A proposed Hamiltonian operator for Riemann Hypothesis

1. Jul 20, 2010

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 fucntion

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

2. Jul 20, 2010

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.

3. Jul 20, 2010

zetafunction

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 ,

Last edited: Jul 20, 2010
4. Jul 20, 2010

CRGreathouse

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

5. Jul 20, 2010

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

Last edited: Jul 20, 2010
6. Jul 21, 2010

HallsofIvy

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

7. Jul 21, 2010

jostpuur

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

8. Jul 22, 2010

zetafunction

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 $$\pi n(E)=Arg\xi(1/2+iE)$$

9. Jul 31, 2010

zetafunction

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

Last edited by a moderator: Apr 25, 2017
10. Aug 1, 2010

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.

11. Aug 1, 2010

zetafunction

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

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

Last edited: Aug 1, 2010
12. Aug 1, 2010

jostpuur

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.

13. Aug 8, 2010

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

14. Aug 11, 2010

Eidos

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

15. Aug 11, 2010

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

16. Aug 11, 2010

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

17. Aug 11, 2010

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

18. Aug 11, 2010

zetafunction

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 $$Im \frac{\partial}{\partial x}log \xi (1/2+ix)$$

Last edited: Aug 11, 2010
19. Aug 11, 2010

No need to yell.

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.

20. Aug 11, 2010

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

21. Aug 11, 2010

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.

Last edited: Aug 11, 2010
22. Aug 11, 2010

zetafunction

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

23. Aug 11, 2010

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.

Last edited: Aug 11, 2010
24. Aug 11, 2010

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 ?

25. Aug 11, 2010

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