A 2D harmonic oscillator \begin{align} H=p_x^2+p_y^2+x^2+y^2 \end{align} has 4 constants of the motion: $E$ the total energy, $D$ the energy difference between coordinates, $L$ the angular momentum and $K$ the correlation. For example see http://aapt.scitation.org/doi/pdf/10.1119/1.1971258

I often read that a system can have at most $2n-1$ constants of the motion, and such a system is maximally superintegrable. In this case $n=2$, implies the most constants of motions the system can have being 3.

Further a 3D harmonic oscillator has 9 constants of the motion, and $9>5$.

How is this reconciled?


The four invariants $$ 2E_x=p_x^2+x^2, \qquad 2E_y=p_y^2+y^2, \qquad L=yp_x-xp_y, \qquad K=xy+p_xp_y, $$ are linearly independent, as per the requirement of that article's linear algebraic method; but they are not algebraically independent quantities, of course, qua hypersurfaces in the 4d phase space.

You may readily check that $$ L^2+K^2=4E_xE_y. $$ This is a good thing, so that there is a trajectory in phase space, and the particle is not frozen: The particle must lie on all constant surfaces, and its trajectory is the intersection of any three. An independent one would intersect it at a point and freeze it!

Recall that the symmetry of the system is SU(2). The 4 invariants represent the 3 generators $L,K,2(E_x-E_y)$ and the identity (Hamiltonian), so, then, the quadratic Casimir invariant thereof.

The analogous dependence is at work, more strikingly, in the 3d oscillator, cf 194768, whose symmetry is SU(3). You might choose the independent set $E_1,E_2,E_3,L_{12},L_{23}$.

But, as a rule, in nd, you have 2 n–1 independent invariants, so a completely specified trajectory as their intersection——superintegrability.

  • $\begingroup$ I realize this may be technical but do you really mean $su(2)$ and not $u(2)$? $\endgroup$ May 26 '17 at 20:51
  • $\begingroup$ No, you got me; technical. Yes, the identity, and also Casimir of the su(2) , has no problem packing together with it to a u(2), Lie algebraically. Reveals my concern to shed invariants, not pile them up...! $\endgroup$ May 26 '17 at 21:50

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