# What is the connection between Poisson brackets and commutators?

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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?

This post imported from StackExchange Physics at 2017-03-13 12:21 (UTC), posted by SE-user 0x90

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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 post imported from StackExchange Physics at 2017-03-13 12:21 (UTC), posted by SE-user David Z
answered Jan 21, 2012 by (660 points)
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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.

This post imported from StackExchange Physics at 2017-03-13 12:21 (UTC), posted by SE-user joseph f. johnson
answered Jan 21, 2012 by (500 points)

The anticommutator also has a classical analogue, the Poisson superbracket!

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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.

This post imported from StackExchange Physics at 2017-03-13 12:21 (UTC), posted by SE-user NikolajK
answered Jan 21, 2012 by (200 points)
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.

This post imported from StackExchange Physics at 2017-03-13 12:21 (UTC), posted by SE-user Qmechanic
@Qmechanic: true, thx.

This post imported from StackExchange Physics at 2017-03-13 12:21 (UTC), posted by SE-user NikolajK
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According to the topic of deformation quantization, the first few entries in the dictionary between

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

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

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

$$\tag{3} \text{Commutator}\quad [\hat{f},\hat{g}] \quad\longleftrightarrow\quad\text{Poisson bracket}\quad i\hbar\{f,g\}_{PB} + {\cal O}(\hbar^2).$$

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 ${\cal O}(\hbar^2)$ in the identification (3).

Example 1: (Fundamental CCR) $$\tag{4} [\hat{q},\hat{p}]~=~i\hbar{\bf 1}\quad\longleftrightarrow\quad \{q,p\}_{PB}~=~1.$$

Example 2:
$$\tag{5} [\hat{q}^2,\hat{p}^2]~=~4[\hat{q},\hat{p}] (\hat{q}\hat{p})_W\quad\longleftrightarrow\quad \{q^2,p^2\}_{PB}~=~4\{q,p\}_{PB} qp,$$ where $(\ldots)_W$ stands for Weyl-symmetrization of operators. See also e.g. this Phys.SE post.

Example 3:
$$\tag{6} [\hat{q}^3,\hat{p}^3]~=~9[\hat{q},\hat{p}] (\hat{q}^2\hat{p}^2)_W+\frac{3}{2}[\hat{q},\hat{p}]^3\quad\longleftrightarrow\quad \{q^3,p^3\}_{PB}~=~9\{q,p\}_{PB} q^2p^2.$$ Note that there are higher-order quantum corrections ${\cal O}(\hbar^3)$ in eq. (6) even after Weyl-symmetrization.

This post imported from StackExchange Physics at 2017-03-13 12:21 (UTC), posted by SE-user Qmechanic
answered Oct 7, 2014 by (3,110 points)
Indeed, eqn (3) describes the celebrated Moyal bracket (MB) and the $\hbar$ multiplying the PB is highly significant. That is, as $\hbar \rightarrow 0$

This post imported from StackExchange Physics at 2017-03-13 12:21 (UTC), posted by SE-user Cosmas Zachos
.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.

This post imported from StackExchange Physics at 2017-03-13 12:21 (UTC), posted by SE-user Cosmas Zachos
.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.

This post imported from StackExchange Physics at 2017-03-13 12:21 (UTC), posted by SE-user Cosmas Zachos
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I don't know any link between Poisson bracket and anti-commutator, but I do know the link between Poisson bracket and commutator. $$[\hat a,\hat b]=i\hbar\{a,b\}_\text{Poison}$$

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

This post imported from StackExchange Physics at 2017-03-13 12:21 (UTC), posted by SE-user Siyuan Ren
answered Jan 21, 2012 by (20 points)
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 $[\hat a,\hat b]~\longleftrightarrow~i\hbar\{a,b\}_{PB}+{\cal O}(\hbar^2)$.

This post imported from StackExchange Physics at 2017-03-13 12:21 (UTC), posted by SE-user Qmechanic
@Qmechanic: With the edit, there won't be any higher order corrections.

This post imported from StackExchange Physics at 2017-03-13 12:21 (UTC), posted by SE-user Siyuan Ren
For instance $\hat{a}=\hat{x}^3$ and $\hat{b}=\hat{p}^3$ would lead to ${\cal O}(\hbar^3)$ corrections.

This post imported from StackExchange Physics at 2017-03-13 12:21 (UTC), posted by SE-user Qmechanic
@C.R. just try computing/calculating those commutators yourself and you will see the $\hbar^3$ term arise.
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