A proposed Hamiltonian operator for Riemann Hypothesis

[zetafunction]

HERE http://vixra.org/pdf/1007.0005v1.pdf

is my proposed proof of an operator whose Eigenvalues would be the Imaginary part of the zeros for the Riemann Hypothesis

the ideas are the following* for semiclassical WKB evaluation of energies the number of levels N(E) is related to the integral of pdq along the classical orbits

* for one dimensional systems, in the WKB approach the inverse of the potential V(x) is proportional to the Half-derivative of N(E)

* For the case of Riemann Zeros the N(E) is given by the argument of Riemann Xi function

Xi(s)= s(s-1)G(s)Z(s) G=gamma function Z= zeta function

evaluated on the critical line $$Arg\xi(1/2+iE)$$

If we combine these points then the inverse of the potential V(x) inside the Hamiltonian H=p^2 +V(x) would be proportional to the HALF INTEGRAL of the logarithmic derivative of

$$\xi(1/2+iE)$$ as i express in the attached .PDF see formulae 3 4 and 5 inside the paper

Also as a final NUMERICAL test, i think that the functional determinant of this operator $$H=p^2 +V(x)$$ is related to the Xi function, since the Hadamard product applied to Xi function can be used to express the Xi function as a product of the eigenvalues of some operator

in brief, the INVERSE function of the potential V(x) is given by the half-integral of the logarithmic derivative of the Xi function evaluated on the critical line 1/2+is

also the FUNCTIONAL DETERMINANT of the operator $$H+m^2$$is the Riemann Xi-function

 

[HallsofIvy]

Why cast everything in terms of physics terminology? If you really do have a proof then you should be able to translate it into purely mathematics terminology which would be much better. The Riemann Hypothesis is, after all, a mathematics problem, not physics.

 

[zetafunction]

Why cast everything in terms of physics terminology? If you really do have a proof then you should be able to translate it into purely mathematics terminology which would be much better. The Riemann Hypothesis is, after all, a mathematics problem, not physics.

translation would be

HAMILTONIAN : linear differential operator of the form $$H=-y''(x)+f(x)y(x)$$

the boundary condition imposed in general are that the solution is on an $$L^{2} (R)$$ space

ENERGIES = eigenvalues of the operator

$$f^{-1}(x)= A \frac{d^{-1/2}g(x)}{dx^{-1/2}}$$ is valid ONLY for one dimensional system and is am asymptotic approximation , in the same way the Prime counting function is asymptotic to x/lnx

$$g(x)= \sum_{n} \delta (E-E_{n})$$ dirac delta distributions over the eigenvalues of the operator H

for our case $$g(x)= \frac{dArg\xi(1/2+ix)}{\pi dx}$$

the inverse of f(x) is NOT exact is valid as an ASYMPTOTIC formula ,

 

[CRGreathouse]

$$f^{-1}(x)= A \frac{d^{-1/2}g(x)}{dx^{-1/2}}$$ is valid ONLY for one dimensional system and is am asymptotic approximation , in the same way the Prime counting function is asymptotic to x/lnx

So don't write = unless you're going to put an error term on the right.

 

[zetafunction]

umm sorry in physics is used to write = even for asymptotic, i think this would be better

$$f^{-1} (x)\sim 2\sqrt \pi \frac{d^{-1/2}g(x)}{dx^{-1/2}}$$

in physics one use this approximation we approximate a sum over eigenvalues , for a double integral in (p,q) for example

$$\sum_{n}^{\infty}e^{iuE_{n}} \approx \iint_{R^{2}} dpdqe^{iup^{2}+iuV(q)}$$

p is the momentum and q is the position

 

[HallsofIvy]

So what is happening is that your phyics notation is leading you into inaccuracies. I was afraid of that. How are you going to go from you approximation, or asymptotic limit, to an exact value?

 

[jostpuur]

Do you believe that you have found the relevant $V(x)$, or that you have only proven its existence?

 

[zetafunction]

So what is happening is that your phyics notation is leading you into inaccuracies. I was afraid of that. How are you going to go from you approximation, or asymptotic limit, to an exact value?

