TheRiemannZetaFunctionandtheconnectiontoHamiltoniansinPhysics UniversityofGroningen

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University of Groningen

Bachelor Thesis

The Riemann Zeta Function and the connection to Hamiltonians in Physics

Author:

D.D. Rouwhorst

Supervisors:

dr. D. Roest prof. dr. J. Top

9th July 2014

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2

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1. INTRODUCTION

In November 1859, Bernhard Riemann published his one and only paper named

“ ¨Uber die Anzahl der Primzahlen unter einer gegebenen Gr¨oße”, translated this means “On the Number of Primes Less Than a Given Magnitude”. As the title suggests Riemann was looking for an estimate of the prime numbers. In his search he used the function ζ(s) =P∞

n=0n^−s, which is only convergent for s with Re(s) > 1. But Riemann used a tool from complex analysis, called meromorphic extension, to extend the function to the entire complex plane. This function now is called the Riemann zeta function.

In his paper Riemann only mentioned briefly that he suspects all the non-real zeroes of his function to lie on the line {s ∈ C | Re(s) = 1/2}, called the critical line. Those non-real zeroes are also called non-trivial zeroes. This now is better known as the Riemann hypothesis. Still this hypothesis is neither proved nor disproved. But, since the proof or disproof of this hypothesis was not Riemann’s purpose of his paper — he only wanted an estimate for the prime numbers —, he proceeded without mentioning the hypothesis again. In 1900, David Hil- bert nominated this hypothesis as the eighth problem on his list of problems in mathematics. Whoever proves of disproves the Riemann hypothesis receives one

million dollars from the Clay Mathematics Institute. ^See^http:

//www.claymath.org/

millenium-problems/

riemann-hypothesis.

Up until now, a lot of people have been searching for a proof using different techniques, as well was and George P´olya. In the early days of quantum mechanics he suggested a physical way to verify the Riemann hypothesis.

On 3rd January 1982, George P´olya sent a letter ^See

http://www.dtc.umn.

edu/~odlyzko/polya/

for the scanned letter.

to Andrew Odlyzko. In his letter he wrote the following.

Dear Mr Odlyzko,

Many thanks for your letter of December 8. I can only tell you what happened to me.

I spent two years in G¨ottingen ending around the begin of 1914. I tried to learn analytic number theory from Landau. He asked me one day: “You know some physics. Do you know a physical reason that the Riemann hypothesis should be true?” This would be the case, I answered, if the non-trivial zeroes of the ξ-function were so connected with the physical problem that the Riemann hypothesis would be equivalent to the fact that all the eigenvalues of the physical problem are real.

I never published this remark, but somehow it became known and it is still remembered.

With best regards.

Your sincerely,

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2 1. Introduction

George P´olya

The ξ-function in this letter is the Riemann xi function and is an adapted form of the Riemann zeta function, as we will see in this thesis.

There are two main indications that the non-trivial zeroes of the Riemann zeta function have such a spectral origin, seeBourgade and Keating (2013). The first one emerged in the 1950’s in the resemblance between the Selberg trace formula and the Weil explicit formula. The second indication came in 1972 from Mont- gomery’s. The non-trivial zeroes exhibit the same repulsion as the eigenvalues of special large unitary matrices.

In 2004, already 10¹³ non-trivial zeroes were checked and the all were on the critical line,Gourdon.

Not only the proof of the Riemann hypothesis might have a physical found- ation, the Riemann zeta function itself appears in different areas of physics. It has connections with classical mechanics, quantum mechanics, nuclear physics, condensed matter physics and statistical physics, as summarized by Schumayer and Hutchinson(2011).

In this thesis we will give an introduction to the mathematical properties of the Riemann zeta function. We will state the Riemann hypothesis and the Hilbert-Pólya conjecture, which implies the Riemann hypothesis. We will look at two Hamiltonians that could provide a possible solution for the Hilbert-Pólya conjecture. We will derive some properties of the Hamiltonian that can solve the Hilbert-Pólya conjecture, by comparing the counting function for the number of non-trivial zeroes of the Riemann zeta function with the counting function for the number of energy eigenstates of physical systems.

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2. PRELIMINARIES

2.1 Sets

For clarity, we specify some sets and their notation that will appear in this thesis.

Notation 2.1 (Natural numbers). This notation agrees

with ISO 80000-2 clause 6.1.

The set of natural numbers is denoted as

N = {0, 1, 2, 3, . . .}.

Notation 2.2 (Integers). This notation agrees

The set of integers is denoted as Z = {. . . , −2, −1, 0, 1, 2, . . .}.

Notation 2.3 (Prime numbers). This notation agrees

The set of prime numbers is denoted as

P = {2, 3, 5, 7, 11, 13, 17, . . .}.

2.2 Notations

Notation 2.4 (Big-O). This notation agrees

The notation f (x) = O(g(x)), f (x) is big-O of g(x), means

f (x) g(x)

is bounded from above in the limit implied by the context.

Notation 2.5 (Little-o). This notation agrees

The notation f (x) = o(g(x)), f (x) is little-o of g(x), means

f (x) g(x) → 0.

in the limit implied by the context.

The symbol “=” in the notations is used for historical reasons and does not have the meaning of equality, because transitivity does not apply.

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4 2. Preliminaries

2.3 Complex exponential and logarithmic functions

Definition 2.1 (Natural logarithm).

This notation agrees with ISO 80000-2 clause 12.5, which also states:

“log(x) shall not be used in place of ln(x).”

The natural logarithmic function, often abbreviated as natural logarithm, ln : R>0 → R, is the function defined by

ln(x) :=

Z x 1

1 t dt.

Definition 2.2 (Argument). The argument of a complex number s is denoted as arg(s) and is defined to be the set

arg(s) := tan⁻¹ Im(s) Re(s)

which contains infinitely many values.

Definition 2.3 (Natural logarithm (continued)). If s 6= 0 is a complex number, then we define ln(s) to be the set

ln(s) :=

Z |s|

1

t dt + i arg(s),

such that it is equal to Definition 2.1 when s is restricted to R^>0.

Definition 2.4. If s is a complex number, then the complex-valued function exp : C → C is implicitly defined by

ln(exp(s)) = s.

One can also write e^s instead of exp(s).

