Hamiltonian brackets & symplectic coordinates

When can a change of variables preserve Hamilton’s equations, and why do conserved quantities need a separate commutativity check? The canonical Poisson bracket answers both questions. This bridge teaches you to calculate the bracket, test a proposed coordinate pair, and exhibit two conserved functions that do not commute. The examples use finite-dimensional, smooth classical dynamics; the final exercise returns these tools to the open Toda chain.

Required background. Differentiate functions of several real variables and use the chain rule. The entry check repairs the signs and the meaning of a partial derivative before any Hamiltonian calculation.

Helpful background. The Toda introduction explains why these tests enter an integrability argument. No differential geometry is required for the calculations below.

The canonical bracket and Hamilton’s equations

Let $U$ be an open region of $\mathbb R^{2n}$, with real coordinates $(q_1,\ldots,q_n,p_1,\ldots,p_n)$. A point in this phase space specifies all coordinates and momenta at one instant. Take a smooth, time-independent Hamiltonian $H:U\to\mathbb R$. For smooth real functions $f,g$ on $U$, define

$$\{f,g\}=\sum_{j=1}^n \left( \frac{\partial f}{\partial q_j}\frac{\partial g}{\partial p_j} -\frac{\partial f}{\partial p_j}\frac{\partial g}{\partial q_j} \right).$$

Here a partial derivative holds all other phase-space coordinates fixed. In particular,

$$\{q_i,q_j\}=0,\qquad \{p_i,p_j\}=0,\qquad \{q_i,p_j\}=\delta_{ij},$$

where $\delta_{ij}$ is one when $i=j$ and zero otherwise. These are the canonical bracket relations. Exchanging the arguments reverses the sign: $\{g,f\}=-\{f,g\}$.

Hamilton’s equations in this convention are

$$\dot q_i=\{q_i,H\}=\frac{\partial H}{\partial p_i}, \qquad \dot p_i=\{p_i,H\}=-\frac{\partial H}{\partial q_i}.$$

They specify a vector field, hence local evolution from each initial point. We make claims only for times when the solution remains in its stated domain. These signs agree with Cannas da Silva, January 2006 revision, § 18.2, p. 107, PDF. The canonical relations and the additional requirement of commuting integrals also appear in Torrielli 2016, § 2.1, pp. 3–4, equations (2.1)–(2.3), PDF.

All explicit examples below use dimensionless coordinates, momenta, Hamiltonians and time. Physical units can be restored for the harmonic oscillator; the bracket convention itself is unchanged.

Entry check and repair

For one pair $(q,p)$, take $f=q^2p$ and $g=p^2$. Find $f_q,f_p,g_q,g_p$, then calculate $\{f,g\}$. Is the answer the same as $\{g,f\}$? Try the calculation before opening the repair.

Repair: differentiate first, then subtract

The four derivatives are

$$f_q=2qp,\quad f_p=q^2,\quad g_q=0,\quad g_p=2p.$$

Therefore $\{f,g\}=(2qp)(2p)-(q^2)(0)=4qp^2$, whereas $\{g,f\}=-4qp^2$. The symbol $p$ is an independent coordinate when differentiating with respect to $q$; do not replace it by a possible trajectory $p(t)$ in this step.

For a short retry, $\{q^2,p\}=2q$ and $\{p,q^2\}=-2q$. These checks fix the sign used in $\dot p=-H_q$.

Evolution and conservation are bracket calculations

Along a solution, the ordinary chain rule gives

$$\begin{aligned} \frac{d}{dt}f(q(t),p(t)) &=\sum_j\left(f_{q_j}\dot q_j+f_{p_j}\dot p_j\right)\\ &=\sum_j\left(f_{q_j}H_{p_j}-f_{p_j}H_{q_j}\right) =\{f,H\}. \end{aligned}$$

Thus a time-independent function $f$ is conserved along every local solution in $U$ exactly when $\{f,H\}=0$ throughout $U$. One direction follows by substitution. Conversely, each point can be used as an initial condition, so conservation for all such solutions forces the derivative to vanish at each point. This is the coordinate form of Cannas da Silva, § 18.4, p. 109, Theorem 18.9, PDF.

For the unit oscillator,

$$H=\frac12(p^2+q^2),\qquad \dot q=p,\qquad \dot p=-q.$$

Its kinetic energy $T=p^2/2$ and potential energy $V=q^2/2$ obey

$$\{T,H\}=-pq,\qquad \{V,H\}=pq.$$

Each term can change while their sum stays constant: $\{H,H\}=0$. At $q=0$, both displayed derivatives vanish at that instant. This does not make $T$ or $V$ conserved functions: neither bracket vanishes throughout phase space.

