On Critical Stability of Three Quantum Charges Interacting Through Delta Potentials

Abstract

We consider three one-dimensional quantum, charged and spinless particles interacting through delta potentials. We derive
sufficient conditions which guarantee the existence of at least one bound state.

Full text

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arXiv:math-ph/0604003v1 1 Apr 2006
Few-Body Systems 0, 1–6 (2008)
Few-
Body
Systems
c ? by Springer-Verlag 2008
Printed in Austria
On critical stability of three quantum
charges interacting through delta potentials
H.D.Cornean1∗, P. Duclos2 ∗∗, B. Ricaud2 ∗∗∗
1Dept. of Math., Aalborg University, Fredrik Bajers Vej 7G, 9220 Denmark
2Centre de Physique Th´ eorique UMR 6207 - Unit´ e Mixte de Recherche du CNRS et des Uni-
versit´ es Aix-Marseille I, Aix-Marseille II et de l’ Universit´ e du Sud Toulon-Var - Laboratoire
affili´ e ` a la FRUMAM, Luminy Case 907, F-13288 Marseille Cedex 9 France
Abstract. We consider three one dimensional quantum, charged and spinless
particles interacting through delta potentials. We derive sufficient conditions
which guarantee the existence of at least one bound state.
1 Introduction
Denote by xi,mi,Zie, i = 1,2,3, the position, mass and charge of the i-th par-
ticle. Our system is formally described by the Hamiltonian
?
adjoint operator associated to the quadratic form with domain H1(R3):
?
?3
i=1−?2
2mi∂2
xi+
1≤i<j≤3ZiZje2δ(xi−xj)acting in L2(R3) which is defined as the unique self-
3
?
i=1
?2
2mi?∂xiψ?2+
?
1≤i<j≤3
ZiZje2
xi=xj
|ψ(σi,j)|2dσi,j,ψ ∈ H1(R3).
Here σi,jdenotes a point in the plane xi= xj. We will consider the cases m1=
m2=: m > 0,m3=: M > 0Z1= Z2= −1,
the question: for what values of m/M and Z does this system possess at least
one bound state after removing the center of the mass?
There is a huge amount of literature on 1-d particles interacting through
delta potentials either all repulsive or all attractive, but rather few papers deal
with the mixed case. We mention the work of Rosenthal, [7], where he considered
M = ∞. The aim of this paper is to make a systematic mathematical study of
the Rosenthal results and extend them to the case M < ∞. It has been shown
in [1] and [2] that these delta models serve as effective Hamiltonians for atoms
Z3=: Z > 0 and answer to
∗E-mail address: cornean@math.aau.dk
∗∗E-mail address: duclos@univ-tln.fr
∗∗∗E-mail address: ricaud@cpt.univ-mrs.fr

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2 Critical stability of three quantum charges with delta self-interactions
M
m
1
∞
θ2,3
2π/3
3π/4
θ1,2
2π/3
π/2
α2
3/4
1/4
ν(α)
1
1/√2
A1
θ12
θ13
A2
A3
Figure 1. Left: table with corresponding values of angles and masses, right: support of the
delta potentials with the unit vectors Ai’s.
in intense magnetic fields or quasi-particles in carbon nanotubes. As one can see
in ([4], [5], [6],[3]), they also seem to be relevant for atomic wave guides, nano
and leaky wires.
2 The spectral problem
Removing the center of mass.
y := x3− (m1x1+ m2x2)/(m1+ m2) and z :=?
relative motion formal Hamiltonian? H = −?2
be the Jacobian of the coordinate change (x′,y′) = {2ν(α)?2/(mZe2)}(x,αy),
and define the unitary (U−1f)(x,y) =
of R2given by A1 :=
ν(α)
2
Define A⊥
i
as Ai rotated by π/2 in the positive sense. Then U? HU−1=
H := −1
Using the Jacobi coordinates: x := x2−x1,
imixi/?
imiwe get the 2-d
y+e2δ(x)−Ze2δ(y−
?1/4 + α2. Let J
m∂2
x−2m+M
4mM?2∂2
x
2) − Ze2δ(y +x
2). Define α2:= (M + 2m)/4M and ν(α) :=
√Jf(x′,y′). Consider three unit vectors
?, A2 :=
1
?α,−1
1
ν(α)
?−α,−1
2
?, and A3 := (0,1).
{mZ2e4}/{2?2ν(α)2} H, where:
x−1
2∂2
2∂2
y− δ(A⊥
1.(x,y)) − δ(A⊥
2.(x,y)) + λδ(A⊥
3.(x,y)),λ :=ν(α)
Z
.
We denote by θi,jthe angle between the vectors Aiand Aj. We give some typical
values of all these parameters (see fig. 1).
The skeleton
τA: H1(R2) → L2(R) defined as (τAψ)(s) := ψ(sA) and if we let τ : H1(R2) →
?3
H as H0+τ⋆gτ where 2H0stands for the free Laplacian and g is the 3×3 diagonal
matrix with entries (−1,−1,λ). Denoting R0(z) := (H0−z)−1and R(z) := (H −
z)−1the resolvents of H0and H, one derives at once, with the help of the second
resolvent equation, the formula for any z in the resolvent sets of H0and H:
Let A be unit vector in R2. If one introduce the ”trace” operator
i=1L2(R) be defined as τ := (τA1,τA2,τA3), we may rewrite the Hamiltonian
R(z) = R0(z) − R0(z)τ∗(g−1+ τR0(z)τ∗)−1τR0(z).
Using the HVZ theorem (see [8] for the case with form-bounded interactions),
we can easily compute the essential spectrum: σess(H) = [−1
is given by the infimum of the spectrum of the subsystem made by the positive
charge and one negative charge.
(1)
2,∞). Its bottom