Definition 2.5. If α is a complex constant and s 6= 0 a complex number, then s^α is defined by

s^α := e^{α ln(s)}.

Definition 2.6. If f (x) = g(x)+ih(x) is a complex-valued function of a real variable x and the real and imaginary parts g(x) and h(x) are differentiable functions of x, then the derivative of f (x) with respect to x is defined to be

f⁰(x) := g⁰(x) + ih⁰(x).

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2.4. Complex analysis 5

2.4 Complex analysis

Definition 2.7 (Holomorphic). The word holomorphic

derives from the Greek words ολος (holos), meaning “whole”, and μορφη (morphe), meaning “form” or

“appearance”.

A complex-valued function f is said to be holomorphic on an open set O if at every point s₀ of O, the limit

lim

Δs→0

f (s₀+ Δs) − f (s0) Δs

exists. This limit is called the derivative of f at s0.

Another frequently used word for holomorphic is analytic and sometimes it is called complex differentiable.

Theorem 2.1 (Cauchy-Riemann condition). Let f (s) be a complex-valued function defined in some open set O. If the first partial derivatives of f (s) with respect to Re(s) and Im(s) exist, are continuous and satisfy the Cauchy-Riemann condition

∂ f (s)

∂ Re(s) = 1 i

∂ f (s)

∂ Im(s) at all points of O, then f is holomorphic in O.

Proof. See Saff and Snider(2003) for a proof.

Definition 2.8 (Meromorphic). The word meromorphic

derives from the Greek words μερος (meros), meaning “part”, and μορφη (morphe), meaning “form” or

“appearance”.

A complex-valued function f is said to be meromorphic in a domain D (an open and connected set) if at every point of D the function is either holomorphic on neighbourhood of that point or has a pole at that point.

A complex-valued function f is said to have a pole (of order k) at s₀ ∈ D if the function g(s) = (s − s₀)^kf (s) is holomorphic in D and g(s₀) 6= 0 for a k ∈ N.

Theorem 2.2. If f is holomorphic and non-zero at each point of a simple closed positively oriented contour C and is meromorphic inside C, then

1 2πi

I

C

f⁰(s)

f (s) ds = N − P,

where N and P are, respectively, the number of zeroes and the number of poles, included multiplicity, of f inside C.

Proof. See Saff and Snider(2003) for a proof.

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6 2. Preliminaries

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3. MATHEMATICS

3.1 The Riemann zeta function

It was Euler who studied the properties of the seriesP∞

n=11/n^s for integer values of s. Later, Dirichlet looked at this series for real s greater than one. But Riemann went even further than that, he allowed s to attain complex values.

Definition 3.1 (Riemann zeta function). The Riemann zeta

function is a special type of the Dirichlet series. The Dirichlet series attached to a complex valued function χ defined on the natural numbers is given by

D(χ, s) :=

∞

X

n=1

χ(n) n^s , so ζ(s) = D(1, s).

Most results in this thesis can be generalized to other Dirichlet series.

The Riemann zeta function for s > 1.

1 2 3 4

0 2 4 6 8

s

ζ(s)

The Riemann zeta function, ζ : {s ∈C | Re(s) > 1} → C, is defined by

ζ(s) :=

∞

X

n=1

1 n^s.

This series converges when Re(s) > 1, because it converges absolutely. To show this, we will use the widely used notation σ := Re(s) and t := Im(s), so s = σ + it. With this notation we have |n^−s| =

n^−σ−it

= |n^−σ| n^−it

=

|n^−σ|

e^{−it ln n}

= |n^−σ| = n^−σ. And we know that the series P∞

n=1n^−σ converges when σ > 1.

The Riemann zeta function is now only defined for input values with real part greater than one. But Riemann meromorphically extended this function to the entire complex plane. This means he found a meromorphic function in C such that this function is equal to ζ(s) when s is restricted to {s ∈ C | Re(s) > 1}.

This meromorphic function turns out to be a holomorphic function, except at the point s = 1, there is has a simple pole, that is, a pole of order one. Moreover, this meromorphic extension is unique, that means if f (s) and g(s) are meromorphic functions in C such that they are equal to ζ(s) when s is restricted to {s ∈ C | Re(s) > 1}, then they are equal everywhere in the complex plane. This is a property of all meromorphic functions and it is called the identity theorem.

Instead of looking at two functions, we could also look at their difference.

Theorem 3.1 (Identity theorem). Let f be a holomorphic function on a domain D. If there exists a point s₀ in D such that f⁽ⁿ⁾(s₀) = 0 for all n ∈ N, then f (s) = 0 for all s in D.

Proof. Let A = s ∈ D | f⁽ⁿ⁾(s) = 0 ∀n ∈ N , which is non-empty by assump- tion. We will show that A and D \ A are open sets, which means A = D because A 6= ∅. And then we are done, because A = D implies f⁽ⁿ⁾(s) = 0 for all n ∈ N and all s ∈ D, and which implies f (s) = 0 for all s ∈ D.

(Proof that A is open) Let a ∈ A be arbitrary. Since D is open, there exists a radius R > 0 such that B(a, R) ⊆ D and

f (s) =

∞

X

n=0

f⁽ⁿ⁾(a)

n! (s − a)ⁿ

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8 3. Mathematics

for ks − ak < R. Since a ∈ A, each f⁽ⁿ⁾(a) = 0, so f is identically zero on B(a, R).

This means that B(a, R) ⊆ A. Because a ∈ A was arbitrary is holds for all points in A and since the union of open sets is again open, A is an open set.

(Proof that D \ A is open) Let b ∈ D \ A be arbitrary. Then there exists an n₀ ∈ N with f⁽ⁿ⁰⁾(b) 6= 0. Since f⁽ⁿ⁰⁾ is a continuous function, there exists an r > 0 such that f⁽ⁿ⁰⁾(s) 6= 0 on B(b, r). This means B(b, r) ⊆ D \ A. Because b ∈ D \ A was arbitrary is holds for all points in D \ A and since the union of open sets is again open, D \ A is an open set.

So D = A = s ∈ D | f⁽ⁿ⁾(s₀) = 0 ∀n ∈ N

⊆ {s ∈ D | f (s₀) = 0}. This proves the theorem.

This extended function is called the Riemann zeta function. In the next section we will show how the function in Definition 3.1 can be meromorphically extended to C with its single pole at s = 1.

