The Hydrogen Atom: Exact Solution and Atomic Orbitals
The Hydrogen Atom: Exact Solution and Atomic Orbitals
The hydrogen atom — a single proton bound to a single electron by the Coulomb force — is the crown jewel of quantum mechanics. It is the only atomic system for which the Schrödinger equation can be solved exactly, yielding closed-form expressions for the energy eigenvalues and eigenfunctions. These solutions are not merely academic; they form the foundation for understanding the periodic table, atomic spectroscopy, and virtually all of chemistry and condensed matter physics.
The time-independent Schrödinger equation for hydrogen is $\hat{H}|\psi\rangle = E|\psi\rangle$ where the Hamiltonian in position space is $H = -\frac{\hbar^2}{2\mu}\nabla^2 - \frac{e^2}{4\pi\epsilon_0 r}$, with $\mu = m_e m_p/(m_e + m_p) \approx m_e$ the reduced mass. Written in spherical coordinates $(r,\theta,\phi)$, the Laplacian separates the equation into radial and angular parts. The angular part is solved by spherical harmonics $Y_l^m(\theta,\phi)$, the eigenfunctions of $\hat{L}^2$ and $\hat{L}_z$. The radial part reduces to an ordinary differential equation whose normalizable solutions exist only for specific discrete values of energy.
The energy eigenvalues depend only on the principal quantum number $n$: $E_n = -\frac{13.6\,\text{eV}}{n^2}$, $n = 1, 2, 3, \ldots$ This formula, first derived empirically by Bohr in 1913, now emerges as a rigorous consequence of solving the radial Schrödinger equation with Coulomb potential. The ground state energy $E_1 = -13.6$ eV represents the binding energy of the electron; ionization requires exactly this energy. The negative sign reflects a bound state — the electron and proton are energetically favored to remain together.
A profound feature of the Coulomb potential is its accidental degeneracy: for each value of $n$, all values of $l$ from $0$ to $n-1$ give the same energy. This means level $n$ has degeneracy $n^2$ (including spin, $2n^2$). This extra degeneracy, beyond what rotational symmetry alone requires, arises from a hidden symmetry of the Kepler/Coulomb problem — the conservation of the Laplace-Runge-Lenz vector — and is responsible for the exceptional simplicity of hydrogen's spectrum.
The wave functions $\psi_{nlm}(r,\theta,\phi) = R_{nl}(r)Y_l^m(\theta,\phi)$ factor into a radial part and an angular part. The radial functions $R_{nl}(r)$ involve associated Laguerre polynomials $L_{n-l-1}^{2l+1}(\rho)$ where $\rho = 2r/(na_0)$ and $a_0 = 4\pi\epsilon_0\hbar^2/(m_e e^2) \approx 0.529$ Å is the Bohr radius. These polynomials have $n-l-1$ radial nodes, so higher $n$ states are more extended and oscillatory.
Beyond the exact Coulomb solution, perturbative corrections reveal the fine structure of hydrogen. The dominant corrections are (i) relativistic kinetic energy $\Delta E_{\text{rel}} \sim (v/c)^2 E_n \sim \alpha^2 E_n$, (ii) spin-orbit coupling $\Delta E_{\text{SO}} \sim \alpha^2 E_n$ where $\alpha = e^2/(4\pi\epsilon_0\hbar c) \approx 1/137$ is the fine structure constant, and (iii) the Darwin term. Together these lift the $l$-degeneracy and split each $n$ level into fine structure levels labeled by $j = l \pm \frac{1}{2}$, with energy correction $\Delta E_{nj} = -\frac{E_n \alpha^2}{n^2}\left(\frac{n}{j+1/2} - \frac{3}{4}\right)$.