If $f$ depends explicitly on time, include that dependence:

$$\frac{df}{dt}=\{f,H\}+\frac{\partial f}{\partial t}.$$

For example, for the free Hamiltonian $H=p^2/2$, the function $f=q-pt$ is constant along trajectories because $\{f,H\}=p$ and $\partial_t f=-p$. The criterion $\{f,H\}=0$ alone applies to functions with no explicit time dependence.

Canonical coordinates preserve the bracket

Suppose $Q=Q(q,p)$ and $P=P(q,p)$ are a smooth, time-independent change of variables with a smooth inverse on the region being used. For one degree of freedom, the chain rule shows that

$$\{F(Q,P),G(Q,P)\}_{q,p} =\left(F_QG_P-F_PG_Q\right)\{Q,P\}_{q,p}.$$

The subscripts identify the original variables in which the bracket is calculated. Consequently $\{Q,P\}_{q,p}=1$ makes the bracket take the same canonical form in $(Q,P)$. For $n$ pairs, the corresponding test is the full set

$$\{Q_i,Q_j\}=0,\qquad \{P_i,P_j\}=0,\qquad \{Q_i,P_j\}=\delta_{ij}.$$

Expanding by the chain rule in all variables proves the same statement: the brackets of coordinate functions are the coefficients of the transformed bracket. Checking only the $n$ diagonal relations misses possible cross terms. A globally valid coordinate system additionally needs a globally one-to-one map on the stated domain; bracket identities alone do not establish that property.

Write the Hamiltonian in the new variables as $K(Q,P)=H(q(Q,P),p(Q,P))$. For a canonical, time-independent change,

$$\dot Q_i=\frac{\partial K}{\partial P_i}, \qquad \dot P_i=-\frac{\partial K}{\partial Q_i}.$$

The function describing the energy has changed its expression, but the dynamics and physical time have not changed.

Why the term symplectic appears

The geometric object preserved by canonical changes is the symplectic form

$$\omega=\sum_i dq_i\wedge dp_i.$$

For one pair, the wedge symbol records oriented infinitesimal area. Expanding the differentials gives

$$dQ\wedge dP =\left(Q_qP_p-Q_pP_q\right)dq\wedge dp =\{Q,P\}\,dq\wedge dp.$$

So the one-pair bracket test preserves this area with its orientation. For several pairs the canonical test preserves the full form $\omega$, a stronger condition than preserving total phase-space volume.

If you encounter the vector-field notation, our convention is $\iota_{X_H}\omega=dH$, meaning that inserting $X_H$ into the first slot of $\omega$ gives $dH$. It yields $X_Hf=\{f,H\}$ and the equations already derived. This matches Cannas da Silva, §§ 18.1–18.3, pp. 105–108, PDF. Torrielli instead writes $\sum_i dp_i\wedge dq_i$ and $X_F(g)=\{F,g\}$ in equations (2.8)–(2.10): both signs change together. Translate these conventions together when comparing formulas.

Worked example: square the coordinate, adjust the momentum

On the half-plane $q\gt0$, define

$$Q=q^2,\qquad P=\frac{p}{2q}.$$

The derivatives are $Q_q=2q$, $Q_p=0$, $P_p=1/(2q)$ and $P_q=-p/(2q^2)$. Therefore $\{Q,P\}=1$. The smooth inverse is

$$q=\sqrt Q,\qquad p=2\sqrt Q\,P, \qquad Q\gt0,\quad P\in\mathbb R.$$

For the unit oscillator, the transformed energy and equations are

$$\begin{aligned} K(Q,P)&=2QP^2+\frac Q2,\\ \dot Q&=4QP,\qquad \dot P=-2P^2-\frac12. \end{aligned}$$

Check these independently in the old variables:

$$\begin{aligned} \dot Q&=2q\dot q=2qp=4QP,\\ \dot P&=\frac{\dot p}{2q}-\frac{p\dot q}{2q^2} =-\frac12-\frac{p^2}{2q^2} =-\frac12-2P^2. \end{aligned}$$

The map is undefined at $q=0$. Extending its formula to both signs of nonzero $q$ would also identify $(q,p)$ with $(-q,-p)$, losing a unique inverse. An oscillator may cross $q=0$ during entirely regular physical motion; this coordinate chart then ceases to apply. A coordinate failure need not be a singularity of the dynamics.