The Riemann hypothesis states that all non-real zeroes of the Riemann zeta function lie on the the line {s ∈ C | Re(s) = 1/2}. The non-real zeroes are called non-trivial zeroes because the real zeroes are easy to calculate.

Conjecture 3.1 (Riemann hypothesis).

The location of the non-trivial zeroes in the complex plane if the Riemann hypothesis is true is represented by the gray line.

−1 1 2

−2

−1 1 2

Re(s) Im(s)

All non-trivial zeroes of ζ(s) have the form

ρ = 1 2 + it, where t is a real number.

3.2 The zeroes of the Riemann zeta function

The goal of this section is to get some insight in the locations of the zeroes of the Riemann zeta function. It turns out that all the non-trivial zeroes are located in the vertical strip {s ∈ C | 0 6 Re(s) 6 1}. We are going to prove this in parts.

First we will show that there are no zeroes in the plane {s ∈ C | Re(s) > 1}.

Theorem 3.2 (Euler product formula). The Riemann zeta function is equal to

ζ(s) = Y

p∈P

1 1 − p^−s, for Re(s) > 1.

Proof. ByDefinition 3.1, the Riemann zeta function is ζ(s) =

∞

X

n=1

1

n^s = 1 + 1 2^s + 1

3^s + 1 4^s + 1

5^s + · · · for Re(s) > 1. Now, we repeat the following kind of operations

1

2^sζ(s) = 1 2^s + 1

4^s + 1 6^s + 1

8^s + 1 10^s + 1

12^s + · · ·

1 − 1

2^s

ζ(s) = 1 + 1 3^s + 1

5^s + 1 7^s + 1

9^s + 1

11^s + · · · 1

3^s

1 − 1

2^s

ζ(s) = 1 3^s + 1

9^s + 1 15^s + 1

21^s + 1

27^s + · · ·

1 − 1

3^s

1 − 1 2^s

ζ(s) = 1 + 1 5^s + 1

7^s + 1 11^s + 1

13^s + 1

17^s + · · ·

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3.2. The zeroes of the Riemann zeta function 9 to obtain

· · ·

1 − 1

11^s

1 − 1 7^s

1 − 1 5^s

1 − 1 3^s

1 − 1 2^s

ζ(s) = 1

as a consequence of the fundamental theorem of arithmetic, the unique factoriz- ation of integers into prime numbers. Rewriting this equation gives

ζ(s) =

1 1 − ₂¹s

1 − ₃¹s

1 −₅¹s

1 −₇¹s

1 −₁₁¹s · · · or equivalently

ζ(s) = Y

p∈P

1 1 − p^−s for Re(s) > 1. This proves the theorem.

This equation is the famous relation between the Riemann zeta function and the prime numbers.

Proposition 3.1. The blue shaded area is

the area without zeroes.

−1 1 2

−2

−1 1 2

Re(s) Im(s)

The Riemann zeta function has no zeroes in the complex plane where Re(s) > 1. In other words, ζ(s) 6= 0 for Re(s) > 1.

Proof. Again, we will use the notation σ := Re(s) and t := Im(s), so s = σ + it.

Now we have |p^−s| = p^−σ. By Theorem 3.2 we have

|ζ(s)| =Y

p∈P

1

|1 − p^−s| >Y

p∈P

1

1 + |p^−s| =Y

p∈P

1

1 + p^−σ = exp −X

p∈P

ln 1 + p^−σ

!

= exp X

p∈P

−p^−σ +1

2p^−2σ− 1

3p^−3σ + O p^−4σ

!

> exp −X

p∈P

p^−σ

!

> 0,

because P

p∈Pp^−σ converges for σ > 1. In the fifth step, we used the Taylor expansion of ln(1 + x) around the point x = 0:

ln(1 + x) =

∞

X

n=0

ln⁽ⁿ⁾(1)

n! xⁿ= x − 1

2x²+ 1

3x³ + O x⁴,

which converges when |x| < 1. Now, |ζ(s)| > 0 or equivalently ζ(s) 6= 0, for Re(s) > 1. This proves the proposition.

To meromorphically extend the Riemann zeta function to the complex plane, we will use the so called Gamma function.

Definition 3.2 (Gamma function). The Gamma function

for s > 1.

0 2 4

0 10 20 30

s

Γ(s)

The Gamma function, Γ : {s ∈C | Re(s) > 0} → C, is defined by

Γ(s) :=

Z ∞ 0

x^s−1e^−xdx.

This notation is introduced by Legendre. Riemann used another notation, he used the notation introduced by Gauß:

Π(s) :=

Z ∞ 0

x^se^−xdx,

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10 3. Mathematics

for Re(s) > −1. Although the definition of Π(s) looks more natural, today almost everyone uses the definition of Γ(s) instead of Π(s). Therefore, we will use it here too.

Let’s proof that the Gamma function converges. We show that it even converges absolutely. Setting σ := Re(s), the absolute value of the integrand equals

|x^s−1e^−x| = x^σ−1e^−x. Splitting the integral at the point s = 1, gives Z ∞

0

x^se^−x dx =

Z ∞ 0

x^σ−1e^−xdx = Z 1

0

x^σ−1e^−xdx + Z ∞

1

x^σ−1e^−xdx

<

Z 1 0

x^σ−1dx + Z ∞

1

x^σ−1e^−xdx.

The first integral converges if σ > 0, and the second integral converges because its a product of a polynomial and e^−x.

Proposition 3.2. The Gamma function defines a holomorphic function if Re(s) >

0.

Proof. We use the Cauchy-Riemann condition, Theorem 2.1, to proof this. We have by definition of the Gamma function, Definition 3.2

∂ Γ(σ + it)

∂σ = ∂

∂σ lim

N →∞

Z N 0

x^σ+it−1e^−xdx = lim

N →∞

∂

∂σ Z N

0

x^σ+it−1e^−xdx

= lim

N →∞

Z N 0

∂ x^σ+it−1e^−x

∂σ dx = lim

N →∞

Z N 0

x^σ+it−1e^−xln(x) dx

= lim

N →∞

Z N 0

∂(it) dx = 1 i lim

N →∞

Z N 0

∂t dx

= 1 i lim

N →∞

∂

∂t Z N

0

x^σ+it−1e^−xdx = 1 i

∂

∂t lim

N →∞

Z N 0

x^σ+it−1e^−xdx

= 1 i

∂ Γ(σ + it)

∂t ,

where in steps three and seven the differentiations and integrations are interchanged by the Leibniz rule. In steps two and eight the derivatives and limits could be interchanged because of pointwise convergence of the integrals and uni- form convergence of their derivatives.

