--- title: Solving the Radial Equation in Full module: The Quantum Hydrogen Atom moduleNumber: 2 lessonNumber: 3 order: 203 summary: > The hydrogen radial equation is solved from the differential equation up. The substitution u = rR turns it into a one-dimensional problem with a centrifugal barrier; matching the asymptotic behaviour at the origin and at infinity peels off the factors r^(ℓ+1) and e^(−r/na₀); a Frobenius series for the remainder must terminate, and that termination condition yields the quantization n ≥ ℓ+1 with E = −Z²Ry/n². The surviving polynomials are the associated Laguerre functions, whose degree n−ℓ−1 counts the radial nodes. topics: [The Quantum Hydrogen Atom] sources: - book: Griffiths & Schroeter ref: "Ch. 4 — Quantum Mechanics in Three Dimensions; §4.2 The Hydrogen Atom" - book: Bransden & Joachain ref: "Ch. 3 — The Hydrogen Atom; §3.2–3.3 Solution of the Radial Equation" - book: Foot ref: "Ch. 2 — The Hydrogen Atom; §2.1–2.2" draft: false --- The [separation of variables](/atomic-physics/quantum-hydrogen-atom/schrodinger-3d-hydrogen) delivered the angular factors as spherical harmonics and left the radial factor $R_{n\ell}(r)$ defined by an ordinary differential equation that carries the Coulomb potential and the energy. That equation was quoted, not solved. Solving it is where the principal quantum number, the restriction $\ell \le n-1$, the $n^2$ degeneracy, and the exponential-times-polynomial form of the wave functions all originate. The method is the standard one for a Schrödinger eigenvalue problem: strip the asymptotic behaviour, expand the remainder in a power series, and force the series to terminate so the solution stays normalizable. ## The radial equation as a one-dimensional problem For a central Coulomb potential $V(r) = -kZe^2/r$ with $k$ the Coulomb constant, the radial equation for $R(r)$ reads[^gs-42] $$ -\frac{\hbar^2}{2\mu}\frac{1}{r^2}\frac{\d}{\d r}\!\left(r^2\frac{\d R}{\d r}\right) + \left[-\frac{kZe^2}{r} + \frac{\hbar^2\ell(\ell+1)}{2\mu r^2}\right]R = ER, $$ with $\mu$ the electron–proton reduced mass. The first-derivative term is a nuisance. It disappears under the substitution $$ u(r) = r\,R(r), \qquad R = \frac{u}{r}, \qquad \frac{1}{r^2}\frac{\d}{\d r}\!\left(r^2\frac{\d R}{\d r}\right) = \frac{1}{r}\frac{\d^2 u}{\d r^2}. $$ Substituting and multiplying by $r$ gives an equation identical in form to a one-dimensional Schrödinger equation for $u(r)$ on the half-line $r \ge 0$: $$ -\frac{\hbar^2}{2\mu}\frac{\d^2 u}{\d r^2} + \underbrace{\left[-\frac{kZe^2}{r} + \frac{\hbar^2\ell(\ell+1)}{2\mu r^2}\right]}_{V_{\text{eff}}(r)}u = Eu. $$ > **Definition (Effective radial potential).** The bracketed term > $V_{\text{eff}}(r) = -kZe^2/r + \hbar^2\ell(\ell+1)/2\mu r^2$ is the potential > felt by the one-dimensional radial motion. It adds to the attractive Coulomb > well a repulsive **centrifugal barrier** $\hbar^2\ell(\ell+1)/2\mu r^2$ that > grows with the angular momentum and dominates near the origin for every > $\ell \ge 1$. The barrier is why higher-$\ell$ states are held away from the nucleus. For $\ell = 0$ the effective potential is the bare Coulomb well and $u$ may be finite and nonzero as $r \to 0$; for $\ell \ge 1$ the $1/r^2$ wall forces $u$ to zero there. The boundary conditions on $u$ are $u(0) = 0$ (so that $R = u/r$ stays finite) and $u(r) \to 0$ as $r \to \infty$ (so that the state is bound and normalizable, $\int_0^\infty |u|^2\,\d r < \infty$). $$ % caption: The effective radial potential for ℓ = 0, 1, 2 (labelled s, p, d). % The centrifugal barrier lifts the ℓ ≥ 1 curves near the origin and pushes the % well minimum outward as ℓ grows. \begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (6.2,0) node[right, black!70] {$r$}; \draw[->, black] (0,-2.8) -- (0,2.4) node[above, black!70] {$V$}; % l=0 pure Coulomb -1/r \draw[acc, very thick] (0.42,-2.6) .. controls (0.9,-1.6) and (1.8,-0.9) .. (3.0,-0.55) .. controls (4.2,-0.35) and (5.4,-0.22) .. (6.0,-0.17); \node[acc, anchor=west] at (0.95,-2.15) {s}; % l=1: barrier + well, minimum around r=1.5 \draw[black, very thick, dashed] (0.55,2.2) .. controls (0.85,0.2) and (1.2,-1.15) .. (1.8,-1.2) .. controls (2.6,-1.2) and (4.0,-0.55) .. (6.0,-0.2); \node[black, anchor=west] at (2.1,-1.35) {p}; % l=2: higher barrier, shallower/farther minimum \draw[black, very thick, densely dotted] (0.75,2.3) .. controls (1.1,0.9) and (1.7,-0.55) .. (2.5,-0.62) .. controls (3.4,-0.62) and (4.6,-0.35) .. (6.0,-0.2); \node[black, anchor=west] at (1.75,1.2) {d}; \draw[black, dashed] (0,0) -- (6.0,0); \node[black, anchor=south east] at (6.0,0.05) {$E=0$}; \end{tikzpicture} $$ ## Dimensionless form and asymptotics Bound states have $E < 0$. Set $$ \kappa = \frac{\sqrt{-2\mu E}}{\hbar}, \qquad \rho = \kappa r, \qquad \rho_0 = \frac{2\mu kZe^2}{\hbar^2\kappa} = \frac{2Z}{\kappa a_0}, $$ where $a_0 = \hbar^2/\mu ke^2$ is the Bohr radius. In these variables every dimensioned quantity is absorbed and the equation becomes[^gs-42] $$ \frac{\d^2 u}{\d\rho^2} = \left[1 - \frac{\rho_0}{\rho} + \frac{\ell(\ell+1)}{\rho^2}\right]u. $$ The eigenvalue now lives entirely in the single number $\rho_0$; finding the allowed $\rho_0$ is finding the allowed energies. Two limits fix the shape of $u$ before any series is written. - **Large $\rho$.** The $1/\rho$ and $1/\rho^2$ terms vanish, leaving $\d^2 u/\d\rho^2 \approx u$, with solutions $e^{\pm\rho}$. Normalizability discards $e^{+\rho}$, so $u \sim e^{-\rho}$ far out. - **Small $\rho$.** The $1/\rho^2$ term dominates, leaving $\d^2 u/\d\rho^2 \approx \ell(\ell+1)\rho^{-2}u$, whose solutions are $\rho^{\ell+1}$ and $\rho^{-\ell}$. Finiteness of $R = u/r$ discards $\rho^{-\ell}$, so $u \sim \rho^{\ell+1}$ near the origin. Peeling both factors off, write the exact solution as their product with an unknown function $v(\rho)$: $$ u(\rho) = \rho^{\ell+1}\,e^{-\rho}\,v(\rho). $$ The two extracted factors already satisfy the boundary conditions; $v$ must not spoil them, which means $v$ may grow no faster than a power of $\rho$. ## The Frobenius series and its recursion Differentiating $u = \rho^{\ell+1}e^{-\rho}v$ twice and substituting reduces the equation to a single ODE for $v(\rho)$:[^gs-42][^bj-33] $$ \rho\frac{\d^2 v}{\d\rho^2} + 2(\ell+1-\rho)\frac{\d v}{\d\rho} + \big[\rho_0 - 2(\ell+1)\big]v = 0. $$ Expand $v$ as a power series about the (regular singular) origin, $$ v(\rho) = \sum_{j=0}^{\infty} c_j\,\rho^{j}. $$ Insert the series, collect the coefficient of $\rho^{j}$, and set it to zero. Each power of $\rho$ gives one relation between consecutive coefficients — the **recursion relation** $$ c_{j+1} = \frac{2(j+\ell+1)-\rho_0}{(j+1)(j+2\ell+2)}\,c_j. $$ One free constant $c_0$ (fixed later by normalization) generates the whole series. The recursion is the engine of the solution, and its large-$j$ behaviour decides everything. $$ % caption: The coefficient recursion generates each c_(j+1) from c_j; a single % seed c₀ propagates the whole series, and the numerator can vanish at one j. \begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth, cell/.style={draw, minimum width=11mm, minimum height=8mm}] \definecolor{acc}{HTML}{4A6FA5} \node[cell, draw=acc, text=acc, thick] (c0) at (0,0) {$c_0$}; \node[cell] (c1) at (1.9,0) {$c_1$}; \node[cell] (c2) at (3.8,0) {$c_2$}; \node[cell] (c3) at (5.7,0) {$c_3$}; \node[cell, draw=black] (cd) at (7.6,0) {$\dots$}; \draw[acc, ->] (c0) -- (c1); \draw[acc, ->] (c1) -- (c2); \draw[acc, ->] (c2) -- (c3); \draw[acc, ->] (c3) -- (cd); \node[black, anchor=north] at (0.95,-0.6) {ratio}; \node[black, anchor=north] at (2.85,-0.6) {ratio}; \node[black, anchor=north] at (4.75,-0.6) {ratio}; \node[align=center, black!70] at (3.8,1.35) {each step multiplies by the\\ recursion ratio at index $j$}; \end{tikzpicture} $$ ## Termination and quantization If the series never stops, the ratio of successive coefficients settles to $$ \frac{c_{j+1}}{c_j} \xrightarrow{\;j\to\infty\;} \frac{2}{j}, $$ which is the coefficient ratio of the Taylor series of $e^{2\rho}$. An infinite $v$ therefore grows like $e^{2\rho}$, making $u \sim \rho^{\ell+1}e^{+\rho}$ — exponentially divergent, not normalizable. The only escape is termination: some $c_{j+1}$ must vanish while $c_j \ne 0$, after which every later coefficient is zero and $v$ is a polynomial. From the recursion, the numerator vanishes at $j = j_{\max}$ when $$ 2(j_{\max}+\ell+1) = \rho_0. $$ Define the integer on the left as the **principal quantum number** $$ n \equiv j_{\max} + \ell + 1, \qquad \rho_0 = 2n. $$ Because $j_{\max} \ge 0$, the polynomial degree is $j_{\max} = n-\ell-1 \ge 0$, which is precisely the restriction $$ \ell = 0, 1, 2, \ldots, n-1. $$ > **Theorem (Radial quantization of hydrogen).** A normalizable solution of the > Coulomb radial equation exists only when $\rho_0 = 2n$ for a positive integer > $n$, with $\ell \le n-1$. The bound-state energies follow from > $\rho_0 = 2Z/\kappa a_0 = 2n$, giving $\kappa = Z/na_0$ and > $$ > E_n = -\frac{\hbar^2\kappa^2}{2\mu} = -\frac{Z^2}{n^2}\,\frac{\hbar^2}{2\mu a_0^2} > = -\frac{Z^2}{n^2}\,\mathrm{Ry}, > \qquad \mathrm{Ry} = \frac{\mu k^2 e^4}{2\hbar^2} \approx 13.6~\text{eV}. > $$ The energy depends on $n$ alone. The angular momentum $\ell$ entered the radial equation through the centrifugal term and left through the termination count $n-\ell-1$, but it cancelled out of the eigenvalue. That cancellation is special to the $1/r$ potential and is examined as a [symmetry](/atomic-physics/quantum-hydrogen-atom/symmetry-degeneracy-runge-lenz) in the next lesson. The Rydberg energy scales as $Z^2$ because the length scale contracts as $a_0/Z$ while the potential deepens as $Z$; the two combine to $Z^2$. $$ % caption: A terminating series (top) gives a polynomial times e^(−ρ) that decays % and is normalizable; a non-terminating series behaves like e^(+ρ) and diverges. \begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (6.2,0) node[right, black!70] {$r$}; \draw[->, black] (0,0) -- (0,3.0) node[above, black!70] {$u(r)$}; % terminating: rises, one node, decays to zero \draw[acc, very thick] (0,0) .. controls (0.6,1.9) and (1.1,2.4) .. (1.6,2.3) .. controls (2.4,2.1) and (3.4,0.7) .. (4.4,0.28) .. controls (5.2,0.12) and (5.8,0.05) .. (6.1,0.03); \node[acc, anchor=west] at (1.6,2.45) {terminates: normalizable}; % non-terminating: turns up and diverges \draw[black, very thick, dashed] (0,0) .. controls (0.7,0.7) and (1.6,0.9) .. (2.4,0.95) .. controls (3.4,1.0) and (4.2,1.6) .. (4.9,2.4) .. controls (5.3,2.9) and (5.5,3.1) .. (5.65,3.2); \node[black, anchor=south east] at (5.6,3.2) {diverges}; \end{tikzpicture} $$ ## The associated Laguerre polynomials The terminating $v(\rho)$ is, up to normalization, an **associated Laguerre polynomial**. In the convention of Griffiths & Schroeter,[^gs-42] $$ v(\rho) \propto L_{n-\ell-1}^{2\ell+1}(2\rho), \qquad L_{q}^{p}(x) = \sum_{k=0}^{q}(-1)^{k}\binom{q+p}{q-k}\frac{x^{k}}{k!}, $$ a polynomial of degree $q = n-\ell-1$. The lower index is the polynomial degree, the upper index $2\ell+1$ is set by the centrifugal factor. The first few are $$ L_0^{p}(x) = 1, \qquad L_1^{p}(x) = -x + (p+1), \qquad L_2^{p}(x) = \tfrac{1}{2}x^2 - (p+2)x + \tfrac{1}{2}(p+1)(p+2). $$ $$ % caption: The associated Laguerre polynomials L_q^1(x) for degrees q = 0, 1, 2. % The degree equals the number of positive real roots, hence the radial nodes. \begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (6.2,0) node[right, black!70] {$x$}; \draw[->, black] (0,-1.6) -- (0,2.2) node[above, black!70] {$L$}; \draw[black] (0,0) -- (6.0,0); % L0 = 1 constant \draw[acc, very thick] (0,1.0) -- (6.0,1.0); \node[acc, anchor=south west] at (4.6,1.0) {$L_0$}; % L1 = -x + 2, root at x=2 \draw[black, very thick, dashed] (0,1.6) -- (5.4,-1.4); \node[black, anchor=south east] at (1.1,1.35) {$L_1$}; \fill[black] (3.6,0) circle (1.6pt); % L2 = x^2/2 - 3x + 3, roots ~1.27, 4.73 (scaled down to fit) \draw[black, very thick, densely dotted] (0,1.5) .. controls (0.9,0.2) and (1.6,-0.55) .. (2.4,-0.72) .. controls (3.2,-0.88) and (3.9,-0.55) .. (4.6,0.15) .. controls (5.1,0.65) and (5.5,1.1) .. (5.9,1.55); \node[black, anchor=south] at (2.4,-0.95) {$L_2$}; \fill[black] (1.15,0) circle (1.6pt); \fill[black] (4.25,0) circle (1.6pt); \end{tikzpicture} $$ Assembling $R = u/r = \rho^{\ell}e^{-\rho}v/\kappa$ and restoring $\rho = Zr/na_0$ (so $2\rho = 2Zr/na_0$) gives the normalized hydrogenic radial function $$ R_{n\ell}(r) = -\sqrt{\left(\frac{2Z}{na_0}\right)^{3}\frac{(n-\ell-1)!}{2n\,(n+\ell)!}} \left(\frac{2Zr}{na_0}\right)^{\ell} e^{-Zr/na_0}\, L_{n-\ell-1}^{2\ell+1}\!\left(\frac{2Zr}{na_0}\right). $$ The three ingredients are visible: a power $r^{\ell}$ from the centrifugal asymptote, an exponential $e^{-Zr/na_0}$ from the large-$r$ asymptote whose range $na_0/Z$ grows with $n$, and a degree-$(n-\ell-1)$ polynomial supplying the oscillations. The overall sign is convention. > **Worked example.** Verify the ground state from the general formula. For > $n=1$, $\ell=0$: the polynomial is $L_0^{1}=1$, the power $r^{\ell}=1$, and > $$ > R_{10}(r) = -\sqrt{\left(\frac{2Z}{a_0}\right)^{3}\frac{0!}{2\cdot 1!}}\;e^{-Zr/a_0} > = 2\left(\frac{Z}{a_0}\right)^{3/2}e^{-Zr/a_0}. > $$ > For $n=2$, $\ell=0$: $L_1^{1}(x) = -x+2$ with $x = 2Zr/2a_0 = Zr/a_0$, so > $L_1^1 = 2 - Zr/a_0$ and > $$ > R_{20}(r) = \frac{1}{2\sqrt{2}}\left(\frac{Z}{a_0}\right)^{3/2}\!