# Can all symplectic-form preserving canonical transformations generated by generating functions

This question is related to this fascinating post and this post and this post, but more limited in scope in discussing the practical definition canonical transformations.

## Canonical transformation from one of four generating functions

Traditionally, canonical transformations are generated by derivatives of a generating function. For example, for $F_2(q, P)$, we have \begin{align} p &= \frac{ \partial F_2 } { \partial q }, \tag{1}\\ Q &= \frac{ \partial F_2 } { \partial P }. \tag{2} \end{align}

## Canonical transformation that preserve symplectic form

More recent literature, e.g., David Tong's lecture notes, tend to define a canonical transformation through the symplectic condition on the Jacobian from $x = (q, p)$ to $X = (Q, P)$: \begin{align} U = \frac { \partial{ X } }{ \partial x }. \tag{3} \end{align} As long as \begin{align} U \, J \, U^T = J, \tag{4} \end{align} we say the transformation is canonical, where \begin{align} J = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right). \end{align} This is because \begin{align} \frac{dX}{dt} &= \frac{\partial X}{\partial x} \frac{d x}{dt} \\ &= \frac{\partial X}{\partial x} J \frac{\partial H}{\partial x} \\ &= \frac{\partial X}{\partial x} J \frac{\partial X}{\partial x} \frac{\partial H}{\partial X}\\ &= U \, J \, U^T \frac{\partial H}{\partial X}, \tag{5} \end{align} where we have assumed that the Hamiltonian of $X$ is the same of that of $x$, which is not necessarily the case, see e.g., this post.

## Questions

Now the first question is

1. Can all canonical transformations preserving the symplectic form, Eq. (4), be generated by one of the generating functions?

If so, it would be nice if we can explicitly construct such a generating function (that is, in a practical way).

The converse is

1. Do all canonical transformations generated by one of the generating functions preserve the sympletic form?

If one of the statements is wrong, do we have some simple counterexamples?

## Attempts

Now if we write \begin{align} U = \left( \begin{array}{ccc} A & B \\ C & D \end{array} \right), \end{align} Eq. (4) means \begin{align} BA^T &= AB^T, \tag{6.1} \\ CD^T &= DC^T, \tag{6.2}\\ AD^T - BC^T &= 1, \tag{6.3} \\ DA^T - CB^T &= 1. \tag{6.4} \end{align} Note that Eq. (6.4) is only the transverse of Eq. (6.3).

It is readily show that Eqs. (1) and (2) satisfies (6), at least in one dimension. Define $Y = (q, P)$, then \begin{align} U &= \left( \frac{ \partial X }{ \partial Y } \right) \left( \frac{ \partial Y }{ \partial x } \right) = \left( \frac{ \partial X }{ \partial Y } \right) \left( \frac{ \partial x }{ \partial Y } \right)^{-1} \\ &= \left( \frac{ \partial (Q, P) }{ \partial (q, P) } \right) \left( \frac{ \partial (q, p) }{ \partial (q, P) } \right)^{-1} \\ &= \left( \begin{array}{cc} \left(\frac{\partial Q}{\partial q}\right)_P & \left(\frac{\partial Q}{\partial P}\right)_q \\ 0 & 1 \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\ \left(\frac{\partial p}{\partial q}\right)_P & \left(\frac{\partial p}{\partial P}\right)_q \end{array} \right)^{-1} \\ &= \left( \begin{array}{cc} \left(\frac{\partial Q}{\partial q}\right)_P & \left(\frac{\partial Q}{\partial P}\right)_q \\ 0 & 1 \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\ - \left(\frac{\partial p}{\partial q}\right)_P \left(\frac{\partial p}{\partial P}\right)_q^{-1} & \left(\frac{\partial p}{\partial P}\right)_q^{-1} \end{array} \right) \\ &= \left( \begin{array}{cc} \left(\frac{\partial Q}{\partial q}\right)_P - \left(\frac{\partial Q}{\partial P}\right)_q \left(\frac{\partial p}{\partial q}\right)_P \left(\frac{\partial p}{\partial P}\right)_q^{-1} & \left(\frac{\partial Q}{\partial P}\right)_q \left(\frac{\partial p}{\partial P}\right)_q^{-1} \\ - \left(\frac{\partial p}{\partial q}\right)_P \left(\frac{\partial p}{\partial P}\right)_q^{-1} & \left(\frac{\partial p}{\partial P}\right)_q^{-1} \end{array} \right) \\ &= \left( \begin{array}{cc} \frac{\partial^2 F_2}{\partial q \partial P} - \frac{\partial^2 F_2}{\partial P^2} \frac{\partial^2 F_2}{\partial q^2} \left(\frac{\partial^2 F_2}{\partial q \partial P}\right)^{-1} & \frac{\partial^2 F_2}{\partial P^2} \left(\frac{\partial^2 F_2}{\partial q \partial P}\right)^{-1} \\ - \frac{\partial^2 F_2}{\partial q^2} \left(\frac{\partial^2 F_2}{\partial q \partial P}\right)^{-1} & \left(\frac{\partial^2 F_2}{\partial q \partial P}\right)^{-1} \end{array} \right). \tag{7} \end{align} It can be verified that Eq. (6.3), hence Eq. (4), is satisfied.