The Gamma function satisfies a special property called the Reduction For- mula, sometimes also called the functional equation of the factorial function.

Theorem 3.3 (Reduction Formula).

The Gamma function is a generalization of the factorial. If n ∈ N then

n! = Γ(n + 1) = n Γ(n)

= n(n − 1) Γ(n − 1)

= · · ·

= n(n − 1) · · · 1 Γ(1)

=

n

Y

i=1

i,

since by definition

Γ(1) = Z ∞

0

e^−xdx

= −e^−x

∞ x=0= 1.

The Gamma function satisfies the Re- duction Formula

Γ(s + 1) = s Γ(s) for Re(s) > 0.

Proof. Using integration by parts, we can see that

Γ(s) = Z ∞

0

x^s−1e^−xdx = 1 s

Z ∞ 0

e^−xdx^s= 1 sx^se^−x

∞

x=0

+1 s

Z ∞ 0

x^se^−xdx = 1

sΓ(s + 1) for Re(s) > 0, since x^s|_x=0 is only well-defined for Re(s) > 0. This implies the Reduction Formula.

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3.2. The zeroes of the Riemann zeta function 11 Yet, the Gamma function is only defined for Re(s) > 0, but we need to extend it to the entire complex plane in order to use it for the extension of the Riemann zeta function. Looking at the Reduction Formula, we see that we can calculate Γ(s + 1) using Γ(s). But we also could use this Reduction Formula to go backwards. Therefore, for −1 < Re(s) 6 0 we can define Γ(s) by

Γ(s) = Γ(s + 1)

s .

This is allowed, since Γ(s + 1) is perfectly defined for −1 < Re(s) 6 0. From this definition we see that Γ(s) has a pole at s = 0. Continuing in this way, we can define the Gamma function in the left side of the complex plane as follows.

Definition 3.3 (Gamma function (continued)). The extended Gamma function for s real.

−4 −2 0 2 4

−10

−5 0 5 10

s

Γ(s)

The Gamma function, Γ : C → C, is defined for −(n + 1) < Re(s) 6 −n, where n ∈ N, by

Γ(s) = Γ(s + n + 1) Qn

m=0(s + m).

Therefore, the Gamma function can be meromorphically extended to C, where it has all its poles at s ∈ Z60, those poles are simple poles. It turns out that the extended Gamma function is never zero.

Theorem 3.4. The Riemann zeta function in Definition 3.1 admits a meromorphic extension to C which satisfies the equation

ξ(s) = ξ(1 − s) (3.1)

where ξ(s) is the so called Riemann xi function defined by ξ(s) :=

1

2s(s − 1)π⁻^s² Γ

s 2

ζ(s). (3.2)

Proof. Substituting x = n²πy into the Gamma function gives Γ

s 2

= Z ∞

0

x^s²⁻¹e^−xdx = Z ∞

0

n²πy₂^s⁻¹

e⁻ⁿ²^πyn²π dy

= n^sπ

s 2

Z ∞ 0

y^s²⁻¹e⁻ⁿ²^πydy

for Re(s) > 0. Substitution of this result into the definition of ξ(s) gives for Re(s) > 1

2

s(s − 1)ξ(s) = π⁻

s 2 Γ

s 2

ζ(s) =

∞

X

n=1

π⁻

s 2 Γ

s 2

1 n^s =

∞

X

n=1

Z ∞ 0

y^s²⁻¹e⁻ⁿ²^πydy

= Z ∞

0

y^s²⁻¹

∞

X

n=1

e⁻ⁿ²^πy

! dy =

Z ∞ 0

y^s²⁻¹ω(y) dy,

where ω(y) :=P∞

n=1e⁻ⁿ²^πyand the sum and integral are exchanged for Re(s) > 1 because of absolute convergence. The Poisson summation formula implies the following functional equation

∞

X

n=−∞

e⁻ⁿ²^π/y =√ y

∞

X

n=−∞

e⁻ⁿ²^πy,

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12 3. Mathematics

see for exampleEdwards (1974), hence 2ω(1/y) + 1 =√

y(2ω(y) + 1), or equivalently

ω 1 y

=√

yω(y) +

√y 2 − 1

2.

The equation for ξ(s) yields, by splitting the integral at y = 1 and making a change of variables y 7→ 1/y in the first integral,

2

s(s − 1)ξ(s) = Z 1

0

y^s²⁻¹ω(y) dy + Z ∞

1

y^s²⁻¹ω(y) dy

= Z ∞

1

y⁻^s²⁻¹ω 1 y

dy +

Z ∞ 1

y^s²⁻¹ω(y) dy

= Z ∞

1

y⁻^s²⁻¹√

yω(y) +

√y 2 − 1

2

dy +

Z ∞ 1

y²^s⁻¹ω(y) dy

= Z ∞

1

y⁻^s²⁻¹ √y 2 − 1

2

dy +

Z ∞ 1

y⁻^s²⁻¹² + y²^s⁻¹

ω(y) dy,

ξ(s) = 1

2 +s(s − 1) 2

Z ∞ 1

y⁻^s²⁻¹² + y^s²⁻¹

ω(y) dy (3.3)

for Re(s) > 1. The right hand side is properly defined on C \ {1}, because ω(y) = O(e^−πy) as y → ∞. Furthermore, this equation is invariant under the change of variables s 7→ 1 − s, therefore ξ(s) = ξ(1 − s).

So, we have, combining (3.1) with (3.2), the following relation for the Riemann zeta function

π⁻

s 2 Γ

s 2

ζ(s) = π⁻

1−s

2 Γ

1 − s 2

ζ(1 − s), (3.4)

and (3.3) in combination with (3.2) gives

ζ(s) = π

s 2

Γ ^s₂

1

s(s − 1) + Z ∞

1

x⁻^s²⁻¹² + x^s²⁻¹

ω(x) dx

. (3.5)

One can see from this equation that ζ(s) is properly defined on C \ {0, 1}. Let’s take a closer look at the points s = 0 and s = 1.

