Hamilton-Jacobi equation: Exercises — Physics deep dive

Before we start, one shared vocabulary reminder, because every problem uses it:

Definition The symbols we reuse everywhere

$q$ — a generalized coordinate: the abstract "position" variable of the system. In each concrete exercise below $q$ wears a physical name — for 1D horizontal motion we write $q = x$ (a length along a line), for vertical free fall $q = z$ (a height), and in the 2D problem the two coordinates are $q_{1} = x$ and $q_{2} = y$ . These are the same kind of object as $q$ ; only the letter changes to match the picture.
$H$ — the old Hamiltonian, the machine that takes a position $q$ and momentum $p$ and returns the total energy.
$L$ — the Lagrangian, kinetic minus potential energy, $L = T - V$ . It is the quantity whose time-integral is the action; we only need it to say what $S$ is physically (see the next bullet). More in Action and Hamilton's Principle.
$K$ — the new Hamiltonian (the "Kamiltonian") after we change variables $(q, p) \to (Q, P)$ . Setting $K = 0$ is the whole trick: it makes the new coordinates $Q$ and new momenta $P$ constant. See Canonical Transformations.
$S (q, t)$ — Hamilton's principal function, the unknown we solve for. It equals the action $\int L d t$ along the true path. It is the generating function linking old and new variables: $K = H + \partial S / \partial t$ .
$W (q)$ — Hamilton's characteristic function: the time-stripped part of $S$ . It exists only when $H$ has no explicit time (energy conserved), and then $S (q, t) = W (q) - E t$ . Its slope gives the momentum too: $p = \partial W / \partial q$ . It answers "what does the action look like once we factor out the steady $- E t$ drift?" We use it in almost every exercise.
$\frac{\partial S}{\partial q}$ — read "how fast $S$ changes if I nudge $q$ while freezing time". The HJ rule says this equals the momentum $p$ .
$\frac{\partial S}{\partial t}$ — "how fast $S$ changes as time ticks with $q$ frozen". With $K = 0$ this equals $- H$ .
Notation guard: lowercase $p$ = old momentum; uppercase $P$ = new (constant) momentum; $α$ = a separation constant, which we then choose to identify with a new momentum $P$ . So $α$ and $P$ are the same object wearing two hats.

The whole game: $H (q, \frac{\partial S}{\partial q}, t) + \frac{\partial S}{\partial t} = 0$ .

Level 1 — Recognition

Exercise 1.1 (L1)

For a 1D system with $H = \frac{p ^{2}}{2 m} + V (x)$ (here the coordinate is $q = x$ ), write the Hamilton-Jacobi equation without solving it.

Recall Solution

What to do: every $p$ becomes $\partial S / \partial x$ , and we append $\partial S / \partial t$ . Why: those come from the generating-function relations $p = \partial S / \partial x$ and $K = H + \partial S / \partial t$ , together with the defining demand $K = 0$ (the new Hamiltonian $K$ was defined in the vocabulary box above). Killing $K$ leaves: $\frac{1}{2 m} (\frac{\partial S}{\partial x})^{2} + V (x) + \frac{\partial S}{\partial t} = 0.$ That is the answer. No integration needed — recognition only.

Exercise 1.2 (L1)

State which is the principal function and which is the characteristic function, and the one equation that links them.

Recall Solution

Principal $= S (q, t)$ — carries time.
Characteristic $= W (q)$ — no time inside it (defined in the vocabulary box above).
Link (only valid when $H$ is time-independent, so energy $E$ is conserved): $S (q, t) = W (q) - E t .$ Why the minus $E t$ : if $\partial S / \partial t = - E$ then $H (q, \partial W / \partial q) = E$ , exactly the conserved-energy statement.

Exercise 1.3 (L1)

The HJ equation is a PDE of what order and is it linear or nonlinear? Give the reason in one sentence.

Recall Solution

It is first order (only first derivatives of $S$ appear) and nonlinear (the momentum $\partial S / \partial q$ enters $H$ typically squared, e.g. $(\partial S / \partial x)^{2}$ , and a square is not linear).

Level 2 — Application

Exercise 2.1 (L2)

Free particle, $H = \frac{p ^{2}}{2 m}$ with coordinate $q = x$ . Solve the time-independent HJ equation for $W (x)$ and write full $S (x, t)$ .

