Hamilton-Jacobi equation

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1. The setup: canonical transformations (WHAT we lean on)

We use the type-2 generating function $F_2(q,P,t) \equiv S(q,P,t)$, which depends on old coordinates $q$ and new momenta $P$. Its defining relations are $p_i = \frac{\partial S}{\partial q_i}, \qquad Q_i = \frac{\partial S}{\partial P_i}, \qquad K = H + \frac{\partial S}{\partial t},$ where $K(Q,P,t)$ is the new Hamiltonian (the "Kamiltonian").

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2. Deriving the HJ equation from scratch (HOW)

Goal: choose $S$ so that the new Hamiltonian vanishes: $K \equiv 0$.

Why does $K=0$ help? Hamilton's equations in the new variables are $\dot Q_i = \frac{\partial K}{\partial P_i}, \qquad \dot P_i = -\frac{\partial K}{\partial Q_i}.$ If $K=0$ everywhere, then $\dot Q_i = 0$ and $\dot P_i = 0$: all new coordinates and momenta are constants. Motion solved.

Why this step? We are deliberately "buying" trivial dynamics by demanding the strongest possible condition on $K$.

Now impose $K=0$ in $K = H + \partial S/\partial t$: $H\!\left(q_1,\dots,q_n,\ \frac{\partial S}{\partial q_1},\dots,\frac{\partial S}{\partial q_n},\ t\right) + \frac{\partial S}{\partial t} = 0.$

Why this step? We substituted $p_i = \partial S/\partial q_i$ (the type-2 relation) into $H$, turning every momentum into a derivative of $S$. The result is the celebrated equation:

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3. What is $S$, physically?

Why this step? We used $\partial S/\partial q_i = p_i$ and $\partial S/\partial t = -H$, then recognized $\sum p_i\dot q_i - H = L$ (Legendre transform). This is the deep payoff: HJ unifies action, canonical transformations, and PDEs.

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4. Time-independent case: separation of $t$

If $H$ has no explicit time dependence, energy is conserved. Try the separation $S(q,P,t) = W(q,P) - E\,t.$

Why this step? Plugging in, $\partial S/\partial t = -E$, so the HJ equation becomes $H\!\left(q,\frac{\partial W}{\partial q}\right) = E.$

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5. Worked Example A — Free particle (1D)

Hamiltonian: $H = \dfrac{p^2}{2m}$.

Step 1. Write HJ: $\dfrac{1}{2m}\left(\dfrac{\partial S}{\partial x}\right)^2 + \dfrac{\partial S}{\partial t} = 0.$ Why? Replace $p\to \partial S/\partial x$.

Step 2. Separate $S = W(x) - E t$: $\dfrac{1}{2m}(W')^2 = E \Rightarrow W' = \sqrt{2mE}.$ Why? Energy conserved, so peel off the time part.

Step 3. Integrate: $W = \sqrt{2mE}\,x$, so $S = \sqrt{2mE}\,x - Et.$

Step 4. New coordinate is constant: take $E$ as the new momentum $P=E$. Then $Q = \frac{\partial S}{\partial E} = \sqrt{\frac{m}{2E}}\,x - t = \text{const} \equiv -t_0.$ Why? $Q$ is constant by construction ($K=0$).

Step 5. Solve for $x$: $x = \sqrt{\dfrac{2E}{m}}\,(t - t_0)$ — uniform velocity $v=\sqrt{2E/m}$. ✅ Exactly the free particle.

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6. Worked Example B — Harmonic oscillator (1D)

$H = \dfrac{p^2}{2m} + \dfrac12 m\omega^2 x^2 = E.$

Step 1. $W' = \sqrt{2m\big(E - \tfrac12 m\omega^2 x^2\big)}$. Why? Solve $H(x,W')=E$ for $W'=\partial W/\partial x$.

Step 2. $S = \int \sqrt{2m\big(E-\tfrac12 m\omega^2 x^2\big)}\,dx - Et$.

