Bachelor thesis
The Schrödinger operator with Morse potential
Supervisor:
prof.dr. H.T. Koelink
Author:
Bas Jordans
Radboud University Nijmegen
11 July 2011
Contents
Introduction and Summary
v
1 Preliminaries
1
2 The
2.1
2.2
2.3
Schrödinger operator with Morse potential
Properties of the solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The discrete spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The asymptotic behaviour of the eigenvalues . . . . . . . . . . . . . . . . . . . . .
7
7
13
20
3 The Hypergeometric Differential Equation
3.1 Properties and solutions of the hypergeometric differential equation . . . . . . . . .
3.2 The discrete spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
29
33
References
37
iii
iv
Introduction and Summary
The main part of this thesis is about the Schrödinger operator with Morse potential. This operator
has applications in both mathematics and physics. In mathematics because for example it can be
used to describe some of the zeros of the Riemann zeta function and in physics for example to
model the potential function of certain molecules consisting of two atoms. In this thesis we will
not treat these applications, we only focus on the theory.
The aim of this thesis is to understand a part of the article of J.C. Lagarias [3] which deals
about the Schrödinger operator with Morse potential. We try to determine the eigenvalues of the
Schrödinger operator with Morse potential. Because if the eigenvalues of an operator are known,
we have almost every information of this operator. Unfortunately most of the times it is impossible
to do determine the eigenvalues (or in general the spectrum) of an operator explicitly. This is also
the case with the Schrödinger operator with Morse potential. However it is possible to give an
estimation of the eigenvalues.
In section one we will introduce the required definitions and theorems using the harmonic oscillator.
It is possible to explicitly describe the eigenvalues of this operator. This will be done at the end
of that section. In the second section we consider the Schrödinger operator with Morse potential
on the right half line. We show that the spectrum is discrete and give the asymptotic behaviour
of the eigenvalues of this operator. In the last section we take a look at the hypergeometric
differential equation which has a link with the Schrödinger operator with Morse potential and
again we describe the spectrum of this operator.
Acknowledgement
I would like to thank Erik Koelink for being my supervisor. Without his advice and ideas it would
be impossible for me to write this thesis.
v
1
Preliminaries
In this section we introduce some of the basics of functional analysis. We use the harmonic
oscillator as an example to explain the definitions, theorems etc. In this section we state the
theorems, propositions and lemma’s without proof.
Definition 1.1 Let E be a vector space over C. A function ∥ · ∥ : E → R, x 7−→ ∥x∥ is called a
seminorm if it satisfies
1. ∥x∥ ≥ 0 ∀x ∈ E
2. ∥λx∥ = |λ| ∥x∥
∀x ∈ E and ∀λ ∈ C
3. ∥x + y∥ ≤ ∥x| + ∥y∥ ∀x, y ∈ E (triangle inequality).
∥ · ∥ is called a norm if it is a seminorm and
4. ∥x∥ = 0 ⇐⇒ x = 0.
E is a normed vector space if there exists a norm ∥ · ∥ : E → R.
Definition 1.2 A Banach space is a normed vector space which is complete relative to the topology induced by its norm.
Definition 1.3 Let E be a vector space. ⟨·, ·⟩ : E × E → C is called a semi-inner product if
1. x 7→ ⟨x, a⟩ is linear ∀a ∈ E
2. ⟨x, y⟩ = ⟨y, x⟩
∀x, y ∈ E
3. ⟨x, x⟩ ∈ [0, ∞) ∀x ∈ E.
⟨·, ·⟩ : E × E → C is an inner product if it is a semi-inner product and
4. ⟨x, x⟩ = 0 ⇒ x = 0.
E is an inner product space if there exists an inner product ⟨·, ·⟩ : E × E → C.
Remark 1.4 We can define a norm on an inner product space by: ∥x∥ :=
particular inner product spaces are normed vector spaces.
√
⟨x, x⟩. So in
Definition 1.5 A Hilbert space is an inner product space which is complete relative to the topology induced by its inner product norm.
• Cn with inner product: ⟨x, y⟩ := x1 y1 + . . . + xn yn is a inner product space.
∫∞
• L2 (R) := {φ : R → C : φ measuarable, −∞ |φ(x)|2 dx < ∞} with inner product: ⟨φ, ψ⟩ :=
∫∞
φ(x)ψ(x) dx is a semi-inner product, because: ⟨φ, φ⟩ = 0 ⇐⇒ φ = 0 a.e. (almost
−∞
everywhere).
Example 1.6
• Define the equivalence relation ∼ on L2 (R) by φ ∼ ψ ⇔ φ(x) = ψ(x) a.e. . Then: L2 (R) :=
L2 (R)/ ∼ with the same inner product as L2 (R) is a inner product space.
Definition 1.7 Let E, F be vector spaces. Define the following vector spaces of linear maps
L(E, F ) := {T : E → F : T linear}
L(E) := L(E, E)
E ′ := L(E, C).
1
1 PRELIMINARIES
If E and F are normed vector spaces, we can define a norm ∥ · ∥ on L(E, F ) by
∥T ∥ :=
sup
x∈E, ∥x∥E =1
∥T x∥F =
∥T x∥F
.
x∈E, x̸=0 ∥x∥E
sup
We call T : E → F a bounded operator if ∥T ∥ < ∞. Denote by BL(E, F ) the bounded linear maps
from E to F
BL(E, F ) := {T ∈ L(E, F ) : ∥T ∥ < ∞}.
Proposition 1.8 Let E and F be normed vector spaces, let T ∈ L(E, F ) then T is bounded if
and only if T is continuous.
Proof. Can be found in [5].
Definition 1.9 Let H be a Hilbert space, D(T ) ⊂ H be a linear subspace. We call T an unbounded
operator on H if
1. T ∈ L(D(T ), H)
2. @M ∈ R such that ∥T (x)∥ ≤ M ∥x∥
∀x ∈ D(T ).
The subspace D(T ) is called the domain of T . (T, D(T )) is densely defined if D(T ) is dense in H.
In this thesis we will only consider densely defined operators. Because if an operator is not densely
defined we can consider the Hilbert subspace H′ := D(T ) in which obviously D(T ) ⊂ H′ dense.
Example 1.10 Define the harmonic oscillator
H :=
d2
− x2
dx2
D(H) := L2 (R) ∩ C 2 (R).
We use this domain, because it’s necessary to differentiate twice and for the inner product it is
necessary that the functions are square integrable. It is easy to verify that H is linear, because
differentiation and multiplying by x2 is linear. We now show that H is unbounded. Let φn (x) :=
2
e−nx .
The function φn ∈ C ∞ (R) so in particular φn ∈ C 2 (R). And for every n ∈ N it is an exponentially
fast decreasing function. Hence φn is quadratic integrable. Thus φn ∈ D(T ). Then
dφn
(x) =
dx
d 2 φn
(x) =
dx2
Hence we obtain
∫
∥Hφn ∥2 =
2
d −nx2
e
= −2nxe−nx = −2nxφn (x)
dx
2
2
2
d
− 2nxe−nx = −2ne−nx + (2nx)2 e−nx = (−2n + (2nx)2 )φn (x).
dx
∞
−∞
Hφn (x)Hφn (x) dx
)2
d 2 φn
2
(x)
−
x
φ
(x)
dx
n
dx2
−∞
∫ ∞
(
)2
=
(−2n + 4n2 x2 − x2 )φn (x) dx
−∞
∫ ∞
)
(
)
)
(
(
2 2
2 2
2 2
dx
+ (4n2 x2 − x2 )2 e−nx
− 4n(4n2 x2 − x2 ) e−nx
=
4n2 e−nx
−∞
∫ ∞
∫ ∞(
)
(
)
2 2
2 2
dx
dx + (4n − 16n3 )
x2 e−nx
e−nx
= 4n2
−∞
−∞
∫ ∞
)
(
2 2
dx.
+ (16n4 − 8n2 + 1)
x4 e−nx
∫
∞
(
=
−∞
2
Recall that
∫
∞
(
e
−∞
∞
∫
−x2
(
x4 e−x
−∞
√
)2
dx =
2
)2
π
2
3
dx =
16
∫
∞
2
(
x
√
e
−x2
)2
−∞
1
dx =
4
√
π
2
π
.
2
By the substitution rule we obtain
√
∫ ∞(
)
2 2
1
π
e−nx
dx = √
n 2
−∞
√
∫ ∞
(
)
2
2
π
3
√
.
x4 e−nx
dx =
2 n
2
16n
−∞
∫
∞
(
)
2 2
x2 e−nx
dx =
−∞
1
√
4n n
√
π
2
So
√
√
√
(
) √
1
π
4n
16n3 1 π
16n4 3 π
3
π
2
√ − √
√
∥Hφn ∥ = 4n √
+
+ 2√
+ (−8n + 1)
2
n 2
n n n n 4 2
n n 16 2
16n n 2
√
√
√
√
√
√
√
√
√
π
π
π
π
π
π
1
3
3
√
= 4n n
− 4n n
+ 3n n
+√
− √
+
2
2
2
n 2
2 n 2
16n2 n 2
√
√
√
√
√
π
1
π
3
π
3
π
√
+√
− √
+
.
= 3n n
2
n 2
2 n 2
16n2 n 2
2
2
And then
∥H(φn )∥2
=
∥φn ∥2
√
√
√ ) (
√ )
√
(
√
π
π
π
π
π
1
3
3
1
√
3n n
+√
− √
+
/ √
2
n 2
2 n 2
16n2 n 2
n 2
= 3n2 + O(1),
which is unbounded as n → ∞. So H is an unbounded operator.
Definition 1.11 Let H =
̸ {0} be a Hilbert space. Let T : D(T ) → H be a linear operator with
dense domain D(T ) ⊂ H. Let λ ∈ C, define: Tλ := T − λI : D(T ) → H where I is the identity
operator on D(T ). If Tλ has a bounded inverse, we denote it by Rλ (T ). So
Rλ := (T − λI)−1 ∈ BL(H)
if it exists. If for λ ∈ C the operator Rλ exists, it is called the resolvent (operator) and λ is a
regular value of T .
Define the resolvent set ρ(T ) of T by
ρ(T ) := {λ ∈ C : λ is a regular value of T }.
Define the spectrum σ(T ) of T by
σ(T ) := C \ ρ(T ).
Now we split up the spectrum in three different types
• The discrete spectrum or point spectrum
σp (T ) := {λ ∈ σ(T ) : Tλ : D(T ) → H is not injective}.
• The continuous spectrum
σc (T ) := {λ ∈ σ(T ) : Tλ : D(T ) → H is injective, not surjective and Im(Tλ ) ⊂ H dense}.
• The residual spectrum
σr (T ) := {λ ∈ σ(T ) : Tλ : D(T ) → H is injective, not surjective and Im(Tλ ) ⊂ H is not dense}.
3
1 PRELIMINARIES
If λ ∈ σp (T ) we call λ an eigenvalue of T .
Remark 1.12
• If λ ∈ σp (T ) then Rλ does not exist. And ∃x ̸= 0, x ∈ ker(Tλ ) so T x = λx.
• If λ ∈ σc (T ) then Rλ is unbounded.
• If λ ∈ σc (T ) then Rλ is not defined on a set which is dense in H.
Remark 1.13 If E = Cn is a finite dimensional vector space and T : E → E a linear map,
then we know from linear algebra that Rλ does not exist if and only if ker(T − λI) ̸= {0}.
Hence if and only if λ ∈ σp (T ). Thus the spectrum of T consists only of discrete spectrum, i.e.
σ(T ) = σp (T ) and σc (T ) = σr (T ) = ∅.
Example 1.14 H has a pure discrete spectrum. To prove this we first calculate the eigenfunc1 2
tions. Let p be a polynomial, then e− 2 x p(x) ∈ C 2 (R) ∩ L2 (R).
)
)
1 2
d 2 ( − 1 x2
d ( − 1 x2
e 2 p(x) =
e 2 p(x) − xe− 2 x p(x)
2
dx
dx
1 2
1 2
1 2
1 2
1 2
= e− 2 x p′′ (x) − xe− 2 x p′ (x) − xe− 2 x p′ (x) − e− 2 x p(x) + x2 e− 2 x p(x)
)
1 2 (
= e− 2 x p′′ (x) − 2xp′ (x) + (x2 − 1)p(x) .
( 1 2
)
1 2
H e− 2 x p(x) = e− 2 x (p′′ (x) − 2xp′ (x) − p(x)) .
Thus
It is known [7, equation 6.34] that the n’th Hermite polynomial Hn = (−1)n ex
the differential equation
Hn′′ (x) − 2xHn′ (x) + 2nHn (x) = 0.
2
dn −x2
dxn e
satisfies
So if we set: p(x) := Hn (x) , λ = −2n − 1, then
0 = e− 2 x (p′′ (x) − 2xp′ (x) − (1 + λ)p(x)) = H(e− 2 x p(x)) − λe− 2 x p(x).
1
2
1
2
1
2
Thus e− 2 x Hn (x) is an eigenfunction with eigenvalue −(2n + 1). It is also known [7, section 6.2]
that the polynomials Hn are orthogonal. Hence we define the collection of orthonormal functions
{yn }n∈N
1 2
e− 2 x Hn (x)
yn (x) := − 1 x2
.
∥e 2 Hn (x)∥
1
2
By the Euclidean algorithm it is obvious that every polynomial can be written as an finite sum of
1 2
Hn ’s. So the functions yn form an orthogonal basis for the space: {p(x)e− 2 x : p is a polynomial}.
From [5, Lemma 9.11] we known that this set is dense in L2 (R). So the collection {yn }n∈N also
form a basis for L2 (R).
For f ∈ L2 (R) we can write, with convergence in L2 norm,
f=
∞
∑
where fn := ⟨f, yn ⟩.
fn y n
n=0
Now we extend the domain of H
D(H)ext := {f ∈ L2 (R) : f =
∞
∑
fn yn and −
n=0
= {f ∈ L2 (R) : f =
∞
∑
∞
∑
fn (2n + 1)yn ∈ L2 (R)}
n=0
fn yn and
n=0
∞
∑
n=0
4
|fn |2 (2n + 1)2 < ∞}.
Continue the operator H : D(H)ext → L2 (R) on the extended domain D(H)ext by
Hf := −
∞
∑
fn (2n + 1)yn
f ∈ D(H)ext .
n=0
Let λ ∈ C \ {−(2n + 1) : n ∈ N} then we obtain
(H − λI)−1 f =
∞
∑
fn (H − λI)−1 yn =
n=0
∞
∑
fn
n=0
yn
.
