Proof Of Rh From Hurwitz Eigenstates

Under my previous (QM-on-graphs) blog postal service nearly the Riemann Hypothesis, Dilaton was forgiven for having brought us about cute internet banalities ;-)



spell Akhmeteli pointed out a newspaper that seems fifty-fifty to a greater extent than promising than my most recent specific attacks:

Hamiltonian for the zeros of the Riemann zeta function

In their PRL paper, Carl M. Bender, Dorje C. Brody, in addition to Markus P. Müller (BBM) truly constructed a Hamiltonian whose eigenvalues seem to hold upwards the zeroes of the zeta percentage in addition to that seems to hold upwards Hermitian, afterwards about straightforward alter of the metric. How does it work?




The Hilbert-Pólya programme wants to assay the Riemann Hypothesis inward this trend – to uncovering a overnice operator \(S\), easily proven to hold upwards Hermitian in addition to easily proven to convey the spectrum linked to the zeroes of the zeta-function. Influenza A virus subtype H5N1 subprogram of this program, the Berry-Keating conjecture, claims that the operator \(S\) that tin give the axe fulfill this amazing role of proving the Riemann Hypothesis is a quantum deformation of the operator \(-ixp=-x\partial_x\) inward quantum mechanics. It makes about feel because \[

-ixp=\frac 12-i\frac{xp+px}{2}

\] hence the existent percentage equal to \(1/2\) is classify of naturally explained.




If in that place is the operator \(S\) amongst the required eigenvalues, in that place should also hold upwards the required eigenstates. For every zip of the zeta function, in that place should hold upwards a percentage – or a collection of probability amplitudes – that should hold upwards affiliated amongst the eigenvalue (the zip of the zeta function) but however depend on about additional observables. So the eigenstates should hold upwards about functions closely related to the Riemann zeta percentage – or their deformations.

At to the lowest degree since 1994, I considered the Hurwitz zeta function to hold upwards the most natural extra-parameter generalization of the Riemann zeta function. In fact, inward 1994 or so, I independently "reinvented it", used it to calculate that \[

1+2+3+\dots = -\frac{1}{12},

\] in addition to the values of the zeta percentage for all other negative integers, in addition to published these derivations inward a student journal (Pictures of Yellow Roses; the spider web version of the mag is to a greater extent than "recent", from 1997; I couldn't convey edited the page or delete it from 1999 or so).

What is the Hurwitz zeta function? It's the Riemann zeta percentage amongst an adjustable "delay" \(q\):\[

\zeta(s,q) = \sum_{n=0}^\infty \frac{1}{(n+q)^{s}}.

\] In the normal zeta function, the initiatory off term of the amount is \(1/1^s\), hither is is \(1/q^s\), in addition to hence yous growth \(q\) past times \(1\) inward each step. So for \(q=1\), the Hurwitz zeta percentage reduces to the Riemann zeta function:\[

\zeta(s,1) = \zeta(s).

\] My convention was shifted inward the 1990s, the normal zeta was obtained for \(q=0\). But now, the flim-flam of BBM – inward my interpretation, the "main" flim-flam of BBM – is to usage these Hurwitz functions every bit functions of \(q\) playing the role of \(x\) inward \(\psi(x)\) every bit the eigenstates. The declaration \(s\) inserted to the Hurwitz zeta percentage is the same that yous desire to associate amongst the eigenvalues of the operator.

I suspect that it could hold upwards useful to usage the alternating Riemann, Hurwitz zeta functions – I hateful the Dirichlet eta function in addition to its Hurwitz-Euler eta generalization – to brand things to a greater extent than convergent inward the critical strip instead of the uniform-sign functions.

OK, hence my version of BBM would start past times proverb that the required Hilbert-Pólya eigenstates are the Hurwitz zeta functions\[

\psi_s(x) = \zeta(s,x+1), \quad x\in \RR^+

\] If yous convey such a percentage of \(x\gt 0\) for a fixed \(s\), which is – permit me remind you\[

\zeta(s,x+1) = \sum_{n=1}^\infty \frac{1}{(n+x)^{s}},

\] how practice yous extract the eigenvalue \(s\) from that function? BBM don't boot the bucket along inward this trend at all – what I am doing would likely hold upwards contrary applied scientific discipline from their viewpoint – but this mental attitude is trend to a greater extent than natural for me. My derivation that the percentage is an eigenstate of an operator is arguably much simpler, too.



The Hurwitz functions of \(x\) await beautiful in addition to "oscillating" enough. Here yous convey a colorful representation of a moving ridge percentage for \(s=3+4i\).

