Poisson brackets — definition, properties, connection to commutators: Exercises

Before we start, one picture of the whole machine so you always know which knob you are turning.

Figure — Poisson brackets — definition, properties, connection to commutators

Level 1 — Recognition

You are only asked to read the definition correctly: pick the right partial derivatives, put them in the right slots, and respect the minus sign.

Problem 1.1

Using the definition with a single coordinate pair $(q, p)$ , compute ${q, p}$ and ${p, q}$ .

Recall Solution 1.1

Here $f$ and $g$ are just the coordinates themselves, so their partial derivatives are $0$ or $1$ . For ${q, p}$ : set $f = q, g = p$ .

$\frac{\partial f}{\partial q} = \frac{\partial q}{\partial q} = 1$ , $\frac{\partial g}{\partial p} = \frac{\partial p}{\partial p} = 1$ .
$\frac{\partial f}{\partial p} = 0$ , $\frac{\partial g}{\partial q} = 0$ . ${q, p} = (1) (1) - (0) (0) = 1.$ For ${p, q}$ : swap the roles. ${p, q} = \frac{\partial p}{\partial q} \frac{\partial q}{\partial p} - \frac{\partial p}{\partial p} \frac{\partial q}{\partial q} = (0) (0) - (1) (1) = - 1.$ So ${q, p} = 1$ and ${p, q} = - 1$ . The minus sign is not decoration — it is antisymmetry made flesh.

Problem 1.2

State which of these are $0$ by the antisymmetry property alone, without computing: ${f, f}$ , ${H, H}$ , ${L_{z}, L_{z}}$ .

Recall Solution 1.2

Antisymmetry says ${f, g} = - {g, f}$ . Put $g = f$ : ${f, f} = - {f, f}$ , so $2 {f, f} = 0$ , giving ${f, f} = 0$ . The same argument uses nothing about what $f$ is. So all three are zero: ${f, f} = {H, H} = {L_{z}, L_{z}} = 0$ .

Problem 1.3

Given constants $a, b$ and the bilinearity rule, simplify ${3 q + 5 p, p}$ .

Recall Solution 1.3

Bilinearity: ${a f + b g, h} = a {f, h} + b {g, h}$ . Here $a = 3$ , $f = q$ , $b = 5$ , $g = p$ , $h = p$ . ${3 q + 5 p, p} = 3 {q, p} + 5 {p, p} = 3 (1) + 5 (0) = 3.$

Level 2 — Application

Now turn the crank on real functions. Same definition, more partial derivatives to track.

Problem 2.1

For one degree of freedom, compute ${q^{2}, p}$ and ${q, p^{2}}$ .

Recall Solution 2.1

${q^{2}, p}$ : $f = q^{2}$ , so $\frac{\partial f}{\partial q} = 2 q$ , $\frac{\partial f}{\partial p} = 0$ . And $g = p$ , so $\frac{\partial g}{\partial p} = 1$ , $\frac{\partial g}{\partial q} = 0$ . ${q^{2}, p} = (2 q) (1) - (0) (0) = 2 q .$ ${q, p^{2}}$ : $f = q$ gives $\frac{\partial f}{\partial q} = 1, \frac{\partial f}{\partial p} = 0$ ; $g = p^{2}$ gives $\frac{\partial g}{\partial p} = 2 p, \frac{\partial g}{\partial q} = 0$ . ${q, p^{2}} = (1) (2 p) - (0) (0) = 2 p .$ Sanity check via Leibniz: ${q^{2}, p} = {q \cdot q, p} = q {q, p} + {q, p} q = q + q = 2 q$ . ✓

Problem 2.2

A particle of mass $m$ in a potential: $H = \frac{p ^{2}}{2 m} + V (q)$ . Compute ${q, H}$ and ${p, H}$ , and identify what physical equations they reproduce.

Recall Solution 2.2

${q, H}$ : $f = q$ , so $\frac{\partial q}{\partial q} = 1$ , $\frac{\partial q}{\partial p} = 0$ . ${q, H} = \frac{\partial q}{\partial q} \frac{\partial H}{\partial p} - \frac{\partial q}{\partial p} \frac{\partial H}{\partial q} = (1) \frac{p}{m} - (0) V^{'} (q) = \frac{p}{m} .$ This is $\overset{q}{˙} = p / m$ — velocity equals momentum over mass. ${p, H}$ : $f = p$ , so $\frac{\partial p}{\partial q} = 0$ , $\frac{\partial p}{\partial p} = 1$ . ${p, H} = \frac{\partial p}{\partial q} \frac{\partial H}{\partial p} - \frac{\partial p}{\partial p} \frac{\partial H}{\partial q} = (0) \frac{p}{m} - (1) V^{'} (q) = - V^{'} (q) .$ This is $\overset{p}{˙} = - V^{'} (q)$ — Newton's second law, force $= - d V / d q$ .