−(2n + 1) − λ
1
Define C := min{−(2n+1)−λ:n∈N}
< ∞. Then for λ ∈
/ {−(2n + 1) : n ∈ N} the operator H − λI is
invertible and bounded, because
∥(H − λI)−1 f ∥2 =
∞
∑
fn
n=0
yn
−(2n + 1) − λ
2
≤ C2
∞
∑
|fn |2 · 1 = C 2 ∥f ∥2 .
n=0
So C \ {−(2n + 1) : n ∈ N} ⊂ ρ(H) thus σ(H) ⊂ {−(2n + 1) : n ∈ N}. And if we let f = yn we
see that {−(2n + 1) : n ∈ N} ⊂ σ(H). Thus the spectrum of H is discrete and σ(H) = σp (H) =
{−(2n + 1) : n ∈ N}.
Definition 1.15 Let H be a Hilbert space, T ∈ BL(H, H). The adjoint of T , notation T ∗ is the
map T ∗ : H → H defined by
⟨T x, y⟩ = ⟨x, T ∗ y⟩ ∀x, y ∈ H.
If T = T ∗ then T is called self-adjoint.
Proposition 1.16 Let H be a Hilbert space, S, T ∈ BL(H, H) then
1. T ∗ is a linear operator
2. T ∗ is bounded, moreover: ∥T ∥ = ∥T ∗ ∥
3. (T ∗ )∗ = T
4. (αT + βS)∗ = αT ∗ + βS ∗
∀α, β ∈ C
5. (T S)∗ = S ∗ T ∗ .
Proof. Can be found in [5].
Definition 1.17 Let H be a Hilbert space, (T, D(T )) densely defined unbounded operator on H
Define
D(T ∗ ) := {x ∈ H : ∃x∗ ∈ H such that ⟨T y, x⟩ = ⟨y, x∗ ⟩ ∀y ∈ D(T )}.
For x ∈ D(T ∗ ), we define T ∗ (x) := x∗ . The map T ∗ is called the adjoint of T .
Lemma 1.18 Let (T, D(T )) be a densely defined operator on H. Then T ∗ : D(T )) → D is a
linear map.
Definition 1.19 Let T be a densely defined operator on an Hilbert Space H. We call T a
symmetric operator if
⟨T x, y⟩ = ⟨x, T y⟩ ∀x, y ∈ D(T ).
We call T a self-adjoint operator if
D(T ) = D(T ∗ ) and T x = T ∗ x ∀x ∈ D(T ).
5
1 PRELIMINARIES
Example 1.20 The harmonic oscillator is symmetric when restricted to the domain C 2 (R) ∩
L2 (R). C02 (R) is dense in C 2 (R) ∩ L2 (R) so we prove the claim for this dense subset. Let
φ(x), ψ(x) ∈ C02 (R), then
∫ ∞
⟨Hφ(x), ψ(x)⟩ =
Hφ(x)ψ(x) dx
−∞
∫ ∞
=
−φ′′ (x)ψ(x) + x2 φ(x)ψ(x) dx
−∞
∫ ∞
′
∞
= −[φ′ (x)ψ(x)]∞
+
[φ(x)ψ
+
−φ(x)ψ ′′ (x) + φ(x)x2 ψ(x)dx
(x)]
−∞
∞
−∞
= −0 + 0 + ⟨φ(x), Hψ(x)⟩.
It is possible to prove (but we will not do it) that (H, D(H)ext ) is even a self-adjoint operator.
6
2
The Schrödinger operator with Morse potential
This section is based on the article [3] of Jeffrey C. Lagarias. The main result is to prove that the
spectrum of the Schrödinger operator with Morse potential has a discrete spectrum and to give
the asymptotic behaviour of the eigenvalues. These results can be found in Corollary 2.34 and
Theorem 2.41.
2.1
Properties of the solutions
In this subsection we give some properties of the solutions of the Schrödinger operator with Morse
potential. And we impose restrictions to the solutions to obtain a one dimensional linear subspace
of the vector space of solutions, see Theorem 2.19.
Definition 2.1 A (one-dimensional) Schrödinger operator is an operator of the form
L := −
d2
+ V (u) where V is a complex valued function.
du2
(2.1)
The Morse potential is the potential function Vk (u) := 14 e2u + keu , with parameter k ∈ R.
Notation 2.2 In section 2 we will consider the Schrödinger operator with Morse potential on
the right half-line [u0 , ∞) u0 ∈ R. We will use the notation Vk (u) to denote the Morse potential
1 2u
d2
1 2u
+ keu and Lk to denote the Schrödinger operator with Morse potential, Lk := − du
+
2 + 4e
4e
keu .
Remark 2.3
300
200
100
0
1
2
3
x
Figure 1: Plot of the Morse potential
In figure 1 is a plot of the Morse potential Vk . The green line in the case k = −5; the red line in
the case k = 1.
Lemma 2.4 The Schrödinger operator with the Morse potential on the right half-line is, under a
translation, equivalent to the Schrödinger operator with potential function VA,B (v) := Ae2v +Bev .
Proof. Use the translation v = u − log( A4 ), then
( )−1
1
A
A
1
Ae2v = Ae2(u− 2 log( 4 )) = A
e2u = e2u
4
4
( )− 21
1
A
A
2B
Bev = Be(u− 2 log( 4 )) = B
eu = √ eu .
4
A
So VA,B (v) = 14 e2u +
2B u
√
e .
A
Now set k =
2B
√
A
and u0 := v0 + log( A4 ).
7
2
THE SCHRÖDINGER OPERATOR WITH MORSE POTENTIAL
Lemma 2.5 The Morse potential Vk (u) is continuous and bounded from below. In particular for
k < 0 it has a unique minimum of −k 2 , attained at u1 = log(−2k). Furthermore Vk is convex on
R if k ≥ 0 and on the interval [log(−k), ∞) if k < 0. And limu→∞ Vk (u) = ∞.
Proof. Continuity is obvious.
If k ≥ 0, then Vk (u) ≥ 0. Hence Vk is bounded from below. Let k < 0. The function x 7→ ex is a
continuous strictly increasing function, so define x = eu . Then Vk (u) = 14 x2 + kx. Minimizing by
differentiation gives: 12 x + k = 0, thus x = −2k. So u1 = log(−2k) and Vk (u1 ) = −k 2 .
Vk′′ (u) = e2u + keu = x2 + kx = x(x + k). If k ≥ 0 then Vk′′ (u) > 0 ∀u ∈ R. If k < 0 then
x(x + k) ≥ 0 if and only if x ≤ 0 or x ≥ −k. The case x < 0 is impossible because then log(x) is
undefined. So for u > log(−k) the function Vk′′ (u) ≥ 0. Hence Vk is convex on R if k ≥ 0 and on
the interval [ log(−k), ∞) if k < 0.
If Vk is convex, it increases at a rate which is at least linear, so limu→∞ Vk (u) = ∞.
Lemma 2.6 Lk is an unbounded linear operator on C 2 [u0 , ∞) ∩ L2 [u0 , ∞).
Proof. Let φ, ψ be twice continuous differentiable functions, then
d2
Lk (cφ(u) + dψ(u)) = − 2 (cφ(u) + dψ(u)) + Vk (u) (cφ(u) + dψ(u))
(du 2
)
(
)
d
d2
= c − 2 + Vk (u) φ(u) + d − 2 + Vk (u) ψ(u)
du
du
= c Lk φ(u) + d Lk ψ(u).
So Lk is a linear operator. We will leave proof that Lk is an unbounded operator as an exercise
for the reader. This proof is analogous to Example 1.10. Consider for example the sequence
n
(φx (x))∞
n=0 with φn (x) = x e
x2
2
∫∞
. And use that
0
∫
2
x2n+4 e−x dx
∞
0
x2n e−x2
dx
=
Γ(n+ 52 )
Γ(n+ 12 )
∼ n2 → ∞ as n → ∞.
But also from Theorem 2.41 we can conclude that the operator is unbounded, because the eigenvalues of Lk tend to infinity.
Remark 2.7 An n’th order linear differential operator L has an n-dimensional vector space of
solutions of the equation Lφ = 0.
Proposition 2.8 Consider the Whittaker’s differential equation
(
( 2
1
2 ))
1 κ
d
4 −µ
+
−
+
+
f (x) = 0,
dx2
4 x
x2
(2.2)
and the Schrödinger equation
Lk ψ(u) = Eψ(u).
Then for κ = −k and µ2 = −E, the function f is a solution of (2.2) if and only if φ(u) := e
is a solution of (2.3).
Notation 2.9 Denote by EE,k the solution space of the Schrödinger equation (2.3) i.e.
{
}
EE,k := φ ∈ C 2 [u0 , ∞) : Lk φ = Eφ .
Proof. First we calculate
d2
du2 φ(u)
(
)
d
d −u
d2
2 f (eu )
φ(u)
=
e
du2
du du
(
)
u
1 u
d
− e− 2 f (eu ) + e− 2 eu f ′ (x)
=
du
2
u
1 −u
1 u
1 u
= e 2 f (eu ) − e− 2 eu f ′ (eu ) + e 2 f ′ (eu ) + e 2 eu f ′′ (eu )
4
2
2
3
1 −u
u
u ′′ u
2
2
= e f (e ) + e f (e ).
4
8
(2.3)
−u
2
f (eu )
2.1
Properties of the solutions
Inserting this equality in (2.3) and substitution x = eu gives
d2 φ
1
(u) + e2u φ(u) + keu φ(u) − Eφ(u)
du2
4
3
u
u
u
1
1 −u
= − e 2 f (eu ) − e 2 u f ′′ (eu ) + e2u e− 2 f (eu ) + keu e− 2 f (eu ) − Ee− 2 f (eu )
4 (
4
)
3
1 −2u
1
u
′′ u
u
u
−u
u
2 −2u
u
2
= −e
f (e ) + e
f (e ) − f (e ) + κe f (e ) − µ e
f (e )
4
4
( 2
)
1
2
3
d f
κ
1
4 −µ
f
(x)
+
f
(x)
+
= −x 2
(x)
−
f
(x)
.
dx2
4
x
x2
−
So if f is a solution of the Whittaker differential equation (2.2) then φ(u) := e− 2 f (eu ) is a solution
of (2.3). Because both equations are second order differential equations it follows from Remark
2.7, that both have a two dimensional vector space of solutions. By Vw denote the solution space of
u
the Whittaker equation (2.2). From the previous calculation follows: {e− 2 f (eu ) : f ∈ Vw } ⊂ EE,k .
And because both vector spaces have equal dimension the claim follows.
u
Definition 2.10 For a ∈ C define the rising factorial or Pochhammer symbol by
∫ ∞
Γ(a + n)
(a)n :=
,
with Γ(z) :=
tz−1 e−t dt z ∈ C, ℜ(z) > 0.
Γ(a)
0
Then (a)n = a(a + 1) . . . (a + n − 1), (a)0 = 0.
For p, q ∈ N∗ the generalized hypergeometric function is the function
p Fq (a1 , . . . , ap ; b1 , . . . , bq ; z) :=
∞
∑
(a1 )n . . . (ap )n z n
(b1 )n . . . (bq )n n!
n=0
bi ∈
/ {n ∈ Z : n ≤ 0}.
If p = 2 and q = 1, the function 2 F1 (a, b; c; z) is called the hypergeometric function. If p = q = 1,
the function 1 F1 (a; b; z) is called the confluent hypergeometric function.
Lemma 2.11 The generalized hypergeometric function p Fq has, unless the series of p Fq terminates, radius of convergence


0 if p > q + 1
1 if p = q + 1 .


∞ if p < q + 1
Proof. Let cn := ((a1 )n · · · (ap )n )/((b1 )n · · · (bq )n · n!) Now apply the ratio test on the sequence
(cn )∞
n=0
(
)
(a1 )n ···(ap )n
(b1 )n ···(bq )n ·n!
cn
(a1 )n
(ap )n (b1 )n+1
(bq )n+1 (n + 1)!
)=
=(
···
···
(a1 )n+1 ···(ap )n+1
cn+1
(a1 )n+1
(ap )n+1 (b1 )n
(bq )n
n!
(b1 )n+1 ···(bq )n+1 ·(n+1)!
=
1
1
···
(b1 + n) · · · (bq + n)(n + 1).
a1 + n
ap + n
b +n
bi +n
cn
cn
q
n+1
= ab11+n
Let p = q+1, then: cn+1
+n · · · ap−1 +n ap +n . Observe that limn→∞ ai +n = 1. So limn→∞ cn+1 =
1. And by the ratio test the radius of convergence is 1.
The cases p > q + 1 and p < q + 1 are analogous. Use the fact that limn→∞ ap1+n = 0 and
limn→∞ n+1
1 = ∞.
It is known [10, Chapter 16] that the solution space of the Whittaker differential equation (2.2) is
spanned by the Whittaker functions Mκ,µ (z) and Wκ,µ (z). These Whittaker functions are defined
in the next definition.
9
2
THE SCHRÖDINGER OPERATOR WITH MORSE POTENTIAL
Definition 2.12 The Whittaker M and W functions are defined by
(
)
1
− z2 12 +µ
Mκ,µ (z) :=e z
− κ + µ; 2µ + 1; z
1 F1
2
)
(
z
1
Γ(−2µ)z 2 +µ e− 2
1
−
κ
+
µ;
2µ
+
1;
z
Wκ,µ (z) :=
F
1
1
2
Γ( 12 − κ − µ)
)
(
z
1
−µ
−
Γ(2µ)z 2 e 2
1
+
− κ − µ; −2µ + 1; z .
1 F1
2
Γ( 21 − κ + µ)
∑∞
−n
Definition 2.13 Let F : C → C or R → R be a function,
be a formal power
n=0 an z
∑N
−n
series. Define the N ’th partial sum SN (z) := n=0 an z
and the remaining term RN (z) :=
F (z) − SN (z). Let Ω ⊂ R or C be an unbounded region. Assume that the following relation holds
(
)
RN (z) = O z −N
for |z| → ∞, z ∈ Ω.
∑∞
Then the series n=0 an z −n is an asymptotic expansion near ∞ of F . We denote
F (z) ∼
∞
∑
an z −n , |z| → ∞, z ∈ Ω.
n=0
Proposition 2.14 The confluent hypergeometric function 1 F1 (a; c; z) has the asymptotic expansion
∞
Γ(c) z a−c ∑ (c − a)n (1 − a)n −n
π
e z
z
for z → ∞, | arg(z)| < .
