\subsection{Rigid Rotator and Particle on a Sphere}

This subsection treats the simplest two-degree-of-freedom Hamilton--Jacobi problem: a particle constrained to the surface of a sphere. The rigid rotator is the classical prototype for spherical angular motion, appearing in the dynamics of diatomic molecules, symmetric rotating tops, and any system whose configuration is naturally described by two angles rather than Cartesian coordinates. It connects directly to material in Unit 5 m5-4 (the moment of inertia $I = mr^2$ for a point mass at distance $r$ from the axis) and Unit 6 m6-1 (the rotational kinetic energy $\tfrac12 I\omega^2$).

\nt{Opening motivation}{The rigid rotator is the simplest genuine two-degree-of-freedom system. Unlike coupled oscillators or the Kepler problem, it has no radial degree of freedom to simplify away, yet it remains analytically solvable on the spot. The geometry of the sphere introduces a curvature-dependent kinetic energy, and the resulting Hamilton--Jacobi equation demonstrates how separability survives in curvilinear coordinates. Physically the model describes the end-to-end rotation of a diatomic molecule, the spinning of a symmetric top with fixed nutation angle, and the angular part of the free-particle Schrödinger equation.}

\dfn{From Lagrangian to Hamiltonian for the rigid rotator}{
Consider a particle of mass $m$ constrained to move on a sphere of fixed radius $r$ with no potential energy. The kinetic energy in spherical coordinates with fixed $r$ is
\[
T = \tfrac12 m r^2 \dot{\theta}^2 + \tfrac12 m r^2 \sin^2\theta\;\dot{\phi}^2.
\]
Because $V = 0$ the Lagrangian is $\mcL = T$. The two generalized coordinates are the polar angle $\theta \in [0,\pi]$ and the azimuthal angle $\phi \in [0,2\pi)$. The conjugate momenta are
\[
p_\theta = \pdv{\mcL}{\dot{\theta}} = m r^2 \dot{\theta},
\qquad
p_\phi = \pdv{\mcL}{\dot{\phi}} = m r^2 \sin^2\theta\;\dot{\phi}.
\]
Invert these relations to express the velocities in terms of momenta:
\[
\dot{\theta} = \frac{p_\theta}{m r^2},
\qquad
\dot{\phi} = \frac{p_\phi}{m r^2 \sin^2\theta}.
\]
The Hamiltonian is the Legendre transform $\mcH = p_\theta \dot{\theta} + p_\phi \dot{\phi} - \mcL$, which for a velocity-quadratic Lagrangian with no explicit time dependence simply equals the total energy:
\[
\mcH = \frac{p_\theta^2}{2m r^2} + \frac{p_\phi^2}{2m r^2 \sin^2\theta}.
\]
Introducing the moment of inertia $I = m r^2$ gives the compact form
\[
\mcH = \frac{p_\theta^2}{2I} + \frac{p_\phi^2}{2I \sin^2\theta}.
\]
The Hamiltonian has no explicit time dependence, so energy is conserved and $\mcH = E$ is a constant.}

\nt{Cross-reference to AP core material}{Notice how the two terms in the Hamiltonian are each of the form $p^2/(2I_{\text{eff}})$. The $p_\phi^2$ term carries an extra $\sin^2\theta$ in the denominator because the effective radius of the circular $\phi$-orbit is $r\sin\theta$, not $r$. When the particle sits on the equator $\theta = \pi/2$, the effective moment of inertia for azimuthal rotation is $I = mr^2$, matching the Unit 5 formula $I = mr^2$ for a point mass. When the particle moves near a pole, the effective radius shrinks and the same canonical momentum $p_\phi$ produces a faster angular speed $\dot{\phi}$, exactly as the Unit 6 relation $K_{\text{rot}} = \tfrac12 I\omega^2$ predicts when $I$ decreases while kinetic energy stays fixed.}

