# 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?

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.

• This answer is not very sophisticated, in my opinion. The Poisson bracket and commutator both satisfy the same algebraic relations and generate time evolution, but the Poisson bracket in classical Hamiltonian mechanics has a definite formula (just like the commutator). How does one reproduce this starting from the axioms of QM? Also see Groenewold's theorem. – thedoctar Apr 8 '19 at 13:57

According to the topic of deformation quantization, the first few entries in the dictionary between

$$\text{Quantum Mechanics}\quad\longleftrightarrow\quad\text{Classical Mechanics}\tag{0}$$

$$\text{Operator}\quad\hat{f}\quad\longleftrightarrow\quad\text{Function/Symbol}\quad f,\tag{1}$$

$$\text{Composition}\quad\hat{f}\circ\hat{g} \quad\longleftrightarrow\quad\text{Star product}\quad f\star g ,\tag{2}$$ 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.

• Indeed, eqn (3) describes the celebrated Moyal bracket (MB) and the $\hbar$ multiplying the PB is highly significant. That is, as $\hbar \rightarrow 0$ – Cosmas Zachos Sep 27 '15 at 19:09
• .cont': not only do the higher order corrections vanish, but also the scale of the commutator shrinks, and the uncertainty principle evaporates to none for the PB; it would be there for PBs, if only that rescaling were not a crucial part of the correspondence! But it is: cf. chapter on uncertainty principle in T Curtright, D Fairlie, & C Zachos, A Concise Treatise on Quantum Mechanics in Phase Space, World Scientific, 2014. – Cosmas Zachos Sep 27 '15 at 19:18
• .cont'. Since the noncommutative product (2) is the average of a commutator and an anticommutator, its classical limit is led by half the anticommutator's limit. So the Classical analog of an anticommutator is a bland "twice the classical product", and there is no need to muddy the waters with discussions of anticommutators: mere commutators capture the essence of non-commutativity in quantum deformation. – Cosmas Zachos Sep 27 '15 at 20:45

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.

• Could you be more precise to describe what you mean by "new quantum phenomenon with no classical analogue"? – linello Sep 25 '19 at 19:27

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.

• Comment to the answer(v1): It seems that $q$, $p$ in the last equation are supposed to denote general elements of the function algebra. Notationally, this is a bit unfortunate, as $q$, $p$ usually denote the basic canonical variables. – Qmechanic Jan 22 '12 at 12:23
• @Qmechanic: true, thx. – Nikolaj-K Jan 25 '12 at 9:28

I don't know any link between Poisson bracket and anti-commutator, but I do know the link between Poisson bracket and commutator. $$[\hat a,\hat b]=i\hbar\{a,b\}_\text{Poisson}$$

## Subtleties

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 $\frac{1}{2}\left(\hat{p}\hat{x}+\hat{x}\hat{p}\right)$.

• Comment to the answer(v1): There could in general be higher-order corrections ${\cal O}(\hbar^2)$ in Planck constant on the rhs. of the identification $[\hat a,\hat b]~\longleftrightarrow~i\hbar\{a,b\}_{PB}+{\cal O}(\hbar^2)$. – Qmechanic Jan 21 '12 at 11:51
• @Qmechanic: With the edit, there won't be any higher order corrections. – Siyuan Ren Jan 21 '12 at 13:55
• For instance $\hat{a}=\hat{x}^3$ and $\hat{b}=\hat{p}^3$ would lead to ${\cal O}(\hbar^3)$ corrections. – Qmechanic Jan 21 '12 at 23:03
• @Qmechanic: Elaborate, please. – Siyuan Ren Jan 24 '12 at 17:26
• @C.R. just try computing/calculating those commutators yourself and you will see the $\hbar^3$ term arise. – Justin L. Jun 21 '13 at 18:42