1

This might be a silly question but I failed to get it. In Ostrogradsky Instability, we deduced that Lagrangian of higher-order derivatives leads to Hamiltonian linear to canonical momenta, and thus, Hamiltonian is unbounded from below. However, a particle in potential: $V(r)=-\frac{1}{r}$ also has a Hamiltonian unbounded from below as $r$ approaches to zero, so why doesn’t it lead to an instability similar to Ostrogradsky instability? What makes them different?

Qmechanic
  • 201,751
Aimikan
  • 57

1 Answers1

3

What is important is that the spectrum of the Hamiltonian operator $\hat{H}$ is bounded from below, and that it has a ground state; not that the classical Hamiltonian $H$ is bounded from below in the classical phase space. E.g. quantum mechanically the hydrogen atom has a ground state despite the Coulomb potential is unbounded from below.

Qmechanic
  • 201,751
  • However, could spectrum be bounded from below, if $\hat{H}$ was linear in $\hat{p}$? – Ján Lalinský Dec 05 '23 at 19:40
  • Yes, e.g. if the phase space is compact. – Qmechanic Dec 05 '23 at 20:06
  • 1
    Any concrete example? – Ján Lalinský Dec 05 '23 at 20:37
  • @JánLalinský a proof is given here on page 9, but perhaps it is not satisfactory. – Aimikan Dec 06 '23 at 14:44
  • @Qmechanic Thanks a lot, the answer is very helpful. But something else confuses me: what is the physical interpretation of Ostrogradskian Hamiltonian in the simplest system(after choosing $Q_{1}=q, P_{1}= \frac{\partial L}{\partial \dot q} -\frac{d}{dt}\frac{\partial L}{\partial \ddot{q} } ,Q_{2}=\dot{q}, P_{2}= \frac{\partial L}{\partial \ddot q} $) or why does it make sense to regard it as the total energy of the system? – Aimikan Dec 06 '23 at 14:58
  • Hi @Aimikan. Thanks for the feedback. You might have a look at e.g. this related Phys.SE post. – Qmechanic Dec 06 '23 at 15:47