Exercises — Canonical transformations — generating functions

$F_3=F_3(p,Q,t)$: old momentum $p$ aur new coordinate $Q$. $q_i=-\frac{\partial F_3}{\partial p_i},P_i=-\frac{\partial F_3}{\partial Q_i},K=H+\frac{\partial F_3}{\partial t}.$ Humne kya kiya: Type 3 ka table row padha. Do minus signs kyun hain: $F_3$ tak master relation se ek Legendre swap ke through pahuncha jaata hai $q$ mein (jo $pdq$ ko $-qdp$ mein badal deta hai), jo $q$ ke aage sign flip kar deta hai; aur $Q$ "$F$-only" side par baitha hai, isliye uska conjugate $P$ bhi minus carry karta hai (same reason as Type 1).

Haan — yeh identity hai, trivially Hamilton's equations preserve karta hai. Yeh $F_2={\sum }_i{q}_i{P}_i$ se aata hai: tab $p_i=\partial F_2/\partial q_i=P_i$ aur $Q_i=\partial F_2/\partial P_i=q_i$. $F_2$ kyun, $F_1$ kyun nahi: $F_1(q,Q)$ identity ko describe hi nahi kar sakta, kyunki identity ke liye $q$ aur $Q$ same variable hain — woh independent nahi hain, isliye $F_1$ ill-defined ho jaata hai. Identity ek aise type ko force karti hai jo $q$ ko new momentum $P$ ke saath mix kare.

Sach. $K=H+\partial F/\partial t$. Agar $F=F(q,Q)$ (koi explicit $t$ nahi), toh $\partial F/\partial t=0$, isliye $K=H$. Physically iska matlab: phase space ka ek time-independent relabelling energy inject ya remove nahi kar sakta; yeh sirf points rename karta hai, isliye ek point par Hamiltonian ki numerical value unchanged rehti hai.

Type 1: $p=\partial F_1/\partial q=Q$ aur $P=-\partial F_1/\partial Q=-q$. Isliye $\boxed{Q=p,P=-q}$. Kaisa dikhta hai: $(q,p)$ plane mein points $90^{\circ }$ rotate hote hain (neeche figure dekho): old momentum axis new coordinate axis ban jaati hai. Position aur momentum interchangeable hain.

Type 2 relations:

$p=\partial F_1/\partial q=m\omega q\cot Q$. $P=-\partial F_1/\partial Q=-\frac{1}{2}m\omega q^2\cdot (-{\csc }^{2}Q)=\frac{m\omega q^2}{2{\sin }^{2}Q}$, $\frac{d}{dQ}\cot Q=-{\csc }^{2}Q=-1/{\sin }^{2}Q$ use karke. $q^2$ ke liye solve karo: $q^2=\frac{2P{\sin }^{2}Q}{m\omega }$, isliye $q=\sqrt{\frac{2P}{m\omega }}\sin Q$. ✓ $\cot Q$ kyun, kuch aur kyun nahi: oscillator ke phase-space orbits ellipses hain. $\cot Q$ precisely woh function hai jiske Type-1 partials $(q,p)$ ko polar-like $(Q,P)$ mein badal dete hain jo un ellipses ko $P$ se set radius ke circles mein straighten kar deta hai.

$(q,p)$ ke do functions ka Poisson bracket hai $\{A,B\}=\frac{\partial A}{\partial q}\frac{\partial B}{\partial p}-\frac{\partial A}{\partial p}\frac{\partial B}{\partial q}.$ Yahan $\partial Q/\partial q=\cos \alpha$, $\partial Q/\partial p=-\sin \alpha$, $\partial P/\partial q=\sin \alpha$, $\partial P/\partial p=\cos \alpha$. Toh $\{Q,P\}=\cos \alpha \cdot \cos \alpha -(-\sin \alpha )\cdot \sin \alpha ={\cos }^2\alpha +{\sin }^2\alpha =1.$ Kyunki $\{Q,P\}=1$ (aur trivially $\{Q,Q\}=\{P,P\}=0$), map har $\alpha$ ke liye canonical hai. Kaisa dikhta hai: poore $(q,p)$ plane ko rotate karna areas preserve karta hai — exactly wahi jo Liouville's theorem demand karta hai. $90^{\circ }$ case $\alpha =\pi /2$ Ex. 2.1 reproduce karta hai ($Q=-p$… wahan choose ki gayi sign convention ke hisaab se).

$Q=\lambda q,P=p$ ke liye: $\{Q,P\}=\frac{\partial (\lambda q)}{\partial q}\frac{\partial p}{\partial p}-\frac{\partial (\lambda q)}{\partial p}\frac{\partial p}{\partial q}=\lambda \cdot 1-0=\lambda \ne 1$ jab tak $\lambda =1$ na ho. Canonical nahi. $Q=\lambda q,P=p/\lambda$ ke liye: $\{Q,P\}=\lambda \cdot (1/\lambda )-0=1$. Canonical. Generating function (Type 2): humein chahiye $Q=\partial F_2/\partial P=\lambda q$ aur $p=\partial F_2/\partial q=\lambda P$. $F_2=\lambda qP$ try karo: tab $\partial F_2/\partial P=\lambda q=Q$ ✓ aur $\partial F_2/\partial q=\lambda P=p$ ⟹ $P=p/\lambda$ ✓. Toh $\boxed{F_2=\lambda qP}$. Momentum kyun shrink karna chahiye: positions ko $\lambda$ se stretch karna phase-space area ko $\lambda$ se stretch karta hai; area fixed rakhne ke liye (canonicity) momenta ko same factor se contract karna padega.