WKB approximations are just this approximation, i have tested my formula with 2 cases

Harmonic Oscillator $$f= (wx)^{2}/4$$ the result is EXACT

POtential well $$f=0$$ then i get the approximate value $$2/\pi$$

i have not 'invented' my formula is just a consequence of the approximate formula used in WKB semiclassical quantum mechanics

$$\pi n(E) \approx 2\int_{a}^{b}dx (E-V(x))^{1/2}$$

http://en.wikipedia.org/wiki/WKB_method

http://en.wikipedia.org/wiki/Bohr–Sommerfeld_theory (see section: one dimensional potential)

here N(E) means how many energy levels are there with energy less than a given E for the Riemann Zeros considered as Energy levels then $$\pi n(E)=Arg\xi(1/2+iE)$$

 

[zetafunction]

Do you believe that you have found the relevant $V(x)$, or that you have only proven its existence?

no i have PROVED and obtained it via the WKB approximation, equations

http://en.wikipedia.org/wiki/WKB http://en.wikipedia.org/wiki/Old_quantum_theory

i really do not understand mathematician, an approximate result does not mean that it is NOT valid , i guess that if you calculate the energies of my operator you will obtain the imaginary part of the zeros

 

[jostpuur]

Either you have a definition for the $V(x)$, or then you don't.

If you have the definition, you can write it here for others to see.

If you cannot write it here for others to see, then you don't have the definition.

It is amazing that you have deceived yourself into believing that you have the definition, even though you have no idea about what the definition could be, and have never even seen it.

 

[zetafunction]

Either you have a definition for the $V(x)$, or then you don't.

If you have the definition, you can write it here for others to see.

If you cannot write it here for others to see, then you don't have the definition.

It is amazing that you have deceived yourself into believing that you have the definition, even though you have no idea about what the definition could be, and have never even seen it.

http://vixra.org/pdf/1007.0005v2.pdf page 3 , formula (5)

$$\sqrt \pi V^{-1} (x) = \frac{1}{\sqrt i} \frac{d^{-1/2}}{dx^{-1/2}}\frac{ \xi ' (1/2+ix)}{\xi (1/2+ix)}+ \frac{1}{\sqrt -i} \frac{d^{-1/2}}{dx^{-1/2}}\frac{ \xi ' (1/2-ix)}{\xi (1/2-ix)}$$

$$\frac{d^{-1/2}f(x)}{dx^{-1/2}}= \frac{1}{\sqrt \pi} \int_{0}^{x} dt \frac{f(t)}{(x-t)^{1/2}}$$

this definition for the INVERSE of the potential is obtained by solving a non-linear integral equation defining the number of Zeros n(E) and the potential in the semiclassical approximation

$$n(E)= A\int_{0}^{E}dx (E-V(x))^{1/2}= \frac{1}{\pi}Arg\xi(1/2+iE)$$

 

[jostpuur]

http://vixra.org/pdf/1007.0005v2.pdf page 3 , formula (5)

I glanced through the paper in hope of seeing the definition. This equation didn't draw my attention because it had an approximation sign "$\approx$" in it.

 

[zetafunction]

i have checked the first 3 eigenvalues of my Hamiltonian and they satisfy

$$E_{0}= 13.1090$$ , $$E_{1}= 20.1254$$ , $$E_{2}= 30.0679$$

using the semiclassical approximation http://en.wikipedia.org/wiki/WKB_approximation to solve the differential equation

 

[Eidos]

Are those the imaginary parts of the first three non-trivial zeros of the Riemann Zeta function?

 

[adriank]

The first few zeros have imaginary part 14.1347, 21.022, 25.0109, 30.4249, 32.9351, says Wolfram.

 

[zetafunction]

i have used approximate methods to solve the Hamiltonian , this is perhaps why my zeros are not exact , perhaps a numerical evaluation or solution of Shchroedinguer equation will yield to more accurate evaluations

 

[adriank]

I can't even tell what you claim to have proven.

 

[zetafunction]

I can't even tell what you claim to have proven.

i give a DIFFERENTIAL OPERATOR whose EIGENVALUES are the imaginary part of the Zeros of the Riemann Zeta function



http://vixra.org/pdf/1007.0005v3.pdf

see formula (5) for the potential.

the EINGENVALUES of the differential operator $$-D^{2}+ f(x)$$ are the IMAGINARY part of the Riemann zeta zeros , with

$$f^{-1} (x)= 2\sqrt \pi \frac{d^{-1/2}g(x)}{dx^{-1/2}}$$

and g(x) here is $$I am \frac{\partial}{\partial x}log \xi (1/2+ix)$$

 

[adriank]

No need to yell.

I have already attempted to read your article. It looks like a complete mess.

How do you know you hit all of the zeros of the function ξ(1/2 + iE)? If RH is false, I think you don't. Whatever branches of Arg you're using in equation (4), it looks like that would only find the zeros where E is real.

You're using approximations all over. Isn't the first = in (2) an approximation, and you simply got lucky that it's exact for the harmonic oscillator? As far as I can tell, everything afterwards, including the equation defining V, uses this approximation, so nothing is exact.

I don't think anyone is convinced that you've done anything substantial.

