\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\normalsize{}'s theorem.}

\nt{Structural identity with the gravitational Kepler problem}{
The Coulomb HJ equation is structurally identical to the gravitational Kepler problem treated in A.08 (kepler--hj.tex). 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. The three action variables $J_r$, $J_\theta$, and $J_\phi$ retain the exact integral structure from the gravitational case; all results for orbits, action--angle variables, and frequencies carry over with the replacement $GM\mu \to k$.}

\nt{From classical action to quantum numbers}{
In the Hamilton--Jacobi formalism, an action variable is defined as the phase--space integral of a canonical momentum over one complete cycle of its conjugate coordinate: $J = \frac{1}{2\pi}\oint p\,\dd q$. This construction assigns a single number to each degree of freedom that measures the area enclosed by the orbit in phase space, normalized by $2\pi$. Planck introduced a fundamental constant $h = 6.626\times 10^{-34}\,\mathrm{J\!\cdot\!s}$ to explain blackbody radiation. The reduced Planck constant $\hbar = h/(2\pi) = 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s}$ is the natural quantum of angular momentum. Bohr and Sommerfeld proposed that classical action variables should be restricted to integer multiples of $\hbar$:
\[
J_i = n_i\hbar,
\]
where $n_i$ is a positive integer called a quantum number. The physical idea is that phase space is granular at atomic scales: only classical orbits whose action variables satisfy this condition survive in the quantum theory. Why quantize action? The action variable sets the scale of an orbit in phase space, and imposing $J = n\hbar$ is equivalent to requiring an integer number of de Broglie wavelengths to fit along the orbit, producing a standing wave. The simplest action variable to evaluate is the azimuthal action. For any central potential, the canonical momentum $p_\phi = \pdv{\mcS}{\phi} = L_z$ is a constant of motion. Because $L_z$ does not vary with $\phi$, the integral evaluates directly:
\[
J_\phi = \frac{1}{2\pi}\oint p_\phi\,\dd\phi
= \frac{1}{2\pi}\int_{0}^{2\pi} L_z\,\dd\phi
= \frac{1}{2\pi}L_z\int_{0}^{2\pi}\dd\phi
= L_z.
\]
Bohr--Sommerfeld quantization $J_\phi = n_\phi\hbar$ then immediately gives $L_z = n_\phi\hbar$. The $z$-component of angular momentum is quantized in units of $\hbar$, exactly as full quantum mechanics predicts.}

\nt{Action--angle quantization and the hydrogen spectrum}{
For the $1/r$ potential the three action variables are $J_\phi = L_z$, $J_\theta = L - |L_z|$, and $J_r = -L + k\sqrt{\mu/(2|E|)}$. The polar and radial actions $J_\theta$ and $J_r$ have the same integral structure as those derived for the gravitational Kepler problem in A.08, with the only change being the coupling constant $k$. Their sum eliminates the angular--momentum dependence:
\[
J_{\mathrm{tot}} = J_r + J_\theta + J_\phi = k\sqrt{\frac{\mu}{2|E|}}.
\]
Bohr--Sommerfeld quantization requires $J_{\mathrm{tot}} = n\hbar$, 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.}

\nt{The Bohr model}{
In 1913, Niels Bohr proposed a model of the hydrogen atom in which the electron orbits the nucleus in circular paths with quantized angular momentum $L = n\hbar$. Bohr postulated that only certain discrete orbits are allowed and that electromagnetic radiation is emitted or absorbed when the electron jumps between them. His model produced the correct Rydberg formula for hydrogen line spectra but relied on ad hoc quantization rules applied to purely classical orbits. The Bohr--Sommerfeld method extends this picture to elliptical orbits and multiple degrees of freedom using action--angle variables derived from the Hamilton--Jacobi formalism. The calculations below show how the Bohr results emerge systematically from semiclassical quantization of classical action variables.}

\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 $J_{\mathrm{tot}} = L$. 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 $J_{\mathrm{tot}} = n\hbar$ 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
\[
J_{\mathrm{tot}} = L = \hbar = 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s}.
\]
The total action equals the reduced Planck constant $\hbar$. This is consistent with the Bohr--Sommerfeld quantization condition $J_{\mathrm{tot}} = n\hbar$ 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
\[
J_{\mathrm{tot}} = n\hbar.
\]
From the HJ action--angle analysis, the total action is $J_{\mathrm{tot}} = k\sqrt{\mu/(2|E|)}$. Equate the two expressions:
\[
k\sqrt{\frac{\mu}{2|E|}} = n\hbar.
\]
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
\[
E = -2.18\times 10^{-18}\,\mathrm{J} = -13.6\,\mathrm{eV},
\qquad
L = 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s} = \hbar,
\]
\[
J_{\mathrm{tot}} = \hbar = 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s},
\qquad
E_1 = -\frac{\mu k^2}{2\hbar^2} = -13.6\,\mathrm{eV}.
\]