Γ(a)
n!
2
n=0
Proof. First we give an integral representation for the function 1 F1 (a; c; z)
1 F1 (a; c; z) =
∞
∑
z n (a)n
n! (c)n
n=0
=
∞
∑
z n Γ(n + a) Γ(c)
n! Γ(a) Γ(c + n)
n=0
=
∞
∑
Γ(c)
z n Γ(n + a)Γ(c − a)
Γ(a)Γ(c − a) n=0 n! Γ(n + a + c − a)
=
=
=
=
=
=
∞
∑
Γ(c)
zn
B(n + a, c − a)
Γ(a)Γ(c − a) n=0 n!
(∫ 1
)
∞
∑
zn
Γ(c)
tn+a−1 (1 − t)c−a−1 dt
Γ(a)Γ(c − a) n=0 n!
0
)
∞ (∫ 1 n
∑
z n a−1
Γ(c)
c−a−1
t t
(1 − t)
dt
Γ(a)Γ(c − a) n=0 0 n!
)
∫ 1 (∑
∞
Γ(c)
z n n a−1
c−a−1
t t
(1 − t)
dt
Γ(a)Γ(c − a) 0 n=0 n!
)
∫ 1 (∑
∞
Γ(c)
(zt)n a−1
t
(1 − t)c−a−1 dt
Γ(a)Γ(c − a) 0 n=0 n!
∫ 1
Γ(c)
ezt ta−1 (1 − t)c−a−1 dt.
Γ(a)Γ(c − a) 0
(2.4)
(2.5)
(2.6)
Equalities (2.4) and (2.5) hold because the beta-function B(x, y) can be written in the two ways
∫ 1
Γ(x)Γ(y)
tx−1 (1 − t)y−1 dt = B(x, y) =
.
Γ(x + y)
0
10
2.1
Properties of the solutions
Equality (2.6) is obtained by using the Lebesgue’s dominated convergence theorem. Furthermore
z
1 F1 (a; c; z) = e 1 F1 (c − a; c; −z), see [7, equation 7.5]. Hence we obtain:
∫ 1
Γ(c)
z
F
(a;
c;
z)
=
e
e−zt tc−a−1 (1 − t)a−1 dt
1 1
Γ(c − a)Γ(a)
0
∫ 1 −zt a−1
Now we want to apply the Lemma of Watson on 0 e t
(1 − t)c−a−1 dt. For this we define
{
(1 − t)a−1 tc−a−1 t < 1
f (t) =
.
0
t≥1
Then for ℜ(c − a) > 0, ℜ(a) > 0 we obtain the following
1. f is a complex valued function with one discontinuity at t = 1.
2. Using Newton’s generalized binomial theorem we have
)
∞ (
∞
∑
∑
r − 1 a−1−k
(a − 1)k
a−1
(1 − t)
=
1
(−t)k =
(−t)k .
k
k!
k=0
k=0
∑∞
c−a−1
(1−a)k k
Hence around zero we have f (t) = t
k=0
k! t .
∫ ∞ −zt
∫ 1 −zt
3. Observe that 0 e f (t) dt = 0 e f (t) dt. Because for ℜ(c − a) > 0, ℜ(a) > 0 the beta
function B(c∫− a, a) is well defined and t 7→ e−zt f (t) is bounded on the interval [0, 1]. Hence
∞
the integral 0 e−zt f (t) dt exists.
Application of the Lemma of Watson [7, Theorem 2.4] gives
1 F1 (a; c; z)
∼
∞
Γ(c) z ∑ Γ(n + c − a) (1 − a)n −n a−c
e
z z
Γ(a) n=0 Γ(c − a)
n!
=
∞
Γ(c) z a−c ∑ (c − a)n (1 − a)n −n
e z
z
Γ(a)
n!
n=0
for | arg(z)| <
π
, z → ∞,
2
as desired.
Proposition 2.15 The function U (a; c; z) defined by
∫ ∞
1
e−zt ta−1 (1 + t)c−a−1 dt for ℜ(z) > 0,
Γ(a) 0
has for ℜ(a) ≥ 0, ℜ(c − a) ≥ 0 the asymptotic expansion
z −a
∞
∑
(a)n (a − c + 1)n
(−z)−n
n!
n=0
for z → ∞.
Proof. As in Proposition 2.14 we apply the Lemma of Watson. Let f (t) := ta−1 (1 + t)c−a−1 .
Then again
1. f is complex and integrable.
∫∞
2. 0 f (t)e−zt dt is convergent because if ℜ(a) ≥ 1, ℜ(c − a) ≥ 1 the functoin f is a polynomial (with possibly) real coefficients and e−zt decreases exponentially fast to zero as
t → ∞ and ℜ(z) > 0. And in the general case ℜ(a) ≥ 0, ℜ(c − a) ≥ 0, the integral is also
convergent.
∑∞
1
3. f (t) = ta−1 n=0 tn (−1)n (a − c + 1)n n!
Because again using Newton’s generalized binomial theorem we obtain (1 + t)c−a−1 =
∑∞ (c−a−1)k k
t .
k=0
k!
11
2
THE SCHRÖDINGER OPERATOR WITH MORSE POTENTIAL
Application of the Lemma of Watson gives
U (a; c; z) ∼
∞
1 ∑
(−1)n (a − c + 1)n 1
Γ(n + a)
Γ(a) n=0
n!
z n+a
= z −a
∞
∑
(a)n (a − c + 1)n
(−z)−n ,
n!
n=0
as desired.
Remark 2.16 1 F1 (a; c; z) and z 1−c 1 F1 (c−a; c; −z) are two independent solutions of the confluent
differential equation zF ′′ +(c−z)F ′ −aF = 0 [7, section 7.1]. And U (a; c; z) is also a solution of this
differential equation. Thus U is a linear combination of those other two solutions [7, equation 7.14]
U (a; c; z) =
The choice a =
1
2
Γ(1 − c)
Γ(c − 1) 1−c
z
1 F1 (a; c; z) +
1 F1 (a − c + 1; 2 − c; z).
Γ(a − c + 1)
Γ(a)
− κ + µ, c = 1 + 2µ gives
1
U ( − κ + µ; 1 + 2µ; z)
2
Γ(−2µ)
1
Γ(2µ)
1
=
z −2µ 1 F1 ( − κ − µ; 1 − 2µ; z).
1 F1 ( − κ + µ; 1 + 2µ; z) +
2
2
Γ( 12 − κ − µ)
Γ( 12 − κ + µ)
When using Definition 2.12 we obtain
1
1
1
Wκ,µ (z) = e− 2 z z 2 +µ U ( − κ + µ, 1 + 2µ, z).
2
(2.7)
Corollary 2.17 The Whittaker-M function has the asymptotic expansion
Mκ,µ (z) ∼ e 2 z −κ
z
∞
Γ(1 + 2µ) ∑ ( 12 + κ + µ)n ( 21 + κ − µ)n −n
z
n!
Γ( 12 − κ + µ) n=0
z → ∞, z ∈ R.
The Whittaker-W function has the asymptotic expansion
Wκ,µ (z) ∼ e− 2 z z κ
1
∞
∑
( 12 − κ + µ)n ( 12 − κ − µ)n
(−z)n
n!
n=0
z → ∞, z ∈ R.
Proof.
The asymptotic expansion for Mκ,µ (z) follows immediately from Definition 2.12 and
Proposition 2.14. For the asymptotic expansions of Wκ,µ (z) use (2.7) and Proposition 2.15
u
Remark 2.18 In figure 2 is a plot of the function e 2 W5,3i (eu ). Thus the solution of L−5 φ = 9φ
which is in L2 [u0 , ∞).
Theorem 2.19 Let
ψ(u; z 2 , k) := e− 2 W−k,iz (eu ) = e− 2 W−k,−iz (eu )
u
u
χ(u; z 2 , k) := e− 2 M−k,iz (eu ).
u
Then EE,k (see Notation 2.9) is spanned by ψ(u; E, k) and χ(u; E, k). And the linear subspace
{
}
FE,k := φ ∈ EE,k : φ ∈ L2 (u0 , ∞) ,
the solutions of the Schrödinger equation (2.3) that belong to L2 (u0 , ∞), is spanned by ψ(u; E, k).
12
2.2
The discrete spectrum
Figure 2: Plot of a solution in L2 [u0 , ∞).
Proof. From Proposition 2.8 we see that ψ(u; E, k) and χ(u; E, k) solutions are of the Schrödinger
equation 2.3. The Whittaker functions Mκ,µ and Wκ,µ are linear independent, so also ψ(u; E, k)
and χ(u; E, k) are linear independent. From the asymptotic expansion of Wκ,µ and Mκ,µ (see
Corollary 2.17) we obtain the asymptotic expansion of ψ and χ
ψ(u; E, k) ∼ e
1 u
−u
−ku
2 −2e
e
e
√
√
∞
∑
( 12 + k + i E)n ( 21 + k − i E)n
(−eu )n .
n!
n=0
We see that ψ(u; E, k) is square integrable (because of the term e− 2 e ). But
1 u
χ(u; E, k) ∼ e− 2 e 2 e eku
u
1 u
√
√
√
∞
Γ(1 + 2i E) ∑ ( 21 + k + i E)n 12 + k − i E)n
√
(−eu )n .
n!
Γ( 12 + k + i E) n=0
1 u
So χ(u; E, k) is not square integrable (because of the term e 2 e ). Thus FE,k is a one dimensional
vector space and is spanned by ψ(u; E, k).
2.2
The discrete spectrum
2
d
Definition 2.20 Let V be a (potential) function. The Schrödinger operator L := − dx
2 + V (x)
is in the limit circle case at infinity respectively at 0 if for all λ ∈ C all solutions of Lφ = λφ are
square integrable at infinity respectively 0. L is in limit point case at infinity respectively at 0 if
it is not in limit circle case at infinity respectively 0.
Corollary 2.21 The operator Lk is in limit point case at infinity.
Proof. Let λ ∈ C. Then (see the proof of Theorem 2.19) the solution χ(u; λ, k) is not square
integrable at ∞.
Another way to prove this result is by using Corollary 2.25, because Vk is bounded from below
(Lemma 2.5).
Definition 2.22 Let f and g be complex valued differentiable functions. The Wronskian of f
and g is the function
W (f, g) := f ′ g − f g ′ .
Lemma 2.23 Let Q be a continuous complex-valued function on (0, ∞) and φ, ψ be solutions of
f ′′ − Qf = 0. Then W (φ, ψ) is constant and W (φ, ψ) = 0 ⇔ φ and ψ are linear dependent.
13
2
THE SCHRÖDINGER OPERATOR WITH MORSE POTENTIAL
Proof. Differentiation of the Wronskian gives
d
d
W (φ(x), ψ(x)) =
(φ′ (x)ψ(x) − φ(x)ψ ′ (x))
dx
dx
= φ′′ (x)ψ(x) + φ′ (x)ψ ′ (x) − φ′ (x)ψ ′ (x) − φ(x)ψ ′′ (x)
= Q(x)φ(x)ψ(x) − φ(x)Q(x)ψ(x)
= 0.
So the Wronskian is constant.
And φ and ψ are linear dependent if and only if (φ, φ′ ) and (ψ, ψ ′ ) are linear dependent if and
only if
)
(
φ ψ
= 0,
det
φ′ ψ ′
which is exactly the Wronskian.
Lemma 2.24 Let V be a continuous real-valued function on (0, ∞) and suppose there exists a
positive differentiable function M such that
1. V (x) ≥ −M (x) ∀x ∈ (0, ∞)
∫∞
2. 1 √ 1
dx = ∞
M (x)
′
M
M 3/2
3.
is bounded near ∞.
Then V is in limit point case at infinity.
Proof. This lemma is exactly [6, Theorem X.8]. We also need this result in section 3, that is why
we give the proof.
d2
2
Let L := − dx
near ∞. Let c0 > 0 such
2 + V (x) and f be a real solution of Lf = 0 which is in L
∫∞
2
that K1 := c0 f (x)dx < ∞ and c0 < c < ∞. Then
∫ c
∫ c
∫ c ′′
V (x)
f (x)f (x)
−K1 ≤ −
f 2 (x)dx ≤
f 2 (x)
dx =
dx
M
(x)
M (x)
c0
c0
c0
[
]x=c
∫ c
f (x)
f ′ (x)M (x) − f (x)M ′ (x)
′
= f (x)
−
f ′ (x)
dx.
M (x) x=c0
(M (x))2
c0
So we obtain the inequality
[
]x=c
∫ c ′
∫ c ′
f (x)
(f (x))2
f (x)f (x)M ′ (x)
′
K1 ≥ − f (x)
+
dx −
dx.
M (x) x=c0
(M (x))2
c0 M (x)
c0
(2.8)
Let K2 be the boundary given by property 3. Application of the Cauchy-Schwarz inequality gives
∫
c
c0
f ′ (x)f (x)M ′ (x)
dx ≤ K2
(M (x))2
Now suppose
∫∞
c0
f ′ (x)2
M (x)
∫
c
c0
≥
(∫
c
c0
) 12 (∫ c
) 12
(f ′ (x))2
2
dx
(f (x)) dx
.
M (x)
c0
(2.9)
dx = ∞. Then rewriting of (2.8) and using (2.9) gives
f ′ (c0 )f (c0 )
f ′ (c)f (c)
≥
− K1 +
M (c)
M (c0 )
≥
f ′ (x)f (x)
√
dx ≤ K2
M (x)
f ′ (c0 )f (c0 )
− K1 +
M (c0 )
f ′ (c0 )f (c0 )
− K1 +
M (c0 )
∫
∫
∫
c
c0
c
c0
c
c0
(f ′ (x))2
dx −
M (x)
(f ′ (x))2
M (x)
(f ′ (x))2
M (x)
∫
f ′ (x)f (x)M ′ (x)
dx
M (x)2
c0
) 12
(∫ c ′
) 12 (∫ c
(f (x))2
(f (x))2 dx
dx − K2
dx
c0
c0 M (x)
(∫ c ′
) 12
(f (x))2
dx − K2 ∥f ∥L2 [c0 ,∞)
dx
.
c0 M (x)
14
c
2.2
The discrete spectrum
The first two terms are finite, the third tends to ∞ as c → ∞ and the fourth to −∞ as c → ∞.
But because of the square root in the fourth term, eventually the second term dominates and the
left hand becomes positive.
