Boundary Conditions for Radial Wavefunctions

Boundary conditions for radial wavefunctions are the part of a central-potential problem most likely to be hidden by notation. The reduced radial equation resembles a one-dimensional Schrödinger equation, but it lives on the half-line, inherits the three-dimensional measure, and has a special endpoint at the origin.

The general boundary-condition philosophy is in Boundary Conditions. This page records the radial rules needed after deriving the Radial Schrödinger Equation.

Helpful background. Central Potentials to separate rotation-invariant Hamiltonians into angular sectors.

The Two Radial Functions

For a separated central-potential state,

$$\psi(r,\theta,\phi) = R_\ell(r)Y_\ell^m(\theta,\phi),$$

the reduced radial wavefunction is

$$u_\ell(r)=rR_\ell(r).$$

These are not interchangeable objects. If the angular function is normalized on the sphere, then

$$\int_{\mathbb R^3} \lvert\psi\rvert^2\,d^3r = \int_0^\infty \lvert R_\ell(r)\rvert^2r^2\,dr = \int_0^\infty \lvert u_\ell(r)\rvert^2\,dr .$$

Thus $R_\ell$ is the radial factor in the three-dimensional wavefunction, while $u_\ell$ is the half-line wavefunction with ordinary $dr$ measure. Many wrong radial solutions come from normalizing one while interpreting the other.

Standard Boundary Conditions

For ordinary bound states in a nonsingular or Coulomb-like central potential, the standard radial conditions are:

• the full wavefunction is regular at the origin;
• the reduced function is square-normalizable on $0\lt r\lt \infty$;
• the bound-state solution decays at infinity;
• matching conditions are imposed at any finite discontinuities or singular shells;
• singular potentials are treated as domain questions, not by casual analogy with regular potentials.

In compact form, a standard bound state satisfies

$$\int_0^\infty \lvert u_\ell(r)\rvert^2\,dr \lt \infty, \qquad u_\ell(r)\to0 \quad \text{as} \quad r\to\infty,$$

together with the appropriate origin condition described below.

Regularity at the Origin

For potentials that are finite at the origin, or less singular than $1/r^2$, the leading near-origin radial equation is controlled by angular momentum:

$$\frac{1}{r^2} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) - \frac{\ell(\ell+1)}{r^2}R \approx0 .$$

Trying $R(r)\sim r^s$ gives

$$s(s+1)=\ell(\ell+1),$$

so

$$s=\ell \qquad \text{or} \qquad s=-\ell-1 .$$

The regular branch is

$$R_\ell(r)\sim r^\ell, \qquad u_\ell(r)\sim r^{\ell+1}.$$

The singular branch is

$$R_\ell(r)\sim r^{-\ell-1}, \qquad u_\ell(r)\sim r^{-\ell}.$$

In the standard no-contact central-potential problem, the singular branch is rejected because the original three-dimensional wavefunction is not regular at the origin. The practical reduced-function condition is therefore

$$u_\ell(0)=0$$

for the usual regular radial domain.

For $\ell=0$, one must be especially careful. The singular branch behaves as $R_0(r)\sim 1/r$ and $u_0(r)\sim\text{constant}$. It may look harmless in the $u$ equation, but it is not a regular three-dimensional wavefunction for an ordinary nonsingular central potential. Allowing it corresponds to changing the domain, often to include contact physics at the origin.

The Coulomb cusp

Regularity does not require a zero radial slope. For $V(r)=-\kappa/r$ with $\kappa>0$, define $a=\hbar^2/(m\kappa)$; in a two-body problem $m$ is the reduced mass. Write the regular solution as $u=r^{\ell+1}(c_0+c_1r+O(r^2))$. The first uncancelled power in the radial equation gives

$$2(\ell+1)c_1+\frac{2}{a}c_0=0.$$

Thus the smooth radial factor after removing the leading power obeys

$$\left.\frac{d}{dr}\log\!\left(\frac{R_\ell(r)}{r^\ell}\right)\right|_{0^+} =-\frac{1}{(\ell+1)a}.$$

For an $s$ wave this is $R_0'(0^+)=-R_0(0)/a$, the Coulomb cusp. The finite nonzero slope cancels the potential’s $1/r$ singularity in the equation. The energy enters only at the next order, so this relation holds for both bound and regular continuum solutions at finite energy. It agrees with the exact hydrogenic radial functions.