Type 3: $q=-\partial F_3/\partial p=-(-Q)=Q$ aur $P=-\partial F_3/\partial Q=-(-p)=p$. Toh $Q=q,P=p$: identity ka ek aur rasta. Analysis: identity alag-alag generator types ke through reachable hai ($F_2=qP$ yahan vs $F_3=-pQ$) kyunki kaun se variables hold karne hain yeh humara choice hai; physical map same hai. Yeh Legendre-transform freedom disguise mein hai — Legendre transformation dekho.

Free particle mein already $H=p^2/2m$ independent of $q$ hai, isliye $q$ cyclic hai aur $p$ conserved hai. Sabse clean synthesis Hamilton–Jacobi choice hai: $F_2=S(q,P)$ lo jo satisfy kare $\frac{1}{2m}{\left(\frac{\partial S}{\partial q}\right)}^2=P,$ taaki $K=H+\partial F_2/\partial t=P$ (jahan $P$ conserved energy hai). Yeh deta hai $\partial S/\partial q=\sqrt{2mP}$, yaani $S=\sqrt{2mP}q$. Tab $p=\frac{\partial S}{\partial q}=\sqrt{2mP},Q=\frac{\partial S}{\partial P}=\frac{m}{\sqrt{2mP}}q=\sqrt{\frac{m}{2P}}q.$ Ab $K=P$ independent of $Q$ hai ⟹ $Q$ cyclic hai, $\dot{P}=0$ (energy conserved), aur $\dot{Q}=\partial K/\partial P=1$ ⟹ $Q=t+\text{const}$. Interpretation: $Q$ literally time ban gaya, aur $P$ energy — yahi Hamilton–Jacobi equation aur Action–angle variables ka essence hai.

Humein chahiye $\partial F_2/\partial P=q^2$, toh $F_2=q^{2}P$ (plus ek $P$-independent piece, use zero lo). Tab $p=\partial F_2/\partial q=2qP$, jisse milta hai $\boxed{P=\frac{p}{2q}}$ ($q\ne 0$ ke liye). Check $\{Q,P\}$: $\partial Q/\partial q=2q,\partial Q/\partial p=0,\partial P/\partial q=-p/(2q^2),\partial P/\partial p=1/(2q)$, toh $\{Q,P\}=2q\cdot \frac{1}{2q}-0\cdot \left(-\frac{p}{2q^2}\right)=1.✓$ Degenerate case: $q=0$ par map $Q=q^2$ invertible nahi hai ($Q\ge 0$, aur $\pm q$ dono same $Q$ dete hain) aur $P=p/2q$ blow up karta hai. Toh transformation sirf har half-line $q>0$ ya $q<0$ par alag-alag canonical hai — ek reminder ki "canonical" aksar ek local statement hai.

$A$: $Q=p,P=-q$. $B$: $\tilde{Q}=P=-q,\tilde{P}=-Q=-p$. Toh net map hai $\boxed{\tilde{Q}=-q,\tilde{P}=-p}$ — origin ke through point reflection ($180^{\circ }$ rotation). Kyun: har $F_1=qQ$ step phase space ka $90^{\circ }$ rotation hai; do steps se $180^{\circ }$ ban jaata hai. Canonical transformations ek group form karte hain — canonical maps ka composition canonical hai, isliye humein result justify karne ke liye kabhi naya generator nahi chahiye.

Old variables ka bracket new wale ke respect se compute kiya gaya, canonical map ke liye $1$ ke barabar hona chahiye: $\{q,p{\}}_{Q,P}=\frac{\partial q}{\partial Q}\frac{\partial p}{\partial P}-\frac{\partial q}{\partial P}\frac{\partial p}{\partial Q}.$ $q=\sqrt{2P/(m\omega )}\sin Q$ aur $p=\sqrt{2Pm\omega }\cos Q$ ke saath:

$K=\omega P$ aur $\dot{P}=0$ se, $P$ constant hai; aur on-shell $K=H=E$ ka matlab hai $E=\omega P$, toh $\boxed{P=E/\omega }$. Ek period mein $Q$ $0$ se $2\pi$ tak jaata hai jabki $P$ fixed hai, toh $\oint pdq=\oint PdQ=P{\int }_0^{2\pi }dQ=2\pi P=\frac{2\pi E}{\omega }.$ (Cross-terms ek closed loop ke around $F_1$ ke total differential ke roop mein integrate hote hain aur vanish ho jaate hain.) Yeh kya hai: $\oint pdq=2\pi P$ se $P$ action variable $I=E/\omega$ ban jaata hai, aur $Q$ uska conjugate angle — Action–angle variables ka poora content. Oscillator ka phase-space ellipse area $2\pi P=2\pi E/\omega$ enclose karta hai; figure dekho.

Type-1 relations $p=\partial F/\partial q$ aur $P=-\partial F/\partial Q$ mein $q$ aur $Q$ ke respect se derivatives hain. Kyunki $h(t)$ sirf $t$ par depend karta hai, $\partial h/\partial q=\partial h/\partial Q=0$, isliye $p$ aur $P$ (hence poora map) unchanged hain. Sirf $K=H+\partial F/\partial t$ ko $+h^{\prime }(t)$ milta hai: $K\to K+h^{\prime }(t)$. Mastery point: generator ka purely time-dependent shift ek gauge freedom hai — yeh $K$ ki "energy zero" rescale karta hai bina physics touch kiye, mirroring karta hai ki Hamilton's equations sirf $H$ ke derivatives ki parwah karte hain. Isliye master relation "$+dF/dt$" tolerate karta hai: total derivatives physically invisible hain.