advanced35 min studyLesson 392 in Physics

Partial-wave phase shifts

Resolve central-potential scattering into angular momentum channels and extract phase shifts from radial matching.

Before you begin

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The question for today

What you’ll understand

Derive the phase-shift form of a radial asymptotic wave.
Relate a channel phase shift to its unit-modulus S element.
Compute partial cross sections and interpret their unitarity limits.

A central potential preserves angular momentum, turning one three-dimensional scattering problem into independent radial problems. Each radial solution remembers the interaction through a phase shift relative to free motion. The complete angular distribution comes from adding these partial-wave amplitudes coherently. Integrated cross sections are simpler because angular orthogonality removes interference between different angular momenta, but that simplification must not be applied before the angular integration.

Match radial waves outside the interaction

For a finite-range real central potential, separate the wavefunction into a spherical harmonic times a reduced radial function divided by radius. The reduced radial equation contains the centrifugal barrier and the potential. Regularity at the origin fixes the interior solution up to normalization. Outside the interaction range, it is a combination of the regular and irregular free radial functions. Matching both the function and its derivative at the boundary determines their ratio and hence the phase shift.

At large radius, choose the radial convention with a sine of free radial phase plus $\delta_\ell$ . A positive shift advances that spatial sine relative to free motion, while an actual wave-packet time delay depends on its energy derivative rather than its sign alone. Phase shifts are defined modulo pi because adding pi only changes an overall radial sign. When following them with energy, use a continuous branch so an artificial arctangent jump does not masquerade as a physical discontinuity.

\left[-\frac{\hbar^2}{2\mu}\frac{d^2}{dr^2}+\frac{\hbar^2\ell(\ell+1)}{2\mu r^2}+V(r)\right]u_\ell=Eu_\ell,\qquad u_\ell\sim C_\ell\sin\left(kr-\frac{\ell\pi}{2}+\delta_\ell\right)

uℓ is the reduced radial function, ℓ a nonnegative integer and δℓ an angle in radians. The asymptotic sine assumes a short-range real central potential and positive elastic-channel energy.↔ Scroll wide equations horizontally.

Convert the radial phase into an amplitude

Writing the sine as incoming and outgoing exponentials shows that the interaction multiplies the outgoing coefficient relative to the free reference by $e^{2i\delta_\ell}$ . A real phase therefore preserves the magnitude of channel flux while changing its phase. The factor of two is essential: the radial sine shift is half the phase of the channel S element. Free scattering corresponds to a unit S element and a zero scattered amplitude, not to a zero S element.

For an incident plane wave along z, the angular basis reduces to Legendre polynomials. Their coefficients include the angular-momentum multiplicity and a partial amplitude with length units. Different angular momenta can reinforce or cancel at one direction even when every integrated contribution is positive. A measured angular minimum therefore does not establish the absence of an interaction in each channel. Inferring phase shifts from intensity alone can have ambiguities because intensities discard part of the complex-amplitude information.

S_\ell=e^{2i\delta_\ell},\qquad f(\theta)=\sum_{\ell=0}^\infty(2\ell+1)\frac{e^{i\delta_\ell}\sin\delta_\ell}{k}P_\ell(\cos\theta)

Sℓ is dimensionless and the partial amplitude is a length. This expansion assumes spinless distinguishable relative motion, a central potential and one elastic channel; identical-particle restrictions will be imposed separately.↔ Scroll wide equations horizontally.

Use orthogonality and the centrifugal barrier

Integrating the squared amplitude uses orthogonality of Legendre polynomials. Each channel contributes a nonnegative area proportional to its multiplicity and squared sine phase. Since that sine cannot exceed one, a fixed partial wave has a maximum elastic cross section at a phase of pi over two modulo pi. The resulting bound scales with inverse squared wavenumber; it is a wave-flux constraint rather than the physical surface area of a hard target.