\thm{Separation of the rigid-rotator Hamilton--Jacobi equation}{
With $H = p_\theta^2/(2I) + p_\phi^2/(2I\sin^2\theta)$, the time-independent Hamilton--Jacobi equation $H\!\bigl(\theta,\phi,\pdv{\mcS}{\theta},\pdv{\mcS}{\phi}\bigr) = E$ separates as follows. Because $\phi$ is cyclic, $\pdv{\mcS}{\phi} = L_z$, a constant. Setting $\mcS = W_\theta(\theta) + L_z\phi - Et$ reduces the equation to
\[
\left(\der{W_\theta}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = 2IE \equiv L^2,
\]
where $L$ is the total angular momentum and $L^2 = 2IE$ is the second separation constant. The $\theta$-equation is
\[
\der{W_\theta}{\theta} = \pm\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}},
\]
which integrates to
\[
W_\theta(\theta) = \int\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}\;\dd\theta.
\]}

\pf{Derivation of the separated equations}{
The time-independent Hamilton--Jacobi equation is obtained by substituting $p_\theta = \pdv{\mcS}{\theta}$ and $p_\phi = \pdv{\mcS}{\phi}$ into the Hamiltonian:
\[
\frac{1}{2I}\left(\pdv{\mcS}{\theta}\right)^2 + \frac{1}{2I\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 = E.
\]
Multiply both sides by $2I$:
\[
\left(\pdv{\mcS}{\theta}\right)^2 + \frac{1}{\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 = 2IE.
\]
The coordinate $\phi$ is absent from the Hamiltonian, so $\phi$ is cyclic. The contribution of the cyclic coordinate to the characteristic function is linear:
\[
\pdv{\mcS}{\phi} = L_z,
\]
where $L_z$ is the conserved $z$-component of the angular momentum. Substituting this constant into the HJ equation gives
\[
\left(\pdv{\mcS}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = 2IE.
\]
Seeking an additive separation $\mcS = W_\theta(\theta) + L_z\phi$, the partial derivative $\pdv{\mcS}{\theta}$ becomes the ordinary derivative $\der{W_\theta}{\theta}$:
\[
\left(\der{W_\theta}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = 2IE.
\]
Define the separation constant $L^2 = 2IE$, which has the dimensions of angular-momentum squared and equals the square of the total angular momentum. Solving for the derivative:
\[
\der{W_\theta}{\theta} = \pm\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}.
\]
The right-hand side vanishes at the turning points where $L^2 = L_z^2/\sin^2\theta$, or equivalently $\sin\theta = |L_z|/L$. Between the turning points the particle oscillates in $\theta$, tracing a cone on the surface of the sphere. The polar angle sweeps between $\theta_{\min} = \arcsin(|L_z|/L)$ and $\theta_{\max} = \pi - \theta_{\min}$, while $\phi$ advances monotonically.}

\nt{Rational and irrational orbit closure}{The rigid rotator has two frequencies, $\omega_\theta$ for the polar oscillation and $\omega_\phi$ for the azimuthal precession. If the ratio $\omega_\theta/\omega_\phi = p/q$ is rational (with $p$ and $q$ coprime integers), then after exactly $q$ complete cycles of $\theta$ oscillation the orbit returns to its starting phase and retraces itself. If the ratio is irrational, the orbit never closes and the trajectory wraps around the invariant torus in phase space, densely filling a two-dimensional ring on the sphere surface. The irrational case exemplifies quasi-periodic motion: the particle explores an angular band without ever repeating its exact configuration. For the rigid rotator the two frequencies turn out equal in magnitude, so every orbit is degenerate and closed with $p/q = \pm 1$. This degeneracy --- identical frequencies for dynamically independent degrees of freedom --- is a special property of the $1/r^2$ force structure and the spherical symmetry of the configuration space.}

