Radial Schrödinger Equation

The radial Schrödinger equation is the one-dimensional ordinary differential equation left after a three-dimensional central-potential problem has been separated into angular and radial parts. The compact separation step is derived in Angular and Radial Separation; this page is the canonical home for the radial half-line equation, boundary behavior, and normalization conventions used in concrete systems such as the hydrogen atom.

The central subtlety is that the radial problem is not an ordinary one-dimensional line problem. The radial coordinate has measure $r^2\,dr$, the origin is a boundary point, and angular momentum produces a centrifugal term.

Central-Potential Setup

A central potential depends only on the distance from the origin:

$$V(\mathbf r)=V(r), \qquad r=\lvert\mathbf r\rvert.$$

The time-independent Schrödinger equation is

$$\left[ -\frac{\hbar^2}{2m}\nabla^2 +V(r) \right]\psi(\mathbf r) =E\psi(\mathbf r).$$

Because the Hamiltonian is rotationally invariant, energy eigenfunctions may be chosen as simultaneous eigenfunctions of $\hat L^2$ and $\hat L_z$. With spherical harmonics normalized on the unit sphere, write

$$\psi_{E\ell m}(r,\theta,\phi) =R_{E\ell}(r)Y_\ell^m(\theta,\phi),$$

where

$$\ell=0,1,2,\ldots, \qquad m=-\ell,-\ell+1,\ldots,\ell.$$

The angular functions obey

$$\hat L^2Y_\ell^m =\hbar^2\ell(\ell+1)Y_\ell^m, \qquad \hat L_zY_\ell^m =\hbar mY_\ell^m.$$

The radial equation is independent of $m$. For a purely central Hamiltonian, this produces the familiar $(2\ell+1)$ angular degeneracy within each allowed $\ell$ sector, unless additional interactions or boundary conditions break rotational symmetry.

Radial Equation for R

In spherical coordinates, the Laplacian may be organized as a radial piece plus angular momentum:

$$\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2\frac{\partial}{\partial r} \right) - \frac{\hat L^2}{\hbar^2r^2}.$$

Substituting $R(r)Y_\ell^m(\theta,\phi)$ gives the radial equation

$$-\frac{\hbar^2}{2m} \left[ \frac{1}{r^2}\frac{d}{dr} \left( r^2\frac{dR}{dr} \right) - \frac{\ell(\ell+1)}{r^2}R \right] +V(r)R =ER.$$

Equivalently,

$$-\frac{\hbar^2}{2m} \frac{1}{r^2}\frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \left[ V(r)+ \frac{\hbar^2\ell(\ell+1)}{2mr^2} \right]R =ER.$$

This is often called the radial Schrödinger equation in $R$ form. The term

$$\frac{\hbar^2\ell(\ell+1)}{2mr^2}$$

is the centrifugal term. It is not an additional physical force inserted by hand; it is the angular kinetic-energy contribution in spherical coordinates.

Reduced Radial Function

It is often cleaner to remove the first-derivative structure by defining

$$u(r)=rR(r).$$

For $r\gt 0$, the radial equation becomes

$$-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} + \left[ V(r)+ \frac{\hbar^2\ell(\ell+1)}{2mr^2} \right]u =Eu.$$

This looks like a one-dimensional Schrödinger equation on the half-line $0\lt r\lt \infty$, with effective potential

$$V_{\mathrm{eff}}(r) = V(r)+ \frac{\hbar^2\ell(\ell+1)}{2mr^2}.$$

The resemblance to a one-dimensional problem is useful, but incomplete: the domain is a half-line, the boundary condition at $r=0$ matters, and the relation $R=u/r$ must remain physically regular.

Normalization

If the spherical harmonics are normalized as

$$\int_{S^2} \lvert Y_\ell^m(\theta,\phi)\rvert^2\,d\Omega =1,$$

then a bound radial state is normalized by

$$\int_0^\infty \lvert R(r)\rvert^2r^2\,dr =1.$$

In terms of $u=rR$ this becomes the ordinary half-line normalization

$$\int_0^\infty \lvert u(r)\rvert^2\,dr =1.$$

This is why $u$ is convenient. The probability of finding the particle between $r$ and $r+dr$, after integrating over angles, is

$$P(r)\,dr = \lvert R(r)\rvert^2r^2\,dr = \lvert u(r)\rvert^2\,dr.$$

For continuum states, the same measure applies, but normalization is usually delta-normalization in energy or momentum rather than square-integrability.

Boundary Behavior

For nonsingular or mildly singular central potentials such as the Coulomb potential, physical bound-state solutions are regular at the origin. Near $r=0$, the centrifugal term dominates for $\ell\gt 0$, and the regular behavior is

$$R(r)\sim r^\ell, \qquad u(r)\sim r^{\ell+1}.$$

The singular alternative behaves like

$$R(r)\sim r^{-\ell-1}, \qquad u(r)\sim r^{-\ell},$$

and is rejected in the standard central-potential problems because it is not an acceptable finite wavefunction at the origin.

For bound states on an infinite domain, normalizability also requires

$$u(r)\to0 \qquad \text{as} \qquad r\to\infty.$$

For scattering states, the large-$r$ behavior is instead oscillatory and is fixed by the chosen incoming, outgoing, or standing-wave convention.

