Is there a way to prove that all the nontrivial zeroes \(s\) of the zeta function, i.e. values of \(s\) obeying \(\zeta(s)=0\), satisfy \(s=1/2+it\) where \(t\in \RR\)? Riemann thought he could prove that theorem but the proof wasn't ever found and it seems likely now that he didn't have one.

The Hilbert-Pólya conjecture or program postulates that there is a natural enough operator \(P\) such as one in quantum mechanics whose eigenvalues are the numbers \(t\) labeling the imaginary part of the zeroes \(s\) of the zeta function. There also exists an independent proof of the Hermiticity of \(P\) and once you prove that the operator has the desired eigenvalues and it is Hermitian, you have proven the Riemann Hypothesis.

Almost all my efforts to prove the Riemann Hypothesis build on the Hilbert-Pólya paradigm because it's the most physical argument. Someone who has done quantum mechanics for a long time must unavoidably find it the most promising approach.

Within the Hilbert-Pólya approach, there are lots of sub-approaches defined by what kind of operators you are trying to build. It's plausible that there exists an infinite matrix \(P_{mn}\) whose eigenvalues are the numbers \(t\) from \(\zeta(1/2+it)=0\) and the matrix entries for \(m,n\in\ZZ\) may be as simple as the greatest common denominator of \(m,n\), or its logarithm, or something of this kind.

I've seen lots of statements like that which seem to work but I haven't been able to find the exact Hermitian matrix \(P_{mn}\) that would have the right eigenvalues.

But there is a rather precise Ansatz I figured out yesterday – after I tried to heavily simplify some more complex and muddled ideas. Quantum mechanics on graphs. What is it and why could it work?

Note that the averaging spacing between two adjacent primes is around \(\ln(p)\) in average where \(p\) is roughly the size of each of them. On the contrary, the spacing is \(1/2\pi \ln(t)\) for the zeroes of the Riemann zeta function \(t\). So including the factor of \(2\pi\), they look like the inverse relationship between the spacing of positions and momenta.

So the roots of the zeta function look like the allowed momenta – except that their spacing gets finer if the momenta are higher. So it looks like the \(n\)-th root of the zeta function corresponds to some momentum mode that "mostly lives" on a circle whose circumference is \(\ln p_n\) where \(p_n\) is the \(n\)-th prime.

If you want a quantum mechanical system, it looks like you want some compact space – to make the spectrum of momenta or energies discrete – and it could locally look like circles of circumference \(\ln p_n\) where \(p_n\) are primes. And you should combine them in some way. Quantum mechanics on graphs could be a way that works. After I found some reasons to think so, I made a Google search and found a paper that used very different – and less specific – arguments but also found some evidence that quantum mechanics on graphs could mimic the Riemann zeta zeroes, perhaps exactly.

Why could the spectrum of the momentum or energy on a graph exactly coincide with the zeroes of the Riemann zeta function?

Imagine that you pick circles of circumference \(\ln 2,\ln 3,\ln 5\), and all the other logarithms of primes. And you join them at one point. Can quantum mechanics describe particle that propagates on this graph, including the singular point where infinitely many circles meet? It could be able to do so. But you must define some boundary conditions at the vertex.

This discussion tells you some basics about the allowed boundary conditions. You still want to produce Hermitian operators for the momentum or the energy. Recall that the Hermiticity of the momentum in quantum mechanics boils down to the integration by parts. But the boundary terms \([uv]\) must vanish. They typically vanish because of vanishing of all fields at infinity; or because of periodicity; or because of Neumann or Dirichlet boundary conditions.

Periodic boundary conditions give you a circle, the vanishing of conditions at infinity is trivial. Dirichlet boundary conditions would be bad because the particle wouldn't be allowed to penetrate through the vertex. The only new interesting condition that makes the boundary terms vanish involves the Neumann boundary conditions\[

\sum_{e\sim v} f'(v) = 0

\] You sum over all edges that terminate at the given vertex, evaluate the derivative of the wave function in the outgoing direction, and sum over all derivatives. If the wave function itself is continuous on the graph – if \(f(v)\) has the same value regardless of the edge on which you approach the vertex – then the momenta may be Hermitian (you must make the edges oriented in that case) and/or the kinetic energy is Hermitian at any rate.

The funny new idea I realized before I saw that paper – and as far as I can see, the authors of that paper don't realize that great news – is that this Neumann boundary condition for the vertex or vertices of the quantum graph could be

*equivalent*to \(\zeta(t)=0\). Recall that in the critical strip, the Riemann zeta function may be calculated from the converging series for the Dirichlet eta function, to pick a nice example:\[

\zeta(s) = \frac{ \frac{1}{1^s}-\frac{1}{2^s}+ \frac{1}{3^s}-\frac{1}{4^s}+\dots }{ 1 - 2^{1-s} }

\] Each term in the numerator could correspond to one contribution \(f'(v)\) to the Neumann boundary conditions for quantum mechanics on the graph. They look similar. For example, \(1/3^s = \exp(-(1/2+it)\ln 3)\) so it looks like some change of the phase over the circle of circumference \(\ln 3\) if the momentum on that circle is \(t\). I don't know how I get the real part \(1/2\) yet.

If one fine-tuned the details so that the conditions would exactly match, you would be basically done. On the interpretation of quantum mechanics on graphs, the condition would guarantee Hermiticity i.e. the reality of the values \(t\). But the condition could be seen to be equivalent to \(\zeta(1/2+it)=0\).

Good luck with fine-tuning the details. I will surely try to make the picture more robust or completely robust if possible. ;-)

I have also some suspicion that the circles of circumference \(\ln p\) should really be viewed as parts of a hyperbola. The normal circle, the unit circle, has the circumference \(-2i\ln -1=2\pi\). If you replace the argument \(-1\) by positive ones and erase the \(i\) in front of it, you should switch to the other signature. So this picture should have a natural interpretation on some hyperbola \(xy=1\) with the Minkowski signature on it. The path from \((x,y)=(1,1)\) to \((p,1/p)\) has the Minkowskian proper length (or rapidity) \(\ln p\).

The \(SO(2,1)\sim SL(2,\RR)\) group or even the modular group \(SL(2,\ZZ)\) could play a role here. It could get combined with one of the other ideas of mine that there are representations of \(SL(2,\RR)\) that are labeled by the zeroes \(t\) of the zeta function.

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