In quantum mechanics, when two observables commute, it implies that the two can be measured simultaneously without perturbing each other's measurement results. Or in other words, the uncertainty in their measurements are not coupled.

But in classical mechanics, in an analogous way we have the Poisson brackets instead, where when two functions/variables of the systems have a vanishing Poisson bracket with each other, we say they are in "involution".

I understand that when a variable $j$ has a vanishing Poisson bracket with the Hamiltonian of the system, means that $j$ is a constant of motion, conserved in time. But for the general case:

If the Poisson bracket of $f$ and $g$ vanishes ($\{f,g\} = 0$), then $f$ and $g$ are said to be in involution.

  • What does the above mean physically? Is it in some sense related to measurement results as it is with commutators in QM, meaning two variables in involution are completely independent of each other?

  • Finally, is there a simple classical example, where one can see 2 observables are in involution, but 2 other (same system) may not be (i.e. counter-example for involution)? I think it would help better understand what in "involution" means.


1 Answer 1


The classical poisson bracket with the generator of any transformation gives the infinitesimal evolution with respect to that transformation. The familiar

$$ \partial_t f = \{H,f\}$$

means nothing else than the time evolution of any observable is given by its Poisson bracket with the Hamiltonian, which is the generator of time translation. More generally, given any element $g$ of the Lie algebra of observables on the phase space, the infinitesimal evolution under the transformation that it generates (parametrized by an abstract "angle" $\phi$) is given by

$$ \partial_\phi f = \{g,f\}$$

What is meant by this is that every observable $f$ gives rise (by Lie integration) to a symplectomorphism $\mathrm{exp}(\phi f)$ on the phase space (since the true Lie integration of observables projects down surjectively onto the symplectomorphisms). More precisely, the statement above therefore reads

$$ \partial_\phi( f \circ \mathrm{exp}(\phi g))\rvert_{\phi = 0} = \{g,f\}$$

so that vanishing Poisson bracket implies $f \circ \mathrm{exp}(\phi g) = f$, i.e. invariance of the observable under the induced symplectomorphism.

Thus, if the Poisson bracket of $f$ and $g$ vanishes, that means that they describe an infinitesimal transformation that is a symmetry for the other.

For example, an observable is invariant under rotation if its Poisson bracket with all $L^i = \epsilon^{ijk}x^jp^k$ (components of $\vec L = \vec x \times \vec p$) vanishes.

For $x$ and $p$, the non-vanishing $\{x,p\}$ means the rather trivial insight that $x$ is not invariant under the translations generated by the momentum.

Fun fact: Wondering what the actual Lie integration of these infinitesimal symmetries might be leads directly to the quantomorphism group, and is a natural starting point for geometric quantization, as discussed in this answer.

Responding to the tangential question in a comment:

Considering the invariant under generation idea, so when trying to apply Noether's theorem, one only has to apply the Poisson bracket of the Lagrangian (e.g.) with $x$, and if it vanishes momentum is conserved for that system?

No, it means that the "Lagrangian" (you cannot really take the Lagrangian, because it is not a function on the phase space) is invariant under varying the momentum of the system (since $x$ generates "translations" in momentum). The Hamiltonian equivalent of Noether's theorem is simply that $f$ is conserved if and only if

$$ \{H,f\} = 0$$

since conservation means invariance under time evolution.

  • $\begingroup$ @user929304: Updated the answer. $\endgroup$
    – ACuriousMind
    Nov 4, 2014 at 16:14
  • $\begingroup$ @user929304: The center of a Lie algebra is the set of all $x$ for which $[x,a] = 0$ for all $a$. To say it is constant for the algebra of observables means that there are no non-trivial, i.e. non-constant, functions on the phase space that would be in involution to all others. It is a fact that for the algebra on a symplectic phase space, the center must be constant. One can generalize Hamiltonian mechanics to arbitrary "phase spaces" and Poisson algebra on them, but I'm no expert on that. (If you can make a good question out of this, ask it!) $\endgroup$
    – ACuriousMind
    Nov 4, 2014 at 16:43
  • $\begingroup$ Thanks, yeah I see, at least now I have some idea what it means. Wow with your answer to this post, and the intuition you offered I feel so much better. Now I see better what is meant by observables that are in involution (invariant under each other's infinitesimal generation), and how this implies the Hamiltonian system can be decomposed into $N$ independent sub-systems, hence integrable (assuming N observables in involution). What is the right word in classical mechanics instead of "observable"? that one's taken from QM I reckon. $\endgroup$
    – user929304
    Nov 4, 2014 at 16:54
  • $\begingroup$ @user929304: I actually learned to call the differentiable functions on the phase space classical observables, but you could just settle for smooth functions. $\endgroup$
    – ACuriousMind
    Nov 4, 2014 at 16:56
  • $\begingroup$ Alright, thanks. I appreciate you taking the time to also answer to my comments, extremely kind and considerate of you! Thanks again, cheers. $\endgroup$
    – user929304
    Nov 4, 2014 at 17:02

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