advanced35 min studyLesson 398 in Physics

Scattering length and effective range

Extract low-energy s-wave parameters with explicit sign conventions and a controlled threshold expansion.

Before you begin

On this page

The question for today

What you’ll understand

Define scattering length from the zero-energy radial wave.
Use the effective-range expansion while keeping elastic unitarity exact.
Identify near-threshold and long-range limits of low-energy universality.

When a de Broglie wavelength is much longer than a short-range interaction region, a small number of parameters can summarize scattering. The leading s-wave parameter is the scattering length; the next correction is the effective range. Neither is simply a geometric radius, and their signs can carry information about threshold states. Writing the expansion for the inverse amplitude keeps the outgoing-flux term explicit and avoids predicting an unphysical unlimited cross section at fixed energy.

Read a length from the zero-energy exterior

For an s wave at zero energy outside a finite-range potential, the reduced radial equation has a linear solution. Write it as proportional to radius minus a constant. The extrapolated zero of that exterior line defines the scattering length. That zero can lie outside the interaction region or at a negative radius; it is an extrapolation parameter, not a literal hard boundary. For a hard sphere it happens to equal the positive sphere radius.

With the radial sine convention already fixed, the phase approaches minus wavenumber times scattering length as energy tends to zero, provided the product remains small. A negative scattering length can describe an attractive potential too weak to bind, while a positive value can occur for a repulsive core or near a shallow bound state. Therefore the sign alone does not uniquely identify attraction or repulsion. Near a threshold pole, the scattering length can diverge while the finite-energy amplitude remains bounded by unitarity.

u_0(r)\propto r-a\quad(E=0,\ r>R),\qquad a=-\lim_{k\to0}\frac{\tan\delta_0(k)}k,\qquad f_0=\frac1{k\cot\delta_0-ik}

a and interaction range R have length units; k has inverse length. The minus sign fixes the scattering-length convention. The exact elastic inverse amplitude includes −ik for an outgoing wave.↔ Scroll wide equations horizontally.

Expand the real inverse amplitude

For sufficiently short-range interactions with an appropriate analytic threshold expansion, the real quantity wavenumber times cotangent of the phase admits an even-power series. Its constant is minus inverse scattering length, and the quadratic coefficient defines effective range. Expanding the inverse amplitude rather than the amplitude is advantageous: retain the exact imaginary outgoing term while approximating the real part. The resulting elastic amplitude automatically has the correct optical-theorem relation at the retained level.

The truncation is not universal. Higher shape parameters matter outside the low-energy interval, and narrow resonances can introduce a large effective-range scale that shrinks that interval. Unscreened Coulomb interactions need a Coulomb-modified expansion; other slowly decaying tails can produce nonanalytic threshold terms. Even for atomic van der Waals interactions, details of the remainder require their own analysis. Restrict the displayed power-series remainder here to finite-range or sufficiently rapidly decaying interactions for which that form exists.

k\cot\delta_0=-\frac1a+\frac12r_e k^2+O(k^4),\qquad f_0\simeq\frac1{-a^{-1}+r_e k^2/2-ik}

The effective range re has length units. The displayed O(k⁴) form assumes an ordinary short-range analytic effective-range expansion and no open inelastic channel; it is not the general Coulomb or arbitrary long-tail formula.↔ Scroll wide equations horizontally.

Take limits in the right order

Ignoring effective range gives a scattering amplitude whose magnitude depends on both scattering length and incident wavenumber. If the product of those scales is small, the total distinguishable-particle s-wave cross section approaches four pi times the squared scattering length. At fixed nonzero wavenumber, however, sending the scattering length to infinity approaches the unitarity value instead of an infinite area. The zero-energy and infinite-scattering-length limits therefore require careful ordering.

