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

2.1.15 · D4 · Physics › Analytical Mechanics › Poisson brackets — definition, properties, connection to com

Intuition Is page ko kaise use karein

Har problem ko pehle khud solve karo, uske baad hi solution kholna. Problems ek ladder ki tarah chad'te hain: L1 Recognition (kya tum is machine ko padh sakte ho?) → L2 Application (kya tum crank ghuma sakte ho?) → L3 Analysis (kya tum dekh sakte ho kyun ek bracket zero hota hai ya nahi hota?) → L4 Synthesis (kya tum results combine karke nayi physics bana sakte ho?) → L5 Mastery (kya tum invent aur defend kar sakte ho?). Yahan sab kuch parent note (the topic note) ki ek definition par tika hai: ${f, g} \equiv \sum_{i = 1}^{n} (\frac{\partial f}{\partial q _{i}} \frac{\partial g}{\partial p _{i}} - \frac{\partial f}{\partial p _{i}} \frac{\partial g}{\partial q _{i}}) .$ Ise aise padho: " $f$ ka $q$ ke saath slope, times $g$ ka $p$ ke saath slope, minus wahi cheez jisme $q$ aur $p$ ke roles flip ho jaate hain." Neeche har solution bas yahi sum hai, carefully track kiya gaya.

Shuru karne se pehle, poori machine ki ek picture — taaki tum hamesha jaano ki tum kaun sa knob ghuma rahe ho.

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

Level 1 — Recognition

Tumse sirf itna kaha ja raha hai ki definition ko sahi se padho: sahi partial derivatives chuno, unhe sahi jagah daalo, aur minus sign ka dhyan rakho.

Problem 1.1

Ek single coordinate pair $(q, p)$ ke saath definition use karke, ${q, p}$ aur ${p, q}$ compute karo.

Recall Solution 1.1

Yahan $f$ aur $g$ sirf coordinates hain, isliye unke partial derivatives $0$ ya $1$ hain. ${q, p}$ ke liye: $f = q, g = p$ rakho.

$\frac{\partial f}{\partial q} = \frac{\partial q}{\partial q} = 1$ , $\frac{\partial g}{\partial p} = \frac{\partial p}{\partial p} = 1$ .
$\frac{\partial f}{\partial p} = 0$ , $\frac{\partial g}{\partial q} = 0$ . ${q, p} = (1) (1) - (0) (0) = 1.$ ${p, q}$ ke liye: roles swap karo. ${p, q} = \frac{\partial p}{\partial q} \frac{\partial q}{\partial p} - \frac{\partial p}{\partial p} \frac{\partial q}{\partial q} = (0) (0) - (1) (1) = - 1.$ Toh ${q, p} = 1$ aur ${p, q} = - 1$ . Minus sign decoration nahi hai — yeh antisymmetry ka saakar roop hai.

Problem 1.2

Bina compute kiye, sirf antisymmetry property se bolo ki inme se kaun se $0$ hain: ${f, f}$ , ${H, H}$ , ${L_{z}, L_{z}}$ .

Recall Solution 1.2

Antisymmetry kehti hai ${f, g} = - {g, f}$ . $g = f$ rakho: ${f, f} = - {f, f}$ , toh $2 {f, f} = 0$ , yaani ${f, f} = 0$ . Yahi argument $f$ ke baare mein kuch bhi use nahi karta. Toh teeno zero hain: ${f, f} = {H, H} = {L_{z}, L_{z}} = 0$ .

Problem 1.3

Constants $a, b$ aur bilinearity rule se, ${3 q + 5 p, p}$ simplify karo.

Recall Solution 1.3

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

Level 2 — Application

Ab real functions par crank ghumao. Same definition, bas zyada partial derivatives track karne hain.

Problem 2.1

Ek degree of freedom ke liye, ${q^{2}, p}$ aur ${q, p^{2}}$ compute karo.

