Differential and Total Cross Sections

A cross section translates scattered flux into an effective area. It is the central observable of scattering theory because detectors count outgoing particles in directions, energies, and channels, not abstract wavefunction coefficients.

For single-channel elastic scattering with asymptotic form

$$\psi_{\mathbf k}^{(+)}(\mathbf r) \sim e^{i\mathbf k\cdot\mathbf r} + f(\theta,\phi)\frac{e^{ikr}}{r},$$

the differential cross section is

$$\frac{d\sigma}{d\Omega} = |f(\theta,\phi)|^2.$$

The total elastic cross section is

$$\sigma_{\mathrm{tot}} = \int d\Omega\, |f(\theta,\phi)|^2$$

when elastic scattering is the only open channel.

Differential Cross Section

The differential cross section is defined by

$$d\sigma = \frac{ \text{particles scattered per unit time into }d\Omega }{ \text{incident flux} }.$$

For the standard elastic convention, the scattered radial current at large $r$ is (with $m$ the particle or reduced mass)

$$j_r^{\mathrm{sc}} \approx \frac{\hbar k}{m} \frac{|f(\theta,\phi)|^2}{r^2}.$$

The scattered rate through the surface element $r^2d\Omega$ is

$$j_r^{\mathrm{sc}}r^2d\Omega \approx \frac{\hbar k}{m} |f(\theta,\phi)|^2d\Omega.$$

Dividing by the incident flux $\hbar k/m$ gives

$$d\sigma = |f(\theta,\phi)|^2d\Omega.$$

Total Cross Section

Integrating over all outgoing directions gives

$$\sigma = \int d\Omega\, \frac{d\sigma}{d\Omega}.$$

For distinguishable particles with azimuthally symmetric scattering,

$$d\Omega = 2\pi\sin\theta\,d\theta,$$

so

$$\sigma = 2\pi \int_0^\pi \sin\theta\, |f(\theta)|^2 \,d\theta.$$

This integral may diverge for long-range interactions such as unscreened Coulomb scattering. Short-range scattering is technically simpler because the asymptotic form separates cleanly into an incident plane wave and outgoing spherical wave.

For a short-range central interaction, Partial-Wave Cross Sections derives the elastic, reaction, and total cross sections as sums over angular-momentum channels.

Example: Integrating an Angular Pattern

If $f(\theta)=a+b\cos\theta$, then

$$\sigma=2\pi\int_{-1}^{1}|a+bx|^2\,dx =4\pi\left(|a|^2+\frac{|b|^2}{3}\right).$$

The interference term integrates to zero because it is odd in $x=\cos\theta$. It still affects individual angular bins. A total cross section can therefore hide differences that an angular distribution reveals.

Units and Scale

Since $f$ has dimensions of length,

$$\left[ \frac{d\sigma}{d\Omega} \right] = \text{area}, \qquad [\sigma]=\text{area}.$$

The unit of solid angle is dimensionless, so both differential and total cross sections carry area units. A large cross section means the target is effective at redirecting incident flux; it need not equal a literal geometric area.

Elastic and Inelastic Channels

If the outgoing channel differs from the incoming channel, one must include the ratio of outgoing to incoming speeds. A common nonrelativistic convention gives

$$\frac{d\sigma_{\beta\leftarrow\alpha}}{d\Omega} = \frac{v_\beta}{v_\alpha} |f_{\beta\alpha}(\theta,\phi)|^2.$$

For elastic scattering of a single particle from a fixed potential,

$$v_\beta=v_\alpha,$$

and the velocity factor is one.

When several final channels are open, the total cross section is a sum over channels and an integral over the corresponding final variables.

Inelastic Scattering Preview develops this formula for internal-state changes, including threshold Q-values, degeneracy sums, golden-rule normalization, and inclusive observables.

Identical-Particle Caveat

For identical particles, direct and exchange assignments can produce the same unordered final event. Their amplitudes must then be added before squaring, and the final pair must be counted only once. The spatial sign depends on the symmetry of the complete spin–space state.

Identical-Particle Scattering derives the direct–exchange amplitudes, the hemisphere versus $1/2!$ counting convention, the even/odd partial-wave rules, and the singlet/triplet spin averages.