Problem 2.3

With $H = \frac{p ^{2}}{2 m}$ (free particle) and $f = q p$ , compute ${f, H}$ and interpret.

Recall Solution 2.3

$f = q p$ : $\frac{\partial f}{\partial q} = p$ , $\frac{\partial f}{\partial p} = q$ . $H = \frac{p ^{2}}{2 m}$ : $\frac{\partial H}{\partial p} = \frac{p}{m}$ , $\frac{\partial H}{\partial q} = 0$ . ${q p, H} = (p) \frac{p}{m} - (q) (0) = \frac{p ^{2}}{m} .$ Since $\frac{d ( q p )}{d t} = {q p, H} = \frac{p ^{2}}{m} = 2 \cdot \frac{p ^{2}}{2 m} = 2 H$ , the quantity $q p$ grows steadily — it is not conserved, and its growth rate is twice the kinetic energy. (This is the classical seed of the virial theorem.)

Level 3 — Analysis

Now the question is why: which brackets vanish, and what that tells you about conservation.

Problem 3.1

For $H = \frac{p _{x}^{2} + p _{y}^{2}}{2 m} + V (r)$ with $r = x^{2} + y^{2}$ , show that $L_{z} = x p_{y} - y p_{x}$ is conserved, i.e. ${L_{z}, H} = 0$ . Explain geometrically why it must vanish.

Recall Solution 3.1

We need ${L_{z}, H}$ . Split $H = T + V$ with $T = \frac{p _{x}^{2} + p _{y}^{2}}{2 m}$ . Kinetic part. $L_{z} = x p_{y} - y p_{x}$ has $\frac{\partial L _{z}}{\partial x} = p_{y}$ , $\frac{\partial L _{z}}{\partial y} = - p_{x}$ , $\frac{\partial L _{z}}{\partial p _{x}} = - y$ , $\frac{\partial L _{z}}{\partial p _{y}} = x$ . And $T$ has $\frac{\partial T}{\partial p _{x}} = \frac{p _{x}}{m}$ , $\frac{\partial T}{\partial p _{y}} = \frac{p _{y}}{m}$ , $\frac{\partial T}{\partial x} = \frac{\partial T}{\partial y} = 0$ . ${L_{z}, T} = \sum_{i = x, y} (\frac{\partial L _{z}}{\partial i} \frac{\partial T}{\partial p _{i}} - \frac{\partial L _{z}}{\partial p _{i}} \frac{\partial T}{\partial i}) = p_{y} \frac{p _{x}}{m} + (- p_{x}) \frac{p _{y}}{m} - 0 - 0 = 0.$ Potential part. $V$ depends on $r$ only, so $\frac{\partial V}{\partial x} = V^{'} (r) \frac{x}{r}$ , $\frac{\partial V}{\partial y} = V^{'} (r) \frac{y}{r}$ , and $\frac{\partial V}{\partial p _{x}} = \frac{\partial V}{\partial p _{y}} = 0$ . ${L_{z}, V} = - \frac{\partial L _{z}}{\partial p _{x}} \frac{\partial V}{\partial x} - \frac{\partial L _{z}}{\partial p _{y}} \frac{\partial V}{\partial y} = - (- y) V^{'} \frac{x}{r} - (x) V^{'} \frac{y}{r} = \frac{V ^{'}}{r} (x y - x y) = 0.$ So ${L_{z}, H} = 0$ : angular momentum is conserved. Geometric why: a central potential $V (r)$ looks identical after any rotation about the $z$ -axis. $L_{z}$ is precisely the generator of those rotations. A quantity that generates a symmetry of $H$ commutes with $H$ — that is the Poisson-bracket face of Noether's theorem. See the figure: the potential's circular symmetry is what kills the bracket.

Problem 3.2

For $H = \frac{p ^{2}}{2 m}$ and the explicitly time-dependent quantity $G = q - \frac{p}{m} t$ , show that $G$ is conserved by using the full equation of motion.

