Quantum Mechanics and Number Theory: an Introduction ...jacobi.fis.ucm.es/metodos/Apuntes/libro piergiulio.pdf · Quantum Mechanics and Number Theory: an Introduction (preliminary

Quantum Mechanics and Number Theory: an

Introduction

(preliminary version)

Giuseppe MussardoInternational School for Advanced Studies (SISSA)

via Beirut 2-4, 34014 Trieste, Italyand

Piergiulio TempestaDepartamento de Fısica Teorica II,Metodos y modelos matematicos,

Facultad de Fısicas, Universidad Complutense,28040 – Madrid, Spain

June 3th, 2010

2

PREFACE

These Lecture notes are a first version of a book in progress on QuantumMechanics and Number Theory. This is a living book: the present versionis still preliminary and will be periodically updated.

The figures will be provided in a separate file.

Contents

I Quantum mechanics 7

1 Mathematical Introduction 91.1 Formulas for π . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.1.1 Leibnitz formula . . . . . . . . . . . . . . . . . . . . . 91.1.2 Viete’s formula . . . . . . . . . . . . . . . . . . . . . . 121.1.3 Other formulas for π . . . . . . . . . . . . . . . . . . . 13

1.2 The quantum propagator . . . . . . . . . . . . . . . . . . . . 151.3 spectrum depending on one quantum number . . . . . . . . . 221.4 The Gamma Function . . . . . . . . . . . . . . . . . . . . . . 241.5 Poisson summation formula . . . . . . . . . . . . . . . . . . . 291.6 Spectral Functions . . . . . . . . . . . . . . . . . . . . . . . . 33

2 WKB Approximation 372.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.2 The connection formulas . . . . . . . . . . . . . . . . . . . . . 392.3 Infinite well . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.4 Airy function . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.5 WKB for radial motion . . . . . . . . . . . . . . . . . . . . . 442.6 Square–well potential . . . . . . . . . . . . . . . . . . . . . . . 442.7 connection formulas . . . . . . . . . . . . . . . . . . . . . . . 472.8 Transmission through a barrier . . . . . . . . . . . . . . . . . 502.9 Metastable states . . . . . . . . . . . . . . . . . . . . . . . . . 502.10 Potential V (x) = λ |x|a . . . . . . . . . . . . . . . . . . . . . 51

2.10.1 WKB quantization of V (x) = λ |x|a . . . . . . . . . . 522.11 semiclassical estimate of 〈xα〉n . . . . . . . . . . . . . . . . . 532.12 Exact WKB quantization . . . . . . . . . . . . . . . . . . . . 56

2.12.1 Some ”numerology” . . . . . . . . . . . . . . . . . . . 592.13 Estimate of the ground state energy . . . . . . . . . . . . . . 602.14 Uncertainty relation and lower bound . . . . . . . . . . . . . 62

3

4 CONTENTS

3 TBA and differential equations 653.1 Spectral determinant . . . . . . . . . . . . . . . . . . . . . . . 653.2 Hadamard Factorization Theorem . . . . . . . . . . . . . . . 723.3 Bethe Ansatz Equations . . . . . . . . . . . . . . . . . . . . . 74

II Analytic Number Theory: an Introduction 77

4 Prime numbers 794.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4.2 Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 814.3 Prime Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 824.4 Measure of Complexity . . . . . . . . . . . . . . . . . . . . . . 834.5 A Primality Test in Quantum Mechanics . . . . . . . . . . . . 844.6 Factorization Problem in Quantum Mechanics . . . . . . . . . 844.7 Elementary facts about prime numbers . . . . . . . . . . . . . 864.8 Fermat’s little theorem . . . . . . . . . . . . . . . . . . . . . . 864.9 Prime numbers and polynomials . . . . . . . . . . . . . . . . 884.10 Distribution of prime numbers . . . . . . . . . . . . . . . . . 894.11 Probabilistic methods . . . . . . . . . . . . . . . . . . . . . . 92

4.11.1 Coprime probability . . . . . . . . . . . . . . . . . . . 924.11.2 Square–free probability . . . . . . . . . . . . . . . . . 924.11.3 Merten’s formula . . . . . . . . . . . . . . . . . . . . . 93

4.12 Mersenne numbers . . . . . . . . . . . . . . . . . . . . . . . . 934.13 Arithmetical functions . . . . . . . . . . . . . . . . . . . . . . 954.14 Dirichlet characters . . . . . . . . . . . . . . . . . . . . . . . . 95

5 The Riemann ζ function 975.1 Relation between ζ (s) and

∏(x) . . . . . . . . . . . . . . . . 98

5.2 Moebius function . . . . . . . . . . . . . . . . . . . . . . . . . 995.3 The Mangoldt function Λ (n) . . . . . . . . . . . . . . . . . . 1005.4 Analyticity properties of ζ (s) . . . . . . . . . . . . . . . . . . 101

5.4.1 Consequences of the functional equation . . . . . . . . 1045.5 Some consequences of Euler’s formula . . . . . . . . . . . . . 1055.6 Dirichlet series . . . . . . . . . . . . . . . . . . . . . . . . . . 107

6 Numerical Spectral Methods 1096.1 Basic formulas for the quantum harmonic oscillator . . . . . . 1096.2 Truncated Hilbert space method . . . . . . . . . . . . . . . . 1116.3 Tight–Binding Method . . . . . . . . . . . . . . . . . . . . . . 114

CONTENTS 5

6.3.1 Particle in a box . . . . . . . . . . . . . . . . . . . . . 114

6 CONTENTS

Part I

Quantum mechanics

7

Chapter 1

Mathematical Introduction

In this first Lecture, we will revise some basic aspects of Elementary NumberTheory.

1.1 Formulas for π

1.1.1 Leibnitz formula

The following result, due to Leibnitz, holds:

π

4= 1− 1

3+

15− 1

7+ ... =

∞∑

n=0

(−1)n

2n + 1.

The proof is immediate, since

π

4= arctg1 =

∫ 1

0

dx

1 + x2=

∞∑

n=0

∫ 1

0(−1)n x2ndx =

∞∑

n=0

(−1)n

2n + 1.

The above formula is completely equivalent to the one of Lord Bronncker

4π

= 1 +12

2 + 32

2+ 52

2+ 72

...

.

This remarkable formula links π with the integers in a far more striking waythat does the decimal expansion of π, which appears to exhibit no regularityin the sequence of its digits.

9

10 CHAPTER 1. MATHEMATICAL INTRODUCTION

The equivalence between the two formulas is based on the fact that a con-vergent infinite series of the form

γ1 + γ1γ2 + γ1γ2γ3 + γ1γ2γ3γ4 + ...

is equivalent to the continued fraction

γ1

1− γ2

1+γ2− γ31+γ3−

.

Other remarkable continued fractions.

1 +1

1 + 11+ 1

1+...

=√

5 + 12

.

11 + e−2π

1+ e−4π

1+...

=

√5 +

√5

2−√

5 + 12

e

2π5 .

The Leibnitz formula is however slowly convergent. Better formulas are, forinstance

π2

6=

∞∑

n=1

1n2

,π4

90=

∞∑

n=1

1n4

.

As we will see, we also have the general formula

ξ (2k) =∞∑

n=1

1n2k

=(2π)2k

2 (2k!)(−1)k−1 B2k,

where B2k are the (even) Bernoulli numbers, defined as the coefficients ofthe expansion

x

ex − 1=

∞∑

n=0

Bnxn

n!.

They satisfy the recursive equation

Bn+1 =n+1∑

k=0

(n + 1k

)Bk,

which is one of the ways for calculating them. Their first values are

B0 = 1, B1 = −12, B2 =

16, B4 = − 1

30...

1.1. FORMULAS FOR π 11

with B2k+1 = 0 for k = 1, 2, 3, ...

ExerciseProve that

7π4

720=

∞∑

n=1

(−1)n

n4.

Another relevant formula, due to Wallis, is

π

4=

23· 43· 45· 65· 67· 87· 89· ...

It can be proved in different ways.1) The first proof is based on the infinite product expansion of sinx

sinx = x∞∏

k=1

[1−

( x

kπ

)2]

.

Substituting x = π2 , we have

2π

=1 · 32 · 2

3 · 54 · 4

5 · 76 · 6 ...

2) Another proof is based on the integral

I (n) =∫ 1

0

(1− x2

)ndx.

It is easy to see that it satisfies the recursive equation

I (n) =2n

2n + 1I (n− 1) .

Hence, if n is an integer, we have

I (1) =23I (0) =

23, I (2) =

45I (1) =

2 · 43 · 5 ,

...

I (n) =2 · 4 · 6 · ... · 2n

3 · 5 · 7 · ... · (2n + 1).

Consider now the sequence of I (n) for half integers

n =12,32,52, ...


where

I

(12

)=

∫ 1

0

√1− x2dx =

π

4,

...

I

(n− 1

2

)=

1 · 3 · 5... (2n− 1)2 · 4 · 6...2n

π

2.

But it is easy to prove that

I (n) ≤ I

(n− 1

2

)≤ I (n + 1) ,

i.e.

2 · 4 · 6 · ... · 2n

3 · 5 · 7 · ... · (2n + 1)≤ 1 · 3 · 5... (2n− 1)

2 · 4 · 6...2n

π

2≤ 2 · 4 · 6 · ... · (2n− 2)

3 · 5 · 7 · ... · (2n + 1),

so that one obtains Wallis’ result for n →∞.

1.1.2 Viete’s formula

Another beautiful formula is

2π

=√

22

√2 +

√2

2

√2 +

√2 +

√2

2...

This formula was found by Viete (2540–160..) at the end of the 16th century.It is the first numerical formula for π and the first one expressed as an infiniteproduct. The proof is based on the duplication formulas of trigonometricfunctions:

sin θ = 2 sinθ

2cos

θ

2= 4 sin

θ

4cos

θ

4cos

θ

2= ... = 2n sin

θ

2ncos

θ

2cos

θ

4... cos

θ

2n.

Sincelim

n→∞ 2n sinθ

2n= θ,

taking θ = π2 and using the formula cos x

2 =√

12 (1 + cosx), we arrive to

Viete’s formula.Another proof is based on the recursive formula

a2n =√

2−√

4− a2n,

which relates the length an of the side of a regular n–gon inscribed in theunit circle to the side a2n of the 2n–gon.

1.1. FORMULAS FOR π 13

1.1.3 Other formulas for π

In the formula

π =∞∑

n=0

[4

8n + 1− 2

8n + 4− 1

8n + 5− 1

8n + 6

](116

)n

the factor 116 appears. Since this turns out to be the anomalous dimension of

the spin–field of the Ising Model, on would speculate on the field theoreticalorigin of this number.The following surprising formula is due to Ramanujan:

1π

=√

89801

∞∑

n=0

(4n)! [1103 + 26390n](n!)4 (396)4n .

The fastest convergent formula is the one found by Gauss, and related to thearithmetic–geometric mean (AGM) of two numbers (a, b). This is definedas the common limit of the sequences

{an+1 = (an+bn) /2 a0 = abn+1 =

√anbn b0 = b

.

It is easy to see that {an} decreases whereas {bn} increases, as it comes fromthe inequalities

an+1 − bn+1 =12

(√an −

√bn

)2≥ 0,

an − an+1 =12

(an − bn) ≥ 0,

bn+1 − bn =√

bn

(√an −

√bn

)≥ 0.

Hence these sequences converge. From the first equality, we have in fact

an − bn ≤ 2−n,

i.e. the limit is reached exponentially fast. The AGM gives rise to thehighly efficient method to compute π. Jonathan and Peter Borwein havefound surprising formulas, like

π = 2[AGM

(√2, 1

)]2/

(1−

∑n

2nεn

), εn = a2

n − b2n.


There is also a formula which links π to the Fibonacci numbers, defined by

Fn+2 = Fn+1 + Fn, F1 = F2 = 1.

There is a close expression for these numbers given by the Binet formula:

Fn =1√5

[(√5 + 12

)n

−(

1−√52

)n],

simple to prove. In fact, from the ansatz Fn = xn , the recursive equationgives

x2 = x + 1 ⇒ x =1±√5

2,

henceFn = Axn

+ + Bxn−,

where the two constants A and B can be fixed by the initial conditions:A = −B = 1√

5.

The Fibonacci numbers are generated by(

1 11 0

)n

=(

Fn+1 Fn

Fn Fn−1

),

from which we inferFn−1Fn+1 − F 2

n = (−1)n .

Notice that the Fibonacci numbers admit a generating function, given by

x

1− x− x2=

∞∑

n−1

Fnxn

(Generating function of a partition

∞∏

n=1

11− qn

=∞∑

n=0

P (n) qn,

with P (n) ' 14n√

3eπ√

2n3 . Finally, the formula connecting Fn with π is

π

4=

∞∑

n=1

arctan1

F2n+1.

1.2. THE QUANTUM PROPAGATOR 15

1.2 The quantum propagator

A basic quantity in Quantum Mechanics is given by the quantum propagator.This is defined as the amplitude for a particle that is in a position q0 at thetime t0 to reach the point q at the time t:

K (q, t; q0, t0) = 〈qt| q0t0〉 .In the following, we will treat for simplicity the one dimensional case, butthe relevant formulas can be easily generalized to higher dimensions (as amatter of fact, they can be generalized to Quantum Field Theory as well).We will assume that at any given time, the coordinate operator q possessesa complete set of eigenvalues, so that

q (t) |q, t〉 = q |q, t〉 ,with the completeness condition

∫dq |q, t〉〈q, t| = 1.

There are two ways of expressing the quantum propagator. The first oneemploys wave functions, the second one the concept of “path integral”.First of all, the translation in time is dictated by the Hamiltonian, so that

K (q, t; q0, t0) = 〈q| e− i~H(t−t0) |q0〉 .

Inserting now a complete set {|En〉}of orthogonal eigenstates of H, we have

K (q, t; q0, t0) =∑n,m

〈q| En〉〈En| e−i~H(t−t0) |Em〉〈En| q0〉 =

∑n

ψn (q) ψ∗n (q0) e−i~En(t−t0),

where〈q| En〉 = ψn (q) , 〈En| q〉 = ψ∗n (q) .

where we have used the orthogonality of the eigenstates. The above is thefirst expression of the quantum propagator. To get the other one, let’sdivide initially the time interval T = (t− t0) in n small intervals (n →∞),and let’s insert (n− 1) times the completeness relation:

〈qt| q0t0〉 =∫

dq1...dqn 〈qt| qn−1tn−1〉〈qn−1tn−1| qn−2tn−2〉 ... 〈q1t1| q0t0〉 .(1.1)


Let us consider each term in this expression.

〈qktk| qk−1tk−1〉 = 〈qk| e−iδt~ H |qk−1〉 =

∫dpk−1 〈qk| pk−1〉〈pk−1| e−

iδt~ H |qk−1〉 , (1.2)

where we have introduced a completeness relation in term of the momentumeigenstates: ∫

dp |p〉〈p| = 1.

Suppose that the Hamiltonian is given by

H =p2

2m+ V (q) .

Using the Baker–Campbell–Hausdorff formula at the leading order in δt, wehave

〈pk−1| e−iδt~ H |qk−1〉 ' 〈pk−1| e−

iδt~

p2

2m e−iδt~ V (q) |qk−1〉 =

〈pk−1| qk−1〉 e− iδt~

[p2

2m+V (qk−1)

]

.

Since〈q| p〉 =

1√2π~

eipq/~.

For the expression (1.2) we have

〈qktk| qk−1tk−1〉 ' 12π~

∫dpk−1e

− iδt~

p2k−12m

+i(qk−qk−1)pk−1

~ e−iδt~ .

In the infinitesimal time δt, we may assume a constant velocity·qk−1, such

that qk − qk−1 ' ·qk−1δt, and we can perform the Gaussian integral. The

result is very simple:

〈qktk| qk−1tk−1〉 =√

m

2πi~δtexp

[δt

~L

(q,·q)]

,

where L(q,·q)

is the Lagrangian of the theory:

L =m·q2

2− V (q) .


Putting now this result into (1.1), we arrive to the interesting result for thequantum propagator:

k (q, t; q0, t0) =∫Dqe

i/~∫ t

t0L

(q,·q)dτ , (1.3)

where the measure is formally defined by

Dq = limn→∞

( mn

2πi~T

)n/2dq1...dqn−1.

The meaning of the above expression is obvious.. One can go from thepoint (q0, t0) to the point (q, t) by following all possible paths. These paths,however, are weighted with the oscillating phase of the action of these paths.In the semiclassical limit ~→ 0, the above integral can be evaluated in thesaddle point approximation, i.e. there exists one path which dominates theexpression. This is given by the condition

δ

∫ t

t0

L(q,·q)

dτ = 0,

i.e.

∫ t

t0

[∂L

∂qδq +

∂L

∂·q

δ·q

]dt =

∫ t

t0

[∂L

∂q− d

dt

∂L

∂·q

]δqdt + pδq |tt0 . (1.4)

For variational paths which leave the end points fixed, the last term in (17)vanishes. Therefore we see that in the semiclassical limit the Lagrangiansatisfies the classical equation of motion

∂L

∂q− d

dt

∂L

∂·q

= 0.

It is now immediate to see that K (q, t, q0, t0) is the Green function of thetime–dependent Schrodinger operator, i.e. the propagator of the theory. Infact, from the time evolution of the states

|ψ (t)〉 = e−i~H(t−t0) |ψ (t0)〉

we can “sandwich” it with 〈q|, obtaining

〈q| ψ (t)〉 = ψ (q, t) =∫

dq′ 〈qk| e−

i~H(t−t0)

∣∣∣q′⟩⟨

q′∣∣∣ ψ〉 =

∫dq

′K

(q, t, q

′0, t

′0

)ψ

(q′, t0

).


The function K (q, q0, t− t0) satisfies the composition law

(q, q

′, t− t

′)=

∫dq

′′K

(q, q

′′, t− t

′′)K

(q′′, q′, t′′ − t

′),

and the differential equation(−i~

∂

∂t+ H

)K

(q, q

′, t− t

′)= −i~δ

(q − q

′).

We want to consider now the density of states of the system. This is definedas

g (E) =∑

n

δ (E − En) gn,

where gn is the degeneracy of the energy levels. We suppose gn = 1. Theintegral of this expression is called the “number staircase function” N (E)and counts the number of levels (including their degeneracy) up to a givenenergy E

N (E) =∫ E

0g (E) dE.

The density of states enters several important formulas and, as we will seeimmediately, is related to the quantum propagator. First of all, a Laplaciantransform of g (E) yields the canonical partition function Z (β) :

Z (β) = Lβ [g (E)] =∫ ∞

0e−βEg (E) dE =

∑n

e−βEn .

In order to relate g (E) to the propagator, let us consider initially the Fouriertransform of K with respect to the time difference t ≡ t

′ − t:

G(q, q

′, E

)≡ − i

~

∫ ∞

0K

(q, q

′, t

)e−

i~Et = − i

~∑

n

ψ∗n(q′)

ψn (q)∫ ∞

0e

i~ (E−En)tdt.