Recall Solution

Step 1 — restricted HJ. With $S = W - E t$ : $\frac{1}{2 m} (W^{'})^{2} = E$ . Step 2 — solve for slope. $W^{'} = \pm 2 m E$ (take the $+$ root for rightward motion; the $-$ root gives leftward motion — both are legal branches). Step 3 — integrate. $W = 2 m E x + C$ . Why we drop $C$ : $C$ is a pure additive constant. Since only derivatives of $S$ ever enter the physics ( $p = \partial S / \partial x$ , $Q = \partial S / \partial E$ ), adding a constant to $S$ changes nothing measurable. We set $C = 0$ without loss of generality — the solution stays fully general. Step 4 — assemble. $S = 2 m E x - E t$ . Matches the parent note's Example A.

Exercise 2.2 (L2)

Continuing 2.1, identify $E$ with the new momentum $P = E$ . Compute the constant new coordinate $Q = \partial S / \partial E$ and solve for $x (t)$ .

Recall Solution

Step 1. $Q = \frac{\partial S}{\partial E} = \frac{\partial}{\partial E} (2 m E x - E t) = \frac{m}{2 E} x - t .$ Why: $\partial_{E} 2 m E = 2 m \cdot \frac{1}{2 E} = m /2 E$ . Step 2. $Q$ is constant (because $K = 0 \Rightarrow \dot{Q} = 0$ ); call it $- t_{0}$ , so $m /2 E x = t - t_{0}$ . Step 3. $x (t) = \frac{2 E}{m} (t - t_{0})$ — uniform velocity $v = 2 E / m$ . Sanity: $\frac{1}{2} m v^{2} = E$ . ✅

Exercise 2.3 (L2)

A particle in free fall, $H = \frac{p ^{2}}{2 m} + m g z$ , where the coordinate is the height $q = z$ . Write and separate the HJ equation to get $W^{'} (z)$ .

Recall Solution

Step 1 — HJ: $\frac{1}{2 m} (\partial_{z} S)^{2} + m g z + \partial_{t} S = 0$ . Step 2 — separate $S = W (z) - E t$ . Why this separation is allowed: the Hamiltonian $H = \frac{p ^{2}}{2 m} + m g z$ contains no explicit $t$ , so energy $E$ is conserved. Whenever $H$ is time-independent the whole $t$ -dependence of $S$ can only be a steady drift $- E t$ (that is precisely what $\partial_{t} S = - E$ says). Peeling it off leaves a purely spatial function $W (z)$ — this is the same time-separation move used in every exercise here and in Separation of Variables. Substituting $\partial_{t} S = - E$ : $\frac{1}{2 m} (W^{'})^{2} + m g z = E .$ Step 3 — solve for slope: $W^{'} (z) = \pm 2 m (E - m g z)$ . Notice $W^{'} \to 0$ at the turning height $z = E / (m g)$ where all energy is potential — the momentum vanishes there, exactly as physics demands. Integrating $W^{'}$ again introduces an additive constant which we drop (only derivatives matter, as in 2.1).

Level 3 — Analysis

Exercise 3.1 (L3)

Harmonic oscillator, $H = \frac{p ^{2}}{2 m} + \frac{1}{2} m ω^{2} x^{2} = E$ (coordinate $q = x$ ). Starting from $W^{'} = \pm 2 m (E - \frac{1}{2} m ω^{2} x^{2})$ , derive $x (t)$ explicitly, showing the substitution.

