
0
we estimate the integral
Za
2
eiv dv =
R Re √2x/3
0
2
eiv dv in (2.81) by
1 √ iπ/4
πe
+ O(a−1 )
2
(a → ∞) .
(2.85)
0
With the substitution v := a − ibt we obtain
a−ib
Z
2
eiv dv
R1
2
2 2
= −ib ei(a −t b −2itab) dt
(2.86)
0
a
and thus
Z
a−ib
R1
2
eiv dv ≤ b e2abt dt =
a
0
1
2a
(e2ab − 1) ≤
1
2a
(2.82) 1
2a
e2ab =
|e2ix/3 | .
(2.87)
31
Note that, by (2.82) and (2.54),
√
p
1
3f
= O( f )
0 < a−1 = √
3/4
2 Re (z )
(f ↓ 0) ,
uniformly for z ∈ M .
Finally combining (2.87), (2.85), (2.82), (2.81), (2.80) and (2.78) gives
1 r πf
√
∧
3/2
2
iπ/4 ϕ
e
(−
z)
+
O(f
)
e−i 3f z + O(f )
I4 (z, f ) =
1/2
2 z
(2.88)
(2.89)
as f ↓ 0, uniformly for z ∈ M .
√
Integration near − z: Define the integral
Z
3
− fi k3 −kz ∧
ϕ (k) dk (z ∈ M , f > 0)
I5 (z, f ) :=
e
(2.90)
e−
Γ
e − near k = −√z. We write k = ζ − √z which gives
along a particular path Γ
k3
ζ3 √ 2iz 3/2
−i
− kz = −i
− zζ 2 .
3
3
3
(2.91)
Using the method of steepest descents we want ζ to be on a curve, Γ− , satisfying
Re
ζ3
3
−
√
zζ 2 = 0.
(2.92)
With the abbreviations (2.64) we want −δ2 ≤ x ≤ 0 for some sufficiently small δ2 and y ≤ 0,
and we obtain for ζ = x + iy on Γ−
y = γ −1 (|z|1/2 + ν)x +
∞
X
en xn
A
(x ∈ [−δ2 , 0])
(2.93)
n=2
en ∈ R. On this curve we have
for some A
−i
ζ3
3
−
√
∞
X
en xn ,
zζ 2 = −eb2 x2 +
B
(2.94)
n=3
where
eb2 := 2γ −2 |z|(|z|1/2 + ν)
(2.95)
en is real. The power series in (2.93) and (2.94) converge for |x| ≤ γ. One has
and B
Z
Z
3
3 √
√
2 3/2
(2.91)
− i k −kz ∧
− i ζ − zζ 2 ∧
ϕ (k) dk =
ϕ (ζ − z) dζ ,
ei 3f z
e f 3
e f 3
(2.96)
e−
Γ
Γ−
32
where Γ− is given by (2.93). Substituting (2.93) and (2.94) into (2.96) and expanding
∧
√
ϕ (ζ − z) in a power series we obtain
r
Z
3 √
√
√
1
πf iπ/4 ∧
− fi ζ3 − zζ 2 ∧
ϕ (− z) + O(f )
ϕ (ζ − z) dζ =
e
e
(2.97)
1/2
2 z
Γ−
as f ↓ 0, uniformly for z ∈ M , and thus
1 r πf
√
∧
2 3/2
iπ/4 ϕ
I5 (z, f ) =
e
(−
z)
+
O(f
)
e−i 3f z
1/2
2 z
(f ↓ 0) ,
(2.98)
uniformly for z ∈ M .
e − to infinity: We give an exact expression for the path Γ− , defined in (2.93),
Connecting Γ
as follows:


s
2
γ − x3
ν
ν
.
y = x
+
(2.99)
+
γ−x
γ−x
γ−x
We continue the path Γ− from its end point ζ(−δ2 ) = −t0 − iΘ0 (some t0 , Θ0 > 0), so that
ζ(t) = −t − iΘ0 , t ≥ t0 . For this ζ(t) (t ≥ t0 ) we calculate
3
t
Θ0 ζ (t) √ 2
2
2
− zζ (t)
+γ +i ν−
Re −i
= Re −i(t − Θ0 + 2itΘ0 ) −
3
3
3
t
Θ
0
+γ
= (t2 − Θ20 ) ν −
− 2tΘ0
3
3
Θ0 = −t2 (Θ0 − ν) − 2γtΘ0 − Θ20 ν −
3
=: −H(t)
(2.100)
and
H 0 (t) = 2t(Θ0 − ν) + 2γΘ0 ,
H 00 (t) = 2t(Θ0 − ν).
(2.101)
From the theory behind steepest descents it follows that H(t0 ) > 0 for t0 > 0. If Θ0 > ν,
then H(t) is positive and increasing for t ≥ t0 .
Equation (2.99) with y = −Θ0 and x = −t0 gives


s
2
γ + t30
ν
ν
.
Θ0 = t 0 
+
+
(2.102)
γ + t0
γ + t0
γ + t0
From (2.102) it follows (after some calculations) that t0 > γν|z|−1/2 implies Θ0 > ν. (Note
that γν|z|−1/2 < ν for z ∈ M .) Furthermore, since (γ + t30 )(γ + t0 )−1 > 1/3, it follows that
t0
Θ0 > √ .
3
(2.103)
33
We now have to pick t0 and Θ0 such that Θ0 − ν > 0 and the path −t − iΘ0 −
∧
is in the region of analyticity of ϕ.