 

[zetafunction]

calculate the EIGENVALUES of the operator and then compare to the known zeros, this is the best proof to check if i am right or not

 

[adriank]

You've done that and they're quite off from the actual zeros. Not to mention that numerical evidence is not proof.

What the heck is the Gelfand-Yaglom theorem? You cite reference [2], which I have searched (thank you Internet) and doesn't mention Gelfand or Yaglom.

Please address the two questions in my previous post.

 

[zetafunction]

You've done that and they're quite off from the actual zeros. Not to mention that numerical evidence is not proof.

What the heck is the Gelfand-Yaglom theorem? You cite reference [2], which I have searched (thank you Internet) and doesn't mention Gelfand or Yaglom.

Please address the two questions in my previous post.

in case you calculate the functional determinant of $$det(z-H)det(z+H)$$ this MUST be equal to the Xi function so ALL the zeros are real since z-H and z+H are HERMITIAN operator, you can calculate this functional determinant if you want to check i am right

 

[adriank]

No, the burden is on you to prove to us that you are right.

You've calculated the eigenvalues of your operator, and they're clearly not the (imaginary parts of the) zeros of the zeta function. And then you're claiming that the functional determinant det(H + z²)/det(H) from this Hamiltonian arising from the approximate potential (as I understand it, equation (11) is approximate as it comes from the WKB approximation (2)) is exactly ξ(1/2 + z)/ξ(1/2)? You still haven't addressed my second question.

Do you claim you have proven the Riemann hypothesis? (It seems like it to me.) It looks far more likely to me that you've obscured things so much that you (and everyone else) are confused.

 

[zetafunction]

my question here is .. an APPROXIMATE result is NOT A VALID result ??

i mean do mathematician USE approximate formulae for the calculations, or whenever a formula is approximate they DO NOT use it ?

 

[Eidos]

Before the mudslinging contest ensues.

Your operator missed the third zero (around 25), why do you think that is?

The other values are fairly close though ^^

If your operator can get them exact, how do you find any other zeros in the RH plane/ prove that aren't any?

Not bad for an approximation I say

 

[zetafunction]

If your operator can get them exact, how do you find any other zeros in the RH plane/ prove that aren't any?

Not bad for an approximation I say

this is one of the problems..

the idea is that $$\frac{ \xi(1/2+iz)}{\xi (1/2)}= \frac{det(H-z)}{detH} \frac{det(H+z)}{detH}$$

MUST hold , for one dimensional system the functional determinant can be calculated without knowing any of its Eigenvalues

 

[adriank]

So prove that it does!

And you're not even being consistent: in your article you write $\frac{\det(H + z^2)}{\det(H)} = \frac{\xi(1/2 + z)}{\xi(1/2)}$, while now you're writing $\frac{\det(H + z)}{\det(H)} \frac{\det(H + z)}{\det(H)} = \frac{\xi(1/2 + iz)}{\xi(1/2)}$. Which one is it!? I'm guessing it's the latter, because otherwise the zeros of $\det(H + z^2)$ are not the zeros of $\xi(1/2 + z)$.

Give a detailed proof. I've already asked: what is the Gelfand-Yaglom theorem? What does it state? As I said above, the book [2] you cited for this theorem does not mention these names.

Once again, as far as I can tell, you're still using an approximation to find V, and then the (approximate) H you get has eigenvalues which you have calculated, and they are different from the zeta zeros, and yet you claim that this same operator satisfies the above equation, implying that the eigenvalues are exactly the zeros. Something's not right.

 

[zetafunction]

So prove that it does!

And you're not even being consistent: in your article you write $\frac{\det(H + z^2)}{\det(H)} = \frac{\xi(1/2 + z)}{\xi(1/2)}$, while now you're writing $\frac{\det(H + z)}{\det(H)} \frac{\det(H + z)}{\det(H)} = \frac{\xi(1/2 + iz)}{\xi(1/2)}$. Which one is it!? I'm guessing it's the latter, because otherwise the zeros of $\det(H + z^2)$ are not the zeros of $\xi(1/2 + z)$.

Give a detailed proof. I've already asked: what is the Gelfand-Yaglom theorem? What does it state? As I said above, the book [2] you cited for this theorem does not mention these names.

