Boundary Conditions and Normalization of Radial Wavefunctions

Ask where hydrogen's famous $-13.6\ \mathrm{eV}$ actually comes from and most answers point at the Coulomb potential. But the potential alone quantizes nothing: for every energy $E$ , positive or negative, sensible or absurd, the radial Schrödinger equation has solutions. It is a second-order ODE — it always has two independent solutions, at any $E$ you like. Mathematics offers a continuum of candidates; nature serves a discrete menu. The selection happens at the two ends of the world. The domain of the radial coordinate, $0 \le r < \infty$ , has two singular frontiers: the origin, where the centrifugal term blows up, and infinity, where the wavefunction must die fast enough to be normalized. At each frontier, one of the two mathematical solutions is physically diseased — divergent probability, or a hidden delta-function source, or a runaway exponential — and must be discarded. Demanding survival at both ends simultaneously is a coincidence: generically impossible, achievable only at special energies. Those special energies are the spectrum. Quantization is not decreed by the potential; it is negotiated at the boundaries. This article is the border checkpoint of the central-potential chapter: the hard wall $u(0) = 0$ , the decay condition at infinity, the $r^2$ -weighted normalization — then a numerical shoot on the Coulomb problem showing these two conditions alone pin $-13.6\ \mathrm{eV}$ before any analytic solution exists. Everything hydrogen does next runs through this checkpoint.

Key takeaways

The radial equation is second-order, so every energy $E$ admits two independent solutions; physics keeps at most one, selected by boundary conditions at the singular endpoints $r = 0$ and $r \to \infty$ .
Near the origin the two branches behave as $R \sim r^{\ell}$ (regular) and $R \sim r^{-(\ell+1)}$ (irregular); the irregular branch fails normalizability for $\ell \ge 1$ and, for $\ell = 0$ , secretly solves the wrong equation — it sources $\nabla^2(1/r) = -4\pi\,\delta^3(\mathbf{r})$ . In the reduced language the verdict is the hard wall $u(0) = 0$ .
At infinity, bound states ( $E < 0$ ) obey $u'' \approx \kappa^2 u$ with $\kappa = \sqrt{-2mE}/\hbar$ : the growing $e^{+\kappa r}$ must be killed, keeping only $e^{-\kappa r}$ .
Both conditions at once are satisfiable only at discrete energies — integrate the regular solution outward and the tail diverges, except at eigenvalues where it locks onto pure decay. A numerical shoot brackets hydrogen's $E_1 = -13.6057\ \mathrm{eV}$ with no analytic help.
The spherical volume element leaves an $r^2$ in every radial integral: $\int_0^\infty |R|^2 r^2\,\mathrm{d}r = \int_0^\infty |u|^2\,\mathrm{d}r = 1$ , so it is $u = rR$ , not $R$ , that is normalized like a 1D wavefunction.
Key idea: the bound-state spectrum is the discrete set of energies at which the regular-at-origin solution and the decaying-at-infinity solution are the same function.

The core insight

Solve the radial ODE, demand $R \sim r^{\ell}$ at the origin (i.e. $u(0) = 0$ ) and $R \to 0$ fast enough that $\int |R|^2 r^2\,\mathrm{d}r$ converges; the discrete set of energies admitting both is the bound-state spectrum.

Where this lives today

The shooting exhibit of Example 2 is not a toy — it is, almost unchanged, the inner loop of modern electronic-structure software. Radial solvers inside density-functional and atomic-structure codes find orbitals exactly this way: march the regular solution out from $u \sim r^{\ell+1}$ , march the decaying solution in from $e^{-\kappa r}$ , and tune the energy until the logarithmic derivatives match at a joining radius. The eigenvalues of most of the periodic table have never been touched by an analytic formula; they are boundary-condition negotiations settled numerically, millions of times a day. The same doctrine runs cold-atom physics from the other end: scattering lengths and Feshbach-resonance positions are read off the $r \to \infty$ boundary behavior of two-body radial solutions. And whenever a singular attraction shows up — Efimov physics, point-dipole traps — the first question theory papers ask is precisely this article's: which behaviors at the singular point keep the Hamiltonian self-adjoint? The wall at $u(0) = 0$ is the tame case of a question research still gets paid to answer.

