---
title: "How a scattering amplitude becomes a detector count"
description: "Derive differential cross sections from probability current and use interference to see why amplitudes are added before probabilities."
canonical_url: "https://sandboxphysics.com/stories/scattering-amplitudes/flux-and-amplitude/"
language: "en"
page_type: "story"
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  zh-CN: "https://sandboxphysics.com/zh/stories/scattering-amplitudes/flux-and-amplitude/"
last_modified: "2026-10-27"
---

[Scattering amplitudes](https://sandboxphysics.com/stories/scattering-amplitudes/)

L1 · Chapter 1

# How a scattering amplitude becomes a detector count

Sandbox Physics · 2026-10-27

A detector counts particles. A calculation first produces a complex arrow with a length and an angle. Add the relevant arrows and square the resulting length to predict relative counts. A cross section expresses that probability as an area relative to the incident flux.

## Before you begin

High-school algebra, trigonometry and the area of a sphere. Picture a complex number as a planar arrow. Probability current, solid angle and differential cross section are introduced here; derivatives appear only in the current formula.

- Derive a differential cross section from incident and outgoing current.
- Explain interference with complex amplitudes and distinguish an amplitude from a rate.

## Count particles through a small area

Send a beam toward a small target. A distant detector usually records different rates when it points in different directions. This angular pattern is the phenomenon we want to explain. To compare experiments, first divide out beam intensity, the number of target particles and the angular size of the detector.

Our starting model is nonrelativistic scattering of a particle of mass $M$ by a short-range central potential. The incident wave travels along positive $z$ with wave number $k$. Far from the target the wave is an incident plane wave plus an outgoing spherical wave. Its amplitude $f(\vartheta)$ has units of length; $\vartheta$ is the angle measured from the incident beam.

$$
\psi(\mathbf r)\simeq e^{ikz}+f(\vartheta)\frac{e^{ikr}}{r}
$$

The outgoing amplitude falls inversely with distance, so its probability density falls inversely with distance squared.

## Obtain an area from probability current

Probability current measures how much probability crosses a unit area per unit time. For a plane wave it is speed times probability density. For the distant outgoing wave, retain the leading radial contribution. The formulas below show both currents with incident amplitude set to one.

An area $dA$ at radius $r$ subtends solid angle $d\Omega=dA/r^2$. Divide the outgoing current through that area by the incident current. The distance cancels. The remaining effective area defines the differential cross section. This derivation assumes a fixed target, elastic scattering and equal incoming and outgoing speeds.

$$
\mathbf j=\frac{\hbar}{M}\operatorname{Im}(\psi^*\nabla\psi),\qquad j_{\rm in}=\frac{\hbar k}{M},\qquad j_{\rm out}=\frac{\hbar k}{M}\frac{|f(\vartheta)|^2}{r^2}
$$

Only the far-field radial current of the spherical wave is written in the final expression.

$$
\frac{d\sigma}{d\Omega}=\frac{j_{\rm out}r^2}{j_{\rm in}}=|f(\vartheta)|^2
$$

This effective area measures scattering into a direction and depends on the wave dynamics and target interaction.

Source: David Tong

## Two paths can darken one direction

Suppose two indistinguishable paths reach the same detector. Either path alone produces counts. Opening both paths can reduce the rate because their complex arrows can point against each other.

Let the path amplitudes be $f_1=a$ and $f_2=ae^{i\phi}$, with positive length $a$ and relative phase $\phi$. Multiplying their sum by its complex conjugate produces an interference term. If the environment distinguishes the paths, its states must enter the calculation; this coherent sum is then insufficient.

Worked example: choose $a=1\,\mathrm{nm}$. Equal phases give an amplitude of twice that length and a differential cross section of $4\,\mathrm{nm}^2$. At relative phase $2\pi/3$ the resultant arrow has length $1\,\mathrm{nm}$, giving $1\,\mathrm{nm}^2$. Opposite arrows cancel completely.

$$
|f_1+f_2|^2=2a^2(1+\cos\phi)
$$

Adding the two path probabilities separately discards the phase-dependent cross term.

## Move to a relativistic normalization

Relativistic calculations usually use an invariant amplitude $\mathcal M$, with a different normalization from $f$. In natural units $\hbar=c=1$, the expression below applies to a distinguishable two-body final state in the center-of-mass frame, with conventional relativistic state normalization. Spin averages and sums must be specified separately.

The gluon chapters study fixed helicities and a fixed cyclic color order. Recombining colors, treating spins and including final-state phase space are further steps toward a collision cross section. A compact partial amplitude alone does not specify an event rate.

$$
\frac{d\sigma}{d\Omega}=\frac{1}{64\pi^2s}\frac{|\mathbf p_{\rm out}|}{|\mathbf p_{\rm in}|}\,|\mathcal M|^2
$$

An integrated identical-particle final state requires the appropriate symmetry factor. The squared total energy is a Mandelstam invariant.

Source: Lance J. Dixon

## Try the reasoning in an experiment

### [Predict a dark direction before changing the potential](https://sandboxphysics.com/labs/partial-wave-scattering/)

1. Keep wave number and potential fixed. Record the strongest and weakest directions in the default angular distribution.
2. Save the default angular scan, change only the potential strength and scan again. Compare the angular distribution and partial-wave points, and explain why adding separate intensities would give a different result.
3. This experiment models nonrelativistic central-potential scattering. Explain its pattern using flux, then identify the additional steps needed for gluon collisions.

## Exercises and explanations

1. Move a detector twice as far away with its area fixed. What happens to the count? What if its area also becomes four times larger?

   Read the explanation

   The fixed area subtends one quarter of the solid angle, so far-field counts fall by four. Increasing the area fourfold restores the solid angle and count. The cross section itself is independent of detector distance.

2. Two equal paths differ in phase by $\pi/2$. What is the differential cross section, and why does it equal the sum of the separate probabilities here?

   Read the explanation

   The cosine vanishes, leaving $2a^2$. The interference term happens to vanish at this phase. Changing the phase generally restores it.

[Next: How a circle limits the strength of scattering](https://sandboxphysics.com/stories/scattering-amplitudes/partial-wave-unitarity/)

## Sources and reproduced scope

Worked examples and schematic figures are authored for this course. Numerical values are calculated model references. The published comparison recalculates a specified analytic CDD curve; it does not copy experimental data, source artwork or the paper’s numerical optimization points. Source locations and reproduced scope are listed below.

1. [David Tong, Topics in Quantum Mechanics, Chapter 10, Scattering Theory](https://www.damtp.cam.ac.uk/user/tong/aqm/aqmten.pdf)

   Sections 10.2.1–10.2.3: three-dimensional asymptotic waves, cross sections and partial waves. We derive the current-to-area relation and unitarity-circle examples for the course; the linked square-well experiment retains its own model and units.

2. [Lance J. Dixon, A brief introduction to modern amplitude methods, arXiv:1310.5353v1 (2013)](https://arxiv.org/html/1310.5353v1)

   Section 2, equation (5), color decomposition; section 3.1, equations (14), (19), (21)–(27), spinors and all-outgoing conventions; section 3.3, equations (41)–(46), polarizations; section 4.1, equations (55)–(57), soft factors; section 4.3, equations (68)–(70), MHV and collinear factors; sections 4.5 and 5.1, equations (74)–(84), complex three-point seeds and recursion. Our conserving scan and four-point residue example are derived extensions.