In this question, it is discussed why, in Lagrangians we usually stick to first derivatives and quadratic terms we never see higher derivatives.
The selected answer shows that, if a Lagrangian $L(q, \dot{q},\ddot{q})$ where chosen, the Hamiltonian $H$ would be $$ H = P_1 Q_2 + P_2 \ddot{q}\left(Q_1, Q_2, P_2\right) - L\left(Q_1, Q_2,\ddot{q}\right), $$
where $Q_1 = q, Q_2 = \dot{q}, P_1 = \frac{\partial L}{\partial \dot q} -\frac{d}{dt}\frac{\partial L}{\partial \ddot q}, \ P_2 = \frac{\partial L}{\partial \ddot q}.$
It is argued that:
However, we now have a problem: $H$ has only a linear dependence on $P_1$, and so can be arbitrarily negative. In an interacting system this means that we can excite positive energy modes by transferring energy from the negative energy modes, and in doing so we would increase the entropy — there would simply be more particles, and so a need to put them somewhere. Thus such a system could never reach equilibrium, exploding instantly in an orgy of particle creation.
Can someone explain this paragraph?
If $H \propto P_1$ then we need negative momenta, but why would this mean negative energies? Where does particle creating come into play (QFT?) ?
