I have two Hamiltonians, $H(p_1, q_1, \dots, p_n, q_n)$ and $H'(p_{n+1}, q_{n+1}, \dots, p_m, q_m)$. They are non-interacting, indeed they have different variables. Moreover, I know that they are both non-integrable. Taken together, the two non-interacting systems are described by the Hamiltonian: $$ H_T(p_1, q_1, \dots, p_m, q_m) = H(p_1, q_1, \dots, p_n, q_n) + H'(p_{n+1}, q_{n+1}, \dots, p_m, q_m) $$

My question is: is it possible that $H_T$ is integrable? I guess "no", but I would like to have a proof.

At first, it could appear as a simple question. Intuitively, "integrable" means "soluble", thus, if the two systems together ($H_T$) are soluble, then a single system alone ($H$) is also soluble.

But "integrable" does not really mean "soluble": it means that it has $m$ constants of motion. So the question is not that simple: it is still possible that $H_T$ has $m$ constants of motion, but a constant of motion of $H_T$ could be not constant for $H$.

More formal statement of the problem

This is a more rigorous formulation of the theorem that I want to prove. Please notice that it is expressed as the contrapositive of what I wrote above.

Two Hamiltonians are given, $H(p_1, q_1, \dots, p_n, q_n)$ and $H'(p_{n+1}, q_{n+1}, \dots, p_m, q_m)$. The Hamiltonian $H_T$ is defined as: $$ H_T(p_1, q_1, \dots, p_m, q_m) = H(p_1, q_1, \dots, p_n, q_n) + H'(p_{n+1}, q_{n+1}, \dots, p_m, q_m) $$

Proposition to be proved: if $H_T$ has at least $n+m$ independent constants of motion, of which $n+m$ in involution, then $H$ has at least $n$ independent constants of motion, of which $n$ in involution.

I would be happy as well to prove much less: if $H_T$ is completely superintegrable (it has $2(n+m)-1$ independent constants of motion, of which $n+m$ in involution), then $H$ has a constant of motion besides $H$ itself. Belive me, also this is tricky.

Important additional information

It is worth noting that, in general, a constant of motion of $H_T$ is not a constant of motion of $H$. It is easy to show an example. Take the equations of a forced (nonlinear) pendulum. It can be written as an autonomous system by adding a degree of freedom: $$ H(p_1, q_1, p_2, q_2) = \frac{p_1^2}{2} + \cos(q_1) + \epsilon q_1 \sin(\omega q_2) + p_2 $$ Clearly, $q_2$ plays the role of time, $q_2=t$. Now, take the same form for $H'$: $$ H'(p_3, q_3, p_4, q_4) = \frac{p_3^2}{2} + \cos(q_3) + \epsilon q_3 \sin(\omega q_4) + p_4 $$

It is easy to show that $H_T=H+H'$ has three constants of motion: $H$, $H'$ and $f=q_4-q_2$. The first two are evident. For the third: $$ \left\{ f, H_T\right\} = \left\{ q_4, H'\right\} - \left\{ q_2, H\right\}= 1-1 = 0 $$ However, this third constant of motion, $f$, is not a constant of motion separately for $H$ and $H'$: $$ \left\{ f, H\right\} = \left\{ q_4-q_2, H\right\} = -1 $$

We can also notice the following. From the knowledge on the non-autonomous forced pendulum, we know that its autonomous version is not integrable. Thus $H$ and $H'$ are non-integrable; since they have only 2 degrees of freedom, this means that they only have one constant of motion, namely $H$ and $H'$, respectively.

Concluding: the answers to this question cannot use the property that a constant of motion of $H_T$ is also a constant of motion of $H$, because this property does not hold in general.

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    $\begingroup$ To be clear, for, e.g., n=2, m=4 , how many constants of the motion qualify for your integrability, as opposed to Liouville, or superintegrability? $\endgroup$ May 30, 2019 at 22:26
  • $\begingroup$ @CosmasZachos : I would like to prove that, if $H_T$ has $m$ independent constants of motion in involution, then $H$ has $n$ constants of motion. It should be the Liouville integrability. But I would be happy to prove much less, e.g. that if $H_T$ is superintegrable (anyway) then $H$ has two independent constants of motion (one besides $H$ itself). $\endgroup$ May 31, 2019 at 7:51
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    $\begingroup$ I see. You need another conserved quantity. Try to descend to $n=2, ~~ m=3$, instead.... $\endgroup$ May 31, 2019 at 18:34
  • $\begingroup$ Yes, also the case $n=2$ and $m=3$ could be interesting! $\endgroup$ May 31, 2019 at 20:28

1 Answer 1


The coupled system will not be Liouville integrable. Because, for integrability the constants of motions will have to be independent and in involution i.e their Poisson bracket has to be 0. If you construct some additional constant of motion for the total system (as you did in additional information), they will either be dependent or will not be in involution ($q_4-q_2$ is not in involution with $H$ and $H’$)

  • $\begingroup$ Good point and interesting approach! But I still have some question. In the example, the three (independent) constants are not in involution, but maybe it is possible to find three functions of them that are in involution. In general, the Liouville integrable $H_T$ could have constants of motion in involution, that do not include $H$ and $H'$. $\endgroup$ May 31, 2019 at 20:32
  • $\begingroup$ Is not any constant of motion of individual systems is also constant of motion for the total system? So $H$ and $H’$ are always constant of motion for $H_T$. $\endgroup$ May 31, 2019 at 20:43
  • $\begingroup$ Consider the following argument. Suppose $H$ has $k<n$ and $H’$ has $l<m$ conserved quantities. Then you already have $k+l$ conserved quantities of total system. For integrability you must have some other conserved quantities involving coordinates of both $H$ and $H’$. Let’s say one such quantity is $f(\vec{p}, \vec{q}) + f’(\vec{p’}+\vec{q’})$. To be independent $f$ and $f’$ must not be constant of motion for $H$ and $H’$. But if you want $f+f’$ to be in involution with $H$ and $H’$ separately that contradicts the previous statement. $\endgroup$ May 31, 2019 at 21:03
  • $\begingroup$ Something is still not clear. $H_T$ has $m$ constants of motion in involution, $J_i$. But in principle it could have additional, independent constant of motion, $K_i$ (For example, superintegrable systems have additional $K_i$). Are we sure that all the $K_i$ are also in involution with the other $J_i$? If not, it is possible that $H=K_1$. $\endgroup$ Jun 1, 2019 at 12:37
  • $\begingroup$ I am not sure if I understand your question. In order to be integrable, $H_T$ need to have $n+m$ constants of motion, not $m$, right? I am saying that if $H, H’$ are not integrable then $H_T$ will not have enough conserved quantities to be integrable. And if it is not integrable, then how can it be superintegrable? $\endgroup$ Jun 1, 2019 at 15:20

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