\left(2-\frac{Zr}{a_0}\right)e^{-Zr/2a_0}. > $$ > Both reproduce the tabulated > [hydrogen wave functions](/atomic-physics/quantum-hydrogen-atom/hydrogen-wave-functions). ## Counting nodes and states The polynomial degree $n-\ell-1$ equals the number of positive roots of $L_{n-\ell-1}^{2\ell+1}$, and each root is a value of $r$ where $R_{n\ell}$ changes sign — a **radial node**. The full node count of $\psi_{n\ell m}$ splits cleanly: - **radial nodes:** $n-\ell-1$, the spheres where $R_{n\ell}(r)=0$; - **angular nodes:** $\ell$, the nodal cones and planes of $Y_{\ell m}$; - **total nodes:** $(n-\ell-1)+\ell = n-1$, independent of $\ell$. A state of fixed $n$ always has $n-1$ nodal surfaces, redistributed between radial and angular as $\ell$ runs from $0$ to $n-1$. The circular state $\ell = n-1$ has no radial nodes and a single-hump $R$; the penetrating state $\ell = 0$ puts all $n-1$ nodes in the radial direction. The degeneracy of the level $E_n$ counts the distinct $(n,\ell,m)$ triples at fixed $n$. For each $\ell$ there are $2\ell+1$ values of $m$, so $$ g_n = \sum_{\ell=0}^{n-1}(2\ell+1) = n^2, $$ doubling to $2n^2$ once electron spin is included. The $\ell$-sum being exactly $n^2$ is the arithmetic face of the accidental degeneracy. | $n$ | allowed $\ell$ | radial nodes $n-\ell-1$ | spatial states $n^2$ | | --- | --- | --- | --- | | $1$ | $0$ | $0$ | $1$ | | $2$ | $0,1$ | $1,0$ | $4$ | | $3$ | $0,1,2$ | $2,1,0$ | $9$ | | $4$ | $0,1,2,3$ | $3,2,1,0$ | $16$ | ## Hydrogenic scaling with nuclear charge Every result carries $Z$ in a fixed pattern, because the Coulomb problem has a single length and a single energy scale that absorb it. Rescaling $r \to Zr$ maps the $Z$-charged problem onto hydrogen: - **length:** the wave function depends on $r$ only through $Zr/a_0$, so all radii contract as $a_0/Z$; the most probable radius of the ground state is $a_0/Z$; - **energy:** $E_n = -Z^2\,\mathrm{Ry}/n^2$, quadratic in $Z$; - **expectation values:** $\langle r\rangle \propto n^2 a_0/Z$ and $\langle 1/r\rangle \propto Z/n^2 a_0$, the subject of a [later lesson](/atomic-physics/quantum-hydrogen-atom/expectation-values-virial). The scaling makes every one-electron ion — $\mathrm{He}^+$ ($Z=2$), $\mathrm{Li}^{2+}$ ($Z=3$), and the highly charged ions of laboratory spectroscopy — a rescaled copy of hydrogen, with the same wave functions and a spectrum stretched by $Z^2$. The exact solvability ends the moment a second electron is added, or the $1/r$ potential is perturbed by relativistic and [fine-structure](/atomic-physics/fine-structure-and-the-dirac-atom/spin-orbit-thomas-precession) corrections, both of which are computed by feeding these exact wave functions into perturbation theory. [^gs-42]: **Griffiths & Schroeter**, _Introduction to Quantum Mechanics_, 3rd ed., §4.2 — the substitution $u=rR$, the dimensionless radial equation, asymptotic peeling, the Frobenius recursion, the termination condition $\rho_0=2n$, and the associated-Laguerre form of $R_{n\ell}$ with its normalization. [^bj-33]: **Bransden & Joachain**, _Physics of Atoms and Molecules_, 2nd ed., §3.2–3.3 — the series solution of the hydrogen radial equation and the identification of the polynomial factor with the associated Laguerre functions.