Recall Solution 2.1

${q^{2}, p}$ : $f = q^{2}$ , toh $\frac{\partial f}{\partial q} = 2 q$ , $\frac{\partial f}{\partial p} = 0$ . Aur $g = p$ , toh $\frac{\partial g}{\partial p} = 1$ , $\frac{\partial g}{\partial q} = 0$ . ${q^{2}, p} = (2 q) (1) - (0) (0) = 2 q .$ ${q, p^{2}}$ : $f = q$ deta hai $\frac{\partial f}{\partial q} = 1, \frac{\partial f}{\partial p} = 0$ ; $g = p^{2}$ deta hai $\frac{\partial g}{\partial p} = 2 p, \frac{\partial g}{\partial q} = 0$ . ${q, p^{2}} = (1) (2 p) - (0) (0) = 2 p .$ Leibniz se sanity check: ${q^{2}, p} = {q \cdot q, p} = q {q, p} + {q, p} q = q + q = 2 q$ . ✓

Problem 2.2

Mass $m$ ka particle ek potential mein: $H = \frac{p ^{2}}{2 m} + V (q)$ . ${q, H}$ aur ${p, H}$ compute karo, aur identify karo ki yeh kaun si physical equations reproduce karte hain.

Recall Solution 2.2

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

Problem 2.3

$H = \frac{p ^{2}}{2 m}$ (free particle) aur $f = q p$ ke saath, ${f, H}$ compute karo aur interpret karo.

Recall Solution 2.3

$f = q p$ : $\frac{\partial f}{\partial q} = p$ , $\frac{\partial f}{\partial p} = q$ . $H = \frac{p ^{2}}{2 m}$ : $\frac{\partial H}{\partial p} = \frac{p}{m}$ , $\frac{\partial H}{\partial q} = 0$ . ${q p, H} = (p) \frac{p}{m} - (q) (0) = \frac{p ^{2}}{m} .$ Kyunki $\frac{d ( q p )}{d t} = {q p, H} = \frac{p ^{2}}{m} = 2 \cdot \frac{p ^{2}}{2 m} = 2 H$ , quantity $q p$ steadily badhti hai — yeh conserved nahi hai, aur iska growth rate kinetic energy ka double hai. (Yeh virial theorem ka classical seed hai.)

Level 3 — Analysis

Ab sawaal yeh hai ki kyun: kaun se brackets vanish hote hain, aur yeh conservation ke baare mein kya batate hain.

Problem 3.1

$H = \frac{p _{x}^{2} + p _{y}^{2}}{2 m} + V (r)$ ke liye jahan $r = x^{2} + y^{2}$ , dikhao ki $L_{z} = x p_{y} - y p_{x}$ conserved hai, yaani ${L_{z}, H} = 0$ . Geometrically explain karo ki yeh kyun vanish karna chahiye.

Recall Solution 3.1

Humein ${L_{z}, H}$ chahiye. $H = T + V$ mein split karo jahan $T = \frac{p _{x}^{2} + p _{y}^{2}}{2 m}$ . Kinetic part. $L_{z} = x p_{y} - y p_{x}$ ke liye $\frac{\partial L _{z}}{\partial x} = p_{y}$ , $\frac{\partial L _{z}}{\partial y} = - p_{x}$ , $\frac{\partial L _{z}}{\partial p _{x}} = - y$ , $\frac{\partial L _{z}}{\partial p _{y}} = x$ . Aur $T$ ke liye $\frac{\partial T}{\partial p _{x}} = \frac{p _{x}}{m}$ , $\frac{\partial T}{\partial p _{y}} = \frac{p _{y}}{m}$ , $\frac{\partial T}{\partial x} = \frac{\partial T}{\partial y} = 0$ . ${L_{z}, T} = \sum_{i = x, y} (\frac{\partial L _{z}}{\partial i} \frac{\partial T}{\partial p _{i}} - \frac{\partial L _{z}}{\partial p _{i}} \frac{\partial T}{\partial i}) = p_{y} \frac{p _{x}}{m} + (- p_{x}) \frac{p _{y}}{m} - 0 - 0 = 0.$ Potential part. $V$ sirf $r$ par depend karta hai, toh $\frac{\partial V}{\partial x} = V^{'} (r) \frac{x}{r}$ , $\frac{\partial V}{\partial y} = V^{'} (r) \frac{y}{r}$ , aur $\frac{\partial V}{\partial p _{x}} = \frac{\partial V}{\partial p _{y}} = 0$ . ${L_{z}, V} = - \frac{\partial L _{z}}{\partial p _{x}} \frac{\partial V}{\partial x} - \frac{\partial L _{z}}{\partial p _{y}} \frac{\partial V}{\partial y} = - (- y) V^{'} \frac{x}{r} - (x) V^{'} \frac{y}{r} = \frac{V ^{'}}{r} (x y - x y) = 0.$ Toh ${L_{z}, H} = 0$ : angular momentum conserved hai. Geometric kyun: ek central potential $V (r)$ $z$ -axis ke ird-gird kisi bhi rotation ke baad identical lagta hai. $L_{z}$ precisely unhi rotations ka generator hai. Ek quantity jo $H$ ki symmetry generate karti hai woh $H$ ke saath commute karti hai — yeh Noether's theorem ka Poisson-bracket roop hai. Figure dekho: potential ki circular symmetry hi bracket ko zero karti hai.

