# An example of symplectomorphism that is not a canonical transformation

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 $$M\to M$$?

• 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. Commented Jan 11 at 6:09
• 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. Commented Jan 18 at 23:40