University Quantum Mechanics II · Quantum Scattering Theory · 15.6
Hard spheres at low energy and the scattering length
Slow particles cannot resolve the shape of a short-range potential: every detail of it reaches them through one length. Here you find that length, first for an impenetrable sphere, which scatters four times its classical area, then for an attractive well, where it changes sign and diverges as a bound state is born.
Build the model
Connect the measurement to the mechanism.
When kR ≪ 1 the wavelength dwarfs the range: every partial wave with l ≥ 1 is held by its centrifugal barrier beyond about (l + ½)/k, far outside the target, and only the s wave, which has no barrier, reaches it. Matching the inside solution to the free one at r = R turns that into a threshold law: with the small-argument forms of jₗ and nₗ, tan δₗ is proportional to (kR)(2l+1) for any potential of range R, so every wave but the s wave fades and the amplitude becomes one isotropic number. The zero-energy wave fixes that number.
Beyond R it obeys u″ = 0, so it is a straight line, u ∝ r − aₛ, and all the interior contributes is where that line crosses zero: the scattering length aₛ, with f → −aₛ and σ → 4πaₛ². An impenetrable sphere pins the zero at R, giving 4πR², four times the classical πR², because a wave much longer than the target meets it as a boundary condition on its whole surface, not as a stream of trajectories to be blocked. An attractive square well tilts the line: aₛ = R[1 − tan(KR)/(KR)] is small and negative for a weak well, where it matches the first Born estimate, and passes through ±∞ at KR = π/2, where the line turns flat and a bound state appears at zero energy.
The one-length description fades as kR approaches 1, where the effective range and then the p wave enter, and it fails outright for a Coulomb tail, which leaves no finite aₛ at all.
- Simple definition
- The scattering length aₛ = −lim tan δ₀/k as k → 0 is the one length that describes a short-range potential to slow particles: outside the potential the zero-energy wave is proportional to r − aₛ, the amplitude tends to −aₛ and σ tends to 4πaₛ².
- Example
- Lead's slow-neutron scattering length is listed as 9.41 fm, so σ → 4π(9.41 fm)² = 1.11 × 10³ fm² = 11.1 b, the measured value, and seven times the classical πR² = 1.58 b of a nucleus of radius 7.1 fm.
Derived from u(R) = 0: the outside wave is sin k(r − R), the free wave with its origin pushed out to R, so σ₀ = (4π/k²) sin²kR.
k = √(2mE)/ħ in m⁻¹ with m the reduced mass in kg; R the sphere's radius in m; δ₀ in rad
Derived for kR ≪ 1 from the quoted small-x forms of jₗ and nₗ. It makes σₗ/σ₀ scale as (kR)(4l), so only the s wave survives slow collisions.
βₗ = R Rₗ′(R)/Rₗ(R), the inside logarithmic derivative, dimensionless; (−1)!! = 1; a hard sphere has βₗ → ∞
Defined here, signed so that a hard sphere has aₛ = R. It stands in for the whole interior: the amplitude tends to f(θ) → −aₛ at every angle.
aₛ in m; at E = 0 the outside solution is u = rR ∝ r − aₛ, a straight line crossing zero at r = aₛ
Derived from σ₀ = (4π/k²) sin²δ₀ with δ₀ ≈ −kaₛ. It depends on aₛ², so the zero-energy cross-section alone cannot reveal the sign of aₛ.
σ in m², 1 b = 10⁻²⁸ m²; needs kR ≪ 1 and k|aₛ| ≪ 1. A hard sphere gives 4πR², four times πR²
Derived by matching sin Kr to r − aₛ at R. A weak well gives aₛ ≈ −2mV₀R³/(3ħ²), first order in V₀ and equal to the first Born value of Topic 15.08.
V₀ in J, m the reduced mass in kg, K in m⁻¹; aₛ → ±∞ at KR = π/2, 3π/2, …, where a bound state appears
Quoted from effective-range theory. The second term sets the energy at which aₛ alone stops being enough.
rₑ, the effective range, in m; k cot δ₀ in m⁻¹; for a hard sphere, expanding −k cot kR gives rₑ = 2R/3
The hard sphere: δ₀ = −kR at every energy
Quote Topic 15.04: beyond the range, Rₗ(r) = A[cos δₗ jₗ(kr) − sin δₗ nₗ(kr)], which tends to A sin(kr − lπ/2 + δₗ)/kr far out. An impenetrable sphere keeps the particle out of r < R, so Rₗ(R) = 0 and tan δₗ = jₗ(kR)/nₗ(kR), exactly and at every energy. For l = 0 the ratio is sin kR/(−cos kR), so δ₀ = −kR, derived, and u = rR ∝ sin k(r − R): the free s wave with its origin moved out to R. The negative sign is the general mark of repulsion, a wave pushed outward whose phase lags. The s-wave cross-section is σ₀ = (4π/k²) sin²kR, already 0.919 of its kR → 0 value 4πR² at kR = 0.5. The same condition for l = 1 gives tan δ₁ = (sin x − x cos x)/(−cos x − x sin x) with x = kR, which at kR = 0.5 is −0.0364: δ₁ = −2.08° beside δ₀ = −28.6°.
