Well, the Possion bracket:

$ \{ A(q,p),B(q,p) \} \equiv \sum_{s} \left( \dfrac{\partial A}{\partial q_{s}} \dfrac{\partial B}{\partial p_{s}} - \dfrac{\partial A}{\partial p_{s}} \dfrac{\partial B}{\partial q_{s}} \right) $

is a coordinate transformation according to the words of this Wikipedia page: (link: http://en.wikipedia.org/wiki/Poisson_bracket)

...".It places mechanics and dynamics in the context of coordinate-transformations: specifically in coordinate planes such as canonical position/momentum, or canonical-position/canonical transformation."...

The Jacobian of, for example, $ A(q,p) $ and $ B(q,p) $ has the same form as the terms in the parenthesis. From what to what coordinates does the bracket transform? Does it make any sense to make comparison (Poisson bracket and the Jacobian)?


4 Answers 4


Poisson brackets are closely related also to transformations of a system.

Consider a "generator" $\delta G$ and some quantity $A$. (All the quantities mentioned are functions of $p_i$, $q_i$, and do not depend on time explicitly. So I'll not write the arguments unless it's necessary). The small transformation, generated by $\delta G$ is:

$A \to A+\delta A,\quad\quad \delta A = -\{\delta G, A\} $

One have to be careful with the signs in the definition. I use the one from Wikipedia.

Example 1 -- momentum
$\delta G = \epsilon_i p_i \quad \delta A = -\epsilon_i\{p_i, A\} = \epsilon_i\frac{\partial A}{\partial q_i},$
$\quad \Rightarrow \quad A(p_i,q_i)\to A(p_i,q_i)+\epsilon_i\frac{\partial A}{\partial q_i} = A(p_i,q_i+\epsilon_i)$
The momentum is the generator of translations.

Example 2 -- angular momentum
$\delta G = \epsilon_i e_{ijk} p_jq_k \quad \delta A = -\epsilon_i e_{ijk} \{p_jq_k, A\}$

Here $e_{ijk}$ -- is a Levi-Civita symbol. Expanding:
$\delta A = -\epsilon_i e_{ijk} \sum_\alpha \left(\frac{\partial p_jq_k}{\partial q_\alpha}\frac{\partial A}{\partial p_\alpha}-\frac{\partial p_jq_k}{\partial p_\alpha}\frac{\partial A}{\partial q_\alpha}\right) = -\epsilon_i e_{ijk}\left(p_j\frac{\partial A}{\partial p_k}+q_j\frac{\partial A}{\partial q_k}\right)$

$\quad \Rightarrow \quad A(p_i,q_i)\to A(p_i,q_i)-\epsilon_i e_{ijk}\left(p_j\frac{\partial A}{\partial p_k} + q_j\frac{\partial A}{\partial q_k}\right) = A(R_{ij}p_j,R_{ij}q_j)$
For infinitesimal rotations $R_{ij}$

So the angular momentum is the generator of rotations.

Example 3 -- energy
By using Hamilton equation one can calculate to which transformations the energy corresponds to (it is essentially a "reverse" of what is written here):

$\delta G = \epsilon H \quad \delta A = -\epsilon\{H, A\} = \epsilon\frac{d A}{d t},$

$\quad \Rightarrow \quad A(p_i(t),q_i(t))\to A(p_i(t),q_i(t))+\epsilon\frac{d A}{d t} = A(p_i(t+\epsilon),q_i(t+\epsilon))$

The energy is the generator of time evolution.

From here one can directly see the relation between symmetries and conservation laws.

If a Hamiltonian is symmetric under certain transformation (and, again, doesn't depend on time explicitly), then the bracket with the corresponding "generator" must vanish $\{\delta G, H\} = 0$. But this also means that the value of the "generator" doesn't change with time.

  • $\begingroup$ how did the minus sign between the two terms in $\delta A$ change to plus sign? or do you want to exchange the indices? $\endgroup$
    – user7757
    Jul 26, 2012 at 6:59
  • $\begingroup$ The last statement about symmetries is only true when the generator does not depend on time explicitly. For instance, in Lorentz invariant theories the boost generator does not Poisson-commute with the Hamiltonian. And this makes sense because the boost change the time $t$ and therefore the Hamiltonian which is the generator of time translations. $\endgroup$ Jul 27, 2012 at 1:14
  • $\begingroup$ @ramanujan_dirac, you are right. I've made a mistake with a general sign and also forgot to change the sign when swapped indices. Tried to fix that. $\endgroup$
    – Kostya
    Jul 27, 2012 at 11:25
  • $\begingroup$ @Kostya. Somebody proposed an edit to your answer. Could you please review it? $\endgroup$
    – Qmechanic
    Jul 29, 2012 at 19:18
  • $\begingroup$ @Qmechanic The edit is just a comment by drake included in the main answer. It is a nice comment but it doesn't fit in the answer. I'm clearly saying at the very beginning that nothing depends on time in my case explicitly. And I don't want to clutter my answer with this details. $\endgroup$
    – Kostya
    Jul 29, 2012 at 19:29

There is no specific coordinate transformation being discussed when defining the Poisson bracket of two general observables, $A$ and $B$. Obviously, the phase space has $2N$ coordinates, $q_i$ and $p_j$, so two coordinates $A,B$ wouldn't be enough.