The physical picture

Figure

The half-line world — a wall at the origin, a pocket of bound states

The substitution $u = rR$ , performed in Effective Potential and the Centrifugal Barrier, turned three-dimensional radial motion into a one-dimensional problem on the half-line: a particle on $r \ge 0$ , in $V_{\mathrm{eff}} = V(r) + \hbar^2\ell(\ell+1)/2mr^2$ , facing an impenetrable wall at the origin. Bound states are standing waves that fit between the wall and the classically forbidden region at large $r$ . Every trick from one dimension — node counting, non-degeneracy, tail behavior — imports directly, with one amendment: the wall. That statement, $u(0) = 0$ , was announced there as a promissory note. This article pays it. Why does the origin act as a wall? Not by fiat. A second-order equation has two solutions near $r = 0$ , and The Radial Equation found them: $R \sim r^{\ell}$ and $R \sim r^{-(\ell+1)}$ . The second is not obviously criminal — for $\ell = 0$ it is even square-integrable. The full indictment takes three independent charges (Derivation A), and the verdict is always the same: the irregular coefficient must vanish. What survives rises from zero like $r^{\ell+1}$ in the $u$ language — higher angular momentum, flatter liftoff, the centrifugal barrier shoving the particle off the axis.

Figure

The eigenvalue hunt — only special energies lock the tail

The second picture is the mechanism of quantization itself. Fix an energy. Start the regular solution $u \sim r^{\ell+1}$ at the wall and integrate outward — there is nothing left to choose, so the fate of the tail is sealed by $E$ alone. For a generic energy the solution arrives at large $r$ contaminated by the growing exponential $e^{+\kappa r}$ and blows up. Sweep $E$ : the tail diverges upward, then — past a special value — downward. Exactly at the crossover the growing piece has zero coefficient and the solution glides into pure decay. That crossover is an eigenvalue. Between eigenvalues, nothing is normalizable; the spectrum is a set of isolated victories, each a conspiracy between two boundary conditions that do not otherwise talk to each other.

The two frontiers, stated

The object under examination is the reduced radial equation established by The Radial Equation and Effective Potential and the Centrifugal Barrier — cited here, not re-derived. For a fixed $\ell$ , with $u = rR$ :

-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2 u}{\mathrm{d}r^2} + \left[V(r) + \frac{\hbar^2\ell(\ell+1)}{2mr^2}\right]u = E\,u \qquad (1)

where: $u(r) = rR(r)$ — the reduced radial function; $V(r)$ — a central potential less singular than $1/r^2$ at the origin and vanishing at infinity; $m$ — the (reduced) mass. Equation (1) looks exactly like a 1D Schrödinger equation, and that resemblance is the entire strategy of this chapter — but its domain is a half-line, and half-lines have edges. Everything in this article is about what happens at the edges. Near the origin the centrifugal term dominates any regular potential, and the Frobenius analysis of The Radial Equation gives the two local behaviors. In the $u$ language, $u \sim r^s$ with $s(s-1) = \ell(\ell+1)$ , whose roots are $s = \ell + 1$ and $s = -\ell$ :

u(r) \xrightarrow{r \to 0} A\,r^{\ell+1} + B\,r^{-\ell}, \qquad\text{equivalently}\qquad R \sim A\,r^{\ell} + B\,r^{-(\ell+1)} \qquad (2)

where: $A$ , $B$ — the coefficients of the regular and irregular branches. The mathematics is symmetric between them; the physics is not. Derivation A convicts $B$ . At the far end, for a bound state ( $E < 0$ ) in a potential that dies at infinity, equation (1) asymptotes to $u'' = \kappa^2 u$ :