Why only the s wave survives: tan δₗ ∝ (kR)(2l+1)
Let any potential of range R act. Whatever it does inside, it hands the outside one number per l, the logarithmic derivative βₗ = R Rₗ′(R)/Rₗ(R). For kR ≪ 1 the inside barely notices the energy, since k² is negligible beside 2m|V|/ħ², so βₗ settles to a constant. Matching gives tan δₗ = [x jₗ′(x) − βₗ jₗ(x)]/[x nₗ′(x) − βₗ nₗ(x)] at x = kR. Insert the small-x forms quoted in Topic 15.03, jₗ ≈ xˡ/(2l + 1)!! and nₗ ≈ −(2l − 1)!!/x(l+1), for which x jₗ′ = l jₗ and x nₗ′ = −(l + 1)nₗ, and tan δₗ ∝ (kR)(2l+1) follows, derived, with the coefficient on the formula card. The hard sphere, βₗ → ∞, gives −x, −x³/3 and −x⁵/45. Since σₗ ∝ (2l + 1) sin²δₗ, σₗ/σ₀ goes as (kR)(4l): the hard sphere's exact σ₁/σ₀ at kR = 0.1 is 3.30 × 10⁻⁵, against (kR)⁴/3 = 3.33 × 10⁻⁵. The cause is the centrifugal barrier, which holds wave l ≥ 1 out beyond about (l + ½)/k; the s wave alone has none. The law fails only if βₗ = −(l + 1), a state sitting exactly at zero energy.
Four times the classical area, and no return to it
At low energy the hard sphere's amplitude is f = e(iδ₀) sin δ₀/k → −R at every angle, so dσ/dΩ = R² and σ = 4πR², the sphere's whole surface area rather than its silhouette. Topic 15.01 found the classical hard sphere isotropic too, with dσ/dΩ = R²/4, so the quantum result is four times larger at every angle. There is no paradox: trajectories need a wavelength short beside R, and at kR ≪ 1 the reverse holds, so the wave meets the sphere as a boundary condition over its whole surface, the regime of diffraction rather than rays. Raising the energy does not bring πR² back. Summing the exact δₗ = arctan[jₗ(kR)/nₗ(kR)] numerically gives σ/πR² = 3.38 at kR = 1, 2.40 at kR = 10 and 2.09 at kR = 100. The limit, quoted from the large-l asymptotics of that sum, is 2πR²: πR² reflected as rays would be, and πR² diffracted into a forward peak about 1/kR wide, the shadow the optical theorem of Topic 15.05 demands.
The scattering length: one intercept for the whole interior
The threshold law makes tan δ₀ linear in k for any short-range potential, so its slope defines a length, aₛ = −lim tan δ₀/k as k → 0, signed so that the hard sphere has aₛ = R. Then sin δ₀ ≈ −kaₛ, the amplitude tends to f → −aₛ, the same at every angle and every low energy, and σ → 4πaₛ². The meaning is in the zero-energy wave. At E = 0 the outside equation is u″ = 0, so u is a straight line, and it is the k → 0 limit of sin(kr + δ₀) ≈ k(r − aₛ): u ∝ r − aₛ. The scattering length is where that outside line, run back through the potential if need be, crosses zero. Repulsion pushes the zero out to positive r; weak attraction pulls the wave in and puts the zero at negative r, so aₛ < 0 while δ₀ > 0. Every detail of the interior reaches a slow particle only through this intercept, so two potentials with the same aₛ cannot be told apart at low energy. The figure draws the line for a hard core inside an attractive shell.
