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

ˆH=(1−e−iˆp)−1(ˆxˆp+ˆpˆx)(1−e−iˆp)

where p=−i∂x is the momentum operator in ℏ=1 units.The authors show that eigenfunctions of ˆH vanishing at infinity must be be in the form of Hurwitz theta function, ψz=−ζ(z,x+1), so that ˆHψz=i(2z−1)ψz

Imposing a boundary condition ψz(0)=0, by the virtue of ζ(z,1)=ζ(z), they show that all non-trivial zeros of the Riemann ζ function must be eigenvalues of ˆH 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 ˆH 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**

I tried to put my own argument on this blog recently but I failed. Let me try again.

1)- The original definition of ζ(z,x) is

ζ(z,x)=∞∑n=11(n+x)z

Then ζ(z,x)−ζ(z,x−1)=−1/xz. 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 ℜz>0. Let ζz denote the map x→ζ(z,x). Then, using the generator A of the dilation operator, it is easy to check that the eigenvalues have the form 1/2+ıE. For A=(pq+qp)/2=−ıxd/dx−ı1/2. This gives A(1/xz)=ı(z−1/2)(1/xz). Going from A to H is done using the operator Δ, leading to the formal equation Hζ1/2+ıE=Eζz. The Dirichlet b.c. at the origin forces E to satisfy ζ(1/2+ıE)=0.

2)- The first problem I see is that ζz does not seem to belong to the Hilbert space H=L2(0,∞). Using a classical method by Hadamard

∫∞0|ζ(z,x)|2dx=11−2ℜzζ(2ℜz−1)

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

3)- The other problem is the definition of Δ. Let S denote the translation operator, defined non rigorously by Sf(x)=f(x−1). If restricted to H, it is not unitary, because it is defined only for x>1. We can defined it by imposing Sf(x)=0 for x∈[0,1]. Then it is only a partial isometry. For S∗f(x)=f(x+1) leading to S∗S=I, but SS∗=P is the projection onto L2(1,∞). Using these notations Δ=I−S=(S∗−I)S. But we get a problem here: the function ζz 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−ıˆp used by the authors?

4)- If ˆp is the usual operator

ˆp=−ıddx

then there is a problem with its domain of definition. On L2(R), it is selfadjoint as can be seen by using Fourier transform. But on 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 L2 functions with L2 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 fz(x)=ezx, then ˆp∗fz=−ızfz, while fz∈H for ℜz<0. The same argument shows that +ı cannot be an eigenvalue. This means that the "defect indices", namely the dimension of the eigenspaces with eigenvalues ±ı, are not equal. Then, the von Neumann theorem show that the operator ˆp 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?

**Attribution***Source : Link , Question Author : Slava Kashcheyevs , Answer Author : Jean V Bellissard*