Instead, the Poisson brackets are relevant for the so-called


canonical transformations which are transformations that preserve the Poisson bracket, e.g. among the positions and the momenta: $$ x_a\to X_a(x_a): \quad \forall a,b, \quad \{X_a,X_b\} = \{x_a,x_b\} $$ where $X_a$ is either $Q_i$ or $P_j$, and $x_a$ is either $q_i$ or $p_j$. In quantum mechanics, canonical transformations become unitary transformations - because those preserve the commutators between the observables.

It is a kind of coincidence that a single term in the Poisson bracket has the form of a Jacobian. Jacobians occur at many places in this discussion but for many degrees of freedom, the Jacobian would be a determinant of a large matrix. The Poisson bracket is a sum (which is different) of many determinants of $2\times 2$ matrices (only).

Of course, it is not a "complete" coincidence that the $2\times 2$ determinant appears here. Mathematically, we say that the canonical transformations preserve the symplectic structure - given by the antisymmetric prescription for all the "fundamental" Poisson brackets. For 2 variables, a single $q$ and a single $p$, the symplectic structure is given by $\epsilon_{ab}$, an antisymmetric invariant, which may also be thought of as a "volume invariant". In other words, $Sp(2,R)\equiv SL(2,R)$: the symplectic transformations are the same ones as the unimodular ones. This isomorphism doesn't generalize to many coordinates.


What the Poisson brackets determine is the symplectic structure(*). The symplectic structure in turn determines the time-evolution of the system's trajectory in phase space. Two sets of phase-space co-ordinates describe the same physics as long as they have the same Poisson bracket. These two sets are then said to be related by a canonical transformation - a transformation that preserves the symplectic structure on the phase space.

The R.H.S. of the Poisson bracket is nothing more than the Jacobian of the transformation as you point out.

The Poisson bracket structure is a powerful guide in transitioning from classical to quantum mechanics. In fact, the simplest prescriptions for "quantizing" involves replacing the phase-space co-ordinates of the classical description (say $A(p,q)$ and $B(p,q)$) with operators ($\hat A(p,q)$, $\hat B(p,q)$), on a Hilbert space, whose commutator is given by:

$$ \{ A(p,q), B(p,q) \} = C(p,q) \rightarrow [\hat A(p,q), \hat B(p,q)] = i \hbar \hat C(p,q) $$

Given two different sets of classical phase-space co-ordinates, which have the same Poisson bracket and are thus related by a canonical transformation, the corresponding sets of operators in the quantum theory are related by a unitary transformation. In other words the symplectic structure determines whether two different representations of the canonical commutation relations (CCRs) yield the same quantization.

This is a non-trivial problem especially in the context of quantum gravity. For more background on these topics look up the Stone-von Neumann theorem.

(*) Given a function $f$ on a phase space we can obtain a vector field $X(f)$:

$$ X(f)_a = \Omega_a^b \nabla_b f $$

where $\Omega_{ab}$ is the symplectic anti-symmetric two form. Lie dragging (roughly speaking "parallel transporting") $\Omega$ along $X(f)$ leaves it ($\Omega$) invariant. $f$ is the "Hamiltonian" and $X(f)_a$ is the "Hamiltonian" vector field which generates time-evolution of a point in phase space.

  • 1
    $\begingroup$ Good, except your second sentence "The symplectic structure in turn determines the time-evolution of the system's trajectory in phase space". The symplectic structure determines what kind of system this is, and you still have to choose the Hamiltonian to discuss time evolution.As Witten likes to point out, the problem of quantization has to do with the former structure - what system this is - and does not necessitate the choice of Hamiltonian. $\endgroup$
    – user566
    Feb 24, 2011 at 14:59
  • 1
    $\begingroup$ "What the Poisson brackets determine is the symplectic structure" -> hm, not really. Poisson manifold can have degenerate symplectic two-form (in contrast with symplectic manifolds) and so it's a bit imprecise to say that it determines the structure. $\endgroup$
    – Marek
    Feb 27, 2011 at 23:41

I do not give you an answer to your questions but an addition to the previous answers, so do not judge me too severely.

One may perform any coordinate change at one's convenience. Preserving this or that form of equations is not obligatory. The variable change should just be reversible. Often the equations in new coordinates may be made simpler to solve. If so, do them without hesitation - nobody can oblige you to perform only "canonical" transformations ;-).

Description of the same physics is possible in terms of quite different coordinates, not only in terms of "canonical" ones. I do not see any physical sense in "canonical" transformations, to tell the truth. Take, for example, the pairs $p, q$ and $a, a^+$. Their CCRs are different. So what? Who cares?

  • 2
    $\begingroup$ "So what? Who cares?" -> those who understand that preservation of structure grants you extra tools that you wouldn't be able to use otherwise ;) $\endgroup$
    – Marek
    Feb 27, 2011 at 23:44
  • $\begingroup$ Marek, you are a clever guy. Now tell me, if the original structure is so "extra excellent", then why do you need to do the coordinate transformations? Solve the original equations in the original variables and be happy with their structure. $\endgroup$ Feb 28, 2011 at 0:13

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