\subsection{Action-Angle Variables}

This subsection develops action-angle variables for integrable periodic systems. We present the canonical transformation that reduces any periodic system to trivial dynamics where the momenta are constant and the angles advance uniformly. The central insight is geometric: the action variable measures the area enclosed by the orbit in phase space, and the angle variable measures where along that orbit the system currently stands.

\nt{Recap of separation of variables}{
The characteristic function $W(q_1,\dots,q_n)$ found by separation in A.02 contains the complete solution to the time--independent Hamilton--Jacobi equation. Once $W$ is known, Jacobi's theorem gives every trajectory. But $W$ alone does not make the periodicity of the motion immediately visible. Here we repackage the information contained in $W$ into action--angle variables, where the periodic structure and oscillation frequencies emerge directly.}

\dfn{Action and angle variables}{
For a periodic degree of freedom with generalized coordinate $q_i$ and conjugate momentum $p_i$, the orbit in the $(q_i,p_i)$ phase plane forms a closed curve. The action variable $J_i$ is defined as the area enclosed by this curve, divided by $2\pi$:
\[
J_i = \frac{1}{2\pi}\oint p_i\,dq_i.
\]
The integral is taken counter--clockwise around one complete closed orbit. Geometrically, $J_i$ is proportional to the phase--space area of one cycle: it equals the enclosed area divided by $2\pi$. For a harmonic oscillator the orbit is an ellipse and the area is a straightforward ellipse computation; for an infinite well the orbit is a rectangle. The angle variable $w_i$ is the canonically conjugate coordinate to $J_i$, defined from Hamilton's characteristic function by
\[
w_i = \pdv{W}{J_i}.
\]
The angle variable $w_i$ is measured in radians and ranges over $[0,2\pi)$. It increases by exactly $2\pi$ during one complete period of the motion. Together, $(w_i,J_i)$ form a set of canonical coordinates obtained from $(q_i,p_i)$ by a canonical transformation.}

\nt{The angle variable as a phase clock}{
The angle variable $w_i$ tracks progress through one complete cycle of periodic motion, much like the hand of a clock. At $w_i = 0$ the system sits at some chosen reference point on its orbit --- for example, the maximum positive displacement. As time advances, $w_i$ sweeps through $[0,2\pi)$ and reaches $2\pi$ precisely when the system returns to its starting phase-space point and the cycle repeats. Because $w_i = \pdv{W}{J_i}$ and the Hamiltonian expressed in action variables is time--independent, Hamilton's equations give $\dot{w}_i = \pdv{\mcH}{J_i} = \omega_i$, a constant. The angle therefore advances uniformly, just as a clock hand rotates at constant angular speed. This makes action--angle variables the natural language for discussing periodic oscillation.}

\ex{Phase-space ellipse for the simple harmonic oscillator}{
The simplest illustration of the geometric definition comes from the harmonic oscillator of mass $m$ and natural frequency $\omega_0$. The energy is $E = p^2/(2m) + \tfrac{1}{2}m\omega_0^2 x^2$. Rearranging, the energy-level curve in phase space is
\[
\frac{x^2}{A^2} + \frac{p^2}{p_{\max}^2} = 1,
\qquad\text{where}\qquad
A = \sqrt{\frac{2E}{m\omega_0^2}},
\quad
p_{\max} = m\omega_0 A.
\]
This is an ellipse in the $(x,p)$ phase plane with semiaxes $A$ along the position axis and $p_{\max} = m\omega_0 A$ along the momentum axis. The area of an ellipse is $\pi$ times the product of its semiaxes:
\[
\text{Area} = \pi A\cdot p_{\max}
= \pi A\cdot m\omega_0 A
= \pi m\omega_0 A^2.
\]
Substituting $A^2 = 2E/(m\omega_0^2)$ gives
\[
\text{Area} = \pi m\omega_0\cdot\frac{2E}{m\omega_0^2}
= \frac{2\pi E}{\omega_0}.
\]
The action variable is the area divided by $2\pi$:
\[
J = \frac{1}{2\pi}\cdot\frac{2\pi E}{\omega_0}
= \frac{E}{\omega_0}.
\]
Inverting, $E(J) = \omega_0 J$, and the oscillation frequency is
\[
\omega = \pdv{E}{J} = \omega_0.
\]
The frequency equals the natural angular frequency $\omega_0$, measured in radians per second, and the period is $T = 2\pi/\omega_0$. This is a direct consequence of isochrony: all oscillations of a harmonic oscillator have the same period regardless of energy.}