Recall Solution 3.2

The master law is $\frac{d G}{d t} = {G, H} + \frac{\partial G}{\partial t}$ . Do not forget the last term. Bracket part. $G = q - \frac{p}{m} t$ : $\frac{\partial G}{\partial q} = 1$ , $\frac{\partial G}{\partial p} = - \frac{t}{m}$ . And $\frac{\partial H}{\partial p} = \frac{p}{m}$ , $\frac{\partial H}{\partial q} = 0$ . ${G, H} = (1) \frac{p}{m} - (- \frac{t}{m}) (0) = \frac{p}{m} .$ Explicit-time part. $\frac{\partial G}{\partial t} = - \frac{p}{m}$ (differentiate treating $q, p$ as fixed). $\frac{d G}{d t} = \frac{p}{m} + (- \frac{p}{m}) = 0.$ $G$ is conserved. Physically $G$ is the initial position: $q (t) - \frac{p}{m} t = q (0)$ for free motion. The conservation only appears when both terms are kept.

Problem 3.3

Show ${L_{z}, p_{x}} = p_{y}$ and ${L_{z}, x} = - y$ , then explain the pattern: $L_{z}$ rotates vectors in the $x y$ -plane.

Recall Solution 3.3

$L_{z} = x p_{y} - y p_{x}$ . ${L_{z}, p_{x}}$ : $\frac{\partial L _{z}}{\partial x} = p_{y}, \frac{\partial p _{x}}{\partial p _{x}} = 1$ ; other surviving pieces zero. ${L_{z}, p_{x}} = \frac{\partial L _{z}}{\partial x} \frac{\partial p _{x}}{\partial p _{x}} - \frac{\partial L _{z}}{\partial p _{x}} \frac{\partial p _{x}}{\partial x} = (p_{y}) (1) - (- y) (0) = p_{y} .$ ${L_{z}, x}$ : $\frac{\partial L _{z}}{\partial p _{x}} = - y, \frac{\partial x}{\partial x} = 1$ . ${L_{z}, x} = \frac{\partial L _{z}}{\partial x} \frac{\partial x}{\partial p _{x}} - \frac{\partial L _{z}}{\partial p _{x}} \frac{\partial x}{\partial x} = (p_{y}) (0) - (- y) (1) = - y .$ Pattern: brackets with $L_{z}$ act like $\frac{d}{d ϕ}$ , an infinitesimal rotation. ${L_{z}, x} = - y$ and ${L_{z}, y} = x$ are exactly $(\overset{x}{˙} \overset{y}{˙}) = (01 - 1 0) (x y)$ , the rotation generator. This is why $L_{z}$ is called the generator of rotations, linking to Angular Momentum Algebra.

Level 4 — Synthesis

Combine properties (Leibniz, Jacobi, bilinearity) to build results you could not get by brute force alone.

Problem 4.1 (Poisson's theorem in action)

Given that $L_{x}$ and $L_{y}$ are both conserved ( ${L_{x}, H} = {L_{y}, H} = 0$ ), prove without touching $H$ 's explicit form that $L_{z}$ is conserved, using ${L_{x}, L_{y}} = L_{z}$ and the Jacobi identity.

Recall Solution 4.1

Jacobi with $f = L_{x}, g = L_{y}, h = H$ : ${L_{x}, {L_{y}, H}} + {L_{y}, {H, L_{x}}} + {H, {L_{x}, L_{y}}} = 0.$ The first two brackets contain ${L_{y}, H} = 0$ and ${H, L_{x}} = - {L_{x}, H} = 0$ , so they vanish: ${H, {L_{x}, L_{y}}} = 0.$ But ${L_{x}, L_{y}} = L_{z}$ , so ${H, L_{z}} = 0$ , i.e. ${L_{z}, H} = 0$ . $L_{z}$ is conserved. Punchline: if two components of angular momentum are conserved, the third is forced to be — you manufactured a new conservation law purely from the algebra. That is Poisson's theorem.

Problem 4.2 (Leibniz to shortcut a hard bracket)

Compute ${x^{2} + y^{2}, L_{z}}$ using the product/Leibniz and bilinearity rules, given ${x, L_{z}} = y$ , ${y, L_{z}} = - x$ (note: these are $-$ of Problem 3.3 by antisymmetry).