Problem 3.2

$H = \frac{p ^{2}}{2 m}$ aur explicitly time-dependent quantity $G = q - \frac{p}{m} t$ ke liye, dikhao ki $G$ conserved hai — full equation of motion use karke.

Recall Solution 3.2

Master law hai $\frac{d G}{d t} = {G, H} + \frac{\partial G}{\partial t}$ . Last term mat bhoolna. Bracket part. $G = q - \frac{p}{m} t$ : $\frac{\partial G}{\partial q} = 1$ , $\frac{\partial G}{\partial p} = - \frac{t}{m}$ . Aur $\frac{\partial H}{\partial p} = \frac{p}{m}$ , $\frac{\partial H}{\partial q} = 0$ . ${G, H} = (1) \frac{p}{m} - (- \frac{t}{m}) (0) = \frac{p}{m} .$ Explicit-time part. $\frac{\partial G}{\partial t} = - \frac{p}{m}$ ( $q, p$ ko fixed maanke differentiate karo). $\frac{d G}{d t} = \frac{p}{m} + (- \frac{p}{m}) = 0.$ $G$ conserved hai. Physically $G$ initial position hai: $q (t) - \frac{p}{m} t = q (0)$ free motion ke liye. Conservation tabhi dikhti hai jab dono terms rakhe jaate hain.

Problem 3.3

Dikhao ki ${L_{z}, p_{x}} = p_{y}$ aur ${L_{z}, x} = - y$ , phir pattern explain karo: $L_{z}$ $x y$ -plane mein vectors ko rotate karta hai.

Recall Solution 3.3

$L_{z} = x p_{y} - y p_{x}$ . ${L_{z}, p_{x}}$ : $\frac{\partial L _{z}}{\partial x} = p_{y}, \frac{\partial p _{x}}{\partial p _{x}} = 1$ ; baaki 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: $L_{z}$ ke saath brackets $\frac{d}{d ϕ}$ ki tarah act karte hain, yaani ek infinitesimal rotation. ${L_{z}, x} = - y$ aur ${L_{z}, y} = x$ exactly $(\overset{x}{˙} \overset{y}{˙}) = (01 - 1 0) (x y)$ hain, rotation generator. Isliye $L_{z}$ ko generator of rotations kaha jaata hai, jo Angular Momentum Algebra se jodta hai.

Level 4 — Synthesis

Properties combine karo (Leibniz, Jacobi, bilinearity) aur aisi results banao jo brute force se nahi aati.

Problem 4.1 (Poisson's theorem in action)

Given ki $L_{x}$ aur $L_{y}$ dono conserved hain ( ${L_{x}, H} = {L_{y}, H} = 0$ ), $H$ ki explicit form chhuye bina prove karo ki $L_{z}$ conserved hai — ${L_{x}, L_{y}} = L_{z}$ aur Jacobi identity use karke.

Recall Solution 4.1

Jacobi $f = L_{x}, g = L_{y}, h = H$ ke saath: ${L_{x}, {L_{y}, H}} + {L_{y}, {H, L_{x}}} + {H, {L_{x}, L_{y}}} = 0.$ Pehle do brackets mein ${L_{y}, H} = 0$ aur ${H, L_{x}} = - {L_{x}, H} = 0$ hain, toh woh vanish ho jaate hain: ${H, {L_{x}, L_{y}}} = 0.$ Lekin ${L_{x}, L_{y}} = L_{z}$ , toh ${H, L_{z}} = 0$ , yaani ${L_{z}, H} = 0$ . $L_{z}$ conserved hai. Punchline: agar angular momentum ke do components conserved hain, teesra force ho kar conserved ban jaata hai — tumne sirf algebra se ek naya conservation law manufacture kar liya. Yahi Poisson's theorem hai.

Problem 4.2 (Leibniz se ek hard bracket shortcut karo)

${x^{2} + y^{2}, L_{z}}$ compute karo — product/Leibniz aur bilinearity rules use karke — given ${x, L_{z}} = y$ , ${y, L_{z}} = - x$ (note: yeh Problem 3.3 ke antisymmetry se $-$ hain).

