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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?
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.
According to the topic of deformation quantization, the first few entries in the dictionary between
$$ \text{Quantum Mechanics}\quad\longleftrightarrow\quad\text{Classical Mechanics}\tag{0}$$
read
$$ \text{Operator}\quad\hat{f}\quad\longleftrightarrow\quad\text{Function/Symbol}\quad f,\tag{1}$$
$$ \text{Composition}\quad\hat{f}\circ\hat{g} \quad\longleftrightarrow\quad\text{Star product}\quad f\star g ,\tag{2}$$ and
$$ \text{Commutator}\quad [\hat{f},\hat{g}] \quad\longleftrightarrow\quad\text{Poisson bracket}\quad i\hbar\{f,g\}_{PB} + \color{red}{{\cal O}(\hbar^2)}. \tag{3}$$
Note that the correspondence (0) depends on which symbols one uses, e.g. Weyl symbols, and that there could in general be higher-order quantum corrections $\color{red}{{\cal O}(\hbar^2)}$ in the identification (3).
Example 1: (Fundamental CCR) $$ [\hat{q},\hat{p}]~=~i\hbar{\bf 1}\quad\longleftrightarrow\quad \{q,p\}_{PB}~=~1.\tag{4} $$
Example 2:
$$ [\hat{q}^2,\hat{p}^2]~=~4[\hat{q},\hat{p}]~ (\hat{q}\hat{p})_W\quad\longleftrightarrow\quad
\{q^2,p^2\}_{PB}~=~4\{q,p\}_{PB} ~qp, \tag{5}$$
where $(\ldots)_W$ stands for Weyl-symmetrization of operators. See also e.g. this Phys.SE post.
Example 3:
$$ [\hat{q}^3,\hat{p}^3]~=~9[\hat{q},\hat{p}]~ (\hat{q}^2\hat{p}^2)_W + \color{red}{\frac{3}{2}[\hat{q},\hat{p}]^3}\quad\longleftrightarrow\quad
\{q^3,p^3\}_{PB}~=~9\{q,p\}_{PB}~ q^2p^2.\tag{6} $$
Note that there are higher-order quantum corrections $\color{red}{{\cal O}(\hbar^3)}$ in eq. (6) even after Weyl-symmetrization.
Both the commutator (of matrices) and the Poisson bracket satisfy the Jacobi identity, $[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0$.
This is why Dirac was inspired by Heisenberg's use of commutators to develop a Hamilton-Jacobi dynamics style of Quantum Mechanics which provided the first real unification of Heisenberg's matrix mechanics with Schroedinger's wave mechanics. The Jacobi identity is also the basic law of Lie algebras, which are useful for symmetry groups in quantum theory.
In classical mechanics, the dynamical variables are the functions $f$ on phase space, and they get a non-trivial algebraic structure from the Poisson bracket. They are the classical « observables ». In quantum mechanics, the observables are matrices, these are the dynamical variables, but they receive a similar algebraic structure from the commutator.
As already pointed out, the anti-commutator is not analogous to the Poisson bracket, it is a distinctly new quantum phenomenon with no classical analogue.
Regarding the significance of the observables momentum and position there are many similarities between Classical and Quantum mechanics. Some of the algebraic relations have been pointed out.
In the end, there is still an important difference, which is obvious by the fact that the function algebra generated by classical quantities is commutative $$q·p=p·q,$$ and the other is not $$Q\ P\ne P\ Q=Q\ P-[Q,P\ ].$$ One might ask if there is a structure for the classical function algebra of $q$ and $p$ with a product, which resembles the quantum mechanical algebra $Q$ and $P$. I.e. is there a product, let's denoted it by $\star\ $, for which $$q\star p-p\star q=[q\ \overset{\star}{,}\ p]\ \ \Longleftrightarrow\ \ [Q,P\ ]=Q\ P-P\ Q.$$
More on questions in this spirit can be found under Weyl quantization.
