Poisson brackets

Start with one spring state

Supply a classical point-particle model on one fixed inertial line, with right positive. Its constant mass is m = 2 kg. Position q is measured from the equilibrium location in m; momentum p = mv is measured in kg m/s, where v is the signed velocity. The restoring force is −kq, with k = 8 N/m. These are exact teaching assumptions, not measured spring data.

At one instant choose q = 0.5 m and p = 2 kg m/s. The particle moves right at p/m = 1 m/s, while its momentum changes at −kq = −4 N. The signs answer different questions: where it moves and how its momentum changes. Its squared position q² is increasing at 2qv = 1 m²/s.

A Poisson bracket will package these rate calculations into one rule. First follow this small state; then use the comparison table before returning to the formal time rule if derivatives are new. Fixed-variable partial derivatives are explained locally below.

Supply the inputs and the Hamiltonian

For this model, q and p form a canonical pair: the Hamiltonian description uses them as independent state inputs with the equations dq/dt = ∂H/∂p and dp/dt = −∂H/∂q. Calling two quantities canonical includes this structural choice; it is not established merely by giving them names. A dot, as in q̇, also means a time derivative.

Choose the Hamiltonian H(q, p) = p²/(2m) + kq²/2. Here it equals kinetic plus potential energy, with potential zero at q = 0. Each term has units J. At the selected state they are 1 J each, so H = 2 J. This energy interpretation belongs to the stated classical model.

The coordinate and momentum inputs may take any real values in this smooth model. A partial derivative changes one input while holding the other and any explicit time fixed. Thus H_p = ∂H/∂p = p/m and H_q = ∂H/∂q = kq. The subscripts name the differentiated input; H_p is not a momentum value. Substitution into the supplied Hamilton equations gives q̇ = p/m and ṗ = −kq.

Define the bracket and keep its order

For smooth scalar functions A(q, p, t) and B(q, p, t), use the convention {A, B} = A_q B_p − A_p B_q. The braces name a Poisson bracket. All four partials are evaluated at the same q, p and t, with explicit time held fixed. The order matters: exchanging A and B reverses the sign.

For q itself, q_q = 1 and q_p = 0. For p, p_p = 1 and p_q = 0. Therefore {q, H} = H_p = p/m and {p, H} = −H_q = −kq. This is why the bracket order {observable, H} matches our stated rate convention. A source using the opposite bracket sign would need a matching change in its evolution rule.

Try the observable A = q², a quantity with units m². Its partials are A_q = 2q and A_p = 0. Hence {q², H} = (2q)(p/m) − 0(kq) = 2qp/m. At q = 0.5 m and p = 2 kg m/s this is +1 m²/s, the rate found in the opening case. The zero product is part of the calculation; it does not require q or p to remain fixed during motion.

Follow an observable through time

An observable here is a smooth scalar quantity calculated from the state, possibly with an explicit time argument. Along a solution q(t), p(t), the ordinary chain rule is dA/dt = A_q q̇ + A_p ṗ + A_t. In A_t, hold q and p fixed. The total derivative follows the changing state and also includes any explicit time dependence.

Insert the supplied Hamilton rates: dA/dt = A_q H_p − A_p H_q + A_t = {A, H} + A_t. This rule is conditional on those Hamilton equations in the stated canonical variables. It is not an additional force law inferred from the bracket definition.

For A = q² there is no explicit time argument, so A_t = 0 and dA/dt = {q², H}. For an observable with explicit time, keep the extra A_t term. The last section shows a concrete case in which that term cancels a nonzero bracket at one instant.

Compare four separate states

The table supplies four alternative instantaneous states. S1 and S2 have the same positive position but opposite momenta; S3 and S4 have the same negative position and opposite momenta. Each has kinetic energy 1 J and potential energy 1 J, yet their signed rates differ. These labels do not order events in time.

In the first row of the diagram, changing the momentum sign changes q̇. In the second, changing the position sign changes ṗ: the restoring rate points toward q = 0. In the third, the sign of d(q²)/dt depends on the product qp. A particle moving toward q = 0 can have a decreasing q² even when its q velocity is positive.

Each bar extends from its row’s zero line to its signed local rate. The rows have different units and scales; equal drawn heights do not mean equal physical quantities. Solid outlines and circles, dashed outlines and diamonds, and dotted outlines and triangles identify the three quantities. There are no connecting trajectories or finite time steps.

Check units and simple identities

The bracket need not be dimensionless. If A and B have quantity units [A] and [B], both terms have units [A][B]/([q][p]); the square brackets here denote units, not Poisson brackets. In this model qp has units kg m²/s = J s. Thus {A, H} has the units of A per second, as a time rate should.

The fundamental bracket {q, p} = 1 is dimensionless because its units cancel. The self-brackets {q, q} and {p, p} both have numerical value zero, but their quantity units are s/kg and kg/s respectively. A zero value does not make every kind of quantity dimensionless.

