I want to check my understanding on the difference between symplectomorphism and canonical transformation. This is a follow-up of my previous post.

  • (A) A map $(q,p)$ to $(Q,P)$ is called a symplectomorphism if it preserves the symplectic two-form: $dp\wedge dq=dP\wedge dQ$.

  • (B) A map $(q,p)$ to $(Q,P)$ is called a canonical transformation if it satisfies $(pdq-H(q,p,t)dt)-(PdQ-K(Q,P,t)dt)=dF$ for a function $F$.

Let $M=\mathbb{R}^2\setminus\{(0,0)\}$ be the phase space. As an example, let us consider the following map: $$Q(q,p)=q\sqrt{1+\frac{2\epsilon}{p^2+q^2}},\quad P(q,p)=p\sqrt{1+\frac{2\epsilon}{p^2+q^2}},$$ where $\epsilon\geq0$ is a parameter. The inverse transformation reads $$q(Q,P)=Q\sqrt{1-\frac{2\epsilon}{P^2+Q^2}},\quad p(Q,P)=P\sqrt{1-\frac{2\epsilon}{P^2+Q^2}}.$$

This map is a symplectomorphism because $$\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial P}{\partial q}\frac{\partial Q}{\partial p}=1,$$ implying that $dp\wedge dq=dP\wedge dQ$.

This map is not a canonical transformation because the following candidate of $F$ and $K$ satisfy $(pdq-H(q,p,t)dt)-(PdQ-K(Q,P,t)dt)=dF$, but this $F$ is not globally defined. $$F=-\epsilon\Big(\frac{pq}{q^2+p^2}+\text{arctan}(q/p)\Big),\quad K(Q,P,t)=H(q,p,t).$$

The infinitesimal version of this transformation reads $$Q(q,p)=q+\epsilon\frac{q}{p^2+q^2}+O(\epsilon^2),\quad P(q,p)=p+\epsilon\frac{p}{p^2+q^2}+O(\epsilon^2),$$ which is suggested in this post.

I thought this is a good example, but it has a problem: the inverse function is only defined for $P^2+Q^2\geq2\epsilon$. Can we improve this map in such a way that it maps $M\to M$?

  • $\begingroup$ Can this be simpler? Let $M$ be a cylinder $S^1\times\mathbb{R}$. This would be a phase space of a point particle moving on a ring. We consider a map from $(\theta,p)$ to $(\Theta,P)$ defined by $\Theta(\theta,p)=\theta$ and $P(\theta,p)=p+\epsilon$. Then $d\Theta\wedge dP=d\theta\wedge dp$ and $(pd\theta-Hdt)-(Pd\Theta-Kdt)=dF$ with $F=-\epsilon\theta$ and $K=H$. In this case $M$ is mapped to $M$ itself. $\endgroup$
    – watahoo
    Commented Jan 11 at 6:09
  • $\begingroup$ Yes. This simplified example works. Its quantum version states that the operator $\hat{U}=e^{i\epsilon\hat{\theta}}$ does not exist unless $n$ is an integer, implying that the fractional part of the flux threading the ring cannot be changed by any unitary transformation, while the integer part can be changed by a large gauge transformation. $\endgroup$
    – watahoo
    Commented Jan 18 at 23:40


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