# A Hamiltonian with smooth term exact to the Riemann zeros

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..

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