Riemann hypothesis: is Bender-Brody-Müller Hamiltonian a new line of attack?

Question

There is a beautiful paper in Physical Review Letters [PRL 118, 130201 (2017), DOI:10.1103/PhysRevLett.118.130201] by Carl Bender, Dorje Brody, and Markus Müller (BBM) on a Hamiltonian approach to the Riemann Hypothesis. The paper is surprisingly easy to follow for a physicist.

BBM define a Hamiltonian $$\hat{H} = (1- e^{-i \hat{p}} )^{-1} ( \hat{x} \hat{p} + \hat{p} \hat{x} ) (1- e^{-i \hat{p}})$$ where $p=- i \partial_x$ is the momentum operator in $\hbar=1$ units.

The authors show that eigenfunctions of $\hat{H}$ vanishing at infinity must be be in the form of Hurwitz theta function, $\psi_z= - \zeta(z,x+1)$, so that $$\hat{H} \psi_z = i (2 z -1)\psi_z $$

Imposing a boundary condition $\psi_z(0)=0$, by the virtue of $\zeta(z,1)=\zeta(z)$, they show that all non-trivial zeros of the Riemann $\zeta$ function must be eigenvalues of $\hat{H}$ with the imposed boundary conditions.

BBM call this result a "a complex extended version of the Berry-Keating conjecture" and go on to provide heuristic arguments that all eigenvalues of $\hat{H}$ are real.

How promising is this new development in the context of solving the Riemann Hypothesis?

Steven Strogatz seems optimistic.

Update [19.10.2020]: The authors of [1] have published additional comments [3 , 4], including a response [4] to @Jean Bellissard's comment [5] that grew out of his answer below.

Answer 1

I tried to put my own argument on this blog recently but I failed. Let me try again. 1)- The original definition of $\zeta(z,x) $ is

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

Then $\zeta(z,x)-\zeta(z,x-1)=-1/x^z$. Then, using the integral representation, this relation extends by analiticity to the maximal domain in which both sides are defined and holomorphic. For our purpose, it is enough to consider $x>0$ and $\Re{z}>0$. Let $\zeta_z$ denote the map $x\to \zeta(z,x)$. Then, using the generator $A$ of the dilation operator, it is easy to check that the eigenvalues have the form $1/2+\imath E$. For $A=(pq+qp)/2=-\imath xd/dx-\imath 1/2$. This gives $A(1/x^z)=\imath (z-1/2)(1/x^z)$. Going from $A$ to $H$ is done using the operator $\Delta$, leading to the formal equation $H\zeta_{1/2+\imath E}=E\zeta_z$. The Dirichlet b.c. at the origin forces E to satisfy $\zeta(1/2+\imath E)=0$.

2)- The first problem I see is that $\zeta_z$ does not seem to belong to the Hilbert space ${\mathcal H}=L^2(0,\infty)$. Using a classical method by Hadamard

$$\int_0^\infty |\zeta(z,x)|^2 dx=\frac{1}{1-2\Re{z}}\zeta(2\Re{z}-1)$$

which converges for $\Re{z}>1$. Using the integral representation, a similar firmula can be obtained, but I failed to show that square integrability hold for $\Re{z}=1/2$. Hence I see a problem here.

3)- The other problem is the definition of $\Delta$. Let $S$ denote the translation operator, defined non rigorously by $Sf(x)=f(x-1)$. If restricted to $\mathcal H$, it is not unitary, because it is defined only for $x>1$. We can defined it by imposing $Sf(x)=0$ for $x\in [0,1]$. Then it is only a partial isometry. For $S^\ast f(x)=f(x+1)$ leading to $S^\ast S=I$, but $S S^\ast =P$ is the projection onto $L^2(1,\infty)$. Using these notations $\Delta= I-S=(S^\ast-I)S$. But we get a problem here: the function $\zeta_z$ is defined for $x>-1$, and the extension to the interval $(-1,0]$ is explicitly used in 1)-. So we cannot use $S$. But then what is the operator $e^{-\imath \hat{p}}$ used by the authors?

4)- If $\hat{p}$ is the usual operator

$$\hat{p}=-\imath \frac{d}{dx}$$

then there is a problem with its domain of definition. On $L^2(\mathbb R)$, it is selfadjoint as can be seen by using Fourier transform. But on $\mathcal H$, it is not. This is a classical exercise found in the very old book of Courant-Hilbert. Namely one can always define it in the set of $L^2$ functions with $L^2$ derivative, vanishing at $x=0$. Then it is symmetric. If so its adjoint is defined on the same space but without the vanishing at $x=0$. Not only the adjoint is not symmetric but its set of eigenvalues is the open lower half plane. This is because if $f_z(x)=e^{zx}$, then $\hat{p}^\ast f_z=-\imath z f_z$, while $f_z\in \mathcal{H}$ for $\Re{z}<0$. The same argument shows that $+\imath$ cannot be an eigenvalue. This means that the "defect indices", namely the dimension of the eigenspaces with eigenvalues $\pm\imath$, are not equal. Then, the von Neumann theorem show that the operator $\hat{p}$ has no selfajoint extension. If not selfadjoint, the definition of its exponential becomes a problem, because the functional calculus is nit defined in general.

