1
$\begingroup$

Given Hamiltonian system with $2n$-dim phase space, if there exist $k\ge n$ independent integrals of motions then we call it integrable Hamiltonian system. The largest number of independent integrals of motions should be $2n-1$. As we know, if a Hamiltonian system is integrable, we can have action-angle variables with $m=2n-k$ independent frequencies $\omega_i$ with $i=1,\cdots, m$.

So we see if for $\forall \ i,j$, $\omega_i/ \omega_j \in \mathbb{Q}$ then the bounded orbit is closed. And certainly if there is only one independent frequency,i.e. maximal integrable $k=2n-1$, then all bounded orbits are closed.

My question:

  1. At first glance, it seems that "a system with all bounded orbits closed" implies that "a system is maximal integrable($k=2n-1$)". But it's not true, I can come up an example $$H=\frac{p_x^2}{2} +\frac{p_y^2}{2} + \frac{1}{2}\omega_1^2 x^2 + \frac{1}{2}\omega_2^2 y^2 \tag{1}$$ with $\omega_1/ \omega_2 \in \mathbb{Q}$. I think this example is not maximal integrable if $\omega_1 \neq \omega_2$ since it have two independent frequencies. (Is this argument right?)

But this example is a little trivial, does there exist an example
$$H= \frac{\mathbf{p}^2}{2}+V(\mathbf{r}) \tag{2}$$ with $V(\mathbf{r})$ not necessary central potential, such that it has two independent frequencies $\omega_{1,2}(I_1,I_2,\cdots)$ which are not constant but the radio is always a rational number.

With action-angle variable, I can come up an example $H=\frac{1}{2} I_1^2+ 2 I_1 I_2+ 2 I_2^2$ with $\omega_2 = 2 \omega_1 = 4 I_2 + 2 I_1$ not a constant. But how to canonical transform to the form of $(2)$? Or any Hamitonian of form $(2)$ with above requirement must be anisotropy harmonic oscillator?

PS: It has no relation with Bertrand's theorem which require central force.

  1. Another system also has all bounded orbits closed is a charged particle in $2$-dim plane with constant magnetic field perpendicular to the plane. All orbits are circle so it can be imagined that this system must be maximal integrable. I'm curious about what the three independent integrals of motions?
$\endgroup$
0
$\begingroup$

Comments to the post (v3):

  1. Any quadratic Hamiltonian that can be separated in normal modes: $$H~=~ \sum_{k=1}^n H_k, \qquad H_k~:=~~\frac{p_k^2}{2m_k}+\frac{1}{2} m_k\omega_k^2q_k^2~=~ \omega_k h_k,$$ $$h_k~:=~\frac{1}{2}(P_k^2+Q_k^2)~=~ \frac{1}{2} Z_k Z^{\ast}_k,$$ $$ Q_k~:=~m_k\omega_k q_k, \qquad P_k~:=~\frac{p_k}{m_k\omega_k}, $$ $$ Z_k ~:=~ P_k+iQ_k, \qquad \{Z_k,Z^{\ast}_{\ell}\}~=~2i\delta_{k\ell}, \qquad k,\ell~\in~\{1,\ldots,n\},\tag{1a} $$ is maximally superintegrable$^1$, i.e. it has $2n-1$ integrals of motion. To see this consider the $n$ complex constants of motion $$C_k ~:=~{\rm Ln}(Z_k)-i\omega_kt, \qquad k~\in~\{1,\ldots,n\},\tag{1b}$$ which depend explicitly on $t$. By eliminating the $t$-variable, it is possible to construct $2n-1$ real integrals of motion.

  2. The 2D Landau Hamiltonian $$ H~=~\frac{p_x^2}{2m}+ \frac{1}{2m}\left( p_y - qBx \right)^2 \tag{2a}$$ is maximally superintegrable, i.e. it has 3 integrals of motion: $$ H, \qquad p_y , \qquad p_x - qBy. \tag{2b}$$

--

$^1$ For terminology, see also this Phys.SE post.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.