University Quantum Mechanics II · Quantum Scattering Theory · 15.3
Free radial solutions and the partial-wave expansion
Every partial-wave calculation rests on two facts built here: outside the potential every radial wave is a mix of one fixed pair of functions, and a plane wave is a sum of them, one per angular momentum. Learn to write both down, and to read from kR how many terms a problem really needs.
Build the model
Connect the measurement to the mechanism.
Outside the range R of a central potential the particle is free, so there the radial equation loses V and depends on r only through x = kr, with k = √(2mE)/ħ for the reduced mass m. Its two solutions are universal: the spherical Bessel function jₗ, finite at the origin, and the Neumann function nₗ, which diverges there. For l = 0 they follow in one line as sin x/x and −cos x/x, and that minus sign fixes the convention for every l; higher orders come from quoted generating formulas, and all of them settle at large x into sin(x − lπ/2)/x and −cos(x − lπ/2)/x, which makes jₗ + i nₗ a purely outgoing wave.
Whatever a potential does is then recorded, channel by channel, in how much nₗ it mixes into jₗ beyond R. The incident wave needs no mixture: e(ikz) is free everywhere and finite at r = 0, so it is a sum of jₗ(kr)Pₗ(cos θ) alone, and Legendre orthogonality with the lowest power of kr fixes the weights at iˡ(2l + 1). Averaged over a sphere, the square of that sum gives wave l the share (2l + 1)jₗ(kr)² of the beam at radius r, tiny inside kr ≈ l because the centrifugal barrier holds wave l out beyond its impact parameter (l + ½)/k.
So a potential of range R touches only l up to about kR. The picture ends where there is no outside: a Coulomb tail never lets the free forms take over, and l ≲ kR is a soft edge to be checked, not a sharp cut.
- Simple definition
- The partial-wave expansion writes an incident plane wave as a sum of free spherical waves of definite angular momentum, e(ikz) = Σ iˡ(2l + 1) jₗ(kr) Pₗ(cos θ), each of which a central potential scatters on its own.
- Example
- For 4.0 MeV neutrons on protons, kR = 0.31 for the 1.41 fm range of the nuclear force: at r = R the s wave carries 97% of the incident beam and the p wave 3%, so one term, l = 0, describes the scattering, and it is isotropic.
Derived: with V = 0 the radial equation depends on r only through kr. The ratio B/A is the only trace V leaves outside; the next topic writes it as −tan δₗ.
k = √(2mE)/ħ in m⁻¹, m the reduced mass in kg, E in J; r and R in m; jₗ and nₗ are dimensionless
The minus sign is a convention, chosen so that nₗ → −cos(x − lπ/2)/x for every l. Tables differ on it, and a flipped sign flips every phase shift.
From u″ + k²u = 0 for u = rR; x is dimensionless. j₀(0) = 1, while n₀ diverges as −1/x
Quoted from Bessel-function theory and checked at l = 1 by substitution. Near x = 0 it gives jₗ ≈ xˡ/(2l + 1)!!, which is why jₗ is tiny for x well below l.
nₗ is the same with sin x/x replaced by −cos x/x. At l = 1: j₁ = sin x/x² − cos x/x and n₁ = −cos x/x² − sin x/x
Derived from jₗ ≈ sin(x − lπ/2)/x and nₗ ≈ −cos(x − lπ/2)/x. A scattered wave is built from hₗ alone; jₗ is half outgoing and half incoming.
