\subsection{Charged Particle in Coulomb Potential}

This subsection treats a charged particle moving in the Coulomb potential of a fixed point charge through the Hamilton-- Jacobi formalism, demonstrating its identical structure to the gravitational Kepler problem and using action-- angle variables to recover the Bohr-- Sommerfeld energy levels of the hydrogen atom.

\dfn{Coulomb Hamilton-- Jacobi equation}{
Consider a particle of reduced mass $\mu$ and charge $q$ moving in the electrostatic potential of a fixed source charge $Q$. The coupling constant is $k = qQ/(4\pi\varepsilon_0)$, with the potential $V(r) = -k/r$ for attractive interaction. For the electron-- proton system, $q = -e$, $Q = +e$, so $k = e^2/(4\pi\varepsilon_0)$. In spherical coordinates the Hamiltonian is
\[
\mcH = \frac{p_r^2}{2\mu} + \frac{p_\theta^2}{2\mu r^2} + \frac{p_\phi^2}{2\mu r^2\sin^2\theta} - \frac{k}{r}.
\]
Substituting $p_i = \pdv{\mcS}{q_i}$ into the Hamilton-- Jacobi equation $\mcH + \pdv{\mcS}{t} = 0$ gives
\[
\frac{1}{2\mu}\left(\pdv{\mcS}{r}\right)^2 
+ \frac{1}{2\mu r^2}\left(\pdv{\mcS}{\theta}\right)^2 
+ \frac{1}{2\mu r^2\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 
- \frac{k}{r} + \pdv{\mcS}{t} = 0.
\]
Because the scalar potential is time-- independent, energy $E = \mcH$ is conserved and the time variable separates as $\mcS = W(r,\theta,\phi) - Et$ with $W$ the Hamilton characteristic function.}

\thm{Orbit equation and eccentricity for the Coulomb problem}{
With $V(r) = -k/r$ the trajectory is a conic section
\[
r(\phi) = \frac{\ell}{1 + \varepsilon\cos(\phi - \phi_0)},
\]
where the semilatus rectum $\ell = L^2/(\mu k)$ and the eccentricity $\varepsilon = \sqrt{1 + 2EL^2/(\mu k^2)}$ are determined by the energy $E$ and the total angular momentum $L$. For bound orbits ($E < 0$, $\varepsilon < 1$) the semimajor axis is $a = -k/(2E)$ and the binding energy $E = -k/(2a)$. A circular orbit occurs at $\varepsilon = 0$ with $L^2 = \mu k a$.}

\pf{Separated Hamilton-- Jacobi equations for the Coulomb problem}{
Set $\mcS = W_r(r) + W_\theta(\theta) + W_\phi(\phi) - Et$ and substitute into the time-- independent HJ equation $\mcH(q,\pdv{W}{q}) = E$:
\[
\frac{1}{2\mu}\left(\der{W_r}{r}\right)^2 
+ \frac{1}{2\mu r^2}\left(\der{W_\theta}{\theta}\right)^2 
+ \frac{1}{2\mu r^2\sin^2\theta}\left(\der{W_\phi}{\phi}\right)^2 
- \frac{k}{r} = E.
\]
The azimuthal coordinate $\phi$ is cyclic, so $\der{W_\phi}{\phi} = L_z$. Multiply by $2\mu r^2$ and rearrange:
\[
r^2\left(\der{W_r}{r}\right)^2 + 2\mu kr - 2\mu Er^2
= -\left(\der{W_\theta}{\theta}\right)^2 - \frac{L_z^2}{\sin^2\theta}.
\]
The left side depends only on $r$, the right only on $\theta$; each equals the separation constant $L^2$. The polar equation is
\[
\left(\der{W_\theta}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = L^2,
\]
and the radial equation is
\[
\left(\der{W_r}{r}\right)^2 = 2\mu E + \frac{2\mu k}{r} - \frac{L^2}{r^2}.
\]
These match the gravitational Kepler equations exactly, with $k$ playing the role of $GM\mu$. The three constants $E$, $L$, and $L_z$ form a complete set required by Jacobi's theorem.}

\nt{Structural identity with the gravitational Kepler problem}{
The Coulomb HJ equation is structurally identical to the gravitational Kepler problem. The only difference lies in the coupling constant: gravity has $k_{\text{grav}} = GM\mu$ while electrostatics has $k_{\text{Coul}} = qQ/(4\pi\varepsilon_0)$. Because the Coulomb interaction is a scalar potential with $\vec{A} = 0$, the minimal coupling is trivial --- the canonical momentum equals the kinetic momentum, $\vec{p} = \mu\dot{\vec{r}}$, and no vector-- potential corrections appear in the Hamiltonian. The separation in spherical coordinates proceeds identically, yielding the same separated radial, polar, and azimuthal equations shown above. All results for orbits, action-- angle variables, and frequencies carry over with the replacement $GM\mu \to k$.}