Now try $\widetilde Q=q^2$, $\widetilde P=p$ on the same half-plane. This map is invertible, but $\{\widetilde Q,\widetilde P\}=2q=2\sqrt{\widetilde Q}$, so it is not canonical. In these variables,

$$\widetilde K=\frac{\widetilde P^2+\widetilde Q}{2}, \qquad \dot{\widetilde Q}=2\sqrt{\widetilde Q}\,\widetilde P, \qquad \dot{\widetilde P}=-\sqrt{\widetilde Q}.$$

Using the canonical rule without the bracket factor would give the wrong velocities. An invertible change of variables is useful, but it does not automatically create canonical coordinates.

Conserved functions need not commute with one another

Two functions $F,G$ are in involution when $\{F,G\}=0$. Conservation instead asks whether each commutes with the chosen Hamiltonian. These are different tests.

On $\mathbb R^4$, consider two uncoupled unit oscillators with equal frequency:

$$\begin{aligned} H&=E_1+E_2,\\ E_1&=\frac12(p_1^2+q_1^2),\qquad E_2=\frac12(p_2^2+q_2^2),\\ L&=q_1p_2-q_2p_1. \end{aligned}$$

The separate energies commute because they involve different canonical pairs, hence $\{E_1,H\}=0$. Direct differentiation of $L$ along the motion gives

$$\dot L=p_1p_2-q_1q_2-p_2p_1+q_2q_1=0.$$

Nevertheless,

$$\begin{aligned} \{E_1,L\} &=(q_1)(-q_2)-(p_1)(p_2)\\ &=-q_1q_2-p_1p_2, \end{aligned}$$

which is not the zero function. At $q_1=q_2=1$ and $p_1=p_2=0$, it is $-1$. Thus $E_1$ and $L$ are both conserved under $H$ but are not mutually in involution. Equal frequencies matter to the conservation of this particular $L$; no assertion about a coupled or unequal-frequency system is being made.

The system still has a commuting choice, $E_1,E_2$. Their differentials are independent wherever neither oscillator is at its origin. In a Liouville argument, a suitable set must satisfy both conditions: mutual involution and independence. Adding every conserved quantity to that set is unnecessary and can destroy involution. See the Toda independence lesson for an explicit rank test, and the Liouville–Arnold reference for the further hypotheses needed to obtain invariant tori and action–angle coordinates.

Exercises

Guided: an instantaneous zero is not conservation

For the unit oscillator, take $D=qp$. Complete $D_q=\underline{\phantom{p}}$ and $D_p=\underline{\phantom{q}}$, then calculate $\{D,H\}$. At the initial point $(q,p)=(1,1)$, $\dot D=0$. Does that make $D$ conserved? Compare with $H^2$: is it conserved, and is it independent of $H$?

Hint

Check the bracket away from $(1,1)$, or differentiate $\dot D$ once more there. For independence, compare $d(H^2)$ with $dH$.

Solution

The derivatives are $D_q=p$ and $D_p=q$, so

$$\dot D=\{D,H\}=p^2-q^2.$$

This vanishes at the chosen point but not throughout phase space. More strongly, along the solution starting there,

$$\ddot D=2p\dot p-2q\dot q=-4pq=-4 \quad\text{at }t=0.$$

Hence $D$ changes immediately beyond its stationary instant. By the product rule, $\{H^2,H\}=2H\{H,H\}=0$, so $H^2$ is conserved. But $d(H^2)=2H\,dH$: it adds no independent integral, even where the two functions have different values.

Independent: logarithmic canonical coordinates

On $q\gt0$, set $Q=\log q$ and $P=qp$. Prove that this is a canonical change onto $\mathbb R^2$, find its inverse, and express the unit-oscillator Hamiltonian as $K(Q,P)$. Derive both transformed equations and check them against the chain rule. At $q=2,p=3$, calculate $Q,P,\dot Q,\dot P$.

Hint

Use $Q_q=1/q$ and $P_p=q$. Inverting the first equation determines $q$ uniquely on this domain, then $p$ follows from the second.

Solution

The bracket is $(1/q)q=1$. The inverse is $q=e^Q$, $p=e^{-Q}P$, with arbitrary real $Q,P$, so the map and inverse are smooth on their stated domains. Substitution gives

$$\begin{aligned} K&=\frac12\left(e^{-2Q}P^2+e^{2Q}\right),\\ \dot Q&=e^{-2Q}P,\\ \dot P&=e^{-2Q}P^2-e^{2Q}. \end{aligned}$$

Independently, $\dot Q=\dot q/q=p/q$ and $\dot P=\dot q\,p+q\dot p=p^2-q^2$, which give exactly these expressions after substitution. At the specified point,

$$Q=\log2,\quad P=6,\quad \dot Q=\frac32,\quad \dot P=5.$$

The factor $e^{-2Q}$ depends on $Q$: treating it as a constant when finding $K_Q$ would lose the kinetic contribution to $\dot P$. As in the squared-coordinate example, this chart does not extend through $q=0$.