We fix ε > 0 sufficiently small and first consider the case ν < ε:
√
t0 := 2ε 3 .
√
z (t ≥ t0 )
If ν < ε, set
(2.104)
Then
Θ0 − ν
(2.103)
>
t
√0 − ν
3
(2.104)
=
2ε − ν > ε > 0.
(2.105)
Furthermore, since γ ≥ ε2 (by the definition of M in Theorem 1.7), equation (2.102) implies
√
(2.106)
Θ0 < 2ε(1 + 3).
Then, if ε > 0 is small compared to k0 and much smaller than ε2 , the path −t−iΘ0 −(γ −iν)
∧
(t ≥ t0 ) is in the region of analyticity of ϕ.
We now consider the case ν ≥ ε:
If ν ≥ ε, we set
γ = αν ,
(2.107)
where the number α = γ/ν is bounded above and below (by the definition of M in Theorem
1.7 and because of ν ≥ ε). We claim that there exist a λ > 0 such that for
t0 := λγ
(2.108)
one has Θ0 = ν(1 + ε). (Then, clearly, Θ0 − ν > 0.) Indeed, dividing (2.102) by ν and then
inserting (2.108) and (2.107) gives
s
2
1 + λ3
Θ0
λ
λ
=
+ (αλ)2
=: g(λ) .
ν
1+λ
1+λ
1+λ
Note that g is continuous, g(0) = 0 and limλ→∞ g(λ) = ∞. Thus, by the mean value
theorem, there exists a λ > 0 for which g(λ) = 1 + ε.
Choosing ε > 0 such that Θ0 = ν(1 + ε) < k0 keeps −t − iΘ0 − (γ − iν) (t ≥ t0 ) in the region
∧
of analyticity of ϕ.
√
Now, by choosing t0 and Θ0 in the way explained above, the path −t − iΘ0 − z (t ≥ t0 ,
∧
z ∈ M ) is in the region of analyticity for ϕ, H(t) (t ≥ t0 ) is positive and increasing, and we
obtain, for z ∈ M and f > 0,
2
I6 (z, f ) := e−i 3f z 3/2
−∞−iΘ
Z 0
e
− fi
ζ3
3
√
− zζ 2 ∧
ϕ (ζ −
√
z) dζ ,
(2.109)
−t0 −iΘ0
Z∞
(2.100)
√
∧
|I6 (z, f )| ≤
e−H(t)/f | ϕ (−t − iΘ0 − z)| dt .
t0
(2.110)
34
Since H 00 (t) > 0 (t ≥ t0 ), we have
H(t) > H(t0 ) + (t − t0 )H 0 (t0 ) .
(2.111)
Then, by use of (2.111) and (2.101) in (2.110), we get
|I6 (z, f )| ≤ e−H(t0 )/f
√
∧
1
sup | ϕ (−t − z − iΘ0 )|
2γΘ0 t≥t0
(z ∈ M , f > 0) .
(2.112)
P∞ e n
We see that from the method of steepest descents the exponent eb2 x2 − n=3 B
n x from
2
e
(2.94) is increasing while b ≥ 2γ so that H(t0 ) is bounded below independent of ν. Then
|I6 (z, f )| ≤ e−c/f
(2.113)
for some c > 0 (independent of f ) and all f > 0, uniformly for z ∈ M .
We thus obtain (2.48) by combining (2.113), (2.98) and (2.89). Estimate (2.49) easily
2 3/2
follows from (2.48) by noticing that |e−i 3f z | ≤ 1 (z ∈ M, f > 0).
We now consider the resolvent matrix element (ϕ , (p2 + f x − z)−1 ϕ)L2 for z ∈ C− :
Proposition 2.10
Let ϕ and M be as in Theorem 1.7. Then
(ϕ , (p2 + f x − z)−1 ϕ)L2 = (ϕ , (p2 − z)−1 ϕ)L2
i
−
f
Z0
e
− fi
k3
3
−zk
∧
ϕ (k) dk
−∞
1
+
z
Z0
e
− fi
k3
3
ϕ(−k) dk
−∞
Z0
e
− fi
k3
3
−zk ∧
∧
ϕ(−k) dk ϕ (+0) +
−∞
+ O(f )
−zk ∧
Z0
e
− fi
k3
3
−zk
∧
∧
ϕ (k) dk ϕ(−0)
−∞
(f ↓ 0) ,
(2.114)
uniformly for z ∈ M . In particular,
(ϕ , (p2 + f x − z)−1 ϕ)L2 = (ϕ , (p2 − z)−1 ϕ)L2
√ ∧ √
4
π ∧
+ √ ϕ (− z)ϕ( z)e−i 3f
z
z 3/2
p
+ O( f )
(2.115)
as f ↓ 0, uniformly for z ∈ M .
Remark 2.11
∧
For ϕ ∈ Dθ0 , ϕ is not necessarily continuous at zero; cf. Remark 1.8.1.
∧
∧
However, by (1.13), the left and right limits ϕ (∓0) and ϕ(∓0) exist.