An attractive well: aₛ = R[1 − tan(KR)/(KR)]
Take V = −V₀ for r < R. At E = 0 the inside equation is u″ = −K²u with K = √(2mV₀)/ħ, and u(0) = 0 gives u = sin Kr. Equal u′/u at R gives K cot KR = 1/(R − aₛ), so aₛ = R[1 − tan(KR)/(KR)], derived: a length times a pure number. For a weak well, tan x ≈ x + x³/3 gives aₛ ≈ −2mV₀R³/(3ħ²), first order in V₀ and equal to the first Born value (2m/ħ²)∫V r² dr quoted from Topic 15.08; the exact aₛ is 3.7% larger in magnitude at KR = 0.3 and 67% larger at KR = 1. As KR → π/2 the inside wave reaches its crest at R, the outside line turns flat and aₛ → −∞; just beyond, aₛ → +∞ and the well holds its first bound state. A shallow bound state's outside wave e(−κr) has u′/u = −κ, and the inside barely changes while EB ≪ V₀, so −κ ≈ 1/(R − aₛ): for aₛ ≫ R, κ ≈ 1/aₛ and EB ≈ ħ²/(2maₛ²), derived to leading order in R/aₛ. Away from the divergences a deep well acts as a hard sphere, aₛ = R − tan(KR)/K lying typically within about 1/K of R: gold and bismuth have slow-neutron scattering lengths of 7.63 fm and 8.53 fm, radii near 7.0 fm.
The effective range, and where one length is not enough
The next term in energy is quoted from effective-range theory: k cot δ₀ = −1/aₛ + ½rₑk² + O(k⁴), with rₑ of the order of the range. For the hard sphere it checks in a line, since −k cot kR = −1/R + k²R/3 + …, so rₑ = 2R/3. A large |aₛ| opens a window R ≪ 1/k ≲ |aₛ| in which only the s wave scatters yet σ is no longer 4πaₛ²: keeping only −1/aₛ gives σ = 4πaₛ²/(1 + k²aₛ²). For the neutron–proton spin-singlet channel, measured as aₛ = −23.7 fm and rₑ = 2.77 fm, that halves σ by a centre-of-mass energy of 74 keV, and the rₑ term reaches 10% of 1/|aₛ| at 0.13 MeV. The description also needs a potential that dies fast: a tail falling as r⁻ⁿ gives a finite aₛ only for n > 3, quoted, and a Coulomb tail gives none, which is why proton–proton scattering uses a Coulomb-modified length, measured as −7.8 fm. So no power of α appears in this topic: the one place α would enter is exactly where aₛ ceases to exist.
Change one variable at a time
Make the relationship visible.
At c = 0 step KR from 1.00 to 1.55: aₛ runs from −0.56R to −30R. At 1.60 it is +22R and a bound state exists; near 4.50 it passes through zero. Then drag c to 1: the shell vanishes and aₛ = R, the hard sphere.
SCATTERING LENGTH aₛ-0.56 R
σ(E → 0) ÷ πR²1.2
SHELL PHASE K(R − c)1.00 rad
BOUND s STATES0
Live interpretationSCATTERING LENGTH aₛ: −0.56 R. σ(E → 0) ÷ πR²: 1.2. SHELL PHASE K(R − c): 1.00 rad. BOUND s STATES: 0
Catch the common trap
Explain before calculating.
Two attractive square wells have the same range R. The first has KR = 1.50 and the second KR = 1.65, where K = √(2mV₀)/ħ. How do their scattering lengths and zero-energy cross-sections compare?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA neutron of kinetic energy 1.00 keV meets a heavy nucleus that you model as a fixed, impenetrable sphere of radius R = 6.0 fm. Find kR, the s-wave phase shift, the cross-section and the p wave's share of it, and compare the cross-section with πR².
- With the nucleus fixed, m is the neutron mass: ħck = √(2mc²E) = √(2 × 939.57 MeV × 1.00 × 10⁻³ MeV) = 1.371 MeV, so k = 1.371 MeV ÷ 197.33 MeV fm = 6.947 × 10⁻³ fm⁻¹ and kR = 0.0417. A real nucleus of mass M and mass number 120, free to recoil, would lower the centre-of-mass k by the factor M/(M + m) ≈ 120/121, only 0.8%.
- Hard sphere: δ₀ = −kR = −0.0417 rad = −2.39°, exactly.
- σ₀ = (4π/k²) sin²kR = 4πR² [sin(kR)/kR]² = 452.4 fm² × 0.99942 = 452.1 fm² = 4.52 b, since 1 b = 100 fm².
- Threshold law: σ₁/σ₀ ≈ (kR)⁴/3 = 1.0 × 10⁻⁶, so the p wave is invisible and the scattering is isotropic.
- Classically σ = πR² = 113 fm² = 1.13 b. The quantum value is 4.00 times larger, the kR → 0 limit reached here to 0.06%. No power of α enters, since the neutron is neutral.
AnswerkR = 0.042 and δ₀ = −2.39°; σ = 4.52 b = 4πR², four times the classical 1.13 b, with the p wave adding a fraction of about 10⁻⁶.
