Hamiltonian for the zeros of the Riemann zeta function Dorje C. - PowerPoint PPT Presentation

Hamiltonian for the zeros of the Riemann zeta function Dorje C. Brody New Trends in Applied Geometric Mechanics – Celebrating Darryl Holm’s 70th Birthday – ICMAT, Madrid, Spain 3–7 July 2017 (Joint work with Carl M. Bender & Markus P. M¨ uller) - 1 -

Madrid, 3 July 2017 Zeros of the Riemann Zeta Function - 2 - Happy Birthday Darryl ! ⃝ DC Brody 2017 New Trends in Applied Geometric Mechanics c

Madrid, 3 July 2017 Zeros of the Riemann Zeta Function - 3 - 1. The asymptotic law of distribution of prime numbers In 1792, Gauss, at the age of 15, conjectured that the asymptotic law of distribution of prime numbers is Λ π (Λ) ∼ log Λ . In 1859, Riemann published his paper On the number of primes less than a given magnitude , in which the zeta function ) − 1 ( 1 1 − 1 ∑ ∏ ζ ( z ) = n z = p z n ≥ 1 p played a prominent role. ⃝ DC Brody 2017 c New Trends in Applied Geometric Mechanics

Madrid, 3 July 2017 Zeros of the Riemann Zeta Function - 4 - 2. Analytic number theory Riemann’s observations are based on the fact that ζ ( z ) can be meromorphically continued (simple pole at z = 1 ), and on the identity: ∫ 2+i ∞ y z 1 { y > 1 } = 1 z d z. 2 π i 2 − i ∞ Consider to start with the counting of the integers: ) z d z ∫ 2+i ∞ { Λ } ( Λ = 1 ∑ ∑ ∑ 1 = n > 1 1 2 π i n z 2 − i ∞ 1 ≤ n ≤ Λ n ≥ 1 n ≥ 1 ∫ 2+i ∞ ζ ( z ) Λ z 1 z d z = Λ − 1 = 2 π i 2 2 − i ∞ ⃝ DC Brody 2017 c New Trends in Applied Geometric Mechanics

Madrid, 3 July 2017 Zeros of the Riemann Zeta Function - 5 - Similarly, we have ) z 1 ∫ 2+i ∞ ( Λ 1 ∑ ∑ ψ (Λ) = log p = log p z d z 2 π i p 2 − i ∞ p p ≤ Λ ∫ 2+i ∞ ∫ 2+i ∞ Λ z ζ ′ ( z ) Λ z 1 log p z d z ≈ 1 ∑ = z d z p z 2 π i 2 π i ζ ( z ) 2 − i ∞ 2 − i ∞ p The derivation of the expression for ψ (Λ) then relies on the study of the properties of ζ ( z ) , in particular, its zeros. The reflection formula ζ ( z ) = 2 z π z − 1 sin( πz/ 2)Γ(1 − z ) ζ (1 − z ) shows that the zeta function vanishes trivially for z = − 2 n ( n = 1 , 2 , . . . ). Riemann conjectured (the Riemann Hypothesis) that the nontrivial zeros of ζ ( z ) all lie on the straight line ℜ ( z ) = 1 2 . ⃝ DC Brody 2017 c New Trends in Applied Geometric Mechanics

Madrid, 3 July 2017 Zeros of the Riemann Zeta Function - 6 - 3. The Hilbert-P´ olya conjecture Assuming that the Riemann hypothesis holds true, and writing z n = 1 2 + i E n , the (real) numbers { E n } should correspond to the eigenvalues of a Hermitian operator (the so-called Riemann operator). 4. The Berry-Keating conjecture In 1989, Berry and Keating conjectured that the Riemann operator should be given by a quantisation of the classical Hamiltonian H = xp. A lot of efforts have been made by various authors to find such Hamiltonian, but without success until now. ⃝ DC Brody 2017 c New Trends in Applied Geometric Mechanics