Since the last integral oscillates and does not converge, we give a smallimaginary part to the energy: E → E + iε, so that

G(q, q

′;E

)=

∑n

ψ∗n(q′)

ψn (q)1

E − En + iε. (1.5)

This Green function satisfies

(E −H) G(q, q

′, E

)= δ

(q − q

′),


and for free theories it can be easily computed

D = 1 G0

(q, q

′, E

)= −

(2m

~2

)i

2kexp

(−ik

∣∣∣q − q′∣∣∣)

,

D = 2 G0

(q, q

′, E

)= −

(2m

~2

)i

4H+

0

(k

∣∣∣q − q′∣∣∣)

,

D = 3 G0

(q, q

′, E

)= −

(2m

~2

) exp[ik

∣∣∣q − q′∣∣∣]

4π |q − q′ | ,

where k =√

2mE. If we now take the trace of (26), i.e.

trG =∫

dqG(q, q

′, E

)=

∑n

1E −En + iε

≡ G (E)

by using the identity

1x + iε

= P

(1x

)− iεπδ (x) ,

we see that

g (E) =∑

n

δ (E − En) = − 1π

ImG (E + iε) = − 1π

∫ImG

(q, q

′, E + iε

)dq.

From the above relation between g (E) and the canonical partition func-tion, we also have that the density of states is given by the inverse Laplacetransform of the latter function

g (E) = L−1E [Z (β)] =

12πi

∫ ε+i∞

ε−i∞eβEZ (β) dβ,

where the integral is to be taken in the complex β–plane along a contourC which is parallel to the imaginary axis. The positive distance ε of thecontour from the imaginary axis has to be chosen large enough so that allpoles of the integrand lie to the left of the contour. If the integrand hasisolated poles we can easily evaluate this integral by the residue theorem.


A very instructive example.Let us consider the harmonic oscillator, with the spectrum given by

En = n (we will take care of the 12 term later). For such system the partition

function is easily done:

Z (β) =∞∑

n=0

e−βn =1

1− e−β=

eβ/2

2 sinhβ/2.

The poles of this function are all simple and all lying on the imaginary axisat the values

βk = 2πik, k = 0,±1,±2, ...

with residue (−1)k. By choosing the contour as in the figure

FIGURE

summing up all residues, and taking into account the extra factors exp (βk/2)as well as exp (Eβk) appearing in the inverse Laplacian transform, we have

g (E) =∑

n

δ (E −En) =+∞∑

k=−∞e2πikE .

As we will see, this is equivalent to the Poisson resummation formula.Since the spectrum is bounded by E = 0, only non–negative values of theenergy are relevant. The above formula can be written

g (E) =

{1 + 2

∞∑

k=1

cos (2πkE)

}.

Note that the constant term, which corresponds to the average level density,comes from the pole at β = 0, whereas the other poles along the imaginaryaxis combine pairwise to make up all Fourier components of the oscillatingpart.It is now trivial to restore the actual dependence En = ~ω

(n + 1

2

)of the

energy levels. In fact, we have

g (E) =1~ω

{1 + 2

∞∑

k=1

cos[2kπ

(E

~ω− 1

2

)]}=

1~ω

{1 + 2

∞∑

k=1

(−1)k cos(

2πkE

~ω

)}.


Comment.Notice that g (E) naturally splits into a smooth part g (E) and an oscillatorypart δg (E):

g (E) = g (E) + δg (E) .

The smooth part is, in the above example, just the constant 1/~ω, namelythere is one energy level per unit ~ω. The oscillating part contains co-sine functions with constant amplitudes; their arguments are multiples of2πE/~ω. This quantity is easily recognized as the classical action: in fact,taking the classical momentum p (x) =

√2m (E − V (x)) and integrating

between the two turning points

x0 = ±√

2E/m

ω,

we obtainS (E) =

∮pdx = 2

∫ x

−x0

p (x) dx =2πE

ω.

For the stair–case function the situation is as in the figure

FIGURE

The previous example is the simplest of a general theory. This states thatthe smooth part is simply given by the corresponding integral on the classicalphase space:

g (E) =1

(2π~)f

∫dqfdpfδ (E −H (q, p)) ,

for a system of f degrees of freedom. This gives reason of the rule that thenumber of eigenstates with energy lass than E equals the number of cells lessthan E. Much more interesting is the oscillatory part, which is controlledby the so–called Gutzwiller trace formula. It reads

δg (E) =∑

Γ∈{ppo}

∞∑

k=1

AΓk(E) cos

[k

~SΓ (E)− σΓk

π

4

],

where Γ counts classes of topologically distinct primitive periodic orbits(ppo) (with energy E as a parameter); k counts the repeated revolutionaround each primitive orbit which yields a series of harmonics; SΓ (E) =∮

pΓdqΓ is the classical action integral along the orbit Γ; and σΓkis the

”Maslov index” of the orbit.


Finally, the amplitudes AΓkdepend on energy, time period and stability of

the orbits and their nature, being isolated or non–isolated. In the lattercase, one must sum over all distinct families of degenerate orbits.The trace formula represents a Fourier decomposition of the oscillating partof the level density. It gives the basis for many interesting interpretations ofquantum phenomena in terms of classical orbits. The reason why classicalperiodic orbits are important can be understood by the path integral repre-sentation of the propagator. Suppose, in fact, that in the formula (1.3), weput q0 = q and integrate on q, in order to obtain the trace of the propagator

FIGURE

In this case the action is a function of the variable q. The paths whichdominate the path integral are those for which pin = pfin, as it can be seenfrom eq. (1.4). Those are just the periodic orbits

1.3 General spectrum depending on one quantumnumber

It is simple to derive in full generality the previous result. Let us assumethat the spectrum En is given by a function f (n) and each level has adegeneracy dn = D (n)

En = f (n) ; dn = D (n) , n = 0, 1, ..

We assume f (n) to be a monotonic function with differentiable inversef−1 (x) ≡ F (x) so that

n = F (En) .

Using the general properties of the δ..function, we have

δ (E −En) = δ (E − f (n)) = δ (n− F (E))∣∣F ′ (E)

∣∣ .

Further defining D (E) = D [F (E)], we have then

g (E) = D (E)∣∣F ′ (E)

∣∣∞∑

n=0

δ (n− F (E)) .

We can apply now the previous formula of the harmonic oscillator (i.e. Pois-son resummation formula) and obtain

g (E) = D (E)∣∣F ′ (E)

∣∣{

1 + 2∞∑

k=1

cos [2πkF (E)]

}.

1.3. SPECTRUM DEPENDING ON ONE QUANTUM NUMBER 23

This formula can be applied to any system where the values of En anddn are known explicitly. This is not restricted to one–dimensional problems:there exists in fact some high–dimensional potentials for which the spectrumcan be written analytically in terms of one single quantum number nwithknown degeneracy. For all these systems the average level density is

g (E) = D (E)∣∣F ′ (E)

∣∣ ,

whereas the oscillating part is

δg (E) = 2D (E)∣∣F ′ (E)

∣∣∞∑

k=1

cos [2πkF (E)] .

Within the periodic orbit theory, 2π~F (E) is the classical action integratedalong an elementary closed orbit. The number k corresponds to the numberof revolutions around the orbit. If several distinct orbits exist, they have tobe summed separately. The amplitudes of the oscillation depend, in thesecases, on the chosen orbit and are, in general, much more difficult to obtain.In fact the Bohr–Sommerfeld quantization rule reads

S (E) =∮

pdx = 2∫ x2

x1

p (x) dx = 2π~n

(the eventual extra constant is irrelevant for the subsequent considerations).Hence, since n = F (En), we see by construction that

F (E) =1

2π~S (E) .

Example 1. Particle in a box of length L.The exact spectrum is given in this case by

En = E0 (n + 1)2 , E0 =~2π2

2mL2, (n = 0, 1...)

In the above notation F (E) =√

EE0− 1, and therefore

g (E) =1

2√

E0E

{1 + 2

∞∑

k=1

cos

[2πk

√E

E0

]}.

Example 2. Spherical harmonic oscillator in D–dimensions.For this system the energy levels are given by one quantum number n witha degeneracy dn which is a polynomial of order D − 1


(ExampleD = 2 dn = n + 1

D = 3 dn =12

(n + 1) (n + 2) .

Through the previous formulas we easily obtain

g (E) =E

(~ω)2

{1 + 2

∞∑

k=1

cos(

2πkE

~ω

)},

for D = 2, and

g (E) =1

2 (~ω)3

[E2 − 1

4(~ω)2

]{1 + 2

∞∑

k=1

(−1)k cos(

2πkE

~ω

)}

for D = 3.

1.4 The Gamma Function

In this Section, some of the basic properties of the Gamma Function will bereviewed, due to its relevance in the subsequent considerations.

Γ (z) is an analytic function of the complex variable z, defined, forRe(z) ≥ 0, by the following integral

Γ (z) =∫ ∞

0dte−ttz−1dt. (1.6)

By making the change of variable t → t2, it can be equivalently expressedas

Γ (z) = 2∫ ∞

0dte−t2t2z−1dt.

When z = 12 ,the above is nothing but the gaussian integral. Therefore

Γ(

12

)=√

π.

The analytic continuation of Γ (z) can be achieved with many different tech-niques. First of all, it is easy to show that Γ (z) has simple poles at z =0,−1,−2, ...,−n. Let us write (1) as

Γ (z) =∫ 1

0dte−ttz−1 +

∫ ∞

1dte−ttz−1.

1.4. THE GAMMA FUNCTION 25

The second integral is convergent for all values of z in the complex plane.About the first one, for Re(z) ≥ 0,where it converges uniformly, we canexpand the exponential and integrate term by term so that

Γ (z) =∫ 1

0dttz−1

∞∑

k=0

(−1)k

k!tk + Γ (z) =

∞∑

k=0

(−1)k

k!

∫ 1

0dttz+k−1 + Γ (z)

=∞∑

k=0

(−1)k

k!1

(z + k)+ Γ (z) .

This expression clearly shows the poles at negative integer values of z.Nearby the poles we have

Γ (z) ' (−1)k

k! (z + k).

Another way of extending analytically its definition in the complex plane isby using its functional equation

Γ (z + 1) = zΓ (z) , (1.7)

which is immediately obtained by integrating by parts eq. (1). By virtue ofeq. (1.7), Γ (z) can be interpreted as a generalization of the factorial functionto complex numbers. Indeed, for integer positive values, we have

Γ (n) = (n− 1)!

Moreover, eq. (1.7) can be used to extend the definition of Γ (z) to othervalues of the complex plane. Since Γ (z) = Γ(z+1)

z , and Γ (z + 1) is definedfor Re(z) ≥ −1, we can use this relation to extend analytically the domainof Γ (z) in the strip −1 ≤ Re(z) ≤ 0. Repeating the same argument, we canextend Γ (z) to all values of the complex plane, except for z = 0,−1,−2.....The Gamma function admits other interesting representations.Infinite limit representation

Γ (z) = limn→∞

1 · 2 · 3 · .. · nz (z + 1) (z + 2) ... (z + n)

nz.

Infinite product representation

1Γ (z)

= zeγz∞∏

n=1

(1 +

z

n

)e−zn

.


To prove the first, let us introduce the function of two variables

F (z, u) =∫ n

0

(1− t

n

)n

tz−1dt.

Since

limn→∞

(1− t

n

)n

= e−t,

we havelim

n→∞F (z, u) = F (z,∞) =∫ ∞

0e−ttz−1dt = Γ (z) . (1.8)

We can evaluate F (z, u) in successive integrations by parts. Take v ≡ tn ,

hence

F (z, u) = nz

∫ 1

0(1− v)n vz−1dv.

Integrating by parts, we get

F (z, u)nz

= (1− v)n vz

z|10 +

n

z

∫ 1

0(1− v)n−1 vzdv.

The integrated part vanishes at both extremes. Repeating the procedure,we have

F (z, u) = nz n (n− 1) ....1z (z + 1) (z + 2) ... (z + n− 1)

∫ 1

0vz+n−1dv

=1 · 2 · 3 · .. · n

z (z + 1) (z + 2) ... (z + n)nz

and using eq. (1.8) we obtain Euler’s result.The Weierstrass result is particularly interesting for our aim: in fact, itresembles the spectral function for the Schrodinger problem! It can be easilyobtained from Euler’s limit form. Let us write it as

Γ (z) = limn→∞

1 · 2 · ... · nz(z + 1)...(z + n)

nz = limn→∞

1z

n∏

m=1

(1 +

z

m

)−1nz.

Inverting this expression and using the relation n−z = e−z ln n, we obtain

1Γ (z)

= z limn→∞ e−z ln n

n∏

m=1

(1 +

z

m

).

1.4. THE GAMMA FUNCTION 27

Multiplying and dividing by

exp[(

1 +12

+13

+ ... +1n

)z

]=

n∏

m=1

ezm

we obtain

1Γ (z)

= z limn→∞ exp

[(1 +

12

+13

+ ... +1n− lnn

)z

]· limn→∞

n∏

m=1

(1 +

z

m

)e−

zm .

Since

limn→∞

( ∞∑

k=1

1k− lnn

)= γ

we finally obtain the Weierstrass result

1Γ (z)

= zeγz∞∏

n=1

(1 +

z

n

)e−

zn .

The Weierstrass definition shows that Γ (z) has simple poles at z = 0,−1,−2, ...and that [Γ (z)]−1 has no poles in the finite complex plane.This formula gives us the opportunity to discuss in general the (infinite)product representation of entire functions. They generalize the concept ofpolynomial expressions

P (z) = anzn + .... + a0 = A (z − z1) (z − z2) ... (z − zn)

which can be always decomposed in terms of the roots in the complex plane.A function analytic for all finite z is called and entire function of z. Thelogarithmic derivative f ′/f is a meromorphic function with a pole expansion.If f (z) has simple zeroes at z = an, we have

f ′ (z)f (z)

=f ′ (0)f (0)

+∞∑

n=1

[1an

+1

zan

]

and integrating both sizes,

lnf ′ (z)f (z)

= zf ′ (0)f (0)

+∞∑

n=1

[ln

(1− z

an

)+

z

an

].

Therefore, we arrive to the Weierstrass infinite product representation forentire functions

f (z) = f (0) exp[zf ′ (0)f (0)

] ∞∏

n=1

(1− z

an

)e

zan .


The Gamma function satisfies a series of remarkable identities. One of theseis

Γ (z) Γ (1− z) =π

sinπz. (1.9)

To prove it, notice that, by virtue of the functional equation, the functionΦ (z) = Γ (z) Γ (1− z) is periodic with period equal to 2,

Φ (z + 2) = Φ (z)

Moreover, it has simple poles at all points z ∈ Z. Hence by multiplying itby

z∞∏

k=1

(1− z2

k2

)

we obtain a function without singularities in the complex plane, i.e. a con-stant equal to 1, as it can be checked by taking the limit z → 0. But

z

∞∏

k=1

(1− z2

k2

)=

1π

sinπz

and eq. (1.9) is proved.Another important result, due to Legendre, is the so called duplication for-mula, relevant in the theory of the Riemann zeta function.

Γ (z) Γ(

z +12

)=√

π21−2zΓ (2z) .

For future applications it is also useful to have the asymptotic expansion ofthe Gamma function. It can be obtained by a saddle point approximation.Performing the change of variable t = xz, we get

Γ (z + 1) = zz+1

∫ ∞

0dxez(log x−x).

The function ϕ (x) = lnx − x goes to −∞both at x → 0 and x → ∞. Itpresents a unique maximum at x = 1. For z → ∞, the above integral isdominated by the region nearby this maximum. Expanding ϕ (x) in seriesnear this point

ϕ (x) = −1− (x− 1)2

2+ ...

1.5. POISSON SUMMATION FORMULA 29

and substituting into (38) we have

Γ (z + 1) = zz+1e−z

∫ ∞

0e−z

(x−1)2

2 dx =√

2πzz+ 12 e−z.

HenceΓ (z + 1) =

√2πzz+ 1

2e−z

ln Γ (z + 1) =(

z +12

)ln z − z +

12

ln 2π + O

(1z

).

Let us mention also the integral representation∫

Ce−ηηzdη =

(e2πiz − 1

)Γ (z + 1)

where the contour is shown in the Figure

FIGURE

This is particularly useful when z is not an integer; in this case the originis a branch point. The above equation can be easily verified by deformingthe contour as in Figure

FIGURE

The integral from ∞ to the origin gives −Γ (z + 1), placing the phaseof the logarithmic branch at the origin. The integral out of ∞ then yieldse2πizΓ (z + 1), whereas the integral around the origin vanishes for z > −1.It is simple to extend the range to include all non–integral values of z : theintegral exists in fact for z < −1, as far as we stay far from the origin.Second, integrating by parts, one gets the usual functional equation

1.5 Poisson summation formula

Let us deal now with a very interesting application of Fourier transform:the Poisson summation formula. Consider the function

S (x) =+∞∑

n=−∞f (x + nT )


where f (x) is an arbitrary function. It is clear that S (x) is periodical withperiod T :

S (x + T ) =+∞∑

n=−∞f (x + (n + 1)T ) =

+∞∑n=−∞

f (x + nT ) = S (x) .

Therefore S (x) can be developed in Fourier series

S (x) =+∞∑

k=−∞Cke

2πikxT

where the coefficients are given by

Ck =1T

∫ T2

−T2

S (x) e−2πikx

T dx.

By substituting the expression for S (x) we have

Ck =1T

∫ T2

−T2

+∞∑n=−∞

f (x + nT ) e−2πikx

T dx.

Making the change of variable y = x + nT , we get

Ck =1T

+∞∑n=−∞

∫ n+T2

n−T2

f (y) e−2πiky

T dy =1T

∫ +∞

−∞f (y) e−

2πikyT dy.

The last term is the Fourier transform of the function f (x) (let us denote itby f (ω)), and then

Ck =2π

Tf

(2πk

T

).

Finally, the function S (x) can be rewritten as

S (x) =+∞∑

n=−∞f (x + nT ) =

2π

T

+∞∑

k=−∞f

(2πk

T

)e

2πikxT .

This is the Poisson Summation Formula. Let us see some applications.

Example 1Compute

S0 =+∞∑

n=0

1a2 + n2

.

1.5. POISSON SUMMATION FORMULA 31

This series can be written as

S0 =1

2a2+

12

+∞∑n=−∞

1a2 + n2

.

Let us apply the PRF to the second term, where

f (v) =1

a2 + v2, T = 1, x = 0.

The Fourier transform of f (v) is

f (v) =12π

∫ +∞

−∞

1a2 + v2

e−ivνdv.

