I am confused by this problem in my homework:

Given a canonical transformation, where $\alpha>0$, $$q\rightarrow \alpha q,$$ $$p\rightarrow \frac{1}{\alpha} p,$$ find the generating function for the corresponding infinitesimal canonical transformation.

Whats in my mind is, lets define $\epsilon=(\alpha-1)/N$ with $N$ being a huge number, then the transformation can be spanned by the following infinitesimal ones $$q\rightarrow q+\epsilon q,$$ $$p\rightarrow p-\frac{\epsilon}{\alpha}p.$$

However, for this infinitesimal transformation to be canonical we need $$1=\frac{1}{\alpha}$$ which says $$\alpha=1.$$

So this infinitesimal transformation doesn't work? We might need to find fancier forms of the infinitesimal transformations, but I cant think of any anymore...


1 Answer 1


Near the identity $\alpha=1+\epsilon$ and $1/\alpha=1-\epsilon$. The infinitesimal canonical transformations generated by a generator $G(q,p)$ are, \begin{equation} q'=q+\epsilon [q,G]_{PB} =q+\epsilon\frac{\partial G}{\partial p}\\ p'=p+\epsilon [p,G]_{PB} =p-\epsilon\frac{\partial G}{\partial q} \ . \end{equation} Now $q'=q+\epsilon q$ and $p'=p-\epsilon p$ so that, \begin{equation} q+\epsilon q=q+\epsilon\frac{\partial G}{\partial p}\\ p-\epsilon p=p-\epsilon\frac{\partial G}{\partial q} \ . \end{equation} The generator is given by the equations, \begin{equation} \frac{\partial G}{\partial p}=q\\ \frac{\partial G}{\partial q}=p \end{equation} which integrate to $G(q,p)=qp$. Now we can write down the infinitesimal canonical transformation again as, \begin{equation} q'=q+\epsilon [q,G]_{PB} =q+\epsilon\frac{\partial G}{\partial p}=q+\epsilon q\\ p'=p+\epsilon [p,G]_{PB} =p-\epsilon\frac{\partial G}{\partial q}=p-\epsilon p \ . \end{equation} giving ODEs, \begin{equation} \frac{dq}{d\epsilon}=q\\ \frac{dp}{d\epsilon}=-p \end{equation} which integrate to $q(\epsilon)=q(0)e^\epsilon$ and $p(\epsilon)=p(0)e^{-\epsilon}$. Now put $\alpha=e^{\epsilon}$ and we get $q'=\alpha q$ and $p'=p/\alpha$ which is the desired finite canonical transformation in the question.


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