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HIlbert-Polya conjecture ¿proof or RH?

  1. Apr 3, 2006 #1
    My question is...could the Hilbert-Polya conjecture if true prove RH (Riemann Hypothesis) i mean let,s suppose we find an operator ( i found a Hamiltonian with a real potential that gave all the roots of [tex] \zeta(1/2+is) [/tex] ) in the form:

    [tex] R=1/2+iH [/tex] with H self-adjoint so all the "eigenvalues" of R are precisely the roots of the Riemann zeta function.... would if mean that Rh is true?..what would happened if we find another operator

    [tex] R^{a}=a+iT [/tex] with T also self-adjoint and a different from 1/2 [/tex] ? or perhaps not so worse, an operator but this time T ISN,T self-adjoint so we are granted that all its eigenvalues wont, be real but...what would happen if T in spite of not being self-adjoint had a real root?..then the Riemann zeta function would have a real root in the form a+it with a and t real and a different from 1/2

    so in what sense is the Hilbert-Polya hypothesis true and is the same as RH?..:grumpy: :grumpy:
  2. jcsd
  3. Apr 3, 2006 #2

    matt grime

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    None of that makes any sense.

    And, no you didn't find a Hamiltonian operator whose roots were those of the zeta function in the critical strip. If you had then you'd have proven the Riemann Hypothesis. As you've not managed to demonstrate you've done this would you mind not implying you had?
  4. Apr 3, 2006 #3
    -i found a Hamiltonian..i still have the paper in .pdf if you want i can submit to you or to your teachers...i think we discussed enough about it at its time don,t you?...

    -teh fact is that according to this H-P conjecture there should be exist an operator:

    [tex] R=1/2+iH [/tex] with H=H+ (self-adjoint) that its eigenvalues are the roots of the zeta function.

    but let,s suppose we managed to find these operators:

    a) [tex] R=a+iT [/tex] T is self-adjoint and a is different from 1/2

    b) [tex] R=a+iT* [/tex] T* is not self adjoint but has a real eigenvalue so

    there would be a root of Zeta function in the form a+it.

    -So perhpas H-P operator existence is a necessary but not sufficient condition to prove RH
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