The definition immediately gives antisymmetry, {B, A} = −{A, B}, and {A, A} = 0 with the appropriate units. It is linear in either input and obeys the product rule {AB, C} = A{B, C} + B{A, C}. Apply a sum only to compatible quantities. For example, {q², p} = 2q, with units m, and reversing the inputs gives {p, q²} = −2q.

Separate an identity from one zero

For the Hamiltonian itself, {H, H} = H_q H_p − H_p H_q = 0, an identity throughout this model’s state domain. Since this H has no explicit time dependence, dH/dt = 0 along a solution. Consequently its energy is constant on a connected time interval on which that solution is defined.

A zero at one state is weaker. At q = 0 and p = 2 kg m/s, the observable A = q² has dA/dt = 0 even though q̇ = 1 m/s. Differentiating its rate once more gives d²A/dt² = 2q̇² + 2q q̈ = 2 m²/s² there. Its squared position is stationary at that instant, not constant on a nearby time interval.

More generally, for an observable without explicit time dependence, a zero bracket throughout the states traversed by a solution gives a constant value along that interval. A bracket identity throughout a region is one sufficient way to establish this for solutions staying in the region. Checking one state, or just several selected states, does not establish that condition.

Retain the explicit time term

Use C(q, p, t) = q − u t with u = 1 m/s, a supplied constant. This is an observable built from our original state, not a replacement canonical frame. Both terms have units m. Holding q and p fixed gives C_t = −u, while C_q = 1 and C_p = 0.

At S1 and t = 0, C = 0.5 m and {C, H} = p/m = 1 m/s. Its total rate is dC/dt = {C, H} + C_t = 1 m/s − 1 m/s = 0. Omitting the explicit time term would give the wrong result. This cancellation is local: its second time rate is q̈ = −kq/m = −2 m/s² there, so C is not constant near that instant.

We have used smooth classical functions, one declared canonical pair and the corresponding Hamilton equations. More canonical pairs use a sum over matched pairs; arbitrary coordinates or constrained systems require the appropriate formulation. Quantum operators are outside this calculation. The next mapped subject, Canonical transformations, asks which changes of variables preserve the canonical structure.

Worked examples

Try it yourself

1. With the stated convention, what are {q, p} and {p, q}? What units do they have?

   Show the answer and reasoning

   {q, p} = +1 and {p, q} = −1, both dimensionless.

   The first has q_q p_p = 1 and the subtracted product is zero. Reversing the entries reverses the sign. The bracket-unit rule cancels [q][p] specifically for these inputs.

2. At S3, find q̇, ṗ and d(q²)/dt. Is the particle moving toward or away from q = 0 at that instant?

   Show the answer and reasoning

   q̇ = +1 m/s, ṗ = +4 N and d(q²)/dt = −1 m²/s. It moves toward q = 0.

   Use p/m, −kq and 2qp/m at q = −0.5 m, p = 2 kg m/s. Positive coordinate rate from negative q moves toward equilibrium, so squared position decreases.

3. At q = 0 and p = 2 kg m/s, is q² conserved just because its first time rate is zero? Is the particle at rest?

   Show the answer and reasoning

   Neither follows. The particle has q̇ = 1 m/s, and q² has second time rate 2 m²/s² at that instant.

   The first rate 2q p/m vanishes because q = 0. Differentiating gives 2q̇² + 2q q̈; only its first term remains. The nonzero second rate excludes local interval constancy, and nonzero q̇ excludes rest.

4. For C = q − u t with u = 1 m/s, what is dC/dt at S2 and t = 0?

   Show the answer and reasoning

   dC/dt = −2 m/s.

   At S2, {C, H} = p/m = −1 m/s and C_t = −1 m/s. Their sum is −2 m/s. The explicit term is added with its own sign.

5. Let D = qp. Its bracket is {D, H} = p²/m − kq². It is zero at all four table states. Is it an identity? Test q = 1 m, p = 0.

   Show the answer and reasoning

   No. At the test state {D, H} = −8 J, so D is not a conserved-function identity throughout the model’s state domain.

   D_q = p and D_p = q give the stated bracket. The test state gives 0 − (8 N/m)(1 m)² = −8 J. Its energy is 4 J, not the four selected states’ 2 J; the test is of a whole-domain identity, not a claim about one particular 2 J trajectory.

6. Both {q, q} and {p, p} have numerical value zero. Must both be dimensionless?

   Show the answer and reasoning

   No. Their units are s/kg and kg/s respectively.

   For {q, q}, the unit rule gives m²/(kg m²/s) = s/kg. For {p, p}, it gives (kg m/s)²/(kg m²/s) = kg/s. Both products cancel numerically, but their quantity kinds and units have not become the same.

What to carry forward

• A Poisson bracket uses a stated canonical pair, fixed-time partial derivatives and an explicit order convention.
• Under the corresponding Hamilton equations, dA/dt = {A, H} + A_t; retain explicit time dependence.
• Equal energy need not determine signed coordinate, momentum or observable rates.
• A bracket identity can establish interval conservation under its conditions; a zero at a selected state cannot.
• Bracket units follow from the units of both functions and the canonical pair, even when the numerical value is zero.