Partial-Wave Cross Sections

For a short-range central interaction, the total elastic cross section separates into nonnegative contributions from independent angular-momentum channels:

$$\sigma_{\mathrm{el}} = \frac{4\pi}{k^2} \sum_{\ell=0}^{\infty} (2\ell+1) \sin^2\delta_\ell.$$

This formula is more than a convenient sum. It identifies which angular momenta dominate, explains threshold suppression, exposes resonance saturation, and gives a controlled convergence test for numerical calculations.

Differential and Total Cross Sections owns the general flux-to-observable construction. Partial-Wave Expansion derives the amplitude decomposition, Phase Shifts owns the meaning and extraction of $\delta_\ell$, and Unitarity owns the allowed amplitude disk and channel bounds. This page is the canonical home for deriving, summing, and using partial-wave cross sections.

Scope and Convention

Assume distinguishable spinless particles in three dimensions, elastic relative momentum $\hbar k$, and a rotationally invariant short-range potential. The scattering amplitude is

$$f(\theta) = \frac{1}{k} \sum_{\ell=0}^{\infty} (2\ell+1) a_\ell P_\ell(\cos\theta),$$

where

$$a_\ell = \frac{S_\ell-1}{2i}.$$

The differential cross section is

$$\frac{d\sigma}{d\Omega} = \lvert f(\theta)\rvert^2.$$

For one-channel elastic scattering,

$$S_\ell=e^{2i\delta_\ell}$$

and

$$a_\ell = e^{i\delta_\ell} \sin\delta_\ell.$$

All formulas below use this normalization. Spin, identical-particle symmetrization, coupled channels, and long-range Coulomb phases require modifications that are stated where they matter. Identical-Particle Scattering owns the direct–exchange interference, event-counting factor, and even/odd partial-wave projections.

Derivation from Legendre Orthogonality

Squaring the partial-wave amplitude gives a coherent double sum:

$$\begin{aligned} \lvert f(\theta)\rvert^2 ={}& \frac{1}{k^2} \sum_{\ell,\ell'=0}^{\infty} (2\ell+1) (2\ell'+1) \\ &\times a_\ell a_{\ell'}^* P_\ell(\cos\theta) P_{\ell'}(\cos\theta). \end{aligned}$$

At a fixed angle, different partial waves interfere. The simplification occurs only after integration over the complete sphere.

Legendre orthogonality is

$$\int_{-1}^{1}dx\, P_\ell(x)P_{\ell'}(x) = \frac{2}{2\ell+1} \delta_{\ell\ell'}.$$

Because $d\Omega=d\phi\,dx$,

$$\int d\Omega\, P_\ell(\cos\theta) P_{\ell'}(\cos\theta) = \frac{4\pi}{2\ell+1} \delta_{\ell\ell'}.$$

Therefore

$$\begin{aligned} \sigma_{\mathrm{el}} &= \int d\Omega\, \lvert f(\theta)\rvert^2 \\ &= \frac{4\pi}{k^2} \sum_{\ell=0}^{\infty} (2\ell+1) \lvert a_\ell\rvert^2. \end{aligned}$$

For elastic phase shifts,

$$\lvert a_\ell\rvert^2 = \sin^2\delta_\ell,$$

which gives

$$\sigma_{\mathrm{el}} = \frac{4\pi}{k^2} \sum_{\ell=0}^{\infty} (2\ell+1) \sin^2\delta_\ell.$$

The dimensions are supplied entirely by $k^{-2}$. The phase shifts and $a_\ell$ are dimensionless.