35
Proof of Proposition 2.10:
Z0
e
− fi
k3
3
−zk
By contour deformation and Proposition 2.8,
r πf
√
∧
2 3/2
ϕ (k) dk =
eiπ/4 ϕ (− z) + O(f ) e−i 3f z + O(f ) ,
1/2
z
∧
−∞
(2.116)
Z0
e
− fi
k3
3
−zk ∧
ϕ(−k) dk =
−∞
Z∞
e
− fi
k3
3
−zk
∧
ϕ (k) dk
0
r πf
∧ √
2 3/2
eiπ/4 ϕ( z) + O(f ) e−i 3f z + O(f )
=
1/2
z
(2.117)
as f ↓ 0, uniformly for z ∈ M . Then (2.115) is proven by inserting (2.116) and (2.117) into
(2.114).
It remains to prove the expansion (2.114): For z ∈ C− the resolvent matrixelement
(ϕ , (p2 + f x − z)−1 ϕ)L2 is given by
1
(ψf , (x − z/f )−1 ψf )L2
f
1 ∧
=
ψ f , ((x − z/f )−1 ψf )∧ L2
f
3
Z Z i k23
k1
− fi
∧
∧
i
f
3 −zk2
3 −zk1
ϕ(k2 ) ϕ (k1 ) dk1 dk2 .
e
=−
e
f
(ϕ , (p2 + f x − z)−1 ϕ)L2 =
k2 >k1
(2.118)
We split up the region of integration into the three components
{(k1 , k2 ) ∈ R × R | k2 ≥ k1 } = {k2 ≥ k1 ≥ 0} ∪ {k1 ≤ 0 ∧ k2 ≥ 0} ∪ {k1 ≤ k2 ≤ 0} .
Then a symmetry argument gives
i
−
f
ZZ
e
i
f
3
k2
3
−zk2
e
− fi
3
k1
3
−zk1
∧
∧
ϕ(k2 ) ϕ (k1 ) dk1 dk2
k2 >k1
i
=−
f
Z0
e
− fi
3
k1
3
−zk1
∧
ϕ (k1 ) dk1
−∞
Z0
e
− fi
3
k2
3
−zk2
∧
ϕ(−k2 ) dk2 + I(z, f )
−∞
(2.119)
for z ∈ C− , where
ZZ
i
I(z, f ) := −
f
e
i
f
3
k2
3
−zk2
e
− fi
3
k1
3
−zk1
ψ(k1 , k2 ) dk1 dk2 ,
(2.120)
k2 >k1 >0
∧
∧
∧
∧
ψ(k1 , k2 ) := ϕ(k2 ) ϕ (k1 ) + ϕ(−k1 ) ϕ (−k2 )
(2.121)
36
for z ∈ C− . (For later use note that ψ(0, k2 ) = lim ψ(k1 , k2 ), k1 , k2 > 0; cf. Remark 2.11.)
k1 ↓0
Next, our main aim is to show that, for z ∈ M , to lowest order in the expression (2.120)
we can drop (k23 − k13 )/3f in the exponent: Let Iε := (γ − ε , γ + ε) for some ε > 0 sufficiently
√
(2.64)
small, where γ = Re z (z ∈ M ), and χ ∈ C0∞ (Iε ) with χ(k) = 1 for k ∈ Iε/2 . Let
χ
e(·) := 1 − χ(·) ,
Θ(·) := 1{k>0} (·) ,
e± (k) := e
± fi
k3
3
−zk
.
(2.122)
By (2.120) and (2.122) one gets
e f ) + J(z, f )
I(z, f ) = J(z,
(z ∈ M , f > 0) ,
(2.123)
where
ZZ
e f ) := − i
J(z,
f
Θ(k2 − k1 )e
χ(k1 )e+ (k2 )e− (k1 )ψ(k1 , k2 ) dk1 dk2 ,
(2.124)
Θ(k2 − k1 )χ(k1 )e+ (k2 )e− (k1 )ψ(k1 , k2 ) dk1 dk2 .
(2.125)
k2 >k1 >0
ZZ
i
J(z, f ) := −
f
k2 >k1 >0
Using
e− (k1 )
(2.122)
=
if
∂
e− (k1 )
k12 − z ∂k1
(2.126)
in (2.124) one finds, for z ∈ M and f > 0,
ZZ
ψ(k1 , k2 ) ∂
e f) =
J(z,
Θ(k2 − k1 )e
χ(k1 )e+ (k2 ) 2
e− (k1 ) dk1 dk2 .
