In order to solve the harmonic oscillator, $$H=\frac{p^2}{2}+\frac{q^2}{2},$$ by using Hamilton-Jacobi theory we have to find a complete solution for the H-J equation, $$\frac{1}{2}\left(\frac{\partial S}{\partial q}\right)^2+\frac{q^2}{2}+\frac{\partial S}{\partial t}=0.$$ We seek for a solution $$S=W(q)+T(t),$$ which leads to the following ODE, $$\frac{dT}{dt}=-\alpha t,$$ $$\frac{dW}{dq}=\sqrt{2\alpha-q^2}.\tag 1$$

It is easy to solve those equations and then write $S(q,\alpha,t)$, from which we can get $$\beta=\frac{\partial S}{\partial \alpha}\Rightarrow q=\sqrt{2\alpha}\sin (t+\beta).$$

Now it comes my issue. When solving for momenta, $$p=\frac{\partial S}{\partial q}=\frac{dW}{dq}=\sqrt{2\alpha-q^2},\tag 2$$ and using the solution found for $q$ we obtain $$p=\sqrt{2\alpha}\sqrt{\cos^2(t+\beta)}=\sqrt{2\alpha}|\cos(t+\beta)|.$$

We know that the solution should be $p=\sqrt{2\alpha}\cos(t+\beta)$. I realize that even Eq. (2) is allowing only for positive momentum. So how to obtain the correct solution without forcing it by knowing the correct answer a priori?

Note that in obtaining (1) we actually should write $$\frac{dW}{dq}=\pm\sqrt{2\alpha-q^2},$$ but yet there is an ambiguity with the plus or minus sign.

  • $\begingroup$ You have two solutions $\pm\sqrt{2\alpha-q^2}$. $\endgroup$
    – Jon
    May 10, 2017 at 14:25
  • $\begingroup$ Exactly, but is there any way of eliminating the ambiguity of the plus or minus sign without knowing the solution a priori? $\endgroup$
    – Diracology
    May 10, 2017 at 15:33

1 Answer 1


By making use of Hamilton's equation to the Hamiltonian given, yeilds: $\dot{q} = \frac{\partial H}{\partial p} = p$, which unambiguously implies $p = \sqrt{2 \alpha} \cos (t + \beta)$.

  • $\begingroup$ So I should use Eq. (1) as $\frac{dW}{dq}=\pm\sqrt{2\alpha-q^2}$ which gives $p=\pm\sqrt{2\alpha-q^2}=\pm\sqrt{2\alpha}|\cos (t+\beta)|$ and choose the correct sign by means of the Hamilton's equation? It works good enough for me, I just thought that once the Hamilton-Jacobi formalism is established, we did not need to trace things back to Hamiltonian formalism. $\endgroup$
    – Diracology
    May 10, 2017 at 16:44
  • $\begingroup$ Well, the equation for q is completely determined by the Hamilton-Jacobi equations and the 2 constants of motion, as you correctly found. But i don't see how to relate this back to momentum p without making use of the Hamilton formalism. If this remains unclear I will edit my answer and elaborate further tomorrow. $\endgroup$ May 10, 2017 at 20:24

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