Exact at every x for l = 0: h₀ = −i e(ix)/x. With e(−iEt/ħ) attached, the phase kr − Et/ħ moves outward
The weights come from Legendre orthogonality and the lowest power of kr. The l = 0 term, sin(kr)/(kr), is the plane wave averaged over directions.
z = r cos θ in m, k in m⁻¹. No nₗ, because e(ikz) is finite at r = 0; only Pₗ, because it has no φ dependence
A semiclassical estimate, not an exact result: wave l carries the share (2l + 1)jₗ(kr)² of the beam at radius r, tiny for kr below about l, so a potential of range R changes only l up to about kR.
bₗ in m, k in m⁻¹; from ħkb = ħ√(l(l + 1)), with l(l + 1) = (l + ½)² − ¼
Outside the range, the potential drops out
Work in the centre-of-mass frame with m the reduced mass and ħk = √(2mE). Quantum Mechanics I's radial equation for u = rR is u″ + [k² − l(l + 1)/r² − 2mV/ħ²]u = 0. Beyond the range R, V = 0, and in the variable x = kr what remains is d²u/dx² + [1 − l(l + 1)/x²]u = 0, which contains nothing about the target. Its two independent solutions are therefore universal functions of kr, the same for every potential and every energy; call them Rₗ = jₗ(kr) and Rₗ = nₗ(kr), so that u = kr jₗ or kr nₗ. Inside R the equation still holds V and is solved with u(0) = 0; outside, the general solution is A jₗ + B nₗ, and matching at r = R fixes one number per l, the ratio B/A. That is the programme of the next four topics. It needs V to vanish beyond R, or at least to fall faster than 1/r so that the free forms take over far out. A Coulomb tail never does: its phase keeps growing as η ln(2kr), with η = Z₁Z₂αc/v, a result quoted here and used in Topic 15.09. For a neutron and a proton, m = 469.5 MeV/c², and a centre-of-mass energy of 2.0 MeV gives k = 0.220 fm⁻¹.
The l = 0 pair, and the sign it fixes
For l = 0 the centrifugal term is absent and d²u/dx² = −u, so u = sin x or u = cos x, and R₀ = u/r. Dividing by k to make them dimensionless gives j₀(x) = sin x/x, which tends to 1 as x → 0, and n₀(x) = −cos x/x, which diverges as −1/x. Finiteness at the origin picks out j₀ and fixes it up to normalisation. The irregular solution is fixed only once it is required to be a pure cosine, a quarter period out of step with j₀, and even then only up to sign; the minus is a convention, chosen so that every nₗ behaves at large x as −cos(x − lπ/2)/x, a quarter period behind jₗ ≈ sin(x − lπ/2)/x: jₗ and nₗ are the spherical versions of sine and minus cosine. Codes and tables differ on this sign, so test n₀ before borrowing any formula, because a flipped sign reverses every phase shift computed with it. Neither function is a bound state. j₀ has nodes at kr = π, 2π, 3π, spaced by half a de Broglie wavelength, and its amplitude falls as 1/(kr), so |R₀|² falls as 1/r², the spreading of a spherical wave.
Higher l from a generating formula, checked at l = 1
For l ≥ 1 this lesson quotes the generating formulas from Bessel-function theory: jₗ(x) = (−x)ˡ[(1/x) d/dx]ˡ(sin x/x), and nₗ(x) the same acting on −cos x/x. One step gives j₁ = sin x/x² − cos x/x and n₁ = −cos x/x² − sin x/x. Check j₁ rather than trust it: u = x j₁ = sin x/x − cos x has u″ = −sin x/x − 2cos x/x² + 2 sin x/x³ + cos x, so u″ + u = (2/x²)(sin x/x − cos x) = 2u/x², the l = 1 equation. Expanding sin and cos gives j₁ ≈ x/3 and n₁ ≈ −1/x² near x = 0, instances of the quoted jₗ ≈ xˡ/(2l + 1)!! and nₗ ≈ −(2l − 1)!!/x(l+1). The large-x form is derived: the leading term comes from letting every derivative fall on sin x, and (−1)ˡ times the l-th derivative of sin x is sin(x − lπ/2). The numbers show where each limit holds: j₁(1) = 0.301 against x/3 = 0.333, and j₁(10) = 0.0785 against sin(10 − π/2)/10 = 0.0839.
