Canonical transformations — generating functions

2.1.16 · Physics › Analytical Mechanics

WHAT hai ek canonical transformation?

WHY ek generating function exist karti hai. Dono — purane aur naye — equations ek variational principle se aate hain (Hamilton's modified principle):

\delta\!\int\!\Big(\sum_i P_i\dot Q_i - K\Big)dt=0.$$ Do integrands same equations of motion dete hain agar aur sirf agar woh kisi function $F$ ki **total time derivative** se differ karein (jiska fixed endpoints par variation zero hota hai): $$\boxed{\;\sum_i p_i\dot q_i - H \;=\; \sum_i P_i\dot Q_i - K \;+\;\frac{dF}{dt}\;}$$ Woh function $==F==$ hi **generating function** hai. Sab kuch is ek equation se nikalta hai. --- ## HOW chaar types aate hain $F$ purane aur naye variables ka mix ho sakta hai. Humein chahiye ki yeh **ek purana + ek naya** variable per degree of freedom par depend kare (taaki transformation fully determined ho). Isse **4 types** milti hain. ### Type 1: $F=F_1(q,Q,t)$ Boxed relation ko $dt$ se multiply karo: $$p_i\,dq_i - H\,dt = P_i\,dQ_i - K\,dt + dF_1.$$ Kyunki $dF_1=\frac{\partial F_1}{\partial q_i}dq_i+\frac{\partial F_1}{\partial Q_i}dQ_i+\frac{\partial F_1}{\partial t}dt$, **independent differentials** $dq_i$, $dQ_i$, $dt$ ke coefficients match karo: > [!formula] Type-1 relations > $$p_i=\frac{\partial F_1}{\partial q_i},\qquad > P_i=-\frac{\partial F_1}{\partial Q_i},\qquad > K=H+\frac{\partial F_1}{\partial t}.$$ > **Why yeh kaam karta hai:** $q_i,Q_i,t$ ko independent treat kiya jaata hai, toh unke coefficients dono sides par *alag-alag* match hone chahiye. ### Type 2: $F=F_2(q,P,t)-\sum_i Q_iP_i$ ($Q$ mein Legendre transform) Humein $(q,P)$ mein generator chahiye. $P_i dQ_i = d(P_iQ_i)-Q_i dP_i$ use karo aur $d(\sum Q_iP_i)$ absorb karo: > [!formula] Type-2 relations > $$p_i=\frac{\partial F_2}{\partial q_i},\qquad > Q_i=\frac{\partial F_2}{\partial P_i},\qquad > K=H+\frac{\partial F_2}{\partial t}.$$ Baaki do bhi same Legendre-swap pattern follow karte hain: | Type | Variables | Relations | |------|-----------|-----------| | $F_1(q,Q)$ | purana $q$, naya $Q$ | $p=\partial_q F_1,\;\;P=-\partial_Q F_1$ | | $F_2(q,P)$ | purana $q$, naya $P$ | $p=\partial_q F_2,\;\;Q=\partial_P F_2$ | | $F_3(p,Q)$ | purana $p$, naya $Q$ | $q=-\partial_p F_3,\;\;P=-\partial_Q F_3$ | | $F_4(p,P)$ | purana $p$, naya $P$ | $q=-\partial_p F_4,\;\;Q=\partial_P F_4$ | Chaaron mein: $K=H+\partial F/\partial t$ (agar $F$ time-independent hai, toh $K=H$). ![[2.1.16-Canonical-transformations-—-generating-functions.png]] --- ## WORKED EXAMPLES > [!example] Example 1 — Identity $F_2=qP$ mein chhupa hai > Lo $F_2=\sum_i q_iP_i$. > - $p_i=\partial F_2/\partial q_i = P_i$ → **Why?** $q_i$ ke w.r.t. derivative $P_i$ laata hai. > - $Q_i=\partial F_2/\partial P_i = q_i$ → toh $Q=q,\;P=p$: **identity** transformation. > > **Lesson:** $F_2=qP$ ek "kuch-nahi-karne-wala" generator hai; isme chhoti additions *infinitesimal* > canonical transformations deti hain (Hamiltonian flow ka seed). > [!example] Example 2 — Coordinates aur momenta swap ($F_1=qQ$) > Lo $F_1=\sum_i q_iQ_i$. > - $p_i=\partial F_1/\partial q_i=Q_i$ → toh $Q_i=p_i$. **Why?