Recall Solution

Step 0 — the two branches. Solving $H = E$ for the slope gives $W^{'} = \pm 2 m (E - \frac{1}{2} m ω^{2} x^{2})$ . The $+$ branch describes the particle moving right (toward $+ A$ ), the $-$ branch moving left. They meet and swap at the turning points $x = \pm A$ where $W^{'} = 0$ . A full oscillation is these two branches stitched together — exactly the L2 lesson. Below we integrate the $+$ branch; the $-$ branch gives the returning half and the two combine into one continuous $sin$ . Step 1 — make the constant coordinate well-posed with an explicit lower limit. First write $W$ as a genuine definite integral from a fixed reference point $x_{0}$ (a constant that does not depend on $E$ ): $W (x, E) = \int_{x_{0}}^{x} 2 m (E - \frac{1}{2} m ω^{2} x^{'2}) d x^{'} .$ Choosing a fixed lower limit is exactly what "dropping the additive constant" means: shifting $x_{0}$ only adds an $x$ -independent constant to $W$ , which never affects any physics (only derivatives of $S$ enter). Because the lower limit $x_{0}$ is $E$ -independent and the upper limit $x$ is $E$ -independent too, differentiation under the integral sign is now legitimate: only the integrand depends on $E$ . With $P = E$ , $Q = \frac{\partial S}{\partial E} = \int_{x_{0}}^{x} \frac{\partial}{\partial E} 2 m (E - \frac{1}{2} m ω^{2} x^{'2}) d x^{'} - t = \int_{x_{0}}^{x} \frac{m d x ^{'}}{2 m ( E - \frac{1}{2} m ω ^{2} x ^{'2} )} - t .$ Why the $\partial_{E}$ pulls down $m$ : $\partial_{E} 2 m (E - \dots) = \frac{2 m}{2 2 m ( \dots )} = \frac{m}{2 m ( \dots )}$ . Step 2 — substitute $x^{'} = \frac{2 E}{m ω ^{2}} sin θ$ . Then the radicand becomes $2 m E cos^{2} θ$ and $d x^{'} = 2 E / m ω^{2} cos θ d θ$ . The integral collapses: $\int \frac{m \cdot 2 E / m ω ^{2} c o s θ d θ}{2 m E c o s θ} = \frac{1}{ω} \int d θ = \frac{θ}{ω} .$ Step 3 — undo the substitution: $θ = arcsin (x \frac{m ω ^{2}}{2 E})$ , so (absorbing the fixed lower-limit contribution into the constant $Q$ ) $Q + t = \frac{1}{ω} arcsin (\dots) + const$ . Step 4 — invert (set $- ω Q + const \equiv ϕ$ ): $x (t) = \frac{2 E}{m ω ^{2}} sin (ω t + ϕ)$ . The amplitude $A = 2 E / (m ω^{2})$ is fixed entirely by the energy $E$ — see the figure below.

The figure below plots this solution so you can see what the algebra produced: the horizontal axis is time $t$ , the vertical axis is position $x (t)$ . The lavender curve is $x (t) = A sin (ω t + ϕ)$ . The two coral dashed lines mark the amplitude $\pm A$ — the classical turning points where the two branches of $W^{'}$ (the $+$ and $-$ roots from Step 0) meet and the momentum $p$ vanishes. The mint dot marks a zero crossing $x = 0$ , where the speed is maximum and all energy is kinetic. Reading the picture: energy $E$ sets how tall the wave is (the amplitude), while $ω$ sets how fast it wiggles.

Figure s01 — HJ solution of the 1D harmonic oscillator. Horizontal axis: time $t$ . Vertical axis: position $x (t) = A sin (ω t + ϕ)$ (lavender). Coral dashed lines: the amplitude $\pm A = \pm 2 E / (m ω^{2})$ , the turning points where momentum $p = 0$ and all energy is potential. Mint dot: a zero crossing $x = 0$ , where speed is maximal and all energy is kinetic. The energy $E$ controls the height of the wave; $ω$ controls its frequency.

Exercise 3.2 (L3)

For that oscillator, verify that the amplitude $A = 2 E / (m ω^{2})$ carries exactly the energy $E$ at maximum displacement.

Recall Solution

At maximum displacement all energy is potential ( $p = 0$ ): $\frac{1}{2} m ω^{2} A^{2} = \frac{1}{2} m ω^{2} \cdot \frac{2 E}{m ω ^{2}} = E . ✅$ The HJ construction is self-consistent: the amplitude the algebra spat out is precisely the classical turning point marked by the coral dashed lines in the figure.

Exercise 3.3 (L3)

Numerical check. Oscillator with $m = 2 kg$ , $ω = 3 s^{- 1}$ , $E = 27 J$ . Find the amplitude $A$ and the maximum speed $v_{m a x}$ .

Recall Solution

Amplitude: $A = \frac{2 E}{m ω ^{2}} = \frac{2 \cdot 27}{2 \cdot 9} = 3 \approx 1.732 m$ . Max speed (all energy kinetic at $x = 0$ ): $\frac{1}{2} m v_{m a x}^{2} = E \Rightarrow v_{m a x} = 2 E / m = 27 = 33 \approx 5.196 m/s$ . Cross-check: $v_{m a x} = ω A = 3 \cdot 3 = 33$ . ✅ Consistent.