The combination that only goes out
Attach the time factor e(−iEt/ħ) and ask which mixture of jₗ and nₗ carries probability outward only. From the large-x forms, jₗ + i nₗ ≈ [sin(x − lπ/2) − i cos(x − lπ/2)]/x = −i e(i(x − lπ/2))/x = (−i)(l+1) e(ix)/x, since e(−ilπ/2) = (−i)ˡ. Its phase kr − Et/ħ is constant on spheres that expand, so this is an outgoing spherical wave, the spherical Hankel function hₗ. For l = 0 the result holds at every r: j₀ + i n₀ = (sin x − i cos x)/x = −i e(ix)/x. The conjugate jₗ − i nₗ is the incoming wave, and jₗ = (hₗ + hₗ*)/2 is an equal mix of the two, a real standing wave that carries no net radial current. So a scattered wave must be built from hₗ alone, and a potential can change only the outgoing half of each jₗ, which the next topic turns into a phase shift. Keep the prefactor (−i)(l+1): it sets the relative phase of different l in the scattering amplitude, and dropping it scrambles their interference.
Rayleigh's formula from orthogonality and the lowest power
A plane wave along z is a free solution everywhere, so each of its partial waves solves the free radial equation; it is finite at r = 0, which rules out every nₗ; and it has no φ dependence, which leaves m = 0 and the angular factor Pₗ(cos θ). Hence e(ikz) = Σ aₗ jₗ(kr) Pₗ(cos θ). Multiply by Pₗ and integrate over x = cos θ; Unit 1's norm 2/(2l + 1) gives aₗ jₗ(kr) = [(2l + 1)/2] ∫ e(ikrx) Pₗ(x) dx from −1 to 1. Compare the lowest power of kr on each side. Powers xⁿ with n ≤ l − 1 in the exponential are orthogonal to Pₗ, so the first survivor is (ikrx)ˡ/l!, and ∫xˡPₗ dx = 2(l+1)(l!)²/(2l + 1)! follows from Pₗ's leading coefficient (2l)!/(2ˡ(l!)²) and its norm. The left side starts aₗ(kr)ˡ/(2l + 1)!!, and (2l + 1)! = 2ˡ l! (2l + 1)!!, so aₗ = iˡ(2l + 1), derived. Checks: at r = 0 only j₀ = 1 survives, as e⁰ = 1 demands; the l = 0 term is the direction average (1/2)∫e(ikrx) dx = sin(kr)/(kr), exactly.
Partial wave l passes at impact parameter (l + ½)/k
Square Rayleigh's formula and average it over a sphere of radius r: orthogonality removes every cross term, and |e(ikz)|² = 1 becomes Σ (2l + 1)jₗ(kr)² = 1, so wave l carries the share (2l + 1)jₗ(kr)² of the beam on that sphere. Inside kr ≈ l the share is tiny, because the centrifugal term makes r ≤ √(l(l + 1))/k classically forbidden to a free particle with angular momentum ħ√(l(l + 1)). Classically the angular momentum is pb = ħkb, so wave l belongs to impact parameter bₗ ≈ (l + ½)/k. Numerically, waves l ≤ L together hold at least 90% of the beam out to kr = L + ½ for every L from 0 to 6, while l = 3 holds only 0.06% at kr = 1. So a potential of range R changes only l up to about kR, a soft edge rather than a cut. Thermal neutrons on a nucleus have kR ≈ 10⁻⁴ and scatter as a pure s wave, which is why neutron diffraction uses one angle-independent scattering length per nucleus, measured as 6.65 fm for carbon; 14 MeV neutrons on nickel have kR ≈ 3.9 and need five waves. When kR runs into the tens the sum grows long, and, where the potential is weak enough to meet its validity condition, the Born approximation of Topic 15.08 becomes the better tool.
Change one variable at a time
Make the relationship visible.