First, we take a look at s = 0. We take the limit of (3.5) as s → 0. Because 1/ Γ(s) = s + o(s) as s → 0, it holds that

1 Γ ^s₂ '

s 2,

as s → 0. Substituting this into (3.5) yields the following limit lims→0ζ(s) = lim

s→0

π

s 2s 2

1

s(s − 1) + Z ∞

1

x⁻²^s⁻¹² + x^s²⁻¹

ω(x) dx

= π

s 2

2

1 s − 1 + s

Z ∞ 1

x⁻²^s⁻¹² + x^s²⁻¹

ω(x) dx

s=0

= 1

2(−1 + 0) = −1 2.

And now we take a look at s = 1. Evaluating (3.4) at this point we obtain the following relation

ζ(1) = π

1

2 Γ(0) ζ(0) Γ(1/2)

.

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3.2. The zeroes of the Riemann zeta function 13 Because Γ(0) is a simple pole and all other terms at the right hand side have a finite value, ζ(1) has to be a simple pole too.

We now have a meromorphical extension of the Riemann zeta function in C, since we can set ζ(0) = −1/2. The only pole of the Riemann zeta function, which is a simple pole, lies at s = 1.

We have already proved that there are no zeroes in the plane {s ∈ C | Re(s) > 1}.

Now we will show what the locations of the zeroes in the plane {s ∈ C | Re(s) < 0}

are.

Proposition 3.3. The blue shaded area is

the area without non-trivial zeroes.

−1 1 2

−2

−1 1 2

Re(s) Im(s)

The symbol “•”

represents the simple pole.

The only zeroes of the Riemann zeta function in the complex plane where Re(s) < 0 are located at all even negative integers.

Otherwise stated, ζ(s) 6= 0 for Re(s) < 0, except in the points where s ∈ 2Z<0.

Proof. Let us look at Re(s) < 0. All the factors on the right hand side of (3.4) are then non-zero, but the factor Γ(s/2) on the left hand side has poles at all s ∈ 2Z60. This means that, since π^−s/2 is never zero, ζ(s) has to be zero. So, for Re(s) < 0, ζ(s) = 0 if and only if s ∈ 2Z<0. Those zeroes are called the trivial zeroes. This proves the proposition.

So, we have already scored out a lot of possible positions of zeroes in the complex plane. But what about the strip {s ∈ C | 0 6 Re(s) 6 1}, can we say something about that?

First, we show that the Riemann zeta function has no zeroes in the interval 0 < s < 1.

Lemma 3.1. The notation bxc means

floor x, the least integer greater then or equal to the real number x (ISO 8000-2 clause 9.17).

For Re(s) > 0 the Riemann zeta function is equal to

ζ(s) = s s − 1 − s

Z ∞ 1

x − bxc

x^s+1 dx. (3.6)

Proof. Observe that n = bxc if

x ∈ (n, n + 1), where n ∈ N.

ByDefinition 3.1 we have for Re(s) > 1

ζ(s) =

∞

X

n=1

1 n^s =

∞

X

n=1

n 1

n^s − 1 (n + 1)^s

=

∞

X

n=1

n −1 x^s

n+1

x=n

= s

∞

X

n=1

n Z n+1

n

1 x^s+1dx

= s Z ∞

1

bxc

x^s+1dx = s

Z ∞ 1

1 x^sdx −

Z ∞ 1

x − bxc x^s+1 dx

= s

s − 1 − s Z ∞

1

x − bxc x^s+1 dx.

The second step is maybe a little clearer when observing the first few terms:

∞

X

n=1

1 n^s = 1

1^s + 1 2^s + 1

3^s + · · ·

= 1 1^s +

−1 2^s + 2

2^s

+

−2 3^s + 3

3^s

+ · · ·

= 1 1 1^s − 1

2^s

+ 2 1 2^s − 1

3^s

+ 3 1 3^s − 1

4^s

+ · · ·

=

∞

X

n=1

n 1

n^s − 1 (n + 1)^s

.

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14 3. Mathematics

But we see that the right hand side of the first equation is also valid for Re(s) > 0.

This proves the lemma.

From (3.6) we can see that for 0 < s < 1

ζ(s) −

s s − 1

= s

Z ∞ 1

x − bxc x^s+1 dx

<

s

Z ∞ 1

1 x^s+1 dx

=

1 x^s

∞

x=1

= 1, or equivalently

−1 < ζ(s) − s

s − 1 < 1.

Therefore,

ζ(s) < 1 + s

s − 1 = 2s − 1 s − 1 .

And for 1/2 < s < 1, we have (2s − 1)/(s − 1) < 0, so ζ(s) 6= 0 in the interval 1/2 < s < 1. From (3.4) we see that the zeroes of ζ(s) are symmetrically located around the line {s ∈ C | Re(s) = 1/2} in the region 0 < Re(s) < 1, since the factors in front of the Riemann zeta functions cannot be zero in this region.

Thus, ζ(s) has no zero in the interval 0 < s < 1.

3.3 Properties of the Riemann xi function

The Riemann xi function has some special properties and is often more useful than the Riemann zeta function.

Lemma 3.2.

According to ISO 80000-2 clause 14.6 the notations s and s^? both mean the complex conjugate of s, where s is meanly used in mathematics and s^? is meanly used in physics and engineering.

A holomorphic function f in C satisfies f (s) = f (s) for all s ∈ C if and only if f is real-valued when restricted to the real values.

Proof. A holomorphic function can always be expanded as a power series around zero, that is for all s ∈ C it holds that

f (s) =

∞

X

n=0

a_nsⁿ,

where an∈ C for all n ∈ N. Therefore, f (s) = f (s) ∀s ∈ C ⇔

∞

X

n=0

a_nsⁿ=

∞

X

n=0

a_nsⁿ∀s ∈ C ⇔ an= a_n∀n ∈ N

⇔ a_n ∈ R ∀n ∈ N ⇔

∞

X

n=0

a_nsⁿ∈ R ∀s ∈ R ⇔ f(s) ∈ R ∀s ∈ R,

which proves the lemma.

A list of useful properties.

• We already saw that the Riemann xi function is holomorphic in C \ {1}.