Individual Angular-Momentum Channels

Define the elastic partial cross section

$$\sigma_\ell^{\mathrm{el}} = \frac{4\pi}{k^2} (2\ell+1) \lvert a_\ell\rvert^2.$$

For purely elastic scattering, equivalent forms are

$$\begin{aligned} \sigma_\ell^{\mathrm{el}} &= \frac{4\pi}{k^2} (2\ell+1) \sin^2\delta_\ell, \\ &= \frac{\pi}{k^2} (2\ell+1) \lvert S_\ell-1\rvert^2. \end{aligned}$$

The full elastic result is an incoherent sum of these integrated contributions:

$$\sigma_{\mathrm{el}} = \sum_{\ell=0}^{\infty} \sigma_\ell^{\mathrm{el}}.$$

The factor $2\ell+1$ reflects angular-momentum multiplicity and the Legendre addition theorem. It is already included in the plane-wave decomposition; it must not be inserted a second time.

Each $\sigma_\ell^{\mathrm{el}}$ is nonnegative. The cumulative sum

$$\Sigma_L = \sum_{\ell=0}^{L} \sigma_\ell^{\mathrm{el}}$$

therefore grows monotonically toward the total elastic cross section. This makes $\Sigma_L$ useful for numerical convergence tests.

Differential Interference and Detector Acceptance

The disappearance of cross terms is a full-solid-angle statement. A detector with an angular cut measures

$$\sigma_{\mathcal A} = \int_{\mathcal A}d\Omega\, \left| \frac{1}{k} \sum_\ell (2\ell+1) a_\ell P_\ell(\cos\theta) \right|^2,$$

and the restricted-domain Legendre integrals are not generally diagonal.

Suppose only $s$ and $p$ waves contribute:

$$f(\theta) = \frac{1}{k} \left( a_0+3a_1\cos\theta \right).$$

Then

$$\begin{aligned} \frac{d\sigma}{d\Omega} = \frac{1}{k^2} \Big[ & \lvert a_0\rvert^2 + 9\lvert a_1\rvert^2\cos^2\theta \\ &+ 6\operatorname{Re} \left( a_0a_1^* \right) \cos\theta \Big]. \end{aligned}$$

The interference term is odd in $\cos\theta$. It integrates to zero over the full sphere but generates a forward–backward asymmetry. The forward and backward hemispheres give

$$\begin{aligned} \sigma_{\mathrm F} ={}& \frac{2\pi}{k^2} \Big[ \lvert a_0\rvert^2 + 3\lvert a_1\rvert^2 \\ &+ 3\operatorname{Re} \left( a_0a_1^* \right) \Big], \\ \sigma_{\mathrm B} ={}& \frac{2\pi}{k^2} \Big[ \lvert a_0\rvert^2 + 3\lvert a_1\rvert^2 \\ &- 3\operatorname{Re} \left( a_0a_1^* \right) \Big]. \end{aligned}$$

Adding them recovers

$$\sigma_{\mathrm F} + \sigma_{\mathrm B} = \frac{4\pi}{k^2} \left( \lvert a_0\rvert^2 + 3\lvert a_1\rvert^2 \right).$$

Thus partial cross sections diagnose the total strength by channel, but they do not reconstruct acceptance-limited data unless the coherent angular amplitude is retained.

Elastic, Reaction, and Total Sums

When inelastic channels are open, the diagonal elastic element can be written

$$S_\ell = \eta_\ell e^{2i\delta_\ell}, \qquad 0\le\eta_\ell\le1.$$

The elastic cross section remains

$$\sigma_{\mathrm{el}} = \frac{\pi}{k^2} \sum_{\ell=0}^{\infty} (2\ell+1) \lvert S_\ell-1\rvert^2.$$

The reaction cross section, meaning loss from the observed elastic channel into all other open channels, is

$$\sigma_{\mathrm{reac}} = \frac{\pi}{k^2} \sum_{\ell=0}^{\infty} (2\ell+1) \left( 1-\lvert S_\ell\rvert^2 \right).$$

Equivalently,

$$\sigma_{\mathrm{reac}} = \frac{\pi}{k^2} \sum_{\ell=0}^{\infty} (2\ell+1) \left( 1-\eta_\ell^2 \right).$$

The total cross section is

$$\begin{aligned} \sigma_{\mathrm{tot}} &= \sigma_{\mathrm{el}} + \sigma_{\mathrm{reac}} \\ &= \frac{2\pi}{k^2} \sum_{\ell=0}^{\infty} (2\ell+1) \left( 1-\operatorname{Re}S_\ell \right). \end{aligned}$$

These expressions satisfy the optical theorem because

$$\operatorname{Im}a_\ell = \frac{ 1-\operatorname{Re}S_\ell }{2}.$$

The channel-by-channel probability interpretation and the bounds implied by $0\le\eta_\ell\le1$ are developed at Unitarity.

