quantum mechanics - What is the connection between Poisson brackets and commutators?

The Poisson bracket is defined as:

$$\{f,g\} ~:=~ \sum_{i=1}^{N} \left[ \frac{\partial f}{\partial q_{i}} \frac{\partial g}{\partial p_{i}} - \frac{\partial f}{\partial p_{i}} \frac{\partial g}{\partial q_{i}} \right]. $$

The anticommutator is defined as:

$$ \{a,b\} ~:=~ ab + ba. $$

The commutator is defined as:

$$ [a,b] ~:=~ ab - ba. $$

What are the connections between all of them?

Edit: Does the Poisson bracket define some uncertainty principle as well?

Answer

Poisson brackets play more or less the same role in classical mechanics that commutators do in quantum mechanics. For example, Hamilton's equation in classical mechanics is analogous to the Heisenberg equation in quantum mechanics:

$$\begin{align}\frac{\mathrm{d}f}{\mathrm{d}t} &= \{f,H\} + \frac{\partial f}{\partial t} & \frac{\mathrm{d}\hat f}{\mathrm{d}t} &= -\frac{i}{\hbar}[\hat f,\hat H] + \frac{\partial \hat f}{\partial t}\end{align}$$

where $H$ is the Hamiltonian and $f$ is either a function of the state variables $q$ and $p$ (in the classical equation), or an operator acting on the quantum state $|\psi\rangle$ (in the quantum equation). The hat indicates that it's an operator.

Also, when you're converting a classical theory to its quantum version, the way to do it is to reinterpret all the variables as operators, and then impose a commutation relation on the fundamental operators: $[\hat q,\hat p] = C$ where $C$ is some constant. To determine the value of that constant, you can use the Poisson bracket of the corresponding quantities in the classical theory as motivation, according to the formula $[\hat q,\hat p] = i\hbar \{q,p\}$. For example, in basic quantum mechanics, the commutator of position and momentum is $[\hat x,\hat p] = i\hbar$, because in classical mechanics, $\{x,p\} = 1$.

Anticommutators are not directly related to Poisson brackets, but they are a logical extension of commutators. After all, if you can fix the value of $\hat{A}\hat{B} - \hat{B}\hat{A}$ and get a sensible theory out of that, it's natural to wonder what sort of theory you'd get if you fixed the value of $\hat{A}\hat{B} + \hat{B}\hat{A}$ instead. This plays a major role in quantum field theory, where fixing the commutator gives you a theory of bosons and fixing the anticommutator gives you a theory of fermions.