But from (3.2) and the fact that ζ(1) is a simple pole, we can see that that the Riemann xi function is holomorphic in C, since the simple zero from s − 1 cancels the simple pole from ζ(s).

• From (3.1) we can see that the Riemann xi function is symmetric in the complex plane around the line {s ∈ C | Re(s) = 1/2}.

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3.4. The counting function 15

−5 −4.5 −4 −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0.5 1 1.5 2 zero-free region

pole

trivial zeroes non-trivial zeroes

Re(s) Im(s)

Fig. 3.1: The location of zeroes of the Riemann zeta function. There can be non-trivial zeroes off the line. Here four such zeroes are depicted. But still such zeroes are not found.

−5 −4.5 −4 −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0.5 1 1.5 2 zero-free region

non-trivial zeroes

Re(s) Im(s)

Fig. 3.2: The location of zeroes of the Riemann xi function. There can be zeroes off the line. Here four such zeroes are depicted. But still such zeroes are not found.

• By (3.4), the trivial zeroes of ζ(s) are cancelled by Γ(s/2). Therefore, the zeroes of the Riemann xi function are the non-trivial zeroes of the Riemann zeta function.

•

1

−40

−20 20 40

Re(s) Im(s)

The first six non-trivial zeroes of the Riemann xi and zeta function.

From (3.6), it is clear that the restriction of the Riemann xi function to the real numbers is real-valued, and therefore satisfies ξ(s) = ξ(s). As result of this, the zeroes of the Riemann xi function, and by (3.2) also of the Riemann zeta function, are located symmetrically around the line {s ∈ C | Im(s) = 0}, the real axis.

3.4 The counting function

We need a function which counts the number of zeroes on the line {s ∈ C | Re(s) = 1/2}

of the Riemann zeta function between 0 and a given height it.

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16 3. Mathematics

Definition 3.4 (Counting function). We define the counting function, N : {T ∈ R | T > 0} → Q, by

N (T ) :=

∞

X

n=1

θ(t − t_n),

where θ : R → Q is the Heaviside step function, also knows as the unit step function, given by

θ(x) :=







0 if x < 0;

1

2 if x = 0;

1 if x > 0,

and t_n is the nth zero of the Riemann zeta function above the real line.

We can use Theorem 2.2 to calculate the number of zeroes in the region R = {s ∈ C | − 6 Re(s) 6 1 + , 0 6 Im(s) 6 T }. To calculate this number, we will use the Riemann xi function instead of the Riemann zeta function, for the following reasons. The Riemann xi function is

ξ(s) = 1

2s(s − 1)π⁻^s² Γ

s 2

ζ(s).

The term s − 1 is there to remove the simple pole of ζ(s) at s = 1, the term s makes ξ(s) symmetric and the factor 1/2 is there for historical reasons. This makes the Riemann xi function a holomorphic function in the entire complex plane. Furthermore, it has the same number of zeroes in R as the Riemann zeta function when 0 6 < 1. So, the number of zeroes of the Riemann zeta function in R is given by

N (T ) = 1 2πi

I

∂R

ξ⁰(s) ξ(s)

ds,

where ∂R is the boundary of R oriented in the usual counterclockwise direction.

We can split the contour ∂R up into two curves γ₁ and γ₂.

1 0.5 iT

−iT

Re(s) Im(s)

The curves γ₁(blue) and γ₂ (red).

The curve γ₁ consists of three straight line segments between the points (0.5, iT ), (−, iT ), (−, 0), and (0.5, 0). The curve γ₂ consists of three straight line segments between the points (0.5, 0), (1 + , 0), (1 + , iT ), and (0.5, iT ). And now we have

N (T ) = 1 2πi

Z

γ1

ξ⁰(s) ξ(s)

ds + 1 2πi

Z

γ2

ξ⁰(s) ξ(s)

ds

= − 1 2πi

Z

γ1

ξ⁰(1 − s) ξ(1 − s)

ds + 1 2πi

Z

γ2

ξ⁰(s) ξ(s)

ds

= 1 2πi

Z

γ3

ξ⁰(s) ξ(s)

ds + 1 2πi

Z

γ2

ξ⁰(s) ξ(s)

ds,

where in the second step we used (3.1) and in the third step we made a change of variables 1 − s 7→ s.

1 0.5 iT

−iT

Re(s) Im(s)

The curves γ₃(blue) and γ₂ (red).

When we changed the variables, the curve γ₁ changed to another curve, which we call γ₃. So the curve γ₃ consists of three straight line segments between the points (0.5, −iT ), (1 + , −iT ), (1 + , 0), and (0.5, 0).

By the chain rule we have

dln(ξ(s))

ds = ξ⁰(s) ξ(s)

.

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3.4. The counting function 17 By using the fundamental theorem of calculus (six times) and letting the argument of the logarithm lie in the interval (−π, π], we can evaluate the integral as follows

N (T ) = 1

2πiln(ξ(s))

1 2+iT

1 2−iT

= 1

2πiln|ξ(s)|

1 2+iT

1 2−iT

+ 1

2πarg(ξ(s))

1 2+iT

1 2−iT

= 1 2πarg

ξ

1 2 + iT

− 1 2πarg

ξ

1 2− iT

= 1 π

arg

ξ

1 2 + iT

= 1 π

arg 1 2+ iT

−1 2+ iT

+ 1 π

arg

π⁻

1 4−^iT₂

+1 π

arg

Γ

1 4+ iT

2

+ 1 π

arg

ζ

1 2 + iT

= 1 − T

2πln(π) +1 π

arg

Γ

1 4+ iT

2

+1 π

arg

ζ

1 2 + iT

. To computation arg(Γ(1/4 + iT /2)), we use the expansion of ln(Γ(s))

ln(Γ(s)) = 1

2ln(2π) +

s − 1

2

ln(s) − s + 1

12s − 1

360s³ + 1

1260s⁵ − · · · which is often called Stirling’s series and can be derived from the asymptotic series of Γ(s). The computation is as follows

arg

Γ

1 4 +iT

2

= Im

ln

Γ

1 4 +iT

2

= Im

−1 4+ iT

2

ln 1

4+ iT 2

− 1 4 +iT

2

+ 1

12 ¹₄ +^iT₂

− 1

360 ¹₄ +^iT₂3 + · · ·

!

