core35 min studyLesson 374 in Physics

The radial equation and centrifugal barriers

Separate a central potential, choose the radial measure correctly, and identify origin conditions and angular-momentum barriers.

On this page

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 μ\mu and central potential V(r)V(r), choose simultaneous eigenfunctions of HH, L2L^2, and LzL_z. Write ψ=Rℓ(r)Yℓm(θ,ϕ)\psi=R_\ell(r)Y_{\ell m}(\theta,\phi) with the spherical harmonic normalized over solid angle. Its angular eigenvalues are ℏ2ℓ(ℓ+1)\hbar^2\ell(\ell+1) and mℏm\hbar, with integer ℓ≥0\ell\ge0 and −ℓ≤m≤ℓ-\ell\le m\le\ell. The radial equation depends on ℓ\ell but not on mm when no direction-selecting interaction is present.

The volume element is r2dr dΩr^2dr\,d\Omega. Defining uℓ=rRℓu_\ell=rR_\ell 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 ∣uℓ∣2dr|u_\ell|^2dr. This does not make ∣Rℓ∣2dr|R_\ell|^2dr 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.

ψ=uℓ(r)rYℓm,∫∣Yℓm∣2dΩ=1,∫0∞∣uℓ(r)∣2dr=∫0∞r2∣Rℓ(r)∣2dr=1\psi=\frac{u_\ell(r)}rY_{\ell m},\qquad \int |Y_{\ell m}|^2d\Omega=1,\qquad \int_0^\infty|u_\ell(r)|^2dr=\int_0^\infty r^2|R_\ell(r)|^2dr=1
For a normalized three-dimensional state, ψ has units length⁻³ᐟ², R has the same units, and u has units length⁻¹ᐟ². The spherical harmonic is dimensionless. The r = 0 behaviour must be imposed through the radial domain rather than through this formal division alone.↔ Scroll wide equations horizontally.

The barrier is an angular kinetic-energy term

The reduced equation contains ℏ2ℓ(ℓ+1)/(2μr2)\hbar^2\ell(\ell+1)/(2\mu r^2) 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 u∝rℓ+1u\propto r^{\ell+1} and u∝r−ℓu\propto r^{-\ell}. The regular three-dimensional Hamiltonian selects the first. In particular, the s-wave condition is u(0)=0u(0)=0 even though R(0)R(0) 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.

−ℏ22μuℓ′′+[V(r)+ℏ2ℓ(ℓ+1)2μr2]uℓ=Euℓ,uℓ(r)∝rℓ+1(r→0, regular V)-\frac{\hbar^2}{2\mu}u_\ell''+\left[V(r)+\frac{\hbar^2\ell(\ell+1)}{2\mu r^2}\right]u_\ell=Eu_\ell,\qquad u_\ell(r)\propto r^{\ell+1}\quad(r\to0\text{, regular }V)
Energy E and V use the same units, r is a length, and μ is the reduced mass. The origin behaviour refers to the ordinary regular central-potential Hamiltonian without a point interaction. The equation is a half-line spectral problem with both origin and infinity conditions.↔ Scroll wide equations horizontally.

A radial level has angular multiplicity

For each allowed radial level at fixed ℓ\ell, the 2ℓ+12\ell+1 magnetic labels have equal energy in a spinless central problem. This rotational degeneracy does not mean there are 2ℓ+12\ell+1 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.

Read a centrifugal barrier quantitatively

A central model has C=ℏ2/(2μ)=1.0 eV nm2C=\hbar^2/(2\mu)=1.0\,\mathrm{eV\,nm^2} and a potential regular near the origin. For ℓ=2\ell=2, find the centrifugal contribution at r=2.0 nmr=2.0\,\mathrm{nm}, the leading regular powers of u and R, and the angular multiplicity of each radial level.

  1. The angular factor is ℓ(ℓ+1)=2(3)=6\ell(\ell+1)=2(3)=6. At 2 nm the barrier energy is C 6/r2=(1.0)(6)/4=1.5 eVC\,6/r^2=(1.0)(6)/4=1.5\,\mathrm{eV}. This is a kinetic contribution, so it adds to whatever central V is specified.

  2. Regularity gives u∝rℓ+1=r3u\propto r^{\ell+1}=r^3. Since R=u/rR=u/r, the ordinary radial function behaves as r2r^2. The shell probability density therefore vanishes as ∣u∣2∝r6|u|^2\propto r^6, rather than as r4r^4.

  3. The allowed magnetic labels are m=−2,−1,0,1,2m=-2,-1,0,1,2, giving 2ℓ+1=52\ell+1=5 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 result

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.

Explore the model

Interactive lab

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.

Track a power with its measureThis 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. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

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.

Check your understanding

When the angular harmonic is normalized and u(r) = rR(r), which quantity gives the probability in a thin spherical shell of thickness dr?

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 u(r)=Asin⁡(πr/a)u(r)=A\sin(\pi r/a) for 0<r<a0<r<a. Find A, its energy in terms of C=ℏ2/(2μ)C=\hbar^2/(2\mu), and the probability of r<a/2r<a/2.

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 ∣A∣2∫0asin⁡2(πr/a)dr=∣A∣2a/2|A|^2\int_0^a\sin^2(\pi r/a)dr=|A|^2a/2, giving A=2/aA=\sqrt{2/a} up to phase. With ℓ=0\ell=0, the radial equation is −Cu′′=Eu-Cu''=Eu, so E=Cπ2/a2E=C\pi^2/a^2. The sine-squared density is symmetric about a/2a/2, and exactly half the norm lies below that radius. Although u(0)=0u(0)=0, the limit R(0)=Aπ/aR(0)=A\pi/a is finite and nonzero. The infinite wall imposes u(a) = 0, a separate condition from origin regularity.

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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