For outward radial integration, the corresponding starting behavior is

$$\frac{u'(r)}{u(r)}=\frac{\ell+1}{r}-\frac{1}{(\ell+1)a}+O(r), \qquad r\to0^+.$$

This refines the regular branch; it is not an additional freely chosen boundary parameter. In particular, imposing $R_0'(0)=0$ would exclude the physical Coulomb $s$ states.

Normalizability at Infinity

For a bound state, square normalizability requires

$$\int_0^\infty \lvert u_\ell(r)\rvert^2\,dr \lt \infty .$$

When $V(r)\to0$ and $E\lt 0$, the large-$r$ reduced equation usually gives exponential decay:

$$u_\ell(r)=e^{-\kappa r}\times\text{a subexponential factor}, \qquad \kappa=\frac{\sqrt{-2mE}}{\hbar}.$$

The factor matters for long-range tails: a Coulomb bound state also has a power of $r$ multiplying the exponential. For finite-range potentials the asymptotic factor tends to a constant.

For continuum states with $E\gt 0$, normalizability is replaced by a continuum or flux convention. One then imposes oscillatory asymptotic behavior rather than decay. The boundary condition at infinity is part of the physical question: bound state, standing wave, incoming scattering state, outgoing wave, or phase-shift convention.

Matching at Finite Radii

If $V(r)$ has a finite jump at $r=a\gt 0$, the radial wavefunction and its derivative match across the jump. In reduced form,

$$u(a^-)=u(a^+), \qquad u'(a^-)=u'(a^+).$$

Equivalently, because $a$ is nonzero, $R$ and $R'$ are continuous at the same point. This is the radial version of ordinary matching across a finite step.

For a delta-shell potential,

$$V(r)=\lambda\delta(r-a), \qquad a\gt 0,$$

the reduced wavefunction is continuous, but its derivative jumps:

$$u(a^-)=u(a^+),$$

and

$$u'(a^+)-u'(a^-) = \frac{2m\lambda}{\hbar^2}u(a).$$

This condition follows by integrating the reduced radial equation across a small interval around $a$. It should not be confused with the origin condition.

Singular Potentials

The simple power law above assumes the potential is less singular than $1/r^2$. The Coulomb potential,

$$V(r)=-\frac{\kappa}{r},$$

is singular at the origin, but it is still milder than $1/r^2$. The usual regular behavior remains the correct bound-state condition.

Potentials with inverse-square behavior require separate analysis:

$$V(r)\sim -\frac{g}{r^2}.$$

Such a term competes directly with the centrifugal term in the effective radial potential. Depending on the sign and size of $g$, the radial Hamiltonian may require extra boundary data at the origin. Strong attractive singularities can produce the fall-to-the-center pathology if no short-distance physics is supplied.

The safe rule is: when the potential is as singular as $1/r^2$ or worse, do not import the regular-potential boundary condition without checking the operator domain.

Current and Self-Adjointness

Boundary conditions are not only aesthetic. They control probability conservation and the self-adjointness of the radial Hamiltonian. In the $u$ representation, the boundary term obtained by integration by parts has the form

$$\left[ v^*u' - {v'}^*u \right]_{0}^{\infty}.$$

For a closed radial bound-state problem, the domain should make this boundary term vanish for allowed functions $u$ and $v$. Regularity at the origin and decay at infinity accomplish this for the standard bound-state domain.

In current language, the reduced half-line current is

$$J_u = \frac{\hbar}{2mi} \left( u^*u'-u{u'}^* \right).$$

An isolated bound-state problem should not leak probability through $r=0$ or $r=\infty$. Scattering problems instead choose asymptotic current conventions deliberately.

