Is at that topographic point a agency to show 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 show that theorem just the proof wasn't always establish in addition to it seems probable straightaway that he didn't accept one.
The Hilbert-Pólya conjecture or plan postulates that at that topographic point is a natural plenty operator \(P\) such every bit 1 inwards quantum mechanics whose eigenvalues are the numbers \(t\) labeling the imaginary component of the zeroes \(s\) of the zeta function. There every bit good exists an independent proof of the Hermiticity of \(P\) in addition to in 1 trial yous show that the operator has the desired eigenvalues in addition to it is Hermitian, yous accept proven the Riemann Hypothesis.
Almost all my efforts to show the Riemann Hypothesis construct on the Hilbert-Pólya image because it's the most physical argument. Someone who has done quantum mechanics for a long fourth dimension must unavoidably discovery it the most promising approach.
Within the Hilbert-Pólya approach, at that topographic point are lots of sub-approaches defined past times what sort of operators yous are trying to build. It's plausible that at that topographic point exists an infinite matrix \(P_{mn}\) whose eigenvalues are the numbers \(t\) from \(\zeta(1/2+it)=0\) in addition to the matrix entries for \(m,n\in\ZZ\) may travel every bit unproblematic every bit the greatest mutual denominator of \(m,n\), or its logarithm, or something of this kind.
I've seen lots of statements similar that which seem to piece of job just I haven't been able to discovery the exact Hermitian matrix \(P_{mn}\) that would accept the correct eigenvalues.
But at that topographic point is a rather precise Ansatz I figured out yesterday – afterwards I tried to heavily simplify some to a greater extent than complex in addition to muddled ideas. Quantum mechanics on graphs. What is it in addition to why could it work?
Note that the averaging spacing betwixt ii following primes is around \(\ln(p)\) inwards 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 role \(t\). So including the constituent of \(2\pi\), they await similar the inverse human relationship betwixt the spacing of positions in addition to momenta.
So the roots of the zeta role await similar the allowed momenta – except that their spacing gets finer if the momenta are higher. So it looks similar the \(n\)-th origin of the zeta role corresponds to some momentum fashion that "mostly lives" on a circle whose circumference is \(\ln p_n\) where \(p_n\) is the \(n\)-th prime.
If yous wish a quantum mechanical system, it looks similar yous wish some compact infinite – to brand the spectrum of momenta or energies discrete – in addition to it could locally await similar circles of circumference \(\ln p_n\) where \(p_n\) are primes. And yous should combine them inwards some way. Quantum mechanics on graphs could travel a agency that works. After I establish some reasons to intend so, I made a Google search in addition to establish a newspaper that used rattling dissimilar – in addition to less specific – arguments just every bit good establish some evidence that quantum mechanics on graphs could mimic the Riemann zeta zeroes, mayhap exactly.
Why could the spectrum of the momentum or loose energy on a graph precisely coincide amongst the zeroes of the Riemann zeta function?
Imagine that yous selection circles of circumference \(\ln 2,\ln 3,\ln 5\), in addition to all the other logarithms of primes. And yous bring together them at 1 point. Can quantum mechanics clitoris particle that propagates on this graph, including the singular betoken where infinitely many circles meet? It could travel able to practise so. But yous must define some boundary atmospheric condition at the vertex.
This discussion tells yous some basics near the allowed boundary conditions. You even thus wish to hit Hermitian operators for the momentum or the energy. Recall that the Hermiticity of the momentum inwards quantum mechanics boils downward to the integration past times parts. But the boundary price \([uv]\) 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 atmospheric condition plough over yous a circle, the vanishing of atmospheric condition at infinity is trivial. Dirichlet boundary atmospheric condition would travel bad because the particle wouldn't travel allowed to penetrate through the vertex. The alone novel interesting status that makes the boundary price vanish involves the Neumann boundary conditions\[
\sum_{e\sim v} f'(v) = 0
\] You core over all edges that plough over the sack at the given vertex, evaluate the derivative of the moving ridge role inwards the outgoing direction, in addition to core over all derivatives. If the moving ridge role itself is continuous on the graph – if \(f(v)\) has the same value regardless of the border on which yous approach the vertex – in addition to thus the momenta may travel Hermitian (you must brand the edges oriented inwards that case) and/or the kinetic loose energy is Hermitian at whatsoever rate.
The funny novel thought I realized earlier I saw that paper – in addition to every bit far every bit I tin see, the authors of that newspaper don't realize that smashing tidings – is that this Neumann boundary status for the vertex or vertices of the quantum graph could travel equivalent to \(\zeta(t)=0\). Recall that inwards the critical strip, the Riemann zeta role may travel calculated from the converging serial for the Dirichlet eta function, to selection a dainty 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 inwards the numerator could correspond to 1 contribution \(f'(v)\) to the Neumann boundary atmospheric condition for quantum mechanics on the graph. They await similar. For example, \(1/3^s = \exp(-(1/2+it)\ln 3)\) thus it looks similar some alter of the stage over the circle of circumference \(\ln 3\) if the momentum on that circle is \(t\). I don't know how I instruct the existent component \(1/2\) yet.
If 1 fine-tuned the details thus that the atmospheric condition would precisely match, yous would travel basically done. On the interpretation of quantum mechanics on graphs, the status would guarantee Hermiticity i.e. the reality of the values \(t\). But the status could travel seen to travel equivalent to \(\zeta(1/2+it)=0\).
Good luck amongst fine-tuning the details. I volition sure attempt to brand the pic to a greater extent than robust or completely robust if possible. ;-)
I accept every bit good some suspicion that the circles of circumference \(\ln p\) should actually travel viewed every bit parts of a hyperbola. The normal circle, the unit of measurement circle, has the circumference \(-2i\ln -1=2\pi\). If yous supercede the declaration \(-1\) past times positive ones in addition to erase the \(i\) inwards front end of it, yous should switch to the other signature. So this pic should accept a natural interpretation on some hyperbola \(xy=1\) amongst 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 \(SO(2,1)\sim SL(2,\RR)\) grouping or fifty-fifty the modular grouping \(SL(2,\ZZ)\) could play a role here. It could instruct combined amongst 1 of the other ideas of mine that at that topographic point are representations of \(SL(2,\RR)\) that are labeled past times the zeroes \(t\) of the zeta function.
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