advanced35 min studyLesson 400 in Physics

Identical-particle scattering symmetrization

Add direct and exchange amplitudes before squaring and count indistinguishable final pairs once.

Before you begin

On this page

The question for today

What you’ll understand

Construct symmetric and antisymmetric spatial scattering amplitudes.
Apply spin symmetry and even/odd angular-momentum restrictions.
Integrate over a nonduplicated final-state domain with correct statistical weights.

Two identical particles can reach the same detected final pair through alternatives that differ only by exchanging their labels. If those labels are not physically available, the alternatives are coherent and their amplitudes must be added with the symmetry required by the complete state. This produces exchange interference in the angular distribution. It also changes state counting: a full-sphere integral of pair directions counts each indistinguishable final pair twice unless a compensating factor is included.

Identify indistinguishable alternatives

Work in the center-of-mass frame for a spin-independent central interaction and equal-mass identical particles. A final momentum direction for one nominal label fixes the opposite direction for the other. Exchanging the labels replaces angle theta by pi minus theta relative to the incoming relative momentum. Let the distinguishable-particle relative-motion amplitude be f. The symmetric spatial amplitude is its direct-plus-exchange sum, and the antisymmetric amplitude is the corresponding difference.

The sign follows the spatial symmetry sector, which need not equal the sign of the total particle statistics when spin is present. Spin-zero bosons have symmetric spatial motion. Identical spin-half fermions in a spin singlet have antisymmetric spin and therefore symmetric spatial motion; a spin triplet instead requires antisymmetric spatial motion. Distinguishable internal states can supply which-particle information, in which case the coherent exchange prescription and spin analysis must be matched to what is prepared and detected rather than assumed automatically.

F_\pm(\theta)=f(\theta)\pm f(\pi-\theta),\qquad \frac{d\sigma_\pm}{d\Omega}=|F_\pm(\theta)|^2

f and F have length units. The ± sign denotes symmetric or antisymmetric spatial motion, respectively. This differential convention resolves an unordered outgoing pair through one chosen hemisphere, or through a detector direction with explicit counting convention.↔ Scroll wide equations horizontally.

Project onto allowed angular momenta

Legendre polynomials change by a parity sign when their argument is reversed. The direct-plus-exchange amplitude therefore removes all odd angular momenta, while the difference removes all even angular momenta. This is a symmetry selection rule, not an approximation about the potential strength. For a sufficiently short-range low-energy interaction, s waves usually dominate when allowed. Spin-polarized identical fermions cannot use that symmetric spatial channel, so their leading ordinary short-range scattering is suppressed by a higher-angular-momentum threshold law.

At a right angle the direct and exchange angles coincide. A symmetric spatial amplitude is doubled there, giving four times the distinguishable differential intensity for the same unsymmetrized amplitude. An antisymmetric one vanishes exactly in this ideal spin-independent elastic setting. This cancellation does not imply the interaction disappeared: other angles or allowed odd partial waves can still scatter. External fields, channel-changing interactions or measurements that distinguish internal labels require a more detailed treatment of the appropriate final states.

P_\ell(-x)=(-1)^\ell P_\ell(x),\qquad F_+=2\sum_{\ell\ {\rm even}}(2\ell+1)f_\ell P_\ell,\quad F_-=2\sum_{\ell\ {\rm odd}}(2\ell+1)f_\ell P_\ell

The angular argument in the sums is cos θ and fℓ is the same partial amplitude used earlier. Even/odd projection assumes a central spin-independent relative-motion problem and a fixed spatial exchange sector.↔ Scroll wide equations horizontally.

Count pairs and average incoherent spin sectors

Integrating a pair cross section over one hemisphere counts each opposite-momentum final pair once. Equivalently integrate over the full sphere and divide by two. For isotropic allowed s-wave scattering, the symmetrized amplitude is twice the distinguishable amplitude; its square is four times as large, while the nonduplication factor halves the full integral. The resulting total for identical spin-zero bosons is twice the distinguishable s-wave total under otherwise matched conditions.