At low energy for sufficiently short-range nonsingular interactions, centrifugal suppression usually leaves the s wave dominant, except near special higher-wave resonances or symmetry restrictions. A rough semiclassical estimate connects angular momentum to impact parameter, explaining why many channels become relevant when the wavelength is short compared with the range. It is an estimate rather than a sharp angular-momentum cutoff. The phase comparison in the wave lab isolates coherent addition but does not produce radial matching, Legendre coefficients or phase shifts for a potential.

\sigma_\ell=\frac{4\pi}{k^2}(2\ell+1)\sin^2\delta_\ell\le\frac{4\pi}{k^2}(2\ell+1),\qquad \sigma_{\rm el}=\sum_\ell\sigma_\ell

These elastic bounds use the stated distinguishable-particle convention. k is in inverse length, so σℓ has area units; equality requires a real phase shift with sin²δℓ = 1.↔ Scroll wide equations horizontally.

Work it through

Two active angular momenta

At $k=0.50\,\mathrm{nm^{-1}}$ , suppose $\delta_0=\pi/6$ and $\delta_1=\pi/2$ , with all higher phase shifts zero. Find both partial cross sections, their sum and the fraction contributed by the p wave.

The common factor is $4\pi/k^2=16\pi\,\mathrm{nm^2}$ . The s-wave multiplicity is one and $\sin^2(\pi/6)=1/4$ , giving $\sigma_0=4\pi\,\mathrm{nm^2}$ .
The p-wave multiplicity is three and its squared sine is one. Thus $\sigma_1=48\pi\,\mathrm{nm^2}$ , which saturates the p-wave elastic unitarity limit at this wavenumber.
Add the integrated contributions to obtain $52\pi\,\mathrm{nm^2}$ . The p-wave fraction is $48/52=12/13\approx92.3\%$ . An angular differential cross section would still contain interference and cannot be obtained by this simple positive sum.

The result

The s and p contributions are 4π and 48π nm², totaling 52π nm²; the p wave contributes 12/13 of the integrated area.

The living notebook

Explore the model

Interactive lab

Interactive lab · waves

Phase-sensitive addition before angular integration

Reset and set BOTH Phase difference = 0° and Amplitude ratio = 1. Predict resultant amplitude 2. Reset and set Phase difference = 120° and Amplitude ratio = 1; predict resultant amplitude 1. These are two equal fixed-frequency sine waves, not radial incoming/outgoing waves or Legendre partial waves. The lab only illustrates that coherent amplitudes depend on relative phase.

SumFirst componentSecond component

Sum at x = 0: 0.866
Resultant amplitude: 1.732
Phase difference: 60°

Phase difference60°Amplitude ratio1

t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). 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 forInterference at a direction must be retained until the appropriate angular integration removes it.

Pause & reason

Check your understanding

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Read the checkpoint explanation

The scattered partial amplitude is (Sℓ − 1)/(2ik), so it vanishes when Sℓ = 1. A zero S element instead means complete loss from that elastic outgoing channel.

Your turn

Try it yourself

An s-wave hard core of radius $R=2.0\,\mathrm{nm}$ enforces $u_0(R)=0$ . At $k=0.050\,\mathrm{nm^{-1}}$ , find the continuous small-k phase shift and its s-wave cross section. Compare with the zero-energy limit.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Outside the core u0 is proportional to sin(kr + δ0). Choose the branch that approaches zero as k approaches zero.

Compare with the solution

The boundary gives $\delta_0=-kR=-0.10$ radians modulo pi. Hence $\sigma_0=(4\pi/0.050^2)\sin^2(0.10)=50.1\,\mathrm{nm^2}$ , rounded to three significant figures. As $k\to0$ , using $\sin(kR)\sim kR$ gives $\sigma_0\to4\pi R^2=16\pi\,\mathrm{nm^2}\approx50.3\,\mathrm{nm^2}$ . This is the s-wave result; at the given small kR, higher partial waves are strongly suppressed, but the exact total hard-sphere answer would include them.

Keep the idea

Take it with you

A phase shift follows from radial boundary matching and is defined modulo pi.
The elastic S phase is twice the radial phase shift.
Partial cross sections add only after angular orthogonality has been used.

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