k1 − z ∂k1
(2.127)
k2 >k1 >0
Then partial integration (against k1 ∈ [0, k2 ]) gives
Z
Z
χ
e(0)
ψ(k2 , k2 )
e
dk2 +
J(z, f ) =
χ
e(k2 ) 2
e+ (k2 )ψ(0, k2 ) dk2
k2 − z
z
k2 >0
k2 >0
ZZ
∂ χ
e(k1 )ψ(k1 , k2 ) −
Θ(k2 − k1 )e+ (k2 )e− (k1 )
dk1 dk2 ,
∂k1
k12 − z
(2.128)
k2 >k1 >0
where χ
e(0) = 1. Integrating by parts once more in the last term of (2.128) gives
ZZ
i
e
J(z, f ) = −
Θ(k2 − k1 )e
χ(k1 )e+ (k2 )e− (k1 )ψ(k1 , k2 ) dk1 dk2
f
k2 >k1 >0
Z
Z
ψ(k2 , k2 )
1
=
χ
e(k2 ) 2
dk2 +
e+ (k2 )ψ(0, k2 ) dk2 + O(f )
k2 − z
z
(2.129)
k2 >0
k2 >0
as f ↓ 0. Thus by use of (2.129) in (2.123) we obtain
Z
Z
ψ(k, k)
1
I(z, f ) = J(z, f ) +
χ
e(k) 2
dk +
e+ (k)ψ(0, k) dk + O(f )
k −z
z
k>0
k>0
(2.130)
37
as f ↓ 0. Combining (2.122), (2.125), (2.126) and (2.130) yields
ZZ
i
I(z, f ) = −
Θ(k2 − k1 )χ(k1 )χ(k2 )e+ (k2 )e− (k1 )ψ(k1 , k2 ) dk1 dk2
f
k2 >k1 >0
ZZ
χ(k1 )e
χ(k2 ) ∂
e
(k
)
e− (k1 )ψ(k1 , k2 ) dk1 dk2
−
Θ(k2 − k1 )
+
2
k22 − z
∂k2
k2 >k1 >0
Z
Z
1
ψ(k, k)
dk +
e+ (k)ψ(0, k) dk + O(f )
(2.131)
+
χ
e(k) 2
k −z
z
k>0
k>0
as f ↓ 0. Integrating by parts twice in the second term on the r.h.s. of (2.131) gives
Z
Z
1
ψ(k, k)
2
dk +
I(z, f ) = Iχ (z, f ) +
(1 − χ (k)) 2
e+ (k)ψ(0, k) dk + O(f ) (2.132)
k −z
z
k>0
k>0
as f ↓ 0, where
i
f
Iχ (z, f ) := −
ZZ
Θ(k2 − k1 )χ(k1 )χ(k2 )e+ (k2 )e− (k1 )ψ(k1 , k2 ) dk1 dk2 .
(2.133)
k2 >k1 >0
In (2.133) we make the change of variables
k
k
k1 + k2
such that k1 = y −
and k2 = y + ,
2
2
2
and a further change k 7→ f k. We obtain, for z ∈ M and f > 0,
k := k2 − k1 ,
y :=
Z Z∞
Iχ (z, f ) = −i
ei(y
2
−z)k ik3 f 2 /12
e
F (y, f k) dy dk ,
(2.134)
Ry k=0
where
k k k
k
F (y, k) := χ y −
χ y+
ψ y − ,y +
2
2
2
2
(y, k ∈ R) .
(2.135)
Since χ ∈ C0∞ (R), one has F ∈ C ∞ (R2 , C) with
supp F ⊂ {ξ ∈ R2 | |ξ| ≤ R}
for some R > 0 sufficiently large.
(2.136)
Thus
|F (y, f k)| ≤ kF (·, ·)k∞ < ∞
(y ∈ R, k ∈ R, f > 0)
(2.137)
and similarly
|∂2 F (y, f k)| ≤ k∂2 F (·, ·)k∞ < ∞
(y ∈ R, k ∈ R, f > 0) ,
(2.138)
where ∂2 denotes the partial derivative with respect to the second entry. We iteratively use
the identity
2
2
eiky
1 ∂
n + 1 eiky
=
+
(n ∈ N)
yn
2ik ∂y
y
y n+1
38
to get
2
1 ∂
1 ∂
2 ∂
3 eiky
.
+
+
+
(2ik)3 ∂y y ∂y y ∂y y y 3
2
eiky =
(2.139)
Then inserting (2.139) into (2.134) gives, for z ∈ M and f > 0,
Z Z1
Iχ (z, f ) = −i
ei(y
2
−z)k ik3 f 2 /12
e
F (y, f k) dk dy
Ry k=0
Z Z∞
−i
2
3 2
∂
1 ∂
2 ∂
3 ei(y −z)k
eik f /12
F
(y,
f
k)
dk dy.
+
+
+
(2ik)3
∂y y ∂y y ∂y y
y3
Ry k=1
(2.140)
By use of (2.137) and (2.138), Taylor’s Theorem with an expansion in f k around zero gives
eik
3
f 2 /12
F (y, f k) = F (y, 0) + O(f )
(f ↓ 0) ,
(2.141)
uniformly for k ∈ [0, 1] and y ∈ R. Analogously,
3
2
1
eik f /12
F (y, f k) =
F (y, 0) + O(f )
3
(2ik)
(2ik)3
(f ↓ 0) ,
(2.142)
uniformly for y ∈ R and k ∈ (1, ∞). For z ∈ M , inserting (2.142) and (2.141) into (2.140)
yields
Z Z1
Iχ (z, f ) = −i
ei(y
2
−z)k
F (y, 0) dk dy + O(f )
R k=0
Z Z∞
−i
2
ei(y −z)k 1 ∂
2 ∂
3
∂
+
−
+
−
+
F (y, 0) dk dy + O(f )
−
y 3 (2ik)3
∂y y
∂y y
∂y y
R k=1
Z Z1
= −i
e
i(y 2 −z)k
Z Z∞
F (y, 0) dk dy − i
R k=0
ei(y
2
−z)k
F (y, 0) dk dy + O(f ) (2.143)
R k=1
as f ↓ 0, uniformly for z ∈ M . Since
Z∞
ei(y
2
−z)k
dk =
y2
i
−z
(z ∈ C− , y ∈ R) ,
0
it follows from (2.143) that
Z
Z
F (y, 0)
χ2 (y)ψ(y, y)
(2.135)
Iχ (z, f ) =
dy
+
O(f
)
=
dy + O(f )
2
y −z
y2 − z
R
y>0
(2.144)
39
as f ↓ 0, uniformly for z ∈ M . Then combining (2.144) and (2.132) gives
Z
Z
ψ(k, k)
1
dk +
k2 − z
z
I(z, f ) =
k>0
(f ↓ 0) ,
e+ (k)ψ(0, k) dk + O(f )
(2.145)
k>0
uniformly for z ∈ M , where ψ(0, k) = limk1 ↓0 ψ(k1 , k) and
Z
ψ(k, k)
dk
k2 − z
∧
∧
∧
Z∞ ∧
Z∞ ∧
ϕ(k) ϕ (k) + ϕ(−k) ϕ (−k)
| ϕ (k)|2
dk =
dk
2
k −z
k2 − z
(2.121)
=
−∞
0
k>0
−1
2
(ϕ , (p − z)
=
(z ∈ C− ) .