Recall Solution 4.2

Pehle given brackets antisymmetry se check karo: 3.3 se, ${L_{z}, x} = - y \Rightarrow {x, L_{z}} = y$ ; aur ${L_{z}, y} = x \Rightarrow {y, L_{z}} = - x$ . ✓ Leibniz ${f g, h} = f {g, h} + {f, h} g$ har square par use karo: ${x^{2}, L_{z}} = x {x, L_{z}} + {x, L_{z}} x = 2 x {x, L_{z}} = 2 x y,$ ${y^{2}, L_{z}} = 2 y {y, L_{z}} = 2 y (- x) = - 2 x y .$ Bilinearity se: ${x^{2} + y^{2}, L_{z}} = 2 x y - 2 x y = 0.$ Interpretation: $x^{2} + y^{2} = r^{2}$ rotationally invariant hai, toh rotation generator $L_{z}$ ke saath uska bracket zero hona hi chahiye. Algebra geometry ko confirm karta hai.

Problem 4.3 (Poori angular-momentum algebra build karo)

Given ${L_{x}, L_{y}} = L_{z}$ , antisymmetry aur cyclic relabelling se teeno brackets ${L_{x}, L_{y}}$ , ${L_{y}, L_{z}}$ , ${L_{z}, L_{x}}$ likho, phir inme se ek ko directly components se verify karo.

Recall Solution 4.3

Parent note ka compact rule hai ${L_{i}, L_{j}} = ϵ_{ij k} L_{k}$ . Cyclically likha gaya: ${L_{x}, L_{y}} = L_{z}, {L_{y}, L_{z}} = L_{x}, {L_{z}, L_{x}} = L_{y} .$ ${L_{y}, L_{z}} = L_{x}$ ka direct check. $L_{y} = z p_{x} - x p_{z}$ , $L_{z} = x p_{y} - y p_{x}$ ke saath: ${L_{y}, L_{z}} = \sum_{i} (\frac{\partial L _{y}}{\partial q _{i}} \frac{\partial L _{z}}{\partial p _{i}} - \frac{\partial L _{y}}{\partial p _{i}} \frac{\partial L _{z}}{\partial q _{i}}) .$ 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: $(p_{x}) (0) - (- x) (0) = 0$ . Ruko — $p_{x}$ / $x$ pairing carefully recompute karo. Saare surviving cross terms collect karo: $\frac{\partial L _{y}}{\partial x} \frac{\partial L _{z}}{\partial p _{x}} - \frac{\partial L _{y}}{\partial p _{x}} \frac{\partial L _{z}}{\partial x} = (- p_{z}) (- y) - (z) (p_{y}) = y p_{z} - z p_{y},$ aur baaki saare index pairs $0$ dete hain. Toh ${L_{y}, L_{z}} = y p_{z} - z p_{y} = L_{x} . ✓$ Yeh closed algebra classical $S O (3)$ hai, jiska quantum image ( $\frac{1}{i ℏ}$ se multiply karke) $[\hat{L}_{i}, \hat{L}_{j}] = i ℏ ϵ_{ij k} \hat{L}_{k}$ hai — dekho Commutators in Quantum Mechanics.

Level 5 — Mastery

Invent karo, generalize karo, aur defend karo. Yeh structure samajhne ko reward karte hain, sirf crank-turning ko nahi.

Problem 5.1 (Quantization consistency)

Classically ${q, p} = 1$ aur ${q, p^{2}} = 2 p$ (Problem 2.1). Dikhao ki yeh Dirac's rule ${f, g} \to \frac{1}{i ℏ} [\hat{f}, \overset{g}{^}]$ ke under quantum images ke saath consistent hain — yaani verify karo $[\overset{q}{^}, \overset{p}{^}] = i ℏ$ aur $[\overset{q}{^}, \overset{p}{^}^{2}] = 2 i ℏ \overset{p}{^}$ .