An uncomplicated – peradventure almost defining – belongings of the Hurwitz zeta function's dependence on \(x\) is that if yous growth \(x\) past times one, it's equivalent to omitting the initiatory off term (the power). So:\[

\exp(\partial_x) \zeta(s,x+1) = \sum_{n=2}^\infty \frac{1}{(n+x)^{s}}

\] Calculate the divergence betwixt the final equation in addition to the previous 1 to get:\[

[\exp(\partial_x) - 1] \zeta(s,x+1) = \frac{1}{(1+x)^s}

\] That's cool: if yous apply this divergence operator (the shift of \(x\) past times one, minus the identity operator), yous teach a simple moving ridge percentage that behaves every bit a might of \(x\). How practice yous extract \(s\) from this moving ridge function? Well, human action amongst the aforementioned Berry-Keating operator \(-x\partial_x\). This operator, when acting on the might \((1+x)^{-s}\), picks the exponent \(s\) every bit a factor, reduces the exponent past times one, in addition to hence returns it to the master copy value. So\[

-x\partial_x[\exp(\partial_x) - 1] \zeta(s,x+1) = \frac{s}{(1+x)^s}.

\] The correct paw side exclusively differs past times the extra \(s\) inward the numerator. We convey almost obtained what nosotros wanted. But to convey an operator amongst its eigenstates, nosotros must truly obtain a multiple of the master copy moving ridge function, i.e. of \(\zeta(s,x+1)\), on the correct paw side. How practice nosotros provide from the simple might on the correct paw side of the previous displayed formula to the total amount defining the Hurwitz zeta function? Well, nosotros must amount over the values of \(x\) betwixt \(x=0\) in addition to \(x=\infty\) again.

So our operator (or the previous displayed equation) has to hold upwards multiplied from the left past times the extra operator\[

1+\exp(\partial_x) + \exp(2\partial_x)+\dots = \frac{1}{1-\exp(\partial_x)}.

\] Once nosotros usage the formula for geometric series, nosotros run into that it's the inverse operator of the operator nosotros started with. I truly practice believe that my derivation of the fact that the operator does what it should is much to a greater extent than straightforward in addition to simpler than theirs. So when the dust settles, our BBM operator of selection is simply\[

south = \frac{1}{1-\exp(\partial_x)} \cdot (-x\partial_x) \cdot [1-\exp(\partial_x)]

\] The eigenvalues should hold upwards just the values of \(s\) for which \(\zeta(s)=0\). Up to sign conventions in addition to a linear redefinition of \(s\) (a factor of \(2i\) in addition to a shift past times \(1/2)\), the operator \(S\) is the same every bit their operator \(H\) inward equation 1 of the BBM paper.

But just attempt to appreciate how far BBM convey gotten: they convey natural moving ridge functions that should hold upwards the eigenstates corresponding to the Riemann zeta function's zeroes; in addition to they tin give the axe write downward a natural operator – a composition of the divergence operator, its difference, in addition to the \(-x\partial_x\) differential operator inward betwixt – that, when acting on the eigenstates, truly produces the eigenstate equation amongst the required eigenvalue.

Now, recollect that nosotros had \(x\gt 0\). The moving ridge percentage must teach to zip for \(x\to 0^+\) – this is just similar for particles inward a potential good (the moving ridge percentage has to vanish where the good starts) – which is equivalent to \(\zeta(s)=0\). And what nearly the Hermiticity? First, the eigenfunctions corresponding to the picayune zeroes – amongst existent negative integer \(s\) – volition arrive at non-normalizable moving ridge functions. On the other hand, nosotros desire the non-trivial zeroes to prevarication on the axis \(1/2+it\) which way that on the relevant Hilbert space, \(S-1/2\) should hold upwards anti-Hermitian.

As written down, the operator \(S\) isn't anti-Hermitian. But if yous define a novel metric on the Hilbert space\[

\langle \varphi,\psi\rangle = \bra\varphi [1-\exp(-\partial_x)][1-\exp(\partial_x)] \ket\psi,

\] hence the denominator at the showtime of \(S\) volition cancel in addition to the extra factor volition only hold upwards the Hermitian conjugate of the final factor inward \(S\), hence nether this modified inner product, \(S-1/2\) volition hopefully hold upwards anti-Hermitian.

(I am a fleck uneasy nearly the divergence operator calculated inward the opposite direction. Also, it's possible that yous truly desire to cancel the numerators in addition to "double" the denominators instead. There is about doubt nearly which selection of these details truly industrial plant best.)

Some comments nearly the normalizability of the moving ridge functions nether the novel inner production await to a greater extent than ambiguous in addition to complicated than what yous would similar hence a select handgrip of may hold upwards hiding here. Also, it seems possible to me that the correct moving ridge percentage shouldn't hold upwards defined for all values of \(x\) but exclusively for integer values of \(x\). All the kernels should hold upwards explicitly rewritten inward the \(x\)-representation, in addition to hence on.

But I think that I convey never made a estimate that the probability that a especial proof of the Riemann Hypothesis is gonna before long hold upwards considered "almost complete" was greater than 50%. This structure looks extremely specific in addition to promising to me.