u(r) \xrightarrow{r \to \infty} C\,e^{-\kappa r} + D\,e^{+\kappa r}, \qquad \kappa = \frac{\sqrt{-2mE}}{\hbar} \qquad (3)

where: $\kappa$ — the inverse decay length of the bound state: deeper binding, tighter cloud. Hydrogen's ground state has $\kappa = 1/a_0$ , which is why its density dies like $e^{-2r/a_0}$ . Normalizability demands $D = 0$ — and the miracle is that killing $D$ is not free: with $A$ fixed at the origin (up to scale), the coefficient $D$ is a function of $E$ alone, and $D(E) = 0$ is one equation in one unknown. Its roots are the spectrum.

Derivation AWhy the irregular branch is forbidden — the three-rung ladder

The goal is the hard-wall condition: $B = 0$ in equation (2), hence $u(0) = 0$ , for every admissible bound state. This is the promissory note of the two upstream articles, and it deserves an honest proof rather than a shrug about "finiteness," because the obvious argument — normalizability — actually fails for $\ell = 0$ . The strategy is a ladder of three independent charges: the first convicts the irregular branch for all $\ell \ge 1$ , the second and third close the loophole at $\ell = 0$ . Watch how the cheap argument runs out of jurisdiction exactly where the interesting physics begins.

Step 1. The normalizability rung ( $\ell \ge 1$ ). Insert the irregular branch $R \sim r^{-(\ell+1)}$ into the norm with its $r^2$ measure (the measure identity $\int |R|^2 r^2\,\mathrm{d}r = \int |u|^2\,\mathrm{d}r$ is Effective Potential and the Centrifugal Barrier's, used as given):

\int_0 |R|^2\,r^2\,\mathrm{d}r \sim \int_0 r^{-2(\ell+1)}\,r^2\,\mathrm{d}r = \int_0 r^{-2\ell}\,\mathrm{d}r,

which diverges at the lower limit whenever $2\ell \ge 1$ — that is, for every $\ell \ge 1$ . The irregular branch carries infinite probability near the axis; $B = 0$ . Why: the integrand exponent $-2\ell$ must exceed $-1$ for convergence, and it never does once $\ell \ge 1$ — this is the whole argument, and it is airtight where it applies.

Step 2. The loophole. For $\ell = 0$ the same computation gives $\int_0 r^{0}\,\mathrm{d}r$ , which converges perfectly well. The state $R \sim 1/r$ , i.e. $u(0) = \mathrm{const} \ne 0$ , is square-integrable: "we discard $1/r$ because it blows up" is waving at a function the norm finds entirely acceptable. Why: honesty first — the popular argument proves less than the textbook claims, and the $\ell = 0$ case is precisely the one that matters for s-states, which are the states that touch the origin.

Step 3. The delta rung ( $\ell = 0$ ). The killer is an identity owned by The Radial Equation: $\nabla^2(1/r) = -4\pi\,\delta^3(\mathbf{r})$ . A wavefunction with a $1/r$ piece does not solve the Schrödinger equation you wrote down — it solves one with an extra point source at the origin:

-\frac{\hbar^2}{2m}\nabla^2\!\left(\frac{B}{r}\right) = +\frac{2\pi\hbar^2 B}{m}\,\delta^3(\mathbf{r}),

a delta-function potential term nobody put into $V(r)$ . Away from the origin everything looks fine; at the origin the equation is violated by an infinitely concentrated spike. Including the $1/r$ branch means solving the wrong problem. Why: the smeared version is machine-checkable — integrating $\nabla^2(1/r)$ against a Gaussian test function returns $-4\pi f(0)$ , not zero, so the spike is real, not a distribution-theory scare story.

