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Consider the ${\rm 2D}$ isotropic oscillator. The hamiltonian is $$H=\frac{1}{2}(p_x^2+p_y^2+x^2+y^2)$$ and the phase space is $4$ dimensional. In this case, the set of all linear canonical transformations that preserve the form of Hamilton's equations form a group ${\rm Sp}(4,{\rm R})$. On the other hand, the group of orthogonal transformations ${\rm SO(4)}$, in the phase space, leaves the hamiltonian $H$ invariant. Now, there can be a subset of canonical transformations that leaves the Hmiltnian invariant in addition to preserving the form of Hamilton's equations.

  • Will it be correct to assert that ${\rm SO(4)}$ is a subgroup of ${\rm Sp}(4,{\rm R})$?
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No, $SO(2n,\mathbb{F})\subseteq Sp(2n,\mathbb{F})$ [understood via their standard embedding into $GL(2n,\mathbb{F})$] is only true for $n=1$. This fact can be proved by considering the corresponding Lie algebras.

Specifically, $Sp(4,\mathbb{R})$ is (the double cover of) the restricted anti de Sitter group $SO^+(3,2;\mathbb{R})$, cf. e.g. my Math.SE answer here.

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  • $\begingroup$ So that would mean, there are some ${\rm SO(4)}$ transformations in the phase space that are not canonical transformations? $\endgroup$ – mithusengupta123 Jun 27 at 18:05
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic Jun 27 at 18:08

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