Low-Energy Dominance of the s-Wave

For a sufficiently short-range nonsingular potential away from exceptional threshold behavior, the threshold law is

$$\delta_\ell(k) = O\left( k^{2\ell+1} \right).$$

At small phase shift,

$$\sin\delta_\ell \sim \delta_\ell.$$

Hence

$$\sigma_\ell^{\mathrm{el}} = O\left( k^{4\ell} \right).$$

The first channels scale as

$$\begin{array}{c|c|c} \ell & \delta_\ell & \sigma_\ell^{\mathrm{el}} \\ \hline 0 & O(k) & O(k^0) \\ 1 & O(k^3) & O(k^4) \\ 2 & O(k^5) & O(k^8) \end{array}$$

This is why the $s$ wave normally dominates low-energy distinguishable-particle scattering.

With scattering length $a_s$,

$$\delta_0(k) \sim -ka_s,$$

so

$$\sigma_0^{\mathrm{el}} \longrightarrow 4\pi a_s^2$$

when both $kR\ll1$ and $k\lvert a_s\rvert\ll1$. Keeping the unitarizing denominator but neglecting effective-range corrections gives

$$\sigma_0^{\mathrm{el}} \approx \frac{ 4\pi a_s^2 }{ 1+k^2a_s^2 }.$$

Near a threshold pole, $\lvert a_s\rvert$ can be much larger than $R$ and the second condition is essential. Scattering Length and Effective-Range Expansion own that regime.

For identical spinless bosons, exchange symmetry permits only even $\ell$; for spin-polarized identical fermions, only odd $\ell$ survive, so the $p$ wave can become the leading channel. Identical final states also require a consistent angular integration domain or symmetry factor. The distinguishable-particle formula must not be reused unchanged; the complete derivation is in Identical-Particle Scattering.

Long-range potentials can alter threshold powers. The Coulomb interaction is not covered by the short-range law above.

Unitarity Limit

Elastic unitarity constrains

$$\lvert a_\ell\rvert\le1.$$

Therefore

$$\sigma_\ell^{\mathrm{el}} \le \frac{4\pi}{k^2} (2\ell+1).$$

For a purely elastic channel, the bound is saturated when

$$\delta_\ell = \frac{\pi}{2} \pmod{\pi}.$$

At that point,

$$S_\ell=-1, \qquad a_\ell=i.$$

The $s$-wave limit is

$$\sigma_0^{\mathrm{el}} \le \frac{4\pi}{k^2}.$$

This bound grows as $k\to0$, but it is a bound at fixed momentum, not a prediction that every low-energy cross section diverges. Saturation requires nonperturbative dynamics such as a nearby threshold pole or an elastic resonance.

Unitarity alone bounds each channel, not the infinite sum. A finite-range or analyticity argument is needed to determine how many partial waves can contribute appreciably.

Impact Parameter and the Black-Disk Sum

At large angular momentum, the semiclassical correspondence is

$$b_\ell \simeq \frac{\ell+1/2}{k}.$$

Associate partial wave $\ell$ with a transverse annulus bounded approximately by

$$\frac{\ell}{k} \le b\le \frac{\ell+1}{k}.$$

Its area is

$$\begin{aligned} \Delta A_\ell &= \pi \left[ \frac{(\ell+1)^2-\ell^2}{k^2} \right] \\ &= \frac{\pi}{k^2} (2\ell+1). \end{aligned}$$

This is exactly the maximum reaction cross section carried by one partial wave. The multiplicity factor has a geometric impact-parameter interpretation.

