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?


6 Answers 6


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.

  • 8
    $\begingroup$ 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. $\endgroup$
    – thedoctar
    Commented Apr 8, 2019 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.

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    $\begingroup$ Indeed, eqn (3) describes the celebrated Moyal bracket (MB) and the $\hbar$ multiplying the PB is highly significant. That is, as $\hbar \rightarrow 0$ $\endgroup$ Commented Sep 27, 2015 at 19:09
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    $\begingroup$ .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. $\endgroup$ Commented Sep 27, 2015 at 19:18
  • 1
    $\begingroup$ .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. $\endgroup$ Commented Sep 27, 2015 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.

  • 2
    $\begingroup$ Could you be more precise to describe what you mean by "new quantum phenomenon with no classical analogue"? $\endgroup$
    – linello
    Commented Sep 25, 2019 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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Qmechanic
    Commented Jan 22, 2012 at 12:23
  • $\begingroup$ @Qmechanic: true, thx. $\endgroup$
    – Nikolaj-K
    Commented Jan 25, 2012 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. $$[\hat a,\hat b]=i\hbar\{a,b\}_\text{Poisson}$$


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

  • 1
    $\begingroup$ 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 $[\hat a,\hat b]~\longleftrightarrow~i\hbar\{a,b\}_{PB}+{\cal O}(\hbar^2)$. $\endgroup$
    – Qmechanic
    Commented Jan 21, 2012 at 11:51
  • $\begingroup$ @Qmechanic: With the edit, there won't be any higher order corrections. $\endgroup$
    – Siyuan Ren
    Commented Jan 21, 2012 at 13:55
  • $\begingroup$ For instance $\hat{a}=\hat{x}^3$ and $\hat{b}=\hat{p}^3$ would lead to ${\cal O}(\hbar^3)$ corrections. $\endgroup$
    – Qmechanic
    Commented Jan 21, 2012 at 23:03
  • $\begingroup$ @Qmechanic: Elaborate, please. $\endgroup$
    – Siyuan Ren
    Commented Jan 24, 2012 at 17:26
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    $\begingroup$ @C.R. just try computing/calculating those commutators yourself and you will see the $\hbar^3$ term arise. $\endgroup$
    – Justin L.
    Commented Jun 21, 2013 at 18:42

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:

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

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
(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.


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