# Complete set of observables in classical mechanics

I'm reading "Symplectic geometry and geometric quantization" by Matthias Blau and he introduces a complete set of observables for the classical case:

The functions $q^k$ and $p_l$ form a complete set of observables in the sense that any function which Poisson commutes (has vanishing Poisson brackets) with all of them is a constant.

I wonder why is it so? That is why do we call it a complete set of observables? As I understand it means the functions satisfying the condition above form coordinates on a symplectic manifold, but I don't see how.

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 I've posted related question at Math.SE math.stackexchange.com/questions/194110/… – Yrogirg Sep 11 '12 at 12:22

Any observable $H$ in classical mechanics defines a flow of states by regarding it as a Hamiltonian. This flow acts on observables $f$ by $df/dt = \{H,f\}$ (this is Hamilton's equation). The idea of a complete set of observables is that it is a set for which any observable with constant flow for all members of the set (ie. Poisson commute with the set) is constant. Intuitively, these flows move all over the phase space, so if $f$ is nonconstant, the flow of $f$ along one of the observables in the complete set can detect this.

These functions don't have to form coordinates.

EDIT: To complement QMechanic's counterexample, here is a compact counterexample: consider the 2-sphere with its ordinary symplectic form and the functions $\mathrm{cos} 2\theta$, $\mathrm{sin}\phi$, and $cos \phi$, where $\theta$ is the polar angle, and $\phi$ is the azimuthal angle. These are symmetric accross the equator, so they don't distinguish points, but it is pretty clear that they are a complete set.

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"I don't think that this means the set of observables can uniquely determine states" The one (but strong) reason I can think of to support this is that the contrary would mean the manifold could be covered with only one map. – Yrogirg Sep 11 '12 at 9:39
@Yrogirg That's a good point. It is of course possible to have a minimal complete set which is larger than the dimension of the phase space. Then that the functions separate points just means they define an embedding into some higher dimensional Euclidean space. For example, the ps and qs are really only defined in their coordinate charts, so to turn them into globally defined functions, we can use a partition of unity so that they are zero outside of their charts. Then it takes several coordinate charts to cover the phase space in general. – user404153 Sep 11 '12 at 17:45
Ok, final clarifications, so you don't see any classical physical meaning in a classical complete set of observables? And in QM $\hat q^2$ and $\hat p$ still form a complete set for a 1D particle? – Yrogirg Sep 12 '12 at 8:14
I have tried to give a description free of any mention of operators, Hilbert spaces, or anything else from the quantum picture. And yes, those do form a complete set for a 1D particle. – user404153 Sep 12 '12 at 8:41

Blau called them a 'complete set' in analogy to the quantum mechanical picture, where a observable commuting (read Poission-commuting in the classical case) with a complete set of commuting observables is proportional to the unit, i.e. a 'constant'. This is called (first) Schur's lemma.

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@Yrogirg indeed another way to say 'complete set of observerables' is to say that these ps and qs form an irreducible representation of the Heisenberg algebra (which is equivalent by Schur's lemma to saying that the only observables which Poisson commute with all the ps and qs are constant). – user404153 Sep 10 '12 at 21:33
@user404153 so specifying the values of all the classical observables from the complete set won't specify a unique classical state? – Yrogirg Sep 11 '12 at 4:23
@Yrogirg That's not quite what I mean. Certainly if one specifies the position and the momentum of a state, one specifies the classical state completely (after all, positions and momenta are supposed to be the coordinates of the space of classical states). What I mean is that if you specify an observable's Poisson brackets with all the ps and qs, then one has specified the observable up to the addition of a constant. Equivalently, an observable which Poisson commutes with all the ps and qs is a constant. – user404153 Sep 11 '12 at 4:59
@user404153 ok, but what has it to do with classical physics alone, without referencing to QM? That's just math interpretation, if that was what I was looking for, I'd asked the question at Math.SE. – Yrogirg Sep 11 '12 at 5:13
@Yrogirg I am confused by your question then. Are you asking why the ps and qs form a complete set of observables? You can show {p,f} = df/dq and {q,f} = df/dp for any function f(p,q), so certainly if {p,f} = {q,f} = 0, f is constant. – user404153 Sep 11 '12 at 8:27

I) Here is my interpretation of OP's question(v1). The mentioned quote is from p.11 in Section 2.2 just below eq.(2.22).

