# What is the process of finding a good canonical transformation to describe a system? How do I choose the correct generating function?

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?

• Solving PDEs is an art. Aug 10, 2020 at 5:42

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 $$f(P)$$.

Meta-problem

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

• "By inspection" isn't this just dividing p by q !? Nov 25, 2022 at 12:51
• @Cheng that's true, what's the problem here? Nov 29, 2022 at 12:39
• Nah it's not a problem, just in case anyone was wondering... And I'm not sure what method did you use when you meant 'by inspection' Nov 30, 2022 at 0:00
• Ahh I see, thanks for the clarification Dec 1, 2022 at 7:01