\nt{Action-- angle quantization and the hydrogen spectrum}{
For the $1/r$ potential the three action variables are $J_\phi = 2\pi L_z$, $J_\theta = 2\pi(L - |L_z|)$, and $J_r = 2\pi(-L + k\sqrt{\mu/(2|E|)})$. Their sum eliminates the angular-- momentum dependence:
\[
J_{\mathrm{tot}} = J_r + J_\theta + J_\phi = 2\pi k\sqrt{\frac{\mu}{2|E|}}.
\]
Bohr-- Sommerfeld quantization requires $J_{\mathrm{tot}}/2\pi = n\hbar$, where $n$ is the principal quantum number. Setting $k\sqrt{\mu/(2|E|)} = n\hbar$ and solving for energy:
\[
|E| = \frac{\mu k^2}{2n^2\hbar^2},
\qquad
E_n = -\frac{\mu k^2}{2\hbar^2 n^2}.
\]
This expression coincides exactly with the ground-- state energy formula from the Schrodinger equation for hydrogen. The separability of the HJ equation in both spherical and parabolic coordinates reflects the hidden $SO(4)$ dynamical symmetry of the $1/r$ potential that makes the hydrogen spectrum depend on a single quantum number.}

\qs{Electron in the Coulomb field of a proton using the HJ action-- angle formalism}{
For an electron bound to a proton, the electrostatic coupling constant is $k = e^2/(4\pi\varepsilon_0) = 2.307\times 10^{-28}\,\mathrm{J\!\cdot\!m}$ and the reduced mass $\mu \approx m_e = 9.11\times 10^{-31}\,\mathrm{kg}$.

\begin{enumerate}[label=(\alph*)]
\item For a bound orbit with semimajor axis $a_0 = 0.529\times 10^{-10}\,\mathrm{m}$ (the Bohr radius), find the orbital energy $E = -k/(2a_0)$ from the HJ action-- angle formalism. Express the result in both joules and electron volts.
\item Find the angular momentum $L = \sqrt{\mu k a_0}$ for this circular orbit and compute the total action $J_{\mathrm{tot}} = 2\pi L$. Compare the energy found in part (a) to the quantum $n=1$ energy of $-13.6\,\mathrm{eV} = -2.18\times 10^{-18}\,\mathrm{J}$.
\item Using the Bohr-- Sommerfeld quantization $J_{\mathrm{tot}} = n h$ with $n=1$, verify that the quantized energy $E_1 = -\mu k^2/(2\hbar^2)$ matches $-13.6\,\mathrm{eV}$.
\end{enumerate}}

\sol \textbf{Part (a).} The HJ action-- angle formalism for any $1/r$ potential gives the energy of a bound orbit in terms of the semimajor axis. The binding energy follows from the virial relation $2T + V = 0$ for a $1/r$ potential, giving
\[
E = -\frac{k}{2a_0}.
\]
Substitute the given numerical values:
\[
E = -\frac{2.307\times 10^{-28}\,\mathrm{J\!\cdot\!m}}{2(0.529\times 10^{-10}\,\mathrm{m})}
= -\frac{2.307\times 10^{-28}}{1.058\times 10^{-10}}\,\mathrm{J}.
\]
Divide:
\[
E = -2.18\times 10^{-18}\,\mathrm{J}.
\]
Convert to electron volts using $1\,\mathrm{eV} = 1.602\times 10^{-19}\,\mathrm{J}$:
\[
E = -\frac{2.18\times 10^{-18}}{1.602\times 10^{-19}}\,\mathrm{eV}
= -13.6\,\mathrm{eV}.
\]
This is precisely the binding energy of the hydrogen atom in its ground state.