The most investigated star product is the Moyal product, which per definition fulfills $$[f\ \overset{\star}{,}\ g]=i\hbar\ \{f,g\}+\mathcal O(\hbar^2).$$
Fields medals are won for this kind of stuff.
I don't know any link between Poisson bracket and anti-commutator, but I do know the link between Poisson bracket and commutator. $$[\hat a,\hat b]=i\hbar\{a,b\}_\text{Poisson}$$
As the operator $\hat a$ and $\hat b$ are counterparts to classical dynamical variable, they must be ①functions of canonical coordinates and momenta (ruling out spin, which cannot be put in a Poisson bracket) ②Hermitian operators (try $[\hat{x}\hat{p},\hat{p}\hat{x}]$).
In addition, the equality sign isn't really an equality, because r.h.s. are commutative numbers while l.h.s are non-commutative operators, so you must be careful relating two sides. For example, the quantum analogy of $xp$ is neither $\hat{x}\hat{p}$ or $\hat{p}\hat{x}$, but $\frac{1}{2}\left(\hat{p}\hat{x}+\hat{x}\hat{p}\right)$.
Algebraically, the relation $[a,b] = iħ\{a,b\}$ isn't just a "correspondence" or something that holds "only in the limit", but is true as is. More precisely, when taken with the definition $a·b ≡ ½(ab + ba)$, the result is a decomposition $$ab = a·b + \frac{iħ}{2} \{a,b\},$$ and an algebra satisfying the following properties: $$ a·b = b·a, \hspace 1em \{a,b\} = -\{b,a\},\\ a·(b·c) - (a·b)·c = \left(\frac{ħ}{2}\right)^2 \{b,\{a,c\}\},\\ \{a,b·c\} = \{a,b\}·c + b·\{a,c\},\\ \{a,\{b,c\}\} = \{\{a,b\},c\} + \{b,\{a,c\}\}. $$ Apart from the defect in the associativity property by the small factor $-(½ħ)^2$, this is just the same kind of algebra as for a classical Poisson manifold; i.e. a Poisson algebra. So, the operator algebra is a deformation of a Poisson algebra. You could just as well write everything algebraically in this way - without the use of complex numbers at all. The rendering of it as an associative, but now non-commutative, complex algebra is just a way of packing together the structure of a real commutative, but non-associative product with a Lie bracket. The only place where the quantum nature is set apart from the classical nature is in the identity for the associativity defect.
This is connected to the Weyl "operator ordering" ($W[⋯]$ = the average taken over all permutation of the basic operators in "⋯") as follows: $$ W[ab] ≡ \frac{ab + ba}{2} = \frac{a·b + b·a}{2},\\ W[abc] ≡ \frac{abc + acb + bac + bca + cab + cba}{6} = \frac{a·(b·c) + b·(c·a) + c·(a·b)}{3},$$ and the cubic combinations of basic operators $a$, $b$, $c$, may be expressed as $$abc = \frac{(a·b)·c + a·(b·c)}{2} + iħ \frac{a·\{b,c\} - b·\{c,a\} + c·\{a,b\}}{2} - ħ^2 \frac{\{\{a,b\},c\} + \{a,\{b,c\}\}}{8}.$$
A limited form of associativity, for powers of the same factor, holds: $$\left(a·b^m\right)·b^n - \left(a·b^n\right)·b^m = \frac{ħ^2}{4} \{\{b^m,b^n\},a\} = \frac{h^2}{4} \{0,a\} = 0 \hspace 1em (m, n = 0, 1, 2, ⋯).$$
It also bears to note that the Lie bracket also satisfies the product rule for the complex product, itself: $$\begin{align} \{a,bc\} &= \{a,b·c\} + \frac{ih}{2} \{a,\{b,c\}\}\\ &= \{a,b\}·c + b·\{a,c\} + \frac{ih}{2} (\{\{a,b\},c\} + \{b,\{a,c\}\})\\ &= \left(\{a,b\}·c + \frac{ih}{2} \{\{a,b\},c\}\right) + \left(b·\{a,c\} + \frac{ih}{2} \{b,\{a,c\}\}\right)\\ &= \{a,b\}c + b\{a,c\}. \end{align}$$
Jordan-Lie and Jordan-Banach Algebras:
https://planetmath.org/JordanBanachAndJordanLieAlgebras
Another sense in which the correspondence to the Poisson bracket holds true is that the underlying Hilbert space is, itself, already a Poisson manifold, with the imaginary part of the inner product connected to the Poisson bracket. But I'm still not entirely clear, yet, on what the exact relation is between its Poisson bracket and the above-mentioned Lie bracket.
