# Contradiction in classical analysis of the hamiltonian $\mathcal{H}=xp$?

I am writing an essay on the Berry Keating article proposing to use the $\mathcal{H}=xp$ hamiltonian to get a correspondence between the nontrivial riemann zeros and the eigenvalues of an Hermitian operator. Now classically, the time evolution of the system as follows from the Hamilton equations is:

$$x(t)=x(0)e^t$$ $$p(t) = p(0)e^{-t}$$

It seems as if the particle is accelerating (since $\frac{dx}{dt}$ clearly increases), while $p$ obviously decreases. Moreover I do not immediately see how this system is chaotic, although clearly the distance as a result of different initial conditions between particles increases over time. Could you please shed some light on this?

• Where is the contradiction? $p$ is not the classical momentum $p=m \dot{x}$ here. Indeed Hamilton equations just say that $\dot{x} = x$, differently from the case ${\cal H}= p^2/2m + U(x)$ where $\dot{x}= p/m$. The Hamiltonian decides what is the relation between $x$ and $p$... – Valter Moretti Dec 9 '14 at 10:58
• ok thank you, i already thought something like that. But that also means that the momentum quantum operator is also different (it is different from $\frac{d}{dx}$ at least)? – user1043065 Dec 9 '14 at 12:12
• also, do you know in which sense the system is chaotic then (equivalently what definition of chaos should be applied here?) – user1043065 Dec 9 '14 at 12:13
• No it does not mean that. Regarding the second question, I do not know, since I do not know the definition of chaotic. However the Hamiltonian flow does not seem very chaotic'' to me in this case, as the maximal solutions are the trivial ones you wrote! – Valter Moretti Dec 9 '14 at 12:49
• – Qmechanic Feb 4 '18 at 12:13