Singular potentials require more care. If $V(r)$ is more singular than the centrifugal term, the self-adjoint domain of the radial Hamiltonian may require additional boundary data. That is a mathematical-domain issue, not a license to ignore the origin.

For a compact checklist of origin, infinity, matching, and singular-potential cases, see Boundary Conditions for Radial Wavefunctions.

Physical Interpretation

The centrifugal term suppresses low-radius probability for $\ell\gt 0$. Classically, nonzero angular momentum makes it difficult to reach the origin; quantum mechanically, the radial wavefunction develops the same qualitative avoidance through the $\ell(\ell+1)/r^2$ term.

The radial equation also explains why central-potential spectra organize into angular-momentum sectors. Each $\ell$ value defines a different half-line spectral problem. The magnetic quantum number $m$ labels orientation within the same angular-momentum multiplet and does not alter the radial equation for a rotationally invariant Hamiltonian.

For the Coulomb potential, the radial equation has enough additional symmetry that the bound-state energy depends only on the principal quantum number $n$, not on $\ell$ or $m$. That stronger degeneracy is special to the $1/r$ potential and is explained in the hydrogen atom page.

Common Mistakes

• Normalizing $R(r)$ with $dr$ instead of $r^2\,dr$.
• Forgetting that $u(r)=rR(r)$ and $R(r)$ have different dimensions and boundary behavior.
• Treating the $u$ equation as a full-line one-dimensional problem rather than a half-line problem.
• Allowing the singular near-origin solution in ordinary central-potential bound states.
• Assuming that $m$ enters the radial equation for a rotationally invariant potential.
• Calling $V_{\mathrm{eff}}$ a new physical potential rather than a compact way to combine $V(r)$ with angular kinetic energy.

Worked Example: Free Radial Motion

For $V(r)=0$, the reduced radial equation is

$$\frac{d^2u}{dr^2} + \left[ k^2-\frac{\ell(\ell+1)}{r^2} \right]u =0, \qquad k^2=\frac{2mE}{\hbar^2}.$$

The regular solutions are proportional to

$$u_\ell(r)=rj_\ell(kr),$$

where $j_\ell$ is a spherical Bessel function. Thus

$$R_\ell(r)=j_\ell(kr).$$

For $\ell=0$, $j_0(kr)=\sin(kr)/(kr)$, so $u_0(r)\propto\sin(kr)$. The boundary condition $u(0)=0$ is the radial remnant of regularity at the origin.

Exercises

1. Starting from the $R$ equation, derive the $u=rR$ equation and identify where the first derivative cancels.

Solution

Use $R=u/r$. Then

$$\frac{dR}{dr} = \frac{u'}{r}-\frac{u}{r^2},$$

so

$$r^2\frac{dR}{dr} = ru'-u.$$

Differentiating once more gives

$$\frac{d}{dr} \left( r^2\frac{dR}{dr} \right) = ru''.$$

Therefore

$$\frac{1}{r^2}\frac{d}{dr} \left( r^2\frac{dR}{dr} \right) = \frac{u''}{r}.$$

Multiplying the $R$ equation by $r$ gives the stated half-line equation for $u$.

2. Show that the $R$ and $u$ normalizations are equivalent when the spherical harmonics have unit angular norm.

Solution

The full normalization is

$$\int_{\mathbb R^3} \lvert R(r)Y_\ell^m(\theta,\phi)\rvert^2\,d^3r = \int_0^\infty \lvert R(r)\rvert^2r^2\,dr \int_{S^2} \lvert Y_\ell^m\rvert^2\,d\Omega.$$

The angular integral is $1$. Since $u=rR$,

$$\lvert R(r)\rvert^2r^2 = \lvert u(r)\rvert^2.$$

Thus the radial normalization may be written as either

$$\int_0^\infty \lvert R(r)\rvert^2r^2\,dr=1 \quad \text{or} \quad \int_0^\infty \lvert u(r)\rvert^2\,dr=1.$$

3. For a nonsingular potential near $r=0$, neglect $V(r)$ and $E$ compared with the centrifugal term. Use a trial form $R(r)\sim r^s$ to find the two possible powers.

Solution

Substituting $R\sim r^s$ into the radial differential operator gives

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

The coefficient must vanish, so

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

The two roots are

$$s=\ell, \qquad s=-\ell-1.$$

The regular solution is $R\sim r^\ell$.

Where This Is Used

• Hydrogen Atom solves the Coulomb radial equation and obtains the Balmer energy spectrum.
• Continuum States of the Coulomb Problem uses the positive-energy radial equation and continuum boundary conventions.
• Central Potentials explains the rotational symmetry and quantum-number labels used here.
• Effective Radial Potential interprets the centrifugal term, radial turning points, and node ordering in fixed-$\ell$ sectors.
• Boundary Conditions for Radial Wavefunctions summarizes the endpoint and matching conditions used by radial solutions.
• Angular and Radial Separation derives the separation constant and the effective radial potential.
• Spherical Coordinates gives the coordinate convention, volume element, and Laplacian used here.
• Spherical Harmonics gives the angular eigenfunctions.
• Orbital Angular Momentum explains the operators that label the angular sectors.
• Normalization Conventions reviews square-integrable and delta-normalized states.
• Sturm–Liouville Theory supplies the mathematical framework for radial eigenvalue problems.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.