= T 2 Re

ln 1

4 +iT 2

− 1 4Im

ln 1

4 +iT 2

− T 2 + −^T₂

12 ₁₆¹ +^T₄² − Im

1

4 − ^iT₂3 360 ₁₆¹ +^T₄²³ + · · ·

= T 2 ln

s

T 2

2

1 + 1 4T²

− 1 4

π

2− arctan

1 4 T 2

!!

− T 2

− 1

6T 1 + _4T¹2

−

T³

8 + 3 −^T₂

−¹₄2

360 ^T₄²2

1 + _4T¹2

3 + · · ·

= T 2 ln T

2

+ T

4 ln

1 + 1 4T²

−π 8+ 1

4arctan

1 2T

−T 2

− 1 6T

1 + 1 4T²

⁻¹

− 1

45T³

1 + 1 4T²

⁻³

+ 1

60T⁵

1 + 1 4T²

⁻³ + · · ·

= T 2 ln T

2

+ T

4 1 4T² − 1

2

1 4T²

2

+ · · ·

!

− π 8

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18 3. Mathematics

+ 1 4

1 2T −1

3

1 2T

3

+ · · ·

!

− T 2 − 1

6T

1 − 1

4T² + · · ·

− 1

45T³

1 − 3

4T² + · · ·

+ 1

60T⁵

1 + 1 4T²

−3

+ · · · This gives finally

arg

Γ

1 4+ iT

2

= T 2 ln T

2

− T 2 − π

8 + O 1 T

.

When we substitute this into our obtained formula for the counting function, we get

N (T ) = T

2πln T 2π

− T 2π+ 7

8+ O 1 T

+ 1

π arg

ζ

1 2+ iT

.

Often, this counting function function is separated into a smooth part and a fluctuating part

N (T ) = hN (T )i + N_fl(T ), (3.7) where

hN (T )i := T

2πln T 2π

− T 2π +7

8 + O 1 T

, (3.8)

is the smooth part and

N_fl(T ) := 1 π

arg

ζ

1 2 + iT

(3.9) is the fluctuating part.

To rewrite (3.9) into a different form, we use the Euler product formula, (3.2), disregarding the fact that this does not converges at these values, and the Taylor expansion of the function ln(1 + s) around the point s = 0, which is given by

ln(1 + s) =

∞

X

m=0

ln^(m)(1)

m! s^m = s − 1 2s²+1

3s³− · · · =

∞

X

m=1

(−1)^m+1 m s^m. This series converges whenever |s| < 1. Now, the fluctuating part becomes

N_fl(T ) = 1 π

arg Y

p∈P

1 1 − p⁻¹²^−iT

!

= −1 π

ImX

p∈P

ln

1 − p⁻¹²^−iT

= −1 π

X

p∈P

Im ln

1 − e(⁻¹²^−iT)^ln(p)

= −1 π

X

p∈P

Im

∞

X

m=1

(−1)^m+1 m

−e(⁻¹²^−iT)^ln(p)m!

= −1 π

X

p∈P

∞

X

m=1

(−1)^2m+1e⁻¹²^{m ln(p)}

m Im −e^{−imT ln(p)}

= −1 π

X

p∈P

∞

X

m=1

−e⁻¹²^{m ln(p)}

m sin(−mT ln(p)) to conclude with

N_fl(T ) = −1 π

X

p∈P

∞

X

m=1

e⁻¹²^{m ln(p)}

m sin(mT ln(p)). (3.10)

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3.4. The counting function 19

(a) The black line is the exact computation of N_fl(T ) from (3.9). The red line is the approximation of N_fl(T ) from (3.10) using the first hundred prime numbers.

(b) The black line is the approximation of hN (T )i from (3.8). The red line is the sum of this black line and the red line fromFigure 3.3a.

(c) The black line is the exact computation of N (T ). The red line is the same as the red line from Figure 3.3b.

Fig. 3.3: The counting function, N (T ), of the number of zeroes of the Riemann zeta function.

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20 3. Mathematics

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4. PHYSICS

4.1 Classical mechanics

4.1.1 Lagrange’s equation of motion

Definition 4.1 (Lagrangian). If the kinetic energy of a particle is denoted by T and the potential energy of that particle by U , then the Lagrangian is defined by

L(x, ˙x) := T ( ˙x) − U (x),

where x is the position of the particle and ˙x is the derivative of the position of the particle with respect to time, called the velocity.

The time is denoted by t. Both position and velocity are time-dependent, that is, x = x(t) and ˙x = ˙x(t). The Lagrangian can be generalized to a more particle system by summing over all particles. Hamilton’s principle states that the path a particle follows from x(t₁) to x(t₂) is that one that minimizes the time integral of the Lagrangian:

δ Z t2

t1

L(x, ˙x) dt = 0.

Note that this path starts at x(t₁) and ends at x(t₂). Actually this equation corresponds to an extremum, but luckily in most cases this is equal to a minimum.

One may interchange integration and differentiation by the Leibniz rule. So

0 = δ Z t2

t1

L(x, ˙x) dt = Z t2

t1

δL(x, ˙x) dt = Z t2

t1

∂L

∂xδx + ∂ L

∂ ˙xδ ˙x

dt

= Z t2

t1

∂L

∂xδx − d dt

∂ L

∂ ˙xδx

dt + ∂ L

∂ ˙xδx

t2

t1

= Z t2

t1

∂L

∂x − d dt

∂ L

∂ ˙x

δx dt,

(4.1)

where in the fourth step we used partial integration. The integrated term vanishes because the begin and endpoints of all possible paths are equal by definition, and thus the differences between the paths are zero: δx(t₁) = 0 and δx(t₂) = 0.

Because the integral in (4.1) must vanish for all possible paths, the integrand itself must vanish:

∂ L

∂x − d dt

∂ L

∂ ˙x = 0. (4.2)

This equation is knowns as Euler’s equation or, when applied to mechanical systems, as the Euler-Lagrange equation.

One can show that Lagrange’s equation of motion (4.2) is equivalent to New- ton’s equation of motion

−∂ U

∂x =: F = m¨x.