Transfer: Toda variables and a missing coordinate

For two open Toda particles, use the dimensionless Hamiltonian

$$H=\frac12(p_1^2+p_2^2)+e^{q_1-q_2}.$$

The variables $a=e^{(q_1-q_2)/2}$, $b_1=p_1$, $b_2=p_2$ are useful for the Lax matrix. Compute $\{a,b_1\}$, $\{a,b_2\}$ and $\{b_1,b_2\}$. Explain why these three variables cannot be a canonical coordinate system on the original four-dimensional phase space.

Then retain the missing center coordinate by introducing

$$Q=q_1-q_2,\quad P=\frac{p_1-p_2}{2},\quad C=\frac{q_1+q_2}{2},\quad \Pi=p_1+p_2.$$

Check that $(Q,P)$ and $(C,\Pi)$ are canonical pairs with vanishing cross brackets. Express $H$ in these variables and derive their four equations of motion. What does this say about the role of $a=e^{Q/2}$?

Hint

Since $\{f,p_i\}=\partial_{q_i}f$, the first brackets need only two derivatives. A simultaneous shift of $q_1,q_2$ leaves $a,b_1,b_2$ unchanged. For the complete transformation, invert the two sums and differences before substituting in $H$.

Solution

The brackets are

$$\{a,b_1\}=\frac a2,\qquad \{a,b_2\}=-\frac a2,\qquad \{b_1,b_2\}=0.$$

They are not the bracket relations of two canonical pairs. Moreover, three variables cannot give an invertible coordinate chart on four-dimensional phase space: $a$ retains only the relative displacement. Both $q_i\mapsto q_i+c$ give the same $(a,b_1,b_2)$.

For the full transformation, the nonzero fundamental brackets are

$$\{Q,P\}=\frac12+\frac12=1, \qquad \{C,\Pi\}=\frac12+\frac12=1.$$

All cross brackets vanish; for example, $\{Q,\Pi\}=1-1=0$ and $\{C,P\}=1/4-1/4=0$. The inverse is global on $\mathbb R^4$:

$$q_1=C+\frac Q2,\quad q_2=C-\frac Q2,\quad p_1=\frac\Pi2+P,\quad p_2=\frac\Pi2-P.$$

Therefore

$$K=\frac{\Pi^2}{4}+P^2+e^Q,$$

and the canonical equations are

$$\dot Q=2P,\qquad \dot P=-e^Q,\qquad \dot C=\frac\Pi2,\qquad \dot\Pi=0.$$

The center moves freely while the interaction depends only on $Q$. The useful exponential variable satisfies $\dot a=(a/2)\dot Q=aP$; it can be evolved by the chain rule without pretending that $(a,b_1,b_2)$ is a canonical chart. A useful Lax parametrization and a canonical coordinate system serve different purposes.

Return to Toda or solve the oscillator

You can now test a coordinate change by its bracket relations and inverse, distinguish conservation from involution, and recognize a domain where an otherwise correct chart fails. Return to Derive the open Toda equations to use Hamilton’s rule with exponential forces, or to Test independence and Poisson commutativity to apply the separate tests to spectral invariants. For a complete trajectory calculation in two coordinate systems, continue to Solve an oscillator two ways.

The calculations here concern canonical coordinates on explicit finite-dimensional domains. They do not assert global coordinates across singular sets or establish action–angle variables for a general Hamiltonian system.

References

• Cannas da Silva, Ana. Lectures on Symplectic Geometry. Lecture Notes in Mathematics 1764. Springer, 2001. DOI: 10.1007/978-3-540-45330-7. Author’s January 2006 revision, PDF, §§ 18.1–18.4, pp. 105–110; especially Definition 18.5 and Theorem 18.9.
• Torrielli, Alessandro. Lectures on Classical Integrability. Lecture notes for the Durham Young Researchers Integrability School, July 2015. arXiv:1606.02946v1 [hep-th], 9 June 2016. Version record. Open PDF, § 2.1, pp. 3–5, equations (2.1)–(2.3) and (2.8)–(2.10).