At large $\ell$, partial wave $\ell$ samples $b\simeq(\ell+1/2)/k$. The annulus between $\ell/k$ and $(\ell+1)/k$ has area $\Delta A_\ell=\pi(2\ell+1)/k^2$, matching the reaction-channel unitarity bound.

In the idealized black-disk model,

$$S_\ell = \begin{cases} 0, & 0\le\ell\le L,\\ 1, & \ell\gt L. \end{cases}$$

Every intercepted partial wave is completely absorbed. Since

$$\sum_{\ell=0}^{L} (2\ell+1) = (L+1)^2,$$

the reaction and elastic-diffraction cross sections are

$$\sigma_{\mathrm{reac}} = \sigma_{\mathrm{el}} = \frac{\pi}{k^2} (L+1)^2.$$

The total is

$$\sigma_{\mathrm{tot}} = \frac{2\pi}{k^2} (L+1)^2.$$

With $L+1\simeq kR$,

$$\begin{aligned} \sigma_{\mathrm{reac}} &\simeq \pi R^2, \\ \sigma_{\mathrm{el}} &\simeq \pi R^2, \\ \sigma_{\mathrm{tot}} &\simeq 2\pi R^2. \end{aligned}$$

The reaction piece is the geometric shadow. The equal elastic piece is diffraction from that shadow. This factor-of-two extinction result is wave physics, not an extra absorptive area.

Examples

Low-energy hard sphere

For an impenetrable sphere of radius $R$, the boundary condition $u_\ell(R)=0$ gives

$$\tan\delta_\ell = \frac{ j_\ell(kR) }{ n_\ell(kR) }.$$

Thus

$$\sin^2\delta_\ell = \frac{ j_\ell^2(kR) }{ j_\ell^2(kR)+n_\ell^2(kR) }.$$

The exact partial-wave sum is

$$\sigma_{\mathrm{el}} = \frac{4\pi}{k^2} \sum_{\ell=0}^{\infty} (2\ell+1) \frac{ j_\ell^2(kR) }{ j_\ell^2(kR)+n_\ell^2(kR) }.$$

For $kR\ll1$, the $s$ wave dominates and

$$\delta_0=-kR \pmod{\pi}.$$

Therefore

$$\sigma_{\mathrm{el}} \simeq \frac{4\pi}{k^2} \sin^2(kR) \longrightarrow 4\pi R^2.$$

The low-energy quantum cross section is four times the geometric area $\pi R^2$. It should not be confused with the high-energy classical shadow. Hard-Sphere Scattering follows the exact sum from this $4\pi R^2$ threshold limit to the $2\pi R^2$ high-energy diffraction limit and derives the classical result separately.

Isolated elastic resonance

If a negligible background and an isolated resonance produce

$$\sin^2\delta_\ell(E) \approx \frac{ \Gamma^2/4 }{ (E-E_R)^2+\Gamma^2/4 },$$

then

$$\sigma_\ell^{\mathrm{el}}(E) \approx \frac{4\pi}{k^2} (2\ell+1) \frac{ \Gamma^2/4 }{ (E-E_R)^2+\Gamma^2/4 }.$$

At $E=E_R$, the channel reaches its elastic unitarity limit if $k$ varies negligibly across the width. Breit–Wigner Form owns the pole-factor derivation and explains how background phases, inelasticity, thresholds, and overlapping poles deform this simple profile.

Reading a numerical channel budget

Suppose a phase-shift solver returns, at one energy,

$$\delta_0=0.35, \qquad \delta_1=0.08, \qquad \delta_2=0.01.$$

In units of $4\pi/k^2$, the channel weights are approximately

$$\begin{aligned} w_0 &= \sin^2(0.35) \approx0.118, \\ w_1 &= 3\sin^2(0.08) \approx0.0192, \\ w_2 &= 5\sin^2(0.01) \approx0.0005. \end{aligned}$$

The total weight is about $0.137$, so the $s$ wave supplies roughly $86\%$, the $p$ wave $14\%$, and the $d$ wave less than $1\%$. These percentages describe the angle-integrated cross section. Even the small $p$ wave can have a visible effect on angular asymmetry through interference with the $s$ wave.