\textbf{Part (b).} For a circular orbit the angular momentum follows from the zero-- eccentricity condition $\varepsilon = 0$, which gives $L^2 = \mu k a$. The angular momentum for the orbit at the Bohr radius is
\[
L = \sqrt{\mu k a_0}.
\]
Compute the product under the square root:
\[
\mu k a_0 = (9.11\times 10^{-31})(2.307\times 10^{-28})(0.529\times 10^{-10})\,\mathrm{kg\!\cdot\!J\!\cdot\!m^2}.
\]
The mantissa product is
\[
(9.11)(2.307)(0.529) = 11.11,
\]
and the exponent is $-31 + (-28) + (-10) = -69$. Thus
\[
\mu k a_0 = 11.11\times 10^{-69}\,\mathrm{kg\!\cdot\!J\!\cdot\!m^2}
= 1.111\times 10^{-68}\,\mathrm{J^2\!\cdot\!s^2}.
\]
Taking the square root:
\[
L = \sqrt{1.111\times 10^{-68}}\,\mathrm{J\!\cdot\!s}
= 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s}.
\]
This equals the reduced Planck constant $\hbar = 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s}$. The total action is
\[
J_{\mathrm{tot}} = 2\pi L = 2\pi\hbar = h = 6.63\times 10^{-34}\,\mathrm{J\!\cdot\!s}.
\]
The total action equals Planck's constant $h$. This is consistent with the Bohr-- Sommerfeld quantization condition $J_{\mathrm{tot}} = n h$ at $n=1$.

Comparing energies: part (a) yielded $E = -2.18\times 10^{-18}\,\mathrm{J} = -13.6\,\mathrm{eV}$, which is exactly the stated quantum $n=1$ energy. The classical HJ action-- angle energy at the Bohr radius coincides numerically with the quantum ground-- state energy.

\textbf{Part (c).} The Bohr-- Sommerfeld quantization condition reads
\[
J_{\mathrm{tot}} = n h = n\cdot 2\pi\hbar.
\]
From the HJ action-- angle analysis, the total action is $J_{\mathrm{tot}} = 2\pi k\sqrt{\mu/(2|E|)}$. Equate the two expressions:
\[
2\pi k\sqrt{\frac{\mu}{2|E|}} = 2\pi n\hbar,
\]
\[
k\sqrt{\frac{\mu}{2|E|}} = n\hbar.
\]
Square both sides:
\[
k^2\frac{\mu}{2|E|} = n^2\hbar^2,
\qquad
|E| = \frac{\mu k^2}{2n^2\hbar^2}.
\]
For $n=1$ the quantized energy is
\[
E_1 = -\frac{\mu k^2}{2\hbar^2}.
\]
Evaluate numerically. First compute the numerator:
\[
k^2 = (2.307\times 10^{-28})^2\,\mathrm{J^2\!\cdot\!m^2}
= 5.322\times 10^{-56}\,\mathrm{J^2\!\cdot\!m^2}.
\]
\[
\mu k^2 = (9.11\times 10^{-31})(5.322\times 10^{-56})
= 48.48\times 10^{-87}
= 4.848\times 10^{-86}\,\mathrm{kg\!\cdot\!J^2\!\cdot\!m^2}.
\]
The denominator is
\[
2\hbar^2 = 2(1.055\times 10^{-34})^2\,\mathrm{J^2\!\cdot\!s^2}
= 2(1.113\times 10^{-68})\,\mathrm{J^2\!\cdot\!s^2}
= 2.226\times 10^{-68}\,\mathrm{J^2\!\cdot\!s^2}.
\]
Therefore,
\[
|E_1| = \frac{4.848\times 10^{-86}}{2.226\times 10^{-68}}\,\mathrm{J}
= 2.178\times 10^{-18}\,\mathrm{J}.
\]
Rounding the coupling constant slightly upward to $k = 2.3071\times 10^{-28}\,\mathrm{J\!\cdot\!m}$ reproduces the conventional value:
\[
E_1 = -2.18\times 10^{-18}\,\mathrm{J}
= -13.6\,\mathrm{eV}.
\]
This matches the quantum ground-- state energy $-13.6\,\mathrm{eV}$ found from solving the Schrodinger equation for hydrogen. The Bohr-- Sommerfeld semiclassical quantization of the HJ action variable therefore predicts the correct hydrogen energy spectrum in its dependence on $n$ and reproduces the ground-- state energy to the precision of the given parameters.

Therefore, the orbital energy, angular momentum, and quantized energy are
\[
E = -2.18\times 10^{-18}\,\mathrm{J} = -13.6\,\mathrm{eV},
\qquad
L = 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s} = \hbar,
\]
\[
J_{\mathrm{tot}} = h = 6.63\times 10^{-34}\,\mathrm{J\!\cdot\!s},
\qquad
E_1 = -\frac{\mu k^2}{2\hbar^2} = -13.6\,\mathrm{eV}.
\]