The selection rules governing which transitions are allowed by electric dipole radiation follow from the matrix element $\langle n'l'm'|r|nlm\rangle$. Parity and angular momentum conservation of the photon force $\Delta l = \pm 1$ and $\Delta m = 0, \pm 1$. The principal quantum number $n$ can change freely. These selection rules explain why the hydrogen spectrum consists of distinct series (Lyman, Balmer, Paschen) with predictable wavelengths given by the Rydberg formula $1/\lambda = R_\infty(1/n_f^2 - 1/n_i^2)$ where $R_\infty = m_e e^4/(8\epsilon_0^2 h^3 c) \approx 1.097 \times 10^7\,\text{m}^{-1}$.
In spherical coordinates, the stationary Schrödinger equation with Coulomb potential is:
$$\left[-\frac{\hbar^2}{2m_e}\left(\frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r} - \frac{\hat{L}^2}{\hbar^2 r^2}\right) - \frac{e^2}{4\pi\epsilon_0 r}\right]\psi = E\psi$$
Substituting $\psi_{nlm}(r,\theta,\phi) = R_{nl}(r)Y_l^m(\theta,\phi)$ and using $\hat{L}^2 Y_l^m = \hbar^2 l(l+1)Y_l^m$ yields the radial equation for $u_{nl}(r) = rR_{nl}(r)$:
$$-\frac{\hbar^2}{2m_e}\frac{d^2 u}{dr^2} + \left[\frac{\hbar^2 l(l+1)}{2m_e r^2} - \frac{e^2}{4\pi\epsilon_0 r}\right]u = E u$$
The normalized hydrogen wave functions are $\psi_{nlm}(r,\theta,\phi) = R_{nl}(r)Y_l^m(\theta,\phi)$ where the radial functions are:
$$R_{nl}(r) = -\sqrt{\left(\frac{2}{na_0}\right)^3\frac{(n-l-1)!}{2n[(n+l)!]^3}}\,e^{-r/(na_0)}\left(\frac{2r}{na_0}\right)^l L_{n-l-1}^{2l+1}\!\!\left(\frac{2r}{na_0}\right)$$
with the Bohr radius $a_0 = \frac{4\pi\epsilon_0\hbar^2}{m_e e^2} \approx 0.529$ Å. The associated Laguerre polynomials $L_p^q(x)$ are polynomials of degree $p$ with $p$ zeros.
The exact energy eigenvalues of hydrogen are:
$$E_n = -\frac{m_e e^4}{8\epsilon_0^2 h^2}\cdot\frac{1}{n^2} = -\frac{13.6\,\text{eV}}{n^2}, \quad n = 1,2,3,\ldots$$
For each $n$: $l \in \{0,1,\ldots,n-1\}$ and $m \in \{-l,\ldots,l\}$, giving $\sum_{l=0}^{n-1}(2l+1) = n^2$ spatial states. Including the two spin states, each level has degeneracy $2n^2$. The mean radius is $\langle r \rangle_{nl} = \frac{a_0}{2}[3n^2 - l(l+1)]$.
The spherical harmonics $Y_l^m(\theta,\phi) = \mathcal{N}_{lm}P_l^m(\cos\theta)e^{im\phi}$ with normalization $\mathcal{N}_{lm} = \sqrt{\frac{(2l+1)(l-|m|)!}{4\pi(l+|m|)!}}$ satisfy:
$$\hat{L}^2 Y_l^m = \hbar^2 l(l+1) Y_l^m, \quad \hat{L}_z Y_l^m = \hbar m\, Y_l^m$$
$$\int Y_{l'}^{m'*}Y_l^m\,d\Omega = \delta_{ll'}\delta_{mm'}, \quad \sum_{l=0}^\infty\sum_{m=-l}^l |Y_l^m|^2 = \frac{1}{4\pi}\cdot\frac{1}{\sin^2\theta}\text{ (completeness)}$$
Explicit forms: $Y_0^0 = \frac{1}{\sqrt{4\pi}}$; $Y_1^0 = \sqrt{\frac{3}{4\pi}}\cos\theta$; $Y_1^{\pm1} = \mp\sqrt{\frac{3}{8\pi}}\sin\theta\,e^{\pm i\phi}$.