′
(x)f (x)
So f M
is positive near ∞. But then f ′ and f have the same sign near ∞. This implies that f
(x)
is eventually monotonic
increasing to ∞, monotonic decreasing to −∞ or tends to a finite non-zero
∫∞
limit. But then c0 |f (x)|2 dx = ∞, which contradicts the fact that f is square integrable near
∫∞ ′ 2
∞. Hence c0 fM(x)
(x) dx < ∞.
Now suppose we have two independent solutions φ, ψ of Lf = 0 which are in L2 near ∞. Without
loss of generality we may assume they are normalized and real such that W (φ, ψ) = 1. Real,
because the potential V is real. So if Lf = 0 then Lf = 0, thus also L(ℑf ) = L(ℜf ) = 0.
Normalized, because if Lf = 0 then also for µ ∈ R : L(µf ) = 0.
′
′2
φ′ ψ
Then by the previous result: φM is L1 near ∞, so √φM is L2 near ∞. Thus √
is L1 near ∞.
M
Analogous
φψ ′
√
M
is L1 near ∞. And
1
φψ ′
φψ ′
√ =√ −√ .
M
M
M
So
√1
M
is in L1 [1, ∞) which contradicts assumption 2.
Corollary 2.25 Let V be a continuous real-valued function on (a, ∞) which is bounded from
below. Then V is in limit point case at infinity.
Proof. Let N be such that V (x) ≥ N ∀x ∈ (a, ∞). And set
{
|N | if N ̸= 0
′
N :=
.
1
if N = 0
Use the translation x 7→ x − a to translate the interval (a, ∞) to (0, ∞). Define W (x) := V (x + a)
and M : (0, ∞) → R, x 7→ N ′ . Then W (x) ≥ −M (x) ∀x ∈ (0, ∞). And M (x) = N ′ > 0 hence
∫∞
′
√1
dx = ∞. And M ′ = 0 so MM3/2 = 0 which is bounded near ∞. From Lemma 2.24 we
1
M (x)
obtain that W is in limit point case at infinity.
d2
Using the inverse translation x 7→ x + a and the fact that if dx
2 φ(x) + V (x) = λφ(x), then
d2
dx2 φ(x + a) + V (x + a) = λφ(x + a), we see that V also is in the limit point case at infinity.
Definition 2.26 Let I ⊂ R be an interval, f : I → R is an absolutely continuous function if
∀ε
∑n> 0 ∃δ > 0 such that for every finite collection of disjoint intervals (xk , yk ), k = 0, . . . , n with
k=0 (yk − xk ) < δ the following inequality holds
n
∑
|f (yk ) − f (xk )| < ε.
k=0
{
}
Denote by AC(I) := f : I → R : f absolutely continuous, f ′ (defined a.e.) is in L2 (I)
It is known that f ∈ AC([a, b]) if and only if f ′ exists a.e., f ′ ∈ L(I) and f (x) =
Remark
∫ x 2.27
′
f (a) + a f (x) dx.
Proposition 2.28 The Schrödinger operator Lk with boundary conditions
cos(α)f (u0 ) + sin(α)f ′ (u0 ) = 0
α ∈ [0, 2π) fixed,
(2.10)
is self-adjoint in L2 [u0 , ∞) when restricted to the domain
{
}
Dα (Lk ) := f ∈ C 1 [u0 , ∞) ∩ L2 [u0 , ∞) : f ′ ∈ AC[u0 , ∞], Lk (f ) ∈ L2 [u0 , ∞), f satisfies 2.10
(2.11)
15
2
THE SCHRÖDINGER OPERATOR WITH MORSE POTENTIAL
Proof. We prove a weaker statement of this proposition, namely that Lk is symmetric on the set
L2 [u0 , ∞) ∩ C 2 [u0 , ∞). The rest of the proposition is too difficult to prove and beyond the scope
of this thesis. See [6, p. 144] to conclude that Lk has indeed as domain for self-adjointness the set
(2.11).
)
∫ ∞(
d2
− 2 φ(u) + Vk (u)φ(u) ψ(u) du
⟨Lk φ, ψ⟩ =
du
u0
)
∫ ∞(
∫ ∞
d2
=
− 2 φ(u) ψ(u) du +
Vk (u)φ(u)ψ(u) du
du
u
u
)
]∞
[ 0 (
)]∞
(
)
[(0
∫ ∞
d2
d
d
+ φ(u)
ψ(u)
+
φ(u) − 2 ψ(u) du
=
− φ(u) ψ(u)
du
du
du
u0
u=u0
u=u0
∫ ∞
+
φ(u)Vk (u)ψ(u) du
u0
′
= [−φ (u)ψ(u) + φ(u)ψ
′
∞
(u)]u=u0
∫
∞
+
u0
(
)
d2
φ(u) − 2 ψ(u) + Vk (u)ψ(u) du
du
= 0 + ⟨φ, Lk ψ⟩
∞
The term [−φ′ (u)ψ(u) + φ(u)ψ ′ (u)]u=u0 is 0, because −φ′ (u)ψ(u) + φ(u)ψ ′ (u) is the Wronskian,
which is constant by Lemma 2.23 for any two solutions of the differential equation (2.3).
Lemma 2.29 Let H be a complex Hilbert Space. Let T : D(T ) → H be a symmetric operator.
Then the discrete spectrum σp (T ) of T is real. And let φ, ψ be eigenfunctions by eigenvalues
λ respectively µ. If λ ̸= µ then ⟨φ, ψ⟩ = 0. Thus eigenfunctions of different eigenvalues are
orthogonal.
Proof. Let λ ∈ σp (T ) and φ be a eigenfunction of λ. Then
λ⟨φ, φ⟩ = ⟨λφ, φ⟩ = ⟨T φ, φ⟩ = ⟨φ, T φ⟩ = ⟨φ, λφ⟩ = ⟨λφ, φ⟩ = λ ⟨φ, φ⟩ = λ⟨φ, φ⟩
Hence λ = λ, so the spectrum is real. And
λ⟨φ, ψ⟩ = ⟨Lk φ, ψ⟩ = ⟨φ, Lk ψ⟩ = ⟨φ, µψ⟩ = µ⟨φ, ψ⟩ = µ⟨φ, ψ⟩
Thus if λ ̸= µ we have ⟨φ, ψ⟩ = 0.
Definition 2.30 Let T ∈ L(D(T ), H) be a linear operator. The spectrum σp (T ) of T is called
simple if for all λ ∈ σp (T ) it satisfies the following condition.
If there exist φ, ψ ∈ D(T ) such that T φ = λφ and T ψ = λψ then φ and ψ are linearly dependent.
Theorem 2.31 (Sturm comparison theorem) Consider two continous functions g and h on
the interval [a, b], such that g(x) > h(x) ∀x ∈ (a, b). Define the Schrödinger operators L :=
d2
d2
2
2
− dx
2 + g(x) and K := − dx2 + h(x). Both on the domain L [a, b] ∩ C [a, b]. Let φ and ψ be
2
solutions (not necessarily in L [a, b]) of respectively Lφ = 0, Kψ = 0, with boundary conditions
φ(a) = ψ(a) = sin(α),
φ′ (a) = ψ ′ (a) = cos(α),
α ∈ [0, 2π) fixed.
Then between any two consecutive zeros of φ there is at least one zero of ψ. Furthermore
#{x ∈ (a, b) : φ(x) = 0} ≤ #{x ∈ (a, b) : ψ(x) = 0}
And the n’th zero x′0 > a of ψ is smaller then the n’th zero x0 > a of φ.
Proof. Because φ and ψ are solutions we obtain
φ′′ (x)ψ(x) − φ(x)ψ ′′ (x) = (g(x) − h(x))φ(x)ψ(x).
16
(2.12)
2.2
The discrete spectrum
Let x1 , x2 be two consecutive zeroes of φ, then integration of (2.12) gives
∫ x2
′
′
′
′
x=x2
φ (x2 )ψ(x2 )−φ (x1 )ψ(x1 ) = [φ (x)ψ(x)−φ(x)ψ (x)]x=x1 =
(g(x)−h(x))φ(x)ψ(x) dx. (2.13)
x1
Now suppose that ψ does not have any zeroes in the interval (x1 , x2 ). We assume that φ and ψ
are positive in (x1 , x2 ) (the other three cases are similar). Then
φ′ (x1 ) ≥ 0
φ′ (x2 ) ≤ 0
ψ(x1 ) ≥ 0
ψ(x2 ) ≥ 0.
But then the left-hand of (2.13) is less or equal to 0, while the right-hand is strictly positive. A
contradiction.
Let x0 be the smallest zero of φ in the open interval (a, b). The only thing left to prove is that ψ
has a zero in (a, x0 ). Again we assume that φ, ψ are positive in (a, x0 ). Suppose that ψ has no
zero in (a, x0 ). Then integration of (2.12) on the interval (a, x0 ) gives
φ′ (x0 )ψ(x0 ) = φ′ (x0 )ψ(x0 ) − 0 + sin(α) cos(α) − sin(α) cos(α)
∫ x0
0
= [φ′ (x)ψ(x) − φ(x)ψ(x)]x=x
=
(g(x) − h(x))φ(x)ψ(x) dx.
x=a
a
And again we obtain the same contradiction.
Lemma 2.32 Let g, h be two continuous functions on I = [a, ∞) or [a, b] with g(x) > h(x) ∀x ∈ I.
d2
d2
Denote L := − dx
2 +g(x) and K := − dx2 +h(x), both with boundary conditions 2.10 on the domain
C 2 (I) ∩ L2 (I). Then the n’th eigenvalue of L is greater than the n’th eigenvalue of K counted
from the bottom of the spectrum.
Proof. We will omit this proof, because it uses techniques and knowledge which is beyond the
scope of this text. For the proof see [9, section 16.4-5]
Lemma 2.32 will be used in Theorem 2.33. In Theorem 2.33 we prove the discreteness of the
spectrum of a certain class of Schrödinger operators. We do so by comparing this operator with
an operator of which we explicitly know the eigenfunctions and the spectrum. Using Lemma 2.32
we can can obtain results about the spectrum of the original operator.
Theorem 2.33 Let q be a continuous potential on the interval [0, ∞). Suppose ∃c ∈ R such
d2
that q(x) ≥ τ ∀x > c. Consider the Schrödinger operator L := − dx
2 + q(x) on the domain
C 2 [0, ∞) ∩ L2 [0, ∞) with boundary condition φ(0) = 0. Then the spectrum of L is discrete in the
interval (−∞, τ ) and the spectrum is empty in (−∞, inf x∈[0,∞) q(x)).
This is the theorem of [8, section 5.5]. We will give the proof, because Theorem 2.33 is essential
in this thesis (It will be used two times).
Proof. Because q is continuous on [0, c] it is bounded (from below) on [0, c]. Let σ ∈ R such that
q(x) ≥ σ if 0 ≤ x ≤ c. Define
{
σ if x ∈ [0, c]
q0 (x) =
.
(2.14)
τ otherwise
First observe that q0 (x) < q(x) ∀x ≥ 0. Define the function φ0 ( · ; λ) : [0, ∞) → R by
{ sin(x−√λ−σ)
if x ∈ [0, c]
− √λ−σ
√
√
√
√
.
φ0 (x; λ) :=
sin(c λ−σ) cos((x−c) λ−τ )
cos(c λ−σ) sin((x−c) λ−τ )
√
√
−
−
otherwise
λ−σ
λ−τ
Some calculations, which are left to the reader, show that φ0 ( . ; λ) ∈ C 2 [0, ∞) ∩ L2 [0, ∞) and that
φ0 satisfies the equation
(
)
d2
− 2 + q0 (x) ψ = λψ(x).
(2.15)
dx
17
2
THE SCHRÖDINGER OPERATOR WITH MORSE POTENTIAL
Let b > c. Now we consider the eigenvalue problem
)
{(
d2
− dx
ψ = λψ(x)
2 + q0 (x)
x ∈ [0, b]
.
(2.16)
ψ(0) = ψ(b) = 0
Hence φ0 is an eigenfunction of (2.16) if and only if φ0 (b; λ) = 0. Using the identity tanh(x) =
1
i tan(ix) we can rewrite φ0 (b; λ) = 0 to
√
√
tan(c λ − σ)
tanh((b − c) τ − λ)
√
√
=−
σ < λ < τ.
λ−σ
τ −λ
Now
d tan(cx)
d sin(cx)
cx cos(cx) cos(cx) − cos(cx) sin(cx) + cx sin(cx) sin(cx)
=
=
dx
x
dx x cos(cx)
x2 cos2 (cx)
cx − 12 sin(2cx)
=
> 0 if x > 0
x2 cos2 (cx)
d tanh(cx)
d sinh(cx)
cx cosh(cx) cosh(cx) − cosh(cx) sinh(cx) − cx sinh(cx) sinh(cx)
=
=
dx
x
dx x cosh(cx)
x2 cosh2 (cx)
cx − 21 sinh(2cx)
=
< 0 if x > 0.
x2 cosh2 (cx)
Hence for λ > σ the function
√
tan(c λ − σ)
√
λ 7→
,
λ−σ
is strictly monotonic increasing from −∞ to ∞ in every interval ((n − 12 ) πc , (n + 12 ) πc ), n ∈ Z.
And for λ < τ the function
√
tanh((b − c) τ − λ)
√
λ 7→
,
τ −λ
is strictly monotonic
decreasing on the interval (σ, τ ).
√
Let N := ⌊ πc τ − σ − 12 ⌋. The fact that σ < λ < τ implies that for 1 ≤ n < N there exists exactly
one µn,b in the interval ((n − 12 ) πc , (n + 12 ) πc ) and at most one µn,b ∈ ((N − 1 − 21 ) πc , τ ) such that
√
√
tan(c µn,b − σ)
tanh((b − c) τ − µn,b )
=−
.
√
√
µn,b − σ
τ − µn,b
There are no eigenvalues of (2.16) in (−∞, σ) because then we would have
√
√
tanh(c σ − λ)
tanh((b − c) τ − λ)
√
√
=−
,
σ−λ
τ −λ
which has no solutions λ ∈ (σ, τ ).
So the number of eigenvalues of (2.16) in (−∞, τ ) is equal to the number of eigenvalues of (σ, τ )
and those are bounded by N which is independent of b.
