Supposedly, canonical transformations are used to provide a general procedure to transform a Hamiltonian such that all coordinates in the new frame are cyclic. I have done the proofs and derivations, but during my course I didn't actually get any practice on finding the right canonical transformations or the generating functions, and I'm feeling sort of in a limbo state where I can prove all the theorems but I can't apply them.

For example, consider the Harmonic oscillator:

$$H=\frac{p^2}{2m} + \frac{1}{2}\omega^2q^2$$

Goldstein provides the following $F_1$ generating function: $$F_1(q,Q)=\frac{m\omega^2q^2}{2} \cot Q $$

under which the hamiltonian transforms to $$H=\omega P$$

via the $F_1$ transformation equations:

$$ \begin{align*} p&=\partial_q F_1 \\ P&=-\partial_Q F_1 \end{align*}$$

I understand where the transformation equations came from and the theory behind the formalism, but I don't understand how to construct a canonical transformation myself or choose a correct generating function ($F_1$ to $F_4$). Basically, it seems that everyone is just pulling them out of the void.

Can anyone provide a "recipe" to transform/solve these problems? Are they just a matter of trial and error? Divine intervention?

  • 4
    $\begingroup$ Solving PDEs is an art. $\endgroup$
    – Qmechanic
    Aug 10, 2020 at 5:42

1 Answer 1


Problem-specific Solution

I stumbled upon the exact same question while studying the same material (Goldstein), and after a while I have it figured out.

Since we're trying to get the expression $f(P)$ now, we should choose a generating function that does not include the variable $P$, therefore it is safe to choose either $F_1 $ or $F_3$.

from $$p = f(P)\cos Q\\q = \frac{f(P)}{m\omega} \sin Q$$ by inspection, we can get $$p = m\omega q \cot Q\tag{1}$$ and that is $$p = p (q,Q)$$

From $\frac{\partial F_1}{\partial q} = p$, we can get the expression of $F_1$. $$F_1 = \frac{1}{2}m\omega q^2 \cot Q$$ From $\frac{-\partial F_1}{\partial Q}=P$, we get $$P = \frac{m\omega q^2}{2 \sin^2Q}\implies q=q(P,Q)$$ therefore we have $$q = \sqrt{\frac{2P}{m\omega}}\sin Q$$

plug this back in to $(1)$, we get $p = p(P,Q)$ $$p = \sqrt{2Pm\omega}\cos Q$$ compare this with $p = f(P)\cos Q$, we get the expression of $f(P)$.


I think what Qmechanic put in the comment section is right, it's about the art of solving PDEs.


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