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Hilbert-Polya conjecture:QM and number theory

  1. Sep 19, 2005 #1
    Now i would like to solve RH by using quantum physics,the problem is equivalent to find an operator [tex]H=aD^{2}+V(x)[/tex] with a real or complex potential V so the eigenvalues of H are precisely the roots of the function (a is a real constant) [tex]\zeta(b+is)[/tex],for b=1/2 the Riemann functional equation:


    tells us that if s is an energy of H for the case a=1/2 also s* is another energy for the H,then taking the expectation value of the Hamiltonian with [tex]\phi_{n}[/tex] and [tex]\phi_{k}[/tex] being the eigenfunctions associated to the eigenvalues s and s* then we would have:

    [tex](<\phi_{n}|H|\phi_{n}>)*=(<\phi_{k}|H|\phi_{k}>) [/tex]

    s=E_{n} and s*=E*_{n}=E_{k} from this last equation we get that for b=1/2 the potential is real.

    The cases b<>1/2 we have complex "energies" in the form s*+(2a-1)i (just apply R Equation) so the potential for these roots must be complex and a complex potential can,t have real energies as <b> (the complex part of the potential is not 0 [H,b]=d being d a non-zero operator.)

    Using perturbation theory we can calculate the potential by solving an integral equation:


    with [tex]E^{0}_{n},\psi [/tex] the eigenvalues and eigenfunctions of H0=P^2/2m ( a free particle moving on an infinite potential well)

    The last equation is an integral equation,we can solve it by Resolvent Kernel method getting finally a formula for V in the form:

    [tex]V(x)=\int_{-\infty}^{\infty}dnR(n,x)\delta{E(n)}[/tex] valid for whatever b is so we could use it to prove RH is any desired form.....as we have found a Hamiltonian whose energies are the roots of [tex]\zeta(b+is)[/tex]. for every a 0<b<1
    Last edited: Sep 20, 2005
  2. jcsd
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