Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

what would happen if one found a Hamiltonian with an smooth level density in the form

$$ N(E)= \frac{E}{2\pi}\log\left(\frac{E}{2\pi e}\right)$$

which is exactly the density of the RIemann zeros..

this means that the energy levels of such operator would be asymptotically exact to the Riemann zeros , and also the level spacing would be on average the same of the Zeros so $$ E_{n} = \frac{2\pi n}{\log n} $$

and $$ E_{n}-E_{n+1} \to 0 $$ as $n \to \infty $

just as montgomery conjecture predicts..

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.