ϕ)
L2
(2.146)
By (2.122), (2.121) and a change of variables k 7→ −k, one finds
Z0
Z
e+ (k)ψ(0, k) dk =
− fi
e
k3
3
−zk ∧
∧
ϕ(−k) dk ϕ (+0)
−∞
k>0
Z0
+
e
− fi
k3
3
−zk
∧
∧
ϕ (k) dk ϕ(−0) .
(2.147)
−∞
Then combining (2.147), (2.146), (2.145), (2.119) and (2.118) proves (2.114).
For the continued resolvent matrix element one has:
Proposition 2.12
Let ϕ and M be as in Theorem 1.7. Then
−1
2
(ϕ , (p + f x − z)
c,Ω
ϕ)L2 f
i
=
f
Z
e
− fi
k3
3
−kz ∧
ϕ(−k) dk
Γ0
−
i
f
Z
e
− fi
k3
3
ϕ(−k) dk
k<0
1
+
z
e
− fi
k3
3
−kz
∧
ϕ (k) dk
Γ0
−kz ∧
Z
Z
e
− fi
k3
3
−kz
∧
ϕ (k) dk
k<0
Z
e
− fi
k3
3
−zk ∧
∧
ϕ(−k) dk ϕ (+0) +
k<0
Z
e
− fi
k3
3
−zk
∧
∧
ϕ (k) dk ϕ(−0)
k<0
+ (ϕ , (p2 − z)−1 ϕ)L2 + O(f )
(f ↓ 0),
(2.148)
uniformly for z ∈ M , where the contour Γ0 is the distorted C− ∪ C+ as used in the
proof of Proposition 2.8. (C− ∪ C+ is defined in (2.46).)
Proof of Proposition 2.12:
Combining (2.31), (2.33), (2.50) and (2.114) proves (2.148).
40
Proof of Theorem 1.7: By Proposition 1.3, the resonances of Hϕ (f ) are the zeros of
c,Ω
Ff,ϕ f in the open lower half complex plane. By (2.31), these are the solutions in C− of
c,Ω
c,Ωf
Ff,ϕ f (z) = 1 − z − (ϕ , (p2 + f x − z)−1 ϕ)L2
= 0.
(2.149)
From the expansion (2.148), (2.50), the estimates given in Corollary 2.9, (2.116) and (2.117),
it follows that for z = r
π ∧ √ ∧ √
p 4 3/2
2 3/2
c,Ω
(ϕ , (p2 + f x − z)−1 ϕ)L2 f = √ ϕ ( z)ϕ(− z) + O( f ) ei 3f z + O(1)ei 3f z ,
z
(2.150)
as f ↓ 0, uniformly for z ∈ M , where we have used that (ϕ , (p2 − z)−1 ϕ)L2 = O(1) (f ↓ 0).
Thus (because of (1.14)), if −Im z > c0 f for c0 > 0 sufficiently large, the equation (2.149)
cannot have a solution z = r in the region M considered in the theorem. This finishes the
proof of Theorem 1.7.
The following corollary is a consequence of (2.50), Corollary 2.9, (2.116), (2.117) and
Proposition 2.12:
Corollary 2.13
Let ϕ and M be as in Theorem 1.7. Then
c,Ωf
(ϕ , (p2 + f x − z)−1 ϕ)L2
4 3/2
π ∧ √ ∧ √
= √ ϕ ( z)ϕ(− z)ei 3f z
z
√ ∧ √ ∧
iπ ∧ √ ∧ √
ϕ ( z)ϕ( z)+ ϕ (− z)ϕ(− z)
+√
z
p
+ (ϕ , (p2 − z)−1 ϕ)L2 + O( f ) (f ↓ 0) ,
(2.151)
uniformly for z ∈ M .
Remark 2.14
We have written (2.148) with a view toward computation (cf. Section
3).√It is better than (2.151) in the sense that the error is of order O(f ) rather than
O( f ).
3
Numerical analysis of the Friedrichs model
The numerical data in this section have been obtained by use of Mathematica 8.
Let
ϕ(x) := µ e−x
2
/2
(x ∈ R, µ > 0 sufficiently small).
(3.1)
Obviously, ϕ ∈ S(R). (S(R) denotes the Schwartz space.) Furthermore, ϕ is a fixed point
∧
of the Fourier transform, i.e., ϕ =ϕ. Thus ϕ ∈ Tk0 for all k0 > 0. For this particular ϕ the
regions defined in (1.11) are
Ωf = C−
(f ≥ 0) .