An unpolarized pair of spin-half fermions has singlet weight one quarter and triplet weight three quarters when its incoming spins are independent and equally populated. Those orthogonal spin sectors are averaged as probabilities, not coherently summed. At low energy with only an s wave, only the singlet contributes. The wave lab shows the algebra of adding two equal-frequency amplitudes with relative sign, but it cannot impose particle statistics, couple spin to angular momentum or perform the final-state pair counting.

\sigma_\pm=\frac12\int_{4\pi}|F_\pm|^2\,d\Omega,\qquad \frac{d\sigma_{\rm unpol}}{d\Omega}=\frac14|F_+|^2+\frac34|F_-|^2

The 1/2 factor belongs to total unordered-pair counting in the full-sphere convention. The unpolarized weights assume two independent equally mixed spin-half particles, spin-independent dynamics and no spin analysis of the final pair.↔ Scroll wide equations horizontally.

Work it through

Identical bosons at an s-wave unitarity point

For a short-range interaction at $k=0.50\,\mathrm{nm^{-1}}$ , let the distinguishable s-wave amplitude be $f=2i\,\mathrm{nm}$ . Find the identical spin-zero-boson differential cross section and the total cross section counting each final pair once. Compare with distinguishable particles.

The amplitude is isotropic, so direct and exchange terms are equal. Bosonic spatial symmetry gives $F_+=2f=4i\,\mathrm{nm}$ and $|F_+|^2=16\,\mathrm{nm^2/sr}$ .
Integrate this constant over one hemisphere of solid angle $2\pi$ , or multiply its full-sphere integral by one half. The total unordered-pair cross section is $32\pi\,\mathrm{nm^2}$ .
The distinguishable total is $4\pi|f|^2=16\pi\,\mathrm{nm^2}$ . Thus the bosonic differential intensity is four times the unsymmetrized value at one direction, while its correctly counted total is only twice the distinguishable total.

The result

The boson differential area is 16 nm²/sr and its pair-counted total is 32π nm², twice the distinguishable total.

The living notebook

Explore the model

Interactive lab

Interactive lab · waves

Direct-plus-exchange algebra only

Reset and set BOTH Phase difference = 0° and Amplitude ratio = 1; predict a doubled amplitude. Reset and set Phase difference = 180° and Amplitude ratio = 1; predict cancellation. These equal fixed-frequency classical waves illustrate the plus and minus algebra only. They are not quantum particle paths and do not calculate exchange symmetry, spin sectors, partial waves or identical-pair cross sections.

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 forIndistinguishable alternatives require coherent amplitude addition before final-state counting.

Pause & reason

Check your understanding

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Read the checkpoint explanation

The spin singlet is antisymmetric. The complete fermionic state must be antisymmetric, so the spatial factor is symmetric; one of the four equally weighted two-spin states is a singlet.

Your turn

Try it yourself

At sufficiently low energy let the unsymmetrized s-wave amplitude have squared magnitude $|f|^2=9\,\mathrm{nm^2}$ . For independent unpolarized identical spin-half fermions and spin-independent dynamics, find the differential and pair-counted total cross sections when higher waves are negligible.

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

Give me a hint

The singlet has F+ = 2f and weight 1/4; the triplet has no s wave. Integrate the resulting differential intensity over a single hemisphere.

Compare with the solution

The singlet differential cross section is $4|f|^2=36\,\mathrm{nm^2/sr}$ , and the triplet s-wave contribution is zero. Spin averaging gives $(1/4)(36)=9\,\mathrm{nm^2/sr}$ . The pair-counted total is $2\pi(9)=18\pi\,\mathrm{nm^2}$ . Integrating this differential value over all directions without the factor one half would double-count the same outgoing pairs. The weights would change for a polarized or correlated incoming spin state, so the unpolarized assumption is part of the numerical answer.

Keep the idea

Take it with you

Exchange acts on amplitudes and depends on the complete spin-spatial symmetry.
Symmetric and antisymmetric spatial sectors select even and odd partial waves.
Total identical-pair cross sections require a nonduplicated angular domain or a factor one half.

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