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H. Cornean, P. Duclos, B. Ricaud3
From this and formula (1) it is standard to prove the following lemma:
Lemma 1.
is a discrete eigenvalue of H if and only if ker(g−1+τR0(E)τ∗) ?= {0}. Note that
up to a scaling this is the same as kerS ?= {0}. Moreover, mult(E) = dim(kerS).
The spectral analysis is thus reduced to the study of S, a 3 × 3 matrix of
integral operators each acting in L2(R). We call S the skeleton of H. Let us
denote by TA,B:= τAR0(−1)τ⋆
vectors A and B, and by?TA,B the Fourier image of TA,B. Then the kernel of
Let k >
1
√2. Define S := kg−1+τR0(−1)τ∗. Then E = −k2< −1
2
B, T0:= TA,A, by θA,Bthe angle between two unit
?TA,Bwhen θA,B?∈ {0,π}, and of?T0read as:
?TA,B(p,q) =
1
2π|sin(θA,B)|
1
?p2−2cos(θA,B)pq+q2
2sin2(θA,B)
+ 1
?,
?T0(p,q) =δ(p − q)
?p2+ 2. (2)
Then?T0is a bounded multiplication operator, and?TA,Bonly depends on |θA,B|.
Reduction by symmetry.
H and S enjoy various symmetry properties which
follow from the fact that two particles are identical. Let π : L2(R) → L2(R) be the
parity operator, i.e. {πϕ}(p) = ϕ(−p) and denote by π1:= π⊗1 and π2:= 1⊗π
the unitary symmetries with respect to the x and y axis. One verifies that for
all i,j ∈ {1,2}, we have [πi,H] = 0 and [πi,πj] = 0. Thus if we denote by πα
α = +,− the eigenprojectors of πion the even, respectively odd functions we
may decompose H into the direct sum
Consequently we denote in the sequel TAi,Ajby Tθi,jor Ti,j.
i,
H =
?
α∈{±}, β∈{±}
Hα,β,Hα,β:= πα
1πβ
2H.
Similarly let Π,σ : L2(R3) → L2(R3) defined by (Πψ)(−p) := ψ(−p) and σψ =
σ(ψ1,ψ2,ψ3) := (ψ2,ψ1,ψ3). They both commute with S, and also [Π,σ] = 0.
Let Παand σα, α = +,−, denote the eigenprojectors of Π and σ symmetric
and antisymmetric resp.. Then we can write S =?
ΠασβS.
From the expression of?Tθ(p,q) one also sees that [π,Tθ] = 0 so that Tθ
operators acting in L2(R+). Due to these symmetry properties we have kerS =
?
of a single operator acting in L2(R+) that we call effective skeleton. We gather
in the following table the four effective skeletons we have to consider with their
corresponding subspaces in L2(R2):
α∈{±},β∈{±}Sα,β,
Sα,β:=
decomposes into T+
θ⊕ T−
θwhere T±
θ:= π±Tθ. As usual we shall consider T±
θas
α,βkerSα,β, and each individual null-space can be expressed as the null-space
Sα,β
++
−+
+−
−−
effective skeleton
1,2+ 2T+
1,2+ 2T−
k − T0+ T+
k − T0+ T−
subspace in L2(R2)
Ranπ+
Ranπ+
Ranπ−
Ranπ−
k − T0− T+
k − T0− T−
2,3(T0+ kλ−1)−1T+
2,3(T0+ kλ−1)−1T−
2,31π+
1π−
1π−
1π+
2
2,32
1,22
1,2
Table 1.
2

