Short version:

I've been reading through some notes on integrable systems/Hamiltonian dynamics, and am stuck on a problem: Show that the coordinate transformation derived via the generating function method gives you a canonical transformation.

Long version:

A coordinate change $(\vec{q},\vec{p})\to (\vec{Q},\vec{P})$ is called canonical if it leaves Hamilton's equations invariant, i.e. the equations in the original coordinates $$\dot{\vec{q}}=\frac{\partial H}{\partial\vec{p}},\quad \dot{\vec{p}}=-\frac{\partial H}{\partial\vec{q}}$$ are equivalent to $$\dot{\vec{Q}}=\frac{\partial\tilde{H}}{\partial\vec{P}}, \quad \dot{\vec{P}}=-\frac{\partial\tilde{H}}{\partial\vec{Q}}$$ where $\tilde{H}(\vec{Q},\vec{P})=H(\vec{q},\vec{p})$.

The generating function method:

Suppose we have a function $S:\mathbb{R}^{2n}\to\mathbb{R}.$ Write its arguments $S(\vec{q},\vec{P})$. Now set $$\vec{p}=\frac{\partial S}{\partial \vec{q}}, \quad \vec{Q}=\frac{\partial S}{\partial \vec{P}}.$$ The first equation lets us to solve for $\vec{P}$ in terms of $\vec{q},\vec{p}$. The second equation lets us solve for $\vec{Q}$ in terms of $\vec{q},\vec{P}$, and hence in terms of $\vec{q},\vec{p}$. The new coordinates $\vec{Q}$, $\vec{P}$ we find this way will give a canonical transformation. Checking this is just a careful application of the chain rule.

My Problem:

So I decided to try and work out this 'careful application of the chain rule', i.e. prove that the transformation obtained via this generating function method is canonical. I have been unable to do so, and help with this problem would be greatly appreciated.

****my progress****

e.g. try to prove $\dot{\vec{P}}=-\frac{\partial\tilde{H}}{\partial\vec{Q}}$. Thinking of $H$ as $H(\vec{q}(\vec{Q},\vec{P}),\vec{p}(\vec{Q},\vec{P}))$, and using the chain rule, $$\frac{\partial\tilde{H}}{\partial Q_i}=\frac{\partial H}{\partial q_j}\frac{\partial q_j}{\partial Q_i}+\frac{\partial H}{\partial p_j}\frac{\partial p_j}{\partial Q_i}=-\dot{p}_j\frac{\partial q_j}{\partial Q_i}+\dot{q}_j\frac{\partial p_j}{\partial Q_i}$$ using the original Hamilton equations.

Meanwhile, $$-\dot{P}_i=-\frac{\partial P_i}{\partial q_j}\dot{q}_j-\frac{\partial P_i}{\partial p_j}\dot{p}_j,$$ so it will suffice to show $$\frac{\partial P_i}{\partial p_j}=\frac{\partial q_j}{\partial Q_i},\quad \frac{\partial P_i}{\partial q_j}=-\frac{\partial p_j}{\partial Q_i}.$$ I've been able to show $$\frac{\partial p_j}{\partial P_i}=\frac{\partial Q_i}{\partial q_j}$$ using chain rule and symmetry of partial derivatives. This then gives us the first desired equality, by inverting the Jacobian matrix: $$\left[\frac{\partial p_j}{\partial P_i}\right]^{-1} =\left[\frac{\partial Q_i}{\partial q_j}\right]^{-1} \implies \left[\frac{\partial P_i}{\partial p_j}\right] =\left[\frac{\partial q_j}{\partial Q_i}\right]$$

However, I'm not sure how to show $$\frac{\partial P_i}{\partial q_j}=-\frac{\partial p_j}{\partial Q_i}.$$


2 Answers 2

  1. Recall that a coordinate transformation $$(q^i,p_i)~~\mapsto~~ \left(Q^i(q,p,t),P_i(q,p,t)\right)\tag{1}$$ with generating function $F$ is of the form$^1$ $$\lambda\underbrace{(\sum_{i=1}^np_i\mathrm{d}q^i-H\mathrm{d}t)}_{~=:~\mathbb{L}_H} ~=~\underbrace{(\sum_{i=1}^nP_i\mathrm{d}Q^i -K\mathrm{d}t)}_{~=:~\mathbb{L}_K} ~+~\mathrm{d}F,\tag{2}$$ where $\lambda\neq 0$ is a non-zero constant.

  2. Since the 2 Lagrangian 1-forms $\mathbb{L}_H$ and $\mathbb{L}_K$ are the same off-shell up to a total derivative and an over-all multiplicative constant $\lambda$, they yield the same Euler-Lagrange (EL) equations, which clearly are Hamilton's equations in both cases. Hence the coordinate transformation (2) leaves Hamilton's equations form-invariant.