Denote by λn,b the n’th eigenvalue of the problem
)
{(
d2
+
q(x)
ψ = λψ x ∈ [0, b]
− dx
2
,
ψ(0) = ψ(b) = 0
this is just L restricted to the interval [0, b]. Then by Lemma 2.32 we see that λn,b ≥ µn,b . Hence
the number of eigenvalues of L on [0, b] is also bounded independently of b. From [9, section 14.5]
we see that λn,b is non-increasing as b → ∞ and tending to a limit λ′n . The eigenvalues of L
on [0, ∞) are among these numbers. From [8, Theorem 5.5] we see that ∀ε > 0 the number of
18
2.2
The discrete spectrum
eigenvalues in (−∞, τ − ε) is finite, so the spectrum of L is discrete in (−∞, τ ).
Because ∀n > 0, ∀b > 0 the eigenvalue µn,b > σ and λn,b ≥ µn,b we have λ′n > σ ∀n ∈ N.
If we set σ = inf x∈[0,∞) q(x) we still have q(x) ≥ σ ∀x > 0. Hence the spectrum is empty in
(−∞, σ) = (−∞, inf x∈[0,∞) q(x)).
To give a asymptotic behaviour of the eigenvalues of the Schrödinger operator with Morse potential
we need to have point tot start counting and the spectrum must be countable. Thus it is necessary
that the spectrum is discrete and bounded from below. Corollary 2.34 shows that this is the case.
Corollary 2.34 Lk , with domain C 2 [u0 , ∞) ∩ L2 [u0 , ∞) and boundary condition (2.10) with
α = 0, has pure discrete simple real spectrum bounded from below. In particular
{
0
if k > 0
E0 (u0 , k) ≥
,
(2.17)
2
−k if k ≤ 0
where En (u0 , k) is the n’th eigenvalue of the eigenvalue problem Lk φ = λφ.
Proof. Vk is bounded from below and continuous (Lemma 2.5). So from Corollary 2.25 we obtain
that the operator Lk is in limit point case at infinity. Since we imposed the boundary conditions
(2.10) at u0 , the spectrum is simple.
Because limu→∞ Vk (u) = ∞ we see from Theorem 2.33 that the spectrum is discrete in (−∞, τ ) ∀τ ∈
R. Hence the spectrum is discrete in R. Lk is symmetric by Proposition 2.28, hence by Lemma
2.29 the spectrum of Lk is real. So the spectrum of Lk is purely discrete real and simple.
From Lemma 2.5 we see that Vk is bounded from below by
{
0
if k > 0
,
2
−k if k ≤ 0
hence using Theorem 2.33 we obtain the lower bound (2.17) of the spectrum.
The result that the spectrum of Lk is bounded from below, can also be proved by comparing Lk
with a scaled and translated version of the harmonic oscillator. This will be done in the proof of
Proposition 2.35.
Proposition 2.35 Consider the eigenvalue problem Lk φ = λφ with domain C 2 [u0 , ∞)∩L2 [u0 , ∞)
and boundary condition (2.10) with α = 0. Denote by En (u0 , k) the n’th eigenvalue of the
eigenvalue problem. Then
{
0
if k > 0
E0 (u0 , k) >
.
2
−k if k ≤ 0
In particular the spectrum of Lk is bounded from below.
Proof. If k > 0 then Vk (x) > V0 (x) ∀x ∈ [u0 , ∞). By lemma 2.32 E0 (u0 , k) ≥ E0 (u0 , 0). So it is
sufficient to prove the claim for k ≤ 0.
Define the potential
VA,u0 ,E (u) := A(u − u0 )2 − E.
Observe that this function is just a scaled and translated version of the potential of the harmonic
oscillator on the interval [0, ∞), see Example 1.10 for the definition of the harmonic oscillator.
Let ε > 0 and set E := k 2 + ε. We want to choose A > 0 such that
VA,u0 ,E (u) = A(u − u0 )2 − k 2 − ε ≤ Vk (u) ∀x ∈ [u0 , ∞),
because then it is possible to apply Lemma 2.32. Observe that for A > 0 and ε > 0 the minimum
value of VA,u0 ,E is −k 2 − ε < −k 2 which is the minimum value of Vk . Set u1 := log(−2k) + 1
19
2
THE SCHRÖDINGER OPERATOR WITH MORSE POTENTIAL
(recall that the minimum of Vk is attained at log(−2k)). Because [u0 , u1 ] is a bounded interval
we can choose A′ > 0 such that VA′ ,u0 ,E (u) < Vk (u) ∀u ∈ [u0 , u1 ]. Note that
′
VA,u
(u) = 2A(u − u0 )
0 ,E
1
Vk′ (u) = e2u + keu
2
′′
VA,u
(u) = 2A
0 ,E
Vk′′ (u) = e2u + keu .
Now choose A′′ > 0 such that VA′ ′′ ,u0 ,E (u1 ) ≤ 12 Vk′ (u1 ) and VA′′′′ ,u0 ,E (u1 ) ≤ Vk′′ (u1 ). This is possible
because Vk′ and Vk′′ both increase exponential as u → ∞ and Vk′ (u), Vk′′ (u) > 0 for u > u1 . Then
VA′ ′′ ,u0 ,E (u) < Vk′ (u) ∀u > u1 . Thus set A := min{A′ , A′′ }.
Application of Lemma 2.32 shows that
E0 ≥ E˜0 − k 2 − ε.
(2.18)
Where E0 = E0 (u0 , k) is the minimum eigenvalue of Vk and E˜0 is the minimum eigenvalue of
VA,0,0 (u) = Au2 on the interval [0, ∞). In the case A=1 we obtain from Example 1.14 that the
spectrum of V1,0,0 is {2n + 1 : n ∈ N}. So the minimal eigenvalue of V1,0,0 is 3. By substitution
1
u 7→ √
u we see that the minimal eigenvalue of VA,0,0 is √3A > 0. So ∀ε > 0 : E0 ≥ √3A −k 2 −ε ≥
4
A
−k 2 − ε. Thus E0 ≥ −k 2 .
Proposition 2.36 Denote by En,α (u0 , k) the n’th eigenvalue from the bottom of the (discrete)
spectrum of Lk with boundary conditions (2.10) on the half-line [u0 , ∞). Then for each n ≥ 0 the
following hold:
1. For fixed n, α and k, with k ≥ 0, En,α (u0 , k) is a increasing function of u0 .
2. For fixed n, α and k, with k < 0, En,α (u0 , k) is a increasing function of u0 when u0 ≥
log(2|k|).
3. For fixed n, α and u0 , En,α (u0 , k) is a increasing function of k.
Proof. Because the spectrum is bounded from below (Proposition 2.35) and the spectrum is
discrete (Corollary 2.34) we can apply Lemma 2.32.
1. Let u0 < u′0 . Define Ṽk (x) := Vk (x − u0 + u′0 ). Then Vk (x) < Ṽk (x) ∀x ∈ (u0 , ∞), because Vk
′
(u0 , k) denote the n’th eigenvalue from the bottom
is a strictly increasing function. By En,α
2
d
′
of the discrete spectrum of − dx
2 + Ṽk (x). Then by lemma 2.32 En,α (u0 , k) < En,α (u0 , k) =
′
En,α (u0 , k).
2. Let log(−2k) ≤ u0 < u′0 . The claim follows analogous to the previous case by considering
Ṽk (x) := Vk (x − u0 + u′0 ), because for u0 ≥ log(2|k|) the potential Vk is strictly increasing.
3. Let k < k ′ . Then Vk (x) < Vk′ (x) ∀x ∈ (u0 , ∞). And the claim follows again by application
of Lemma 2.32.
2.3
The asymptotic behaviour of the eigenvalues
From the previous subsection we obtain that the spectrum of the Schrödinger operator with Morse
potential is discrete and bounded from below. Now it is possible to determine the distribution
of the eigenvalues of this operator. To obtain this distribution (Theorem 2.41) We need some
theorems of [8, Chapter 7]. We will give the proof of these theorems.
2
d
Notation 2.37 Let L be the Schrödinger operator − du
2 + V (u) on the interval [u0 , ∞) with
domain D(L) and boundary conditions cos(α)φ(u0 ) + sin(α)φ′ (u0 ) = 0. Assume that L has a
discrete spectrum bounded from below. Define
N (T ; α, u0 ) := #{λ ∈ (−∞, T ) : λ is an eigenvalue of the eigenvalue problem},
the number of eigenvalues of L smaller then T .
20
2.3
The asymptotic behaviour of the eigenvalues
Lemma 2.38 Let Q be a positive continuous differentiable function on the interval [0, a]. Let
d2
2
L := − du
2 + Q (u) and let φ be a solution of Lφ = 0. Define θ : [0, a] → R such that
tan(θ(u)) =
Q(u)φ(u)
φ′ (u)
with (0 ≤ θ(0) < π).
Then θ(t) = mπ if and only if t is the m’th zero of φ in (0, a]. And if m is the number of zeros of
φ, then
∫ a ′
∫ a
|Q (u)|
Q(u) du ≤
mπ −
du + π.
Q(u)
0
0
Proof. First observe that arctan is a multi valued function. In particular arctan(0) = kπ (k ∈ Z).
= 0 if and only if φ(u) = 0, because Q is
So also θ is multi valued. Furthermore we have Q(u)φ(u)
φ′ (u)
positive. Now
dθ
=
du
1
(
1+
=Q+
Qφ
φ′
)2
Q′ φφ′ + Qφ′2 + QφQ2 φ
(Q′ φ + Qφ′ )φ′ − Qφφ′′
=
′2
φ
φ′2 + (Qφ)2
Q′ φφ′
.
+ (Qφ)2
(2.19)
(2.20)
φ′2
2
d
2
Equality (2.19) holds because du
2 φ = Q φ.
dθ
Let s ∈ [0, a] so that φ(s) = 0, then du (s) = Q(s) > 0. Hence θ increases at s, so θ increases at
every zero of φ. Let t0 < t1 be two consecutive zeros of φ. Because θ increases at t0 and t1 we
must have θ(t0 ) < θ(t1 ). Hence if θ(t0 ) = kπ it follows with the observations that θ(t1 ) = (k + 1)π.
And the fact that 0 ≤ θ(0) < π completes the first claim.
If φ has m zeros in (0, a) then it follows that 0 ≤ θ(a) − mπ ≤ π. Because 0 ≤ (φ′ − Qφ)2 =
φ′2 − 2Qφφ′ + Q2 φ2 , the inequality 2Qφφ′ ≤ φ′2 + Q2 φ2 holds. So using this estimate and (2.20)
we obtain
dθ
1 Q′
−Q ≤
.
du
2Q
Integration of this inequality gives
∫ a
∫
′
θ (u) du −
0
0
a
1
Q(u) du ≤
2
∫
a
0
|Q′ (u)|
du,
Q(u)
as desired.
Proposition 2.39 depends heavily on Lemma 2.38. Lemma 2.38 encodes for finite intervals [0, a]
the behaviour of the zeros of φ and thus of the eigenvalues of L. We now use Lemma 2.38 to say
something about the distribution of the eigenvalues in case the interval is infinite.
Proposition 2.39 [8, Theorem 7.5] Assume that q is a real-valued, C 2 and convex function. Let
d2
2
2
L := − du
2 + q(u) on the domain C [0, ∞) ∩ L [0, ∞) with boundary condition φ(0) = 0. Let uT
(for T sufficiently large) be the unique value uT ∈ [0, ∞) with q(uT ) = T . Then
∫
1 uT √
N (T ; α, 0) =
T − q(u) du + O(1) as T → ∞.
(2.21)
π 0
Proof. Because q is convex on (0, ∞), q strictly increasing at least with a linear rate. Hence
limt→∞ q(t) = ∞. So from Theorem 2.33 we obtain that the spectrum of L is discrete. As in
Proposition 2.28 we can conclude that L is symmetric on C 2 [0, ∞)∩L2 [0, ∞). So from Lemma 2.29
we can conclude that the spectrum is real. The potential q is continuous and limt→∞ q(t) = ∞,
hence q is bounded from below. An application of Corollary 2.25 shows that q is in limit point
case at infinity. This in combination with the boundary condition at 0 shows that the spectrum
is simple.
21
2
THE SCHRÖDINGER OPERATOR WITH MORSE POTENTIAL
From [8, section 5.4] we see that N (T ; α, 0) is equal to the number of zeros of the solution of
d2
(− du
2 + q(u) − T )f = 0. So we try to determine the behaviour of the zeros of the solutions as
T → ∞.
Let
√ φ(u) be a solution of Lφ = λφ. Now we try to use the Lemma 2.38, so define Q(u) :=
λ − q(x) and
(
)
φ(u)Q(u)
θ(u) := arctan
, 0 ≤ θ(0) < π.
φ′ (u)
Then this function θ is again multi valued. Some calculations show that
′
Q′ φφ′
dθ
= Q + ′2
du
φ + (Qφ)2
φφ′ Q · −q ′
=Q+
2
2
(Q φ + φ′2 )2(λ − q)
q′
4(λ − q)
=Q−
=
√
λ−q−
=Q+
(2.22)
φ′2 + Q2 φ2
q′
2Qφφ′
=Q−
2
4(λ − q) Q φ2 + φ′2
(
( ))
q ′ sin 2 arctan Qφ
√
φ′
= λ−q−
4(λ − q)
2 Qφ
φ′
Q2 φ2
φ′2
φφ′ 2√−q
λ−q
+1
(2.23)
q ′ sin(2θ)
4(λ − q)
(2.24)
Equality (2.22) is calculated in the proof of Lemma 2.38 and (2.23) follows from the identity
sin(2 arctan(x)) = x22x
+1 .
√
′
q ′ (u) sin(2θ(u))
sin(2θ(u))
′
gives
Thus we have θ (u) = λ − q(u) − q(u)
3
4(λ−q(u)) . Multiplying both sides with
4(λ−q(u)) 2
q ′ (u) sin(2θ(u))
4(λ − q(u))
3
2
θ′ (u) =
q ′ (u) sin(2θ(u)) q ′2 (u) sin2 (2θ(u))
.
−
5
4(λ − q(u))
16(λ − q(u)) 2
(2.25)
Combining (2.24) and (2.25) gives
θ′ (u) =
√
q ′ (u)θ′ (u) sin(2θ(u)) q ′2 (u) sin2 (2θ(u))
λ − q(u) −
−
.
3
5
4(λ − q(u)) 2
16(λ − q(u)) 2
Define
∫
u
I1 (u) :=
q ′ (t)
θ′ (t) sin(2θ(t))
0
3
(λ − q(t)) 2
Then
∫
u
θ(u) = θ(0) +
0
∫
dt
u
I2 (u) :=
q ′2 (t)
0
sin2 (2θ(t))
5
(λ − q(t)) 2
dt.
√
1
1
λ − q(t) dt − I1 (u) − I2 (u).
