Hamiltonian — definition H = Σpᵢq̇ᵢ − L: Worked examples

2.1.11 · D3 · Physics › Analytical Mechanics › Hamiltonian — definition H = Σpᵢq̇ᵢ − L

Intuition Ye page kya hai

Parent note ne tumhe recipe dikhaayi thi: momentum $p = \partial L / \partial \overset{q}{˙}$ nikalо, usse invert karo, aur $H = \sum p_{i} \overset{q}{˙}_{i} - L$ assemble karo. Ye page us recipe par har tarah ke system던ta hai taaki tum koi bhi aisa case kabhi na dekho jо tumne pehle solve na kiya ho. Hum scenarios ki ek matrix se guzarte hain — clean energy cases, wo cases jahan $H \neq = E$ , wo cases jahan $H$ conserved bhi nahi hota, multi-coordinate systems, ek magnetic (velocity-dependent) potential, aur ek exam twist.

Shuru karne se pehle: yaad karo wo do sawaal jo hum hamesha ek Hamiltonian ke baare mein poochhte hain.

Kya $H$ conserved hai? Haan bilkul tab jab $\partial L / \partial t = 0$ ho (Lagrangian mein koi naked $t$ na ho).
Kya $H$ energy $T + V$ hai? Haan sirf tab jab $T$ velocities mein ek pure quadratic ho (neeche refresher mein explain kiya gaya hai) aur $V$ mein koi velocity na ho. Ye do sawaal independent hain — yahi is page ka poora drama hai.

Recall Do refreshers jo tumhe chahiye honge (taaki ye page akele kaam kare)

Euler's theorem, ek line mein. Agar $T$ sirf $\overset{q}{˙}_{j} \overset{q}{˙}_{k}$ jaise terms se bana ho (har term mein exactly do velocity factors hain), tо differentiate karke sum karne par $2 T$ wapas milta hai: $\sum_{i} \overset{q}{˙}_{i} \frac{\partial T}{\partial q ˙ _{i}} = 2 T$ . Socho: do velocity factors mein se har ek "ek baar count hota hai", tо $T$ do baar wapas milta hai. Exactly isliye $\sum p_{i} \overset{q}{˙}_{i} = 2 T$ aur isliye $H = 2 T - (T - V) = T + V$ — lekin sirf tab jab $T$ ke har term mein exactly do velocity factors hon. Jacobi integral, ek line mein. Jab $L$ mein explicit $t$ nahi hota lekin $T$ pure velocity-quadratic nahi hota (jaise ek moving constraint velocity-free piece add kare), tо conserved quantity $H = \sum p_{i} \overset{q}{˙}_{i} - L$ phir bhi exist karta hai lekin energy nahi hoti. Us conserved-lekin-not-energy object ka ek naam hai: Jacobi integral. Isse tum cell C4 mein miloge.

The scenario matrix

Neeche har row ek alag "class" of behaviour hai. Aakhiri do columns upar ke do independent sawaal hain. C8 exam twist hai — iske sach-sachy jawaab hain $H$ conserved: NO, $H = E$ : YES (poora worked out neeche; pahunche se pehle predict karne ki koshish karo).

Cell	Scenario class	Example jo isse hit karta hai	$H$ conserved?	$H = E$ ?
C1	Textbook 1D, $H = E =$ const	(a) free-fall particle	✓	✓
C2	2 coordinates, ek cyclic	(b) 2D central motion, polar	✓	✓
C3	$L$ mein explicit time	(c) driven oscillator	✗	✓ (value drift karti hai)
C4	Moving constraint, $H \neq = E$ lekin conserved	(d) bead on rotating wire	✓	✗
C5	Velocity-dependent potential (magnetic)	(e) charge in magnetic field	✓	✓ (subtle)
C6	Degenerate / zero input	(f) free particle, $V = 0$ , $p = 0$ limit	✓	✓ (trivial)
C7	Sign / direction cases	(g) particle thrown up vs down	✓	✓
C8	Exam twist: time-scaled potential	(h) growing stiffness	✗	✓

Har cell neeche worked out hai aur labelled hai. Jo prerequisites hum use karte hain: Lagrangian Mechanics, Legendre Transform, Hamilton's Canonical Equations, Conservation Laws & Noether's Theorem, aur pictures ke liye Phase Space.