Numerical Workflow

Given computed phase shifts or channel $S$-matrix elements:

1. Match conventions. Confirm the definitions of $S_\ell$, $a_\ell$, $k$, and the angular measure.
2. Compute channel contributions. Evaluate $\sigma_\ell^{\mathrm{el}}$ and, if needed, $\sigma_\ell^{\mathrm{reac}}$.
3. Inspect the channel budget. Plot or tabulate contributions rather than reporting only their sum.
4. Increase the cutoff. Require stability of the cumulative total as $L_{\max}$ grows.
5. Converge the amplitude separately. Differential observables can remain sensitive to a high-$\ell$ tail even when the integrated total looks stable.
6. Check unitarity. Verify $\lvert S_\ell\rvert=1$ for elastic channels or match the deficit to explicit inelastic channels.
7. Vary numerical controls. Test radial box size, matching radius, grid spacing, and potential-tail truncation.

A useful total-cross-section tail diagnostic is

$$\epsilon_L = \frac{ \sum_{\ell=L+1}^{L+\Delta L} \sigma_\ell^{\mathrm{el}} }{ \sum_{\ell=0}^{L+\Delta L} \sigma_\ell^{\mathrm{el}} }.$$

This is an indicator, not a rigorous bound on all omitted $\ell$. A long-range tail can make convergence slow and invalidate the estimate based on a short block of channels.

Common Mistakes

• Dropping interference terms in a differential or acceptance-limited cross section.
• Multiplying by $2\ell+1$ twice.
• Using the elastic phase-shift formula when $\eta_\ell\lt1$.
• Calling $\sigma_{\mathrm{el}}$ the total cross section when reaction channels are open.
• Applying the short-range threshold law to an unscreened Coulomb potential.
• Assuming $s$-wave dominance for spin-polarized identical fermions.
• Summing a fixed number of partial waves at all energies rather than increasing $L_{\max}$ with $kR$.
• Treating the unitarity limit as a prediction instead of an upper bound.
• Comparing a low-energy hard-sphere result $4\pi R^2$ with the geometric area without recognizing diffraction.
• Declaring the differential amplitude converged because the integrated total has stabilized.

Exercises

Derive the elastic sum

Starting from

$$f(\theta) = \frac{1}{k} \sum_{\ell=0}^{\infty} (2\ell+1) a_\ell P_\ell(\cos\theta),$$

derive the total elastic cross section.

Solution

Use

$$\int d\Omega\, P_\ell(\cos\theta) P_{\ell'}(\cos\theta) = \frac{4\pi}{2\ell+1} \delta_{\ell\ell'}.$$

Then

$$\begin{aligned} \sigma_{\mathrm{el}} &= \int d\Omega\,\lvert f\rvert^2 \\ &= \frac{1}{k^2} \sum_{\ell,\ell'} (2\ell+1) (2\ell'+1) a_\ell a_{\ell'}^* \\ &\quad\times \frac{4\pi}{2\ell+1} \delta_{\ell\ell'} \\ &= \frac{4\pi}{k^2} \sum_\ell (2\ell+1) \lvert a_\ell\rvert^2. \end{aligned}$$

For elastic scattering, $\lvert a_\ell\rvert^2=\sin^2\delta_\ell$.

Recover the s–p interference asymmetry

For

$$f(\theta) = \frac{1}{k} \left( a_0+3a_1\cos\theta \right),$$

show that $\sigma_{\mathrm F}-\sigma_{\mathrm B}$ is proportional to $\operatorname{Re}(a_0a_1^*)$.