5)- The previous argument can be rephrased in terms of the operator $S$. Its adjoint admits a lot of eigenvalues, namely the points inside the unit disk.

In conclusion, the sloppyness of the definitions used but the authors leads to a complete mess. Nothing is correct in this paper.

As long as physicists use algebra, or algorithmic arguments, they can find outstanding results. But when it comes to analysis, they may loose their judgment, and grave mistakes show up at the corner. Analysis is not easily amenable to algoritmic descriptions. And this is precisely where the power of Mathematics lies: by manipulating infinities, Mathematics goes way beyond the Church-Turing definition of computability. And what is Analysis if not manipulating infinities, through limits, convergence and the likes?

Answer 2

I didn't get everything written in this paper.

But keep in mind this line of attack has some good chances to be way too simple : all this works the same for $F(s) = \alpha L(s,\chi_5)+\overline{\alpha}L(s,\overline{\chi_5})= 2\sum\limits_{n=1}^\infty \Re(\alpha \;\chi_5(n)) n^{-s}$ where $\chi_5$ is the non-real character modulo $5$ and $\alpha \in \mathbb{C}$.

$F(s)$ has the same kind of integral representation and functional equation as $\zeta(s)$, so we can write for it the same kind of differential operator. But the RH obviously fails for $F(s)$ (it doesn't have an Euler product)

Answer 3

It's an interesting article. The main problem I can see is that their boundary condition invokes the zeta function itself, so this approach may be incorporating the RH itself from the very beginning. The approach doesn't seem to bring in anything deep about the primes nor number theory.

Having worked on this problem over several years, I remain skeptical about the Hilbert-Polya idea. Berry-Keating did some very interesting and important work. However the random matrix theory connection would seem to cast doubts on Hilbert-Polya rather than support it: if the statistics of the zeros can be explained by random matrix theory, it is hard to imagine a simple, non-random hamiltonian that would reproduce this spectrum. But this is not a rigorous counter-argument of course.

Answer 4

I am a number theorist who has thought about the Riemann Hypothesis for many years. This paper has elementary mistakes and there are simple counterexamples to the approach taken. It would not have been published in a mathematics journal, and I am disappointed (but not surprised) it has been published in PRL. Bellissard has listed some of the mistakes in the analysis (see arXiv:1704.02644) and the number theory follies would be clear to any graduate student in the field.

It is interesting sociologically that so many well known physicists are blogging or tweeting about this when they could just consult experts in their universities. For those experts, this is a joke.

A dark day for science.

Answer 5

The paper in question is written at the physical level of rigor, is published in a physics journal, and doesn't seem to make claims that it establishes mathematically rigorous results. However, I'm not convinced by the criticism so far that it's impossible to turn the main idea of the paper into a mathematically well-defined $L^2$ spectral problem whose solution would imply RH.

Perhaps something along the following lines may work.

1) The eigenfunctions $\psi_z(x)$ of $\hat{H}$ are indeed not in $L^2(0, \infty)$ but it seems that one can map the ones that correspond to the non-trivial zeros of $\zeta$ as follows. Let $w(x)$ be a positive smooth function on $(0,\infty)$ that coincides with $1/x$ near $0$ and with $1/x^{\frac{3}{2}}$ near infinity and let $\hat{W}$ be the operator given by multiplication with $w$. Then $\hat{W}\psi_z \in L^2(0, \infty)$ if and only if $z$ is a non-trivial zero of the Riemann zeta function. Indeed, $\hat{W}\psi_z$ is integrable at $0$ if and only if $\psi_z(0)=0$, i.e. if $z$ is a zero of $\zeta$. Further, known asymptotic properties of the Hurwitz zeta function (already mentioned in the BBM paper, namely sublinear growth for $z$ in the critical strip and much faster growth when $z$ is a trivial zero) imply that $\hat{W}\psi_z$ is square-integrable at infinity only when the zero $z$ is non-trivial. (The behavior of $\psi_z(x)$ at infinity is also discussed in Jean Bellissard's Comment paper arxiv.org/abs/1704.02644, using the integral representation of the Hurwitz zeta function.)

2) Now consider the operator $\hat{H}_W=\hat{W}\hat{H}\hat{W}^{-1}$, so that, at least formally, $\hat{W}\psi_z$ are eigenfunctions of $\hat{H}_W$ for every non-trivial zero $z$. To make this rigorous, one has to show that $\hat{H}_W$ is a densely defined closable operator on $L^2(0,\infty)$ and that $\hat{W}\psi_z$ are in the domain of its closure. (That of course would include settling quite a few technicalities regarding the proper definition and domain of $\hat{\Delta}^{-1}$, the inverse of the difference operator considered in the BBM paper, but I don't see why this would be apriori impossible.)