Section 2.2 is devoted to the case where phase space is a $2n$ dimensional real vector space $V$ with $2n$ global coordinates

$$(x^1, \ldots, x^{2n})~=~(q^1, \ldots, q^n; p_1 \ldots, p_n),$$

and canonical Poisson bracket, cf. eq. (2.22). This is a special case of a general symplectic manifold.

An observable $f$ is by definition a smooth function on $V$, i.e., $f\in C^{\infty}(V)$. Or in plain words, $f$ is a smooth function of $x^1, \ldots, x^{2n}$. On the other hand, the $2n$ coordinates $x^1, \ldots, x^{2n},$ form a complete set of generators for the algebra $(C^{\infty}(V),+,\cdot)$.

Let us assume that the function $f$ Poisson commutes (has vanishing Poisson brackets) with all of the $2n$ variables,

$$\forall x\in V \forall I\in\{1,\ldots, 2n\}~:~\{x^I, f(x)\}~=~0.$$

By the definition (2.21) of the canonical Poisson bracket, we deduce that $f$ has vanishing derivatives wrt. all the positions and momenta.

Hence $f$ is just a constant function.

II) On the other hand, let us imagine that we have $2n$ differential functions $g^1, \ldots, g^{2n},$ such that

$$\forall f\in C^{\infty}(V): [ (\forall x\in V\forall I\in\{1,\ldots, 2n\}~:~\{g^I(x), f(x)\}~=~0)~ \Rightarrow ~ f ~\text{is constant} ].$$

OP essentially asks in a comment:

Do $g^1, \ldots, g^{2n},$ locally form a coordinate system? By the word locally we mean: Given a fixed point $x_{(0)}\in V$, does there exist a sufficiently small neighborhood of $x_{(0)}$, such that $g^1, \ldots, g^{2n},$ could serve as coordinate functions there?

Answer: In general the answer is No, but if we e.g. additionally assume that the Jacobian matrix

$$\left(\frac{\partial g^I}{\partial x^J}\right)_{1\leq I,J \leq 2n}$$

is invertible in the fixed point $x_{(0)}$, then the answer is Yes by the inverse function theorem.

Counterexample: Let $n=1$, i.e. the phase space $V$ is $2$-dimensional with coordinates $(x^1,x^2)=(q,p)$. Let the fixed point be $x_{(0)}=(0,0)$. Let $g^1(q,p)=q^2$ and $g^2(q,p)=p$. The map $x\mapsto$ $g(x)$ is not invertible in $x_{(0)}=(0,0)$, so that $g^1$ and $g^2$ cannot serve as coordinate functions. On the other hand, clearly only a constant function $f$ would have (identically) vanishing Poisson brackets with $g^1$ and $g^2$.

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 It's nice, but I meant why do we call it "a complete set of observables"? – Yrogirg Sep 10 '12 at 4:10 oh, what is left in your answer is to show the things backwards, from commuting to the fact that they are coordinates. – Yrogirg Sep 10 '12 at 4:15 or if you agree with my interpretation of what "a complete set of observables" should mean, then it can be moved to Math.SE – Yrogirg Sep 10 '12 at 4:37 I updated the answer. – Qmechanic♦ Sep 11 '12 at 14:47 Nice counterexample. :) – user404153 Sep 11 '12 at 17:50
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Blau's definition is a classical analog of the Schur's lemma. The reasoning behind this definition is the requirement that under a faithful quantization map which carries functions on the phase space to operators on some Hilbert space, the representation of the algebra of the quantum observables is irreducible. The irreducibility requirement has a physical origin as irreducible representations correspond to "single" quantum systems and if a representation is reducible, then it can be reduced to independent subsystems. Of course due to the Groenwold-Van Hove theorem, in general, no such quantization map exists. We usually give up faithfulness for the sake of irreducibility.

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