A system with $n$ degrees of freedom is called \emph{completely integrable} when it possesses $n$ independent constants of motion that are in involution --- their pairwise Poisson brackets all vanish, $\{F_i, F_j\} = 0$ for all $i,j = 1,\dots,n$. Complete integrability guarantees that the $2n$--dimensional phase space is foliated by invariant $n$--dimensional tori, each labeled by constant values of the action variables. On each torus the dynamics reduces to uniform rotation of the angles. For such systems the Hamiltonian depends only on the actions: $\mcH = \mcH(J_1,\dots,J_n)$.

\thm{Hamilton's equations in action-angle variables}{
Let $\mcH = \mcH(J_1,\dots,J_n)$ be the Hamiltonian expressed as a function of the action variables alone. Hamilton's canonical equations in the $(w,J)$ variables are
\[
\dot{J}_i = -\pdv{\mcH}{w_i} = 0,
\qquad
\dot{w}_i = \pdv{\mcH}{J_i} \equiv \omega_i.
\]
The action variables $J_i$ are constant in time because the angle variables do not appear in $\mcH$ and therefore the actions experience no conjugate forces. The angle variables advance linearly:
\[
w_i(t) = \omega_i t + w_i(0),
\]
with constant angular frequencies $\omega_i = \pdv{\mcH}{J_i}$ measured in radians per second. Since the angle variable $w_i$ increases by $2\pi$ during one complete period of the $i$-th degree of freedom, the oscillation period is $T_i = 2\pi/\omega_i$.}

The frequency $\omega_i$ provides direct access to the temporal characteristics of the motion without solving differential equations. For a single degree of freedom, the procedure is mechanical: evaluate $J = \frac{1}{2\pi}\oint p\,dq$ by computing the enclosed phase--space area or by direct integration, invert the resulting relation to express $E = E(J)$, and differentiate to find $\omega = \pdv{E}{J}$. The period follows from $T = 2\pi/\pdv{E}{J}$. When $T$ turns out independent of amplitude the system is isochronous, a property shared by both the harmonic oscillator and the Kepler problem, and one that proves significant in the quantum limit.

\ex{Direct integration for the simple harmonic oscillator}{
We confirm the geometric result of the previous example by direct computation. For the harmonic oscillator $\mcH = p^2/(2m) + \tfrac{1}{2}m\omega_0^2 x^2$, at fixed energy $E$ the momentum is $p = \pm\sqrt{2mE - m^2\omega_0^2 x^2}$ and the turning points are $x = \pm A$ with $A = \sqrt{2E/(m\omega_0^2)}$. The raw phase--space integral around the orbit is
\[
\oint p\,dx = 2\int_{-A}^{A}\sqrt{2mE - m^2\omega_0^2 x^2}\,dx.
\]
Substitute $x = A\sin\phi$, so $dx = A\cos\phi\,d\phi$ and the limits become $\phi = -\pi/2$ to $\pi/2$. The integrand simplifies:
\[
\sqrt{2mE - m^2\omega_0^2 A^2\sin^2\phi}
= \sqrt{2mE - 2mE\sin^2\phi}
= \sqrt{2mE}\cos\phi,
\]
since $m^2\omega_0^2 A^2 = 2mE$. The integral becomes
\[
\oint p\,dx = 2\sqrt{2mE}\cdot A\int_{-\pi/2}^{\pi/2}\cos^2\phi\,d\phi.
\]
Using $\int_{-\pi/2}^{\pi/2}\cos^2\phi\,d\phi = \pi/2$, this gives
\[
\oint p\,dx = 2\sqrt{2mE}\cdot A\cdot\frac{\pi}{2}
= \pi\sqrt{2mE}\cdot A.
\]
Substituting $A = \sqrt{2E/(m\omega_0^2)}$ yields
\[
\oint p\,dx = \pi\sqrt{2mE}\cdot\sqrt{\frac{2E}{m\omega_0^2}}
= \frac{2\pi E}{\omega_0}.
\]
Dividing by $2\pi$, we recover
\[
J = \frac{1}{2\pi}\cdot\frac{2\pi E}{\omega_0}
= \frac{E}{\omega_0},
\qquad
E(J) = \omega_0 J,
\qquad
\omega = \pdv{E}{J} = \omega_0,
\]
in complete agreement with the geometric calculation. The angular frequency equals $\omega_0$ directly, with no separate $2\pi$ factors to track.}

