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advanced_notions:poisson_bracket

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$ \left \{F,G \right \} = \sum_{n = 1}^{N} \left( \frac{\partial F}{\partial q_n} \frac{\partial G}{\partial p_n} - \frac{\partial F}{\partial p_n} \frac{\partial G}{\partial q_n} \right) $

Poisson Bracket

Intuitive

The Poisson bracket is a mathematical tool that allows us to calculate the time evolution in classical mechanics.

Concrete

For two sets of canonical coordinates $ q_1,q_2, … q_N $ and momenta $ p_1, p_2, … p_N $ the Poisson bracket is defined by

$$ \left \{F,G \right \} = \sum_{n = 1}^{N} \left( \frac{\partial F}{\partial q_n} \frac{\partial G}{\partial p_n} - \frac{\partial F}{\partial p_n} \frac{\partial G}{\partial q_n} \right) $$ for any two functions $F$ and $ G $ of the canonical coordinates and momenta.

The Poisson bracket of two observables can be thought of as the rate of change of the first along the flow given by the second. The most famous example is the time evolution. To get the time evolution of some observable $O$ all we have to do is calculate the Poisson bracket of $O$ with the Hamiltonian function.

Properties of the Poisson Bracket
It is linear for both $F$ and $G$. It is antisymmetric $$ \left[ G,F \right] = -\left[ F,G \right] $$ The Posisson bracket also satisfies the Jacobi identity $$ \left[\left[F,G \right],H \right] + \left[\left[G,H \right],F \right] + \left[\left[H,F \right],G \right] = 0. \tag{2.2} $$

The Poisson bracket satisfied

$$ \left\{ F G, H \right\} = F \left\{ G, H \right\} + G \left\{ F, H \right\} .$$

which looks like the Leibniz rule in calculus. This suggests that we interpret the Poisson bracket as derivative of the first argument with respect to the second argument.


The Poisson Bracket describe time evolution

In general, the time derivative of a function $F$ that is a function of generalized position and momentum coordinates, but not a direct function of time (which is often accurate) is

$$\frac{dF}{dt}=\frac{\partial F}{\partial q}\frac{dq}{dt}+\frac{\partial F}{\partial p}\frac{dp}{dt}$$

Hamilton's equations of motion are $$\frac{dq}{dt}=\frac{\partial H}{\partial p}$$ and $$ \frac{dp}{dt}=-\frac{\partial H}{\partial q}.$$

Putting these equations into our general time derivative from above yields

$$ \frac{dF}{dt}=\frac{\partial F}{\partial q}\frac{\partial H}{\partial p}-\frac{\partial F}{\partial p}\frac{\partial H}{\partial q}\equiv \{H,F\},$$

which shows that indeed Poisson brackets describe the time evolution of observables. If the function $F$ explicitly depends on the time we get instead $$ \frac{\mathrm{d}F}{\mathrm{d}t} = \{F,H\} + \frac{\partial F}{\partial t}. $$ Take note how similar this equation is to the Heisenberg equation which describes the time evolution in quantum mechanics $$ \frac{\mathrm{d}\hat F}{\mathrm{d}t} = -\frac{i}{\hbar}[\hat F,\hat H] + \frac{\partial \hat F}{\partial t}.$$ The only difference is that the Poisson brackets have been replaced with the commutator.

This is one way to make the difference between quantum and classical mechanics explicit:

$$ \text{Commutator}\quad [\hat{f},\hat{g}] \quad\longleftrightarrow\quad\text{Poisson bracket}\quad \{f,g\}$$


The Poisson bracket describes general transformations/symmetries

As mentioned above the Poisson bracket of two observables can be thought of as the rate of change of the first along the flow given by the second.

So when we use the Hamiltonian $H$ as the generator, we end up with time evolution. Therefore, as shown above for an observable $A$ without explicit time dependence we have

$\frac{\mathrm{d}A}{\mathrm{d}t} = \{A, H\}.$$

Therefore, $A$ is a constant of motion (= does not change as time passes on) only if $\{A, H\} =0$.

However, we can equally consider the reverse Poisson bracket \{H, A\}. Here $A$ appears in the second slot and therefore this Poisson bracket represents the rate of change of $H$ along the flow generated by $A$. So if $A$ generates a symmetry of a system we expect that $\{H, A\}$ is true. This comes about because if the Hamiltonian remains unchanged by the flow generated by $A$ then also the equations of motion: $\dot q = \frac{\partial H}{\partial p}, \quad \dot p = - \frac{\partial H}{\partial q}$ remain unchanged under the same flow.

  • If $\{H, A\} =0$ which means we have a symmetry of the Hamiltonian generated by $A$, then since the Poisson bracket is antisymmetric we automatically have
  • $\{A, H\} =0$, which states that $A$ is a conserved quantity.

