QUANTUM MECHANICS AND QUANTUM MECHANICS AND RIEMANN HYPOTHESISRIEMANN HYPOTHESIS

We want to prove that the Energies of the Hamiltonian

2

2

( )( ) ( ) ( ) y(0)=0=y( )n

d y xV x y x E y x

dx

Are the square of the imaginary part of the Riemann Zeros 2n nE

So the Energies of the Hamiltonian are the roots of the Riemann Xi-function\Riemann zeta function

/ 2( 1)( )

2 2ss s s

s s

So, we have managed to prove that

2

0

1 1log det

2 2nn

d dm E i H E Arg ik

dE k dk

det 1/ 4 (1 )( )

(0) det 1/ 4

H s ss

H

The Riemann Xi-function can be expressed as a functional determinant on the variAble ‘s’

The Riemann-Weil trace formula can be obtained from the derivative of the folloWing functional determinant

The potential inside the Quantum Hamiltonian is igven implicitly by

1/ 2

11/ 2

0

( ) 2 ( )( ) 2

1/ 2

xd N x d dtN tV x

dx dx x t

Here , N(E) is the Eigenvalue staircase , that counts the number of Eigenvalues/Energies on the interval (0,E)

0

( ) ( )nn

N E H E E

0 0

( )1 x>0

xH x

0

( ) ( )nn

dNd E E E

dE

In case of the Riemann Hypothesis, the Eigenvalue staircase is given by theArgument of the Riemann Xi function defined on the first page

1 1( )

2N E Arg i E

In order to get the proof that our quantum system defined on page first , gives theSquare of the zeros of the Riemann zeta function as Eigenvalues, we need to intro-Duce 2 important formulae , the Riemann-Weil explicit formula and the quantizationRules by Bohr and Sommerfeld

0

( ) 2 ( )a

C

pdq E V x dx n E

BOHR-SOMMERFELD QUANTIZATION RULE

Here, n=n(E) is an integer this formula is an approximation but works pretty well For big values of the Quantum Energies ‘n’ , the pont ‘a’ is a turning point so the Momentum of the particle is p=0 and E=V(a) , in one dimension we can obtain The potential (its inverse) from the eigenvalue staircase as follows

nE E

( ) 13/ 2

0 0

1/ 2 1 1/ 21

1/ 2 1/ 2

( )2 ( ) 2 ( ) 2

( ) ( )2 ( ) 2 ( )

a a E E

n n x

dx dV xn E E V x dx E V D

dV dx

d d dV x d n xn x V x

dx dx dx dx

The constant integration is set to 0 so at x=0 the potential and the n(E) take the Value 0 , from this equation above we have proved that we can recover the inverseOf the potential (and hence the potential) from the half derivative of the EigenvalueStaircase

If we introduce the momentum and wave number in terms of the Energy

E p k in units with 2 1m

This is made in order to simplify the calculations and avoid constants

RIEMANN-WEIL EXPLICIT FORMULA FOR ZEROS AND PRIMES

1

( ) 1 ' 1( ) 2 2 (log ) ( ) (0) log

2 2 4 2n

i n irh h g n drh r g

n

( ) ( ) ( )iuxh u g x e h u

( ) ( )g x g x

There is a duality between the imaginary part of the Riemann Zeros and the primeNumbers

log ( )

0 otherwise

kp n pn

Von Mangoldt function

If we plug the function

Inside the Riemann-Weil explicit formula we obtain

2( , ) ( )h r E E r

2 1 1 ' 1 1 log

2 2 2 2 2

' 1 1 ' 1 1 2 24 2 4 4 2 4 2

regE ip ipp p p

i ip p

ip ip

p p p

To get this formula we have used the following expressions

log

1

( ) ' 1( )

'( ) 2n

n i E nreg

x nn

x x nf x e i E

f x n

2( ) imk

m m

k n e

2

2 22

( )( ) 0 (0) 0 (1)

d y xn y x y y

dx

In OUR case we will choose the branch of the logarithm for the RiemannXi function so

10

2Arg

In this case the density of states is related to a sum over primes, this is not new For example for the infinite potential well the density of Energies can be evaluatedBy an exponential sum over ALL the integers.