Recall Solution 4.2

First check the given brackets by antisymmetry: from 3.3, ${L_{z}, x} = - y \Rightarrow {x, L_{z}} = y$ ; and ${L_{z}, y} = x \Rightarrow {y, L_{z}} = - x$ . ✓ Use Leibniz ${f g, h} = f {g, h} + {f, h} g$ on each square: ${x^{2}, L_{z}} = x {x, L_{z}} + {x, L_{z}} x = 2 x {x, L_{z}} = 2 x y,$ ${y^{2}, L_{z}} = 2 y {y, L_{z}} = 2 y (- x) = - 2 x y .$ By bilinearity: ${x^{2} + y^{2}, L_{z}} = 2 x y - 2 x y = 0.$ Interpretation: $x^{2} + y^{2} = r^{2}$ is rotationally invariant, so its bracket with the rotation generator $L_{z}$ must vanish. The algebra confirms the geometry.

Problem 4.3 (Building the full angular-momentum algebra)

Given ${L_{x}, L_{y}} = L_{z}$ , use antisymmetry and cyclic relabelling to write all three brackets ${L_{x}, L_{y}}$ , ${L_{y}, L_{z}}$ , ${L_{z}, L_{x}}$ , then verify one of them directly from components.

Recall Solution 4.3

The compact rule from the parent note is ${L_{i}, L_{j}} = ϵ_{ij k} L_{k}$ . Written out cyclically: ${L_{x}, L_{y}} = L_{z}, {L_{y}, L_{z}} = L_{x}, {L_{z}, L_{x}} = L_{y} .$ Direct check of ${L_{y}, L_{z}} = L_{x}$ . With $L_{y} = z p_{x} - x p_{z}$ , $L_{z} = x p_{y} - y p_{x}$ : ${L_{y}, L_{z}} = \sum_{i} (\frac{\partial L _{y}}{\partial q _{i}} \frac{\partial L _{z}}{\partial p _{i}} - \frac{\partial L _{y}}{\partial p _{i}} \frac{\partial L _{z}}{\partial q _{i}}) .$ Nonzero partials: $\frac{\partial L _{y}}{\partial x} = - p_{z}$ , $\frac{\partial L _{y}}{\partial z} = p_{x}$ , $\frac{\partial L _{y}}{\partial p _{x}} = z$ , $\frac{\partial L _{y}}{\partial p _{z}} = - x$ ; $\frac{\partial L _{z}}{\partial x} = p_{y}$ , $\frac{\partial L _{z}}{\partial y} = - p_{x}$ , $\frac{\partial L _{z}}{\partial p _{x}} = - y$ , $\frac{\partial L _{z}}{\partial p _{y}} = x$ .

$x$ -term: $(- p_{z}) (0) - (z) (p_{y}) = - z p_{y}$ .
$y$ -term: $(0) (- y) - (0) (- p_{x}) = 0$ .
$z$ -term: $(p_{x}) (0) - (- x) (0) = 0$ . Wait — recompute the $p_{x}$ / $x$ pairing carefully. Collect all surviving cross terms: $\frac{\partial L _{y}}{\partial x} \frac{\partial L _{z}}{\partial p _{x}} - \frac{\partial L _{y}}{\partial p _{x}} \frac{\partial L _{z}}{\partial x} = (- p_{z}) (- y) - (z) (p_{y}) = y p_{z} - z p_{y},$ and all other index pairs give $0$ . So ${L_{y}, L_{z}} = y p_{z} - z p_{y} = L_{x} . ✓$ This closed algebra is the classical $S O (3)$ , whose quantum image (multiply by $\frac{1}{i ℏ}$ ) is $[\hat{L}_{i}, \hat{L}_{j}] = i ℏ ϵ_{ij k} \hat{L}_{k}$ — see Commutators in Quantum Mechanics.

Level 5 — Mastery

Invent, generalize, and defend. These reward understanding the structure, not just crank-turning.

Problem 5.1 (Quantization consistency)

Classically ${q, p} = 1$ and ${q, p^{2}} = 2 p$ (Problem 2.1). Show these are consistent with the quantum images under Dirac's rule ${f, g} \to \frac{1}{i ℏ} [\hat{f}, \overset{g}{^}]$ , i.e. verify $[\overset{q}{^}, \overset{p}{^}] = i ℏ$ and $[\overset{q}{^}, \overset{p}{^}^{2}] = 2 i ℏ \overset{p}{^}$ .

