08 May 2026

Poisson Brackets

Definition, algebraic properties, time evolution, and elementary examples of Poisson brackets.

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Hamiltonian mechanics takes place in phase space, where both coordinates and momenta are independent variables. The Poisson bracket is a compact operation on phase-space functions. It tells how one quantity changes under the flow generated by another quantity.

With the Hamiltonian as the generator, the Poisson bracket gives time evolution. With momentum or angular momentum as the generator, it gives translations or rotations.

Definition

For two functions $f(q_i,p_i,t)$ and $g(q_i,p_i,t)$,

\[\boxed{ \{f,g\}=\sum_i\left( \frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i} \right). }\]

The first term measures how $f$ changes with coordinate while $g$ changes with momentum. The second term subtracts the reverse contribution. This antisymmetric structure is what makes the bracket suitable for canonical mechanics.

Fundamental brackets

For canonical variables,

\[\boxed{ \{q_i,q_j\}=0, \qquad \{p_i,p_j\}=0, \qquad \{q_i,p_j\}=\delta_{ij}. }\]

These relations are the phase-space signature of canonical coordinates.

Properties

The Poisson bracket satisfies:

\[\{f,g\}=-\{g,f\},\] \[\{af+bg,h\}=a\{f,h\}+b\{g,h\},\] \[\{fg,h\}=f\{g,h\}+g\{f,h\},\]

and the Jacobi identity

\[\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}=0.\]

Time evolution

For any dynamical quantity $f(q,p,t)$,

\[\boxed{ \frac{df}{dt}=\{f,H\}+\frac{\partial f}{\partial t}. }\]

If $f$ has no explicit time dependence, then

\[\frac{df}{dt}=\{f,H\}.\]

Thus $f$ is conserved when

\[\boxed{\{f,H\}=0.}\]

Hamilton equations from Poisson brackets

Using the fundamental brackets,

\[\dot q_i=\{q_i,H\}=\frac{\partial H}{\partial p_i},\] \[\dot p_i=\{p_i,H\}=-\frac{\partial H}{\partial q_i}.\]

So Hamilton’s equations are Poisson-bracket equations.

Example: harmonic oscillator

For

\[H=\frac{p^2}{2m}+\frac12kx^2,\]

we get

\[\dot x=\{x,H\}=\frac{p}{m},\] \[\dot p=\{p,H\}=-kx.\]

Therefore

\[\ddot x+\frac{k}{m}x=0.\]

Brackets for one-dimensional motion

For one canonical pair $(x,p)$,

\[\left\{x,\frac{p^2}{2m}\right\} =\frac{\partial x}{\partial x} \frac{\partial}{\partial p}\left(\frac{p^2}{2m}\right) -\frac{\partial x}{\partial p} \frac{\partial}{\partial x}\left(\frac{p^2}{2m}\right).\]

Hence

\[\left\{x,\frac{p^2}{2m}\right\}=\frac{p}{m}.\]

Also,

\[\{p,V(x)\} =0-\frac{\partial p}{\partial p}\frac{dV}{dx} =-\frac{dV}{dx}.\]

Thus for

\[H=\frac{p^2}{2m}+V(x),\]

the Poisson-bracket equations give

\[\dot x=\frac{p}{m}, \qquad \dot p=-\frac{dV}{dx}.\]

Main points

Practice questions

  1. Define the Poisson bracket.
  2. Prove ${q_i,p_j}=\delta_{ij}$.
  3. Derive Hamilton’s equations using Poisson brackets.
  4. Find $\dot x$ and $\dot p$ for the harmonic oscillator.
  5. State the Jacobi identity.

Next: Poisson Theorems and Angular Momentum

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