C1 — Textbook case: 1D free-fall

Worked example (a) Mass

$m$ ka ek particle gravity ke under gir raha hai, height $y$ . $L = \frac{1}{2} m \overset{y}{˙}^{2} - m g y$ . Forecast: aage padhne se pehle $H$ guess karo. Ye $\frac{p ^{2}}{2 m} + m g y$ hona chahiye, aur ye dono conserved aur energy ke equal hona chahiye. Kyun? Kyunki explicit $t$ nahi hai, aur $T = \frac{1}{2} m \overset{y}{˙}^{2}$ ek clean quadratic hai.

Momentum: $p = \frac{\partial L}{\partial y ˙} = m \overset{y}{˙}$ . Ye step kyun? Legendre transform velocity ko uski "slope" $p = \partial L / \partial \overset{y}{˙}$ se swap karta hai; pehle wo slope dhundni zaroori hai.
Invert: $\overset{y}{˙} = p / m$ . Ye step kyun? $H$ sirf $(y, p)$ ka function hona chahiye — har $\overset{y}{˙}$ ko bahar karna hoga.
Assemble: $H = p \overset{y}{˙} - L = p \cdot \frac{p}{m} - (\frac{1}{2} m (p / m)^{2} - m g y) = \frac{p ^{2}}{m} - \frac{p ^{2}}{2 m} + m g y$ . Ye step kyun? Ye definition $H = \sum p_{i} \overset{q}{˙}_{i} - L$ hai jisme inverted velocity plug ki gayi hai. $H = \frac{p ^{2}}{2 m} + m g y = T + V = E .$ Verify: Hamilton's equations ko free-fall reproduce karni chahiye. $\overset{y}{˙} = \partial H / \partial p = p / m$ ✓ (step 1 se match). $\overset{p}{˙} = - \partial H / \partial y = - m g$ , yaani $m \overset{y}{¨} = - m g$ — Newton's law. ✓ Numerically $m = 2 kg$ , $\overset{y}{˙} = 3 m/s$ , $y = 5 m$ , $g = 9.8$ ke liye: $p = 6$ , $H = 36/4 + 2 \cdot 9.8 \cdot 5 = 9 + 98 = 107 J$ .

C2 — Do coordinates, ek cyclic: 2D central motion

Worked example (b) Mass

$m$ ka ek particle central potential $V (r)$ mein, polar coordinates $(r, θ)$ . $L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2}) - V (r)$ . Forecast: ab do momenta hain. $θ$ kabhi $L$ mein appear nahi karta (sirf $\dot{θ}$ karta hai) — tо $θ$ cyclic hai aur iska momentum $p_{θ}$ (angular momentum) conserved hai. $H$ ko $T + V$ hona chahiye.

Momenta: $p_{r} = \frac{\partial L}{\partial r ˙} = m \overset{r}{˙}$ , $p_{θ} = \frac{\partial L}{\partial θ ˙} = m r^{2} \dot{θ}$ . Ye step kyun? Har coordinate ko apna conjugate momentum milta hai; dono slopes chahiye.
Invert: $\overset{r}{˙} = p_{r} / m$ , $\dot{θ} = p_{θ} / (m r^{2})$ . Ye step kyun? Do velocities, do evictions.
Assemble: $H = p_{r} \overset{r}{˙} + p_{θ} \dot{θ} - L$ . $H = \frac{p _{r}^{2}}{m} + \frac{p _{θ}^{2}}{m r ^{2}} - (\frac{p _{r}^{2}}{2 m} + \frac{p _{θ}^{2}}{2 m r ^{2}} - V (r)) = \frac{p _{r}^{2}}{2 m} + \frac{p _{θ}^{2}}{2 m r ^{2}} + V (r) .$ Ye step kyun? Same Legendre assembly; $\dot{θ}$ term neatly fold ho jaata hai kyunki $T$ quadratic hai. $H = \frac{p _{r}^{2}}{2 m} + \frac{p _{θ}^{2}}{2 m r ^{2}} + V (r) = E .$ Verify: Kyunki $θ$ $H$ mein absent hai, $\overset{p}{˙}_{θ} = - \partial H / \partial θ = 0$ — angular momentum conserved, jaisa Conservation Laws & Noether's Theorem rotational symmetry ke liye predict karta hai. ✓ Numerically $r = 2$ , $p_{r} = 1$ , $p_{θ} = 4$ , $m = 1$ , $V = 0$ par: $H = 1/2 + 16/ (2 \cdot 4) = 0.5 + 2 = 2.5$ .

