# Poisson structure comes from hamiltonian?

I am interested in studying quantization, but it seems I am lacking the basics of classical mechanics. Any help would be appreciated.

I would first like to ask what is necessary to have a sympletic/Poisson structure in a classical mechanical system (e.g. an isolated particle moving on some Riemannian manifold $Q$). Do we need an evolution rule to begin with?

For example, if we have some ("nice"?) Lagrangian structure (this gives an evolution rule), we can define the momenta and get to Hamilton's equations, and build a Hamiltonian flow on the phase space $P = T^*Q$. This gives us a Poisson structure for the algebra of smooth observables $C^\infty(P,\mathbf{R})$.

But many books depart from a Poisson manifold $(P, \{\,,\,\})$ and input the Hamiltonian later, or even suggest it can be a parameter, some $h\in C^\infty(P,\mathbf{R})$. This suggests that the Poisson structure is, in fact, more of a kinematic entity, rather than a dynamic one (means it has no a priori relation with the rule of how the system evolves). Is that right? On a practical example, how do we even know what the Poisson structure would be, if we don't even know what are the momenta?

Also, as I see, the fact that one can use a Poisson manifold also comes from the fact that a point in phase space will determine the state (give a solution curve for the motion by initial value formulation) only if we restric ourselves to evolution equations of second order, which means some restriction on the Lagrangian, no?

Lastly: For quantization of a Poisson manifold, do we need the Hamiltonian beforehand? Or: quantization is kinematic or dynamic?

Thank youuu!

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Sympletic structure arises naturally on the cotangent bundle $T^*Q$ of the configuration space $Q$. Nothing needed, it's already there. Not even Riemannian structure (kinetic energy) on $Q$, $Q$ should be just a smooth manifold. That's the naturallity in math. – Yrogirg Jul 24 '12 at 19:16

I too have been trying to understand meaning of quantization, and have had similar kinds of questions as you are asking. So I would like to share my own understanding regarding these.

First of all when we encounter problems in classical mechanics, it is almost never that we are given with a symplectic or Poisson manifold, with a smooth Hamiltonian function on it so that all you have to do is to carry out formal well defined mathematical steps to find equations of motion. Usually finding the phase space itself is part of the problem. And in some cases (good examples I think come from field theories) it could be a nontrivial problem.

On the other hand if some mathematician gives you with a classical mechanics problem s/he will at least provide you with the data {Phase space, Hamiltonian}, because mathematically this data is necessary for defining the problem. Here by phase space I mean a symplectic manifold (or may be a Poisson manifold with a given Poisson structure).

Now coming to your questions :

I would first like to ask what is necessary to have a sympletic/Poisson structure in a classical mechanical system (e.g. an isolated particle moving on some Riemannian manifold Q). Do we need an evolution rule to begin with?

If you want to work on $TQ$ ("velocity space") then here too you can work in Hamiltonian (rather than action) formulation, but your symplectic structure here will depend upon Lagrangian. Also this I think would put some conditions (call them conditions A) on Lagrangian so that you can have a "good" symplectic structure on $TQ$ (i.e. a nice Hamiltonian formulation of the problem on $TQ$) . As of now I don't know what these conditions are but it should not be very difficult to find them from expression of symplectic form in terms of Lagrangian.

If, on the other hand, you want to work on $T^*Q$ then here symplectic structure will not depend upon the Hamiltonian. That is why I think people usually study Hamiltonian formulation of classical mechanics on $T^*Q$ rather than on $TQ$ because here symplectic structure is fixed once and for all and thus to study different physical systems you only need to change your Hamiltonian. In order to go from $TQ$ to $T^*Q$ one needs Legendre transformation and for this to be possible Lagrangian needs to satisfy some conditions (I am not sure what (if any) is the relation of these conditions with conditions A mentioned above; though its likely that they are related).

But many books depart from a Poisson manifold $(P,{,})$ and input the Hamiltonian later, or even suggest it can be a parameter, some $h\in C^\infty(P,R)$. This suggests that the Poisson structure is, in fact, more of a kinematic entity, rather than a dynamic one (means it has no a priori relation with the rule of how the system evolves). Is that right? On a practical example, how do we even know what the Poisson structure would be, if we don't even know what are the momenta?

From what we saw above it should be clear that symplectic structure on Phase space $T^*Q$ is independent of Hamiltonian. On $T^*Q$ we know what position is and what momentum is and that is all that's needed to define a symplectic structure.

Also, as I see, the fact that one can use a Poisson manifold also comes from the fact that a point in phase space will determine the state (give a solution curve for the motion by initial value formulation) only if we restric ourselves to evolution equations of second order, which means some restriction on the Lagrangian, no?

If classical problem is given in Lagrangian formulation then (as mentioned above) there should be some conditions that Lagrangian must satisfy in order for the problem to have Hamiltonian formulation. On the other hand on a phase space mathematically it would be fine to choose any smooth function as your Hamiltonian.

Lastly: For quantization of a Poisson manifold, do we need the Hamiltonian beforehand? Or: quantization is kinematic or dynamic?