Level 4 — Synthesis

Exercise 4.1 (L4)

Separable 2D problem. $H = \frac{p _{x}^{2}}{2 m} + \frac{p _{y}^{2}}{2 m} + \frac{1}{2} m ω^{2} x^{2}$ (oscillator in $x$ , free in $y$ ; the two generalized coordinates are $q_{1} = x$ and $q_{2} = y$ ). Use an additive separation $W = W_{x} (x) + W_{y} (y)$ to split the HJ equation into two independent 1D pieces. Identify the separation constants.

Recall Solution

Step 1 — restricted HJ: $\frac{1}{2 m} (W_{x}^{'})^{2} + \frac{1}{2} m ω^{2} x^{2} + \frac{1}{2 m} (W_{y}^{'})^{2} = E$ . Step 2 — why additive? Because $H$ is a sum of an $x$ -only part and a $y$ -only part. Guessing $W = W_{x} + W_{y}$ makes each $\partial W / \partial x = W_{x}^{'}$ depend on $x$ only (Separation of Variables). Step 3 — regroup: $function of x \frac{1}{2 m} (W_{x}^{'})^{2} + \frac{1}{2} m ω^{2} x^{2} = function of y E - \frac{1}{2 m} (W_{y}^{'})^{2} .$ A function of $x$ equals a function of $y$ for all $x, y$ only if both equal a constant. Call it $α_{x}$ . Step 4 — the two 1D equations: $\frac{1}{2 m} (W_{x}^{'})^{2} + \frac{1}{2} m ω^{2} x^{2} = α_{x}, \frac{1}{2 m} (W_{y}^{'})^{2} = E - α_{x} \equiv α_{y} .$ Separation constants: $α_{x}$ (energy in the $x$ -oscillator) and $α_{y} = E - α_{x}$ (energy in the free $y$ -motion). Following our notation guard, we identify these constants with the new momenta: $P_{1} = α_{x}, P_{2} = α_{y}$ — same objects, two hats. This links directly to Action-Angle Variables, where such constants become the actions.

Exercise 4.2 (L4)

In 4.1, solve the $y$ -piece completely and give $y (t)$ .

Recall Solution

Step 1: $(W_{y}^{'})^{2} = 2 m α_{y} \Rightarrow W_{y}^{'} = \pm 2 m α_{y} \Rightarrow W_{y} = 2 m α_{y} y$ (taking the $+$ branch for rightward drift; additive constant dropped as before). Step 2 — conjugate coordinate: treat $α_{y}$ as the new momentum $P_{2}$ , so $Q_{y} = \partial S / \partial α_{y}$ . The relevant piece of $S$ is $W_{y} - α_{y} t$ (the $y$ -share of $E t$ ), so $Q_{y} = \frac{m}{2 α _{y}} y - t = const .$ Step 3 — invert: $y (t) = \frac{2 α _{y}}{m} (t - t_{0})$ — a free particle drifting at speed $2 α_{y} / m$ , exactly as expected. The oscillator part follows Exercise 3.1 with $E \to α_{x}$ .

Level 5 — Mastery

Exercise 5.1 (L5)

The bridge to quantum mechanics. Insert $ψ = e^{i S /ℏ}$ (the "wave built from the action") into the time-dependent Schrodinger Equation $i ℏ \partial_{t} ψ = - \frac{ℏ ^{2}}{2 m} \partial_{x}^{2} ψ + V ψ$ (coordinate $q = x$ ). Show that as $ℏ \to 0$ you recover the HJ equation, and identify the leftover term.