This integral can be computed using the residue theorem.a) v < 0. Close the contour as in figure..The integral on the semicircle vanishes when the radius goes to infinity, soit remains the contribution of the residue:

I = iRes[

1a2 + v2

e−ivν

]

v=ia

=12a

e−a|v|.

b) v > 0. we get again the same result. Consequently

S0 =1

2a2+

π

2a

+∞∑

k=−∞e−a|2πk| =

12a2

− π

2a+

π

a

11− e−2πa

.

Let us prove as an example the following identity∞∑

n=1

1n2

=π2

6.

Notice that

S0 =∞∑

n=0

1a2 + n2

=1a2

+∞∑

n=1

1a2 + n2

,

therefore ∞∑

n=1

1n2

= lima→0

[S0 − 1

a2

].

Expanding in series S0

S0 ' 12a2

− π

2a+

π

a

(1 + 2πa + (2πa)2

2 + ...)

2πa(1 + 2πa2! + (2πa)2

3! + ...)=


12a2

− π

2a+

12a2

(1 + 2πa +(2πa)2

2+ ...)(1− 2πa

2!− (2πa)2

3!+

(2πa

2!

)2

...) =

12a2

− π

2a+

12a2

+π

2a+

12

[π2 − 4π2

6

]=

π2

6+

1a2

,

hence the result. The Poisson resummation formula is clearly useful fromthe point of view of numerical calculus.

Example 2Consider the series

S (t) =+∞∑−∞

e−k2t.

Suppose that we want compute it at t = 0.01. In this case, to have aprecision of 10−10 we need ∼ 50 terms. Using Poisson formula we have

S (t) =√

π

t

+∞∑−∞

e−π2n2

t .

At t = 0.01, we get

S =√

π

0.01

(1 + 2e−100π2

+ ...)

= . . .

Exercise 1Show that

S1 =∞∑

n=0

1a2 + (2n + 1)2

=π

4a

1− e−πa

1 + e−πa.

Exercise 2Show that

S2 =∞∑

n=0

(−1)n

a2 + n2=

12a2

+π

a

e−πa

1 + e−2πa

Exercise 3Prove that ∞∑

n=0

(−1)n

n4= −7π4

720.

1.6. SPECTRAL FUNCTIONS 33

1.6 Spectral Functions

The previously introduced functions are some representatives of a large classof functions which deal with the spectrum of a system. They will be theobject of our future study. To fix the ideas, let us take a one–dimensionalsystem with Hamiltonian

H =p2

2m+ V (q) , [p, q] = i~ (1.10)

with the eigenvalues which goes like

En∼= nα (1.11)

(we will see later how to compute these quantities). Let us define the fol-lowing spectral functions

• Partition function

Z (t) = Tr exp(−tH) =∞∑

n=0

e−tEn , Ret ≥ 0

• Resolvent Trace

R (E) = Tr (H + E)−1 =∞∑

n=0

1E + En

, E ∈ C−{−�} (1.12)

• Fredholm determinant

∆ (E) = det(1−EH−1

)=

∞∏

n=0

(1− E

En

), E ∈ C. (1.13)

• Zeta function,

ξ (s) = TrH−s =∞∑

n=0

1Es

n

, Res >1a. (1.14)

• Mellin Transform of R (E)

M (s) =∫ ∞

0E−sR(E)dE (1.15)


Few comments(i) Note that we have defined the resolvent trace with (−E), comparing withthe previous formula.(ii) In the Fredholm determinant we assume the absence of Hadamard fac-tors, i.e. we assume ∑ 1

En< ∞.

This formula can be modified consequently if this condition of convergenceis not satisfied.It is easy to see that all these functions are related to each other, a resultwhich is intuitive, since all of these are based on the spectrum {En}. Wehave

R (E) =∫ ∞

0Z (t) e−tEdt

in the region where ReE > 0. Similarly

∆ (−E) =∞∏

n=0

(1 +

E

En

)= exp

[∫ E

0n (λ) dλ

], E ∈ C. (1.16)

Moreover, ∫ ∞

0Z (t) ts−1dt =

∞∑

n=0

∫ ∞

0e−tEnts

dt

t

making the change of variables tEn = xwe have∫ ∞

0Z (t) ts−1dt =

∑ 1Es

n

∫ ∞

0dxe−xxs−1 = Γ (s) ξ (s) .

Finally

ξ (s) =1π

sinπsM (s) , (1.17)

which can be proved as follows.

∑n

∫ ∞

0E−s 1

E + EndE =

∑n

∫ ∞

0dEe−s ln E 1

E + En.

In order to compute this integral, let us choose the contour as in the figure

FIGURE

1.6. SPECTRAL FUNCTIONS 35

In going to the lower branch II, the logarithmic function changes, sinceE → Ee2πi. Therefore, applying the theorem of residues to the above con-tour, with vanishing contribution from the external circle and the one nearthe origin, we have

I(1− e−2πis

)= 2πie−iπs

∑n

E−sn .

Hence we get formula (1.17).


Chapter 2

Semiclassical methods:WKBApproximation

2.1 Introduction

Consider the one dimensional Schrodinger equation

d2Ψdx2

+2m

~2[E − V (x)]Ψ = 0. (2.1)

If the potential doesn’t have a simple form, solving the Schrodinger equationis usually a complicated problem, which requires the use of approximate ornumerical methods, such as

1) Perturbation theory2) Variational methods3) Numerical diagonalization.

One particular method has found its greatest use in the case of one–dimensionalsystems. This is the so called WKB method (after Wentzel, Kramers andBrillouin).The basic idea is very simple and it is similar to the eikonal expansion inoptics.Notice that if V = const, eq. (2.1) has the solutions Ψ± = e±ikx. Thissuggests to look for a solution of the form

Ψ = eiS(x).

Substituting into eq. (2.1), we get

iS′′ − (S′)2 + [K (x)]2 = 0,

37

38 CHAPTER 2. WKB APPROXIMATION

where

K (x) =

√2m~ [E − V (x)], E > V (x)

−i√

2m~ [V (x)−E] E < V (x)

The above equation can be solved recursively, i.e. substituting the n–thapproximation, for the (n + a)–th approximation we have

Sn+1 = ±∫ x √

K2 (x) + iS′′n (x) + Cn+1,

henceS0 = ±

∫ x

K (x) + C0,

S1 = ±∫ x √

K2 (x)± iK ′ (x)dx + C1.

Obviously, to make sense of this procedure, the next approximation shouldbe close to the previous one, which for the above terms is equivalent to say

∣∣K ′ (x)∣∣ ¿ ∣∣K2 (x)

∣∣ (2.2)

In the above equations, both signs have to be chosen equal, since S1 is sup-posed to be an improvement of S0. Hence

S1 = ±∫ x

K (x)

√1± i

K ′

K2dx ' ±

∫ x [K (x)± i

K ′

K

]dx =

±∫ x

K (x) +i

2log K (x) + C.

Therefore, at this level of approximation, for the wave function we have

Ψ (x) ' A√K (x)

exp[±i

∫ x

K (t) dt

].

The condition (2.2) has a very physical meaning. We can define, in theregions where E > V (x), an effective wave length

λ (x) =2π

k (x)

and therefore eq. (2.2) can be written

λ (x)∣∣∣∣dp

dx

∣∣∣∣ ¿ |p (x)| ,

2.2. THE CONNECTION FORMULAS 39

where p (x) = ±~K (x) is the momentum of the particle. Hence, the rangeof the validity of the WKB approximation (at this order) is that the changeof the momentum over a wavelength must be small compared to the squareof its amplitude.

The method falls down if K (x) vanishes or if K (x) varies very rapidly.This certainly happens at the classical turning point for which

V (x) = E,

or whenever V (x) has a very steep behaviour. In both cases a more accuratesolution has to be found. Obviously, the WKB method would not be usefulunless we find a way to analyze such situations.

2.2 The connection formulas

Consider a potential as in the figure

FIGURE

Let us consider initially the case x À x2. In this region the classical mo-mentum is imaginary and therefore one of the solutions blows up. Hence wehave to consider only the exponentially decaying part, i.e.

ψ (x) =c√

|K (x)| exp[−

∫ x ∣∣K ′ (x)∣∣ dx

′]

. (2.3)

In the region x1 < x < x2, on the other hand, far away from the turningpoints, we have

ψ (x) =A√

K (x)exp

[i

∫ x

K(x′)

dx′]

+B√K (x)

exp[−i

∫ x

K(x′)

dx′]

(2.4)Let us consider now more closely the situation near one of the turning point,let’s say x = x2. Suppose that the potential is smooth and admits a seriesexpansion. Keeping only the leading term we have

d2Ψdx2

=2m

~2(x− x2) V

′(x2) Ψ.

By posing2m

~2V′(x2) ≡ α3, y = αx,


the above equation becomes

d2Ψdy2

= y Ψ.

This is the differential equation of the Airy function (see later). Disregardingonce again the exponential growing function, its solution is expressed by theso called Airy function Ai (y),defined by

Ai (y) =1√π

∫ ∞

0dv cos

(v3

3+ vy

).

For negative y, corresponding to the interior of the potential, it is oscillatory,whereas for positive y is exponentially decaying. Its asymptotic expansionsare

Ai (y) =

|y|− 1

4 sin(

23 |y|

32 + π

4

), y → −∞

12y−

14 exp

(−2

3y32

), y → −∞

.

We have to match this behaviour with our previous expressions. We have

B = −A = iCei π4 ,

and therefore for x1 < x < x2 we have now

Ψ =2C√K (x)

sin[i

∫ x2

xK

(x′)

dx′+

π

4

]. (2.5)

Inspecting the expressions (2.3) and (2.4), we see that a phase factor π4 has

been introduced in the wave function at the turning point of a smoothpotential.The WKB condition can be now obtained by demanding single–valuedness ofthe wave function. In fact, repeating now the analysis for the other turningpoint, we arrive to the expression

Ψ =2C√K (x)

sin[i

∫ x2

xK

(x′)

dx′+

π

4

]. (2.6)

The two phase factors in eqs. (2.5) and (2.6) should be the same up toa negative sign, which can be absorbed in the overall function. So, withC = (−1)nC, we have

∫ x

x1

K(x′)

dx′+

π

4= −

∫ x2

x1

K(x′)

dx′ − π

4+ (n− 1)π,

2.2. THE CONNECTION FORMULAS 41

i.e. ∫ x2

x1

K (x) dx =(

n +12

)π, n = 0, 1, 2, ...

By extracting the ~dependence, this becomes the familiar equation of quan-tization

2∫ x2

x1

p (x) dx =∮

p (x) dx =(

n +12

)h.

The left–hand side of this equation is equal to the area enclosed by the curverepresenting the motion in phase space and it is called the phase integral I.On the other hand, it also measure the phase change which the oscillatorywave function Ψ (x) undergoes between the turning points.

∫ x2

x1

K (x) dx =(

n +12

)h.

Dividing this expression by 2π, we see that according to the WKB approx-imation

n

2+

14

of quasi wavelengths fit between x1 and x2. Hence, n represents the numberof modes of the wave function, a fact which helps in visualizing the elusiveψ. According to eq. (23), the area of the phase space between one boundstate and the next one is equal to h. From this observation we see that itcomes the statement often used in statistical mechanics, that each quantumstate occupies a volume h in phase space.In order to find the normalization of the semi–classical wave function, weassume that the contribution beyond the turning points is negligible, so

A2u

∫ x2

x1

dx

p (x)sin2

(∫ x

x1

K(x′)

dx′+

π

4

)' A2

2

∫ x2

x1

dx

p (x),

hence

A2 =2∫ x2

x1

dxp(x)

.

With all the above quantities fixed, it is nice to check how the semi–classicalwave functions match the exact ones. This check can be done easily forthe harmonic oscillator. Obviously at the turning points the wave functionblows up and one need the connecting formulas.


2.3 Infinite well

The quantization condition changes if one of the walls of the potential is aninfinite step. In this case the wave function is zero at the boundary. Supposethen that we have two turning points, one x2 with a smooth behaviour andx1 with an infinite well. In this case, coming from right hand side of x2, wehave for the wave function

Ψ ∼ sin[∫ x2

xK (x) +

π

4

].

But then we have to impose

Ψ (xc) = 0.

Therefore we have the quantization condition∫ x2

x1

K (x) dx +π

4= (n + 1)π

and therefore, in this case∮

p (x) dx =(

n +34

)h.

Since π4 of the phase of the right hand side of the above equation may be

attributed as before to the turning point x2, the extra phase factor π/2 hasto be attributed to the step well.

2.4 Airy function

Consider the Schrodinger equation in a linear potential V = Ax,

d2Ψdx2

=2m

~2[Ax−E] Ψ.

It is obvious that if we shift the variable

x → x +E

A,

we can absorb the eigenvalue E, so if we find a solution for one value, wefind it for all other values. We write then

d2Ψdx2

=2m

~2AxΨ.

2.4. AIRY FUNCTION 43

Introduce the quantities

α3 =(

2mA

~2

),

y = αx.

The above equation becomes

d2Ψdy2

= yΨ.

The Fourier transform of Ψ is

Ψ (y) =∫ +∞

−∞dpΨ(p) eipy.

Then we have the differential equation

id

dpΨ(p) = p2Ψ,

whose solution is

Ψ (p) = exp[−i

p3

3

].

Hence

Ψ (y) =∫ +∞

−∞dp exp

[i

(py − p3

3

)]= 2

√πΦ(y) ,

with

Φ (y) =1√π

∫ ∞

0cos

(yp +

p3

3

)dp.

This function can be written in terms of Bessel functions

Φ (y) =

√y3πK 1

3

(23y

32

), y > 0

13

√πy

[J 1

3

(23y

32

)+ J− 1

3

(23y

32

)], y < 0

.

Hence, it goes exponentially to zero for y → +∞, whereas has an infinitenumber of zeros for y < 0. [Comments. Ideal spectral function! Up toHadamard factors,

∆ (E) =∏

i

(1 +

E

λi

),

with λi zeros of Airy function.


2.5 WKB for radial motion

The previous considerations can be easily extend to deal with quantummechanical systems in a central potential. Let us consider first a two–dimensional problem

− ~2

2m

[∂2

∂r2+

1r

∂

∂r+

1r2

∂2

∂ϕ2

]Ψ(r, ϕ) + V (r)Ψ (r, ϕ) = EΨ(r, ϕ) .

The radial and angular eqs. are separable in this case. Furthermore, bydividing the radial wave function by

√r, we obtain for it an eq. in the

radial variable which is similar to the one–dimensional case, i.e.

Ψ (r, ϕ) =∑

l

vl (r)√r

eilϕ

and

Q2l (r) =

2m

~2

[E − V (r)− ~2

2mr2

(l2 − 1

4

)],

where ~l is the eigenvalue of the angular momentum. The above formulasare indeed similar to the one dimensional case and we are tempted to writedown analogous WKB expressions, with

p (x)~

→ Ql (x) .

To check if this procedure gives the correct asymptotic phase in the wavefunction, let’s analyze a solvable case.

2.6 Square–well potential

Consider V (r) to be a square–well potential of depth V0 Let us also assumel 6= 0. The exact regular solution is Jl (kr), where Jl is the Bessel functionand the wave number

k2 =2m (E − V0)

~2

is a constant. For large r, such that

kr À |l| ,

Jl (kr) '√

2πkr

cos(

kr − lπ

2− π

4

).

2.6. SQUARE–WELL POTENTIAL 45

Let us construct the WKB wave function for this problem. To match theasymptotic form let’s do in Q2

l the replacement(l2 − 1

4

) → s2, where s2 isa parameter to be determined shortly.Then, using the previous formulas, the WKB formula for vl is

vl (r) ' 2√Ql (r)

cos[∫ r

r1

Qc

(r′

)dr

′ − π

4

]

where

Qc (r) ≡ 2m

~2

[E − V0 − ~2

2mr2s2

]

and r1 is the turning point near the origin determined by the condition

k2r21 − s2 = 0.

The integral in (10) can be done explicitly

I =∫ r

r1

[k2r

′2 − s2]1/2 dr

′

r′=

√(kr)2 − s2 − s arccos

( s

kr

).

In the asymptotic region kr À s, it is equal to

I ' kr − sπ

2.

Comparing now with (8), we see that

s2 = er,

in order to have a match of the phases. In turns out that this prescriptionis also valid in the more general case where V (r) varies with r, i.e. we haveto use the substitution

l2 − 14→ h2.

This condition can be justified by the request that the angular part of thewave function is semi–classical as well.We can now derive the WKB quantization condition for the bound states.Define initially the phases

φl (r1, r) ≡∫ r

r1

Ql

(r′)

dr′

φl (r1, r2) ≡∫ r2

r1

Ql

(r′)

dr′.


Since the WKB wave function at r may be written in terms of either one ofthem, it follows that

cos(φl (r1, r)− π

4

)= ± cos

(φl (r1, r2)− π

4

),

and therefore we arrive to the quantization condition

φl (r1, r2) =∫ r2

rQl (r) dr =

(nl +

12

)π.

Defining the action variable

Ir ≡ 12π

Sr =12π

∮prdr,

the quantization condition may be written in the generalized form

Ir =(nr +

µ

4

)~,

where µ is called the Maslov index. In the above example µ counts thenumber of classical turning points, where the amplitude of the wave functiondiverges.Note that in this case the WKB wave function over a complete cycle acquiresa factor ±2π

(nr + µ

4

).

In case the particle encounters a hard wall b times, the wave function goesto zero under Dirichel b.c. at every encounter and picks an extra phase bπfor the b reflections. The total acquired phase may be written then as

2π

(nr +

µ

4+

b

2

).

Under Neumann b.c., on the other hand, there is no change in the phase ofthe wave at the wall and b = 0, in this case. Thus the above quantizationcondition may be written as

Ir =(

nr +µ

4+

b

2

)~.

This formula is important for obtaining Gutzwiller trace formula in theintegrable cases, although it does not apply for the non–integrable ones.Finally, in the 3–dimensional case, the semiclassical formalism is set up bythe substitution

l (l + 1) →(

l +12

)r

,

obtained by the same request of semi–classical behaviour for the angularwave function.

2.7. CONNECTION FORMULAS 47

2.7 Connection formulas for semiclassical wave func-tions

The previous analysis can be done in another way which is very instructive.This provides the connection formula between semi–classical wave functionsin the allowed and forbidden classical region. Let us consider first a turningpoint in the right part, as in figure

FIGURE

In the region I : x > a, far away from the turning point we have

Ψ =C√|p| exp

[−1~

∫ x

a|p| dx

](2.7)

whereas, in the region II : x < a, the wave function is given by a linearsuperposition of the two allowed solutions

Ψ =A√p

exp[

i

~

∫ x

apdx

]+

B√p

exp[− i

~

∫ x

apdx

]. (2.8)

How can we determine the coefficients A and B ?In the vicinity of the turning point we have

E − V (x) ' F0 (x− a) , F0 = −dv

dx|x=a< 0.