** $dq$ coefficient match karo. > - $P_i=-\partial F_1/\partial Q_i=-q_i$ → toh $P_i=-q_i$. > > Result: $Q=p,\;P=-q$. Coordinates aur momenta swap ho jaate hain (sign tak)! Yeh dikhata hai ki $q$ aur $p$ phase space mein **equal footing** par hain — position ke baare mein kuch intrinsically "position-wala" nahi hai. > [!example] Example 3 — Harmonic oscillator, killer app ($F_1=\tfrac12 m\omega q^2\cot Q$) > $H=\dfrac{p^2}{2m}+\dfrac12 m\omega^2 q^2$. Try karo $F_1(q,Q)=\dfrac{m\omega q^2}{2}\cot Q$. > 1. $p=\dfrac{\partial F_1}{\partial q}=m\omega q\cot Q$. **Why?** $p$ ke liye Type-1 rule. > 2. $P=-\dfrac{\partial F_1}{\partial Q}=\dfrac{m\omega q^2}{2\sin^2 Q}$. **Why?** $\frac{d}{dQ}\cot Q=-\csc^2Q$. > 3. (2) se $q$ solve karo: $q=\sqrt{\dfrac{2P}{m\omega}}\,\sin Q$. (1) mein sub karo: $p=\sqrt{2Pm\omega}\,\cos Q$. > 4. $H$ mein plug karo (time-independent hai toh $K=H$): > $$K=\frac{2Pm\omega\cos^2Q}{2m}+\frac12 m\omega^2\frac{2P}{m\omega}\sin^2Q=\omega P(\cos^2Q+\sin^2Q)=\omega P.$$ > Toh $K=\omega P$ — **$Q$ cyclic hai!** Phir $\dot P=-\partial K/\partial Q=0$ ($P$ const = energy/$\omega$), > aur $\dot Q=\partial K/\partial P=\omega$ ⟹ $Q=\omega t+\phi$. Humne oscillator ko bina kisi > differential equation ke solve kar diya — sirf ek clever generator se. (Yeh action–angle variables ka gateway hai.) --- ## Common mistakes (steel-manned) > [!mistake] "Koi bhi smooth change of variables canonical hota hai." > **Why sahi lagta hai:** Lagrangian mechanics mein *koi bhi* invertible point transformation $q\to Q(q)$ > kaam karta hai. **Fix:** *phase* space mein tum $q$ aur $p$ mix karte ho, aur tumhe symplectic form preserve karna hoga — > equivalently Poisson brackets $\{Q_i,P_j\}=\delta_{ij}$, $\{Q,Q\}=\{P,P\}=0$. Zyaadatar random > maps yeh fail kar dete hain. Generating-function machinery ise *automatically guarantee* karti hai. > [!mistake] Time term bhool jaana: "$K=H$ hamesha." > **Why sahi lagta hai:** static examples mein yeh *hota hi* hai. **Fix:** $K=H+\partial F/\partial t$. > Extra term essential hai jab $F$, $t$ par depend kare — exactly yahi Hamilton–Jacobi $K=0$ banata hai. > [!mistake] Signs mix up karna ($P=+\partial_Q F_1$). > **Why sahi lagta hai:** $p$ aur $P$ "symmetric lagte hain." **Fix:** minus sign isliye aata hai kyunki $dQ$ boxed equation ke **right** side par baitha hai: $P_i dQ_i$ sign ke saath move hota hai. Yaad rakho: > *naya variable jo "$F$-only" side par ho, use Types 1 & 3 mein minus milta hai*. --- ## Active recall > [!recall]- Khud test karo (answers chhupao) > - Woh kaun si ek equation hai jo SAARI canonical transformations generate karti hai? > - Kyun dono integrands sirf ek constant se nahin, balki total time derivative se differ karne chahiye? > - $F_2$ ke liye, $\partial_P F_2$ kaun sa variable hai? > - $\partial F/\partial t \ne 0$ $H$ ke saath kya karta hai? #flashcards/physics Canonical transformation kya define karta hai? ::: Ek change $(q,p)\to(Q,P)$ jo kisi naye Hamiltonian $K$ ke liye Hamilton ke equations ka canonical form preserve kare. CTs generate karne wala master relation kya hai? ::: $\sum p_i\dot q_i - H = \sum P_i\dot Q_i - K + \frac{dF}{dt}$. Type-1 generating function relations? ::: $p_i=\partial F_1/\partial q_i,\;P_i=-\partial F_1/\partial Q_i,\;K=H+\partial F_1/\partial t$. Type-2 generating function relations? ::: $p_i=\partial F_2/\partial q_i,\;Q_i=\partial F_2/\partial P_i,\;K=H+\partial F_2/\partial t$. Kaun sa generator identity transformation deta hai? ::: $F_2=\sum_i q_iP_i$ ⟹ $Q=q,\,P=p$. Generator $F_1=qQ$ kya produce karta hai? ::: $Q=p,\;P=-q$ (coords aur momenta ka swap). $K$ aur $H$ mein generally kya relation hai? ::: $K=H+\partial F/\partial t$; barabar sirf tab jab $F$ time-independent ho. Harmonic oscillator ke liye, $F_1=\tfrac12 m\omega q^2\cot Q$ kya achieve karta hai? ::: Yeh $K=\omega P$ deta hai, $Q$ ko cyclic banata hai toh $P$ constant hai aur $Q=\omega t+\phi$. Canonicity ke liye Poisson-bracket test? ::: $\{Q_i,P_j\}=\delta_{ij},\;\{Q_i,Q_j\}=\{P_i,P_j\}=0$. > [!recall]- Feynman: ek 12-saal ke bachche ko samjhao > Socho ek treasure map jahan khazana dhundhna bahut mushkil hai. Ek canonical transformation aise hai > jaise usi map par **naya grid** kheenchna taaki khazana seedha ek grid line par aa jaaye — khazana hila nahi, > tumne sirf cheezon ko relabel kiya taaki raasta ek seedha, boring line ban jaaye. "Generating function" woh > **single recipe** hai jo tum likhte ho jो exactly batati hai ki grid ko dobara kaise kheenchna hai bina game > ke rules todte hue (rules hain Hamilton ke equations). > [!mnemonic] Signs aur variables yaad rakhna > **"1Q 2P, 3Q 4P"** — *kaun sa naya variable appear hota hai*: $F_1,F_3$ naya **Q** use karte hain; $F_2,F_4$ naya **P** use karte hain. > Aur **"purane ko plain milta hai, akele-naye ko minus"**: woh variable jo *sirf* $F$ mein appear kare (conjugate pair argument nahi) use Types 1 & 3 mein minus sign milta hai. ## Connections - [[Hamilton's equations]] — woh form jo CTs preserve karti hain. - [[Poisson brackets]] — CTs ke under invariant; canonicity ka algebraic test. - [[Hamilton–Jacobi equation]] — $F$ choose karo taaki $K=0$ ho. - [[Action–angle variables]] — Example 3 generalized. - [[Symplectic geometry]] — CT = symplectomorphism jo $dp\wedge dq$ preserve kare. - [[Legendre transformation]] — chaar generator types ko relate karta hai. - [[Liouville's theorem]] — phase-space volume ek CT invariant hai. ## 🖼️ Concept Map ```mermaid flowchart TD HE[Hamilton equations canonical form] CT[Canonical transformation] VP[Hamilton modified principle] F[Generating function F] K[New Hamiltonian K] F1[Type 1 F1 q,Q] F2[Type 2 F2 q,P] F3[Type 3 F3 p,Q] F4[Type 4 F4 p,P] REL[Transformation relations] CT -->|preserves| HE CT -->|requires| K VP -->|two integrands differ by dF/dt| F F -->|manufactures| CT F -->|depends on one old + one new var| F1 F -->|Legendre swap| F2 F -->|Legendre swap| F3 F -->|Legendre swap| F4 F1 -->|match differentials| REL F2 -->|match differentials| REL REL -->|gives p,P,Q and| K ```