Extra $\frac{p _{θ}^{2}}{2 m r ^{2}}$ term centrifugal barrier hai. Neeche ka figure aise padho: blue curve raw attractive potential $V (r) = - k / r$ hai (ek well jo andar ki taraf kheenchti hai), pink curve centrifugal term $p_{θ}^{2} / (2 m r^{2})$ hai (ek wall jo $r = 0$ ke paas upar shoot karti hai, kyunki $r^{2}$ se divide karne par blow up hota hai), aur yellow curve unka sum $V_{eff}$ hai. Dhyan do yellow curve ek finite radius par ek minimum tak neeche jaati hai — wo dip ek stable circular orbit hai, wo radius jahan inward pull aur outward barrier balance karte hain. Barrier hi wajah hai ki nonzero $p_{θ}$ waala central-force particle kabhi centre mein nahi girta.

Figure — Hamiltonian — definition H = Σpᵢq̇ᵢ − L

C3 — $L$ mein explicit time: driven oscillator

Worked example (c) Ek 1D oscillator jo external force

$F (t) = F_{0} cos (Ω t)$ se push ho raha hai. $L = \frac{1}{2} m \overset{x}{˙}^{2} - \frac{1}{2} k x^{2} + F_{0} cos (Ω t) x$ . Forecast: term $F_{0} cos (Ω t) x$ mein explicitly $t$ hai, tо $\partial L / \partial t \neq = 0$ — expect karo $H$ conserved nahi. Lekin $T$ phir bhi clean quadratic hai aur $V$ mein koi velocity nahi, tо $H$ ko phir bhi numerically har instant par $T + V$ ke equal hona chahiye.

Momentum: $p = m \overset{x}{˙}$ , invert $\overset{x}{˙} = p / m$ . Ye step kyun? Driving term mein $\overset{x}{˙}$ nahi hai, tо momentum touch nahi hota.
Assemble: $H = p \overset{x}{˙} - L = \frac{p ^{2}}{2 m} + \frac{1}{2} k x^{2} - F_{0} cos (Ω t) x$ . Ye step kyun? Standard assembly; note karo drive minus sign ke saath survive karti hai ( $L$ mein $+ F_{0} cos (Ω t) x$ tha, tо $- L$ ise $- F_{0} cos (Ω t) x$ kar deta hai). $H = \frac{p ^{2}}{2 m} + \frac{1}{2} k x^{2} - F_{0} cos (Ω t) x .$
Time behaviour: $\frac{d H}{d t} = \frac{\partial H}{\partial t} = + F_{0} Ω sin (Ω t) x \neq = 0$ . Ye step kyun? Motion ke saath $H$ ka total time-derivative $\partial H / \partial t$ ke equal hota hai (Hamilton's equations se $q, p$ parts cancel ho jaate hain), aur yahan wo partial nonzero hai. Verify: Kya $H = E$ instant-by-instant hai? $E = T + V = \frac{p ^{2}}{2 m} + \frac{1}{2} k x^{2} - F_{0} cos (Ω t) x$ — haan, identical, kyunki $V$ ab drive term ko include karta hai. Tо $H = E$ lekin dono conserved nahi: energy andar bahar pump hoti hai. Numeric snapshot: $m = 1, k = 4, F_{0} = 3, Ω = 2$ , $t = 0$ par tо $cos = 1, sin = 0$ , $x = 1, p = 2$ : $H = 2 + 2 - 3 = 1$ ; aur $d H / d t = 3 \cdot 2 \cdot 0 \cdot 1 = 0$ is instant par. ✓