Practical Checklist

When solving a radial problem, check the following before trusting a spectrum or wavefunction:

1. Have you decided whether the physical radial function is $R(r)$ or $u(r)$ in each formula?
2. Is the normalization using $r^2\,dr$ for $R$ or $dr$ for $u$?
3. Is the near-origin solution regular as a three-dimensional wavefunction?
4. Does the large-$r$ behavior match bound-state decay or the intended scattering convention?
5. Are finite jumps, delta shells, or hard cutoffs matched correctly?
6. Is the potential singular enough to require a domain analysis beyond the standard rule?

Common Mistakes

• Treating the reduced equation as a full-line one-dimensional problem.
• Requiring $R(0)=0$ for every regular radial state. For $\ell=0$, regular $R$ can be finite and nonzero at the origin.
• Allowing $u(0)\ne0$ in an ordinary no-contact $s$-wave problem.
• Normalizing $R$ with $dr$ instead of $r^2\,dr$.
• Forgetting that continuum radial states are not square-normalized like bound states.
• Assuming Coulomb singularity and inverse-square singularity have the same domain behavior.
• Imposing derivative continuity across a delta-shell potential.

Exercises

1. Check the Coulomb cusp using the $1s$, $2s$, and $2p$ functions in Radial Wavefunctions. Why do the two $s$ states have the same logarithmic slope despite their different energies, and which factor should be differentiated for the $p$ state?

Solution

For $1s$, $R_{10}\propto e^{-r/a}$, so $R_{10}'(0)/R_{10}(0)=-1/a$. For $2s$,

$$R_{20}\propto(2-r/a)e^{-r/(2a)}, \qquad \frac{R_{20}'(0)}{R_{20}(0)}=-\frac{1}{2a}-\frac{1}{2a}=-\frac{1}{a}.$$

The polynomial contributes half the $2s$ slope; using only its exponential would give the wrong answer. Both states must cancel the same local Coulomb singularity. Their different energies affect higher coefficients of the origin expansion.

For $2p$, $R_{21}\propto r e^{-r/(2a)}$. Its logarithmic derivative contains $1/r$ because of the leading angular-momentum power. Remove that power first:

$$\frac{d}{dr}\log(R_{21}/r)=-\frac{1}{2a} =-\frac{1}{(\ell+1)a},\qquad \ell=1.$$

All three states retain $u(0)=0$. The cusp specifies the next term in the regular solution rather than replacing that origin condition.

2. Show that the regular branch implies $u_\ell(0)=0$.

Solution

For the regular branch,

$$R_\ell(r)\sim r^\ell.$$

Since $u_\ell=rR_\ell$,

$$u_\ell(r)\sim r^{\ell+1}.$$

For all allowed orbital angular momenta $\ell=0,1,2,\ldots$, this tends to zero as $r\to0$. Thus the regular radial domain has

$$u_\ell(0)=0.$$

3. Derive the derivative jump condition for a delta-shell potential $V(r)=\lambda\delta(r-a)$ with $a\gt 0$.

Solution

Integrate the reduced radial equation across $[a-\epsilon,a+\epsilon]$:

$$\int_{a-\epsilon}^{a+\epsilon} \left[ -\frac{\hbar^2}{2m}u'' + \lambda\delta(r-a)u \right]dr = \int_{a-\epsilon}^{a+\epsilon} \left[ E - \frac{\hbar^2\ell(\ell+1)}{2mr^2} \right]u\,dr .$$

The right side vanishes as $\epsilon\to0$ for finite $u$. The left side becomes

$$-\frac{\hbar^2}{2m} \left[ u'(a^+)-u'(a^-) \right] + \lambda u(a) =0.$$

Therefore

$$u'(a^+)-u'(a^-) = \frac{2m\lambda}{\hbar^2}u(a).$$

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• National Institute of Standards and Technology, Digital Library of Mathematical Functions, Eqs. 18.39.34–35, hydrogenic radial functions and scales.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• G. Teschl, Mathematical Methods in Quantum Mechanics, 2nd ed., American Mathematical Society, 2014.