  3. Eqs. (2) represent a notion of a canonical transformation (CT), cf. e.g. this Phys.SE post.


$^1$ OP's generator $F=S(q,P,t)-\sum_{i=1}^nP_iQ^i$ is of type 2 and in OP's case $\lambda=1$.


What you have found is basically another way to describe canonical transformations. Let's start by calculating $$\dot{Q}_i = \frac{\partial Q_i}{\partial q_j} \dot{q}_j + \frac{\partial Q_i}{\partial p_j} \dot{p_j} = \frac{\partial Q_i}{\partial q_j} \frac{\partial H}{\partial p_j} - \frac{\partial Q_i}{\partial p_j} \frac{\partial H}{\partial p_j}.$$ This has to be equal to $$\frac{\partial H}{\partial P_i} = \frac{\partial H}{\partial q_j} \frac{\partial q_j}{\partial P_i} + \frac{\partial H}{\partial P_j}\frac{\partial p_j}{\partial P_i}.$$ Comparing these two equations yields $$\frac{\partial H}{\partial p_j} \left( \frac{\partial Q_i}{\partial q_j} - \frac{\partial p_j}{\partial P_i}\right) = \frac{\partial H}{\partial q_j}\left(\frac{\partial q_j}{\partial P_i} + \frac{\partial Q_i}{\partial p_j}\right).$$ Since neither $\partial H / \partial p_j$ nor $\partial H / \partial q_j$ are identical 0, the expressions in the brackets need to equal 0. In your case, you will have to show that the generating function $S$ yields a transformation that abides these equations (a similar argument for $\dot{P}_i$ yields the identiy in your question).

From a more theoretical point of few, a transformation is canonical if the poisson brackets are preserved: $$\{Q_i, Q_j\} = 0, \qquad \{P_i, P_j\} = 0, \qquad \{Q_i, P_j\} = \delta_{ij}.$$ And from a even more abstract point of few, someone might show that a transformation is canonical by checking if it preserves the symplectic structure of Hamilton's equations. If $J$ is the Jacobian matrix and $E$ the $2n\times 2n$ matrix $$ E = \begin{pmatrix} 0 &\delta_{ij} \\ -\delta{ij} & 0 \end{pmatrix} $$ then the symplectic structure is preserved iff $J E J^\top = E$. Incidentally, $J$ is called symplectic in that case.

A canonical transformation induced by the generating function $S$ has to satisfy all three conditions, but since they are equivalent, you just need to check that one of them is true.

  • $\begingroup$ Hi, thanks for the response. When you say "In your case, you will have to show that the generating function $S$ yields a transformation that abides these equations" - this is precisely the bit I am stuck on! More specifically, how do I show the second bracket is equal to zero: $$\frac{\partial q_j}{\partial P_i}+\frac{\partial Q_i}{\partial p_j}=0?$$ $\endgroup$ Feb 5, 2020 at 3:24
  • $\begingroup$ Just plug in the generating function with $Q_i = \frac{\partial S}{\partial P_j}$. Differentiate this expression with respect to $p_j$, which should yield $0$, since $S$ does not depend on $p_j$. $\endgroup$
    – Tobi7
    Feb 5, 2020 at 11:42
  • $\begingroup$ Not sure I agree with that. Strictly speaking if you were to differentiate a function with arguments $(\vec{q}, \vec{P})$ with respect to $p_j$ you would have to re express $\vec{P}$ as $\vec{P}(\vec{q},\vec{p})$. Your derivative then would then involve the chain rule and be nonzero $$\frac{\partial S}{\partial p_j}=\frac{\partial S}{\partial P_k}\frac{\partial P_k}{\partial p_j}.$$ Partial derivatives refer to a derivative w.r.t. a certain slot of the function's arguments. $\endgroup$ Feb 6, 2020 at 11:48
  • $\begingroup$ I agree with you, my argument is regarding the derivative is not right. But i've been thinking of exploiting the symmetry of partial derivatives, i.e $$\frac{\partial}{\partial p_j}\frac{\partial}{\partial P_k}S = \frac{\partial}{\partial P_k}\frac{\partial}{\partial p_j}S$$. What do you think? $\endgroup$
    – Tobi7
    Feb 6, 2020 at 13:43
  • 1
    $\begingroup$ Hey, I found the 'official solution' after some browsing online. It's here if you're interested: damtp.cam.ac.uk/user/acla2/IS_handout1.pdf. It turns out our approach might not have been the right one. Rather, you're meant to use the fact that a transformation $$x\to y(x)$$ is canonical iff the Jacobian matrix $Dy$ is symplectic $$Dy J (Dy)^T=J$$ $\endgroup$ Feb 13, 2020 at 10:17

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