The sum involving the Dirac delta is over the zeros of f(x) E p k The sum involving the Mangoldt function has been zeta-regularized

The sum over the imaginary part of the zeros inside the Riemann-Weil formula isOver ALL the zeros, not just the zeros lying on the critical line Re(s)=1/2

If we integrate over ‘E’ and use ( ) log( )Arg z m z

2

0 0

1 1( ) ( ) ( )

2n nn n

H E H E E Arg i E n E

RIEMANN-VON MANGOLDT FORMULA FOR THE NUMBER OF ZEROS ON THE INTERVAL (0,T)

1 1 1( ) log

2 2 2 2

T T TN T O Arg iT T

T

1/ 21

1/ 2

( )( ) 2

d N xV x

dx

We MUST include ALL the corrections the SMOOTH part and the OSCILLATING part , the SMOOTH Part is given by the first 2 terms inside the Riemann-Mangoldt formula, the OSCILLATING part is Given by the argument of the Riemann Zeta function on the critical line

/ 22 2 / 2

14 ( )

( ) smooth

e A x BV x

2

2 2( ) 4( )smooth

x BV x e

A

11 2

2 0

12 ( ) ( ) 2

( ) log ( )1!log 1

2

kkk

oscn k

kn i

V x n skkn n

0

33 12

2 2 !1

2n

A B en

1 1

1

2( ) log ( ) ( ) ( )

2 2 n n

E E nN E E N E E f n N E

e W ne

1

1

( )( )

!

nn

n

nW x x

n

If we invert the Riemann-Von Mangoldt function, we can get a first asymptotic toThe imaginary part of the Riemann Zeros , given bye the Lambert W-function

ZETA REGULARIZED DETERMINANT AND RIEMANN HYPOTHESIS

Given an operator with a Hadamard product expansion2

0

1n n

E

E

We can define its ‘Zeta regularized determinant’ in the form

2 2

0 0

ˆ(0, ) (0,0) log det log logs s n nn n

Z E Z H E E E

1 2

0( ) ( , ) ( ) expss s E dt t t tE

1

0( ) s ts dtt e

2 20

1,s

nn

s EE E

For a Hamiltonian 2 †( ) H p V x H H We can define an EXACT and an WKB Theta function

2

0 0 0

1( ) exp ( ) exp( ( ))

2

exact WKB definition

stn

n

t tE s dtN t e dq dp tp tV q

How should the potential V(x) be so ( ) ( )WKBt t

Before doing any calculation , we will introduce some properties of theLaplace inverse and direct transform

1( ) ( )

2

c ist kt kt

c i

D f t dsF s e s D e k ei

If two functions are EQUAL so are their Laplace transforms

0 0

( ) ( ) ( ) ( )st stdt f t e dtg t e f t g t

2

0 0 0

12 ( )

0 0

1exp ( ) exp( ( ))

2

1 1 1 ( )exp( ( ))

2 2 2

stn

n

tV q tr

tE s dtN t e dq dp tp tV q

dV rdq dp tp tV q dqe dre

drt t

1 1

0 0 0 0

( ) ( ) 2 ( ) ( )2

st st st stss dtN t e dtV t e s dtN t e dte V t

We can also use the integration by parts for the fractional operator, in the same Way we introduced the fractional differential operator for the Laplace inverse

1/ 2 1/ 2

1/ 2 1/ 20 0

( ) ( )( ) ( )

d d f x d d g xdx g x dx f xdx dx dx dx

From the equation 1

0 0

2 ( ) ( )st sts dtN t e dte V t

We get, that the inverse of the potential (by Laplace transform theory) is Equal to the following 1/ 2

11/ 2

2 ( ) ( )d

N x V xdx

¡¡ But this is precisely the same functional equation we obtained with the BohrSommerfeld quantization rules ¡¡

For the case of the RH and the Riemann Zeros

2

0 0

1 1( ) exp ( ) ( )

2st

nn

t t s dtN t e N x Arg i x

In this case we can define the zeta regularized Hadamard product

2

20

1 11

2 2 n n

zz

For the EXACT Theta function

For the WKB Theta function we have the Zeta function

3/ 2

0 0

2 ( ) ( , ) exp ( )sWKBs s E t dt dq tV q

1/ 21

1/ 2

( )( ) ( ) ( ) 2exact WKB

d n xt t iff V x

dx

For the WKB formulation Zeta regularization gives

2( , ) ( ,0) log det log dets WKB s WKBs E s H E H

So if we put the 2 equations together ( ) ( )WKBt t

2 2 1/ 21

1/ 22

1det ( ) ( )2

( ) 21 det ( )2

x

x

z V x z d n xV x

dxV x

SInce the Hamiltonian is Hermitian the factor 2z HWill only vanish for

n nz i E i

So, ALL the Zeros of the Riemann Xi function 1

2s

Will be PURELY IMAGINARY , and bye a change of variables izThen, ALL the Zeros of the Riemann Xi function , will be REAL

1

2iz

The Riemann Xi function (above) is proportional to the Functional determinant

2 2det ( )x V x s 1/ 2

11/ 2

1( ) 2

2

d ArgV x i x

dx

2 22

2 2( , ) ( , ) 0 +boundary conditionny f x y E f x y

x y

21 1 0

4 2n n nE s Z is

( )

0 .

