Wednesday, March 08, 2017

Proof of RH from Hurwitz eigenstates

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

while Akhmeteli pointed out a paper that seems even more 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, and Markus P. Müller (BBM) actually constructed a Hamiltonian whose eigenvalues seem to be the zeroes of the zeta function and that seems to be Hermitian, after some straightforward change of the metric. How does it work?

The Hilbert-Pólya program wants to prove the Riemann Hypothesis in this way – to find a nice operator \(S\), easily proven to be Hermitian and easily proven to have the spectrum linked to the zeroes of the zeta-function. A subprogram of this program, the Berry-Keating conjecture, claims that the operator \(S\) that can fulfill this amazing role of proving the Riemann Hypothesis is a quantum deformation of the operator \(-ixp=-x\partial_x\) in quantum mechanics. It makes some sense because \[

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

\] so the real part equal to \(1/2\) is sort of naturally explained.

If there is the operator \(S\) with the required eigenvalues, there should also be the required eigenstates. For every zero of the zeta function, there should be a function – or a collection of probability amplitudes – that should be affiliated with the eigenvalue (the zero of the zeta function) but still depend on some additional observables. So the eigenstates should be some functions closely related to the Riemann zeta function – or their deformations.

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

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

\] and the values of the zeta function for all other negative integers, and published these derivations in a student journal (Pictures of Yellow Roses; the web version of the journal is more "recent", from 1997; I couldn't have edited the page or delete it from 1999 or so).

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

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

\] In the normal zeta function, the first term of the sum is \(1/1^s\), here is is \(1/q^s\), and then you increase \(q\) by \(1\) in each step. So for \(q=1\), the Hurwitz zeta function reduces to the Riemann zeta function:\[

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

\] My convention was shifted in the 1990s, the normal zeta was obtained for \(q=0\). But now, the trick of BBM – in my interpretation, the "main" trick of BBM – is to use these Hurwitz functions as functions of \(q\) playing the role of \(x\) in \(\psi(x)\) as the eigenstates. The argument \(s\) inserted to the Hurwitz zeta function is the same that you want to associate with the eigenvalues of the operator.

I suspect that it could be useful to use the alternating Riemann, Hurwitz zeta functions – I mean the Dirichlet eta function and its Hurwitz-Euler eta generalization – to make things more convergent in the critical strip instead of the uniform-sign functions.

OK, so my version of BBM would start by saying that the required Hilbert-Pólya eigenstates are the Hurwitz zeta functions\[

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

\] If you have such a function of \(x\gt 0\) for a fixed \(s\), which is – let me remind you\[

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

\] how do you extract the eigenvalue \(s\) from that function? BBM don't proceed in this way at all – what I am doing would probably be reverse engineering from their viewpoint – but this attitude is way more natural for me. My derivation that the function is an eigenstate of an operator is arguably much simpler, too.

The Hurwitz functions of \(x\) look beautiful and "oscillating" enough. Here you have a colorful representation of a wave function for \(s=3+4i\).

An elementary – maybe almost defining – property of the Hurwitz zeta function's dependence on \(x\) is that if you increase \(x\) by one, it's equivalent to omitting the first term (the power). So:\[

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

\] Calculate the difference between the last equation and the previous one to get:\[

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

\] That's cool: if you apply this difference operator (the shift of \(x\) by one, minus the identity operator), you get a simple wave function that behaves as a power of \(x\). How do you extract \(s\) from this wave function? Well, act with the aforementioned Berry-Keating operator \(-x\partial_x\). This operator, when acting on the power \((1+x)^{-s}\), picks the exponent \(s\) as a factor, reduces the exponent by one, and then returns it to the original value. So\[

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

\] The right hand side only differs by the extra \(s\) in the numerator. We have almost obtained what we wanted. But to have an operator with its eigenstates, we must actually obtain a multiple of the original wave function, i.e. of \(\zeta(s,x+1)\), on the right hand side. How do we return from the simple power on the right hand side of the previous displayed formula to the full sum defining the Hurwitz zeta function? Well, we must sum over the values of \(x\) between \(x=0\) and \(x=\infty\) again.

So our operator (or the previous displayed equation) has to be multiplied from the left by the extra operator\[

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

\] Once we use the formula for geometric series, we see that it's the inverse operator of the operator we started with. I really do believe that my derivation of the fact that the operator does what it should is much more straightforward and simpler than theirs. So when the dust settles, our BBM operator of choice is simply\[

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

\] The eigenvalues should be exactly the values of \(s\) for which \(\zeta(s)=0\). Up to sign conventions and a linear redefinition of \(s\) (a factor of \(2i\) and a shift by \(1/2)\), the operator \(S\) is the same as their operator \(H\) in equation 1 of the BBM paper.

But just try to appreciate how far BBM have gotten: they have natural wave functions that should be the eigenstates corresponding to the Riemann zeta function's zeroes; and they can write down a natural operator – a composition of the difference operator, its difference, and the \(-x\partial_x\) differential operator in between – that, when acting on the eigenstates, really produces the eigenstate equation with the required eigenvalue.

Now, recall that we had \(x\gt 0\). The wave function must go to zero for \(x\to 0^+\) – this is just like for particles in a potential well (the wave function has to vanish where the well starts) – which is equivalent to \(\zeta(s)=0\). And what about the Hermiticity? First, the eigenfunctions corresponding to the trivial zeroes – with real negative integer \(s\) – will produce non-normalizable wave functions. On the other hand, we want the non-trivial zeroes to lie on the axis \(1/2+it\) which means that on the relevant Hilbert space, \(S-1/2\) should be anti-Hermitian.

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

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

\] then the denominator at the beginning of \(S\) will cancel and the extra factor will simply be the Hermitian conjugate of the last factor in \(S\), so under this modified inner product, \(S-1/2\) will hopefully be anti-Hermitian.

(I am a bit uneasy about the difference operator calculated in the opposite direction. Also, it's possible that you actually want to cancel the numerators and "double" the denominators instead. There is some uncertainty about which choice of these details actually works best.)

Some comments about the normalizability of the wave functions under the new inner product look more ambiguous and complicated than what you would like so a catch may be hiding here. Also, it seems possible to me that the right wave function shouldn't be defined for all values of \(x\) but only for integer values of \(x\). All the kernels should be explicitly rewritten in the \(x\)-representation, and so on.

But I think that I have never made a guess that the probability that a particular proof of the Riemann Hypothesis is gonna soon be considered "almost complete" was greater than 50%. This construction looks extremely specific and promising to me.

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