For example, if the Hamiltonian is invariant under translations, we have $\{H, p\}=0$ since $p$ generates translations. In turn, we automatically also have $\{p, H\}=0$ which means that momentum is conserved.

In general, the infinitesimal evolution under the transformation generated by any element $g$ of the Lie algebra of observables on our phase space, (parametrized by an abstract "angle" $\varphi$) is given by

$$ \partial_\phi F = \{G,F\}.$$

This has immediately important consequences because it means that if the Poisson bracket of $F$ and $G$ vanishes, they describe an infinitesimal transformation that is a symmetry for the other. So if the Poisson bracket of an observable with the Hamiltonian vanishes it means that this observable is invariant under time translations, i.e. a constant in time. Similarly, if the Poisson bracket of an observable and the generator of rotations (= angular momentum $L^i = \epsilon^{ijk}x^jp^k$) vanishes it is invariant under rotations.

More precisely, consider a "generator" $\delta G$ and some quantity $A$. The infinitesimal transformation, generated by $\delta G$ is:

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

Momentum
$\delta G = \epsilon_i p_i \quad \delta A = -\epsilon_i\{p_i, A\} = \epsilon_i\frac{\partial A}{\partial q_i},$ $\quad \Rightarrow \quad A(p_i,q_i)\to A(p_i,q_i)+\epsilon_i\frac{\partial A}{\partial q_i} = A(p_i,q_i+\epsilon_i)$ Therefore momentum is the generator of translations.
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 the Levi-Civita symbol. Expanding yeilds $\delta A = -\epsilon_i e_{ijk} \sum_\alpha \left(\frac{\partial p_jq_k}{\partial q_\alpha}\frac{\partial A}{\partial p_\alpha}-\frac{\partial p_jq_k}{\partial p_\alpha}\frac{\partial A}{\partial q_\alpha}\right) = -\epsilon_i e_{ijk}\left(p_j\frac{\partial A}{\partial p_k}+q_j\frac{\partial A}{\partial q_k}\right)$

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

So we conclude that angular momentum is the generator of rotations.

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

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

Therefore energy is the generator of time evolution.

Abstract

The Poisson bracket is a tool that allows us to take the derivative of some function with respect to some other function. This becomes possible because the arena of classical mechanics is the phase space which is a symplectic manifold. A symplectic manifold is a manifold that is endowed with a symplectic form. This symplectic form can be used to map any function on the manifold to a vector field on the manifold.

Each vector field can be used to define (integral) curves. Starting with one point on the manifold, we can use the vector field to get a curve.

Now, the Poisson bracket tells us how we can differentiate a function along one such curve. We do this by comparing the value of the function at one point to its value at another point on the curve which lies infinitesimally close.

In this sense $\{F,G\}$ is a short-hand notation for using the symplectic form to map $G$ into a vector field and then taking the derivative of $F$ along the curves that are described by this vector field.


Any system in Classical Mechanics can be thought of rigorously as a phase space which is more precisely formalized as a symplectic manifold $(X,ω)$, or even more precisely a Poisson Manifold. In words, this means that the algebra of all functions on our phase space $X$, is canonically equipped with a Lie bracket: the Poisson bracket. Formulated differently, dynamics in mechanics are modeled on the cotangent bundle $T^∗M$ which has a canonical symplectic structure.

The Poisson bracket is like a ”Lie derivative” found in differential geometry. The Poisson bracket of two observables can be thought of as as the rate of change of the first along the flow given by the second. This is formally expressed as a Lie derivative:$$\{A, B\} = \mathcal{L}_{X_B} A .$$ In canonical coordinates this takes the form:$$\{A, B\} = \sum_{i=1}^N \frac{\partial A}{\partial q_i} \frac{\partial B}{\partial p_i} - \frac{\partial A}{\partial p_i} \frac{\partial B}{\partial q_i}$$

The Poisson bracket Lie algebra $\mathscr{P}(X,ω)$, which is the algebra of classical observables, is an extension of the Lie algebra of Hamiltonian vector fields $\mathscr{H}_{\mathcal{v}}(X)$ on our phase space $X$ by the "line Lie algebra $\mathbb{R}$" $$ \mathbb{R} \longrightarrow \mathscr{P}(X,ω) \longrightarrow \mathscr{H}_{\mathcal{v}}(X).$$

Why is it interesting?

Poisson brackets are necessary to describe the time evolution of observables in the Hamiltonian formulation of classical mechanics. Formulated differently, the Poisson bracket controls the dynamics in classical mechanics.

Poisson brackets play more or less the same role in classical mechanics that commutators do in quantum mechanics.

FAQ

What is the Lie group that is generated by Poisson brackets?
The Lie group which integrates the Poisson bracket is called the “quantomorphism group”. For a nice discussion, see e.g. https://www.quora.com/How-is-the-Poisson-bracket-different-from-a-commutator
advanced_notions/poisson_bracket.1523186290.txt.gz · Last modified: 2018/04/08 11:18 by jakobadmin