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Recently I have come to know that for a system with $2n$ dimensional phase space, the set of all canonical transformations form a group ${\rm Sp(2n, R)}$. But in contrast to other Lie groups e.g. ${\rm SO(3)}, {\rm SU(2)}$ etc, I find this group to be quite abstract. Let me explain.

Any matrix $M\in{\rm Sp(2n, R)}$ act on the column vector $$\underline{z}=(q_1,...,q_n,p_1,...,p_n)^T$$ and preserve the Poisson bracket structure. But unlike rotations, CTs $(Q,P)\to(q,p)$ are most often not linear. Therefore, in contrast to rotation matrices $R_{\hat{n}}(\vec{\theta})\in {\rm SO(3)}$ where the group elements $R_{\hat{n}}(\vec{\theta})$ depend only on the group parameters $\vec{\theta}$, the matrices $M$ seems to depend on $(q,p)$ variables too. For instance, for the hamiltonian $H=\frac{1}{2}(p^2+q^2)$, a CT like $$q=\sqrt{2P}\cos Q, ~ p=\sqrt{2P}\sin Q$$ cannot be written as $$(Q, P)^T=M ~(q,p)^T$$ with an $M$ that is independent of $(q,p)$.

  • This does not feel correct to me. I am accustomed to seeing linear transformations forming a group, not nonlinear transformations.

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Next, the CTs can be widely different in their appearance. To state more clearly, for systems with ${\rm 2D}$ dimensional phase space, all the transformations below

$Q=p~ \& P=-q$,

$Q=\log \frac{1}{q}\sin p~ \&~ P=q\cot p$,

$Q=\tan^{-1}(q/p) ~ \& ~ P=\frac{1}{2}(p^2+q^2)$

... etc, are examples of CTs (in the sense that all of them preserve canonical PB structure). Therefore, if I am not terribly wrong, all these transformations are elements of ${\rm Sp(2,R)}$. These transformations are wildly different from each other; in no way they one is obtainable from the other by continuous variation of group parameters. This is also quite bizarre.

  • Again, I surely hold serious misunderstanding but can't find it. Can someone point out what's wrong with my line of thought? Thanks!
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  1. The full groupoid of all (possibly non-linear) canonical transformations (CTs) is infinite-dimensional. Infinitesimal CTs are Hamiltonian vector fields $$ \delta z^I ~=~{\{F(z,t),z^I\}}, \qquad I~\in~\{1,\ldots,2n\}.\tag{1}$$

  2. The symplectic group $Sp(2n, \mathbb{R})$ of dimension $n(2n+1)$ is the group of all linear time-independent CTs. The corresponding symplectic Lie algebra $sp(2n,\mathbb{R})$ is the set of all linear time-independent infinitesimal CTs, which have time-independent quadratic generating functions $$F(z)~=~\frac{1}{2}\sum_{I,J=1}^{2n}a_{IJ}z^Iz^J.\tag{2}$$

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  • $\begingroup$ Okay. Therefore, for the 2D phase space all CTs of the form $Q=\alpha q+\beta p, P=\gamma q+\delta p$ ($\alpha,\beta,\delta,\gamma$ are independent of $q,p$) belong to the group ${\rm Sp(2,R)}$. My 2nd and 3rd examples of CTs, which involve logarithmic, trigonometric functions do not belong to ${\rm Sp(2,R)}$. Am I correct? $\endgroup$ – mithusengupta123 Jun 30 at 15:58
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    $\begingroup$ $\uparrow$ Right. $\endgroup$ – Qmechanic Jun 30 at 16:10
  • $\begingroup$ Isn't then the Wikipedia notation $M(z,t)\in Sp(2n,R)$ misleading? en.wikipedia.org/wiki/Symplectic_group#Classical_mechanics $\endgroup$ – mithusengupta123 Jun 30 at 16:43
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    $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic Jun 30 at 17:08

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