C4 — Moving constraint: $H \neq = E$ lekin phir bhi conserved

Worked example (d) Mass

$m$ ka bead ek wire par jo fixed angular rate $ω$ se spin ho raha hai; radial coord $r$ . Constraint $ϕ = ω t$ time-dependent hai. $T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} ω^{2})$ , $V = 0$ : $L = \frac{1}{2} m \overset{r}{˙}^{2} + \frac{1}{2} m r^{2} ω^{2} .$ Forecast: khatre ki ghanti. $T$ mein ek piece $\frac{1}{2} m r^{2} ω^{2}$ hai jo koi $\overset{r}{˙}$ carry nahi karta — ye velocity mein pure quadratic nahi hai. Tо Euler's theorem ka " $\sum p \overset{q}{˙} = 2 T$ " fail hoga aur hame $H \neq = E$ milna chahiye. Lekin $L$ mein explicit $t$ nahi hai ( $ω t$ substitute ho gaya tha), tо $H$ phir bhi conserved hai.

Momentum: $p_{r} = \partial L / \partial \overset{r}{˙} = m \overset{r}{˙}$ , invert $\overset{r}{˙} = p_{r} / m$ . Ye step kyun? Sirf $\overset{r}{˙}$ appear karta hai; $ω$ term velocity-free hai.
Assemble: $H = p_{r} \overset{r}{˙} - L = \frac{p _{r}^{2}}{m} - (\frac{p _{r}^{2}}{2 m} + \frac{1}{2} m r^{2} ω^{2}) = \frac{p _{r}^{2}}{2 m} - \frac{1}{2} m r^{2} ω^{2}$ . Ye step kyun? Velocity-free term Legendre swap se untouched rehta hai, tо ye $H$ mein $- L$ ke sign ke saath survive karta hai: minus. $H = \frac{p _{r}^{2}}{2 m} - \frac{1}{2} m r^{2} ω^{2} .$
Energy se compare: $E = T + V = \frac{p _{r}^{2}}{2 m} + \frac{1}{2} m r^{2} ω^{2}$ . Tо $H = E - m r^{2} ω^{2} \neq = E$ . Ye step kyun? Rotating wire constraint ke through energy feed karta hai; $H$ ek alag conserved bookkeeping quantity hai — upar wale refresher ka Jacobi integral. Verify: conserved? $\partial L / \partial t = 0$ ✓, tо $d H / d t = 0$ . Energy nahi? $r = 1, ω = 2, m = 1, p_{r} = 3$ par: $H = 9/2 - 2 = 2.5$ , jabki $E = 9/2 + 2 = 6.5$ . Different. ✓

C5 — Velocity-dependent potential: charge in a magnetic field

Worked example (e) Charge

$e$ , mass $m$ , uniform field $B = B \overset{z}{^}$ mein, vector potential $A = \frac{1}{2} B \times r$ use karte hue. Magnetic Lagrangian hai $L = \frac{1}{2} m ∣ \dot{r} ∣^{2} + e \dot{r} \cdot A$ . Hum poora vector throughout carry karte hain — vectors bas saath chalte hain, aur end mein numbers sanity-check karne ke liye hum ek scalar component par aayenge. Forecast: yahan "potential" term $e \dot{r} \cdot A$ mein velocity hai, tо naive " $H = T + V$ " argument void hai. Surprise ye hai: kyunki wo term $\dot{r}$ mein linear hai, wo $H$ se cancel ho jaata hai aur hame phir bhi $H =$ kinetic energy milti hai.

Momentum: $p = \frac{\partial L}{\partial r ˙} = m \dot{r} + e A$ . Ye canonical momentum hai, $m v$ nahi. Ye step kyun? Linear-in-velocity term slope ko shift karta hai; $e A$ bhoolna classic magnetic mistake hai.
Invert: $\dot{r} = (p - e A) / m$ . Ye step kyun? Velocity bahar karo; note karo combination $p - e A = m v$ kinetic momentum hai.
Assemble: $H = p \cdot \dot{r} - L$ . Substitute karke aur simplify karne par (linear terms cancel ho jaate hain): $H = \frac{( p - e A ) ^{2}}{2 m} .$ Ye step kyun? $p \cdot \dot{r} = \frac{1}{m} p \cdot (p - e A)$ aur $L = \frac{( p - e A ) ^{2}}{2 m} + \frac{e}{m} A \cdot (p - e A)$ ; subtract karne par cross terms khatam ho jaate hain. Verify: $H = \frac{1}{2} m ∣ v ∣^{2} = T$ — pure kinetic energy, kyunki magnetic force koi kaam nahi karta. Tо $H = E$ sach hai, lekin sirf canonical momentum sahi use karne ke baad. Numeric scalar check (ek component, $A$ matlab $A$ ka relevant component): lo $m = 1, e = 1, A = 2, p = 5$ : kinetic momentum $= 5 - 2 = 3$ , $H = 3^{2} /2 = 4.5$ ; aur $T = \frac{1}{2} m v^{2}$ with $v = 3$ deta hai $4.5$ . ✓ Same. $L$ time-independent hai, tо $H$ bhi conserved hai. ✓