Madrid, 3 July 2017 Zeros of the Riemann Zeta Function - 7 - 5. Outline of the talk We consider the ‘Hamiltonian’ operator 1 ˆ x ) ( 1 − e − iˆ p ) , H = p (ˆ x ˆ p + ˆ p ˆ 1 − e − iˆ which reduces to the classical Hamiltonian function H = 2 xp . It will be shown that with the boundary condition ψ (0) = 0 the eigenvalues { E n } of ˆ { 1 } H satisfy the property that 2 (1 − i E n ) are the zeros of the Riemann zeta function. The Riemann hypothesis follows if all eigenvalues of ˆ H are real. Using the pseudo-Hermiticity of ˆ H , a heuristic analysis will be presented that suggests that this is indeed the case. ⃝ DC Brody 2017 c New Trends in Applied Geometric Mechanics

Madrid, 3 July 2017 Zeros of the Riemann Zeta Function - 8 - 6. The shift operator and its inverse Defining ˆ ∆ ≡ 1 − e − iˆ p , in units ℏ = 1 we have p = − i d ˆ d x so that ˆ ∆ f ( x ) = f ( x ) − f ( x − 1) . ∆ − 1 we have As for ˆ ∞ p ) n − iˆ p ( − iˆ 1 p = 1 p − 1 = 1 ∆ − 1 = ˆ ∑ B n . 1 − e − iˆ e − iˆ iˆ p iˆ p n ! n =0 In particular, if f ( x ) → 0 sufficiently fast, then we have ∞ ˆ ∑ ∆ − 1 f ( x ) = − f ( k + x ) . k =1 ⃝ DC Brody 2017 c New Trends in Applied Geometric Mechanics

Madrid, 3 July 2017 Zeros of the Riemann Zeta Function - 9 - 7. Uniqueness of ˆ ∆ ψ We multiply the eigenvalue equation ˆ Hψ = Eψ on the left by ˆ ∆ . Recall that H = ˆ ˆ x ) ˆ ∆ − 1 (ˆ x ˆ p + ˆ p ˆ ∆ . This gives a first-order linear differential equation ( 2 x d ) x ) ˆ ∆ ψ = E ˆ ˆ (ˆ x ˆ p + ˆ p ˆ ∆ ψ = − i d x + 1 ∆ ψ for the function ˆ ∆ ψ , whose solution is unique and is given by ˆ ∆ ψ = x − z up to a multiplicative constant. Therefore, ψ ( x ) = ˆ ∆ − 1 x − z . ⃝ DC Brody 2017 c New Trends in Applied Geometric Mechanics

Madrid, 3 July 2017 Zeros of the Riemann Zeta Function - 10 - 8. Eigenstates and eigenvalues The eigenstates of ˆ H are given by the Hurwitz zeta function ψ z ( x ) = − ζ ( z, x + 1) on the positive half line R + , with eigenvalues E = i(2 z − 1) . To see this, observe that, up to an additive constant, ∞ p ) n ∆ − 1 x − z = 1 ( − iˆ ˆ ∑ x − z B n iˆ p n ! n =0 ∞ p ) n p ) x 1 − z = 1 ( − iˆ ∑ B n (iˆ iˆ p n ! 1 − z n =0 ∞ p ) n 1 ( − iˆ ∑ x 1 − z . = B n 1 − z n ! n =0 Because iˆ p = ∂ x and Γ( µ + 1) x x µ = ∂ n Γ( µ − n + 1) x µ − n , ⃝ DC Brody 2017 c New Trends in Applied Geometric Mechanics

Madrid, 3 July 2017 Zeros of the Riemann Zeta Function - 11 - setting µ = 1 − z we find ∞ ( − 1) n x 1 − z − n ∆ − 1 x − z = Γ(2 − z ) ˆ ∑ B n Γ(2 − z − n ) , 1 − z n ! n =0 but we have Γ(2 − z ) = (1 − z )Γ(1 − z ) and Γ(2 − z − n ) = 1 1 ∫ d u e u u n + z − 2 , 2 π i C so ∞ ( − u/x ) n ∆ − 1 x − z = Γ(1 − z ) ∫ d u e u u z − 2 ˆ ∑ x 1 − z B n 2 π i n ! C n =0 = Γ(1 − z ) d u e u u z − 2 − u/x ∫ x 1 − z e − u/x − 1 2 π i C d u e u u z − 1 = Γ(1 − z ) ∫ x − z 1 − e − u/x . 2 π i C Now we scale the integration variable according to u/x = t and obtain d u e xt t z − 1 ∆ − 1 x − z = Γ(1 − z ) ∫ ˆ 1 − e − t = − ζ ( z, x + 1) . 2 π i C ⃝ DC Brody 2017 c New Trends in Applied Geometric Mechanics