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4Critical stability of three quantum charges with delta self-interactions
3 Sectors without bound states
Properties of the Tθoperators.
is self-adjoint and has a finite Hilbert-Schmidt norm. The proof of the following
lemma in not at all obvious, but will be omitted due to the lack of space.
From (2) we get 0 ≤ T0≤ 1/√2. Then Tθ
Lemma 2. For all θ ∈ [π/2,π) one has ±T±
θ ?→ ±inf T±
Absence of bound state in the odd sector with respect to y.
the following result:
θ≥ 0 and the mapping [π/2,π) ∋
θis strictly increasing.
We have
Theorem 3. For all Z > 0 and all 0 < M/m ≤ ∞, H has no bound state in the
symmetry sector Ranπ−
2.
Proof. The symmetry sector Ranπ−
2corresponds to the second and third lines
in Table 1. For the third line one uses that T+
k > 1/√2 since we are looking for eigenvalues below σess(H) = [−1
1,2≥ 0 by Lemma 2, and that
2,∞). Hence
k − T0+ T+
1,2≥ k −
1
√2> 0
thus ker(k − T0+ T+
eigenvalues in Ranπ−
T−
1,2) = {0}, and by Lemma 1 this shows that H has no
1π−
2,3(T0+ kλ−1)−1T−
2. By the same type of arguments one has: k − T0−
2,3≥ k −
The above theorem has a simple physical interpretation. Wave
functions which are antisymmetric in the y variable are those for which the
positive charge has a zero probability to be in the middle of the segment joining
the negative charges. A situation which is obviously not favorable for having a
bound state.
1,2+ 2T−
1
√2> 0.
Remark 4.
Absence of bound state in the odd-even sector with respect to x and
y.
Looking at the fourth line of Table 1 we have to consider
?
where?T−
for θ1,2= 2π/3, this will imply that S−,−(2−1
then for all π/2 ≤ θ1,2≤ 2π/3 by the monotonicity of inf T−
as stated in Lemma 2; finally looking at (3) this will show that S−,−(k) > 0 for
all k > 1/√2 and therefore that kerS−,−(k) = {0}. But?T−
one of T−
O?(p2+ q2)?. It turns out that −1 is an eigenvalue of?T−
?
p2+2
p(2p2+ 3)
S−,−(k) := k − T0+ T−
1,2=:
k − T0
?
1 +?T−
1,2(k)
??
k − T0
(3)
1,2(k) := (k − T0)−1
M ≥ m, i.e. π/2 ≤ θ1,2≤ 2π/3. Assume that we can prove that?T−
2T−
1,2(k − T0)−1
2. Here we will only consider the case
1,2(2−1
2) ≥ −1
2) ≥ 0 first for θ1,2= 2π/3 and
1,2with respect to θ
1,2:= T−
1,2(2−1
2) ( for
θ1,2= 2π/3) is Hilbert-Schmidt since its kernel decay at infinity faster than the
1,2and it has the following behavior at the origin:?T−
?1/2
1,2(p,q) ∼ −16√2
1,2with eigenvector
3√3π+
R+∋ p ?→
1
√2−
1
√
p→0
∼
1
32
5
4
+ O(p2)

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H. Cornean, P. Duclos, B. Ricaud5
and since the Hilbert-Schmidt norm of?T1,2 can be evaluated numerically to
proved the
??T−
Theorem 5. For all Z > 0 and all 1 ≤ M/m ≤ ∞, H has no bound state in the
symmetry sector Ranπ−
2.
1,2?HS ≤ 1.02, all the other eigenvalues of?T−
1,2are above −1. Thus we have
1π−
4 The fully symmetric sector
According to Table 1, we need to find under which conditions one has
kerS+,+(k) ?= {0} where
S+,+(k) := k − T0− K(k), withK(k) := T+
1,2− 2T+
2,3(T0+ kλ−1)−1T+
2,3.
The proof of the following lemma is an easy application of Fredholm and analytic
perturbation theory:
Lemma 6.(i) {Rek2> 0} ∋ k ?→ S+,+(k) is a bounded analytic self-adjoint
family of operators.
(ii) If inf σS+,+(2−1
{0}.
Denote by K(p,q) the integral kernel of K(2−1
Theorem 7. For all 0 < M/m ≤ ∞, H has at least one bound state in the
symmetry sector Ranπ+
2if Z is such that K(0,0) > 0.
Proof. We will now look for an upper bound on inf S+,+(2−1
method. Let j ∈ C∞
functions: ∀ǫ > 0, ψǫ(p) := ǫ−1j(pǫ−1), φǫ:=
ψǫconverges as ǫ → 0 to the Dirac distribution. First one has
?
R+
ǫ2
2√2?j?2
?
2)
?
< 0, then there exists k > 1/√2 so that kerS+,+(k) ?=
2). Our last result is:
1π+
2) by the variational
0(R+,R+) so that?
R+j(x)dx = 1 and define two families of
√ǫ
||j||ψǫ, ?φǫ? = 1. We know that
((2−1
2− T0)φǫ,φǫ) =
1
ǫ√2?j?2
[1 − (1 + p2/2)−1/2]j2(p/ǫ)dp
≤
?
R+
p2j(p)2dp.(4)
Then one has (K(2−1
that (S+,+(2−1
ǫ > 0 small enough, provided K(0,0) > 0.
It is possible to compute K(0,0) analytically. It can be shown that there
exists Zub
c(M/m) such that for any Z larger than this value, we have K(0,0) >
0. If we now define the critical Z as Zc(M/m) := inf{Z
H(Z,M/m) has at least one bound state}, it follows from our last theorem that
Zc(M/m) ≤ Zub
The curve Zub
c(M/m) is plotted on figure 2, where we used θ1,2instead of
the ratio M/m.
2)φǫ,φǫ) =
ǫ
?j?2(K(2−1
2 − (T0φǫ,φǫ) − (K(2−1
2)ψǫ,ψǫ) =
ǫ
?j?2(K(0,0) + O(ǫ)) so
2)φǫ,φǫ) will be negative for
2)φǫ,φǫ) = 2−1
> 0, H=
c(M/m).