Relation to One-Dimensional Scattering

In one-dimensional barrier scattering, one usually quotes reflection and transmission probabilities as current ratios:

$$R = \frac{|j_{\mathrm{ref}}|}{j_{\mathrm{inc}}}, \qquad T = \frac{j_{\mathrm{trans}}}{j_{\mathrm{inc}}}.$$

For incidence from the left, $j_{\mathrm{ref}}<0$ because the reflected current points toward decreasing $x$; its magnitude is needed for the nonnegative probability $R$. Those are not three-dimensional cross sections. They are the one-dimensional analogs of flux ratios. The shared principle is the same: compare outgoing flux to incident flux.

One-Dimensional Scattering Revisited organizes these dimensionless probabilities as a two-channel $S$-matrix and connects them to transfer matrices, poles, resonant tunneling, and WKB.

Convention Differences

Nonrelativistic quantum mechanics often defines cross sections through wavefunction asymptotics. Quantum field theory usually defines them through invariant amplitudes, phase space, and flux factors. The physical structure is continuous across the two subjects, but the symbols and normalizations are not identical.

Whenever comparing formulas, check:

• state normalization,
• incoming flux convention,
• final-state density of states,
• whether spin averages or sums are included,
• whether identical-particle factors are present.

Common Mistakes

• Calling $f$ itself a cross section.
• Forgetting to integrate over solid angle to get the total cross section.
• Omitting velocity factors in inelastic scattering.
• Adding probabilities instead of amplitudes for indistinguishable identical-particle alternatives.
• Applying a three-dimensional cross-section formula to a one-dimensional barrier coefficient.

Exercises

1. If $f(\theta,\phi)=a$ is a constant, compute the total cross section.

Solution

The differential cross section is

$$\frac{d\sigma}{d\Omega}=|a|^2.$$

Integrating over the full solid angle,

$$\sigma = \int d\Omega\,|a|^2 = 4\pi |a|^2.$$

2. A detector accepts all azimuths and $x=\cos\theta$ in the interval $x_0-w\le x\le x_0+w$, with $w>0$ and the interval contained in $[-1,1]$. For the angular pattern $f=a+bx$, calculate the cross section averaged over this bin and the count rate at luminosity $\mathcal L$ and uniform efficiency $\epsilon$. Neglect angular smearing. How much does evaluating the cross section at the bin center underestimate the rate when $b=3a\ne0$, $x_0=0$, and $w=1/3$?

Solution

After integrating over azimuth, $d\Omega=2\pi\,dx$ with increasing $x$ as the integration variable, so the bin has solid angle $\Delta\Omega=4\pi w$. Set $y=x-x_0$. The term linear in $y$ integrates to zero, while $\langle y^2\rangle=w^2/3$. Therefore

$$\begin{aligned} \left\langle\frac{d\sigma}{d\Omega}\right\rangle_{\mathrm{bin}} &=\frac{1}{2w}\int_{x_0-w}^{x_0+w}|a+bx|^2\,dx\\ &=|a+bx_0|^2+\frac{|b|^2w^2}{3},\\ R_{\mathrm{bin}} &=\mathcal L\epsilon\Delta\Omega \left(|a+bx_0|^2+\frac{|b|^2w^2}{3}\right). \end{aligned}$$

For the stated values, the average is $4|a|^2/3$, whereas the central value is $|a|^2$. The central-value estimate gives $3/4$ of the actual count rate: it underestimates the rate by $25\%$. Even a bin centered on an amplitude zero can contain events, since the $|b|^2w^2/3$ term remains positive. A finite-bin measurement gives an average; interpreting it as a point value requires a model for the angular variation.

For varying efficiency or finite angular resolution, use the response integral in What Is a Scattering Experiment?.

3. Why does an inelastic channel generally require a speed ratio in the cross section?

Solution

Cross sections are flux ratios. For a wave with the same amplitude convention, the current is proportional to the particle speed. If the outgoing channel has a different kinetic energy or reduced mass, its outgoing current per unit amplitude differs from the incoming current. The factor $v_\beta/v_\alpha$ converts amplitude squared into the correct flux ratio.

Photoelectron Spectroscopy shows how total, differential, and acceptance-integrated cross sections enter a concrete bound-to-continuum measurement.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• MIT OpenCourseWare, Quantum Physics III (8.06), Solution Set 9, 29 April 2005, Problem 2. Incident and scattered currents, differential cross sections, and forward interference.
• R. G. Newton, Scattering Theory of Waves and Particles, 2nd ed., Springer, 1982.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover, 2006.