Madrid, 3 July 2017 Zeros of the Riemann Zeta Function - 12 - As for the eigenvalues, we have ∆ − 1 (ˆ ∆ − 1 x − z = i(2 z − 1) ψ z ( x ) . Hψ z ( x ) = ˆ ˆ x ) ˆ ∆ ˆ x ˆ p + ˆ p ˆ Note that for ℜ ( z ) > 1 we have ∞ d u e xt t z − 1 − ζ ( z, x + 1) = Γ(1 − z ) ∫ 1 ∑ 1 − e − t = − ( x + k ) z . 2 π i C k =1 ⃝ DC Brody 2017 c New Trends in Applied Geometric Mechanics

Madrid, 3 July 2017 Zeros of the Riemann Zeta Function - 13 - 9. The boundary condition We now impose the boundary condition that ψ n (0) = 0 for all n . Because ζ ( z, 1) = ζ ( z ) , this implies that z can only be discrete zeros of the Riemann zeta function: If z = 1 2 (1 − i E ) then i(2 z − 1) = E . Can z be a trivial zero? For the trivial zeros z = − 2 n , n = 1 , 2 , . . . , we have 1 ψ z ( x ) = − 2 n + 1 B 2 n +1 ( x + 1) , ⇒ | ψ z ( x ) | grows like x 2 n +1 as x → ∞ . where B n ( x ) is a Bernoulli polynomial = For the nontrivial zeros ψ z ( x ) oscillates and grows sublinearly. Thus, for the trivial zeros ˆ ∆ ψ z ( x ) blows up and for the nontrivial zeros ˆ ∆ ψ z ( x ) goes to zero as x → ∞ . ⃝ DC Brody 2017 c New Trends in Applied Geometric Mechanics

Madrid, 3 July 2017 Zeros of the Riemann Zeta Function - 14 - 10. Relation to pseudo-Hermiticity If we consider the space-time (PT) inversion on the canonically transformed variables (ˆ x, ˆ p ) → (ˆ p, − ˆ x ) so that PT : (ˆ x, ˆ p, i) − → (ˆ x, − ˆ p, − i) , then we find that i ˆ H is PT symmetric. However, since PT ψ n ( x ) = ψ − n ( x ) , the PT symmetry is broken for all z n ∈ C . “Eigenvalues of i ˆ H are purely imaginary” ⇒ “The Riemann hypothesis holds” p † is symmetric and that To proceed, assume that ˆ 1 H † = ( 1 − e iˆ ˆ p ) (ˆ x ˆ p + ˆ p ˆ x ) p . 1 − e iˆ Then if we define the operator ˆ η according to η = sin 2 1 ˆ 2 ˆ p, which is nonnegative, bounded, and Hermitian under our assumption, we get H † = ˆ ˆ η ˆ η − 1 . H ˆ ⃝ DC Brody 2017 c New Trends in Applied Geometric Mechanics

Madrid, 3 July 2017 Zeros of the Riemann Zeta Function - 15 - Thus, our Hamiltonian ˆ H is pseudo-Hermitian: ρ − 1 = ˆ ρ ˆ ˆ H ˆ h, where ρ † ˆ η = sin 2 1 ˆ ρ = ˆ 2 ˆ p. ⃝ DC Brody 2017 c New Trends in Applied Geometric Mechanics

Madrid, 3 July 2017 Zeros of the Riemann Zeta Function - 16 - 11. Quantisation condition for the Berry-Keating Hamiltonian ρ † ˆ η = sin 2 1 Recall that ˆ ρ = ˆ 2 ˆ p . ρ = ˆ Choosing ˆ ∆ we have the Berry-Keating Hamiltonian h BK = ˆ ˆ x ˆ p + ˆ p ˆ x, with eigenstates and eigenvalues ϕ BK z ( x ) = x − z and E = i(2 z − 1) . The boundary condition ψ (0) = 0 then translates into the quantisation condition for the Berry-Keating Hamiltonian, either as ϕ BK [ ] lim z ( x ) − ζ ( z, x − 1) = 0 x → 0 or alternatively as x → 1 ϕ BK lim z ( x ) = − lim x → 1 ζ ( z, x + 1) . ⃝ DC Brody 2017 c New Trends in Applied Geometric Mechanics