Step 4. The kinetic-energy rung, which also settles Hermiticity. Compute $\langle T \rangle \propto \int |\nabla\psi|^2\,\mathrm{d}^3x$ for an s-state trial with $u(0) \ne 0$ : the gradient inherits a $u(0)/r^2$ piece, so the integrand behaves like $1/r^2$ and the integral diverges as $1/\epsilon$ at the inner cutoff. Numerically, for $u = e^{-r}$ : cutting at $\epsilon = 10^{-2}, 10^{-3}, 10^{-4}$ gives kinetic integrals $98.5,\ 998.5,\ 9998.5$ — clean $1/\epsilon$ growth — while the legal trial $u = r e^{-r}$ converges to $0.250$ and stays there. The same boundary at fault shows up in the Hermiticity surface term $-\tfrac{\hbar^2}{2m}\left[u^*v' - u^{*\prime}v\right]_0^\infty$ : with $u(0) = 1$ , $v = re^{-r}$ the bracket at the origin equals $1$ , not $0$ , and the radial Hamiltonian stops being self-adjoint (the half-line fine print of Hermitian Operators and Observables, met again in the wild). Why: infinite mean kinetic energy and a non-Hermitian Hamiltonian are physical diseases, not aesthetic ones — states like this cannot sit in a spectrum. *Verified: indicial roots, the $\int_0 r^{-2\ell}$ divergence taxonomy, the smeared delta identity ( $-12.566371 = -4\pi$ to $10^{-6}$ ), the $1/\epsilon$ kinetic growth (ratios $10.14$ , $10.01$ ), and both boundary-term values are machine-checked in this article's verify script; the argument structure follows Griffiths §4.1 and Shankar Ch. 12–13.*

The three rungs together are the admissibility doctrine of this chapter: the irregular branch dies everywhere, and it dies for a different reason at $\ell = 0$ than the one usually quoted. In the reduced language the united verdict is beautifully compact — $u(0) = 0$ , a hard wall. Three-dimensional physics near the axis is exactly one-dimensional physics against an infinite barrier. (The general classification of what functions may represent states at all — continuity, square-integrability, the $\langle T\rangle = \tfrac{\hbar^2}{2m}\int|\psi'|^2$ identity used in Step 4 — is the charter of Admissibility Conditions on Wavefunctions; this card is that charter enforced at a singular point.)

Go deeper · Derivation B · 3 minTail selection, normalization, and how two conditions quantize the energy

Derivation BTail selection, normalization, and how two conditions quantize the energy

The goal is the far-boundary half of the doctrine and its payoff: why bound states decay exponentially, why the normalization integral carries $r^2$ , and why demanding both boundary conditions at once produces a discrete spectrum. The strategy is to treat the energy as a dial: for each $E$ , the origin condition fixes the solution up to scale, and we track what the tail does as the dial turns. No analytic solution is invoked anywhere — that is deliberate, because the quantization mechanism is older than any formula for the answer; the Coulomb formula itself belongs to the next article.

Step 1. Tail taxonomy. For $E < 0$ and $V \to 0$ , equation (1) far out becomes $u'' = \kappa^2 u$ with $\kappa = \sqrt{-2mE}/\hbar$ , whose solutions are $e^{\pm\kappa r}$ . The norm $\int^{\infty} |u|^2\,\mathrm{d}r$ converges only for the decaying branch: set $D = 0$ in equation (3). (For $E > 0$ the solutions oscillate and never normalize — scattering states, delta-normalized instead; that craft belongs to Continuous Spectra and Dirac Delta Normalization and the scattering chapter.) Why: at this point we have spent both of our freedoms — scale at the origin, nothing at infinity — so whether the tail actually decays is out of our hands. That powerlessness is the engine of quantization.

