Canonical transformations — generating functions: Question bank

2.1.16 · D5 · Physics › Analytical Mechanics › Canonical transformations — generating functions

Traps se pehle, hum teen ideas rebuild karte hain jinpe poora page tika hai — master relation, chaar generator types, aur phase-space picture — taaki koi bhi symbol use hone se pehle earn ho jaye.

Formula Chaar generator types ("

$F_{1}, \dots, F_{4}$ " ka matlab) $F$ ko ek purane aur ek naye variable par depend karna chahiye per degree of freedom, jisse bilkul chaar choices milti hain. Box ko $d t$ se multiply karo aur independent differentials ke coefficients match karo:

F_{1} (q, Q) : F_{2} (q, P) : F_{3} (p, Q) : F_{4} (p, P) : p_{i} = \frac{\partial F _{1}}{\partial q _{i}}, p_{i} = \frac{\partial F _{2}}{\partial q _{i}}, q_{i} = - \frac{\partial F _{3}}{\partial p _{i}}, q_{i} = - \frac{\partial F _{4}}{\partial p _{i}}, P_{i} = - \frac{\partial F _{1}}{\partial Q _{i}}, Q_{i} = + \frac{\partial F _{2}}{\partial P _{i}}, P_{i} = - \frac{\partial F _{3}}{\partial Q _{i}}, Q_{i} = + \frac{\partial F _{4}}{\partial P _{i}}, K = H + \frac{\partial F _{1}}{\partial t} K = H + \frac{\partial F _{2}}{\partial t} K = H + \frac{\partial F _{3}}{\partial t} K = H + \frac{\partial F _{4}}{\partial t}

$F_{2}, F_{3}, F_{4}$ ko $F_{1}$ se ek Legendre transformation ke zariye reach kiya jaata hai (ek variable ko uske conjugate se swap karo, jaise $P d Q = d (P Q) - Q d P$ ). Pattern dekho: jo variable sirf $F$ ke andar appear karta hai (kabhi "free" argument ke roop mein nahi) usey Types 1 aur 3 mein minus milta hai.

Neeche di gayi picture har aane wale question ki arena hai: phase space, wo flat sheet jiske do axes hain $q$ (horizontal) aur $p$ (vertical). System ki ek state ek dot hai; jaise time chalta hai dot ek curve trace karta hai. Ek CT is sheet par grid ko redraw karta hai bina use faade.

Figure — Canonical transformations — generating functions

Wo ek geometric fact jo CTs ko special banata hai wo ye hai ki ye is sheet par oriented area preserve karte hain. Agla figure ek choti si states ki patch dikhata hai jo ek CT se carry ho rahi hai: ye stretch aur shear kar sakti hai, lekin uski area unchanged rehti hai — ye Poisson bracket test ${Q, P} = 1$ ka visual meaning hai aur Liouville's theorem ka bhi.

True or false — justify

TF1. "Har invertible smooth map

(q, p) \to (Q, P)

canonical hoti hai."

False — invertibility kaafi nahi hai; map ko symplectic (area) structure preserve karni chahiye, equivalently

{Q, P} = 1

. Zyaadatar random smooth maps area ko unevenly stretch karti hain aur Hamilton's equations toot jaate hain.

TF2. "Agar generating function

F

mein koi explicit time dependence nahi hai, to

K = H

phase space par functions ke roop mein hoga."

True — kyunki

K = H + \partial F / \partial t

aur

\partial F / \partial t = 0

, naye Hamiltonian ki value har phase point par purane ke barabar hoti hai (sirf

Q, P

mein re-express hoti hai).

TF3. "Ek canonical transformation kisi point par energy ki numerical value nahi badal sakta."

Saamaanya taur par False — jab

F

t

par depend karta hai,

K \neq = H

, isliye naye Hamiltonian ki value alag hoti hai; energy conservation ek concept ke roop mein survive karta hai lekin usse attached number badal sakta hai.

TF4. "Chaar generator types

F_{1}

–

F_{4}

chaar alag tarah ke canonical transformations describe karte hain."

False — ye wahi CTs ko alag independent variables ke saath describe karte hain; type ek bookkeeping choice hai ki kaun sa old/new pair free treat kiya jaye.

TF5. "

F_{2} = \sum_{i} q_{i} P_{i}

identity transformation generate karta hai."

True — Type-2 rules dete hain

p_{i} = \partial F_{2} / \partial q_{i} = P_{i}

aur

Q_{i} = \partial F_{2} / \partial P_{i} = q_{i}

, yaani kuch nahi badalta; ye "kuch-na-karo" generator woh base point hai jahan se infinitesimal CTs grow karte hain.

TF6. "Kyunki

Q = p

P = - q

canonical hai, position aur momentum interchange kiye ja sakte hain."