C6 — Degenerate / zero input: free particle

Worked example (f) Free particle,

$V = 0$ , aur limit $p \to 0$ . $L = \frac{1}{2} m \overset{x}{˙}^{2}$ . Forecast: koi potential nahi, $H$ = pure kinetic. Degenerate point $p = 0$ par (particle at rest) $H = 0$ . Kuch bhi blow up nahi hona chahiye.

Momentum: $p = m \overset{x}{˙}$ , invert $\overset{x}{˙} = p / m$ . Ye step kyun? Standard; koi potential worry karne waala nahi.
Assemble: $H = p \overset{x}{˙} - L = \frac{p ^{2}}{m} - \frac{p ^{2}}{2 m} = \frac{p ^{2}}{2 m}$ . Ye step kyun? Definition $H = p \overset{x}{˙} - L$ mein inverted velocity plug karo; koi potential nahi tо sirf kinetic piece survive karta hai.
Degenerate check $p = 0$ : $H = 0$ , $\overset{x}{˙} = \partial H / \partial p = 0$ , $\overset{p}{˙} = - \partial H / \partial x = 0$ . Particle hamesha ke liye still baithta hai — Phase Space mein ek fixed point. Ye step kyun? Zero input ka sensible (non-singular) answer aana chahiye; aata hai. $H = \frac{p ^{2}}{2 m} .$ Verify: $\overset{p}{˙} = 0$ har jagah (koi force nahi), tо $p$ conserved hai — translational symmetry se momentum conservation. $p = 0$ par: $H = 0$ . ✓ $p = 4, m = 2$ par: $H = 16/4 = 4$ . ✓

C7 — Sign / direction cases: thrown up vs down

Worked example (g) C1 waala particle phir, lekin velocity ke dono signs examine karo: upar pheka gaya (

$\overset{y}{˙} > 0$ , tо $p > 0$ ) aur neeche pheka gaya ( $\overset{y}{˙} < 0$ , tо $p < 0$ ). $H = \frac{p ^{2}}{2 m} + m g y$ . Forecast: $H$ $p^{2}$ par depend karta hai, tо dono signs same height par same energy dete hain — upar aur neeche same speed par $H$ ke liye indistinguishable hain. $p$ ka sign sirf phase space mein motion ki direction set karta hai.

Up case: $\overset{y}{˙} = + 3 m/s$ , $m = 2$ , tо $p = + 6$ , $y = 5$ par: $H = 36/4 + 2 \cdot 9.8 \cdot 5 = 9 + 98 = 107$ . Ye step kyun? Positive $p$ matlab $\overset{y}{˙} = \partial H / \partial p = p / m > 0$ , upar move kar raha hai.
Down case: $\overset{y}{˙} = - 3 m/s$ , tо $p = - 6$ , same $y = 5$ : $H = (- 6)^{2} /4 + 98 = 9 + 98 = 107$ . Ye step kyun? Negative $p$ matlab neeche move kar raha hai, lekin $p^{2}$ sign erase kar deta hai — same energy. Verify: dono $107 J$ dete hain; phase-space trajectory ek parabola hai jo $p \to - p$ ke under symmetric hai. ✓ Ye mirror symmetry exactly ek pheke gaye ball ki upar-neeche flight hai.

Neeche ka figure aise padho: horizontal axis momentum $p$ hai, vertical axis height $y$ hai, aur yellow curve un saare $(p, y)$ ka set hai jinka same $H = 107 J$ hai. Blue dot (right) "thrown up, $p = + 6$ " hai aur pink dot (left) "thrown down, $p = - 6$ " hai — dono same height $y = 5$ par baithe hain, vertical $p = 0$ axis ke across mirror images. Wo left–right mirror symmetry hi ye fact hai ki $H$ sirf $p^{2}$ dekhta hai: travel ki direction badalne se energy untouched rehti hai.