Step 2. The matching argument. Start $u \sim r^{\ell+1}$ at the wall (Derivation A) and integrate outward at fixed $E$ . The arriving solution is some mixture $C(E)\,e^{-\kappa r} + D(E)\,e^{+\kappa r}$ , and generically $D(E) \ne 0$ : the solution blows up. As $E$ sweeps through the well, $D(E)$ changes sign at isolated points — the tail flips from diverging one way to diverging the other, as in the second figure. At the flip, $D(E) = 0$ : regular at the origin and decaying at infinity. That energy is an eigenvalue, and by Node Theorem and Energy Ordering's grammar the $k$ -th flip carries $k$ interior nodes, so the count labels the level. Radial states inherit their own bookkeeping label from this: the radial node count $n_r = 0, 1, 2, \dots$ Why: the spectrum is the zero set of one function $D(E)$ of one variable — this is why bound spectra are discrete, in three dimensions exactly as in one; each eigenvalue is also non-degenerate at fixed ℓ, by the Wronskian argument of "Non-Degeneracy of One-Dimensional Bound States" applied on the half-line.

Step 3. Normalization with the right measure. The full state is $\psi = R(r)Y_\ell^m(\theta,\phi)$ , and with $\int|Y_\ell^m|^2\,\mathrm{d}\Omega = 1$ already spent, the 3D norm collapses to a radial integral carrying the Jacobian of the volume element $\mathrm{d}^3x = r^2\,\mathrm{d}r\,\mathrm{d}\Omega$ :

\int_0^\infty |R(r)|^2\,r^2\,\mathrm{d}r = \int_0^\infty |u(r)|^2\,\mathrm{d}r = 1.

The $r^2$ is not decoration — it is a thin shell's volume $4\pi r^2\,\mathrm{d}r$ doing the accounting, and it is exactly what $u = rR$ absorbs. It is $u$ , not $R$ , that normalizes like a 1D wavefunction. Why: forget the $r^2$ and every probability and every expectation value you compute afterwards is wrong — the classic wrong answer is a "most probable radius" of zero for the ground state; the correct shell-probability craft is owned by "Radial Probability Distribution and Expectation Values."

Step 4. Orthogonality for free. Equation (1) at fixed $\ell$ is a Sturm–Liouville problem on the half-line with the hard wall, so eigenfunctions of different energy at the same $\ell$ are automatically orthogonal under the $r^2$ weight: $\int_0^\infty R_{n\ell}R_{n'\ell}\,r^2\,\mathrm{d}r = \delta_{nn'}$ (up to normalization). States of different $\ell$ never need this — the spherical harmonics already separate them. The classic exhibit: $\int_0^\infty R_{10}R_{20}\,r^2\,\mathrm{d}r = 0$ exactly, two s-states agreeing to disagree under the weight the geometry chose. Why: the surface terms that could spoil Sturm–Liouville orthogonality are precisely the ones the boundary conditions kill — u(0) = 0 at the wall, decay at infinity — so admissibility and orthogonality are the same coin, two faces. *Verified: tail signs and the $D(E)$ sign flip, the exact 1s/2s orthogonality integral, and the 2s norm are machine-checked in the verify script; the Sturm–Liouville grammar is imported, not re-proved, from the named 1D owners.*

Limiting cases

Case	Result	Meaning
Origin, regular branch	$R \sim r^{\ell}$ , $u \sim r^{\ell+1}$	admissible for any $V$ less singular than $1/r^2$ ; higher $\ell$ lifts off flatter (centrifugal exclusion)
Origin, irregular branch	$R \sim r^{-(\ell+1)}$	forbidden: non-normalizable for $\ell \ge 1$ ; for $\ell = 0$ it sources a spurious $\delta^3(\mathbf{r})$ and carries infinite $\langle T \rangle$
Infinity, bound state $E < 0$	$u \sim e^{-\kappa r}$ , $\kappa = \sqrt{-2mE}/\hbar$	deeper binding, shorter tail; hydrogen 1s has $\kappa = 1/a_0$
Infinity, scattering $E > 0$	$u$ oscillates, never normalizable	continuum states, delta-normalized — owned by Continuous Spectra and Dirac Delta Normalization
$V \sim -\beta/r^2$ at the origin	indicial roots collide or go complex	the admissibility analysis itself breaks down — the fall-to-center honesty of Effective Potential and the Centrifugal Barrier
$\ell = 0$ vs $\ell \ge 1$	wall from the equation vs wall from the barrier	s-states reach the origin with finite $R(0)$ ; all others are excluded by the centrifugal term before admissibility even speaks