Suppose now that we can still use the semiclassical approximation suffi-ciently close to the turning point. In this case, in the region I we have

Ψ =C√

2m |F0|1

(x− a)1/4exp

[−1~

∫ x

a

√2m |F |√x− adx

]. (2.9)

Consider now this function as a function of the complex variable x and thepassage from positive to negative (x− a) must be along a path which isalways sufficiently far from the point x = a, so that the WKB is still valid.Let us consider first the variation of this function from right to left along asemicircle of radius ρ in the upper half–plane.

FIGURE


Along this semicircle

x− a = ρeiϕ,

∫ x

a

√t− adt =

23z

32 =

23ρ

32

(cos

32ϕ + i sin

32ϕ

).

The exponential factor in (2.9), at the beginning(0 < ϕ < 2

3π)increases in

modulus and then decreases to modulus 1. At the end of the semicircle theexponent becomes purely imaginary and equal to

− i

~

∫ x

a

√2m |F0| (a− x)dx = − i

~

∫ x

ap (x) dx.

In the coefficient of the exponential, the change along the semicircle is

(x− a)−14 → (a− x)−

14 e−i π

4 .

Hence the function (2.7) has become the second term in (2.8) with

B = Ce−i π4 .

Why is it possible to determine only the coefficient B passing in the uppercomplex plane? There is a simple explanation of this fact. Suppose wefollow the variation of the function (2.8) along the same semicircle but inthe opposite direction.

FIGURE

In this case we have, as before∫ x

a

√(a− x)dx =

23ρ

32

(cos

32ϕ + i sin

32ϕ

).

Correspondingly, for the exponentials

A exp[−2

3ρ

23 sin

32ϕ

]exp

[i

~23ρ

23 cos

32ρ

]+

B exp[23ρ

23 sin

32ϕ

]exp

[− i

~23ρ

23 cos

32ρ

].

We see that at the very beginning of the path the first term becomes expo-nentially small with respect to the second and it must be discarded! So, inthis way we have lost the information on A. In order to obtain A, we mustcontinue the function (2.7) along a path in the lower half plane.

2.7. CONNECTION FORMULAS 49

FIGURE

Repeating the exercise we find that along this path the exponential hasbecome the one of the term in A, whereas its prefactor has become

(x− a)−14 → (a− x) ei π

4 .

In this way we findA = Cei π

4

and therefore the final solution is

Ψ =

{C√p exp

[− 1~

∫ xa |p| dx

], x > a

2C√p cos

[1~

∫ xa pdx + π

4

], x < a

∣∣∣∣∣ .

What has it been the connection formula for the divergent solution? Namely,suppose that in I we have

Ψ =C√p

exp[1~

∫ x

a|p| dx

].

What is the analytic continuation in x < a ? Repeating the calculations,coming from the path in the upper plane, we find this time

A = Ce−i π4 ,

whereas, along the path in the lower plane

B = Cei π4 .

Hence

C√p

exp[1~

∫ x

a|p| dx

]→ C√

pei π

2 exp[

i

~

∫ x

apdx + i

π

4

]+

C√pe−i π

2 exp[− i

~

∫ x

apdx− i

π

4

]=

− 1√pC sin

[∫ x

apdx +

π

4

].

So, we have the connection formulas

1√p

cos[1~

∫ x

apdx +

π

4

]→ − 1√

pexp

[− i

~

∫ x

a|p| dx

],

1√p

sin[1~

∫ x

apdx +

π

4

]→ − 1√

pexp

[1~

∫ x

a|p| dx

].

From these formulas we see that an “innocent” component sin[

1~

∫ xa pdx + π

4

]in the allowed region produces a dangerous component outside!


2.8 Transmission through a barrier

Suppose we have a barrier potential with a particle hitting from the left thebarrier, with an energy E insufficient (classically) to pass on the right. Wewould like to compute semiclassically the transmission coefficient.

For the semiclassical solution, in the three regions, we have

Ψ =

A√p exp

[i∫ xa kdx

]+ B√

p exp[−i

∫ xa kdx

], x < a

C√p exp

[− ∫ xa |k| dx

]+ C√

p exp[∫ x

a |k| dx], a < x < b

F√p exp

[i∫ xb kdx

]+ G√

p exp[−i

∫ xb kdx

], x > b

∣∣∣∣∣∣∣.

Using the previous connection formula, we can easily establish the linearcombination between the coefficients and the final result is very simple.

(AB

)=

12

(2θ + 1

2θ i(2θ − 1

2θ

)−i

(2θ − 1

2θ

)2θ + 1

2θ

) (FG

),

where

θ ≡ exp∫ b

a|p| dx.

This parameter measures the height and the thickness of the barrier. Forthe transmission coefficient, we have simply (assuming G = 0, i.e. no wavefrom the right)

T =∣∣∣∣F

A

∣∣∣∣2

=4(

2θ + 12θ

)2 .

For a high and broad barrier θ À 1 and

T ' 1θ2

= exp[−2~

∫ b

a|p| dx

].

2.9 Metastable states

Consider now the same problem of computing the transmission amplitudefor a potential as in the figure

FIGURE

i.e., a symmetric potential with respect to the origin and a deep well inside.Repeating the previous calculations, one obtains

T =4(

4θ2 + 14θ2

)cos2 L + sin2 L

,

2.10. POTENTIAL V (X) = λ |X|A 51

where

θ ≡ exp∫ b

a|k| dx,

L ≡∫ a

−ak (x) dx.

The above quantity reaches its maximum when cosL = 0, i.e. when

L =∫ a

−aK (x) dx =

(n +

12

)π,

the quantization condition for the bound states!

FIGURE

This is a very important observation. In fact, suppose we want to determineif a given number belongs or not to a given sequence {En}. If we are ableto find a potential which has {En} as a spectrum, the above question, ofa mathematical nature, becomes a physical question. In fact, once suchpotential is constructed, we round it to make its eigenstates metastableand we perform a scattering experiment, with a beam of particles hittingthe target with the energy equal to the number we want to probe. If theparticle passes with probability 1, we have hit an eigenvalue! Obviously,once we round the potential, its eigenvalues acquire an imaginary part. Infact, there is now a probability to escape from the well and to go at infinity.We have a decay process

En → En − 12iΓn,

with

Γn ≡ 1θ2

(∂L∂E

)E=En

.

2.10 Potential V (x) = λ |x|a

Consider the Schrodinger equation in one variable:

− ~2

2m

d2

dx2Ψ(x) + V (x)Ψ (x) = EΨ(x) .


We want to write initially the Schrodinger equation in homogeneous variable.To this aim, let us introduce the dimensionless quantity

η =x

ξ,

where ξ is a length scale. We have[− ~2

2mξ2

d2

dη2+ λξa |η|a

]ψ = Eψ.

Hence

E0

[−1

2d2

dη2+ b |η|a

]ψ = Eψ,

where

E0 =~2

mξ2, b =

mλ

~2ξa+2. (2.10)

From now on, all the energies will be measured in units of E0. Notice that b isa dimensionless parameter. It will be fixed through the WKB quantization.

2.10.1 WKB quantization of V (x) = λ |x|a

Consider the following potential.

FIGURE

We have: ∮p (x) dx =

(n +

12

)h.

For the left–hand side we have∮

p (x) dx = 4∫ x

0

√2m (E − V (x))dx = 4

∫ x

0

√2m (E − λ |x|a)dx =

√2m4E1/24

∫ x

0

√1− λ |x|a

Edx.

Making the change of variable ya =(

λ1/axE1/a

)a, the above integral becomes

4√

2mE

12+ 1

a

λ1/a

∫ 1

0

√1− yady.

2.11. SEMICLASSICAL ESTIMATE OF 〈Xα〉N 53

Since

I (a) =∫ 1

0

√1− yady =

√π

2Γ

(1 + 1

a

)

Γ(

32 + 1

a

) ,

we finally have for the LHS∮

p (x) dx = 2√

2πmΓ

(1 + 1

a

)

Γ(

32 + 1

a

) Ea+22a

λ1a

.

Equating this formula to the RHS, we deduce the quantization of energylevels:

En = E0

(n +

12

) 2aa+2

,

E0 =

[~√2m

λ1/aΓ(

32 + 1

a

)

Γ(1 + 1

a

) √π

] 2aa+2

.

In order to compare the two results we have to make equal the two energyscales, i.e. E0 and E0. One way of doing this consists in extracting initiallyξ from the second of (2.10)

ξ =(~2b

mλ

) 1a+2

,

and then substituting it into E0, which becomes

E0 =

[~√

2√2m

λ1a

b 1a

] 2aa+2

.

Comparing now this expression with the corresponding one for the semiclas-sical expression, the two ones match if we choose

E0 =

[√2π

Γ(1 + 1

a

)

Γ(

32 + 1

a

)]a

.

2.11 Harmonic oscillator: semiclassical estimateof 〈xα〉n

There is a beautiful application of the semiclassical formula. This consistsin the semiclassical estimate of

〈n|xα |n〉 ≡∫ +∞

−∞dxxαψn (x) ψn (x) dx,


where

ψn (x) =(

1π

) 14 1

2n/2

1√n!

e−x2

2 Hn (x) .

The difficulty of the problem consists in the lacking of suitable approxi-mation of the Hermite polynomials with respect to the variable n. Wecan get around this problem by using the semiclassical approximation. Be-fore doing that, we have to fix a convenient notation. In the units wherem = ω = ~ = 1, the exact energy levels (and the semiclassical ones) areEn = n + 1

2 . The position of the inversion points an

FIGURE

is given by the conditionV (an) = En,

i.e.12a2

n = n +12,

and thereforean =

√2n + 1.

The momentum p (x) is given by

p (x) =√

2m (E − V (x)) =√

a2n − x2.

With the above notation the semiclassical wave function in the allowed re-gion −a < x < a (|x| > a is supposed to give a negligible contribution to allthe following computation) is represented by

ψn (x) ' An√p

cos[∫ x

−ap (t) dt− π

4

].

The normalization constant An is easily fixed. In fact∫ a

−aψ2

ndx = A2n

∫ a

−a

dx

p (x)cos2

[∫ x

−ap (t) dt− π

4

]' A2

2

∫ a

−a

dx

p (x).

Here we have taken the average 12 of the rapid oscillating trigonometric func-

tion. Since∫ a

−a

dx

p (x)= 2

∫ a

−a

dx√a2 − x2

= 2∫ 1

0

dt√1− t2

= π,

2.11. SEMICLASSICAL ESTIMATE OF 〈Xα〉N 55

for the normalization constant A we have

A =

√2π

.

Last observation. We can write the wave function as

ψn =A

a12n

[1−

(xan

)2] 1

4

cos

[2a2

n

∫ x/an

−1

√1− t2dt− π

4

],

i.e.ψn =

A

a12n [1− η2]

14

cos[a2

nf (η)],

where η ≡ xan

. By using the above formula is now easy to compute theaverage of xα. We have

〈xα〉 =∫ an

−an

xαψn (x) ψn (x) dx =A2

an

∫ an

−an

xα 1√1− η2

cos2[a2

nf (η)]dx.

By making the change of variable x → x/an and taking the average of therapid oscillating function, we have

〈xα〉 = A2aαn

∫ 1

0dt

tα√1− t2

.

For the last integral we obtain

I (α) ≡∫ 1

0dt

tα√1− t2

=√

π

α

Γ(

1+α2

)

Γ(

α2

) ,

i.e.

〈n|xα |n〉 =2√π

1α

Γ(

1+α2

)

Γ(

α2

) (2n + 1)α2 .

Simple checks of the above formula are obtained by takingα = 2

⟨x2

⟩=

2√π

12

Γ(

32

)

Γ(

12

) (2n + 1) =12

(2n + 1) = n +12,

which coincides with the exact result.


α = 4⟨x4

⟩=

2√π

14

Γ(

52

)

Γ (2)(2n + 1)2 .

Since Γ(

52

)= 3

4

√π,

⟨x4

⟩=

34

(2n2 + 2n +

12

),

whereas the exact result seems to be (according to Landau)

⟨x4

⟩ex

=34

(2n2 + 2n + 1

).

2.12 Exact WKB quantization

The WKB quantization condition can be actually converted into an algebraicform, which can be used to calculate the eigenvalues to any order. Themethod is due to Dunham and brought to its full glory by Bender, Olanssemand Way.In the following we will use their notation.Starting from the Schrodinger equation written as

[− d2

dx2+ V (x)− E

]ψ = 0, ψ (±∞) = 0

we introduce a small parameter ε and consider the eigenvalue problem

ε2ψ′′

= Q (x) ψ, Q (x) = V (x)− E. (2.11)

The parameter ε helps in organizing the WKB series and it os obviouslyrelated to ~. The WKB solution is expressed as

ψ (x) = exp

[1ε

∞∑

n=0

εnsn (x)

].

Substituting this expression into (2.11) and comparing the equal powers ofε, we obtain

S′(0) = − [a (x)]1/2 ,

2S′0S

′n +

n−1∑

j=1

S′jS

′n−j + S

′′n−1 = 0, n ≥ 1. (2.12)

2.12. EXACT WKB QUANTIZATION 57

The recursion equation in (2.12) is, as a matter of fact, a simple algebraicrule for computing S′n (x) from s

′j for j < n. Straightforward computations

give

S′1 (x) = −Q′ (x)

4Q (x),

S′2 (x) =

5 [Q′ (x)]2

32 [Q (x)]52

− Q′′(x)

8 [Q (x)]32

,

S′3 (x) = −

15[Q′(x)

]3

64 [Q (x)]4+

9Q′(x) Q

′′(x)

32 [Q (x)]3− Q

′′′(x)

16 [Q (x)]2,

and so on. Once the S′n have been found, there is a simple formula which isa generalization of the well–known formula of the WKB quantization

∫ x2

x1

[E − V (x)]12 dx =

(k +

12

)π,

which states the exact quantization of the eigenvalues. This formula issimply

12πi

∮ ∞∑

n=0

S′n (x) = k k = 0, 1, 2... (2.13)

and it expresses, through the logarithmic derivative of the wave function,the fact that this solution possesses k zeros. It is easy to see that the aboveformula reduces to the usual one at the leading order. The leading WKBformula takes into account both S0 (x) and S1 (x) substituting S

′0 (x) into

(2.13) and making [Q (x)]12 single valued by joining the turning points x1

and x2 by a branch cut, we have

− 12πi

∮[V (x)−E]

12 dx =

∫ x2

x1

√E − V (x)dx,

for the term coming from S0, whereas for the term coming from S′1 (x)

− 12πi

∮Q′ (x)4Q (x)

dx = − 18πi

lnQ (z) |C= − 18πi

4πi = −12, (2.14)

Here lnQ (z) has been evaluated once around the contour (it gives 4πi becausethe contour encircles two simple poles of Q (x)).Then, one reproduces the well–known formula

∫ x2

x1

√E − V (x) = k +

12.


An important comment is in order. It is remarkable that the above formula(2.13) takes into account only the basic property of the wave function, i.e.the number of nodes. Any reference, for instance, to the Airy function hasdisappeared. Moreover, there is no reference to ψ (x) or to its boundary con-ditions. The construction is purely algebraic; it involves differentiation andnot integration. It is necessary though to use complex contour integrationinstead of ordinary integration along the real axis between x1 and x2. Thisbecause all the functions S

′i (x) are singular at the turning points.

There is an interesting open question. Since the integrals in (2.13) involveclosed contours, it is possible to add total derivatives to S′n (z) under theintegral without altering the value of the integral. Hence, S′n as generatedby the recursion equation (2.12) is but one element of a large equivalenceclass which we denote by Fn. The elements of Fn are all different but theircontour integral are always the same. Is it possible that for all n there issome elements of Fn which is so simple that the indicate sum in (2.13) may beevaluated in close form? This possibility seems quite remote. Nevertheless,there are some interesting results.

(i) One element of F2n+1 is always zero (for n ≥ 1), because S′2n+1 is

itself a total derivative. For example, S′3 (x) can be written

S′3 (x) =

d

dx

5[Q′(x)

]2

64 [Q (x)]3− Q

′′(x)

16 [Q (x)]2

,

which vanishes once integrated along the close contour. This can be un-derstood by a simple argument. The quantization condition (2.14) is aconstraint on the phase of ψ (x). S

′2n+1 (x) (n ≥ 1) is always real because

it contains no fractional powers of (V −E) and therefore cannot contributeto the phase of ψ (x). It is S′2n instead that becomes imaginary as x crossesinto a classically allowed region and causes the wave function to becomeoscillatory. Hence it is no longer surprising that S

′2n+1 (n ≥ 1) drops from

the quantization condition.(ii) The even terms S′2n (z) , the only ones which survive, can be drasti-

cally simplified by adding and subtracting total derivatives. For instance

S′2 (x) = − Q

′′(x)

48 [a (x)]32

− d

dx

[5Q

′(x)

48 [a (x)]32

].

(iii) There is a close parallelism between the search of the simplest ele-ment of F2n and the construction of the conserved quantities for the non–

2.12. EXACT WKB QUANTIZATION 59

linear wave equation like the KdV

·u = u

′′′ − 6uu′.

For this equation, the conserved quantities are derived from the Miura trans-formation

u (x) = −v2 (x)− v′.

They satisfy a recursive equation, similar to (2.12).

The recursion equation for the elements S′n of the WKB expansion maybe derived from the Miura transformation if we put u = v/ε and let v =∑

εnS′n.

The Miura transformation if nothing but the Riccati equivalent of theSchrodinger equation.

2.12.1 Some ”numerology”

Let us analyze the potential V = xN , with N even. For such potential, theWKB series is a power series in inverse fractional powers of the energy E,i.e.

E1N

+ 12

∞∑

n=0

E−n(1+ 2N )an (N) =

(k +

12

)π.

The coefficients an (N) have the form

an (N) =2√

πΓ(1 + 1−2n

N

)

Γ(

3−2n2 + 1−2n

N

) Pn (N) (−1)n

(2n + 2)!2n,

where Pn (N) is a polynomial in N :

P0 (N) = 1

P1 (N) = 2 (N − 1)

P2 (N) = (N − 3) (N − 1) (2N + 3)

P3 (n) =49

(N − 1) (N − 5)(24N3 + 22N2 − 117N − 139

)

...

All these polynomials contain (N − 1) as a factor, and this explains why forthe harmonic oscillator the WKB quantization is exact.


One should keep in mind that the above series is an asymptotic series.Like the Stirling series for the Γ function, the coefficients get smaller for awhile but eventually grow without bound. This means that for any givenvalue of k, successive approximations to E(k), obtained by truncating theseries, improve to some maximal accuracy and then become worse. Also,since E increases with k, more terms in the series should be required to reachmaximal accuracy as k increases, and the accuracy should also increase withk. This is precisely what happens. For instance, with V = x4,

E0exact = 1.060362090

(WKB)1 = 0.87

(WKB)2 = 0.98

(WKB)4 = 0.95

(WKB)6 = 0.78

(WKB)8 = 1.13

(WKB)10 = 1.40,

E8exact = 37.923001027033

(WKB)1 = 37.904471845068

(WKB)2 = 37.923021140528

...