Recall Solution

Step 1 — compute derivatives of $ψ = e^{i S /ℏ}$ . Using the chain rule (and, for the second $x$ -derivative, the product rule on the two factors it creates): $\partial_{t} ψ = \frac{i}{ℏ} (\partial_{t} S) ψ, \partial_{x} ψ = \frac{i}{ℏ} (\partial_{x} S) ψ,$ $\partial_{x}^{2} ψ = \partial_{x} [\frac{i}{ℏ} (\partial_{x} S) ψ] = [\frac{i}{ℏ} \partial_{x}^{2} S + (\frac{i}{ℏ} \partial_{x} S)^{2}] ψ = [- \frac{1}{ℏ ^{2}} (\partial_{x} S)^{2} + \frac{i}{ℏ} \partial_{x}^{2} S] ψ .$ Notice the two terms in $\partial_{x}^{2} ψ$ : the square from differentiating the exponent twice, and the $\partial_{x}^{2} S$ from differentiating the slope once — this second term is the whole point of the exercise. Step 2 — substitute and divide by $ψ$ (nonzero everywhere): $i ℏ \cdot \frac{i}{ℏ} \partial_{t} S = - \frac{ℏ ^{2}}{2 m} [- \frac{1}{ℏ ^{2}} (\partial_{x} S)^{2} + \frac{i}{ℏ} \partial_{x}^{2} S] + V .$ Step 3 — simplify. Left side: $i ℏ \cdot \frac{i}{ℏ} = i^{2} = - 1$ , so it is $- \partial_{t} S$ . Right side, distribute the $- \frac{ℏ ^{2}}{2 m}$ : $- \partial_{t} S = \frac{1}{2 m} (\partial_{x} S)^{2} - \frac{i ℏ}{2 m} \partial_{x}^{2} S + V .$ Step 4 — rearrange into HJ shape (move everything to one side): $\frac{1}{2 m} (\partial_{x} S)^{2} + V + \partial_{t} S = \frac{i ℏ}{2 m} \partial_{x}^{2} S .$ Step 5 — take $ℏ \to 0$ . The right-hand side is proportional to $ℏ$ , so it vanishes in the limit, leaving $\frac{1}{2 m} (\partial_{x} S)^{2} + V + \partial_{t} S = 0$ — exactly the Hamilton-Jacobi equation. The leftover $\frac{i ℏ}{2 m} \partial_{x}^{2} S$ is the first quantum correction (the seed of the WKB expansion). So HJ is the classical (short-wavelength) limit of Schrödinger's equation — Hamilton anticipated wave mechanics by a century.

Exercise 5.2 (L5)

Design problem. You are told the desired trajectory is $x (t) = a t$ (constant velocity $a$ ) for a free particle. Reverse-engineer the principal function $S (x, t)$ that generates it, and confirm $p = \partial S / \partial x$ gives the right momentum.

Recall Solution

Step 1 — pick energy. Constant speed $a$ means $E = \frac{1}{2} m a^{2}$ . Step 2 — use free-particle $S$ : $S = 2 m E x - E t = (ma) x - \frac{1}{2} m a^{2} t$ , since $2 m E = m^{2} a^{2} = ma$ . Step 3 — check momentum: $p = \partial S / \partial x = ma$ . ✅ Correct: constant momentum $ma$ for speed $a$ . Step 4 — check the constant coordinate: $Q = \partial S / \partial E$ . With $E = \frac{1}{2} m a^{2}$ , better to differentiate the general form: $Q = m /2 E x - t = x / a - t$ . Setting $Q = 0$ gives $x = a t$ . ✅ The design reproduces the target trajectory.

Exercise 5.3 (L5)

Numerical mastery. Free particle, $m = 4 kg$ , prescribed speed $a = 5 m/s$ . Compute $E$ , the momentum $p$ , and evaluate $S$ at $x = 10 m, t = 2 s$ .

Recall Solution

$E = \frac{1}{2} m a^{2} = \frac{1}{2} \cdot 4 \cdot 25 = 50 J$ .
$p = ma = 4 \cdot 5 = 20 kg\cdotm/s$ .
$S = p x - E t = 20 \cdot 10 - 50 \cdot 2 = 200 - 100 = 100 J\cdots$ . All consistent with $2 m E = 2 \cdot 4 \cdot 50 = 400 = 20 = p$ . ✅

Recall Final self-test (hide and answer)

Order and linearity of HJ? ::: First order, nonlinear.
How many separation constants for the 2D oscillator-plus-free system? ::: Two: $α_{x}$ and $α_{y} = E - α_{x}$ .
What term survives Schrödinger→HJ before $ℏ \to 0$ ? ::: $\frac{i ℏ}{2 m} \partial_{x}^{2} S$ .
Amplitude of oscillator with $m = 2, ω = 3, E = 27$ ? ::: $3 \approx 1.732 m$ .

Where does the classical limit of

ψ = e^{i S /ℏ}

live?

ℏ \to 0

, giving the HJ equation exactly.

For the 2D separable system, what are the new momenta?

The separation constants

P_{1} = α_{x}, P_{2} = α_{y} = E - α_{x}