Recall Solution 5.1

Dirac's rule: a classical bracket equal to $C$ maps to a commutator equal to $i ℏ C$ . From ${q, p} = 1$ : predicted $[\overset{q}{^}, \overset{p}{^}] = i ℏ \cdot 1 = i ℏ$ . ✓ (the canonical commutation relation). From ${q, p^{2}} = 2 p$ : predicted $[\overset{q}{^}, \overset{p}{^}^{2}] = i ℏ \cdot 2 \overset{p}{^} = 2 i ℏ \overset{p}{^}$ . Independent quantum check using $[\hat{A}, \hat{B} \hat{C}] = [\hat{A}, \hat{B}] \hat{C} + \hat{B} [\hat{A}, \hat{C}]$ (the commutator Leibniz rule): $[\overset{q}{^}, \overset{p}{^}^{2}] = [\overset{q}{^}, \overset{p}{^}] \overset{p}{^} + \overset{p}{^} [\overset{q}{^}, \overset{p}{^}] = i ℏ \overset{p}{^} + \overset{p}{^} i ℏ = 2 i ℏ \overset{p}{^} . ✓$ The classical Leibniz rule ( ${q, p^{2}} = 2 p {q, p} = 2 p$ ) and the quantum Leibniz rule give the same shape — that structural identity is exactly why Dirac's map works.

Problem 5.2 (Design a conserved quantity)

For a particle under constant force, $H = \frac{p ^{2}}{2 m} - F q$ (so $F$ is the force). Find a time-dependent quantity $G (q, p, t)$ that is conserved, and prove it.

Recall Solution 5.2

Guess from physics: under constant force, $p (t) = p_{0} + F t$ , so $p - F t$ should be constant. Let $G = p - F t$ . $\frac{d G}{d t} = {G, H} + \frac{\partial G}{\partial t} .$ Bracket: $G = p - F t$ has $\frac{\partial G}{\partial q} = 0$ , $\frac{\partial G}{\partial p} = 1$ . And $H$ has $\frac{\partial H}{\partial p} = \frac{p}{m}$ , $\frac{\partial H}{\partial q} = - F$ . ${G, H} = (0) \frac{p}{m} - (1) (- F) = F .$ Explicit time: $\frac{\partial G}{\partial t} = - F$ . So $\frac{d G}{d t} = F + (- F) = 0.$ $G = p - F t$ is conserved: it is the initial momentum $p_{0}$ . This is the classic "explicit-time term is essential" case flagged in the parent note.

Problem 5.3 (Jacobi guards the whole edifice)

The Jacobi identity ${f, {g, h}} + {g, {h, f}} + {h, {f, g}} = 0$ must hold for the bracket to be a legitimate Lie algebra. Verify it explicitly for $f = q, g = p, h = q p$ in one degree of freedom.

Recall Solution 5.3

Compute the three inner brackets first.

${g, h} = {p, q p}$ . By Leibniz: ${p, q p} = q {p, p} + {p, q} p = 0 + (- 1) p = - p$ .
${h, f} = {q p, q}$ . Leibniz: ${q p, q} = q {p, q} + {q, q} p = q (- 1) + 0 = - q$ .
${f, g} = {q, p} = 1$ . Now the three outer brackets:
${f, {g, h}} = {q, - p} = - {q, p} = - 1$ .
${g, {h, f}} = {p, - q} = - {p, q} = - (- 1) = + 1$ .
${h, {f, g}} = {q p, 1} = 0$ (bracket with a constant is $0$ ). Sum: $- 1 + 1 + 0 = 0. ✓$ Jacobi holds. This is not a coincidence — it holds for all $f, g, h$ , and it is what makes Poisson's theorem (Problem 4.1) and the canonical structure work.

Recall Self-test checklist (reveal after you finish)

Ordering ::: ${p, q} = - {q, p}$ — always flip the sign on swap. Time law ::: $\frac{df}{d t} = {f, H} + \frac{\partial f}{\partial t}$ — never drop the last term. Conservation ::: $f$ conserved $⟺ \frac{df}{d t} = 0$ , which needs both terms to cancel. Angular momentum ::: ${L_{i}, L_{j}} = ϵ_{ij k} L_{k}$ (cyclic). Quantization ::: ${f, g} \to \frac{1}{i ℏ} [\hat{f}, \overset{g}{^}]$ .

Connections

Hamiltonian Mechanics — Problems 2.2, 2.3, 5.2 reproduce Hamilton's / Newton's equations.
Noether's Theorem & Conservation Laws — Problem 3.1 is Noether via brackets.
Angular Momentum Algebra — Problems 3.3, 4.1–4.3 build the $S O (3)$ algebra.
Commutators in Quantum Mechanics — Problem 5.1 crosses the Dirac bridge.
Canonical Transformations — the bracket structure Problem 5.3 protects.
Liouville's Theorem — same phase-space flow underlying every problem here.