Step 3. The conserved new coordinate: with $P=E$, $Q = \frac{\partial S}{\partial E} = \int \frac{m\,dx}{\sqrt{2m(E-\tfrac12 m\omega^2 x^2)}} - t = \text{const}.$ Why? Differentiate under the integral; $\partial/\partial E$ pulls down $m$.

Step 4. The integral evaluates (sub $x=\sqrt{2E/m\omega^2}\sin\theta$) to $\frac{1}{\omega}\arcsin\!\left(x\sqrt{\frac{m\omega^2}{2E}}\right) = t + \text{const}.$

Step 5. Invert: $x(t) = \sqrt{\dfrac{2E}{m\omega^2}}\,\sin(\omega t + \phi)$. ✅ The familiar oscillation with amplitude set by $E$.

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7. Common mistakes

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8. Active recall

Recall Quick self-test (hide and answer)

• What condition on the new Hamiltonian $K$ produces the HJ equation? → $K=0$.
• What is $S$ physically? → the action $\int L\,dt$.
• How many constants in a complete integral for $n$ DOF? → $n$ (plus one trivial additive constant).
• What replaces $p_i$ in $H$? → $\partial S/\partial q_i$.

Recall Feynman: explain to a 12-year-old

Imagine you must walk a complicated maze. Newton's way: feel the walls and push step by step. The Hamilton-Jacobi way: first draw a magic map where every point already has an arrow saying "go here next." Once you have that perfect map ($S$), you never have to think — you just follow arrows and the maze solves itself. Building the magic map is hard (it's one big equation), but after that, walking is free. The arrows are the momenta, and the map's "height" is the total effort (action) to reach each point.

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9. Connections

• Canonical Transformations — HJ is the canonical transformation to constant variables.
• Hamilton's Equations — the starting point; HJ replaces $2n$ ODEs with one PDE.
• Action and Hamilton's Principle — $S$ is the action; HJ is the PDE the action obeys.
• Separation of Variables — the practical method to solve HJ for integrable systems.
• Action-Angle Variables — built directly from $W$ for periodic motion.
• Schrodinger Equation — the classical limit of QM reduces to HJ ($\psi\sim e^{iS/\hbar}$).

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Hamilton-Jacobi equation ka core idea bahut elegant hai. Newton bolta hai "force se chalo", Lagrange bolta hai "energy se", aur Hamilton phase space ki geometry use karta hai. HJ ek step aage jaata hai: hum ek aisa generating function $S$ dhoondhte hain jo coordinates ko aise naye coordinates $(Q,P)$ me badal de jo bilkul move hi na karein — yaani $\dot Q=0$, $\dot P=0$. Iske liye hum demand karte hain ki naya Hamiltonian $K=0$ ho. Jaise hi yeh hota hai, saari dynamics ek hi PDE me simat jaati hai: $H(q,\partial S/\partial q,t)+\partial S/\partial t=0$. Yahi hai Hamilton-Jacobi equation.

Sabse khoobsurat baat: yeh $S$ koi random function nahi hai — yeh actually action hai, $S=\int L\,dt$. Jab tum $dS/dt$ nikaalte ho trajectory ke along, toh wo $\sum p\dot q - H = L$ ban jaata hai. Matlab action khud apne differential law ko follow kar raha hai. Yeh wahi cheez hai jo aage chal ke Schrodinger equation se connect hoti hai ($\psi \sim e^{iS/\hbar}$), isliye yeh classical aur quantum ka pul hai.

Practically, jab $H$ time pe depend nahi karta, hum $S=W-Et$ likhte hain, jahan $W$ characteristic function hai aur $E$ energy. Phir $H(q,\partial W/\partial q)=E$ solve karte hain. Free particle aur harmonic oscillator dono is method se 4-5 steps me exact solution de dete hain. Yaad rakho: $n$ degrees of freedom ke liye complete integral me $n$ constants aati hain, jo new momenta ban jaati hain — yahi sabse common galti hai ki log in constants ko ignore kar dete hain.