A proposed Hamiltonian operator for Riemann Hypothesis

[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.