Edit: the second question is answered by Qmechanic in point 2 of this post. The answer appears to be true, in agreement with the observation of Eq. (7).

• The second subquestion (v4) is answered in my Phys.SE post here. – Qmechanic Oct 14 '15 at 13:44
• @Qmechanic. Thank you sir for this thorough explanation of the relationship between different definitions. I really appreciate it! – hbp Oct 14 '15 at 17:35

## 1 Answer

These questions can be trivially answered if you know some basic symplectic geometry. I'll provide explanations here with a tl;dr if you don't cannot follow the details.

Can all canonical transformations preserving the symplectic form, Eq. (4), be generated by one of the generating functions?

tl;dr: If the first cohomology group of the symplectic manifold is trivial, i.e. if ${\cal H}^1 ( M) = 0$, then every canonical transformation can be generated by a function.

Proof: Let $M$ be a symplectic manifold with form $\Omega$. A canonical transformation is a diffeomorphism $\xi$ on this manifold that preserves the symplectic form, i.e. $${\cal L}_\xi \Omega = 0$$ But ${\cal L}_\xi = {\text d} i_\xi + i_\xi {\text d}$ and $\Omega$ is closed. Thus, the equation above can be re-written as $${\text d}( i_\xi \Omega ) = 0$$ Now, if ${\cal H}^1 ( M) = 0$, then every closed one-form is exact. Thus, there exists a function $f$, such that $$i_\xi \Omega = - {\text d} f$$ This $f$ above is precisely the generating function.

Finally, since $\Omega$ is invertible, this formula can be inverted to determine $\xi$ from $f$. Thus, there is a one-to-one correspondence between canonical transformations $\xi$ and generating functions $f$.

Do all canonical transformations generated by one of the generating functions preserve the sympletic form?

tl;dr: Yes. A canonical transformation is by definition a transformation that preserves the symplectic form.

This follows immediately from the discussion above. Any exact form is closed so any transformation obtained from a generating function as shown above is trivially canonical.

• @hbp The formula for $\mathrm{d}f$ in Prahar's answer reads $\xi^i \Omega_{ij} = - \nabla_j f$ in coordinate components which has a formal solution in terms of a line integral $f=-\int \xi^i \Omega_{ij} \mathrm{d}x^j$. – Void Oct 19 '15 at 7:59
• The answer is simple: $H$ is the generating function of the canonical transformation corresponding to time-evolution. You can understand generating functions as Hamiltonians generating particular "time" flows. – Void Nov 23 '15 at 19:00
• I downvoted the answer not because of its content but because of its educational value. If it is clear that the asker is not that familiar with the topic, one should formulate an adequate answer. No offence :-) just trying to be constructive. – mr. curious Dec 18 '20 at 21:02
• @mr.curious - The asker him/herself accepted this answer and did not follow up with any questions or curiosities in the comments. Who are you to judge whether the answer is adequate? – Prahar Mitra Dec 18 '20 at 21:06