Solution

With $x=\cos\theta$,

$$\begin{aligned} \lvert f\rvert^2 ={}& \frac{1}{k^2} \Big[ \lvert a_0\rvert^2 + 9\lvert a_1\rvert^2x^2 \\ &\qquad+ 6\operatorname{Re} \left( a_0a_1^* \right)x \Big]. \end{aligned}$$

Integrating over $0\le x\le1$ and $-1\le x\le0$ gives

$$\begin{aligned} \sigma_{\mathrm F} - \sigma_{\mathrm B} &= \frac{12\pi}{k^2} \operatorname{Re} \left( a_0a_1^* \right). \end{aligned}$$

The interference cancels in the full sum $\sigma_{\mathrm F}+\sigma_{\mathrm B}$ but controls the asymmetry.

Obtain the threshold powers

Assume $\delta_\ell=O(k^{2\ell+1})$. Determine the low-$k$ scaling of $\sigma_\ell^{\mathrm{el}}$ for $\ell=0,1,2$.

Solution

For small phase shift,

$$\sin^2\delta_\ell = O\left( k^{4\ell+2} \right).$$

The prefactor $k^{-2}$ gives

$$\sigma_\ell^{\mathrm{el}} = O\left( k^{4\ell} \right).$$

Hence

$$\begin{gathered} \sigma_0=O(k^0), \qquad \sigma_1=O(k^4), \\ \sigma_2=O(k^8). \end{gathered}$$

Sum the black disk

Take $S_\ell=0$ for $0\le\ell\le L$ and $S_\ell=1$ above $L$. Find the elastic, reaction, and total cross sections.

Solution

For an absorbed channel,

$$\lvert S_\ell-1\rvert^2=1, \qquad 1-\lvert S_\ell\rvert^2=1.$$

Using

$$\sum_{\ell=0}^{L} (2\ell+1) = (L+1)^2,$$

gives

$$\sigma_{\mathrm{el}} = \sigma_{\mathrm{reac}} = \frac{\pi}{k^2} (L+1)^2$$

and

$$\sigma_{\mathrm{tot}} = \frac{2\pi}{k^2} (L+1)^2.$$

The elastic piece is diffraction from the absorbed disk.

Check the low-energy hard sphere

For $\ell=0$, use

$$j_0(x)=\frac{\sin x}{x}, \qquad n_0(x)=-\frac{\cos x}{x}$$

to find the low-energy hard-sphere cross section.

Solution

The boundary condition gives

$$\tan\delta_0 = \frac{j_0(kR)}{n_0(kR)} = -\tan(kR).$$

Choose the branch continuous from zero:

$$\delta_0=-kR.$$

Then

$$\sigma_0 = \frac{4\pi}{k^2} \sin^2(kR) \longrightarrow 4\pi R^2$$

as $kR\to0$.

Locate resonant saturation

For a purely elastic partial wave with

$$\sin^2\delta_\ell(E) = \frac{ \Gamma^2/4 }{ (E-E_R)^2+\Gamma^2/4 },$$

find the peak cross section.

Solution

At $E=E_R$,

$$\sin^2\delta_\ell=1.$$

Therefore

$$\sigma_\ell^{\mathrm{peak}} = \frac{4\pi}{k_R^2} (2\ell+1),$$

where $k_R$ is the relative wave number at the resonance. This is the elastic partial-wave unitarity limit.

References

• B. Zwiebach, “Chapter 7: Scattering,” MIT OpenCourseWare 8.06 Quantum Physics III (2018), PDF.
• J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover, 2006, Chapters 3 and 11.
• R. G. Newton, Scattering Theory of Waves and Particles, 2nd ed., Springer, 1982, Chapters 10 and 11.
• C. J. Joachain, Quantum Collision Theory, 3rd ed., North-Holland, 1983, Chapters 3 and 7.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977, Sections 132–134.
• Particle Data Group, “Kinematics,” Review of Particle Physics (2012), PDF.