\nt{Evaluating the $\theta$ integral}{The integral $W_\theta(\theta) = \int\sqrt{L^2 - L_z^2/\sin^2\theta}\;\dd\theta$ can be rearranged by factoring the square root:
\[
\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}
= \frac{\sqrt{L^2\sin^2\theta - L_z^2}}{\sin\theta}.
\]
A useful substitution is $u = \cos\theta$, for which $\dd\theta = -\dd u/\sqrt{1-u^2}$ and $\sin^2\theta = 1-u^2$. The integral then reads
\[
\int \frac{\sqrt{L^2(1-u^2) - L_z^2}}{\sqrt{1-u^2}}\cdot \left(-\frac{\dd u}{\sqrt{1-u^2}}\right)
= -\int \frac{\sqrt{(L^2-L_z^2) - L^2 u^2}}{1-u^2}\;\dd u.
\]
This has the structure of an elliptic integral in general. The denominator $(1-u^2)$ together with the quadratic radicand $\sqrt{(L^2-L_z^2) - L^2 u^2}$ prevents a simple elementary antiderivative. However, for the purpose of computing the action variable --- a closed-loop integral between turning points --- the explicit antiderivative is not needed. The turning points in $u$-space occur at $u = \pm L_z/L$, where the radicand vanishes. A second substitution $u = (L_z/L)\cos\psi$ converts the definite integral to a standard trigonometric form that evaluates in closed fashion, yielding $J_\theta = |L| - |L_z|$ without computing an indefinite integral.}

\nt{Action-angle variables for the rigid rotator}{The two independent action variables are computed by integrating the conjugate momenta over their respective cycles, each divided by $2\pi$. For the azimuthal coordinate $\phi$, which is $2\pi$-periodic:
\[
J_\phi = \frac{1}{2\pi}\oint p_\phi\,\dd\phi = \frac{1}{2\pi}\int_{0}^{2\pi} L_z\,\dd\phi = L_z.
\]
Physically, $J_\phi = L_z$ equals the vertical spin: the amount of angular momentum aligned with the symmetry axis. A larger $|L_z|$ means the motion stays closer to the equator circle. For the polar coordinate $\theta$, which oscillates between turning points, the full cycle traverses the range twice:
\[
J_\theta = \frac{1}{2\pi}\oint p_\theta\,\dd\theta
= \frac{1}{\pi}\int_{\theta_{\min}}^{\theta_{\max}}\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}\;\dd\theta
= |L| - |L_z|.
\]
Physically, $J_\theta = |L| - |L_z|$ measures the tilt from the equator: the ``missing'' angular momentum that keeps the trajectory from being perfectly planar. When $|L| = |L_z|$ the tilt vanishes, $J_\theta = 0$, and the motion is confined to the equatorial circle. When $L_z = 0$ there is no preferred axis and $J_\theta = |L|$, the full angular momentum goes into the polar oscillation. Inverting gives $L_z = J_\phi$ and $|L| = J_\theta + |J_\phi|$. The Hamiltonian in action variables is
\[
H(J_\theta,J_\phi) = \frac{\bigl(J_\theta + |J_\phi|\bigr)^2}{2I}.
\]}

\nt{From classical action to quantum numbers}{The Bohr--Sommerfeld quantization rule states that each action variable must be an integer multiple of Planck\normalsize{}'s reduced constant: $J_i = n_i\hbar$. Applying this to the rigid rotator, the azimuthal action quantizes as $J_\phi = m\hbar$ with $m$ an integer, giving directly $L_z = m\hbar$. The polar action quantizes as $J_\theta = \ell_\theta\hbar$ with $\ell_\theta$ a non-negative integer. Since $|L| = J_\theta + |J_\phi|$, we obtain $|L| = (\ell_\theta + |m|)\hbar$. Defining the total angular-momentum quantum number $\ell = \ell_\theta + |m|$, the condition $\ell_\theta \ge 0$ becomes $\ell \ge |m|$. Therefore $L^2 = |L|^2 = \ell^2\hbar^2$, and the semiclassical energy is $E = \ell^2\hbar^2/(2I)$. The fully quantum-mechanical result from solving the angular Schrödinger equation replaces $\ell^2$ with $\ell(\ell+1)$, giving $E = \ell(\ell+1)\hbar^2/(2I)$. The factor $\ell(\ell+1)$ rather than $\ell^2$ emerges from the non-commutativity of $L_x$, $L_y$, and $L_z$ --- no quantum state can simultaneously have definite values of all three components, so the total angular momentum magnitude always exceeds $|L_z|$ by a half-integer step. The Bohr--Sommerfeld approach captures the ladder of energy levels, the integer structure of quantum numbers, and the constraint $\ell \ge |m|$ that limits how far the spin axis can tilt. However, it cannot produce the $+\ell$ correction inside $\ell(\ell+1)$. For large $\ell$ the semiclassical and quantum results agree well, since $\ell(\ell+1) \approx \ell^2$ when $\ell \gg 1$. This bridge from classical action variables to quantum angular momentum was the key step in old quantum theory and its subsequent replacement by matrix and wave mechanics.}

