The radial equation and centrifugal barriers
Separate a central potential, choose the radial measure correctly, and identify origin conditions and angular-momentum barriers.
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The question for today
What you’ll understand
- Transform the three-dimensional central problem into a half-line equation.
- Normalize a reduced radial function with the correct measure.
- Check origin regularity and interpret the centrifugal term without confusing it with a new force.
A central potential depends only on distance from the origin, making angular momentum a conserved label. Separating the angles leaves a radial differential equation, but the radial coordinate is not simply an ordinary Cartesian coordinate restricted to positive values. Its integration measure and its endpoint at zero carry physical information. A change of radial unknown makes that information easier to track while exposing an effective angular-momentum barrier.
Separate angles and change the radial unknown
For a constant reduced mass and central potential , choose simultaneous eigenfunctions of , , and . Write with the spherical harmonic normalized over solid angle. Its angular eigenvalues are and , with integer and . The radial equation depends on but not on when no direction-selecting interaction is present.
The volume element is . Defining removes the first derivative from the radial kinetic operator and turns normalization into an ordinary half-line integral. The shell probability between nearby radii is . This does not make the shell probability, nor does it discard the angular distribution for measurements in a particular direction. Each expression must be paired with its own measure.
The barrier is an angular kinetic-energy term
The reduced equation contains in addition to the physical potential. It is nonnegative and rises toward the origin for nonzero angular momentum. This centrifugal barrier comes from angular kinetic energy, not an extra interaction added to the original Hamiltonian. It suppresses low-energy penetration into small radii for higher partial waves. It does not imply that every effective potential has a positive classical turning radius for every energy.
For a potential regular at the origin, the leading radial solutions behave as and . The regular three-dimensional Hamiltonian selects the first. In particular, the s-wave condition is even though may be finite and nonzero. Square-integrability alone can fail to exclude a singular s-wave candidate; absence of a contact interaction and the operator domain also matter. Singular inverse-square interactions require a separate endpoint analysis.
A radial level has angular multiplicity
For each allowed radial level at fixed , the magnetic labels have equal energy in a spinless central problem. This rotational degeneracy does not mean there are different radial solutions. Conversely, states with different angular momentum need not share an energy in a generic central potential. Hydrogen's additional equality across several angular momenta is a special Coulomb feature, rather than a consequence of spherical symmetry by itself.
Bound radial states must decay sufficiently at infinity to have finite norm. Positive-energy partial waves instead describe scattering and use continuum normalization. An effective barrier may also support a resonance that leaks outward rather than a square-integrable stationary eigenstate. A peak trapped behind a barrier in a finite-time picture therefore does not establish a true bound state. The spectral distinction uses the endpoint conditions and energy threshold, not simply the presence of a hump in an effective-potential sketch.
Work it through
Read a centrifugal barrier quantitatively
A central model has and a potential regular near the origin. For , find the centrifugal contribution at , the leading regular powers of u and R, and the angular multiplicity of each radial level.
The angular factor is . At 2 nm the barrier energy is . This is a kinetic contribution, so it adds to whatever central V is specified.
Regularity gives . Since , the ordinary radial function behaves as . The shell probability density therefore vanishes as , rather than as .
The allowed magnetic labels are , giving equal-energy states for each such radial solution. This count excludes spin and does not imply a degeneracy with an s-wave radial level.
The angular barrier contributes 1.5 eV at 2 nm. Regular u and R behave as r³ and r², and every radial level in this ℓ = 2 sector has five magnetic substates in the spinless central model.
The living notebook
Explore the model
Interactive lab · function
Track a power with its measure
Reset and set both controls to a = 1 and x = 0.5; predict y = 0.25. Reset and set both controls to a = 1 and x = 1; predict y = 1. Regard this quadratic only as the near-origin R scaling of the ℓ = 2 example. The lab does not multiply by the radial measure, solve a radial equation, plot a 1/r² barrier, or produce a wavefunction normalization.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA plotted radial function and a probability density can have different powers; the lab shows only its stated quadratic.
Pause & reason
Check your understanding
Your turn
Try it yourself
An s-wave particle is confined by a spherical infinite wall at radius a, with V = 0 inside and no contact interaction at the origin. Take the lowest reduced radial function for . Find A, its energy in terms of , and the probability of .
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Normalize u with dr and use the symmetry of sine squared about the midpoint of the interval.
Compare with the solution
The norm is , giving up to phase. With , the radial equation is , so . The sine-squared density is symmetric about , and exactly half the norm lies below that radius. Although , the limit is finite and nonzero. The infinite wall imposes u(a) = 0, a separate condition from origin regularity.
Keep the idea
Take it with you
- The reduced radial function absorbs the spherical probability measure.
- Angular kinetic energy produces an ℓ-dependent inverse-square barrier.
- Origin domains, angular multiplicity, and infinity conditions all matter for radial spectra.
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