Partial Wave Analysis

For a central potential $V(r)$ — one that depends only on the distance from the target — angular momentum is conserved. That lets us decompose the scattering problem into independent angular-momentum channels, each labeled by $\ell = 0, 1, 2, \dots$ The amplitude $f(\theta)$ becomes a sum over these partial waves, and in each channel the entire effect of the potential is compressed into a single real number, the phase shift $\delta_\ell$ .

Why central potentials simplify

When $V = V(r)$ , the Hamiltonian commutes with $\mathbf{L}^2$ and $L_z$ . The scattering is azimuthally symmetric about the beam axis, so $f$ depends only on $\theta$ , and we may expand everything in Legendre polynomials $P_\ell(\cos\theta)$ — the $m=0$ spherical harmonics. The incident plane wave itself has the exact expansion (the Rayleigh formula)

e^{ikz} = \sum_{\ell=0}^{\infty} (2\ell+1)\, i^{\ell}\, j_\ell(kr)\, P_\ell(\cos\theta),

where $j_\ell$ are spherical Bessel functions. Each term is a partial wave of definite $\ell$ .

The asymptotic anatomy of a partial wave

Far from the target, the free spherical wave $j_\ell(kr)$ behaves as a standing wave built from equal incoming and outgoing spherical pieces:

j_\ell(kr) \;\xrightarrow{\,r\to\infty\,}\; \frac{\sin\!\big(kr - \tfrac{\ell\pi}{2}\big)}{kr} = \frac{1}{2ikr}\Big( e^{i(kr-\ell\pi/2)} - e^{-i(kr-\ell\pi/2)}\Big).

A short-ranged potential cannot create or destroy probability, so it cannot change the amplitude of the outgoing piece — only its phase. The full solution therefore has the same form with the outgoing wave multiplied by a unit-modulus factor $e^{2i\delta_\ell}$ :

R_\ell(r) \;\xrightarrow{\,r\to\infty\,}\; \frac{1}{2ikr}\Big( e^{2i\delta_\ell}\,e^{i(kr-\ell\pi/2)} - e^{-i(kr-\ell\pi/2)}\Big) \;\propto\; \frac{\sin\!\big(kr - \tfrac{\ell\pi}{2} + \delta_\ell\big)}{kr}.

That single real shift $\delta_\ell$ of the asymptotic sine wave is the phase shift for channel $\ell$ . It is the output of solving the radial equation — the next lesson computes one explicitly.

The partial-wave amplitude

Subtracting the incident wave from the full solution and reading off the coefficient of $e^{ikr}/r$ gives the amplitude as a sum over channels:

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

Each channel contributes a partial-wave amplitude $f_\ell = e^{i\delta_\ell}\sin\delta_\ell / k$ .

The cross-section channel by channel

Because the Legendre polynomials are orthogonal, $\int_{-1}^{1} P_\ell P_{\ell'}\,d(\cos\theta) = \frac{2}{2\ell+1}\delta_{\ell\ell'}$ , the cross terms vanish when we integrate $|f|^2$ over angle. The total cross-section becomes a clean sum of independent contributions:

\sigma_{\text{tot}} = \frac{4\pi}{k^2}\sum_{\ell=0}^{\infty}(2\ell+1)\,\sin^2\delta_\ell .

Each channel's contribution is bounded by $\sigma_\ell \le \frac{4\pi}{k^2}(2\ell+1)$ , the unitarity limit, saturated when $\delta_\ell = \pi/2$ (a resonance).

Why this matters

Partial-wave analysis converts a partial differential equation into a tower of ordinary radial equations, one per $\ell$ , each producing a single number $\delta_\ell$ . At low energy the tower truncates after the first term. The remaining task — extracting $\delta_\ell$ from a given $V(r)$ — is exactly what the next lesson does for a concrete potential.