(3.2)
41
Resonances of Hϕ for small µ > 0:
Figure 3.1: Contour deformation used in order to get (3.5)
According to Proposition 1.3, the resonances of Hϕ , defined in (1.7), are precisely the
solutions of
c,Ω0
c,Ω0
=0
(z) = 1 − z − (ϕ , (p2 − z)−1 ϕ)L
F0,ϕ
2
(3.3)
in C− . For the resolvent matrix element one finds
−1
2
(ϕ , (p − z)
ϕ)L2
∧
∧
= (ϕ , (k 2 − z)−1 ϕ)L2 = µ2
Z
2
e−k
dk
k2 − z
R
−k2
Z
e
√
√ dk
(k − z)(k + z)
R
Z
1
2
µ2
1 √ −
√
= √
e−k
dk
2 z
k− z
k+ z
= µ2
(z ∈ C+ ) .
(3.4)
R
√
The integrand in (3.4) has singularities at ± z. Now in (3.4) we deform the integration path
R into a path γ as shown in Figure 3.1 without changing the integral. Then continuation
gives
Z
γ
2
2
c,Ω0
e−k
√ e−k
√ dk
√ ,± z +
= 2πi Res
k∓ z
k∓ z
Z
R
2
e−k
√ dk
k∓ z
(z ∈ C− ) ,
(3.5)
42
where the residues are
Res
e−k2
√ √ , ± z = e−z
k∓ z
(z ∈ C− ) .
(3.6)
Inserting (3.6) into (3.5) yields
e−z
2
0
√ + µ2
(ϕ , (p2 − z)−1 ϕ)c,Ω
=
µ
2πi
2
L
z
2
e−k
dk
k2 − z
Z
(z ∈ C− ) ,
(3.7)
R
and
Z
√
2
−πe−z e−k
erf(i z) √
dk
=
i
+
k2 − z
i
z
(z ∈ C− ) ,
(3.8)
R
where
2
erf(x) := √
π
Zx
2
e−t dt
(x ∈ R) .
(3.9)
0
The error function erf(·) has an entire extension (see, e.g., [O, Chapter 2.4]).
Finally, by combining (3.1), (3.3), (3.7) and (3.8), the resonances of Hϕ are the solutions
of
√
π e−z erf(i z) e−z
2
√
√
−
i+
0=1−z−µ
2πi
(z ∈ C− ).
(3.10)
i
z
z
By Proposition 1.6, for µ > 0 sufficiently small there is exactly one resonance of Hϕ in
a sufficiently small complex neighborhood of 1 (the embedded eigenvalue of H0 defined in
(1.6)). Let r0 (µ), µ > 0 sufficiently small, denote this unique resonance near the embedded
eigenvalue 1 of H0 . Solving1 (3.10) for µ > 0 sufficiently small reveals the behavior of
the resonance r0 (µ) of Hϕ as µ ↓ 0 shown in Figures 3.2 through 3.4. One finds that the
resonance of Hϕ converges to the original embedded eigenvalue 1 of H0 .
For later use we compute
r0 µ =
1
= 1.01905 − 0.0111115 i =: r0 .
10
(3.11)
Resonances of Hϕ (f ):
Fix µ > 0 sufficiently small. Again by Proposition 1.3 the resonances of Hϕ (f ) (f > 0) are
the solutions of
c,Ω
c,Ω
Fϕ,f f (z) = 1 − z − rϕ,f f (z) = 0
(z ∈ C− ) ,
(3.12)
where
c,Ω
(2.31)
c,Ωf
rϕ,f f (z) = (ϕ , (p2 + f x − z)−1 ϕ)L2
1 by
(z ∈ C− ) ,
(3.1)
2
ϕ(x) = µe−x
/2
use of the FindRoot-algorithm of Mathematica 8 (which is basically Newton’s method)
.
(3.13)
43
Re r0 HΜL
2.0
1.8
1.6
1.4
1.2
1.0
Μ
0.0
0.2
0.4
0.6
0.8
Figure 3.2: Real part of the resonance r0 ( · ) of Hϕ
Im r0 HΜL
Μ
0.2
0.4
0.6
0.8
-0.05
-0.10
-0.15
Figure 3.3: Imaginary part of the resonance r0 ( · ) of Hϕ
For this particular ϕ one has
c,Ω
rϕ,f f (z) =
2πi 2 z ψ
+ (ϕ , (p2 + f x − z)−1 ϕ)L2
f f f
(z ∈ C− ) ,
(3.14)
44
Im r0 HΜL
1.2
1.4
1.6
1.8
Re r0 HΜL
-0.05
-0.10
-0.15
Figure 3.4: The resonance r0 (µ) (0 ≤ µ ≤ 0.85) of Hϕ in the complex plane
where ψf is the entire extension of
Z
3
(2.3)
k2
µ
(3.1)
−i k
3f −kx
e
ψf (x) = √
e− 2 dk
2π
(x ∈ R, f > 0).
(3.15)
R
Note that ψf (z/f ) is a (highly) oscillatory integral. In order to numerically solve2
c,Ω
c,Ω
(3.12), we have substantially used the expansion for rϕ,f f (z) = (ϕ , (p2 + f x − z)−1 ϕ)L2 f
given in Section 2.2, Proposition 2.12. It follows from (2.148) that the error produced by
using this expansion in (3.12) is of order O(µ2 f ) = O(10−2 f ) as f ↓ 0. In addition there
is a numerical error produced by Mathematica which is less than 10−a + |r| 10−p , where
a = p = 53 log10 (2) ≈ 16 (machine precision) and r is the computed zero. (The parameters
a and p are called “accuracy” and “precision”.) Note that |r| is very close to one.
