I was solving an exercise of classical mechanics :

Consider the following hamiltonian

$H(p,q,t) = \frac{p^2}{2m} + \lambda pq + \frac{1}{2}m\lambda^2\frac{q^6}{q^4+\alpha^4}$

Where $\lambda,m,\alpha$ are positive parameters.

After having solved the equations of motion, I found something odd, and plotting the phase space (p,q) (which is y,x below) one can find that there are loops where momentum never changes sign, yet the system is localized (loop)

enter image description here

What sort of physical system does such a hamiltonian describe ? I cannot imagine something with an oscillating, never zero momentum, but still localized in space. I know this happens because of the cross-term, but I cannot find an interpretation for it

  • 6
    $\begingroup$ As you said the velocity is not proportional to the momentum in your system, if you compute the velocity around loops you will find that it is oscillating $\endgroup$
    – Ikiperu
    Commented May 14, 2013 at 0:18
  • 1
    $\begingroup$ Stupid me, you are right. Taking the derivative with respect to t of q(t) gives a term proportional to cos(t). In this case, the Lagrangian certainly contained couplings, and the conjugate momentum picks up cross terms when going to the Hamiltonian, so I cannot see it as a velocity. $\endgroup$
    – Mathusalem
    Commented May 14, 2013 at 0:25
  • 3
    $\begingroup$ or directly from hamilton eqs., the velocity is $\dot{q}=p/m +\lambda q$ $\endgroup$
    – Ikiperu
    Commented May 14, 2013 at 0:28

1 Answer 1


The key inside to OP's question has already been provided by Ikiperu in above comments. Here we just want to show that the problem becomes very simple to study in the corresponding Lagrangian formalism.

The Hamiltonian reads

$$\tag{1} H(p,q) ~:=~ \frac{p^2}{2m} + \lambda pq + \frac{m\lambda^2}{2}\frac{q^6}{q^4+\alpha^4}. $$

Since there is no explicit time dependence in (1), the Hamiltonian (= the mechanical energy of the system) is preserved. The velocity can be calculated from Hamilton's equation

$$\tag{2} \dot{q}~=~\frac{\partial H}{\partial p}~=~\frac{p}{m}+ \lambda q. $$

If we eliminate the momentum

$$\tag{3} \frac{p}{m}~=~ \dot{q}-\lambda q $$

in the Hamiltonian (1), we get a surprisingly simple energy function

$$\tag{4} h(q,\dot{q})~=~ \frac{m}{2}\dot{q}^2+V(q). $$

Here the potential $V(q)$ is the double-well

$$\tag{5} V(q)~=~ -\frac{m\alpha^2}{2}\frac{\lambda^2}{\left(\frac{q}{\alpha}\right)^2+\left(\frac{\alpha}{q}\right)^2} ,$$

which has two stable positions

$$\tag{6} q~=~\pm \alpha.$$

In $(q,\dot{q})$ space, there are two stable points

$$\tag{7} (q,\dot{q})~=~(\pm \alpha,0)$$

on the horizontal $q$-axis.

In $(q,p)$ phase space, the two stable points

$$\tag{8} (q,p)~=~\pm \alpha(1,-m\lambda)$$

are shifted by the transformation (3), in accordance with OP's figure.


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