Complete set of observables in classical mechanics

Question

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.

Answer 1

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.

Answer 2

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 $g^2$.

Answer 3

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.

Answer 4

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.