Once again, as far as I can tell, you're still using an approximation to find V, and then the (approximate) H you get has eigenvalues which you have calculated, and they are different from the zeta zeros, and yet you claim that this same operator satisfies the above equation, implying that the eigenvalues are exactly the zeros. Something's not right.

in order to calculate the zeros i have used APPROXIMATION to the eigenfunctions, the idea would be to solve $$-D^{2}+ V(x)$$ by Numerical methods with boundary conditions $$y(0)=y(\infty)$$ an APPROXIMATE formula is NOT an incorrect formula

http://iopscience.iop.org/1751-8121/42/27/272001/fulltext

gelfand YAglom theorem EXPLAINED in order to calculate a functional determinant you do NOT need to know explicitly the Eigenvalues, so if you calculate the functional determinant i propose you could check if you recover the Xi function

 

[CRGreathouse]

in order to calculate the zeros i have used APPROXIMATION to the eigenfunctions, the idea would be to solve $$-D^{2}+ V(x)$$ by Numerical methods with boundary conditions $$y(0)=y(\infty)$$ an APPROXIMATE formula is NOT an incorrect formula

Back that up. What statement do you really prove, and how do you know the approximation is close enough to prove RH?

 

[jackmell]

i have checked the first 3 eigenvalues of my Hamiltonian and they satisfy

$$E_{0}= 13.1090$$ , $$E_{1}= 20.1254$$ , $$E_{2}= 30.0679$$

using the semiclassical approximation http://en.wikipedia.org/wiki/WKB_approximation to solve the differential equation

I do not feel that is a good approximation. The first 10 via Mathematica are:

In[2]:= Table[N[Im[ZetaZero[n]]], {n, 1, 10}]

Out[2]= {14.1347, 21.022, 25.0109, 30.4249, 32.9351, 37.5862, \
40.9187, 43.3271, 48.0052, 49.7738}

That's really just not close enough. Now, if you could compute say at least the first 20 and get them to at least one digit to the right of the decimal place and show your numerical calculations become more accurate as you increase the internal precision of the computations, then that's more encouraging.

 

[zetafunction]

a bit of explanation ..

QUANTUM MECHANICS AND SEMICLASSICAL SOLUTIONS

the schröedinguer equation in units $$\hbar = 2m=1$$ for one dimensional particle under the influence of a potential f(x) is

$$-D^{2} y(x)+f(x)y(x)=E_{n}y(x)$$

here the eigenfunctions are on an $$L^{2} (R)$$ space so $$\int_{-\infty}^{\infty}dx|y(x)|^{2} < \infty$$

of course this equation is complicate to solve EXACTLY depending on the form of f(x) so we still must rely on approximations

$$y(x) \approx e^{iS(X)}$$ , here S(x) is a function that satisfy the first order differential equation (1)

$$E= (\frac {dS}{dx})^{2}+f(x)$$ and here 'E' is the Energy

from the ansatz in (1) we obtain the following APPROXIMATE (is an approximate solution INVALID in mathematics ? ) for the quantization of Energies

$$\pi n(E) = \int_{0}^{a}dx (E-V(x))^{1/2}$$ (2)

here 'a' is TURNING POINT , in other words 'a' is a value of position 'x' so $$E=V(x)$$ as you will notice if the POtential is bigger than the Energy the integrand is COMPLEX

This integral equation in (2) has NO exact solution for V(x) but can be solved to get the INVERSE of the potential function f(x) so $$f^{-1} (x) = 2 \sqrt \pi \frac{d^{1/2}n(x)}{dx^{1/2}}$$

ONCE we have the INVERSE we could solve this equation by numerical methods to get f(x) and then solve NUMERICALLY the Schroedinguer equation $$-D^{2} y(x)+f(x)y(x)=E_{n}y(x)$$

this is WHY quantum mechanic is RELEVANT to solve the RIemann Hypothesis, AGAIN my question for the mathematician is

AN APPROXIMATE EQUATION IS STILL VALID ? , of course perhaps the EIGENVALUES will be ASYMPTOTICS to the zeros , is this still invalid ??

also you can check this formula for the HARMONIC OSCILLATOR or LINEAR POTENTIAL to check that gives CORRECT results

 

[zetafunction]

Also, if we consider the Riesz function , R(x) that under Riemann HYpothesis satisfies

$$R(x)=O(x^{1/2+\epsilon} )$$ for any positive epsilon

this Riesz function satisfy the integral equation

$$e^{-x}-1 = \int_{0}^{\infty} \frac{dt}{t}R(t)frac( ( xt^ {-1})^{1/2})$$

so if we consider $$R(x)=O(x^{1/2+\epsilon} )$$ and plug it into the integral you get that $$e^{-x}-1 \le C \frac{ |\zeta (1/2+2\epsilon)|}{1+2\epsilon}x^{1/4+ \epsilon}$$

 

[zetafunction]

the inverse of the potential is given by

$$AV^{-1}(x)= \sum_{\gamma}H(x-\gamma^{2})(x-\gamma^{2})^{-1/2}$$

using the Riemann-Weyl formula this sum can be turned into a sum over primes and prime powers.