(WKB)10 = 37.923001027043(10−14!

).

2.13 Estimate of the ground state energy

In the following we will discuss some useful tools which allows us to esti-mate the ground state energy, i.e. to have upper and lower bounds for thisquantity.Variational principleThe Hamiltonian can be decomposed as

H =∑

n

En |ψn〉〈ψn| . (2.15)

2.13. ESTIMATE OF THE GROUND STATE ENERGY 61

In the following we assume to have an increasing sequence of eigenvaluesand, for simplicity, to have no degeneracy. Taking the expectation value of(2.15) on a state |ψ〉, we have

〈ψ|H |ψ〉 =∑

n

En |〈ψ| ψn〉|2 ≥ E0

∑n

En |〈ψn| ψ〉|2 = E0 〈ψ| ψ〉 , (2.16)

i.e.E0 ≤ 〈ψ|H |ψ〉

〈ψ| ψ〉 . (2.17)

If the state depends on a parameter, i.e. |ψ〉 = |ψ (β)〉 and it has the samequalitative features of the ground state wave function (i.e. it has no nodesand decreases sufficiently rapidly at infinity), we can obtain an estimate ofE0 by computing (2.17) and determining the value minimizing this quantity.Example

H =p2

2m+ λ |x|a = ε0

[p2

2+ b |η|a

], (2.18)

where we have chosen the natural units, previously introduced, which matchthe semiclassical expression. Let us take now, as a trial wave function,

ψβ (η) =

√β√π

exp(−β2η2

2

).

By using the formula

I (n) =∫ +∞

−∞|x|n e−α2x2

dx =(

1α

)n+1

Γ(

n + 12

),

it is easy to obtain the expectation value of the Hamiltonian. In fact, wehave,

ψ′= −β2ηψ; ψ

′′= β4η2ψ − β2ψ.

Therefore, for the kinetic term we get

〈ψ| p2

2|ψ〉 = −1

2〈ψ| ψ′′⟩

=β2

2− β4

2〈ψ| η2 |ψ〉 =

β2

2− β2

4=

β2

4,

whereas for the potential term we have

〈ψ|V |ψ〉 = b 〈ψ| |η|a |ψ〉 =b

β

Γ(

a+12

)√

π=

B

β,


where

B =b (a)√

πΓ

(a + 1

2

).

Hence

〈ψ|H |ψ〉 = ε0

[β2

4+

B

βa

]. (2.19)

Minimizing with respect to β, we have

∂

∂β〈ψ|H |ψ〉 = ε0

[β

2− a

B

βa+1

]= 0,

i.e.βop = (2aB)

1a+2 .

Substituting now into (2.19), we have

E0 ≤ ε0

[14

(2aB)2

a+2 +B

(2aB)1

a+2

]= ε0

(2aB

4+ B

)1

(2aB)1

a+2

=

ε0

(a + 24a

)(2aB)

2a+2 = ε0

(a + 24a

)[2ab (a)√

πΓ

(a + 1

2

)] 2a+2

,

so, finally

E0 ≤ ε0

(a + 24a

) [2ab (a)√

πΓ

(a + 1

2

)] 2a+2

.

For a = 2, we get correctly E0 = ε02 .

The variational principle can be used to establish the following general result:an attractive potential, i.e. a potential V (x) < 0, has always at least a boundstate.

2.14 Uncertainty relation and lower bound

The previous variational principle permits to obtain an upper bound forthe ground state energy E0. In order to obtain a lower bound, we rely onHeisenberg’s uncertainty relation

∆p∆q ≥ ~2.

For an Hamiltonian as the one in (2.18), we have

E0 ≥ ε0

[18

(1

∆q

)2

+ b |η|a]

.

2.14. UNCERTAINTY RELATION AND LOWER BOUND 63

Requiring that ∆q is of the order of 2η we have

E0 ≥ ε0

[132

(2

∆q

)2

+ b

(∆q

2

)a]≡

ε0

[132

(1x

)2

+ bxa

], (2.20)

where x ≡ ∆q2 . Computing the extremum of the above quantity w.r.t. x we

have

− 116

(1x

)3

+ abxa−1 = 0, ⇒ xop =(

116ab

) 1a+2

.

Substituting this value into (2.20), we have, after some manipulations

E0 ≥ ε0

(a + 232a

)(16ab)

2a+2 .

For the ratio of the uncertainty bound and the one of the variational principlewe have

Eun

Evar=

18

(8√

π

Γ(

a+12

)) 2

a+2

,

and this quantity is correctly always less than 1.


Chapter 3

Thermodynamic BetheAnsatz and OrdinaryDifferential Equations

3.1 Spectral determinant

In this section we will come back to the problem of the spectral determi-nant of the Schrodinger equation. We will see that for a particular class ofpotentials this function satisfies a nonlinear functional equation which, as amatter of fact, is equal to the Bethe Ansatz equations for integrable modelsin Statistical Mechanics! Consider the Schrodinger equation

(− d2

dx2+ P (x)

)ψ (x) = 0 (3.1)

for arbitrary complex values of x, with

P (x) = x2M − E +l (l − 1)

x2.

For non integer values of M we have a branch cut at the origin. Using theresults of the analysis by Sibuya in [?], we state that (3.1) admits a solutiony = y (x,E, l), such that

(i) y is an entire function of x,E, where due to the branch point at x = 0,x must be in general considered to live on a suitable cover of the puncturedcomplex plane;

(ii) y and y′admit the asymptotic expressions

y ∼ x−M2 exp

[− 1

M + 1xM+1

](3.2)

65

66 CHAPTER 3. TBA AND DIFFERENTIAL EQUATIONS

y ∼ −xM2 exp

[− 1

M + 1xM+1

], (3.3)

as x goes to infinity in any closed sector satisfying

|arg x| < 3π

2M + 2.

(Remark. Extra terms appear though for 0 < M ≤ 1 ).(iii) Furthermore, the solution y is uniquely identified by the above in-

formation. From WKB theory, we know that eq. (3.1) has two solutions Φ±given by

Φ± (x) = [P (x)]−14 exp

[±

∫ x

x0

√P (x)dx

].

It is then easy to see that the asymptotic (ii) is, up to a normalizationconstant N , the large x limit of Φ−:

y (x,E) ' NΦ− (x,E)

with x = ρeiθ we have

Φ± (ρ, θ) ' x−π2 exp

[±ρM + 1

M + 1eiθ(M+1)

].

Let us fix now some terminology. A solution tending to zero for large ρ inthe sector

a < θ < b

is called subdominant in that sector. A solution growing for large ρ in thesector

a′< θ

′< b

′

is called dominant in this sector.Let us denote by Sk the sector

∣∣∣∣θ −2kπ

2M + 2

∣∣∣∣ <π

2M + 2.

We see then(i) in S0, y ' NΦ− → 0, i.e. y is subdominant in S0;(ii) at θ = ± π

2M+2 it decays algebraically;(iii) in S±1, y ' NΦ− →∞, i.e. y is dominant in these sectors.FIGURE

3.1. SPECTRAL DETERMINANT 67

Note that in S±1 , with some coefficients C (E)

y (x) = y− (x) + C (E) y+ (x)

and that exactly at θ ≥ 3π2M+2 we have lost control of the exponential asymp-

totic behaviour of y. This is the reason of the above restriction

|θ| = |arg x| < 3π

2M + 2.

To find subdominant solutions on other sectors, we can use the followingtechnique. Consider y (x) = y (ax, E, l) for any constant a. This functionsatisfies the differential eq.

[− d2

dx2+ a2M+2x2M − a2E +

l (l + 1)x2

]y (x) = 0.

Thus, if a2M+2 = 1, P(y

(ax, a−2E, l

))is another solution of (3.1). Setting

ω = eiπ/M+1,

we have therefore the set of solutions

yk ≡ yk (x,E, l) = ωk2 y

(ω−kx, ω2kE, l

)(3.4)

with yk subdominant in Sk and dominant in Sk±1 (the prefactor ωk2 is for

later convenience). The first conclusion of the above analysis is that eachpair

{yk, yk+1}of functions provides a set of linearly independent solutions of (3.1) and anyother solution can be expressed in terms of them. In particular

yk−1 (x,E, l) = Ck (E, l) yk (x,E, l) + Ck (E, l) yk+1 (x,E, l) . (3.5)

The functions Ck and Ck are called the Stokes multipliers for yk−1 w.r.t.yk and yk+1. Clearly, from (3.4), we have

Ck (E, l) = Ck−1

(ω2E, l

)

Ck (E, l) = Ck−1

(ω2E, l

).

From now on, we denote C0 and C0 by C and C. The second result is thatthe Stokes multipliers can be expressed in terms of the Wronskians.


Definition 1. The Wronskian W [f, g] of two functions is defined by

W (f, g) = fg′ − f

′g.

If f and g are both solutions of the Schrodinger equation (3.1), their Wron-skian is a constant, independent of x. In fact

W ′ (f, g) = f′g′+fg

′′−f′′g−f

′g′= fg

′′−f′′g = −P (x,E, l) (fg − gf) = 0.

Consider the Wronskian of (3.5) (with k = 0) with respect to the functionsy1 and y0:

y−1 = C0y0 + C0y1

W [y−1, y1] ≡ W−1,1 = C0W [y0, y1] = C0W01

W [y−1, y0] ≡ W−1,0 = C0W[y1, y0 = C0W1,0 = −C0W01

],

i.e.C =

W−1,1

W01; C = −W−1,0

W0,1. (3.6)

These Wronskians are entire functions of E and l (since the same holdsfor the functions yk). Since y0 and y1 are linearly independent, W nevervanishes. Therefore C and C are also entire functions. As a matter of fact,all Ck are identically equal to −1,

Ck = −1.

This result follows from the relations

Wk1+1,k2+1 = Wk1k2

(ω2E, l

)

W0,1 (E, l) = 2i.

The last condition is obtained by evaluating W0,1 as x →∞ in the sectors S0

or S1, where the asymptotic behaviours of y0 and y1 (and their derivatives)are determined by eqs. (3.2)–(3.3). Since y−1 (x,E, l) = y1 (x∗, E∗, l∗) italso follows from (3.6) that C (E, l) is real whenever E and l are real.Finally, the basic Stokes relation (3.5) at k = 0 can be written in the form

C (E, l) y0 (x,E, l) = y−1 (x,E, l) + y1 (x,E, l)

C (E, l) =12i

W−1,1 (E, l) , (3.7)


i.e.

C (E, l) y (x, E, l) = ω−12 y

(ωx, ω−2E, l

)+ ω

12 y

(ω−1x, ω2E, l

).

Let us analyze this equation. Any solution of (3.1) can be written in termsof another basis of functions

{ψ+, ψ−}where ψ+ (x) is the solution which for x → 0 goes as

ψ+ ∼ xl+1 (3.8)

whereas ψ− (x) behaves asψ− ∼ x−l.

Notice that both of them vanish if l (l + 1) < 0, whereas one of the twodiverges if l (l + 1) > 0.Since l enters eq. (3.1) only in the combination l (l + 1), we can decide tofix ψ+ (x) as the solution which behaves as (3.8) and to define ψ− (x) by itsanalytic continuation, i.e.

ψ− (x,E, l) ≡ ψ+ (x,E,−1− l) .

In analogy with what done previously, we define the shifted solutions ψ±k

ψ±k = ψ±k (x,E, l) = ωk2 ψ±

(ω−kx, ω2kE, l

).

They also solve the original problem (3.1). By considering the limit x → 0,it is easily seen that

ψ±k (x,E, l) = ω∓k(l+ 12)ψ± (x,E, l) .

We also have

W[yk+1, ψ

±k2+1

](E, l) = W

[yk+1, ψ

±k2+1

] (ω2E, l

)

and therefore

W[yk, ψ

±](E, l) = ω±k(l+ 1

2)W[yk, ψ

±](E, l) =

ω±k(l+ 12)W

[y, ψ±

] (ω2kE, l

). (3.9)

We can now take the Wronskian of both sides of (3.7) with ψ±. Defining

D∓ (E, l) ≡ W[y (x, E, l) , ψ± (x,E, l)

]


and using (3.9), the Stokes eq. (3.7) becomes

C (E, l) D∓ (E, l) = ω∓(l+ 12)D∓ (

ω−2E, l)+ ω±(l+ 1

2)DD∓ (ω2E, l

). (3.10)

We will discover soon the meaning and the beauty of this functional equa-tion.First of all, notice that expressing

y = αψ+ + βψ−,

the coefficients of this linear combination are given by the above Wronskians.In fact, using

W[ψ−, ψ+

]= x−l (l + 1)xl + lx−l−1xl+1 = 2l + 1,

we haveW

[y, ψ+

] ≡ D− = β (2l + 1)

W[y, ψ−

] ≡ D+ = −α (2l + 1) ,

i.e.(2l + 1) y = D− (E, l) ψ− −D+ (E, l) ψ+.

Moreover, from the relation

ψ− (x,E, l) = ψ+ (x,E,−1− l)

we deduceD (E, l) ≡ D− (E, l) = D+ (E,−1− l) .

Requiring a non–singular behaviour at the origin for y (x) restricts E to theset of eigenvalues {En} of the spectral problem, i.e.

D (E, l) |E=En= 0, l > −12.

Therefore we arrive at the following remarkable conclusion: D (E, l) is anentire function which has zeros on the real positive axis, in correspondencewith the eigenvalues EM of the Schrodinger equation. This because thesolution y vanishes at infinity (x →∞) and it is regular as x → 0 . HenceD± (E, l) are the Fredholm determinants. We have the following properties.

(i) C and D− are entire functions of E, since the functions enteringtheir definition are entire.

(ii) If l is real and larger than −12 , then the zeros of D− all lie on the pos-

itive real axis of the complex plane E. In fact, we have already commented


that a zero of D− (E, l) signals the existence of a proper eigenfunction. TheHermitian nature of the problem for l > −1

2 then ensures the reality of thesezeros.. Moreover, for l > 0 the potential is everywhere positive. Therefore,by multiplying for ψ∗ and integrating from (0,∞) shows that all eigenvaluesEM must be positive.

(iii) When M > 1, D− (E, l) has the following large E asymptotic

ln D− (E, l) ' a0

2(−E)µ , |E| → ∞, |arg (−E)| < π, (3.11)

where

a0 = − 1√π

Γ(−1

2− 1

2M

)Γ

(1 +

12M

), µ =

M + 12M

.

(iv) If E = 0, then

D− (0, l) = − 1√π

Γ(

1 +2l + 1

2M + 2

)(2M + 2)

2l+12M+2

+ 12 .

Notice that at l = 0, we simply have

D− (E, 0) ≡ y (x,E) |x=0, D+ (E, 0) ≡ y′ (x,E) |x=0 .

In this case, the problem reduces to an ordinary Schrodinger equation, buton a half line. WE can choose Dirichlet b.c. for the solution, i.e.

y (x,E) |x=0= 0,

or Neumann b.c.y′(x,E) |x=0= 0.

The first one are those appropriate for the wave function in the odd–paritysector, as for instance

FIGURE

whereas the second one refers to the even–parity sector, as

FIGURE

In this case D− (E, 0) will have zeroes in correspondence of the energy levelsE−

n with odd parity, whereas D+ (E, 0)will have zeros in correspondence ofE+

n . This is equivalent, in fact, to choose the proper combination whichexpresses y in this case.


For l = 0, it is simple to establish the formula (3.11). In this case, in fact,D− (E, 0) ≡ y (x, E) |x=0= 0, so we have to find the value of this functionat the origin. When E →∞, we have

P ′ (x,E)P (x,E)

→ 0,

and therefore we can rely on WKB. We have, in fact

y (x,E) ' Nφ− (x,E)

and therefore

lny

N∼=

∫ x0

x

(E2M − E

) 12 dE '

x,x0À0

xM+10

M + 1− xM+1

M + 1.

We can drop the first term by choosing

lnN = −∫ x0

0tMdt.

Sending now x0 →∞, we end up with

ln y | x=0.EÀ0

∼∫ ∞

0

[(t2M − E

) 12 − tM

]dt = a0 (−E)M ,

where

a0 =∫ ∞

0

([t2M + 1

] 12 − tM

)dt = − 1√

πΓ

(−M + 1

2M

)Γ

(1 +

12M

),

withµ =

M + 12M

.

The same result can be shown to hold also for l 6= 0. We can now use theHadamard factorization theorem.

3.2 Hadamard Factorization Theorem

An entire function f (z) is said to be of finite order if there is a positivenumber A such that

|f (z)| = O(e|z|

A)

, |z| → ∞.

3.2. HADAMARD FACTORIZATION THEOREM 73

The lower bound δ of the numbers A for which this is true is called the orderof f (z). In our specific example,

lnD (E, l) |−EÀ0∼ a0 (−E)M

which is valid everywhere except for arg (E) = 0, i.e. the real axis, wherethe growth of D is not greater than for arg (E) 6= 0. Hence

order [D] = µ =M + 12M

< 1, M > 1.

The celebrated Hadamard theorem can be stated as follows.

Theorem 1. An entire function f(z) of finite order δ admits a factorizationover his zeros {zn} of the form

f (z) = eQ(z)∞∏

n=0

K

(z

zn, P

),

where

K (v, 0) = 1− v; K (v, p) = (1− v)v+ v2

2+...+ vp

p

are called “primary factors” and Q (z) is a polynomial of order not greaterthan δ. If δ is not an integer, then p = Int [δ]. Consequently, for M > 1,

δ = µ =M + 12M

, p = 0

and

D (E, l) = D (0, l)∞∏

n=0

(1− E

En

).

For M ≤ 1 things are complicated by the possibility of nontrivial Q (z).Exactly at M = 1,

D (E, l) = D (0, l)∞∏

n=0

(1− E

En

)e

EEn .


3.3 Bethe Ansatz Equations

The functional eq. (3.10) can be used to derive the energy levels of theconsidered quantum mechanical problem. Let us consider the one for D− ≡D

C (E, l) D (E, l) = ω−(l+ 12)D

(ω−2E, l

)+ ω(l+ 1

2)D(ω2E, l

).

Setting E = Ei the left–hand side vanishes and we are led to the equation

−1 = ω2l+1 D(ω2Ei, l

)

D (ω−2Ei, l); ω ≡ eiπ/M+1.

For M > 1, we can use the Hadamard factorization theorem and write

−1 = ω2l+1∞∏

n=0

(En − Eiω

2

En − Eiω−2

).