\cor{Equatorial orbit}{
When the total angular momentum equals the absolute value of its $z$-component, $L = |L_z|$, the square root in the $\theta$-equation vanishes everywhere except at $\sin\theta = 1$. The polar momentum $p_\theta = \der{W_\theta}{\theta}$ is zero, so the motion is confined to the equator $\theta = \pi/2$. From Hamilton\normalsize{}'s equations, the azimuthal velocity is
\[
\dot{\phi} = \pdv{H}{p_\phi} = \frac{p_\phi}{I\sin^2\theta} = \frac{L_z}{I}
\]
since $\sin(\pi/2) = 1$. The azimuthal angle advances at constant rate:
\[
\phi(t) = \frac{L_z}{I}\,t + \phi_0,
\]
representing uniform circular motion at angular speed $\omega = |L_z|/I$. This orbit corresponds to $J_\theta = 0$ and is the only truly one-degree-of-freedom subcase of the rigid rotator.}

\ex{Action-angle frequencies for the rigid rotator}{
Using the Hamiltonian in action variables,
\[
H = \frac{\bigl(J_\theta + |J_\phi|\bigr)^2}{2I},
\]
the two frequencies are
\[
\omega_\theta = \pdv{H}{J_\theta}
= \frac{J_\theta + |J_\phi|}{I}
= \frac{|L|}{I},
\qquad
\omega_\phi = \pdv{H}{J_\phi}
= \pm\frac{J_\theta + |J_\phi|}{I}
= \pm\frac{|L|}{I}.
\]
Here the sign conventionally matches the sign of $J_\phi$, so a negative $J_\phi$ gives retrograde azimuthal motion at the same frequency magnitude. The two frequencies are equal in magnitude, confirming the degeneracy noted above: every trajectory on the sphere is a closed orbit with rational frequency ratio $1:1$. For the special case $J_\theta = 0$ (equatorial orbit), the polar frequency is defined by continuity and the motion is purely azimuthal.}

\qs{Diatomic molecule as a rigid rotator}{A diatomic molecule is modeled as a rigid rotator with moment of inertia
\[
I = 1.46\times 10^{-46}\,\mathrm{kg\!\cdot\!m^2}.
\]
The total angular momentum is $L = 2\hbar$, where $\hbar = 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s}$.

\begin{enumerate}[label=(\alph*)]
\item Write the Hamilton--Jacobi equation for the rigid rotator in spherical coordinates with fixed $r$. Identify the cyclic coordinate and state the corresponding conserved quantity.
\item Use the classical action-angle result $E = L^2/(2I)$ to compute the rotational energy of the molecule in SI units. Compare this to the quantum result $E = \hbar^2\ell(\ell+1)/(2I)$ with $\ell = 2$.
\item Find the absolute difference between the classical and quantum energy values and express it as a percentage of the quantum value.
\end{enumerate}}