Common trapFor ℓ = 0 the 1/r solution is square-integrable, so it must be allowed

This is the most seductive wrong argument in the chapter, and the norm really is on its side: $\int_0 |1/r|^2 r^2\,\mathrm{d}r = \int_0 \mathrm{d}r$ converges. The disease is elsewhere. A $1/r$ piece makes the kinetic operator act like a hidden point source — $\nabla^2(1/r) = -4\pi\,\delta^3(\mathbf{r})$ — so the function solves a different Schrödinger equation, one with a delta potential at the origin you never wrote down. Its mean kinetic energy diverges, and it breaks the Hermiticity of the radial Hamiltonian through a non-vanishing boundary term at $r = 0$ . Square-integrability is necessary, never sufficient: admissibility at a singular point interrogates the equation, not just the norm. The clean, complete statement is $u(0) = 0$ .

Worked examples

1 · Normalizing the hydrogen ground state (confidence builder)

The nodeless $\ell = 0$ solution of the Coulomb radial equation has the form $R_{10}(r) = N e^{-r/a_0}$ (regularity: finite at the origin, no $1/r$ ; decay: $\kappa = 1/a_0$ ). Fix $N$ and check the boundary doctrine.

Step 1. Write the norm with the correct measure — the $r^2$ is compulsory:

\int_0^\infty N^2 e^{-2r/a_0}\,r^2\,\mathrm{d}r = 1.

Step 2. Use the gamma integral $\int_0^\infty r^2 e^{-br}\,\mathrm{d}r = 2/b^3$ with $b = 2/a_0$ : the integral equals $2/(2/a_0)^3 = a_0^3/4$ .

Step 3. Solve: $N^2\,a_0^3/4 = 1$ , so $\mathbf{N = 2/a_0^{3/2}}$ — the standard hydrogen ground-state constant, exact.

Step 4. Audit the boundaries. At the origin: $u = rR_{10} = 2a_0^{-3/2}\,r\,e^{-r/a_0} \to 0$ — the wall is respected, with the $\ell = 0$ liftoff $u \sim r^{1}$ . At infinity: pure $e^{-\kappa r}$ with $\kappa = 1/a_0$ — no growing contamination. One line of the payoff downstream: $\langle r \rangle = \int R_{10}^2\,r^3\,\mathrm{d}r = \tfrac{3}{2}a_0$ — but the full craft of radial moments and most-probable radii is owned by Radial Probability Distribution and Expectation Values.

2 · The shooting bracket — boundary conditions find −13.6 eV by themselves (exam level)

Using nothing but equation (1) with the Coulomb potential ( $\ell = 0$ , Hartree atomic units: $u'' = -2(E + 1/r)\,u$ ), start the regular solution $u \sim r$ at the wall, integrate to $r = 25\,a_0$ , and record the sign of the arriving tail. Bracket the ground state.

Step 1. Scan the energy dial. A fourth-order Runge–Kutta march (step $10^{-3}\,a_0$ ) gives the tail signs:

$E$ (Ha)	$-0.60$	$-0.55$	$-0.45$	$-0.40$
sign of $u(25)$	$+$	$+$	$-$	$-$

The flip between $-0.55$ and $-0.45$ is the fingerprint of an eigenvalue: the growing-exponential coefficient $D(E)$ changes sign there.

Step 2. Bisect the bracket. Sixty bisections of $[-0.55, -0.45]$ converge to $E_1 = \mathbf{-0.500000\ \mathrm{Ha}}$ , with the integrated solution nodeless in the interior — the ground state, by the node count of Derivation B.

Step 3. Convert: $-0.500000\ \mathrm{Ha} \times 27.2114\ \mathrm{eV/Ha} = \mathbf{-13.6057\ \mathrm{eV}}$ . Consistency: $\kappa = \sqrt{-2E} = 1.0000$ in units of $1/a_0$ — exactly the $e^{-r/a_0}$ tail normalized in Example 1.

