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A proposed Hamiltonian operator for Riemann Hypothesis

  1. Jul 20, 2010 #1
    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 [tex] Arg\xi(1/2+iE) [/tex]

    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

    [tex] \xi(1/2+iE) [/tex] 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 [tex] H=p^2 +V(x) [/tex] 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 [tex] H+m^2 [/tex]is the Riemann Xi-function
  2. jcsd
  3. Jul 20, 2010 #2


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    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.
  4. Jul 20, 2010 #3
    translation would be

    HAMILTONIAN : linear differential operator of the form [tex] H=-y''(x)+f(x)y(x) [/tex]

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

    ENERGIES = eigenvalues of the operator

    [tex] f^{-1}(x)= A \frac{d^{-1/2}g(x)}{dx^{-1/2}} [/tex] 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

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

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

    the inverse of f(x) is NOT exact is valid as an ASYMPTOTIC formula ,
    Last edited: Jul 20, 2010
  5. Jul 20, 2010 #4


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    So don't write = unless you're going to put an error term on the right.
  6. Jul 20, 2010 #5
    umm sorry in physics is used to write = even for asymptotic, i think this would be better

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

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

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

    p is the momentum and q is the position
    Last edited: Jul 20, 2010
  7. Jul 21, 2010 #6


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    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?
  8. Jul 21, 2010 #7
    Do you believe that you have found the relevant [itex]V(x)[/itex], or that you have only proven its existence?
  9. Jul 22, 2010 #8
    WKB approximations are just this approximation, i have tested my formula with 2 cases

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

    POtential well [tex] f=0 [/tex] then i get the approximate value [tex] 2/\pi [/tex]

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

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


    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 [tex] \pi n(E)=Arg\xi(1/2+iE) [/tex]
  10. Jul 31, 2010 #9
    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
  11. Aug 1, 2010 #10
    Either you have a definition for the [itex]V(x)[/itex], 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.
  12. Aug 1, 2010 #11
    http://vixra.org/pdf/1007.0005v2.pdf page 3 , formula (5)

    [tex] \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)} [/tex]

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

    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

    [tex] n(E)= A\int_{0}^{E}dx (E-V(x))^{1/2}= \frac{1}{\pi}Arg\xi(1/2+iE) [/tex]
    Last edited: Aug 1, 2010
  13. Aug 1, 2010 #12
    I glanced through the paper in hope of seeing the definition. This equation didn't draw my attention because it had an approximation sign "[itex]\approx[/itex]" in it.
  14. Aug 8, 2010 #13
    i have checked the first 3 eigenvalues of my Hamiltonian and they satisfy

    [tex]E_{0}= 13.1090 [/tex] , [tex]E_{1}= 20.1254 [/tex] , [tex]E_{2}= 30.0679 [/tex]

    using the semiclassical approximation http://en.wikipedia.org/wiki/WKB_approximation to solve the differential equation
  15. Aug 11, 2010 #14
    Are those the imaginary parts of the first three non-trivial zeros of the Riemann Zeta function?
  16. Aug 11, 2010 #15
    The first few zeros have imaginary part 14.1347, 21.022, 25.0109, 30.4249, 32.9351, says Wolfram.
  17. Aug 11, 2010 #16
    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
  18. Aug 11, 2010 #17
    I can't even tell what you claim to have proven.
  19. Aug 11, 2010 #18
    i give a DIFFERENTIAL OPERATOR whose EIGENVALUES are the imaginary part of the Zeros of the Riemann Zeta function


    see formula (5) for the potential.

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

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

    and g(x) here is [tex] Im \frac{\partial}{\partial x}log \xi (1/2+ix) [/tex]
    Last edited: Aug 11, 2010
  20. Aug 11, 2010 #19
    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.
  21. Aug 11, 2010 #20
    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
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