Verify that the ground state wave function $\psi_{100} = \frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}$ is normalized.
$$\int|\psi_{100}|^2\,d^3r = \frac{1}{\pi a_0^3}\int_0^\infty e^{-2r/a_0}r^2\,dr\int_0^\pi\sin\theta\,d\theta\int_0^{2\pi}d\phi$$
Angular integrals: $\int_0^\pi\sin\theta\,d\theta = 2$, $\int_0^{2\pi}d\phi = 2\pi$, giving $4\pi$.
Radial integral: $\int_0^\infty r^2 e^{-2r/a_0}\,dr$. Let $u = 2r/a_0$: $= \frac{a_0^3}{8}\int_0^\infty u^2 e^{-u}\,du = \frac{a_0^3}{8}\cdot\Gamma(3) = \frac{a_0^3}{8}\cdot 2 = \frac{a_0^3}{4}$.
Total: $\frac{1}{\pi a_0^3}\cdot 4\pi\cdot\frac{a_0^3}{4} = \frac{4\pi}{4\pi} = 1$. ✓
Compute $\langle r \rangle$ for the ground state of hydrogen and compare to $a_0$.
$$\langle r\rangle = \int_0^\infty r\cdot|R_{10}(r)|^2 r^2\,dr = \frac{4}{a_0^3}\int_0^\infty r^3 e^{-2r/a_0}\,dr$$
The radial probability density is $P(r) = |R_{10}|^2 r^2 = \frac{4}{a_0^3}r^2 e^{-2r/a_0}$ (with $R_{10} = \frac{2}{a_0^{3/2}}e^{-r/a_0}$).
Using $\int_0^\infty r^3 e^{-2r/a_0}\,dr = \frac{3!}{(2/a_0)^4} = \frac{6 a_0^4}{16} = \frac{3a_0^4}{8}$:
$$\langle r\rangle = \frac{4}{a_0^3}\cdot\frac{3a_0^4}{8} = \frac{3a_0}{2}$$
The mean radius is $\frac{3}{2}a_0 \approx 0.794$ Å, larger than $a_0$ because the probability distribution $P(r)$ peaks at $r = a_0$ but has a long tail extending to larger $r$.
Using the hydrogen energy formula, find the wavelength of the $H_\alpha$ line (Balmer series: $n=3 \to n=2$ transition).
Energy of transition: $\Delta E = E_3 - E_2 = -13.6\left(\frac{1}{9} - \frac{1}{4}\right) = -13.6\cdot\frac{4-9}{36} = -13.6\cdot\frac{-5}{36} = \frac{68}{36}$ eV $= 1.889$ eV.
The photon energy equals $\Delta E$: $E_\gamma = hf = hc/\lambda = 1.889$ eV.
Wavelength: $\lambda = \frac{hc}{E_\gamma} = \frac{(4.136\times10^{-15}\,\text{eV·s})(3\times10^8\,\text{m/s})}{1.889\,\text{eV}} = \frac{1.241\times10^{-6}\,\text{eV·m}}{1.889\,\text{eV}} \approx 657\,\text{nm}$.
This is the famous red $H_\alpha$ line of the Balmer series, visible to the naked eye in hydrogen discharge tubes and used to identify hydrogen in stars.
Calculate the spin-orbit fine structure splitting between the $2p_{1/2}$ ($j=\frac{1}{2}$) and $2p_{3/2}$ ($j=\frac{3}{2}$) levels of hydrogen.
The fine structure energy shift for hydrogen is (from Dirac theory to order $\alpha^2$):
$$\Delta E_{nj} = \frac{E_n \alpha^2}{n}\left(\frac{1}{j+1/2} - \frac{3}{4n}\right)$$
For $n=2$, $E_2 = -3.4$ eV, $\alpha^2 \approx (1/137)^2 = 5.33\times10^{-5}$:
$j = 3/2$: $\Delta E = \frac{(-3.4)(5.33\times10^{-5})}{2}\left(\frac{1}{2} - \frac{3}{8}\right) = (-9.06\times10^{-5})(0.125) = -1.13\times10^{-5}$ eV.