This is an equation for the energy levels, equivalent to the Bethe Ansatzequation in Statistical Mechanics! This equation can be converted into anintegral nonlinear equation, which can be solved very fast numerically. Tothis aim, let us define

a (E, l) ω−2l−1 =D

(Eω2

)

D (Eω−2)=

∞∏

n=0

(En − Eω2

En − Eω−2

). (3.12)

When a (E, l) = −1, we get the quantization condition for the energy levels.From eq. (3.12), taking the logarithm we have

lnQ (E, l) = i2l + 1M + 1

π +∞∑

n=0

F

(E

En

),

with

F (E) = ln1− ω2E

1− ω−2E.

Assuming that all En lie on the positive real axis, and that these are the onlypoints in some strip about this axis for which a (E, l) = −1. By Cauchy’stheorem we have

ln a (E, l) = i2l + 1M + 1

π +∫

C

dE′

2πiF

(E

E′

)∂E′ ln

(1 + a

(E′, l

)).

FIGURE

3.3. BETHE ANSATZ EQUATIONS 75

since

∂E′ ln(1 + a

(E′, l

))=

a′ (E′)1 + a (E′)

and then apply the residue theorem. We make the change of variables

E = eθµ

and integrate by parts. We get

ln a (θ) = i2l + 1M + 1

π +∫

Cdθ′R

(θ − θ′

)ln

(1 + a

(θ′

))+

−∫

Cdθ′R

(θ − θ′

)ln

(1 + a

(θ′

)),

where the contour C1 and C2 run from −∞ to +∞, infinitesimally belowand above the real axis, and

R (θ) ≡ i

2π

d

dθF

(e

θµ

).

FIGURE

Using the fact that i ln a is a real analytic function for which

[a (θ)]∗ = [a (θ∗)]−1 ,

and pushing C1 and C2 toward the real axis, we have

ln a (θ) = i2l + 1M + 1

π+∫ +∞

−∞dθ′R

(θ − θ′

){ln

[1 + a

(θ′ − iε

)−1]∗− ln

(1 + a

(θ′ − iε

))}

= i2l + 1M + 1

π +∫ +∞

−∞dθ′R

(θ − θ′

) [ln a

(θ′

)− 2iIm ln(1 + a

(θ′ − iε

))].

Let us define the Fourier transform

f (k) = F (f) (k) =∫ +∞

−∞f (θ) e−ikθdθ

f (θ) = F−1[f](θ) =

12π

∫ +∞

−∞f (k) eiθkdk,


and taking the Fourier transform of the above equation[1− R (k)

]F [ln a] (k) =

2iπ (2l + 1)M + 1

δ (k)− 2iR (k) ImF [ln (1 + a)] (k) .

Transforming back and restoring the integrations over C1 and C2, we have

ln a (θ) = iπ

(l +

12

)− ib0e

θ+

+∫

C1

dθ′ϕ

(θ − θ′

)ln

(1 + a

(θ′

))−∫

C2

dθ′ϕ

(θ − θ′

)ln

(1 + a−1

(θ′

)),

with

ϕ (θ) ≡∫ +∞

−∞

dk

2πeikθ shπ

2 (ξ − 1) k

2shπ2 ξkchπk

2

, ξ =1M

.

b0 = 2 cosπ

2Ma0.

The driving force ib0eθ of the above equation arises from a zero mode, which

can be traced to the zero of

1− R (k) at K = i.

Its value can be fixed by the WKB result

D (E, l) 'E→−∞

a0 (−E)n = a0

(−e

θµ

)n

and from the definition

a (E, l) = ω2l+1 D(Eω2, l

)

D (Eω−2, l).

A first consistency check of this equation is immediate in the large θ limit,the driving term dominates and therefore

a (θ, l) = −1

at the points θn, with

b0eθn ' π

(l +

12

)+ (2n + 1)π, n = 0, 1, . . .

giving

En '[

π

2b0(4n + 2l + 3)

] 1n

, n = 0, 1, . . .

which at M = 1, a potential can be solved exactly, coincides with the exactresult!

Part II

Analytic Number Theory: anIntroduction

77

Chapter 4

Prime numbers

4.1 Introduction

Number theory is one of the most fascinating areas of Mathematics. Thereare many unproved results and conjectures. The most amazing aspect isthat there are problems which can be stated in a very simple way, but verydifficult to prove.Example: (3n + 1) Problem.Take a number N. If it is even divide it by 2, otherwise multiply it by 3 andadd 1.

N →{

N/2, N even3N + 1, N odd

∣∣∣∣ .

Iterate the operation. Does the numbers always end to 1?(Argument based on random walk).Number theory has deep overlaps with calculus and also with geometry.Example: Pitagoric triple.

a2 + b2 = c2, a, b, c ∈ N.

a = 3, b = 4, c = 5,

a = 5, b = 12, c = 13,a = 8, b = 15, c = 17,

......

...

79

80 CHAPTER 4. PRIME NUMBERS

There is an infinite number of them. Geometrically, the problem consists infinding all rational points on a circle

(a

c

)2+

(b

c

)2

= 1.

FIGURE

This happens when the slope is rational! The generalization of this problemconsists in finding rational points on an elliptic curve. According to therecent proof of Fermat’s last theorem, there are no integer solutions to theequation

an + bn = cn, n > 2.

Another classical connection between Number Theory and Geometry isabout the possibility to construct regular polygons by rule and compass.According to the proof by Gauss, this is possible for all those polygons of nsides when n is a product of different Fermat primes and powers of 2:

Fn = 22n+ 1.

At present, only 5 such primes are known for n = 0, 1, 2, . . ., i.e.

F0 = 3, F1 = 5, F2 = 17, F3 = 257, F4 = 65537.

So, regular polygons with

n = 2, 3, 4, 5, 6, 8, 10, 12, 15, 17, 20, 24, 30, . . .

sides are constructible while those with

n = 7, 9, 11, 13, 14, 18, 19, 21, 22, 23, 25, . . .

cannot be constructed.In these lectures, we will cover only a small set of topics of this large subject.In the first lecture we will discuss some beautiful identities, collected andpresented almost in a random order. The other two lectures will deal withone of the most fascinating and mysterious functions of Mathematics, i.e.the Riemann zeta function. We will discuss its main properties and itsrelationship with prime numbers.

4.2. INDUCTION 81

4.2 Induction

One of the strongest properties of integers is their inductive nature: if theassumption that some law holds for n implies that holds for n + 1 as well(and it is proved true for n = 1), then it is always true.A typical result that can be easily proven by induction is

n∑

k=1

k =n (n + 1)

2.

Another, less well known, is the derivation by Newton of the infinite expan-sion of the series

(1 + x)−2 = 1− 2x + 3x2 − 4x3 + 5x4 + ...

(1 + x)−3 = 1− 3x + 6x2 − 10x3 + 15x4 + ...

Consider the Tartaglia triangle for the binomial expression (1 + x)n, with nnonnegative integer:

n = 0 1

n = 1 1 1

n = 2 1 2 1

n = 3 1 3 3 1

n = 4 1 4 6 4 1

Of course, each number is the sum of the two numbers above it, i.e.(

n + 1k

)=

(nk − 1

)+

(nk

).

By applying this rule back–forward, we get

n = −3 1 − 3 6 − 10 15

n = −2 1 − 2 3 − 4 5 − 6

n = −1 1 − 1 1 − 1 1− 1

n = 0 1 0 0 0 0

Let us consider now some limiting formulas√

1 +√

1 +√

1 + ... → 1 +√

52


√7 +

√7 +

√7 + ... → 1 +

√29

2√1 + 2

√1 + 3

√1 + 4√... → 3

√

1 +

√2 +

√3 +

√4+... →

√3

√1 +

√2 +

√...√

pn+... → 1.8507...

4.3 Prime Numbers

Let us consider now the Prime Numbers pi, the building blocks of NumberTheory. The first primes are

pi = 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ...

There are infinitely many prime numbers. Several proofs of this fact areknown; the oldest and simplest one is due to Euclid.Euclid’s proof. Assume that there are only k primes p1, ..., pk. Considerthe number N = p1 · ... · pk + 1. It is not divisible for any pi, so either it isa prime or it has a prime divisor larger than all p1, ..., pk.The basic theorem of Arithmetic states that any natural number n either isa prime or admits an unique decomposition in terms of the primes, i.e.:

n = pα11 · ... · pαk

k .

The mathematical problem of identifying a prime number or to determineits prime decomposition is one of the most challenging of Mathematics.Let us discuss briefly the Primality Problem. The crucial point is that itdoesn’t exist a formula which determines all and only the primes, i.e. arelation of the type Pn = f (n). There are several formulas providing ananswer to the primality problem, but in practice almost useless.Example. Wilson formula

f (n) = sin[π

(n− 1)! + 1n

].

One can prove that this function is zero if and only if n is a prime. Unfortu-nately, this primality test is of no practical advantage, because calculatingf (n) takes longer than the ordinary Eratosthenes’s sieve.Example: 101! ' 10160.

4.4. MEASURE OF COMPLEXITY 83

4.4 Measure of Complexity

Consider an algorithm which applies to a problem made of N data. Thecomplexity of the problem can be related to the time of elaboration, inparticular how this time scales with N . The problem has a polynomialcomplexity if the time of elaboration scales as a power law of N :

τpoly ∼ Nα.

These are the “easy” problems of mathematics. On the other hand, theproblem is said to have an exponential complexity if the time scales fasterthan any power of N

τexp ∼ exp [βNγ ] .

These are the “toughest” problems of mathematics. Among them:

• the problem of salesman,

• the determination of the vacuum states in random systems.

Consider now a number A, expressed, say, in the usual decimal basis. Thenumber of its digits is about log10 A. In this context, a polynomial algorithmscales as

τpoly ∼ (log10 A)α ,

whereas an exponential algorithm scales as the number A itself, since

τexp ∼ exp [β log A] ∼ Aβ.

Let us discuss then the primality test. In its simplest version it consists inthe Eratosthenes’s sieve. This means, in practice, testing the divisibility ofthe number A for all numbers B ≤ √

A, hence

τ ' A1/2,

and we have an exponential algorithm! Only recently, it has been proved byAgrawal et al. that there exists a primality test with polynomial complexity.The same story happens for the factorization problem. In fact, is its naiveimplementation, the factorization algorithm scales as well as τ ' A1/2. Untilnow, it doesn’t still exist a factorization algorithm with polynomial complex-ity.Let us see how Quantum Mechanics may help in this respect.


4.5 A Primality Test in Quantum Mechanics

Suppose we have constructed a potential V (x)which has as eigenvalues theprime numbers sequence (this can be reasonably done in a semi–classicalapproximation). We can construct an algorithm for a primality test withzero complexity, i.e. it does not scale or depend on the number of digits ofthe number A under check!

FIGURE

Let us denote by G (in honour of Gauss) the device furnishing the primalitytest. If a wave of energy E = ~ωA is transmitted, A is a prime. Otherwise,it is composite, and it remains the problem of determining its factors :A = p1 · ... · pk.For simplicity, assume they are all different; the following discussion can beeasily generalized. The question is: how Quantum Mechanics does solve thefactorization problem?

4.6 Factorization Problem in Quantum Mechanics

Quantum Mechanics works well with sums of numbers rather than withproducts. Hence take initially the logarithm of the numbers under checkand promote them to energies. From the unique factorization theorem, weknow that any number can be written in an unique way as a product ofprimes: A = p1 · ... · pk, so

log A =k∑

i=1

log pi.

In the last expression, single out one term (any of them), and write

log A = log pj +k∑

i6=j

log pi ≡ log pj + log Nj ,

where Nj = A/pj .Let us construct now (by the inversion formula) a potential Vlog p (x)whichpossesses as eigenvalues all and only the logarithms of primes. Let us con-struct also another potential Wln N (y)whose eigenvalues, instead, are all

4.6. FACTORIZATION PROBLEM IN QUANTUM MECHANICS 85

the logarithms of the integer numbers. Consider now the 2–dimensionalHamiltonian

H (x, y) =1

2m

[p2

x + p2y

]+ V (x) + W (y) .

With respect to this Hamiltonian, the eigenvalues associated with a num-ber A of k prime factors is k–times degenerate. This is obvious since onecan choose any term ln pj in the previous decomposition. Denote the k–thelement of the basis of this degenerate level as

|pj〉∣∣∣Nj

⟩.

Suppose that we switch on an energy precisely equally to log A into thesystem. According to Von Neumann’s theorem, the system immediatelyafter will be in a linear combination of these k states:

|ψ (t)〉 =k∑

j=1

cj |pj , Nj ; t〉 .

This state, under time evolution, remains always in the original degeneratesubspace.Suppose now that we send on the system an electromagnetic wave, polarizedprecisely along the y–axis. In a dipole approximation the system couples tothe operator y. This operator induces a transition among the levels of theW (y) potential, in particular the most important ones are between the nextneighborhoods

|ln N〉 ↗↘|ln N + 1〉

|ln N + 1〉and the spectrum of emission/absorption of the system becomes the one ofk multiplets.

FIGURE

If the original system was an harmonic oscillator, the splitting of these mul-tiplets would be always the same, since the energy levels of the harmonicoscillator are equally spaced.

FIGURE


But is this case the potential along y has eigenvalues which are log N , hence

∆E± = ln(N ± 1)

N' ± 1

N,

i.e. the separation of the energy levels contains information on the level. Inthis way, we have two basic informations.

1. How many factors the number A is made of, since it is equal to thenumber of multiplets,

2. From their splittings we can derive 1/Ni, taken the largest one (i.e. theone associated to the smallest Ni), one determines Ni. Dividing theoriginal number by Ni, one extracts pi and the problem has reducedits difficulty to a step less.

Continuing in this way, in k measurements one ends up with a completefactorization of the original number A.

4.7 Elementary facts about prime numbers

• Each odd prime number is always of the form

4n± 1.

• All prime numbers of the form 4n + 1 can be written as sum of twosquares, while the other ones cannot.

4.8 Fermat’s little theorem

This is a famous result in elementary number theory.

Theorem 2. If a is any integer and p is a prime, then ap − a is divisibleby p, or

ap−1 ≡ 1 (modp) .

This theorem has been generalized by Euler. He introduced the functionφ (m), called Euler’s φ function or totient function. It counts the numberof positive integers r smaller than m that are coprime to m.Example. m = 10, r = 1, 3, 7, 9, φ (10) = 4.

4.8. FERMAT’S LITTLE THEOREM 87

For a prime number p, each of the previous r = 1, 2, ..., p − 1 is coprime top, hence

φ (p) = p− 1.

It is easy to prove that

φ (pα) = (p− 1) pα−1 = pα

(1− 1

p

).

In fact, among the pα numbers less than the number φ (pα), pα−1 are divisibleby p, hence the result.The Euler function is a multiplicative function, i.e.

φ (n ·m) = φ (n)φ (m) if (n, m) = 1.

Therefore, it is easy to calculate for any number m =∏

i pεii

φ (m) =∏

i

φ (pεii ) =

∏

i

(pi − 1) pεi−1i = m

∏

i

(1− 1

pi

).

Notice that this function depends on the multiplicative properties of theconsidered number so, it has wild jumps

φ (n) = 1, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8, 8, 16, ...

n = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17...

With the above definitions, the Euler theorem states that

bφ(m) ≡ 1 (modm) if (b,m) = 1.

We will estimate later on the behaviour of φ (m) and its average properties.Note that

φ (n)n

=∏

pi|n

(1− 1

pi

).

We can convert this product over primes dividing n into a product over allprimes by a probabilistic argument. Indeed, the probability that any “old”prime will divide n is 1/pi and the probability that it will not is 1 − 1/pi.In the latter case, the prime pi “contributes” the factor 1 to the product.Hence

∏

pi|n

(1− 1

pi

)'

∏pi

[(1− 1

pi

)1pi

+ 1 ·(

1− 1pi

)]=

∏pi

(1− 1

p2i

).

Hence 〈φ (n) /n〉 ' (∑n

1n2

)−1 = 6π2 .


4.9 Prime numbers and polynomials

We mentioned before that it does not exists a close formula for the value ofprime numbers. However, there is, first of all, an excellent estimate of then–th prime, given by

pn ' n log n + n (log log n− 1) + O

(n log log n

lnn

).

This expression can be derived by knowing the function∏

(x) which countsthe primes ≤ x.It is also interesting to notice that there is a remarkable number of polyno-mial formulas which furnish prime numbers for long sequences of consecutiveintegers. Such is, for example, the polynomial

P (x) = x2 + x + 41,

(Euler!) . P (x) is prime for the 40 consecutive values x = 0, 1, 2, ..., 39!Observe that

Lemma 1. A polynomial with integer coefficients cannot take on primevalues for all integral values of the argument.

Proof. Suppose to the contrary that P (n) is prime for every n. Letα be an arbitrary positive integer and put β = P (α). Consider now thesequence

P (α + β) ; P (α + 2β) ;P (α + 3β) ...

By virtue of the binomial theorem, for each n ∈ N

P (α + nβ)− P (α) = integral multiple of β.

Hence P (α + nβ) is also an integral multiple of β and, since this value mustbe a prime, it can be only β. Thus

P (α + nβ)− P (α) = P (α + nβ)− β = 0

for every positive integer n , contradicting the fact that a polynomial canonly have finitely many roots.Other remarkable polynomial expressions are

P (n) = n2 − 79n + 1601,

with P (n) prime for n = 1, 2, ...79.

4.10. DISTRIBUTION OF PRIME NUMBERS 89

P (n) = n2 + n + 17

prime for n = 0, ..., 16,

P (n) = 2n2 + 29,

prime for n = 0, 1, ..., 28.These quadratic expressions plotted on a spiral lattice (square spiral) pro-duce straight lines!

16 15 14 135 4 3 126 1 2 117 8 9 10

.

What happens on a triangular spiral lattice?

4.10 Distribution of prime numbers

How are prime numbers distribute among the integer ones? This questionwas posed for the first time by Gauss and it has given rise to extraordinarydevelopments in analytic number theory, culminating with the work of Rie-mann and the prime number theorem by Chebyshev, de la Vallee–Poussinand Hadamard. The Gauss’s contribution was almost empirical, i.e. he tookthe table of primes known at that time and he simply counted the relativefrequency of primes among the integers. It should be noticed that

(i) the primes become rarer and rarer, larger they get(ii) apart a certain regularity in their mean density, their distribution

seems rather irregular.In few words, prime numbers resemble the behaviour of statistical mechan-ics, i.e. smooth properties of the ensemble can be rather irregular for andparticular representative.The determination of π (x) by Riemann will be the topic of the future lec-tures. Here we will present a simple argument which permits to derive thecorrect asymptotic estimate

π (x) ' x

ln x

It should be noticed that the close formula for the n–th prime numbercould be obtained if one knew how to invert the function π (x), i.e.

pn = π−1 (n) .


FIGURE

The simple proof is based on the observation that the probability that anarbitrary integer is divisible by the prime pi is 1/pi. Said in another way,one number over pi is precisely divisible by it (Eratosthenes’s sieve!). Thus,the probability that it is not divisible by pi is (1− 1/pi). Assuming now thatdivisibility by different primes is not correlated (we will see that, actually,it is not the case), the probability that a number x is not divisible by anyprime below it is expressed as

W (x) '(

1− 12

) (1− 1

3

) (1− 1

5

)... '

∏

pi≤x

(1− 1

pi

).