True —

F_{1} = \sum q_{i} Q_{i}

ke saath Type-1 rules dete hain

p_{i} = \partial F_{1} / \partial q_{i} = Q_{i}

(isliye

Q_{i} = p_{i}

) aur

P_{i} = - \partial F_{1} / \partial Q_{i} = - q_{i}

; minus sign boxed relation ke right-hand

P d Q

ki wajah se forced hai, aur geometrically ye

q

–

p

sheet ka

9 0^{\circ}

rotation hai (rotations area preserve karti hain, isliye ye canonical hai), jo dikhata hai ki phase space mein koi built-in "position" nahi hai (dekho Symplectic geometry).

TF7. "Generating-function method galti se ek non-canonical map produce kar sakta hai."

False — kisi bhi valid generator se derived transformation construction se canonical hoti hai, kyunki ye shared variational principle (boxed master relation) se derive hoti hai jo Hamilton's form fix karta hai.

TF8. "Canonical transformations phase-space volume preserve karti hain."

True — ye figure s02 ka symplectic area preserve karti hain, isliye higher dimensions mein volume bhi (ye Liouville's theorem hai CT lens se dekha gaya).

TF9. "Ek time-dependent CT jo

K = 0

banati hai us system ko bina dynamics ke chhod deti hai."

False —

K = 0

ka matlab hai naye variables saare constants of motion hain; dynamics poori tarah transformation mein encode hai, jo bilkul Hamilton–Jacobi equation strategy hai.

Spot the error

SE1. "Type-1 rule:

p_{i} = \partial F_{1} / \partial q_{i}

aur

P_{i} = + \partial F_{1} / \partial Q_{i}

Sign galat hai:

P_{i} = - \partial F_{1} / \partial Q_{i}

. Minus isliye aata hai kyunki

P_{i} d Q_{i}

master relation ke right par hai, isliye use cross karne par sign flip hota hai.

SE2. "

F_{2} (q, P)

ke liye hume milta hai

Q_{i} = - \partial F_{2} / \partial P_{i}

Phir galat sign:

Q_{i} = + \partial F_{2} / \partial P_{i}

. Type 2 ek Legendre swap (

P d Q = d (P Q) - Q d P

) se banaya gaya tha bilkul isi liye ki ye term positive aaye — jaisa four-types box mein listed hai.

SE3. "Hum freely

F_{1} = F_{1} (q, p, t)

choose kar sakte hain, ek purane coordinate ko uske apne purane momentum ke saath mix karke."

Ek generator ko per degree of freedom ek purane aur ek naye variable par depend karna chahiye;

F_{1} (q, p)

do purane variables use karta hai aur transformation equations close nahi kar sakta.

SE4. "Kyunki

K = H + \partial F / \partial t

F

mein ek constant add karne se

K

badal jaata hai."

Ek constant ka time-derivative aur spatial derivatives dono zero hain, isliye ye na

K

badalta hai na transformation; sirf

q, Q, t

par genuine dependence matter karta hai.

SE5. "Oscillator example

F_{1} = \frac{1}{2} mω q^{2} cot Q

mein, humne

H

ko ek differential equation integrate karke solve kiya."

Koi differential equation solve nahi ki — poora point ye hai ki generator

Q

ko cyclic banata hai (

K = ω P

), isliye motion

Q = ω t + ϕ

P = const

purely algebraically follow karta hai.

SE6. "Ek point transformation

Q = Q (q)

sirf linear

Q (q)

ke liye canonical hai."

Koi bhi invertible point transformation canonical hai agar momenta sahi transform hoti hain (

P = p \partial q / \partial Q

); nonlinearity bilkul theek hai, ye

F_{2} = \sum P_{i} Q_{i} (q)

se generate hoti hai.

SE7. "Kyunki Poisson brackets CTs ke under invariant hain,

{Q, P}

compute karna canonicity kabhi test nahi kar sakta."

Invariance hi wajah hai ki bracket ek test hai: ek proposed map canonical hai iff

{Q, P} = 1

(aur

{Q, Q} = {P, P} = 0

) purane variables mein upar di gayi definition se evaluate ki gayi.

Why questions

WHY1. Dono integrands ek total time derivative

d F / d t

se kyun differ karne chahiye, sirf ek constant se nahi?

Ek total time derivative fixed endpoints par integrate hokar ek boundary term banata hai jiska variation vanish hota hai, isliye equations of motion untouched rehti hain; ek mere constant ek special (trivial) case hai jo nontrivial variable changes generate nahi kar sakta.

WHY2. Generating function ek single scalar function kyun hai aur har naye variable ke liye formulas ka set nahi?

Saare transformation equations ek scalar

F

ke partial derivatives hain, jo automatically wo integrability/consistency guarantee karta hai (mixed partials match hote hain) jo map ko canonical rakhta hai.