MediumModel the neutron–proton interaction in its spin-singlet state as an attractive square well of range R = 2.50 fm and depth V₀ = 15.0 MeV, with ħ²/2m = 41.47 MeV fm² for the reduced mass. Find aₛ and σ as k → 0, compare aₛ with the measured −23.7 fm and with the first Born estimate, and find the depth at which a bound state would appear.
- K = √(2mV₀)/ħ = √(15.0 MeV ÷ 41.47 MeV fm²) = 0.6014 fm⁻¹, so KR = 1.50355 rad, just below π/2 = 1.5708: no bound state yet.
- So close to π/2 the tangent needs every digit: tan(KR) = 14.85 and tan(KR)/(KR) = 9.876, so aₛ = R[1 − tan(KR)/(KR)] = 2.50 fm × (1 − 9.876) = −22.19 fm. That falls 6.4% short of the measured −23.7 fm, and V₀ = 15.08 MeV would reproduce it.
- σ → 4πaₛ² = 4π(22.19 fm)² = 6.19 × 10³ fm² = 61.9 b, 79 times the 0.785 b of a hard sphere of the same radius.
- First Born value: −2mV₀R³/(3ħ²) = −(15.0/41.47) fm⁻² × (2.50 fm)³/3 = −1.88 fm, the right sign but 12 times too small. It is the KR → 0 limit of the exact formula, and here KR sits beside π/2, where tan(KR) runs away.
- A bound state appears at KR = π/2: V₀ = 41.47 MeV fm² × (π/5.00 fm)² = 16.37 MeV, only 9% deeper. At 17.0 MeV, aₛ = +54.8 fm and EB ≈ ħ²/(2maₛ²) = 13.8 keV; the exact bound state of that well is at 14.5 keV.
Answeraₛ = −22 fm and σ = 62 b, against the measured −23.7 fm; the Born estimate is only −1.9 fm; a bound state appears at V₀ = 16.4 MeV, 9% deeper.
HardA model spin-triplet neutron–proton interaction has a hard core of radius c = 0.50 fm inside an attractive shell, V = −V₀ for c < r < R = 2.50 fm, with ħ²/2m = 41.47 MeV fm². Derive aₛ and check its limits, find the depth at which the first bound state appears, and for V₀ = 38.0 MeV compare aₛ and the binding it implies with the measured 5.42 fm and 2.2246 MeV.
- At E = 0 inside the shell u″ = −K²u with u(c) = 0, so u = sin K(r − c); outside u ∝ r − aₛ. Equal u′/u at R gives K cot K(R − c) = 1/(R − aₛ), so aₛ = R − tan[K(R − c)]/K, derived.
- Checks: c = 0 gives R[1 − tan(KR)/(KR)]; as V₀ → 0, tan[K(R − c)]/K → R − c, so aₛ → c, a hard sphere of the core's size; c = R gives aₛ = R. The first bound state appears at K(R − c) = π/2: V₀ = 41.47 MeV fm² × (π/4.00 fm)² = 25.58 MeV, 56% deeper than the 16.37 MeV the same well needs with no core.
- At V₀ = 38.0 MeV: K = √(38.0/41.47) fm⁻¹ = 0.9572 fm⁻¹ and K(R − c) = 1.9145 rad, past π/2, so one bound state. tan 1.9145 = −2.794, so aₛ = 2.50 fm + 2.794/0.9572 fm = 2.50 + 2.919 = 5.42 fm, the measured value, and σ → 4πaₛ² = 369 fm² = 3.69 b.
- Leading-order binding: EB ≈ ħ²/(2maₛ²) = 41.47 MeV fm² ÷ 29.36 fm² = 1.41 MeV, 37% below the measured 2.2246 MeV. The estimate assumes aₛ ≫ R, and here aₛ/R is only 2.2. Solving the model's own bound-state condition, K′ cot K′(R − c) = −κ with K′² = K² − κ², numerically gives 2.76 MeV, 24% too high.
- One length cannot fix the binding. The model's effective range, from its phase shifts, is about 2.3 fm against the measured 1.753 fm. With that rₑ and the model's unrounded aₛ = 5.419 fm, the quoted expansion continued to k = iκ, 1/aₛ = κ − ½rₑκ², gives κ = 0.2315 fm⁻¹ and EB = 2.223 MeV, within 0.1% of the measurement.
Answeraₛ = R − tan[K(R − c)]/K = 5.42 fm, as measured; the first bound state needs V₀ = 25.6 MeV; EB ≈ 1.41 MeV at leading order and 2.76 MeV exactly for the model, against 2.2246 MeV, which the effective range recovers.