4
16
′
Fix λ > max{q(0), q (0)} and let X be the unique value in (0, ∞) such that q(X) = λ. Then
q ′ (u)
3 is a strictly increasing function on the interval (0, X) from a value less then 1 to ∞. It
(λ−q(u)) 2
is strictly increasing because
q ′′ (u)(λ − q(u)) 2 + 32 q ′2 (u)(λ − q(u)) 2
q ′ (u)
d
> 0.
=
du (λ − q(u)) 23
λ − q(u)
3
Hence there exists a unique number Y ∈ (0, X) such that
The function
2
q ′ (u)
λ−q(u)
1
q ′ (Y )
3
(λ−q(Y )) 2
= 1.
is continuous, increasing and bounded on the interval [0, Y ], because q is
C , convex and Y is chosen such that
q ′ (Y )
λ−q(Y )
= 1. The function θ′ (u) sin(2θ(u)) is integrable
22
2.3
The asymptotic behaviour of the eigenvalues
on the interval [0, Y ] since θ is C 2 . Hence by the second mean-value theorem for integration [1,
lemma 6.9] there exists a ξ ∈ (0, Y ) such that
]t=Y
[
∫ Y
q ′ (Y )
1
′
cos(2θ(t))
I1 (Y ) =
θ
(t)
sin(2θ(t))
dt
=
1
·
−
.
3
2
(λ − q(Y )) 2 ξ
t=ξ
So |I1 (y)| ≤ 1. And we also have
∫ Y
q ′ (t)
0 ≤ I2 (Y ) ≤
q ′ (t)
5 dt
(λ − q(t)) 2
0
Since
′
q (Y )
3
(λ−q(Y )) 2
= 1,
′
q (0)
3
(λ−q(0)) 2
[
=
2
q ′ (t)
3 (λ − q(t)) 23
]t=Y
t=0
′′
q (t)
> 0 and q is convex so
2
3
−
3
(λ−q(t)) 2
∫
q ′′ (t)
Y
3
0
(λ − q(t)) 2
dt ≤
2
. (2.26)
3
> 0.
φ does not vanish identically on the interval (0, Y ), because the spectrum is discrete and hence [8,
section 5.4] the solution φ has a finite number of zeros in (0, ∞). Let m be the number of zeros
in the interval (0, Y ). Then from Lemma 2.38
∫
Y
mπ −
∫
√
λ − q(u) du ≤
0
0
Hence
∫
Y
mπ =
∫
′
Y
−q (u)
√
∫ Y
λ−q(u)
q ′ (u)
√
du + π = 2
du + π.
λ − q(u)
0 λ − q(u)
2
(∫
√
λ − q(u) du + O
0
0
Y
=
∫
0
∫
0
Y
Y
=
Y
=
q ′ (u)
du
λ − q(u)
)
+ O(1)
√
λ − q(u) du + O(log(λ − q(0)) − log(λ − q(Y ))) + O(1)
( (
))
√
λ − q(0)
λ − q(u) du + O log
+ O(1)
λ − q(Y )
√
λ − q(u) du + O(1).
(2.27)
0
The
function q is )strictly increasing, so λ − q(Y ) ≥ λ − q(u) ∀u ∈ (Y, X). The solutions of
(
d2
− du
f = 0 are
2 + q(Y ) − λ
√
√
f (u) = c cos( λ − q(Y )) + d sin( λ − q(Y )) (c, d ∈ R fixed).
√
λ−q(Y )
Those solutions have
(X − Y ) + O(1) zeros in (Y, X). The O(1) term comes from the
π
round off. Suppose φ has n − m zeros in (Y, X), then by the Sturm comparison Theorem 2.31
√
(2.28)
(n − m)π ≤ λ − q(Y )(X − Y ) + O(1).
Since λ = q(X), we have
∫
X
X −Y =
Y
q ′ (t)
1
dt ≤ ′
′
q (t)
q (Y )
∫
X
q ′ (t) dt =
Y
λ − q(Y )
.
q ′ (Y )
(2.29)
Combining inequalities (2.28), (2.29) and the choice of Y gives
3
(n − m)π ≤
(λ − q(Y )) 2
+ O(1) = 1 + O(1) = O(1).
q ′ (Y )
Furthermore, using (2.29) and again the choice of Y
∫ X√
3
√
(λ − q(Y )) 2
λ − q(t) dt ≤ λ − q(Y )(X − Y ) ≤
= O(1).
q ′ (Y )
Y
23
(2.30)
(2.31)
2
THE SCHRÖDINGER OPERATOR WITH MORSE POTENTIAL
Combining (2.27), (2.30) and (2.31) gives
1
n = m + (n − m) =
π
∫
Y
0
∫
√
1 X√
λ − q(t) dt + O(1) =
λ − q(t) dt + O(1),
π 0
as desired.
The Morse potential Vk is for k < 0 not convex on the whole real line. Proposition 2.39 is in that
case not sufficient to prove the asymptotic distribution if u0 is too small. But it is possible to
adapt the proof in a way that it is sufficient for the potential to be convex on [X, ∞) for certain
X ∈ R. See the Theorem 2.40 for that result.
Theorem 2.40 Let q be a real-valued C 2 -potential on the interval [u0 , ∞). Suppose there exists
u1 ∈ [u0 , ∞) such that u is increasing and convex on [u1 , ∞). Then the Schrödinger operator
d2
L := − du
2 + q(u) on the interval [u0 , ∞) with boundary condition φ(u0 ) = 0, has a pure discrete
and simple spectrum. Furthermore there is a constant T1 such that for all T ≥ T1 there exists a
unique solution uT ≥ u0 of q(uT ) = T , and the equality
∫
1 uT √
N (T ; α, u0 ) =
T − q(u) du + O(1),
(2.32)
π u0
holds for T1 ≤ T < ∞.
Proof.
As in Proposition 2.39 we can conclude that the spectrum is discrete and simple in
(−∞, ∞).
Let u1 ∈ [u0 , ∞) such that q is increasing and convex on [u1 , ∞). If u1 = u0 we can directly apply
Proposition 2.39. Hence suppose u1 > u0 . Set T1 := max{q(u) : u0 ≤ u ≤ u1 }. Because q is
continuous and convex there exists for λ > T1 a unique uλ ∈ [u0 , ∞) such that q(uλ ) = λ. Hence
{u ≥ u0 : q(u) ≤ λ} = [u0 , uλ ]. Let λ > T1 . We can apply the proof of Proposition 2.39 until the
q ′ (Y )
choice of Y . But there exists a unique Y ∈ [u1 , uλ ] such that
3 = 1. The estimation of
(λ−q(Y )) 2
I1 (Y ) is the same, but the estimation of I2 (Y ) is no longer valid. However
∫
u1
I2 (Y ) =
≤
u0
∫ u1
q ′2 (t)
q ′2 (t)
u0
sin2 (2θ(t))
(λ − q(t))
5
2
sin2 (2θ(t))
(λ − q(t))
5
2
∫
Y
dt +
u1
dt +
q ′2 (t)
sin2 (2θ(t))
5
(λ − q(t)) 2
dt
2
= O(1).
3
The last integral is finite because q is C 2 [u0 , ∞), hence q ′ is bounded on [u0 , u1 ] and u1 < T . The
2
3 comes from (2.26). And the rest of the proof goes the same as the proof of Proposition 2.39. Theorem 2.41 For the Schrödinger operator with Morse potential Vk (u) on the interval [u0 , ∞)
with boundary condition φ(u0 ) = 0, on the domain C 2 [u0 , ∞) ∩ L2 [u0 , ∞), the eigenvalue density
satisfies
N (T ; α, u0 ) =
√
√
1√
1
T log( T ) + (2 log(2) − 1 − u0 ) T + O(1)
π
π
as T → ∞.
The O(1) depends both on k and u0 .
Remark 2.42 This estimate gives the behaviour of the eigenvalues of the Schrödinger operator.
And because of Proposition 2.8 it also gives information of the Whittaker differential equation.
Observe that leading term is independent of u0 , k and α. In the second term only u0 appears.
The dependence of k is just in the factor O(1).
Using this method it is not possible to give a further expansion of the eigenvalue density, because
in lemma 2.38 we already have a factor π which comes back in the O(1) term. And that one still
needs to be taken into account.
24
2.3
The asymptotic behaviour of the eigenvalues
Proof. The Morse potential Vk is increasing and convex if k ≥ 0 or if u > log(−2k) for k < 0
(Lemma 2.5) so we can apply Proposition 2.40. Then we need to estimate the integral
1
N (T ; α, u0 ) =
π
∫
uT
√
u0
1
T − e2u − keu du + O(1).
4
(2.33)
First assume that u0 ≥ log(6|k| + 1). Then Vk is strictly increasing and convex on [u0 , ∞). The
case u0 < log(6|k| + 1) will be dealt later on.
Let T1 be such that ∀T ≥ T1 the equation T = Vk (u) has a unique solution. Let
√ T > T1 . Then
T = Vk (uT ) ≥ Vk (u0 ) = 41 (6|k| + 1)2 + k(6|k| + 1) > k 2 . So we can set T ∗ := T + k 2 and pick
the unique uT ∈ [u0 , ∞) such that Vk (uT ) = T .
)
)2
(
(
Then (T ∗ )2 − 21 eu + k = T + k 2 − 14 e2u + keu + k 2 = T − Vk (u). Hence we need to estimate
1
I :=
π
∫
uT
√
u0
1
1
T − e2u − keu du =
4
π
T∗
Apply the substitution w =
1 u
2 e +k
∫
uT
√
(
(T ∗ )2
−
u0
1 u
e +k
2
)2
du.
(2.34)
. The function u 7→ 21 eu + k is monotone increasing on [u0 , ∞),
so the substitution is single valued. And 12 eu0 + k > 0, because of the choice of u0 . Hence the
substitution is well defined. The integration boundaries and Jacobian become
T∗
=: w0
+k
T∗
T∗
T∗
=1
=√
=√
1 uT
+k
T + k2
Vk (uT ) + k 2
2e
1 u0
2e


∗
(
)
1 u
k 1 eTu +k
e
+
k
−
k
dw
T∗
1 u
k
2
= −(
= −w 1 − 1 u
= −w 1 − 2 ∗ 
)2 e = −w 1 u
1 u
du
T
e +k 2
2e + k
2e + k
2
= −w(1 −
kw
).
T∗
So (2.34) can be written as
∫
1
√
(T ∗ )2
(T ∗ )2 −
w2
I=−
w0
(
1
1 − kw
T∗
)
1
dw = T ∗
w
∫
w0
√
1
1
1− 2
w
(
1
1 − kw
T∗
)
1
dw.
w
(2.35)
Another substitution v = log(w) gives
∗
v0 : = log(w0 ) = log(T ) − log
)
(
1 u
∗
e + k ev .
T =
2
log(1) = 0
ev = w =
T∗
+k
1 u
2e
so
(
1 u0
e +k
2
)
dv
1
=
dw
w
(2.36)
Then the integral (2.35) transforms into
I=T
∗
∫
v0
√
(
1−
e−2v
0
1
v
1 − ke
T∗
)
dv.
If k = 0 then obviously |k|e
T ∗ = 0. Suppose that k ̸= 0. Using that u0 > log(6|k| + 1) and v > 0
gives
kev
|k|
|k|
1
|k|ev
|k|
|k|
0≤
= 1 u
≤
≤ .
=
< 1
<
∗
T
T∗
3|k|
+
k
+
1
2|k|
2
e
+
k
(6|k|
+
1)
+
k
2
2
v
25
2
So it is allowed to write
THE SCHRÖDINGER OPERATOR WITH MORSE POTENTIAL
1
kw
1− T
∗
as a geometric series
∑∞ ( kev )n
n=0
T∗
. Then I = M + R where (use
Lebesgue’s dominated convergence theorem on R)
∫ v0 √
M := T ∗
1 − e−2v dv
R := T ∗
0
∞ (
∑
j=1
k
T∗
)j ∫
v0
ejv
(2.37)
√
1 − e−2v dv.
(2.38)
0
√
Observe that − 1 − e−2v + 1 ≤ 21 e−2v+log(2) ∀v > 0. Because if we write x = e−2v then
√
√
1
e−2v = e−2v+log(2) ≥ − 1 − e−2v + 1
⇐⇒
−x + 1 ≤ 1 − x
⇐⇒
2
x2 − 2x + 1 ≤ 1 − x
⇐⇒
x2 − x ≤ 0
⇐⇒
(2.39)
0≤x≤1
⇐⇒
v≥0
√
And v0 ≥ 0 ⇔ w0 ≥ 1 ⇔ T ∗ ≥ 12 eu0 + k ⇔ T + k 2 ≥ 21 eu0 + k ⇔ T ≥ 14 e2u0 + keu0 which
√
holds. Now use the inequality − 1 − e−2v + 1 ≤ 12 e−2v+log(2) , substution v 7→ v − log(2) and the
definition of v0 (2.36) to obtain
[
]∞
∫ ∞√
∫ ∞
1 −2v
1 −2v
−2v
e
dv = − e
1−e
− 1 dv ≤
(2.40)
4
v0
v0 −log(2) 2
v=log(T ∗ )−log( 21 eu0 +k)−log(2)
)
(
1
.
=O
(T ∗ )2
Using the estimate (2.40) and the definition of v0 (2.36) it is possible to estimate M (2.37) by
∫ v0 √
M = T ∗ v0 + T ∗
1 − e−2v − 1 dv
0
(
(
))
∫ ∞√
1 u0
∗
∗
=T
log(T ) − log
e +k
+ T∗
1 − e−2v − 1 dv + O(T ∗ ).
(2.41)
2
0
√
The identity (2.42) can be obtained using the substitution y = 1 − e−2v .
∫ ∞
√
1 − 1 − e−2v dv = 1 − log(2),
(2.42)
0
Inserting (2.42) in (2.41) gives
1
M = T log(T ) + T (log(2) − 1 − log( eu0 + k)) + O
2
∗
∗
∗
(
1
T∗
)
.
( )
The term O T1∗ depends on u0 and k. This one can also be estimated by O(1).
Now split up R (2.38) in the two parts R1 and R2
)j ∫ v0
)j ∫ v0 √
∞ (
∞ (
∑
∑
k
k
jv
∗
∗
e dv
R2 := T
( 1 − e−2v − 1)ejv dv.
