# Lie Algebra of Classical Observables under Poisson Bracket

I am confused with understanding the fundaments of classical mechanics. All classical observables commute since they are represented by regular functions on phase space. All classical observables form a Lie Algebra under the Poisson Bracket as stated here: Why quantum mechanics?

How should I interpret this: If I have two observables f and g, they commute with each other, but individually, or as a sum or product, they obey the Poisson Bracket too? Or in other words they obey Hamilton's equation of motion?

I guess I am confused by looking at:

$\frac{d}{dt}f(p,q,t) =$ {$f,H$} + $\frac{\partial}{ \partial_t} f$ .

The Hamiltonian is an observable right? So that bracket should be zero right, then according to what is said above.. But that only holds for functions that are constants of motion..

I guess what I also not completely get is: What is the big deal with a Lie-group. Why is the Lie-bracket, or in this case the Poisson bracket, the fundamental property that is needed for things to work?

I am so confused..My apologies for my lack of knowledge on the subject.

• I'm not sure what you're asking. Could you please reformulate your question such that it is more precise. – Alpha001 Mar 14 '17 at 10:58
• Dear Alpha001, Thank you for your reply. What maybe is a concrete question: What does the following statement mean exactly: All classical observables form a Lie Algebra under the Poisson Bracket. The under the poisson bracket part... – Tyson Fasno Mar 14 '17 at 11:04
• Classical observables do not necessarily commute, either at the quantum or classical bracket level: $x$ and $p$ are a case and point. Indeed the quantum or classical bracket of $x$ and $p$ defines canonically conjugate variables. Are you confusing conserved quantities with observables? – ZeroTheHero Mar 14 '17 at 11:34
• What kind of commutator are you thinking of when you "all observables commute since they are represented by regular functions on phase space"? In particular, non-commuting observable mapped to functions as you state but one needs to use the $\star$ product to multiply them, and the star product is not commutative, i.e. $f\star g\ne g\star f$. Is this what you have in mind? – ZeroTheHero Mar 14 '17 at 12:43

A Lie algebra is a vector space $V$ together with a "bracket" $[-,-] : V\times V\to V$ that is linear, antisymmetric and obeys the Jacobi identity. Crucially, nothing in the definition of a Lie algebra says that the elements of $V$ are anything more than vectors, so in the abstract setting it does not make sense to talk about them "commuting" since a vector space carries no native notion of multiplication.
Now, the vast majority of examples of Lie algebras comes from algebras of operators, i.e. linear maps on some vector space. For such operators, you have a natural notion of multiplication (simply carrying the transformations they representation out in succession), and you can talk about the commutator $AB-BA$. If we now take the commutator as a bracket on the space of operators, that space becomes a Lie algebra. But, crucially, nothing forces us to take the commutator as the bracket, in particular, we can have the following:
In classical mechanics we have the algebra of classical observables as smooth $\mathbb{R}$-valued functions $C^\infty(M)$ on phase space $M$. Such functions can naturally be multiplied with each other and therefore also have a notion of "commutator", but it's just always zero since $\mathbb{R}$ is commutative unlike rings of operators. But, nevertheless, we may define a bracket on $C^\infty(M)$ by $$\{f,g\} = \frac{\partial f}{\partial x}\frac{\partial g}{\partial p} - \frac{\partial f}{\partial p}\frac{\partial g}{\partial x},$$ the Poisson bracket. This is non-zero even though the functions $f,g$ commute under multiplication, and therefore gives a non-trivial Lie algebra structure on $C^\infty(M)$. That they commute under naive multiplication has nothing to do with their Poisson bracket. Note that functions with $\{f,g\} = 0$ are sometimes said to "Poisson commute", which means they commute under the in general non-commutative "multiplication" $f\bullet g := \frac{\partial f}{\partial x}\frac{\partial g}{\partial p}$.
The significance of the Lie/Poisson bracket is that given two observables $f,g$ it tells you how one changes under the other infinitesimally. For example: the Hamiltonian is the generator of time translation, so $\partial_t f(t) = \{H,f\}$ (up to a sign). Angular momentum $L = x\times p$ is the generator of rotation, so if $\phi$ is the variable angle of rotation (say, around the z-axis), then $\partial_\phi (R_z(\phi) f) = \{L_z,f\}$, where $R_z(\phi)$ is the operator that acts by rotation on the function $f$, and so on.