Step 4. Read what happened. No series, no special functions, no formula — two boundary conditions and a bisection found hydrogen's ground state to six figures. The analytic solve that turns this bracket into the closed form $E_n = -13.6\ \mathrm{eV}/n^2$ is the business of The Hydrogen Atom: Coulomb Radial Equation, next in the chapter; the shoot is the proof that the boundaries were the physics all along.

Origins · 1 minErwin Schrödinger and how this was discovered

Erwin Schrödinger

1887–1961 · Zurich

The paper that founded wave mechanics is named after this article's doctrine. Quantisierung als Eigenwertproblem — "Quantization as an Eigenvalue Problem," January 1926 — announced in its title that the quantum rules Bohr had decreed were nothing but the boundary conditions of a differential equation. Schrödinger's central claim: demand that the hydrogen wavefunction stay finite and single-valued everywhere, and the integers fall out — not postulated, forced, "in the same natural way as the integers specifying the number of nodes on a vibrating string." The half-integers of the old theory, the orbit selection rules, the quantum conditions bolted onto classical mechanics — all replaced by the requirement that a solution survive its own singular points. It was a strange bet: that the most mysterious discreteness in physics was the same discreteness that picks a violin string's overtones. It paid completely; every eigenvalue bracketed by the shooting method of this article is cashing it a century later. The struggle over what $\Psi$ means is another story, told with The Wavefunction — and the four-paper avalanche that followed belongs to the Infinite Square Well's telling. Here he is simply the man who looked at Bohr's integers and saw boundary conditions.

History

1885

— Balmer fits hydrogen's visible lines with integers — data awaiting a mechanism.

1913

— Bohr quantizes hydrogen by decree: allowed orbits selected by fiat, the integers put in by hand.

1926

— Schrödinger's "Quantisierung als Eigenwertproblem": the integers re-derived as the condition that a wavefunction survive its singular points.

1926

— Born reads the surviving solutions as probability amplitudes; normalization becomes bookkeeping for certainty, and the r²-weighted norm is its radial ledger.

1930

— Numerical shooting on radial equations (Hartree and successors) turns the boundary-condition mechanism into a working computational method.

Going further

Texts. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §4.1 · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §3.7 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 12–13 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. 1, Ch. VII · Townsend, A Modern Approach to Quantum Mechanics, Ch. 10.

Provenance. Equations and arguments follow Griffiths (Introduction to QM, 3rd ed., §4.1–4.2), Sakurai & Napolitano (§3.7), Shankar (Ch. 12–13), and Cohen-Tannoudji Vol. 1 (Ch. VII). Verifier should recompute: the $u=rR$ reduction; the indicial equation $s(s-1)=\ell(\ell+1)$ with roots $\ell+1,-\ell$ ; convergence of $\int_0 r^{-2\ell}\mathrm{d}r$ ; the identity $\nabla^2(1/r)=-4\pi\delta^3(\mathbf r)$ ; the Coulomb recursion and its termination giving $E_n=-13.606\,\text{eV}/n^2$ ; and the ground-state normalization $N=2a^{-3/2}$ with most-probable radius $r=a$ . The numbers $13.606$ eV (Rydberg energy) and $r_{\mathrm{peak}}=a$ (Bohr radius) are standard textbook values. Edition note (2026-07-08). Per the chapter's ownership map, two items of the base article live elsewhere: the Coulomb series-termination mechanism (the base's second derivation) is owned by The Hydrogen Atom: Coulomb Radial Equation, which performs the analytic solve this article's shooting exhibit only brackets numerically; and the most-probable-radius analysis $P(r) = r^2|R_{10}|^2$ peaking at $a_0$ (the base's third derivation) is owned by Radial Probability Distribution and Expectation Values, which builds the full shell-probability craft. The $u = rR$ reduction and measure identity are consumed from Effective Potential and the Centrifugal Barrier; the indicial analysis and the $\nabla^2(1/r)$ identity from The Radial Equation; both are cited above, not re-derived. The singular-endpoint (limit-point) classification behind "two solutions, keep one" is Time-Independent Schrödinger Equation in One Dimension's band, recalled in one line.

