# ${\rm 2D}$ isotropic oscillator: Is ${\rm SO(4)}$ a subgroup of ${\rm Sp}(4,{\rm R})$?

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 $${\rm Sp}(4,{\rm R})$$?

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