Interpretation of the Figures 3.5 – 3.8:
In Figures 3.5 through 3.8 one sees that there exist many resonances of Hϕ (f ) in a neigh(3.11)
borhood of the pre-existing resonance r0 = 1.01905 − 0.0111115 i of Hϕ (0). This is in
good accordance with the result of Corollary 1.5.
Figure 3.7 and Figure 3.8 show that the imaginary part of the resonances r(f ) of Hϕ (f )
in a neighborhood of the pre-existing resonance r0 converges to zero as f ↓ 0. In particular,
r(f ) 6→ r0 as f ↓ 0. Thus the pre-existing resonance r0 is unstable. This behavior confirms
the instability result of Theorem 1.7. According to Figure 3.5 and Figure 3.6, the real part
of the resonances r(f ) seems to converge to the original embedded eigenvalue 1 of H0 as
f ↓ 0. At present it is not clear to the authors how to analytically prove (or disprove) this
convergence of the real part.
A priori, the graphs in Figures 3.7 and 3.8 can be interpreted in two different ways which
are hard to distinguish numerically:
1. In Figure 3.7 and Figure 3.8 each “ray” tending to zero is a “bundle” of imaginary
parts, all close to each other. Analogously, in Figure 3.5 and Figure 3.6 each “ray”
(possibly tending to 1) is a “bundle” of real parts, all close to each other.
2. In Figure 3.7 and Figure 3.8 each “ray” tending to zero is one (highly) oscillatory
function of f , representing the imaginary part of a single resonance. Analogously, in
Figure 3.5 and 3.6 each “ray” (possibly tending to 1) is one oscillatory function of f ,
representing the real part of a single resonance.
2 again
by use of the FindRoot-algorithm of Mathematica 8
45
It is not clear what kind of objects (in terms of f ) the real and imaginary parts of the
resonances are: Branches of multi-valued analytic functions? Strata?
Re zHf L
1.05
1.04
1.03
1.02
1.01
1.00
0.002
0.004
0.006
0.008
0.010
f
0.98
Figure 3.5: Real part of resonances r(f ) of Hϕ (f ) vs. f ; for µ = 1/10
Re zHf L
1.0006
1.0004
1.0002
1.0000
0.9998
f
0.0002
0.0003
0.0004
0.0005
0.0006
Figure 3.6: Real part of resonances r(f ) of Hϕ (f ) vs. f on a finer scale; for µ = 1/10
46
Im zHf L
0.002
0.004
0.006
0.008
0.010
f
-0.0005
-0.0010
-0.0015
-0.0020
-0.0025
-0.0030
-0.0035
Figure 3.7: Imaginary part of resonances r(f ) of Hϕ (f ) vs. f ; for µ = 1/10
Im zHf L
0.0002
0.0003
0.0004
0.0005
0.0006
f
-0.00005
-0.00010
-0.00015
Figure 3.8: Imaginary part of resonances r(f ) of Hϕ (f ) vs. f on a finer scale; for µ = 1/10
47
A
Proof of Corollary 1.5
c,Ω
By Proposition 1.3, the real zeros of Ff,ϕ f (f > 0) are precisely the eigenvalues of the
self-adjoint operator Hϕ (f ). Thus by Proposition 1.4 (and Definition 1.2) it suffices to show
that Hϕ (f ) has at most finitely many eigenvalues:
u
Let f > 0, ϕ 6= 0. Let λ be an eigenvalue of Hϕ (f ). Then, for some H 3
6= 0
c
2
p + f x − λ Mϕ
u
0
=
.
(ϕ, · )L2
1−λ
c
0
In particular,
(p2 + f x − λ)u + cϕ = 0 ,
(ϕ, u)
L2
(A.1)
+ c(1 − λ) = 0 .
(A.2)
Since p2 + f x does not have any eigenvalues, equation (A.1) implies c 6= 0. W.l.o.g. one can
set c = 1. Then (A.1) implies that
u = −(p2 + f x − λ)−1 ϕ
(A.3)
and ϕ is in the domain of (p2 + f x − λ)−1 . Inserting (A.3) into (A.2) (with c = 1) gives
1 − λ − (ϕ , (p2 + f x − λ)−1 ϕ)L2 = 0 .
(A.4)
By Lemma 2.3 and (2.3) one has
λ
1
ψf , (x − )−1 ψf L2 ,
f
f
(ϕ , (p2 + f x − λ)−1 ϕ)L2 =
(A.5)
where the function ψf is entire by Lemma 2.4. Then
k(p2 + f x − λ)−1 ϕk22 =
1
λ
1
k(x − )−1 ψf k22 =
f
f
f
Z
R
|ψf (x)|2
dx < ∞
|x − λf |2
implies that
λ
= 0.
ψf
f
(A.6)
Thus the set of all eigenvalues λ of Hϕ (f ) must be discrete, since otherwise (A.6) would
imply that ψf (x) = 0 for all x ∈ R.
Next we will prove that for eigenvalues λ of Hϕ (f ) the resolvent matrix element (A.5) is
bounded. Then (A.4) implies that there are at most finitely many eigenvalues of Hϕ (f ).
Therefor we write
λ
λ
|(ψf , (x − )−1 ψf )L2 | = f
Z
λ
R\[ λ
f −1, f +1]
+
f +1
Z
λ
f −1
|ψf (x)|2
dx.