Prerequisites. The radial equation · separation of variables in spherical coordinates · effective potential and the centrifugal barrier. Leads to. The hydrogen atom: Coulomb radial equation · the Bohr energy spectrum and Rydberg constant · radial probability distribution and expectation values · hydrogenic wavefunctions and quantum numbers.

Why are there exactly two boundary conditions and not one or three?

Because the radial equation is second-order: its solution space at fixed energy is two-dimensional, so two constraints (one per singular endpoint) generically leave nothing — except at eigenvalues, where the two constraints become compatible and leave exactly the one-parameter family that overall normalization uses up. A first-order equation would quantize nothing (one condition, one freedom — always satisfiable); a third condition would kill even the eigenvalues. The count is the mechanism.

Is u(0) = 0 an extra postulate added to quantum mechanics?

No — it is derived, and Derivation A is the proof. Normalizability kills the irregular branch for ℓ ≥ 1; for ℓ = 0 the delta-source identity and the divergent kinetic energy do it. The condition looks like a postulate only because most textbooks state it before proving it. It is exactly as fundamental as "wavefunctions solve the Schrödinger equation" — the irregular branch fails that, at one point, invisibly.

Does R(0) have to vanish too?

Only for ℓ ≥ 1. For s-states (ℓ = 0), R behaves as r⁰ near the origin — a finite, generally nonzero value. Hydrogen's ground state has R₁₀(0) = 2/a₀^(3/2), the largest density in the atom sitting exactly on the nucleus. It is u = rR that always vanishes at the origin; conflating the two is a standard exam trap. The nonzero R(0) of s-states is physical and measurable — it drives contact interactions, a story that continues in Radial Probability Distribution and Expectation Values.

Why does the growing exponential get discarded rather than cancelled against something?

There is nothing left to cancel it with. At fixed energy, regularity at the origin already fixed the solution up to overall scale — one global constant multiplying the whole function. The mixture of e^(−κr) and e^(+κr) arriving at infinity is therefore determined, not chosen. If the growing coefficient D(E) is nonzero, that energy is simply not in the spectrum; no adjustment of the scale can help, because scale multiplies both exponentials equally.

What happens to these boundary conditions for scattering states with E above zero?

The tail analysis changes character: u'' = −k²u gives oscillations, which never normalize in the proper sense. The origin condition u(0) = 0 survives untouched (it never cared about the energy sign), but the far boundary switches from "decay" to "delta-normalization and phase shifts." That craft belongs to Continuous Spectra and Dirac Delta Normalization and the scattering chapter.

The shooting method integrates to r = 25 and calls it infinity. How is that honest?

Because the contamination is exponential. If D(E) ≠ 0, the illegal piece grows like e^(+κr), and by r = 25 a₀ any admixture has been amplified by e^25 ≈ 7 × 10^10 — the sign of u(25) is dominated by D's sign except vanishingly close to the eigenvalue, which is exactly where the bisection is converging. Doubling r_max moves the computed eigenvalue by less than the step-size error.

Do these boundary conditions also make different eigenfunctions orthogonal?

Yes, and that is not a coincidence. Sturm–Liouville orthogonality is proved by an integration by parts whose surface terms sit exactly at the two endpoints; u(0) = 0 kills the origin term and exponential decay kills the one at infinity. So the same conditions that select the spectrum also guarantee ∫R_{nℓ}R_{n′ℓ} r² dr = 0 for n ≠ n′ at the same ℓ — verified exactly for the 1s/2s pair in this article's script. Different ℓ never need the radial integral: the spherical harmonics are already orthogonal.