If x is not divisible by any prime below it, it is of course a prime. Takingthe logarithm of both terms and expanding it for small 1/pi, we have

ln W (x) '∑pi<x

ln(

1− 1pi

)'

∑

i

1pi

.

The last sum on primes can be expressed as a sum on integers by using thesame function W (x). In fact,

∑

i

1pi'

∑n

W (n)n

'∫ n

1

W (t)t

dt.

Hence, we arrive to the expression

log W (x) ' −∫ x

1

W (t)t

dt.

Deriving both terms we obtain the differential equation

1W 2 (x)

dW

dx= −1

x,

whose solution isW (x) ' 1

log x.

This function expresses the probability that an integer x is a prime. Theirnumber up to x is then

Π (x) =∫ x

1dtW (t) '

∫ x

1

dt

ln t≡ Li (x) ' x

lnx,

4.10. DISTRIBUTION OF PRIME NUMBERS 91

i.e., the result by Gauss.We will say that the exact expression for the smooth part of Π (x) will begiven by

Π (x) ≡ R (x) = Li (x)− 12Li

(√x)− 1

3Li

(3√x

)+ ...

≡∞∑

n=1

µ (n)n

Li(x

1n

),

where µ (n) ,n = p1...pk is the Moebius coefficient, defined by

µ (n) =

1 n = 10 ...

(−1)k otherwise

.

More surprising is the fact that there exists a close exact formula for Π (x),able to follow also its jumps. This formula involves the nontrivial zeros ofthe Riemann zeta function. Indeed

Π (x) = limk→∞

Rk (x) ,

where

Rk (x) ≡ R (x) +k∑

l=−k

R (xρl) ,

and ρl is the l–th zero of the Riemann zeta function

ζ (s) =∞∑

n=1

1ns

.

The first nontrivial zeros have all real part equal to 1/2, whereas their imag-inary part is given by

TABLE


4.11 Probabilistic methods

By using the previous result on the distribution of primes, we can have afair estimate of several questions in number theory. In the following we willdiscuss some of them.

∑

i

1pi→∞

since ∑

i

1pi'

∑n

W (n)n

'∫ Λ

1

dn

n ln n' log log Λ.

4.11.1 Coprime probability

What is the probability that two numbers (a, b) randomly chosen are co–primes?

The probability that one of them is divisible by pi is 1pi

and the proba-bility that both are divisible by the same prime, assuming no correlations,is 1/p2

i . Hence the probability that they are not both divisible by pi is

P =∏

i

(1− 1

p2i

)=

6π2

.

Since

∏

i

(1− 1

p2i

)−1

=∏

i

(1 +

1p2

i

+1p4

i

+ ...

)=

∞∑

n=1

1n2

=π2

6.

4.11.2 Square–free probability

What is the probability that a randomly chosen integer n is square free (i.e.not divisible by a square)? The probability is equal to the above one, i.e.

P =6π2

= 0.608...

In fact, if an integer is square free, it must not be divisible for the sameprime pi more than once. Hence, either it is not divisible by pi or, if it is, itis not divisible again. Thus

Pi '(

1− 1pi

)+

1pi

(1− 1

pi

)= 1− 1

p2i

,

and taking the product on all i, we obtain the above result.

4.12. MERSENNE NUMBERS 93

4.11.3 Merten’s formula

The previous estimate of the product on primes can be refined as

∏pi<x

(1− 1

pi

)' e−γ

ln x,

where γ is the Euler–Mascheroni constant

γ = 0.57721..., e−γ = 0.5614...

The above number gives, somehow, an estimate of the correlation amongprime numbers. This means that in the previous derivation of the “primenumber theorem”, we should take the product on the primes not less thanx (this would be excessive any way), but

∏

pi<xexp(−γ)

(1− 1

pi

)' 1

ln x.

4.12 Mersenne numbers

These are the primes of the form

Mp = 2p − 1,

where p is a prime (observe that if p were not a prime, say p = n ·m, wewould have

2n·m − 1 = (2n − 1)(2n(m−1) + 2n(m−2) + ...1

),

i.e. the number is composite). Moreover, not for any prime p is 2p − 1 aprime number.Notice that any Mersenne number has as a companion a perfect even num-ber, i.e. a number equal to the sum of its divisors:

P = Mp2p−1.

The above formula is a necessary and sufficient condition in order for aneven number to be perfect. It is still an open question whether there existodd perfect numbers.


There exists a remarkable conjecture, due to Wagstaff, concerning thesenumbers. Given a number of the form 2p − 1, the probability that it isprime is about

1ln (2p − 1)

' 1p ln 2

.

Among the numbers of this form, the Mersenne’s ones are quite special.Indeed, it can be shown that if a prime q divides 2p − 1, then q must be ofthe form q = kp + 1 for some k ∈ N. So, no “small” primes can divide anumber of the form 2p − 1. Qualitatively, this implies that such numbershave a bigger probability of being primes than the “prime theorem” estimate.To quantify it, we will argue on the reverse, namely, the fact that for eachprime q < p, the probability that q divides 2p− 1 is zero. This increases the

probability of 2p − 1 to be a prime by(1− 1

q

)−1. Therefore, for all primes

less than p, we have∏q<p

(1− 1

q

)−1

' eγ ln p.

Combining now the two probabilities, together with the one that p is aprime, we obtain for the probability that 2p − 1 is prime,

eγ ln p

ln p ln (2p − 1)' eγ

ln 2p.

Therefore, for the total number of Mersenne primes less than x we have∏

Mers (x) ' eγ

ln 2

∑

p<ln x

1p

=eγ

ln 2ln lnx.

There is a remarkable fit of this formula. This can also be expressed asfollows: If x is the n–th Mersenne number, so that

n =∏

Mers (x) ,

Thenlog log x ' ln 2

eγn.

This conjecture in turn implies, since log log x → ∞, that there exist in-finitely many Mersenne primes. But the quantity of Mersenne numbers lessthan x os approximately π (log x), i.e. the quantity of primes less than log x(because log2 xdiffers only by a constant factor log 2 from log x). Since

π (log x) ' log x

log log x

4.13. ARITHMETICAL FUNCTIONS 95

is much bigger than log log x, the Wagstaff conjecture implies also that thereare infinitely many composite Mersenne numbers.

4.13 Arithmetical functions

Sequences of real or complex numbers play a major role in number theory.They are called arithmetical functions.

Definition 2. An arithmetical function is a function f: N→ R or C.

There are many classical examples of arithmetical functions.

Definition 3. An arithmetical function f (n) is said to be multiplicative if

f (n ·m) = f (n) · f (m) , (n,m) = 1.

The multiplicative function f (n) is said to be completely multiplicativeif

f (n ·m) = f (n) · f (m) , for all n, m ∈ N.

4.14 Dirichlet characters

There are also periodic arithmetical functions, which are sequences of realof complex numbers, periodic with period T :

f (n + T ) = f (n) .

The most important example is provided by the Dirichlet characters.

Definition 4. A Dirichlet character of the conductor T is a function χ (n)satisfying the following axioms:

1) periodicity: χ (n + T ) = χ (n)2)multiplicativity: χ (nm) = χ (n) χ (m)3) χ (n) = 0, if (n, T ) 6= 1.

The principal character χT of conductor T is given by

χT (n) ={

1 if (n, T ) = 10 if (n, T ) 6= 1

.

This definition is motivated by the role played by Dirichet characters ingroup theory.


Chapter 5

The Riemann ζ function

The Riemann ζ (s) function is defined for Res ≥ 1 by the following relation

ζ (s) =∞∑

n=1

1ns

. (5.1)

The series is absolutely convergent for σ > 1, where s = σ+it. The followingresult holds:

∞∑

n=1

1ns

=∏pi

1(1− 1

psi

) . (5.2)

Expanding the factor(1− 1

psi

)−1in the domain σ > 1, we have

∏pi

1(1− 1

psi

) =∏

i

[1 +

1ps

i

+1

p2si

+1

p3si

+ ...

]=

1 +(

1ps1

+ ... +1ps

k

)+

(1

ps1p

s2

+ ...

)+ ...

Each term of the product is an integer, due to the unique factorizationtheorem, hence one obtains eq. (5.2). The above formula (as well as theothers which we will discuss below) is actually a particular example of amore general formula involving multiplicative functions.For these functions we have

f(pα11 pα2

2 ...pαkk

)= f (pα1

1 ) f (pα22 ) ...f

(pαk

k

)

97

98 CHAPTER 5. THE RIEMANN ζ FUNCTION

and the general identity∞∑

n=1

f (n)ns

=∏pi

[1 +

f (pi)ps

i

+f

(p2

i

)

p2si

+f

(p3

i

)

p3si

+ ...

].

The identity (5.1) was already established by Euler, but it was Riemann whohad the idea of studying the ζ function in terms of the complex variable s.It is easy to see that ζ (s) has no zeroes for σ > 1. In fact, for σ > 1, we canuse its product representation of this function and we have(

1− 12s

)(1− 1

3s

)(1− 1

4s

)· ... ·

(1− 1

ps

)ζ (s) = 1 +

1ms

1

+1

ms2

+ ...

where m1, m2,... are all integers whose prime factors exceed P . Hence∣∣∣∣(

1− 12s

)(1− 1

3s

) (1− 1

4s

)· ... ·

(1− 1

ps

)ζ (s)

∣∣∣∣ ≥→ 1− 1(p + 1)σ−...− 1

(p + k)σ > 0

if P is large enough. Therefore |ζ (s)| > 0.Said in another way, a product can vanish if and only if at least one of thefactors vanishes. Since the general expression is

psi

psi − 1

=es ln pi

es ln pi − 1=

eσ ln p [cos (t ln p) + i sin (t ln p)]eσ ln p [cos (t ln p) + i sin (t ln p)]− 1

,

we see that the generic term never vanishes. On the other hand, the points = 1 is singular for ζ (s).

5.1 Relation between ζ (s) and∏

(x)

There is a deep relation between the behaviour of ζ (s) and the functionπ (x), which counts the primes less than x. It can be seen as follows. Forσ > 1 (where Euler’s formula is valid), we have

log ζ (s) = −∑

p

ln(

1− 1ps

)= −

∞∑

n=2

[π (n)− π (n− 1)] ln(

1− 1ns

)=

−∞∑

n=2

π (n)[ln

(1− 1

ns

)− ln

(1− 1

(n + 1)s

)]=

∞∑

n=2

π (n)∫ n+1

n

s

x (xs − 1)dx = s

∫ ∞

2

π (x) dx

x (xs − 1),

i.e.ln ζ (s) = s

∫ ∞

2

π (x) dx

x (xs − 1).

5.2. MOEBIUS FUNCTION 99

5.2 Moebius function

For σ > 1, we define

ZM (s) =1

ζ (s)=

∏p

(1− 1

ps

)=

∞∑

n=1

µ (n)ns

, (5.3)

where

µ (n) =

1 n = 1(−1)k n = p1...pk, pi 6= pj

0 n = p21...

,

i.e. µ (n) is zero each time that the number n contains more than once thesame factor pi. From this respect, prime numbers appear as “fermionic”particles!The arithmetic function µ (n) is known as the Moebius function. It plays afundamental role in analytic number theory, since it rules the convolutionof series. Let us study its main properties.First of all ∑

d|qµ (d) =

{1 q = 10 q > 1

,

This identity follows from the relations

1 = ZM (s) ζ (s) =∞∑

m=1

1ms

∞∑

n=1

µ (n)ns

=∞∑

q=1

1qs

∑

d|qµ (d) .

The Moebius function enters the so called inversion formulas, i.e.

g (q) =∑

d|qf (d)

⇓f (d) =

∑

d|qµ

(q

d

)g (d) .

The above formula can be written in a variety of ways:

g (x) =∞∑

n=1

f(x

n

)↔ f (x) =

∞∑

n=1

µ (n) g(x

n

), (5.4)

H (x) =∞∑

n=1

h (nx) ↔ h (x) =∞∑

n=1

µ (n) H (nx) . (5.5)


Let us see the proof of (5.4). Making the change of variable x = t/m andsumming on m, we get

∞∑

m=1

f

(t

m

)=

∞∑

m=1

∞∑

n=1

µ (n) g

(t

nm

)=

∑q

g

(t

q

) ∑

d|qµ (d) = g (x) ,

since∑

d|q µ (d) = δq,1.

5.3 The Mangoldt function Λ (n)

Taking the logarithm and differentiating eq. (5.2), we have for σ > 1

ζ ′ (s)ζ (s)

= −∑

p

ln p

ps

(1− 1

ps

)−1

= −∑

p

ln p∞∑

m=1

1pms

=

= −∞∑

n=2

Λ (n)ns

,

where

Λ (n) ={

log p ifn = p or a power of f0 otherwise

.

Integrating we have

log ζ (s) =∞∑

n=2

Λ1 (n)ns

,

where Λ1 (n) = Λ (n) / lnn.The Riemann function enters a variety of identities. For instance,

ζ (s− 1)ζ (s)

=∞∑

n=1

φ (n)ns

σ > 2,

where φ (n) is the Euler function. In fact

ζ (s− 1)ζ (s)

=∏p

(1− p−s

1− p1−s

)=

∏p

{(1− 1

ps

)(1 +

p

ps+

p2

p2s+ ...

)}=

∏p

{1 +

(1− 1

p

) (p

ps+

p2

p2s+ ...

)}.

5.4. ANALYTICITY PROPERTIES OF ζ (S) 101

Observe that, if n = pα11 ...pαk

k

φ (n) = n

(1− 1

p1

)...

(1− 1

pk

).

Hence we obtain the above formula.We also have

∑p

1ps

=∞∑

n=1

µ (n)n

log ζ (ns) .

To prove it, notice that

log ζ (s) =∑m

∑p

1mpms

=∞∑

m=1

P (ms)m

,

with P (s) =∑

p p−s.Hence

∞∑

n=1

µ (n)n

log ζ (ns) =∞∑

n=1

µ (n)n

∞∑

m=1

P (mns)m

=

∞∑

r=2

P (rs)r

∑

n|2µ (n) = P (s) .

Exercise. Prove that ∞∏

p=2

p2 + 1p2 − 1

=52.

5.4 Analyticity properties of ζ (s) and its functionalequation

The function ζ (s) is regular for all values of s except s = 1, where there isa simple pole with residue 1. It satisfies the functional equation

ζ (s) = 2sπs−1 sinπs

2Γ (1− s) ζ (1− s) .

This equation can be written in a more symmetric way: introducing thefunction

φ (s) ≡ 12s (s− 1)π−

s2 Γ

(s

2

)ζ (s)


we haveφ (s) = φ (1− s) .

We know that ζ (s) is defined for Res > 1. How can we extend analyti-cally its definition to the complex plane? Consider the Γ function: Γ (s) =∫∞0 dxxs−1e−x . With the substitution x → nx, we have

Γ (s) = ns

∫ ∞

0dxxs−1e−nx,

henceΓ (s)ns

=∫ ∞

0dxxs−1e−nx.

Summing now on n , we have

Γ (s) ζ (s) =∫ ∞

0dx

xs−1

ex − 1.

Consider now the integral

I (s) =∫

C

zs−1

ez − 1dz,

where the contour C starts at infinity on the positive axis, encircles theorigin once in the positive direction excluding the points ±2πi, ±4πi, . . .and returns to the positive axis again.

FIGURE

If Res > 1, we can shrink the circle around the origin to zero, so that

I (s) = −∫ ∞

0

xs−1

ex − 1dx +

∫ ∞

0

(xe2πi

)s−1

ex − 1dx =

(e2πis − 1

) ∫ ∞

0

xs−1

ex − 1dx =

(e2πis − 1

)Γ (s) ζ (s) =

2πieiπs

Γ (1− s)ζ (s) .

Since Γ (s) Γ (1− s) = πsin πs , we obtain

ζ (s) =e−iπsΓ (1− s)

2πi

∫

C

zs−1

ez − 1dz.

This formula has been proved for Res > 1. However, the integral is uni-formly convergent in any finite region of the s–plane and so it is an integral

5.4. ANALYTICITY PROPERTIES OF ζ (S) 103

function of s. Hence, it defines the analytic continuation of ζ (s) for allvalues of s. The only possible singularities are the poles of Γ (1− s), i.e.s = 1, 2, 3, .... We know already that ζ (s) is regular at s = 2, 3, 4... and thisfollows immediately from Cauchy theorem (the contour, in these cases, doesnot encircle any singularities).The only possible singularity is a simple pole at s = 1. In this case

I (1) =∫

C

dz

ez − 1= 2πi,

Γ (1− s) = − 1s− 1

+ ...

Thereforeζ (s) ' 1

s− 1, s → 1.

Notice that if s is any integer (negative or null), the integrand in I (s) isone–valued, and I (s) can be evaluated by the theorem of residues. Since

x

ex − 1=

∞∑

n=0

Bnxn

n!,

we haveζ (0) = −1

2, ζ (−2m) = 0,

ζ (1− 2m) = (−1)m B2m

2m.

In order to derive the functional equation of ζ (s), consider the integral alongthe contour CM

FIGURE

Inside the contour there are now poles at

±2πi,±4πi, ...,±2nπi.

The residues at 2mπi and −2mπi are together

(2mπe

iπ2

)s−1+

(2mπe

3iπ2

)s−1= (2mπ)s−1 eiπ(s−1)2 cos

π (s− 1)2

=

−2 (2mπ)s−1 eiπs sinπs

2.


Hence

I (s) =∫

CM

zs−1

ez − 1dz + 4πieiπs sin

πs

2

n∑

m=1

(2mπ)s−1 .

Consider now Res < 0, and n →∞. The function 1/ (ex − 1) is bounded onthe contour C and zs−1 = 0

(|z|s−1

), hence

∫CM

→ 0, i.e.

I (s) = 4πieiπs sinπs

2

∞∑

m=1

(2mπ)s−1 = 4πieiπs sinπs

2(2π)s−1 ζ (1− s) ,

from which one derives the functional equation.

5.4.1 Consequences of the functional equation

• Link with the Bernoulli numbers:

ζ (2m) = 22m−1π2m B2m

(2m)!

• For the derivative at zero:

ζ ′ (0) = −12

ln 2π.

From the functional equation,

−ζ ′ (1− s)ζ (1− s)

= − ln 2π − 12π tan

πs

2+

Γ′ (s)Γ (s)

+ζ ′ (s)ζ (s)

.

Near s ' 1, 12π tan πs

2 ' − 1s−1 , and Γ′(s)

Γ(s) = −γ, so

ζ ′ (s)ζ (s)

=− 1

(s−1)2+ A + ...

1(s−1) + γ + A (s− 1)

= − 1s− 1

+ γ.