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22 4. Physics

4.1.2 Hamilton’s equations of motion

When the Lagrangian is denoted by L, the momentum is defined by p := ∂ L

∂ ˙x

where x is the position of the particle and ˙x is the derivative of the position of the particle with respect to time, the velocity. The time is denoted by t. Both position and velocity are time-dependent, that is, x = x(t) and ˙x = ˙x(t). The momentum can be generalized to a more particle system by summing over all particles. We can use the definition of momentum to express ˙x in terms of x and p, that is, ˙x = ˙x(x, p). Using (4.2), we can see that the time derivative of p is given by

˙ p = ∂ L

∂x.

Definition 4.2 (Hamiltonian). The Hamiltonian is defined by H(x, p) := p ˙x − L(x, ˙x),

which we observe as a function of only x and p, by using ˙x = ˙x(x, p).

Taking the total differential of the left hand side of the definition of the Hamiltonian, gives

dH = ∂ H

∂x dx + ∂ H

∂p dp. (4.3)

The total differential of the right hand side is

dH = d(p ˙x) − dL(x, ˙x) = ˙x dp + p d ˙x −∂ L

∂x dx − ∂ L

∂ ˙x d ˙x

= ˙x dp + p d ˙x − ˙p dx − p d ˙x = − ˙p dx + ˙x dp.

(4.4)

After identifying the coefficients of dx and dp in (4.3) and (4.4), we find











˙x = ∂ H

∂p ,

˙

p = −∂ H

∂x .

(4.5)

Those equations are called Hamilton’s equations of motion. By construction, they are equivalent to Lagrange’s equation of motion.

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4.2. Quantum mechanics 23

4.2 Quantum mechanics

If Ψ = Ψ (x, t) is the wave function of a particle, then the Schr¨odinger equation reads

i~∂ Ψ

∂t = − ~² 2m

∂²Ψ

∂x² + U Ψ,

where x is the position and m the mass of the particle, t is the time and U the potential energy.

For a particle in the state Ψ = Ψ (x, t), the expectation value of the observable for position x is

hxi = Z ∞

−∞

x|Ψ |²dx = Z ∞

−∞

Ψ^?xΨ dx.

The expectation value for momentum p is hpi =

mdx

dt

= mdhxi dt = m

Z ∞

−∞

x∂

∂t|Ψ |²dx = m Z ∞

−∞

x∂

∂tΨ^?Ψ dx

= m Z ∞

−∞

x ∂Ψ^?

∂t Ψ + Ψ^?∂ Ψ

∂t

dx = i~

2 Z ∞

−∞

x

−∂²Ψ^?

∂x² Ψ + Ψ^?∂²Ψ

∂x²

dx

= i~

2 Z ∞

−∞

x ∂

∂x

−∂ Ψ^?

∂x Ψ + Ψ^?∂ Ψ

∂x

dx = i~

2 Z ∞

−∞

∂Ψ^?

∂x Ψ − Ψ^?∂ Ψ

∂x

dx

= ~ i

Z ∞

−∞

Ψ^?∂ Ψ

∂x dx = Z ∞

−∞

Ψ^?~ i

∂

∂xΨ dx,

where in the eighth and ninth steps we used integration by parts, the integrated term vanishes because the wave function Ψ goes to zero at ±∞. We say that the operator x represents position, and we denote this operator by

x := x,b

and the operator (~/i)(∂/∂x) represents momentum, and this operator is denoted by

p :=b ~ i

∂

∂x.

The expectation value of an observable H(x, p) is expressed as hH(x, p)i =

Z ∞

−∞

ψ^?H(b bx,p)ψ dx =:b D

ψ, bH(bx,p)ψb E .

An operator bQ, constructed from the observable Q(x, p), is called Hermitian

if D

ψ, bQψE

=D

Qψ, ψb E for all ψ(x).

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24 4. Physics

4.3 The Hilbert-P´ olya conjecture

From the letter in the Introduction, we know George P´olya suggested a physical approach to prove the Riemann hypothesis.

Conjecture 4.1 (Hilbert-P´olya conjecture). If the non-trivial zeroes of the Riemann zeta function are written in the form

ρ_n = 1 2 + it_n,

then the t_n’s correspond to the eigenvalues of a Hermitian operator.

Also David Hilbert is associated with this conjecture, although there is no evidence he conjectured it. When we formulate this into a conjecture, we get the following.

Theorem 4.1. The Hilbert-P´olya conjecture,Conjecture 4.1, implies the Riemann hypothesis, Conjecture 3.1.

Proof. Note that if ρ_n = 1/2 + it_n are the non-trivial zeroes of the Riemann zeta function, then the t_n’s do not necessarily have to be real. Suppose t_n= a_n+ ib_n, then

ρn= 1

2+ itn = 1

2 + i(an+ ibn) = 1

2 − bn+ ian.

So ρn can be indeed any non-trivial zero. Now suppose the Hilbert-P´olya conjecture is true. That is, the t_n’s correspond to the eigenvalues of a Hermitian operator, say bH. That means that if a t_n is a eigenvalues corresponding to the eigenfunction ψn of such an operator then

According to ISO 80000-2 clause 14.6 the notations s and s^? both mean the complex conjugate of s, where s is meanly used in mathematics and s^? is meanly used in physics and engineering.

tnhψn, ψni = htnψn, ψni =D

Hψb n, ψn

E

= D

ψn, bHψn

E

= D

Hψb n, ψn

E?

= htnψn, ψni^? = t^?_nhψn, ψni^? = t^?_nhψn, ψni,

so tn = t^?_n, therefore tn has to be real. That means the non-trivial zeroes have the form ρ = 1/2 + it, where t is real. And that is exactly what the Riemann hypothesis states.

4.4 Equivalences between the fluctuating parts of the counting functions

It would be nice if we could find a Hermitian operator that satisfies the conditions for the Hilbert-P´olya conjecture. This Hermitian operator can be obtained by quantizing a Hamiltonian of a dynamical system which spectrum, that is, the set of eigenvalues, matches the non-trivial zeroes of the Riemann zeta function. A necessary condition for this is that the counting function of the number of states of an “energy” between 0 and E > 0, denoted as N (E), equals the counting function of the number of non-trivial zeroes with imaginary part between 0 and T > 0, which we denoted by N(T ), see (3.7).

In mathematical physics the counting function N (E) of the number of states with an energy between 0 and E always contains a fluctuating part. That means N (E) can be written as

N (E) = hN (E)i + N_fl(E).