3) Assuming that was described in 2) can be done, one gets an embedding of the nontrivial zeros into the discrete spectrum of $\hat{H}_W$. Now It's known that the Berry-Keating Hamiltonian $\hat{x} \hat{p} + \hat{p} \hat{x}$ is selfadjoint and hence has real spectrum, so one may try to use the fact that $\hat{H}_W$ is similar to $\hat{x} \hat{p} + \hat{p} \hat{x}$ to obtain estimates on the resolvent operator that imply that the spectrum of $\hat{H}_W$ is contained in the spectrum of $\hat{x} \hat{p} + \hat{p} \hat{x}$.

Finally, of course it may be that $\hat{H}_W$ has lots of other spectrum including complex one, while its discrete spectrum is real, so that RH may still be true but not accessible via this line of attack.

Answer 6

In my previous answer I left out the issue regarding how one can make sense rigorously of the operators $\hat{\Delta}=1-e^{-i \hat{p}}$ and $\hat{\Delta}^{-1}=(1- e^{-i \hat{p}} )^{-1}$ used in the BBM paper. I'd like to elaborate on this a bit.

1) As it's well-known (and explained in Jean Bellissard's answer) the momentum operator $\hat{p}$ doesn't admit selfadjoint extensions on $L^2(0,\infty)$, hence one cannot use (at least not directly) functional calculus to make sense of $\hat{\Delta}$. In the BBM paper $\hat{\Delta}$ is interpreted as a difference operator and $\hat{\Delta}^{-1}$ as its inverse defined via infinite series (apparently, following Euler rather closely), but this approach is not satisfactory for many reasons.

2) It would be much more fitting to the general quantization idea to think of $\hat{\Delta}$ and $\hat{\Delta}^{-1}$ as pseudodifferential operators. Since this cannot be done directly, I suggest to proceed as follows, using appropriate cut-off functions to obtain smooth compactly supported symbols. Let $\kappa_{\varepsilon}(p)$ be a family of functions in $C^{\infty}_{c}(\mathbb{R})$ which is 1 on $[-1/ \varepsilon, 1/ \varepsilon]$ and $0$ away from this interval, so that $\kappa_{\varepsilon} \rightarrow 1 $ as $\varepsilon \rightarrow 0$. Let $\hat{\Delta}_{\varepsilon}$ be the family of pseudodifferential (in fact smoothing) operators with symbols $(1-e^{-i p})\kappa_{\varepsilon}(p)$. Further, let $\chi_{\varepsilon}(p)$ be another family in $C^{\infty}_{c}(\mathbb{R})$ which is $0$ on $[-\varepsilon, \varepsilon]$ as well as outside $[-1/ \varepsilon, 1/ \varepsilon]$ and 1 in between, so that still one has $\chi_{\varepsilon} \rightarrow 1 $ as $\varepsilon \rightarrow 0$. Let $\hat{\Delta}_{\varepsilon}^{-1}$ be the family of pseudodifferential operators with symbols $(1-e^{-i p})^{-1}\chi_{\varepsilon}(p)$. (There is a plenty of research on pseudodifferential operators on the half-line and I suspect some of it may be very relevant to this particular context.)

3) Now an appropriate limit (say, weak graph limit) as $\varepsilon \rightarrow 0$ of these families of operators may or may not exist but even if doesn't not exist, it looks plausible to me that one can show that the formal relations among $\hat{\Delta}$, $\hat{\Delta}^{-1}$ and the Hurwitz zeta function given in the BBM paper hold ''asymptotically'' as $\varepsilon \rightarrow 0$. Thus the same should be true for the formal eigenvalue/eigenfunction equation, so that if one defines $\hat{H}_{\varepsilon}=\hat{\Delta}_{\varepsilon}^{-1}(\hat{x} \hat{p} + \hat{p} \hat{x})\hat{\Delta}_{\varepsilon}$, one would have $$\lim_{\varepsilon \rightarrow 0}\hat{H}_{\varepsilon}\psi_z=i(2z-1)\psi_z $$ as, say, weak limit of distributions.

4) Finally, one can look at the operators $\hat{H}_{\varepsilon,W}=\hat{W}\hat{H}_{\varepsilon}\hat{W}^{-1}$, where $\hat{W}$ was defined in my previous answer. The desired reality of the eigenvalues would then follow from a uniform (in $\varepsilon$) bound on the norms of resolvent operators of $\hat{H}_{\varepsilon,W}$ for every fixed non-real $z$. There are well-known sufficient conditions for the $L^2$-boundedness of a pseudodifferential operator in terms of its symbol.

Answer 7

Some of the main discrepancies regarding the Bender-Brody-Müller (BBM) conjecture are with regard to the domain of the operator, and the convergence of the eigenspectrum. By considering the problem as a Schrödinger operator, one may obtain a convergent eigenspectrum. Such solutions are discussed here, and here.