( ) 1 det (1 )pl s n

n p o

Z s e s s

2 22

2 2y

x y

11 s

p

s p

THE SELBERG ZETA FUNCTION SATISFY RIEMANN HYPOTHESIS, AND IS ALSOPROPORTIONAL TO A FUNCTIONAL DETERMINANT

The Selberg zeta function can be Defined as

The analogy between the Selberg and the RiemannZeta function is clear

logpl p

In both cases the infinite product for The Riemann Zeta over the primes, and for theSelberg Zeta is only convergent for

Re( ) 1s

2 22 2

2 2 21

sinh( )0 (0) 0 (1) 1

n

x xn

x x n

A simpler analogy is given by the hyperbolic sine function and the infinite potenTial well in QM in this case the problem is exactly solvable and gives.

So , we can expect that our method will work, HOWEVER in order to work we NEED to include both, the SMOOTH and oscillating part of the Eigenvalue StairCase N(E) , otherwise the results may be poor

1 1( ) log ( )

2 2 2smooth oscillating

E EN E N E Arg i E

e

Weyl’s law 1/ 2 1( ) ( )

2smoothN E O E d

For x >0 the potential is given by the implicit equation

1/ 21

1/ 2

1( ) 2

2

d ArgV x i x

dx

For x <0 the potential is equal to ( ) 0V x x At the point x=0 there is an infinite wall (similar to the infinite potential wellProblem , this is the reason we have introduced the boundary conditions

(0) 0 ( )y y Since the potential V(x) is positive on the Real line then

2 2 0 n n n nH p V E E E R

If the square of the Riemann zeros are ENERGIES , the imaginary part of the Riemann Zeros play the role of momenta

n n nk p

So, the Riemann Xi-function can be expressed as the quotient of 2 functionalDeterminants involving a Hamiltonian (Hermitian) Operator

1det (1 )

( ) 41(0) det4

H s ss

H

† 2( ) ( ) ( ) ( ) ( ) y(0)=0=y( ) xHy x H y x y x V x y x

The potential is defined implicitly by

1/ 21

1/ 2

1( ) 2

2

dV x Arg i x

dx

2

0

1 1log det

2 2nn

d dm E i H E Arg ik

dE k dk

The Riemann-Weil formula can be derived by taking the logarithmic derivativeOf the Funcitonal determinant

E k 0

The evaluation of the right part yields to

1 1 log 1 ' 1

2 2 2 4 2 2 2 2 2

ik i ie ik e k k

k k k k k

Using the Riemann-Weil formula we can also evaluate

2 20

1 2 1 1 ' 1 '( ) (1 )

1 1( 1) 2 ( 1) ( 1)( 1) ( 1)4 4

nn

drs s

s s s s s ss s r s s

In this case we have used the test function

2 2

1 1 1( , ) and

2 2h r a a s a s

r a

This invariance still holds 1s s

REFERENCES AND FURTHER LECTURES

http://www.claymath.org/millennium/Riemann_Hypothesis/riemann.pdf E. BombieriTalking about Riemann Hypothesis

http://en.wikipedia.org/wiki/Functional_determinant How to evaluate Zeta regulariZed Functional determinants, other methods are explained there too.

http://www.math.binghamton.edu/loya/papers/KirsLoyaAJP.pdf More about func-Tional determinant and Zeta regularization algorithm applied to physics

http://arxiv.org/PS_cache/quant-ph/pdf/0206/0206179v1.pdf WKB introductionIn physics and mathematics

http://vixra.org/abs/1111.0105 My original paper on the subject

http://arxiv.org/PS_cache/arxiv/pdf/1104/1104.1850v3.pdf Berry and KeatingHamiltonian model

http://arxiv.org/PS_cache/arxiv/pdf/1101/1101.3116v1.pdf Physics and RiemannHypothesis: an introduction for pedestrians

http://www.phy.bris.ac.uk/people/berry_mv/the_papers/Berry154.pdf Quantization,Functional determinants, Riemann Hypothesis, Selberg trace and More :D

¡¡¡ THANK YOU FOR WATCHING

http://arxiv.org/PS_cache/arxiv/pdf/1012/1012.4264v1.pdf More physics and Riemann Hypothesis , Berry Hamiltonian and more stuff.

http://euclides.us.es/da/actividades/G_Sierra_HR_Sevilla-09.pdf Riemann Hypo-Thesis and Physics.