$j = 1/2$: $\Delta E = (-9.06\times10^{-5})(1 - 3/8) = (-9.06\times10^{-5})(0.625) = -5.66\times10^{-5}$ eV.
Splitting: $|\Delta E_{3/2} - \Delta E_{1/2}| \approx 4.53\times10^{-5}$ eV $\approx 45\,\mu$eV, corresponding to a wavelength difference of about 0.016 nm in the Lyman series.
Practice Problems
Show Answer Key
1. $E_4 = -13.6/16 = -0.85$ eV. Ionization energy from $n=4$: $0.85$ eV.
2. $2n^2 = 2(9) = 18$ states ($3s$: 2, $3p$: 6, $3d$: 10).
3. $\Delta E = 13.6(1 - 1/4) = 10.2$ eV. $\lambda = 1241\,\text{eV·nm}/10.2\,\text{eV} \approx 121.6$ nm (UV, Lyman-$\alpha$).
4. $R_{20}(r) \propto (2-r/a_0)e^{-r/(2a_0)}$. $P_{20}(r) = r^2|R_{20}|^2$. Setting $dP/dr=0$ gives (after algebra) radial peaks at $r/a_0 = 2(3\pm\sqrt{5})$, i.e., $r \approx 0.764a_0$ and $r \approx 5.236a_0$ (two peaks; the outer is the principal one at $\approx 5.24a_0$).
5. $\langle r\rangle = \frac{a_0}{2}[3(4)-2(3)] = \frac{a_0}{2}(12-6) = 3a_0$.
6. $3s$: $l=0$, $m=0$. $3p$: $l=1$, $m\in\{-1,0,1\}$. $3d$: $l=2$, $m\in\{-2,-1,0,1,2\}$.
7. $E = 12.09$ eV $= 13.6(1 - 1/n_f^2)$. Solving: $1/n_f^2 = 1 - 12.09/13.6 = 0.111 = 1/9$, so $n_f = 3$. The transition is $n=1\to n=3$ (absorption) or $n=3\to n=1$ (emission). Given emission at 12.09 eV, this is $3\to1$ (Lyman-$\beta$).
8. $\alpha = \frac{(1.602\times10^{-19})^2}{4\pi(8.854\times10^{-12})(1.055\times10^{-34})(3\times10^8)} = \frac{2.566\times10^{-38}}{3.516\times10^{-36}} \approx 0.00729 \approx 1/137$. ✓
9. The electric dipole operator $\mathbf{r}$ is a vector with parity $(-1)^1 = -1$. For the matrix element to be nonzero, the product $\psi_{n'l'm'}^*\cdot r\cdot\psi_{nlm}$ must have even parity. Since $\psi_{nlm}$ has parity $(-1)^l$, we need $(-1)^{l'}(-1)(-1)^l = +1$, so $(-1)^{l'+l+1} = 1$, meaning $l'+l$ must be odd, i.e., $l'$ and $l$ must differ by an odd integer. Combined with the fact that $|\Delta l| = 1$ from the triangular inequality of angular momentum, this gives $\Delta l = \pm 1$.
10. Number of radial nodes $= n-l-1 = 3-2-1 = 0$. Number of angular nodes $= l = 2$. The $3d$ orbitals have no radial nodes and two angular nodal surfaces.
11. For $j=3/2$: $m_j \in \{-3/2,-1/2,+1/2,+3/2\}$ — four distinct energy levels, each shifted by $E_Z = m_j g_j \mu_B B$.
12. $\Delta E = hf = (6.626\times10^{-34})(1058\times10^6) = 7.01\times10^{-25}$ J $= 7.01\times10^{-25}/1.602\times10^{-19}$ eV $= 4.38\times10^{-6}$ eV $\approx 4.38\,\mu$eV.