Poisson Brackets and Constants of the Motion
For a quantity with no explicit time dependence, the Poisson bracket with the Hamiltonian gives its rate of change, so such a conserved quantity is one whose bracket vanishes. For the canonical examples, the bracket algebra reappears as the commutator in quantum mechanics.
What a learner can do afterwards
- Computes a Poisson bracket and uses it to test whether a quantity is conserved
- Recovers the canonical equations as brackets with the Hamiltonian
- States the correspondence between a Poisson bracket and a quantum commutator
1 · Read
The Poisson bracket of two phase-space functions A(q, p, t) and B(q, p, t) in one canonical pair (q, p) is the signed combination {A, B} = (∂A/∂q)(∂B/∂p) − (∂A/∂p)(∂B/∂q). For several canonical pairs the bracket sums this expression over every pair: {A, B} = Σᵢ [(∂A/∂qᵢ)(∂B/∂pᵢ) − (∂A/∂pᵢ)(∂B/∂qᵢ)]. The total time derivative of A along a motion is dA/dt = {A, H} + ∂A/∂t. If A has no explicit time dependence, the partial term drops and the bracket alone gives the rate. Then {A, H} = 0 along the motion means A is a constant of the motion.
Worked example: the spring oscillator with q = x and H = p²/(2m) + ½kx². The bracket {x, H} = (∂x/∂x)(∂H/∂p) − (∂x/∂p)(∂H/∂x) = 1 · p/m − 0 · kx = p/m, which is ẋ. The bracket {p, H} = (∂p/∂x)(∂H/∂p) − (∂p/∂p)(∂H/∂x) = 0 · p/m − 1 · kx = −kx, which is ṗ. The canonical equations are brackets with H. Conservation check on H itself: {H, H} = (∂H/∂x)(∂H/∂p) − (∂H/∂p)(∂H/∂x) = kx · p/m − p/m · kx = 0, and since H has no explicit t, the energy stays fixed.
The canonical equations are brackets with H: position changes as {q, H} and momentum as {p, H}. One rule therefore delivers both the motion itself and the list of conserved quantities. Care is needed on two points. Zero bracket means zero everywhere along the relevant trajectory, not merely at one sampled point. And a nonzero bracket at one instant does not by itself prove continuing drift, since an explicit time dependence can cancel it in dA/dt = {A, H} + ∂A/∂t.
The commutator of two operators A and B is AB − BA, and a hat marks an operator. ħ is the reduced Planck constant, δᵢⱼ is 1 when i = j and 0 otherwise. For the stated canonical examples, quantization swaps each bracket for a commutator: [q̂ᵢ, p̂ⱼ] = iħδᵢⱼ, [q̂ᵢ, q̂ⱼ] = 0, and [p̂ᵢ, p̂ⱼ] = 0. So {q, p} = 1 becomes [q̂, p̂] = iħ, and vanishing mutual brackets become compatible observables within these examples. The map is not an exact blanket rule for arbitrary classical functions: operator ordering and quantum corrections can spoil it.
With no explicit t, the bracket with H gives the rate; zero bracket along the whole motion means conserved.
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8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.