\nt{Degeneracy and closed orbits in the Kepler problem}{
The Kepler problem (gravitational or electrostatic $V = -k/r$) provides a remarkable illustration of frequency degeneracy. It has three independent action variables --- $J_r$ for radial motion, $J_\theta$ for polar angle, and $J_\phi$ for azimuthal rotation. The energy in action--angle variables depends only on their sum:
\[
E = -\frac{mk^2}{2(J_r + J_\theta + J_\phi)^2}.
\]
The three frequency derivatives are therefore identical:
\[
\omega_r = \pdv{E}{J_r} = \pdv{E}{J_\theta} = \pdv{E}{J_\phi}.
\]
Every degree of freedom oscillates at the same angular frequency. On the three--dimensional invariant torus this means all three angle variables advance by $2\pi$ simultaneously after one period, so the trajectory retraces itself and forms a closed curve. In contrast, when frequency ratios are irrational, motion on a torus fills the surface densely without ever repeating --- the so--called irrational winding. Because the Kepler frequencies are equal, no such winding occurs: every bound Kepler orbit closes exactly after one period, producing the familiar planetary ellipses. This is the degeneracy singled out by Bertrand's theorem, which proves that only the Kepler potential $V \propto -1/r$ and the harmonic potential $V \propto r^2$ among all central potentials produce closed bounded orbits. Adding a small perturbing term, such as $V = -k/r + \epsilon/r^2$, breaks the degeneracy: the frequencies separate, the orbit no longer closes, and the perihelion precesses.}

\qs{Particle in a one-dimensional infinite potential well}{
A particle of mass $m$ is confined to a region $0 < x < L$ by infinite potential walls, so $V(x) = 0$ for $0 < x < L$ and $V = \infty$ elsewhere. Inside the well the Hamiltonian is $\mcH = p^2/(2m)$ and the total energy is $E$.

\begin{enumerate}[label=(\alph*)]
\item Compute the action variable $J = \frac{1}{2\pi}\oint p\,dx$ for this system, showing that $J = L\sqrt{2mE}/\pi$.
\item Express the energy as $E(J)$ and compute the frequency $\omega = \pdv{E}{J}$ and the period $T = 2\pi/\omega$. Show that $T = 2L\sqrt{m/(2E)}$, which equals the time for the particle to travel the distance $2L$ at speed $v = \sqrt{2E/m}$.
\item For an electron with mass $m = 9.11\times 10^{-31}\,\mathrm{kg}$ confined to a region of width $L = 1.00\times 10^{-10}\,\mathrm{m}$ with total energy $E = 1.00\,\mathrm{eV} = 1.60\times 10^{-19}\,\mathrm{J}$, compute the numerical values of $J$ (in $\mathrm{kg\!\cdot\!m^2/s}$) and $T$ (in seconds).
\end{enumerate}}

\sol \textbf{Part (a).} Inside the well the particle has kinetic energy $E = p^2/(2m)$, so the magnitude of momentum is $|p| = \sqrt{2mE}$ and is independent of position. The particle travels back and forth between the walls at $x = 0$ and $x = L$. During the forward leg the momentum is $p = +\sqrt{2mE}$ and during the return leg $p = -\sqrt{2mE}$.