You don't need to know Hamiltonian for quantizing your phase space. Quantization treats all observables on equal footing. However there are some subtleties due to which not all observables may get quantized. So some of the steps of quantization may require you to refer to your Hamiltonian just to make sure that at least your Hamiltonian function can be assigned to a corresponding quantum operator.

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The last part is just not true. You can't "quantize" a classical phase space independent of extra structure, since the quantum commutation relations are not invariant under general canonical transformations. If [x,p]=i, it isn't true that any other classical functions of x and p whose Poisson bracket is 1 have the same commutation relation, because of operator ordering issues. – Ron Maimon Jul 25 '12 at 3:53
@Ron Maimon Thanks for your correction. Ya, choice of polarization in geometric quantization method may require extra structure on phase space. But still Hamiltonian doesn't have much role to play here. Right ? – user10001 Jul 25 '12 at 4:24
If I was sure, I would have written an answer. The problem is that classical mechanics is more symmetrical than quantum mechanics, and QM picks out certain canonical pairs as "truly canonical" (in that their commutation relation is really a c-number, equal to their Poisson bracket, equal to one), and other variables which look just as canonically conjugate classically are no good as quantum canonical pairs. This is something which makes the program of "quantization" impossible as mathemticians imagine it, although it can be done to leading semiclassical order. – Ron Maimon Jul 25 '12 at 4:30
I agree you don't need a Hamiltonian to write down q and p and identify p as the derivative with respect to q. This is all that you are doing when "quantizing". But you do need to know which way is q and which way is p (although you can interchange the two, or rotate, you can't do a general nonlinear transformation). I will think what this extra structure is, it's like a foliation of some kind. – Ron Maimon Jul 25 '12 at 4:34
@RonMaimon: One can also argue that QM is more symmetric than CM, as the classical commutation relations are not invariant under general unitary mappings. – Arnold Neumaier Jul 25 '12 at 13:32

Yes, the Poisson structure determines part of the kinematic set-up. A choice of relevant observables spanning a Lie algebra under the PB (a Heisenberg algebra for a multiparticle system, $iso(3)$ for rigid motions, etc.) makes up the remainder of the kinematics. In a Lagrangian framework, for Lagrangians at most quadratic in the first derivatives and not containing higher derivatives, the PB is determined by the derivative part of the Lagrangian, hence is far from fixing the dynamics.

One needs the Poisson structure in order that the dynamics is determined by the Hamiltonian.

Quantization is kinematic, too, and has nothing to do with particular Hamiltonians. The quantum-classical correspondence of the Hamiltonian is anyway valid only up to $O(\hbar)$, as the same classical $H$ is the classical limit $\hbar\to 0$ of many quantum $H$s, and as $\hbar$ is a constant in Nature.)

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You are speaking about systems with quadratic Lagrangians only, this is a subset of the systems that interest the OP, but it is the one where the Lagrangian formulation is easiest, and it does include most field theories people study. – Ron Maimon Jul 25 '12 at 16:24
@RonMaimon: Thanks. I edited my answer to reflect that. – Arnold Neumaier Jul 25 '12 at 17:12

The symplectic structure is independent of the Hamiltonian, but often a Hamiltonian gives us a hint to which symplectic structure we should use.

Here's what I mean in more detail. The way I intuit a symplectic structure is as a polarization of the 2n coordinates of the manifold into n positions and n conjugate momenta. This shouldn't depend on the Hamiltonian, which just tells me the energy of a configuration, but the Hamiltonian will be easier to manipulate with a good choice of polarization.

A symplectic structure is specified by a nondegenerate real closed 2-form $\omega$. We can relate this to the Poisson bracket as follows. Given a function $f:M\rightarrow\mathbb{R}$ on the phase space $M$, since $\omega$ is nondegenerate we can find a unique vector field $X_f$ such that $\omega(X_f, -)=df(-)$ as 1-forms. Then given another $g:M\rightarrow\mathbb{R}$, we can define $\{f,g\}=\omega(X_f,X_g)$.

So far everything is time-independent. In this symplectic perspective, the notion of time should be a flow on $M$, so that given a function $f$ on $M$ as before, we can form a 1-parameter family of such functions $f(t)$ such that $\frac{df}{dt}$ depends only on $f$. One way of doing this is to specify a function $H:M\rightarrow\mathbb{R}$ and declare $\frac{df}{dt}=\{H,f(t)\}$. We could also take a 1-parameter family of Hamiltonians $H(t)$ and do the same thing.

Finally, it is important to note that our phase space is often not a cotangent bundle when we have symmetries or constraints. However, there is a theorem of Darboux which says that the symplectic form always looks locally like the natural symplectic form on a cotangent bundle.

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The local statement is obvious, but the global statement eluded me--- what's an example of a global phase space which is not a cotangent bundle globally? – Ron Maimon Jul 27 '12 at 7:10
The trivial example is the two-sphere. Its symplectic structure is given by the volume form, but as $S^2$ is compact it could not be a cotangent bundle. Its the classical (and quantum mechanical, in the language of geometric quantum mechanics) phase space of the spin degrees of freedom. – Tobias Diez Jul 27 '12 at 17:38