\sol \textbf{Part (a).} The Hamiltonian of the rigid rotator is $H = p_\theta^2/(2I) + p_\phi^2/(2I\sin^2\theta)$. The full Hamilton--Jacobi equation follows by replacing $p_\theta$ with $\pdv{\mcS}{\theta}$, $p_\phi$ with $\pdv{\mcS}{\phi}$, and appending the time derivative:
\[
\frac{1}{2I}\left(\pdv{\mcS}{\theta}\right)^2 + \frac{1}{2I\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 + \pdv{\mcS}{t} = 0.
\]
For time-independent motion, the action separates as $\mcS = W(\theta,\phi) - Et$. The Hamiltonian does not depend explicitly on $\phi$, so $\phi$ is the cyclic coordinate. Its conjugate momentum is conserved:
\[
\pdv{\mcS}{\phi} = L_z = \text{const},
\]
which is the $z$-component of the angular momentum.

\textbf{Part (b).} The classical action-angle result for the rigid rotator is $E = L^2/(2I)$. With $L = 2\hbar$, we have $L^2 = 4\hbar^2$. First compute $\hbar^2$:
\[
\hbar^2 = \left(1.055\times 10^{-34}\right)^2\,\mathrm{J^2\!\cdot\!s^2}
= 1.113\times 10^{-68}\,\mathrm{J^2\!\cdot\!s^2}.
\]
The classical energy is
\[
E_{\mathrm{class}} = \frac{4\hbar^2}{2I}
= \frac{2\hbar^2}{I}
= \frac{2(1.113\times 10^{-68})}{1.46\times 10^{-46}}\,\mathrm{J}
= 1.525\times 10^{-22}\,\mathrm{J}.
\]
The quantum energy with $\ell = 2$ is
\[
E_{\mathrm{quant}} = \frac{\hbar^2\,\ell(\ell+1)}{2I}
= \frac{(1.113\times 10^{-68})(6)}{2(1.46\times 10^{-46})}\,\mathrm{J}
= \frac{6.678\times 10^{-68}}{2.92\times 10^{-46}}\,\mathrm{J}
= 2.287\times 10^{-22}\,\mathrm{J}.
\]
The quantum energy exceeds the classical value. The ratio is
\[
\frac{E_{\mathrm{quant}}}{E_{\mathrm{class}}}
= \frac{6}{4}
= 1.50.
\]

\textbf{Part (c).} The absolute difference between the two energies is
\[
\Delta E = E_{\mathrm{quant}} - E_{\mathrm{class}}
= 2.287\times 10^{-22} - 1.525\times 10^{-22}\,\mathrm{J}
= 7.62\times 10^{-23}\,\mathrm{J}.
\]
Expressed as a percentage of the quantum value:
\[
\frac{\Delta E}{E_{\mathrm{quant}}}\times 100\%
= \frac{7.62\times 10^{-23}}{2.287\times 10^{-22}}\times 100\%
= 33.3\%.
\]
Analytically, since $E_{\mathrm{class}} = \ell^2\hbar^2/(2I)$ and $E_{\mathrm{quant}} = \ell(\ell+1)\hbar^2/(2I)$, the fractional difference is
\[
\frac{\Delta E}{E_{\mathrm{quant}}}
= \frac{\ell(\ell+1) - \ell^2}{\ell(\ell+1)}
= \frac{\ell}{\ell(\ell+1)}
= \frac{1}{\ell+1}.
\]
For $\ell = 2$ this gives $1/3 = 33.3\%$, matching the numerical calculation. The discrepancy arises from the quantum $+\ell$ correction in $\ell(\ell+1)$ relative to the classical $\ell^2$.

Therefore,
\[
\frac{1}{2I}\left(\pdv{\mcS}{\theta}\right)^2 + \frac{1}{2I\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 + \pdv{\mcS}{t} = 0,
\qquad
p_\phi = L_z = \text{const};
\]
\[
E_{\mathrm{class}} = 1.53\times 10^{-22}\,\mathrm{J},
\qquad
E_{\mathrm{quant}} = 2.29\times 10^{-22}\,\mathrm{J},
\qquad
\frac{\Delta E}{E_{\mathrm{quant}}} = 33.3\%.
\]