The phase--space integral around one complete cycle is
\[
\oint p\,dx = \int_{0}^{L}\sqrt{2mE}\,dx + \int_{L}^{0}\left(-\sqrt{2mE}\right)\,dx.
\]
Each integral equals $L\sqrt{2mE}$, so the total enclosed area is
\[
\oint p\,dx = L\sqrt{2mE} + L\sqrt{2mE} = 2L\sqrt{2mE}.
\]
Dividing by $2\pi$ gives the action variable:
\[
J = \frac{1}{2\pi}\cdot 2L\sqrt{2mE}
= \frac{L\sqrt{2mE}}{\pi}.
\]

\textbf{Part (b).} Solve the result from part (a) for $E$:
\[
\frac{\pi J}{L} = \sqrt{2mE},
\qquad
\frac{\pi^2 J^2}{L^2} = 2mE,
\qquad
E(J) = \frac{\pi^2 J^2}{2mL^2}.
\]
Differentiate with respect to $J$ to find the angular frequency:
\[
\omega = \pdv{E}{J} = \frac{\pi^2 J}{mL^2}.
\]
The period is $T = 2\pi/\omega$:
\[
T = \frac{2\pi}{\pi^2 J/(mL^2)}
= \frac{2mL^2}{\pi J}.
\]
Substitute $J = L\sqrt{2mE}/\pi$ to express $T$ in terms of $E$:
\[
T = \frac{2mL^2}{\pi\cdot L\sqrt{2mE}/\pi}
= \frac{2mL^2}{L\sqrt{2mE}}
= \frac{2mL}{\sqrt{2mE}}
= 2L\sqrt{\frac{m}{2E}}.
\]
Independently, the particle's speed inside the well is $v = \sqrt{2E/m}$ and the round--trip distance is $2L$. The travel time for one complete cycle is
\[
T = \frac{2L}{v} = 2L\sqrt{\frac{m}{2E}},
\]
which agrees exactly with the action--angle result.

\textbf{Part (c).} The given values are $m = 9.11\times 10^{-31}\,\mathrm{kg}$, $L = 1.00\times 10^{-10}\,\mathrm{m}$, and $E = 1.60\times 10^{-19}\,\mathrm{J}$.

First compute the product $2mE$:
\[
2mE = 2(9.11\times 10^{-31})(1.60\times 10^{-19})\,\mathrm{kg^2\!\cdot\!m^2/s^2}
= 2.92\times 10^{-49}\,\mathrm{kg^2\!\cdot\!m^2/s^2}.
\]
Taking the square root:
\[
\sqrt{2mE} = 5.40\times 10^{-25}\,\mathrm{kg\!\cdot\!m/s}.
\]
Now compute the action variable $J = L\sqrt{2mE}/\pi$:
\[
J = \frac{(1.00\times 10^{-10})(5.40\times 10^{-25})}{\pi}\,\mathrm{kg\!\cdot\!m^2/s}
= \frac{5.40\times 10^{-35}}{\pi}\,\mathrm{kg\!\cdot\!m^2/s}
= 1.72\times 10^{-35}\,\mathrm{kg\!\cdot\!m^2/s}.
\]
Next, the period $T = 2L\sqrt{m/(2E)}$. Compute $m/(2E)$:
\[
\frac{m}{2E} = \frac{9.11\times 10^{-31}}{2(1.60\times 10^{-19})}\,\mathrm{kg/J}
= 2.85\times 10^{-12}\,\mathrm{s^2/m^2}.
\]
Taking the square root:
\[
\sqrt{\frac{m}{2E}} = 1.69\times 10^{-6}\,\mathrm{s/m}.
\]
Multiplying by $2L$:
\[
T = 2(1.00\times 10^{-10})(1.69\times 10^{-6})\,\mathrm{s}
= 3.37\times 10^{-16}\,\mathrm{s}.
\]

Therefore,
\[
J = 1.72\times 10^{-35}\,\mathrm{kg\!\cdot\!m^2/s},
\qquad
T = 3.37\times 10^{-16}\,\mathrm{s}.
\]