I'll have to work through this, here.
More precisely, quantum states are not actually non-zero Hilbert space vectors. Instead, a vector $v ∈ ℌ - \{0\}$ in a Hilbert space $ℌ$ gives rise to a pure state that may be identified uniquely as $W_v = |v〉〈v|/|v|^2$, where $|v| = \sqrt{〈v,v〉}$. This is a projective geometry since $W_v = W_{-v}$ and, in fact, $W_v = W_{λv}$ for any real or complex $λ ≠ 0$. Mixed states are convex unit norm sums or integrals of pure states.
The inner product of two states is $$〈W_v, W_{v'}〉 = \frac{〈v,v'〉}{|v||v'|},$$ which satisfies the identity $$|〈W_v, W_{v'}〉|^2 = Tr(W_v W_{v'}),$$ and can, thus, be considered as a complex square root of the states' overlap.
This extends, bi-linearly to an inner product $〈W, W'〉$ the full space of pure and mixed states, $W$, $W'$.
The Poisson bracket arises, via a Poisson tensor $Ω(W, W')$, as the complex part of the inner product $$〈W, W'〉 = G(W, W') + i Ω(W, W') = 2ħ〈W, W'〉,$$ the real part giving you a metric $G(W, W')$ for a Kähler manifold, with the constraint $G(W, W') = 2ħ$ giving rise to the re-phase symmetry of the Hilbert space $ℌ$, itself, as a gauge symmetry.
The expectation value of an operator $a$ in state $W_v$ is $\widetilde{a}(W_v) = Tr(W_v a) = 〈v|a|v〉/〈v,v〉$, so operators are actually functions over this space, and their Poisson bracket is directly connected to the Lie Bracket as $$\{\widetilde{a}, \widetilde{b}\} = \widetilde{\{a,b\}}.$$
That's close to how N. P. Landsman treats it in his 1998
Mathematical Topics Between Classical and Quantum Mechanics
https://link.springer.com/book/10.1007/978-1-4612-1680-3
in his section 2.5, except his treatment was more cumbersome, because he failed to make the direct identification of $W_v$ as the state for the Hilbert space vector $v$. Ashtekar and Shilding also discussed the underlying geometry in:
Geometric Formulation Of Quantum Mechanics
https://arxiv.org/pdf/gr-qc/9706069v1.pdf
(1997 June 21)
So, the Lie bracket $\{a,b\}$ actually is a Poisson bracket, when treated as a state space function; not merely something that has a Poisson bracket as a classical limit or corresponds to a Poisson bracket. Instead, what you find, here, is that the classical Poisson bracket, is the classical limit of the quantum Poisson bracket: $$\lim_{ħ → 0}\widetilde{\{a,b\}} = \lim_{ħ → 0}\{\widetilde{a},\widetilde{b}\} = \left\{\lim_{ħ → 0}\widetilde{a},\lim_{ħ → 0}\widetilde{b}\right\},$$ and the entire infrastructure of the algebras, manifolds and brackets are passing over into the limit (or in the reverse direction: arising as deformations of their respective classical limits), not just individual objects.