R1 := T
∗
∗
T
T
0
0
j=1
j=1
(2.43)
(2.44)
∑∞ j
Using the power series of log(1 − x) = − n=1 xj , the definition of v0 (2.36) and the identity
− log( xy ) = log(y) − log(x) we obtain the estimation


)j (
)
)j (
) ∑
)j
∞ (
∞ (
∞ (
∑
∑
1
k
1
k
1
k

(ejv0 − 1)
= T∗ 
ejv0 −
R1 = T ∗
∗
∗
∗
T
j
T
j
T
j
j=1
j=1
j=1
(
(
)
(
))
(
(
)
( ))
v0
ke
k
k
1
= T ∗ − log 1 − ∗ + log 1 − ∗
= T ∗ − log 1 − 1 u
+O
0
T
T
T∗
+k
2e
( (
)
(
))
1 u0
1 u0
= T ∗ log
e + k − log
e
+ O(1).
2
2
26
2.3
The asymptotic behaviour of the eigenvalues
The O( T1∗ ) comes from the first term of the power series of log(1 − Tk∗ ).
Using (2.39), the definition of v0 (2.36), we can estimate R2 (2.44) by
)j ∫ v0
∞ (
∑
|k|
∗
e(j−2)v dv
R2 ≤ T
∗
T
0
j=1
)
((
)j−2
( )2
)j
∞ (
∗
∑
|k|
|k|
1
T
|k|
= T ∗ ∗ (1 − e−v0 ) + T ∗
(v0 − 0) + T ∗
−1
1 u0
T
T∗
T∗
j−2
+k
2e
j=3
∑ ( |k| )j ( T ∗ )j−2
|k|2
≤ |k|(1 − e−v0 ) + ∗ v0 + T ∗
1 u0
T
T∗
+k
2e
j=3
)
(
j−2
k2 ∑
|k|
≤ O(1) + ∗
T j=3 12 eu0 + k
= O(1).
Combining the estimates of R1 and R2 gives
(
)
1 u0
∗
R = R1 + R2 = T log
e + k + T ∗ log(2) − T ∗ u0 + O(1).
2
(2.45)
Summing the estimates of M and R (2.41) and (2.45) gives
I = T ∗ log(T ∗ ) + T ∗ (2 log(2) − 1 − u0 ) + O(1).
(2.46)
∗
But this√asymptotic expansion (2.46) is given in terms of T instead of T . The Taylor expansion
of x 7→ T − x around x = 0 gives
(
)
√
√
1
1
√
T −x= T + √ x+O
.
(2.47)
2 T
T T
Hence as T → ∞ the estimate
(
)
√
1
T∗ = T + O √
(2.48)
T
holds. Inserting the expansion 2.48 of T ∗ in 2.46 we obtain the desired estimate
√
√
1√
1
T log( T ) + (2 log(2) − 1 − u0 ) T + O(1),
(2.49)
π
π
which holds as u0 ≥ log(6|k| + 1).
Now suppose u0 < log(6|k| + 1) =: u1 . The value T and hence uT are assumed to be sufficiently
large, in particular uT > u1 . Then we can split the integral (2.33) in the two parts
∫ u1 √
∫ uT √
T − Vk (u) du +
T − Vk (u) du + O(1).
(2.50)
u0
u1
Because u1 = log(6|k| + 1) we can use (2.49) to estimate the second part
∫ uT √
T − Vk (u) du, by
u1
√
√
1√
1
T log( T ) + (2 log(2) − 1 − u1 ) T + O(1).
(2.51)
π
π
The potential √
Vk is continuous and hence bounded in [u0 , u1 ]. Using again the Taylor expansion
(2.47) of x 7→ T − x we obtain
∫ u1 √
∫ u1 √
√
√
T − Vk (u) du =
T − O(1) du = (u1 − u0 ) T − O(1) = (u1 − u0 ) T + O(1). (2.52)
u0
u0
The constant in O(1) depends both on k and u0 . Using the estimations (2.50), (2.51) and (2.52)
we can estimate the integral (2.33) by
√
√
1
1√
T log( T ) + (2 log(2) − 1 − u0 ) T + O(1).
π
π
Which is the desired asymptotic behaviour of the eigenvalues.
27
28
3
3.1
The Hypergeometric Differential Equation
Properties and solutions of the hypergeometric differential equation
Definition 3.1 The hypergeometric differential equation is the differential equation
z(1 − z)f ′′ + (c − (a + b + 1)z)f ′ − ab f = 0.
(3.1)
Notation 3.2 By Ha,b,c denote the hypergeometric differential operator
Ha,b,c f :=
(
)
d2
d
t(1 − t) 2 + (c − (a + b + 1)t) − ab f.
dt
dt
(3.2)
Let u0 < 0, we consider Ha,b,c on the domain C 2 (−∞, u0 ] ∩ L2 (−∞, u0 ] and try to describe the
spectrum of this operator.
Proposition 3.3 The hypergeometric functions 2 F1 (a, b; c; z) and z 1−c 2 F1 (a − c + 1, b − c + 1; 2 −
c; z) (see Definition 2.10) span for c ∈
/ Z the two dimensional vector space of solutions of the
hypergeometric differential equation (3.1).
Proof. First calculate the derivatives of 2 F1 (a, b; c; z).
∞
∑
(a)n (b)n n
z
2 F1 (a, b; c; z) =
(c)n n!
n=0
′
2 F1 (a, b; c; z)
=
∞
∑
(a)n (b)n n−1
nz
(c)n n!
n=1
=
∞
∑
(a + n)(b + n) (a)n (b)n n
z
c+n
(c)n n!
n=0
′′
2 F1 (a, b; c; z)
=
∞
∑
(a)n (b)n
n(n − 1)z n−2
(c)
n!
n
n=2
=
∞
∑
(a + n)(b + n) (a)n (b)n n
n
z
(c + n)
(c)n n!
n=1
We can write z(z − 1) 2 F1′′ + (c − (a + b + 1)z) 2 F1′ − ab 2 F1 as a power series. Then it is sufficient
to prove that the coefficient of z n is 0 for every n. So let n ∈ N, then by using the right forms of
the derivative of 2 F1 , coefficient of z n becomes
−n(n + 1) +
(a + n)(b + n)
(a + n)(b + n)
n+c
− n(a + b + 1) + ab.
c+n
c+n
(3.3)
Multiplying by (c + n) gives
n(1 − n)(c + n) + n(a + n)(b + n) + c(a + n)(b + n) − n(a + b + 1)(c + n) − ab(c + n)
= (cn − cn2 + n2 − n3 ) + (abn + bn2 + an2 + n3 ) + (abc + bcn + acn + cn2 )
− (acn + an2 + bcn + bn2 + cn + n2 ) − (abc + abn)
= 0.
Because c ∈
/ Z the equation (3.3) must be zero. Hence 2 F1 (a, b; c; z) is a solution of (3.1).
For the function z 1−c 2 F1 (a − c + 1, b − c + 1; 2 − c; z) it is possible to do a similar calculation.
That will be left as an exercise for the reader.
In general the hypergeometric functions 2 F1 (a, b; c; z) and z 1−c 2 F1 (a − c + 1, b − c + 1; 2 − c; z)
are independent [7, section 5.3]. This can be seen by considering the behaviour as z → 0 and by
considering an analogue of the Wronskian. So 2 F1 (a, b; c; z) and z 1−c 2 F1 (a−c+1, b−c+1; 2−c; z)
span the solution space of the hypergeometric differential equation.
Remark 3.4 We see from Lemma 2.11 that the radius of convergence of 2 F1 (a, b; c; z) is 1. But
with analytic continuation we can extend 2 F1 outside the unit disc [7, section 5.2].
29
3 THE HYPERGEOMETRIC DIFFERENTIAL EQUATION
2
1
0
x
Figure 3: Plot of three different hypergeometric functions2 F1 (a, b; c; z).
Remark 3.5
In figure 3 is a plot of three different hypergeometric functions. The red line is a plot of
2 F1 (1, 1; 1; z), the green line of 2 F1 (1, 2; 1; z) and the yellow line of 2 F1 (1, 1; 2; z). Observe that
∑∞ (1)n (1)n n ∑∞ n
1
2 F1 (1, 1; 1; z) =
n=0 (1)n n! z =
n=0 z So the analytic continuation of 2 F1 (1, 1; 2; z) = 1−z .
.
And 2 F1 (1, 1; 2; z) has analytic continuation − log(1−z)
z
Lemma 3.6 Let
(
(α,β)
φλ
(t) :=
2 F1
)
1
1
(α + β + 1 − iλ), (α + β + 1 + iλ); α + 1; − sinh2 (t) ,
2
2
(α,β)
be a Jacobi function. Then φλ
(
(t) is a solution of
(
)
)
d2
1
d
2
2
+ (2α + 1)
+ (2β + 1) tanh(t)
+ (α + β + 1) − λ ψ = 0.
dt2
tanh(t)
dt
Proof. Fix α, β and λ in C. For the ease of notion denote
(
f (s) := 2 F1
)
1
1
(α + β + 1 − iλ), (α + β + 1 + iλ); α + 1; s .
2
2
(α,β)
Then φλ (t) = f (− sinh2 (t)). For the readability we will, for the rest of the proof, omit the
(α,β)
indices λ . Then
dφ
(t) = −2 sinh(t) cosh(t)f ′ (− sinh2 (t))
dt
(
)
d2 φ
(t) = −2 cosh2 (t) + sinh2 (t) f ′ (− sinh2 (t)) + 4 sinh2 (t) cosh2 (t)f ′′ (− sinh2 (t)).
2
dt
So
1
dφ
(t)
2 sinh(t) cosh(t) dt
)
1 d2 φ
1(
− sinh2 (t) cosh2 (t)f ′′ (− sinh2 (t)) = −
(t) −
cosh2 (t) + sinh2 (t) f ′ (− sinh2 (t)).
2
4 dt
2
f ′ (− sinh2 (t)) = −
30
(3.4)
(3.5)
3.1
Properties and solutions of the hypergeometric differential equation
The function f is defined as a hypergeometric function, so according to Proposition 3.3 f satisfies
the equation
(
)
0 = − sinh2 (t) cosh2 (t)f ′′ (− sinh2 (t)) + (α + 1) + (α + β + 2) sinh2 (t) f ′ (− sinh2 (t))
(
)
1
1
−
(α + β + 1)2 + λ2 f (− sinh2 (t))
4
4
(
2
) ′
(
)
1d φ
1(
2
2
2
=−
cosh
(t)
+
sinh
(t)
f
(−
sinh
(t))
+
(α + 1) cosh2 (t) − sinh2 (t)
(t)
−
4 dt2
2
(
)
)
1
1 2
2
2
′
2
+ (α + β + 2) sinh (t) f (− sinh (t)) −
(α + β + 1) + λ φ(t).
(3.6)
4
4
Equality (3.6) follows by inserting (3.4) and from the identity cosh2 (t) − sinh2 (t) = 1 ∀t ∈ R.
Multiplying (3.6) by −4 and taking some coefficients together gives
)
d2 φ (
+ (2 − 4α − 4) cosh2 (t) + (2 + 4α + 4 − 4α − 4β − 8) sinh2 (t) f ′ (− sinh2 (t))
dt2 (
)
+ (α + β + 1)2 + λ2 φ(t)
)
(
)
d2 φ (
= 2 − (4α + 2) cosh2 (t) + 4(β + 2) sinh2 (t) f ′ (− sinh2 (t)) + (α + β + 1)2 + λ2 φ(t)
dt
(
)
)
d2 φ
1
dφ (
= 2 + (2α + 1)
+ (2β + 1) tanh(t)
+ (α + β + 1)2 + λ2 φ(t).
(3.7)
dt
tanh(t)
dt
0=
Equality (3.7) followed by inserting (3.5).
In the previous section we developed a theory for Schrödinger operators. The hypergeometric
differential equation however is not a Schrödinger operator, because of the first order derivative.
In the Lemma 3.8 this first order derivative will be removed by a conjugation of the operator. We
conjugate with γ. We choose this function γ such that
(
γ −1 (t) (2α + 1)
)
1
d
+ (2β + 1) tanh(t)
γ(t) = 0.
tanh(t)
dt
This γ is unique up to a constant.
Notation 3.7 By ∆α,β (t) we denote the weight function:
(2 sinh(t))2α+1 (2 cosh(t))2β+1 ,
Lemma 3.8 Let Mα,β :=
∆′α,β (t) d
d2
dx2 + ∆α,β (t) dt
and γα,β (t) := √
t > 0.
1
.
∆α,β (t)
−1
Define Kα,β := γα,β
Mα,β γα,β .
Then:
(
Kα,β f =
(
)
(
)
)
d2
1
1
1
2
2
2
+
−α
+
+
−β
+
tanh
(t)
−
(2αβ
+
2α
+
2β
+
1)
f
dt2
4 tanh2 (t)
4
Proof. For the readability of this proof, we will omit the subscripts
α,β .
2 · − 12 ∆ 2 ∆′
∆′
∆′
∆′
2γ ′
∆′
+
=
=
−
+
= 0.
+
1
γ
∆
∆
∆
∆
∆− 2
3
31
First we calculate
3 THE HYPERGEOMETRIC DIFFERENTIAL EQUATION
′
So: γ ′ = − 12 γ ∆
∆ . Using the chain rule and this identity gives
(
)
∆′ ′
∆′ ′
Kf = γ −1 γ ′′ f + 2γ ′ f ′ + γf ′′ +
γf+
γf
∆
∆
( ′
)
(
)
′
′
2γ
∆
∆ ′
= f ′′ +
+
f ′ + γ −1 γ ′′ +
γ f
γ
∆
∆
((
( ′ )2 )
)
′ ′
1 ∆
1
∆
′′
′
−1
− γ
=f +0·f +γ
− γ
f
2 ∆
2
∆
(
( )2
(
)
( ′ )2 )
1 ∆′
1 ∆′
1 ∆′
1
1 ∆′′ ∆
∆
′′
−1
γ+
γ−
− γ
− γ
f
=f +γ
−
2 ∆2
2 ∆
2∆
2 ∆
2
∆
(
( )2 )
1 ∆′′
1 ∆′
′′
=f + −
+
f.
2 ∆
4 ∆
Further calculations show
∆ = (et − e−t )2α+1 (et + e−t )2β+1
∆′ = (2α + 1)(et − e−t )2α (et + e−t )2β+2 + (2β + 1)(et − e−t )2α+2 (et + e−t )2β
∆′′ = (2α)(2α + 1)(et − e−t )2α−1 (et + e−t )2β+3 + (2α + 1)(2β + 2)(et − e−t )2α+1 (et + e−t )2β+1
+ (2α + 2)(2β + 1)(et − e−t )2α+1 (et + e−t )2β+1 + (2β)(2β + 1)(et − e−t )2α+3 (et + e−t )2β−1 .