Recall Solution 5.1

Dirac's rule: $C$ ke barabar classical bracket ek commutator $i ℏ C$ mein map hota hai. ${q, p} = 1$ se: predicted $[\overset{q}{^}, \overset{p}{^}] = i ℏ \cdot 1 = i ℏ$ . ✓ (canonical commutation relation). ${q, p^{2}} = 2 p$ se: predicted $[\overset{q}{^}, \overset{p}{^}^{2}] = i ℏ \cdot 2 \overset{p}{^} = 2 i ℏ \overset{p}{^}$ . Independent quantum check $[\hat{A}, \hat{B} \hat{C}] = [\hat{A}, \hat{B}] \hat{C} + \hat{B} [\hat{A}, \hat{C}]$ (commutator Leibniz rule) use karke: $[\overset{q}{^}, \overset{p}{^}^{2}] = [\overset{q}{^}, \overset{p}{^}] \overset{p}{^} + \overset{p}{^} [\overset{q}{^}, \overset{p}{^}] = i ℏ \overset{p}{^} + \overset{p}{^} i ℏ = 2 i ℏ \overset{p}{^} . ✓$ Classical Leibniz rule ( ${q, p^{2}} = 2 p {q, p} = 2 p$ ) aur quantum Leibniz rule dono same shape dete hain — yahi structural identity hai jo Dirac's map ko kaam karne deti hai.

Problem 5.2 (Ek conserved quantity design karo)

Constant force wale particle ke liye, $H = \frac{p ^{2}}{2 m} - F q$ (toh $F$ force hai). Ek time-dependent quantity $G (q, p, t)$ dhundo jo conserved ho, aur ise prove karo.

Recall Solution 5.2

Physics se guess: constant force ke under, $p (t) = p_{0} + F t$ , toh $p - F t$ constant hona chahiye. Maano $G = p - F t$ . $\frac{d G}{d t} = {G, H} + \frac{\partial G}{\partial t} .$ Bracket: $G = p - F t$ ke liye $\frac{\partial G}{\partial q} = 0$ , $\frac{\partial G}{\partial p} = 1$ . Aur $H$ ke liye $\frac{\partial H}{\partial p} = \frac{p}{m}$ , $\frac{\partial H}{\partial q} = - F$ . ${G, H} = (0) \frac{p}{m} - (1) (- F) = F .$ Explicit time: $\frac{\partial G}{\partial t} = - F$ . Toh $\frac{d G}{d t} = F + (- F) = 0.$ $G = p - F t$ conserved hai: yeh initial momentum $p_{0}$ hai. Yeh wahi classic "explicit-time term zaroori hai" case hai jo parent note mein flag kiya gaya tha.

Problem 5.3 (Jacobi poori edifice ki raksha karta hai)

Jacobi identity ${f, {g, h}} + {g, {h, f}} + {h, {f, g}} = 0$ bracket ke liye ek legitimate Lie algebra hone ke liye zaroori hai. Ise explicitly verify karo $f = q, g = p, h = q p$ ke liye, ek degree of freedom mein.

Recall Solution 5.3

Pehle teeno inner brackets compute karo.

${g, h} = {p, q p}$ . Leibniz se: ${p, q p} = q {p, p} + {p, q} p = 0 + (- 1) p = - p$ .
${h, f} = {q p, q}$ . Leibniz: ${q p, q} = q {p, q} + {q, q} p = q (- 1) + 0 = - q$ .
${f, g} = {q, p} = 1$ . Ab teeno outer brackets:
${f, {g, h}} = {q, - p} = - {q, p} = - 1$ .
${g, {h, f}} = {p, - q} = - {p, q} = - (- 1) = + 1$ .
${h, {f, g}} = {q p, 1} = 0$ (constant ke saath bracket $0$ hota hai). Sum: $- 1 + 1 + 0 = 0. ✓$ Jacobi holds. Yeh coincidence nahi hai — yeh saare $f, g, h$ ke liye hold karta hai, aur yahi Poisson's theorem (Problem 4.1) aur canonical structure ko kaam karne deta hai.

Recall Self-test checklist (khatam karne ke baad reveal karo)

Ordering ::: ${p, q} = - {q, p}$ — swap par hamesha sign flip karo. Time law ::: $\frac{df}{d t} = {f, H} + \frac{\partial f}{\partial t}$ — last term kabhi mat chhodna. Conservation ::: $f$ conserved $⟺ \frac{df}{d t} = 0$ , jiske liye dono terms ka cancel hona zaroori hai. 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 Hamilton's / Newton's equations reproduce karte hain.
Noether's Theorem & Conservation Laws — Problem 3.1 brackets ke zariye Noether hai.
Angular Momentum Algebra — Problems 3.3, 4.1–4.3 $S O (3)$ algebra build karte hain.
Commutators in Quantum Mechanics — Problem 5.1 Dirac bridge cross karta hai.
Canonical Transformations — bracket structure jo Problem 5.3 protect karta hai.
Liouville's Theorem — same phase-space flow jo yahan har problem ke neeche hai.