C8 — Exam twist: kya explicit time conservation aur energy dono todta hai?

Worked example (h) Ek particle ek aisi potential mein jiska

strength time mein badhti hai: $V (x, t) = \frac{1}{2} k (t) x^{2}$ with $k (t) = k_{0} e^{t / τ}$ . $L = \frac{1}{2} m \overset{x}{˙}^{2} - \frac{1}{2} k_{0} e^{t / τ} x^{2}$ . Forecast (solve karne se pehle guess karo!): (i) Kya $H = E$ hai? $T$ clean quadratic hai, $V$ mein velocity nahi — tо haan, $H = T + V$ . (ii) Kya $H$ conserved hai? $L$ mein explicit $t$ hai ( $e^{t / τ}$ wala) — tо nahi. Trap ye hai ki assume kar lo " $H = E$ " ⟹ " $H$ conserved". Ye alag hain!

Momentum: $p = m \overset{x}{˙}$ , invert $\overset{x}{˙} = p / m$ . Ye step kyun? Time-dependent stiffness $V$ mein baithti hai, $p$ ko touch nahi karti.
Assemble: $H = p \overset{x}{˙} - L = \frac{p ^{2}}{2 m} + \frac{1}{2} k_{0} e^{t / τ} x^{2} = T + V = E$ . ✓ (tо $H = E$ ) Ye step kyun? Clean quadratic par Legendre exactly $T + V$ deta hai; poori time-dependence $V$ ke andar untouched ride kart rahti hai. $H = \frac{p ^{2}}{2 m} + \frac{1}{2} k_{0} e^{t / τ} x^{2} .$
Conservation: $\frac{d H}{d t} = \frac{\partial H}{\partial t} = \frac{1}{2} k_{0} \cdot \frac{1}{τ} e^{t / τ} x^{2} \neq = 0$ . Ye step kyun? Motion ke saath $d H / d t = \partial H / \partial t$ ; stiffness mein explicit $t$ energy andar leak karta hai. Conserved nahi. Verify: $t = 0$ , $τ = 1$ , $k_{0} = 4$ , $m = 1$ , $x = 1$ , $p = 2$ par: $H = 2 + 2 = 4$ (energy). $d H / d t = \frac{1}{2} \cdot 4 \cdot 1 \cdot 1 \cdot 1 = 2 \neq = 0$ . Tо $H = E = 4 J$ abhi hai lekin badh raha hai. ✓ Twist ka jawaab: $H = E$ YES, $H$ conserved NO.

Recap: kaun sa cell kaun sa sawaal answer karta hai?

Recall Do independent switches

Conserved? ::: Haan iff $\partial L / \partial t = 0$ ( $L$ mein koi naked $t$ nahi). Energy ke equal? ::: Haan iff $T$ velocities mein pure quadratic ho AUR $V$ velocity-free ho. Woh cell jahan conserved lekin energy NAHI ::: C4, bead on rotating wire (Jacobi integral). Woh cell jahan energy lekin conserved NAHI ::: C3 aur C8, $L$ mein explicit time. Woh cell jahan DONO hold karte hain ::: C1, C2, C6, C7 (aur C5 canonical momentum use karne ke baad).

Sab kuch ekath karo: aath cells sirf do switches ke char combinations hain, har ek ek concrete system se illustrate kiya gaya hai. C1/C2/C6/C7 happy corner mein baithe hain (conserved aur energy). C3 aur C8 "energy lekin drifting" dikhate hain — explicit time on karo aur $H$ ki value constant rehni band ho jaati hai, chahe har frozen instant par $H$ phir bhi $T + V$ ke equal ho. C4 opposite mismatch dikhata hai — "conserved lekin energy nahi" — jahan moving constraint tumhe $E$ ki jagah Jacobi integral deta hai. C5 happy corner ke roop mein disguised trap hai: ye lagta hai jaisa fail hona chahiye (potential mein velocity) lekin linearity ise bachaa leti hai, baste ho canonical momentum $p = m v + e A$ use karo na ki $m v$ . In char corners ko master karo aur koi bhi exam scenario tumhe surprise nahi kar sakta.