Then we obtain
∆′′
1
= (4α2 + 2α)
+ (4β 2 + 2β) tanh2 (t) + 8αβ + 6α + 6β + 4
∆
tanh2 (t)
∆′
1
= (2α + 1)
+ (2β + 1) tanh(t)
∆
tanh(t)
( ′ )2
1
∆
= (4α2 + 4α + 1)
+ (4β 2 + 4β + 1) tanh2 (t) + (8αβ + 4α + 4β + 1).
∆
tanh2 (t)
Observe that
Mα,β =
d2
1
+ (2α + 1)
+ (2β + 1) tanh(t).
dt2
tanh(t)
This explains why we are interested in the operator M .
And then the potential becomes
( )2
1 ∆′′
1 ∆′
1
−
+ (−2β 2 − β) tanh2 (t) − (4αβ + 3α + 3β + 2)
+
= (−2α2 − α)
2 ∆
4 ∆
tanh2 (t)
)
)
(
(
1
1
1
2
tanh2 (t) + (2αβ + α + β + 1)
+ α2 + α +
+
β
+
β
+
4 tanh2 (t)
4
(
)
(
)
1
1
1
2
= −α2 +
+
−β
+
tanh2 (t) − (2αβ + 2α + 2β + 1),
4 tanh2 (t)
4
as desired.
Remark 3.9 The applied transformations of the hypergeometric differential equation are unitary transformations combined with a translation of weighted function spaces as Hilbert Spaces,
namely (L2 , dx) and (L2 , ∆(x)dx). The spectrum does not change under a unitary transformation.
Because let T be an operator with eigenvalue λ and eigenfunction φ then T φ = λφ. Let S be a
unitary transformation, then we obtain
S −1 T Sφ = ψ ⇐⇒ T (Sφ) = Sψ.
32
3.2
The discrete spectrum
Thus T (Sφ) = λ(Sφ). So the transformations only cause a translation of the spectrum. In the
last proposition the spectrum of the hypergeometric differential equation is calculated by inverting
the translations.
Because of Lemma 3.8 we make the following definition.
Definition 3.10 Let α, β ∈ R, then we define:
(
)
(
)
1
1
1
1
2
2
Vα,β (t) := α −
+ β −
tanh2 (t) − (α2 + β 2 − )
4 tanh2 (t)
4
2
(
)(
) (
)
1
1
1
2
2
= α −
−1 + β −
(tanh2 (t) − 1)
4
4
tanh2 (t)
d2
Lα,β := − 2 + Vα,β
dt
Remark 3.11
(3.8)
(3.9)
25
20
15
10
5
0
1
2
3
4
5
x
6
7
8
9
10
1
Figure 4: Plot µ tanh(t)
+ λ tanh(t)
1
Figure 4 is a plot of two translated potentials of the operator Lα,β , namely the function µ tanh(t)
+
λ tanh(t). The red line is the case µ = λ = 1, the green line in the case µ = 1, λ = −5.
1
Remark 3.12 It is obvious that tanh and tanh
are continuous functions on the open interval
x
−x
(0, ∞). So Vα,β also is a continuous function on this interval. Recall that tanh(t) = eex −e
+e−x . Then
limt→∞ tanh(t) = limt→∞
et −e−t
et +e−t
= limt→∞
lim tanh2 (t) = 1
t→∞
Furthermore
3.2
1
tanh(t)
lim
1−e−2t
1+e−2t
t→∞
= 1. Hence
1
(t) = 1
tanh2
lim Vα,β (t) = 0.
t→∞
has a singularity at x = 0.
The discrete spectrum
In this subsection we calculate the spectrum of the hypergeometric differential equation on the
interval (−∞, u0 ] for a u0 < 0. Because of the results in the previous subsection we consider the
operator Lα,β of Definition 3.10 on the interval [w0 , ∞).
Notation 3.13 The Jacobi transformed hypergeometric differential equation Let w0 > 0
Consider the problem
{
Lα,β ψ(t) = λψ(t)
∀t ∈ [w0 , ∞)
,
ψ(w0 ) = 0
33
3 THE HYPERGEOMETRIC DIFFERENTIAL EQUATION
on the domain L2 [w0 , ∞) ∩ C 2 [w0 , ∞).
Lemma 3.14 Vα,β is in limit point case at infinity.
Proof. From Remark 3.12 we see that limt→∞ Vα,β = 0. Because w0 > 0 the potential Vα,β is
continuous on the interval [w0 , ∞) and hence is bounded from below. Application of Corollary
2.25 shows that the potential is limit point case at infinity.
Theorem 3.15 Lα,β with domain C 2 [w0 , ∞)∩L2 [w0 , ∞) has a finite discrete spectrum in (−∞, 0)
and a continuous spectrum in [0, ∞).
Proof. Combining the fact that limt→∞ Vα,β (t) = 0 and Theorem 2.33 shows that the spectrum
of Lα,β is discrete in (−∞, 0). To show that the discrete spectrum in (−∞, 0) is finite and that the
spectrum is continuous in [0, ∞) we use [4, Lemma 3.2] respectively [4, Theorem 3.2]. So we need
to show that Vα,β ∈ L[w0 , ∞) respectively (1 + x2 ) Vα,β ∈ L[w0 , ∞). First consider the continuous
spectrum. Observe that
d
cosh2 (x) − sinh2 (x)
(x − tanh(x)) = 1 −
= tanh2 (x)
dx
cosh2 (x)
(
)
d
1
sinh2 (x) − cosh2 (x)
1
x−
=1−
=
.
dx
tanh(x)
sinh2 (x)
tanh2 (x)
Thus
∫ ∞
∫
w0
∞
u0
x=∞
tanh2 (x) − 1 dx = [x − tanh(x) − x]x=w0 = tanh(w0 ) − lim tanh(a) = tanh(w0 ) − 1
[
1
1
− 1 dx = x −
−x
2
tanh(x)
tanh (x)
Hence
∫
∞
w0
|Vα,β (x)| dx ≤ α2 −
a→∞
]x=∞
=
x=w0
1
1
1
− lim
=
− 1.
tanh(w0 ) a→∞ tanh(a)
tanh(w0 )
1
1
1
1
+ β 2 − tanh(w0 ) + α2 + β 2 + < ∞.
4 tanh(w0 )
4
2
(3.10)
Using the translation x 7→ x − w0 and application of [4, Theorem 3.2] shows that the interval
(0, ∞) is precisely the continuous spectrum of Lα,β .
To show that the discrete spectrum of Lα,β is finite in (−∞, 0), it is sufficient that
∫ ∞
|(1 + x2 )Vα,β (x)| dx < ∞.
w0
∫∞
∫∞
∫∞
Because 1 > 0 and x2 > 0 it follows that w0 |(1+x2 )Vα,β (x)| dx = w0 |Vα,β (x)| dx+ w0 x2 |Vα,β (x)| dx.
∫∞
The previous calculation showed that w0 |Vα,β (x)| dx < ∞. It is very difficult to calculate
∫∞ 2
x |Vα,β (x)| dx exactly, but it is possible to prove that the integral is finite. For this we
w0
consider the behaviour of x2 Vα,β (x) as x → ∞,
( 2x
)
)
)
(( x
e − 2 + e−2x − (e2x + 2 + e−2x )
e − x−x
2
−
1
=
x
x2 (tanh2 (x) − 1) = x2
ex + e−x
e2x + 2 + e−2x
x2
−4x2
∼ 2x
= x
−x
2
(e + e )
e +1
(
)
(( x
)
)
( 2x
)
−x
1
e +x
e + 2 + e−2x − (e2x − 2 + e−2x )
2
2
x2
(x)
−
1
=
x
−
1
=
x
ex − e−x
e2x − 2 + e−2x
tanh2
x2
4x2
∼ 2x
.
= x
−x
2
(e − e )
e −1
34
3.2
The discrete spectrum
( 1
)
Hence x2 (tanh2 (x) − 1) and x2 tanh
are integrable on the interval [w0 , ∞). So x2 Vα,β is
2 (x) − 1
integrable on [w0 , ∞). And then using inequality (3.10) it follows that
∫ ∞
(1 + x2 )Vα,β (x) dx < ∞.
w0
By substitution x 7→ x−w0 and application of [4, Lemma 3.2] we obtain the fact that the spectrum
of Lα,β is finite in (−∞, 0).
Now we know what the spectrum of the operator Lα,β is, but we started with the hypergeometric
differential equation. In Theorem 3.16 we see what the consequence of Theorem 3.15 is for the
hypergeometric differential equation.
Theorem 3.16 The hypergeometric differential equation has a finite discrete spectrum in the
interval(−3a2 + 4ab − 3b2 − 21 , ∞) and the spectrum is continuous in the interval(−∞, −3a2 +
4ab − 3b2 − 12 ].
Proof. According to Theorem 3.15, the operator Lα,β on the interval [w0 , ∞) has a finite discrete
spectrum in (−∞, 0) and a continuous spectrum in [0, ∞). We make a table to denote the operator,
interval of the domain and on which set the spectrum is discrete.
Operator
Lα,β
2
2
1
1
d2
2
2
− dt
2 + (α − 4 ) tanh +(β − 4 ) tanh
2
2
d2
1
1
2
2
dt2 + ( 4 − α ) tanh +( 4 − β ) tanh
Kα,β
Mα,β
1
d
d2
dt2 + ((2α + 1) tanh + (2β + 1) tanh) dt
2
2
+(α + β + 1) − λ
Ha,b,c
Interval
[w0 , ∞)
[w0 , ∞)
[w0 , ∞)
[w0 , ∞)
[w0 , ∞)
[w0 , ∞)
(−∞, v0 ]
Finite discrete spectrum
(−∞, 0)
(−∞, α2 + β 2 − 12 ]
(−α2 − β 2 + 12 , ∞)
(−α2 − β 2 − 2αβ − 2α − 2β − 21 , ∞)
(−α2 − β 2 − 2αβ − 2α − 2β − 21 , ∞)
(−α2 − β 2 − 2αβ − 2α − 2β − 21
−(α + β + 1)2 + λ2 , ∞)
(−α2 − β 2 − 2αβ − 2α − 2β − 21
−(α + β + 1)2 + λ2 , ∞)
The constant w0 is chosen such that v0 = − sinh2 (w0 ). Furthermore we have
(−α2 − β 2 − 2αβ − 2α − 2β −
1
1
− (α + β + 1)2 + λ2 = −2(α + β + 1)2 + λ2 − .
2
2
And for a, b, c ∈ C, we can choose α, β, λ ∈ C such that
1
(α + β + 1 − iλ)
2
1
b = (α + β + 1 + iλ)
2
c = α + 1.
a=
Namely set α = c − 1, β = a + b − c and λ = i(a − b). Then
−2(α + β + 1)2 + λ2 −
1
1
= −2(c − 1 + a + b − c + 1)2 − (a − b)2 −
2
2
1
= −3a2 + 4ab − 3b2 − .
2
Hence the hypergeometric differential operator (3.2) has a finite discrete spectrum in the interval
(−3a2 + 4ab − 3b2 − 12 , ∞) and a continuous spectrum in the interval (−∞, −3a2 + 4ab − 3b2 − 21 ].
Observe that this result is independent of c.
Remark 3.17 A final remark.
In section two we calculated the spectrum for the Schrödinger operator with Morse potential, see
35
3 THE HYPERGEOMETRIC DIFFERENTIAL EQUATION
Theorem 2.41. The Schrödinger operator with Morse potential is equivalent to the Whittaker
differential equation, see Proposition 2.8. And the Whittaker differential equation is the Liouville
transformation of the confluent hypergeometric differential equation (or Kummer’s equation) [7,
section 7.3].
tg ′′ (t) + (c − t)g ′ (t) − ag(t) = 0.
(3.11)
The confluent hypergeometric differential equation can be obtained from the hypergeometric
differential equation by a limit process. Namely set g(t) = f ( bt ). Then g ′ (t) = 1b f ′ ( bt ) and
g ′′ (t) = b12 f ′′ ( bt ). Let f be a solution of (3.1) then
( (
)) ( ) (
) ( )
( )
t
t
t
t
t
t
0=
1−
f ′′
+ c − (a + b + 1)
f′
− ab f
b
b
b
b
b
b
= t(b − t)g ′′ (t) + (bc − (a + b + 1)t)g ′ (t) − ab g(t).
Dividing by b gives
(
)
) )
(
(
t
a+1
′′
+ 1 t g ′ (t) − ag(t).
0=t 1−
g (t) + c −
b
b
By taking limb→∞ we obtain the confluent hypergeometric differential equation (3.11). We have
also calculated the spectrum of the spectrum of the hypergeometric differential equation (Theorem
3.16). So it might be interesting to investigate what happens to the spectrum during this limit
process.
36
References
[1] D.M. Bressoud, A radical approach to real analysis, The Mathematical Association of America, Washington, 2006.
[2] T.H. Koornwinder, Jacobi Functions and Analysis on Noncompact Semisimple Lie Groups,
Special Functions: Group Theoretical Aspects and Applications, 1-85, D. Reidel Publishing
Company, 1984.
[3] J.C. Lagarias, The Schrödinger operator with Morse potential on the right half-line, Communications in Number Theory and Physics, 3 (2009), No. 2, 323–361.
[4] B.M. Levitan and I.S. Sargsjan Introduction to Spectral Theory: Selfadjoint Ordinary Differential Operators, American Mathematical Society, Providence, 1975
[5] B. de Pagter and A.C.M. van Rooij, An Invitation to Functional Analysis, to appear.
[6] M. Reed and B. Simon Methods of Modern Mathematical Physics, II: Fourier Analysis, SelfAdjointness, Academic Press, New York, 1975.
[7] N.M. Temme, Special Functions, an Introduction to the Classical Functions of Mathematical
Physics, John Wiley & Sons, Inc., New York, 1996.
[8] E.C. Titchmarsh, Eigenfunction Expansions, Associated with Second-order Differential Equations, Part I, Clarendon Press, Oxford, 1962.
[9] E.C. Titchmarsh, Eigenfunction Expansions, Associated with Second-order Differential Equations, Part II, Clarendon Press, Oxford, 1958.
[10] E.T. Whittaker and G.N. Watson, A Course of Modern Analysis, Cambridge University Press,
Cambridge, 1963.
37