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..