λ
x− f
(A.7)
48
Clearly,
Z
λ
R\[ λ
f −1, f +1]
Z
|ψf (x)|2
dx ≤ |ψf (x)|2 dx = kψf k22
λ
x− f
(A.8)
R
and
λ
f +1
Z
λ
f −1
|ψf (x)|2
(A.6)
dx =
λ
|x − f |
Z1 |ψ (x + λ ) − ψ ( λ )|2
f
f f
f
dx .
|x|
(A.9)
−1
One finds
Z
λ ∧
λ
1 λ
√
− ψf
=
eik f (eixk − 1) ψ f (k) dk ψf x +
f
f
2π
R
Z
1
ε
ε ∧
1
≤ c|x|ε/4 hki− 2 − 4 |hki 2 + 2 ψ f (k)| dk
(A.10)
R
for x ∈ R, small f, ε > 0 and some c < ∞. By Schwarz’ inequality one has
Z
1
1
ε
1
ε ∧
ε
1
ε ∧
hki− 2 − 4 |hki 2 + 2 ψ f (k)| dk = h · i− 2 − 4 , |h · i 2 + 2 ψ f | L2 R
1
ε
1
ε
∧
≤ kh · i− 2 − 4 k2 kh · i 2 + 2 ψ f k2 ,
(A.11)
where
1
ε
∧
1
ε
kh · i 2 + 2 ψ f k2 = khpi 2 + 2 ψf k2 .
(A.12)
Then, by (A.11) and (A.10),
Z1 |ψ (x + λ ) − ψ ( λ )|2
f
f f
f
|x|
0
dx ≤ c khpi
1
ε
2+2
ψf k22
−1
Z1
ε
|x|−1+ 2 dx
−1
0
1
ε
4c
khpi 2 + 2 ψf k22 .
ε
Combining (A.7) through (A.13) proves
=
|(ψf , (x −
λ −1
) ψf )L2 |
f
≤
kψf k22 +
1
ε
4c0
khpi 2 + 2 ψf k22
ε
(A.13)
(A.14)
for some c0 < ∞. By (2.3) one has
kψf k2 = kϕk2 ,
1
ε
1
ε
khpi 2 + 2 ψf k2 = khpi 2 + 2 ϕk2 .
Thus, by combining (A.5) and (A.15), the estimate (A.14) is equivalent to
1
ε
1
4c0
|(ϕ, (p2 + f x − λ)−1 ϕ)L2 | ≤
kϕk22 +
khpi 2 + 2 ϕk22 .
f
ε
(A.15)
(A.16)
We remark that the estimate (A.16) also follows using Mourre’s method; see [PSiSim].
49
B
Boundedness of dilation analytic vectors in a sector
Lemma B.1
Let A be the generator of dilations in L2 (R), i.e., the self-adjoint realization of (px + xp)/2 in L2 (R). Let f ∈ D(A2 ). Then f is differentiable on R\{0}
and
p
(B.1)
|f (x)| ≤ (2 |x|)−1 k(A2 + 1)f k2 .
Proof: We define L2 (R+ ) := {I[0,∞] φ | φ ∈ L2 (R)}, where I[0,∞] denotes the indicator
function on the interval [0, ∞]. Note that U (θ) = eiθA (θ ∈ R) maps L2 (R+ ) into itself.
Define
W : L2 (R+ ) → L2 (R) ,
W φ(·) := e(·)/2 φ(e(·) ).
(B.2)
Then W is unitary and
(W U (θ)W −1 g)(u) = g(u + θ)
(g ∈ L2 (R), u, θ ∈ R).
Thus if f ∈ D(A2 ), the function W I[0,∞] f is in H2 (R) (the Sobolev space of second order)
and is thus a differentiable function. It follows that f (x) is differentiable for x > 0. An
analogous proof works for x < 0. This shows the differentiability. Next we will prove (B.1).
We compute for f ∈ L2 (R)
((1 + iA)−1 f )(x) + ((1 − iA)−1 f )(x) = 2((1 + A2 )−1 f )(x)
Z ∞
=
e−|θ| eθ/2 f (eθ x)dθ .
−∞
Thus
2(1 + A2 )−1 f )(x) ≤
Z
∞
e−2|θ| dθ
−∞
Z
1/2 Z
∞
1/2
−∞
∞
=
eθ |f (eθ x)|2 dθ
eθ |f (eθ+u x/|x|)|2 dθ
1/2
(B.3)
−∞
= |x|
−1/2
≤ |x|
−1/2
Z
∞
f (yx/|x|)|2 dy
1/2
−∞
kf k2 ,
where in (B.3) the substitution |x| = eu was used. The lemma thus follows.
Proposition B.2
If f is dilation analytic in angle θ0 in the sense of Definition 1.1,
iφ
then f is analytic in the union of sectors Γ±
θ0 := {z ∈ C | z = ±|z|e , |z| > 0, |φ| < θ0 }
and
p
|f (eiφ x)| ≤ (2 |x|)−1 k(A2 + 1)e|φA| f k2 (|φ| < θ0 , x ∈ R\{0}) .
50
Proof: Using the unitary operator W (defined in (B.2)) it is easily seen that if f is dilation
−
analytic in angle θ0 , f has a version which is analytic in the set Γθ0 := Γ+
θ0 ∪ Γθ0 . In addition
the analytic extension of U (θ)f for |Im θ| < θ0 is given by eθ/2 f (eθ x).
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Acknowledgements: J.R. is grateful to the DFG for making her stay in Charlottesville
possible and to Markus Klein for some helpful discussions!