Effective range provides a measurable energy dependence beyond the leading constant, but two fitted parameters do not reconstruct the complete microscopic potential. Different short-range interactions can share the same low-energy values while differing at higher energy or in additional channels. The simple function lab can illustrate the quadratic dependence of the real inverse-amplitude correction on wavenumber, but it cannot infer scattering parameters, perform threshold matching or generate the imaginary term that enforces flux conservation.

r_e=0:\quad f_0=-\frac{a}{1+ika},\qquad \sigma_0=\frac{4\pi a^2}{1+k^2a^2}\le\frac{4\pi}{k^2}

These equations use a real scattering length and distinguishable elastic s-wave scattering. They assume neglected effective-range and higher terms are small over the selected energy band; identical particles require symmetry factors.↔ Scroll wide equations horizontally.

Work it through

Retain the effective-range correction

Within a valid truncated effective-range model take $a=4.0\,\mathrm{nm}$ , $r_e=2.0\,\mathrm{nm}$ and $k=0.25\,\mathrm{nm^{-1}}$ . Find the s-wave amplitude and cross section, and compare with the same calculation omitting effective range.

The real inverse-amplitude part is $K=-1/a+r_ek^2/2=-0.25+0.0625=-0.1875\,\mathrm{nm^{-1}}$ . The outgoing imaginary term is $-ik=-0.25i\,\mathrm{nm^{-1}}$ .
Invert $K-ik$ by multiplying its conjugate. Since $K^2+k^2=25/256\,\mathrm{nm^{-2}}$ , the amplitude is $f_0=(-1.92+2.56i)\,\mathrm{nm}$ . Its imaginary part is positive, consistent with elastic unitarity.
The cross section is $4\pi/(K^2+k^2)=1024\pi/25\,\mathrm{nm^2}\approx128.7\,\mathrm{nm^2}$ . With $r_e=0$ it is $4\pi/(0.25^2+0.25^2)=32\pi\,\mathrm{nm^2}\approx100.5\,\mathrm{nm^2}$ .

The result

The retained-range model gives f0 = (−1.92 + 2.56i) nm and σ0 = 128.7 nm², compared with 100.5 nm² when the range correction is dropped.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

The quadratic real correction

Reset and set BOTH Coefficient a = 1 and Position x = 0.50; predict y = 0.25. Reset and set Coefficient a = 1 and Position x = 1; predict y = 1. Regard x as a dimensionless wavenumber and y only as a quadratic correction shape. The lab plots y = ax², its slope and area; it does not include the scattering-length offset, complex amplitude, effective-range fitting or unitarity.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 forA quadratic correction can grow quickly with wavenumber even while the leading threshold parameter stays fixed.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The denominator 1 + k²a² regularizes the fixed-energy limit. The cross section saturates the elastic s-wave unitarity limit rather than growing as a² indefinitely.

Your turn

Try it yourself

Neglect effective range and take $a=-3.0\,\mathrm{nm}$ at $k=1/3\,\mathrm{nm^{-1}}$ . Find the amplitude and total s-wave cross section. Compare the cross section with the result for positive $a=3.0\,\mathrm{nm}$ at the same k.

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

Give me a hint

Use f0 = −a/(1 + ika). The cross section depends on a², but the real amplitude retains its sign information.

Compare with the solution

For negative a, $ka=-1$ , giving $f_0=3/(1-i)=(1.5+1.5i)\,\mathrm{nm}$ . Thus $|f_0|^2=4.5\,\mathrm{nm^2}$ and $\sigma_0=18\pi\,\mathrm{nm^2}$ . Positive a gives $f_0=-3/(1+i)=(-1.5+1.5i)\,\mathrm{nm}$ and exactly the same cross section. Intensity at this level cannot distinguish the two signs, although their real phases and pole structures differ. Both results assume real a, purely elastic scattering and negligible higher threshold terms at the stated wavenumber.

Keep the idea

Take it with you

Scattering length is a signed extrapolation parameter, not generally a target radius.
The effective-range expansion approximates the real inverse amplitude while −ik enforces outgoing flux.
At fixed energy a large scattering length reaches a unitarity limit.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.