Hence

−ζ ′ (0)ζ (0)

= − ln 2π.

The functional equation for the Riemann function can be derived in manydifferent ways. One is, for instance, to use the Poisson resummation formula

+∞∑−∞

f (n) =+∞∑−∞

∫ +∞

−∞f (x) cos 2πnxdx.

5.5. SOME CONSEQUENCES OF EULER’S FORMULA 105

Let us sketch the original proof by Riemann.Make the change of variable x → n2πx in the Γ function Γ

(s2

):

∫ ∞

0x

s−12 e−n2πxdx =

Γ(

s2

)

nsπs2

.

For Res > 1, summing on n, we have

Γ(

s2

)ζ (s)

πs2

=∫ ∞

0x

s2−1ψ (x) dx

ψ (x) ≡∞∑

n=1

e−n2πx.

This function is a Jacobi θ–function. It satisfies the basic functional identi-ties

+∞∑−∞

e−n2πx =1√x

+∞∑−∞

e−n2 πx ,

2ψ (x) + 1 =1√x

[2ψ

(1x

)+ 1

].

After some manipulations, we get

Γ(s

2

)ζ (s) π−

s2 =

∫ 1

0x

s2−1ψ (x) dx +

∫ ∞

1x

s2−1ψ (x) dx =

1s (s− 1)

+∫ ∞

1

[x−

s2−1 + x

s2−1

]ψ (x) dx.

The last integral is convergent for all values of s and therefore this formularemains true, by analytic continuation, for all values of s. But the righthand side is invariant under the substitution s → 1− s. Hence

π−s2 Γ

(s

2

)ζ (s) = π−

12+ s

2 Γ(

1− s

2

)ζ (1− s) .

5.5 Some consequences of Euler’s formula

The Euler’s formula reads

ζ (s) =∞∑

n=1

1ns

=∞∏p

(1− 1

ps

)−1

.


The relation with the function π (x):

ln ζ (s) = −∑

p

ln(

1− 1ps

)= −

∞∑

n=2

[π (n)− π (n− 1)] ln(

1− 1ns

)=

−∞∑

n=2

π (n)[ln

(1− 1

ns

)− ln

(1− 1

(n + 1)s

)]=

=∞∑

n=2

π (n)∫ n+1

n

s

x (xs − 1)dx = s

∫ ∞

2

π (x) dx

x (xs − 1).

Also,1

ζ (s)=

∏p

(1− 1

ps

)=

∞∑

n=1

µ (n)ns

.

The Moebius function has the property∑

d|qµ (d) = δq,1.

It is easy to prove:

1 =∞∑

m=1

1ms

∞∑

n=1

µ (n)ns

=∞∑

q=1

1qs

∑

d|qµ (d) .

Moebius inversion formula:

g (x) =∞∑

n=1

f(x

n

)

f (x) =∞∑

n=1

µ (n) g(x

n

).

Indeed:

f( x

m

)=

∞∑

n=1

µ (n) g( x

mn

);

∑m

f( x

m

)=

∑m.n

µ (n) g( x

mn

)=

∑q

∑

d|qµ (q) g

(x

q

)= g

(x

1

)= g (x) .

Finally, observe that

5.6. DIRICHLET SERIES 107

∑p

1ps

=∞∑

n=1

µ (n)n

log ζ (ns) ,

with

ln ζ (s) =∑m

∑p

1mpms

=∞∑

m=1

F (ms)m

,

where F (s) =∑ 1

ps .Hence,

∑ µ (n)n

ln ζ (ns) =∑n,m

µ (n)nm

F (mns) =∞∑

r=1

Γ (rs)s

∑

n|rµ (r) = Γ (s) .

5.6 Dirichlet series

Dirichlet series are one of the most relevant and classical topics in analyticnumber theory. Here we will follow strictly the book by J.–P. Serre [10].Let a (n) be a complex–valued arithmetic function.

Definition 5. A Dirichlet series with exponents {λn} is a series of the form

∞∑

n=1

a (n) e−λnz, (5.6)

where {λn} is a sequence of real numbers tending to infinity, and z ∈ C.

The simplest example is when λn = n for all n. Then, putting e−z = t, theseries (5.6) is a power series in t.Another nontrivial case is λn = log n. This reduces the series (5.6) to anordinary Dirichlet series.We remind that, if the coefficients a (n) are bounded, then ϕ (s) is absolutelyconvergent for Re(s) > 1.

Definition 6. An (ordinary) Dirichlet series with coefficients a (n) is aseries of the form

ϕ (s) =∞∑

n=1

a (n)ns

.


The most known examples of Dirichlet series are the Riemann zeta functionζ (s) =

∑∞n=1

1ns and the L–series

L (s, χ) =∞∑

n=1

χ (n)ns

,

corresponding to a Dirichlet character χ of conductor T . Other interestingexample are

ζ (s− 1)ζ (s)

=∞∑

n=1

φ (n)ns

,

where φ (n) is the totient function, and

ζ (s) ζ (s− a) =∞∑

n=1

σa (n)ns

,

where σa (n) is the divisor function. Also, the logarithm of the zeta functionis expressed, for Re(s) > 1, by the Dirichlet series

log ζ (s) =∞∑

n=2

Λ (n)log (n)

1ns

,

where Λ (n) is the von Mangoldt function. A very simple result holds for thelogarithmic derivative of a Dirichlet series Φ (s) =

∑∞n=1

f(n)ns , corresponding

to a completely multiplicative function f (n):

Φ′ (s)Φ (s)

= −∞∑

n=1

f (n) Λ (n)ns

.

Chapter 6

Numerical Spectral Methods

Most of the times, the Schrodinger equation cannot be solved exactly. Inorder to determine the energy spectrum, one has to rely on some numericalmethods. Here we will discuss, in particular, two different approaches. Eachof them has certain advantages and disadvantages.The first of them is the truncated Hilbert space method, the second one thetight–binding method. Before starting the discussion, it is useful to remindsome basic formulas concerning the harmonic oscillator.

6.1 Basic formulas for the quantum harmonic os-cillator

The Hamiltonian is

H =p2

2m+

12mω2q2 =

mω2

2

(q + i

p

mω

)(q − i

p

mω

)− ~ω

2,

with[q, p] = i~.

Introducing

a =√

mω

2~

(q + i

p

mω

), a+ =

√mω

2~

(q − i

p

mω

)

we get [a, a+

]= 1.

109

110 CHAPTER 6. NUMERICAL SPECTRAL METHODS

Hence

q =

√~

2mω

(a + a+

)=

l√2

(a + a+

),

p = i

√mω~

2(a+ − a

)=

i~l

1√2

(a+ − a

),

where l is the basic length

l =

√~

mω.

In terms of a and a+, the Hamiltonian can be written as

H = ~ω(

a+a +12

).

Denoting by |n〉 the n–th eigenvector of this Hamiltonian, we have

a |n〉 =√

n |n− 1〉 ,a+ |n〉 =

√n + 1 |n + 1〉 ,

The Hilbert space of physical states is finitely generated, since it can beobtained from the vacuum state via the formula

|n〉 =1√n!

(a+

)n |0〉 .

The occupation number operator is defined by

N = a+a,

and for its eigenstates we have

N |n〉 = n |n〉 .Finally, we recall that the Schrodinger representation is defined by

aψ0 (q) =(

q +~

mω

d

dq

)ψ0 (q) = 0,

i.e.

ψ0 (q) =(mω

~π

) 14 exp

(−mωq2

2~

).

The excited states are expressed by

ψn (q) =

√1

2nn!

(mω

~π

) 14e−

mω2~ q2

Hn

(√mω

~q

), (6.1)

where the operator

Hn (x) = (−1)n ex2 dn

dxne−x2

is the generator of the classical Hermite polynomials.

6.2. TRUNCATED HILBERT SPACE METHOD 111

6.2 Truncated Hilbert space method

Let us assume for simplicity that the potential V (x) entering the Schrodingerequation is an even function, unbounded at infinity

FIGURE

The problem is to find the eigenvalues and the eigenvectors of the Hamil-tonian

H =p2

2m+ V (x) .

The THSM, first of all, amounts to choosing a basis |ψn〉 in the Hilbert spaceand computing the matrix elements

〈ψn|H |ψm〉 ≡ Hn,m.

Second, one has to truncate the above basis at a certain level N and toconsider the finite–dimensional Hamiltonian

HN = Hn,m n,m = 0, 1, ..., N

which can be diagonalized numerically. A practical way of implementing thealgorithm consists of using the harmonic oscillator basis (6.1). Its vectorsdepend on the free parameter ω. The strategy then is the following.

(i) Use the variational principle on the ground state wave function toidentify the optimal value of ω, say ω, and use this value to construct theremaining vectors of the basis.

(ii) Compute separately the matrix elements of the kinetic term T = p2

2mand of the potential term V (x).

(iii) On the harmonic oscillator basis, T has matrix elements on thediagonal, and along the next main diagonals only

T =

. . . . . .

. . . . . . . . . 0. . . . . . . . .

. . . . . . . . .

0. . . . . . . . .

. . . . . .


as evident from the representation p.(iv) The potential V (x), in general, has sparse matrix elements, given

by all the overlapping integrals

Vnm =∫

dxψn (x) V (x) ψm.

The main advantages of the method are:1) It is easy to implement.2) It furnishes reasonable estimates of the first eigenvalues.

The main disadvantages are:1) The integrals Vnm can be only poorly estimated for large values of the

quantum numbers n and m, due to the oscillatory nature of the Hermitepolynomials (the same is true for any other basis).

2) The final form of the truncated Hamiltonian HN , as mentioned above,in general possesses nonzero matrix elements in all entries. For large values ofN this may represent a problem for a fast and sufficiently accurate numericaldiagonalization of this matrix.However, for polynomial potentials, one can overcome at least the first ofthe mentioned disadvantages, since the algebraic expressions of the matrixelements can be computed without performing the numerical integration.Let’s see how this is possible.Consider the Hamiltonian

H =p2

2m+ λ |x|n ,

where n is an integer. First of all, we can identify the optimal choice of ω forthis Hamiltonian and, correspondingly, the harmonic oscillator length

l =

√~

mω.

The related representation of the operators x and p reads

x =l√2

(a + a+

),

p =i~l

1√2

(a+ − a

).

On the other hand, the above Hamiltonian can be put into and a–dimensionlessform by using the intrinsic length scale of the problem

ξ =(~2

mλb (n)

) 1n+2

,

6.2. TRUNCATED HILBERT SPACE METHOD 113

where the function b (n), as before, is chosen so that the energy levels match,for large values of the quantum numbers, with their semi–classical expres-sions.We are led to compute the universal ration of the two scales

R =l

ξ,

and by substituting the value of the optimal choice of ω,

ωop =

(2nΓ

(n+1

2

)λ~

n−22

√πm

n2

) 2n+2

,

we obtain

R =

( √π

2nb (n) Γ(

n+12

)) 1

n+2

.

The Hamiltonian can be written as

H = ξ0

[−1

41

R2

(a+ − a

)2 +bRn

2n2

(a + a+

)n]

. (6.2)

The operators a and a+ can be easily computed, since

a =

0√

1 0 0 · · ·0 0

√2 0

0 0 0√

3 00 0 0 0

√4

0 0 0 0 0. . .

a+ =

0 0 0 0 · · ·√1 0 0 0

0√

2 0 0 00 0

√3 0 0

0 0 0√

4 0. . .

.

Obviously, the truncated matrices a(N) and a+(N) do not satisfy the usual

commutation relation [a, a+

]= I.


Instead, they satisfy the relation[a(N), a

+(N)

]= IN − (N + 1) δN,N ,

i.e.

[a(N), a

+(N)

]=

11

11

. . .−N

,

so that the trace of the commutator vanishes, as it happens for any finitedimensional representation. Even with this drawback, it is however evidentthe algebraic advantage of the expression (6.2). In fact, the matrix elementsof this matrix can be computed just by sum and matrix multiplication ofthe known matrices aN and a+

N .It still remains the computational problem of diagonalizing a sparse matrix.This problem can be attacked by using a different method, the so–calledTight–Binding Method.

6.3 Tight–Binding Method

This method reduces the problem of finding the eigenvalues and the eigenvec-tors of the Schrodinger Hamiltonian to the one of diagonalizing numericallya matrix with non–vanishing matrix elements only along the main diago-nal and the upper and lower ones. However, a certain care is needed inorder to interpret correctly the results. To this aim, it is useful to discusspreliminarily the problem of a particle in a box.

6.3.1 Particle in a box

Consider the Schrodinger problem for a free particle in a box of length L:

− ~2

2m

d2

dx2ψ = Eψ,

ψ (0) = ψ (L) = 0.

FIGURE

6.3. TIGHT–BINDING METHOD 115

The solution of the problem is well known. We have

d2ψ

dx2= −2m

~2Eψ,

with

ψ = A sin

(√2mE

~2x + ϕ

).

By imposing the boundary conditions, we end up with ϕ = 0 and sin(√

2mE~2 L

)=

0, i.e. √2mE

~2L = nπ (n = ±1,±2, ...)

En =~2π2

2mL2n2, n = 1, 2, ... (6.3)

The (normalized) wave functions are given by

ψn =

√2L

sin(nπ

Lx)

.

Notice that ψn and ψ−n describe the same physical vector. Suppose nowthat we discretize the previous equation, i.e. first we divide the interval(0, L) in N steps, as in figure

FIGURE

with lattice space

ε ≡ L

N.

The wave functions take now values only on the discrete points pn ≡ nε,with ψ0 = ψN = 0. The discrete version of the second derivative term is

d2ψ (x)dx2

→ ψm+1 + ψm−1 − 2ψm

ε2.

In this way we are led to find the eigenvalues and eigenvectors of the followinglinear problem

ψm+1 + ψm−1 − 2ψm =(−2mε2

~2

)Eψm. (6.4)

This problem can be easily solved by expressing ψm as


ψm =∑

k

eikmεψk,

where the range of k will be clear from the subsequent considerations.Substituting this expression into (6.4), we have

2 cos (kε− 1) ψk = −2mε2

~2Eψk,

i.e.

E (k) =~2

m

1ε2

(1− cos kε) . (6.5)

For an infinitesimal value of εwe shall recover the previous result: expandingthe right hand side of (6.5) we have

E (k) ' ~2

m

k2

2, (6.6)

and comparing with (6.3) we obtain the discrete values of k:

kn =πn

L, n = 0,±1, ...

Since L = nε, the only relevant values are

kn =πn

nε, n = 0,±1, ...± (N − 1) .

However, due to the parity of the expression (6.5) for k → −k, and takinginto account the boundary conditions

ψ0 = ψN = 0,

we can restrict the attention to the values

kn =πn

L, n = 0, 1, ..., N − 1,

so that, finally we get

En =~2

m

N2

L2

(1− cos

πn

N

), n = 1, 2, ..., N − 1.

For the corresponding wave functions we have

ψ(n)mε =

n−1∑

p=0

sin[πpm

N

]ψ(n)

p ,

6.3. TIGHT–BINDING METHOD 117

where

ψ(n)p =

√2

Nεδn,p.

The important thing to notice is that now the spectrum of the discreteSchrodinger equation has additional features, namely the lowest part of thespectrum is similar to the actual Schrodinger equation, whereas the higherpart of the spectrum is dominated by the lattice effects and it shows thetypical form of the case

FIGURE

Consider now the discrete version of the Schrodinger equation in thepresence of a potential

− ~2

2m

1ε2

(ψm+1 + ψm−1 − 2ψm) + Vmψm = Eψm.

For studying the problem, we have to cut the potential on a finite interval(0, L). This introduces new aspects in the problem: with boundary condi-tions

ψ0 = ψL = 0.

The effective potential felt by the particle is V (x) inside the box, but V =∞ at x = 0 and x = L.

FIGURE

It is obvious that if we discretize the Schrodinger equation in terms of Npoints, we will obtain N different energy levels.However, not all of them are those relative to the potential V ! Only thosewhich take values

E > V0 = VL

may have some resemblance to the actual eigenvalues. The higher ones, infact, come just from the infinite well problem, further masked by the latticeeffects. So, typically the spectrum obtained in this way has three differentbehaviours, as shown in figure

FIGURE

The region I, the lowest part of the spectrum, probes the potential andgives an approximation of the true eigenvalues. The region II, where the


eigenvalues are larger that the value of the potential at the edge of theinterval, is simply the spectrum of the infinite well potential, for values ofthe discrete index n which allows the expansion (6.6). The region III is,instead, just a lattice artifact of the spectrum.

In order to separate the above region one has to perform a finite sizescaling, i.e. follow the evolution of the eigenvalues by changing the value ofL (increasing correspondingly the number N of discrete points), the “true”eigenvalues are those which do not change by varying L.

FIGURE

In conclusion, the tight binding method provides an Hamiltonian whichis relatively simple and fast to diagonalize numerically, given its (almost)diagonal form

. . . . . .

. . . . . . . . . 0. . . . . . . . .

. . . . . . . . .

0. . . . . . . . .

. . . . . .

The potential is treated in a local way and it is not necessary to computethe overlapping integrals. However, it is necessary to have certain carein interpreting the result and in identifying the ”true” eigenvalues of theproblem.

Bibliography

[1] T. M. Apostol, Introduction to analytic Number Theory, Springer–Verlag, Berlin, 1976.

[2] C.M. Bender, S. A. Orszag, Advanced mathematical methods for scien-tists and engineers. Springer, Berlin, 1999.

[3] C. Cohen–Tanoudji, B. Diu, F. Laloe, Quantum Mechanics, John Wileyand Sons, New York, 2005.

[4] P. Erdos, J. Suranyi, Topics in the Theory of Numbers, Springer, Berlin,2003.

[5] L. D. Landau, E. M. Lifschitz, Quantum Mechanics: non–relativistictheory, Oxford : Butterworth-Heinemann, 2002.

[6] G. Mussardo, Il modello di Ising : introduzione alla teoria dei campi edelle transizioni di fase, Boringhieri, Torino, 2007.

[7] G. Mussardo, Statistical Field Theory: An introduction to exactly solvedmodels in Statistical Physics, Oxford Graduate Texts, 2009.

[8] A. Ogg, Modular Forms and Dirichlet Series, Mathematics Lecture NoteSeries, W. A. Benjamin, Inc. 1969.

[9] J. J. Sakurai, Advanced Quantum Mechanics, Addison-Wesley, 1967.

[10] J.–P. Serre, A course in Arithmetic, Springer–Verlag, Berlin, 1973.

[11] Frontiers in Number Theory, Physics and Geometry, P. cartier, B. Julia,P. Moussa and P. Vanhove Editors, Springer–Verlag, vol. I, 2006 and vol.II, 2007.

[12] From Number Theory to Physics, M. Waldschmidt, P. Moussa, J.–M.Luck and C. Itzykson Editors, Springer–Verlag, Berlin, 1995.

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