WHY3.

F_{1}

F_{2}

jaane par Legendre transformation kyun appear hota hai?

Free variable ko

Q

se uske conjugate

P

par switch karna exactly "natural variable" ka change hai, jo Legendre transformation karta hai; hum

\sum Q_{i} P_{i}

subtract karte hain

d Q

d P

se trade karne ke liye.

WHY4. Ek clever CT Hamiltonian ke coordinate ko cyclic kyun bana sakta hai, aur wo win kyun hai?

Agar

Q

K

mein appear nahi karta to

\dot{P} = - \partial K / \partial Q = 0

, isliye

P

conserved hai aur

\dot{Q}

constant hai — motion trivial straight-line drift ban jaati hai, jo Action–angle variables ka seed hai.

WHY5. Coefficients match karte waqt

q

aur

p

ko independent differentials kyun treat kiya jaata hai?

Master relation mein

q, Q, t

(Type 1) independent free variables choose ki jaati hain, isliye unke differentials alag-alag vary kiye ja sakte hain; tabhi har coefficient dono sides par match karna chahiye.

WHY6. Generating-function trick fail kyun hoti hai agar hum

F

ko ek conjugate pair ke dono members par simultaneously depend karte hain?

Tab transformation equations degenerate ho jaate hain — tum saare naye variables ko purane ke terms mein solve nahi kar sakte — kyunki pair symplectic constraint ke under independent nahi hai.

Edge cases

EC1. Agar

F_{1}

Q

par independent ho to Type-1 relations ka kya hoga?

Tab

P_{i} = - \partial F_{1} / \partial Q_{i} = 0

, saare naye momenta ko zero force karta hai — ek degenerate, non-invertible map jo valid CT nahi hai.

EC2. Kya swap

Q = p

P = - q

apna khud ka inverse hai?

Ise do baar apply karne par

Q \to - q

P \to - p

milta hai, yaani phase space mein

18 0^{\circ}

rotation, identity nahi — isliye ye order-four hai, ek acchi reminder ki CTs ek group banate hain.

EC3. Identity ke paas "sabse chhota" nontrivial CT kya hai?

Ek infinitesimal wala,

F_{2} = \sum q_{i} P_{i} + ϵ G (q, P)

; first order tak ye variables ko

{\cdot, G}

se shift karta hai,

G

ke Hamiltonian flow ko generate karta hai.

EC4. Agar

H

already zero hai, to kya koi transformation canonical hai?

Dynamics trivial hai (

\overset{q}{˙} = \overset{p}{˙} = 0

), lekin ek map tab bhi canonical hai sirf tab jab ye symplectic form preserve kare; canonicity map ki property hai, jo bhi

H

tum carry karo usse independent hai.

EC5. Kya ek time-dependent CT energy conservation preserve karta hai?

Zaruri nahi —

\partial F / \partial t \neq = 0

ke saath,

K \neq = H

aur

K

khud time par depend kar sakta hai, isliye conserved quantity (agar koi ho)

K

hai, original

H

nahi.

EC6. Oscillator generator

F_{1} = \frac{1}{2} mω q^{2} cot Q

Q \to 0

par limiting behaviour kya hai?

q = 2 P / mω sin Q

use karke, product

q cot Q = 2 P / mω cos Q

finite rehta hai, isliye

p = mω q cot Q \to 2 mω P

(uska maximum) jab

q \to 0

— turning point jahan saari energy kinetic hai, aur

P

poore time finite rehta hai.

Recall One-line self-audit

Kya main master relation memory se state kar sakta hoon aur har sign locate kar sakta hoon? ::: $\sum p \overset{q}{˙} - H = \sum P \dot{Q} - K + d F / d t$ ; $P$ par minus (Types 1,3) $P d Q$ ke right par hone se aata hai.
Kya main canonicity ka ek algebraic test jaanta hoon? ::: Poisson brackets ${Q, P} = 1$ , ${Q, Q} = {P, P} = 0$ .
Kya main explain kar sakta hoon ki cyclic $Q$ goal kyun hai? ::: Tab $\dot{P} = 0$ aur $Q$ linearly drift karta hai — ye Hamilton–Jacobi aur action–angle variables ka poora trick hai.

Connections

Hamilton's equations — wo form jise CTs preserve karte hain.
Poisson brackets — algebraic canonicity test.
Legendre transformation — Types 2–4 ke peeche ka swap.
Action–angle variables — jahan cyclic- $Q$ generators le jaate hain.
Hamilton–Jacobi equation — $K = 0$ time-dependent CT.
Liouville's theorem — CTs ke under volume preservation.
Symplectic geometry — kyun $q$ aur $p$ equal footing par hain.