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In their expository paper, ''Physics of the Riemann Hypothesis arxiv.org/abs/1101.3116v1'', Hutchison and Schumayer suggested the following approach on the Hilbert Polya conjecture, via quantisation of the classical Hamiltonian $H=xp$.

They consider the dilation symmetry of this Hamiltonian $x\mapsto \lambda x, p \mapsto p/\lambda$) which manifests itself in the transformation of the corresponding wave function as

$$\varphi(\lambda x) = \frac{1}{\lambda^{1/2-iE}}\varphi(x)$$

and one might suggest restricting ourselves to λ being a positive integer. They claim that this could be an attractive suggestion, because the wave-packet, generated by the uniform superpositions of all these transformed wavefunctions is

$$\psi(x) =\zeta(1/2 - iE)\varphi(x).$$

where $\zeta$ is the Riemann zeta function and $E$ is real. They then state that however, there is no physical motivation which would require this $\zeta$ pre-factor to vanish. Furthermore this integer-based dilation-symmetry does not form a group, because the multiplicative inverse element (which would be $\lambda = 1/m$) is missing.

But even if one could find some physical motivation for the pre $\zeta$ factor to vanish, wouldn't it only imply that some (not necessarily all) zeros of $\zeta(1/2 -iE)$ are real, which is not suffficient for the Hilbert-Polya conjecture to be true ?

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    $\begingroup$ Isn't the whole problem of these types of Hamiltonians, that no one knows if they are hermitian (or at least $\mathcal{PT}$ inversion symmetric). And that this question is exactly the RH. And it seems to be everyone is exactly as far from proving that as he is from proving the RH. $\endgroup$ Feb 14, 2019 at 12:03
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    $\begingroup$ you can find a recent criticism of this approach based on $H=(xp+px)/2$ in a comment by Jean Bellissard. $\endgroup$ Feb 14, 2019 at 12:42
  • $\begingroup$ The authors addressed the comments. The conjecture is asking about the real part of a complex number, if the root is imaginary then the real part is simply zero. Also, the roots have been bounded in the range 0 to 1 already. So their proof is sufficient. $\endgroup$
    – user149963
    Dec 16, 2019 at 8:22

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