Keep kR = 3 and raise L from 0: the dot at r = R climbs 0.002, 0.361, 0.807, 0.968, 0.997, so this range needs l up to 4, and each new dashed wave rises through its line and peaks just beyond it. Halve kR to 1.5 and L = 2 already holds 0.994.
b / R FOR WAVE L0.83
s-WAVE SHARE AT r = R0.002
p-WAVE SHARE AT r = R0.358
LARGEST l WITH bₗ ≤ R2
Live interpretationb / R FOR WAVE L: 0.83. s-WAVE SHARE AT r = R: 0.002. p-WAVE SHARE AT r = R: 0.358. LARGEST l WITH bₗ ≤ R: 2
Catch the common trap
Explain before calculating.
Thermal neutrons with k = 3.5 × 10¹⁰ m⁻¹ scatter from a nucleus whose potential has range R = 5.0 fm, and the measured angular distribution in the centre-of-mass frame is isotropic. Why?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyNeutrons with lab kinetic energy 4.0 MeV scatter from protons at rest. Taking the range of the nuclear force as the pion's reduced Compton wavelength, R = ħ/(mπ c), find k in the centre-of-mass frame, kR and the p wave's impact parameter, and decide which partial waves the angular distribution needs.
- Centre-of-mass energy and reduced mass: E = 4.0 MeV × mₚ/(mₙ + mₚ) = 4.0 × 938.272/1877.837 MeV = 1.999 MeV, and mc² = mₙmₚc²/(mₙ + mₚ) = 469.46 MeV.
- ħck = √(2mc²E) = √(2 × 469.46 × 1.999) MeV = 43.32 MeV, so k = 43.32 MeV ÷ 197.33 MeV fm = 0.2195 fm⁻¹. Dimensions: MeV divided by MeV fm is fm⁻¹.
- The range, from the measured pion mass 139.57 MeV/c²: R = 197.33/139.57 fm = 1.414 fm, so kR = 0.2195 × 1.414 = 0.310.
- The p wave's impact parameter is b₁ = 1.5/k = 6.83 fm, 4.8 times R. At r = R it carries 3j₁(kR)² = 0.031 of the beam, close to the small-x estimate (kR)²/3 = 0.032, while the s wave carries j₀(kR)² = 0.968.
- So one partial wave, l = 0, does the scattering, and an s wave has no angular dependence: dσ/dΩ is isotropic in the centre-of-mass frame. Measured n–p angular distributions at a few MeV are isotropic to within a few per cent.
- Order in α: none, since the neutron is uncharged. For protons on protons at the same energy η = αc/v = 0.079 is small, but the Coulomb phase grows as ln r, so the free forms would never take over.
Answerk = 0.220 fm⁻¹ and kR = 0.31; b₁ = 6.8 fm lies far outside R, so only l = 0 matters and the distribution is isotropic, as measured.
MediumTest Rayleigh's formula at one point: kr = 1.0 and θ = 60°, so kz = 0.5. Sum the partial waves through l = 4, with jₗ(1) from the explicit forms and Pₗ(0.5) from Bonnet's recursion, and compare with e(0.5i) = cos 0.5 + i sin 0.5. Which l build the real part, and why do five terms suffice?
- iˡ is ±1 for even l and ±i for odd l, while jₗ and Pₗ are real, so even l build the real part and odd l the imaginary part. The targets are cos 0.5 = 0.877583 and sin 0.5 = 0.479426.
- j₀(1) = sin 1 = 0.8414710, j₁(1) = sin 1 − cos 1 = 0.3011687, j₂(1) = 2 sin 1 − 3 cos 1 = 0.0620351 and j₃(1) = 9 sin 1 − 14 cos 1 = 0.0090066, where two numbers near 7.57 nearly cancel. The recursion jₗ₊₁ = (2l + 1)jₗ/x − jₗ₋₁, quoted with the generating formula, gives j₄(1) = 7 × 0.00900658 − 0.06203505 = 0.0010110; carry the extra digit, because the subtraction loses two.
- Pₗ(0.5) = 1, 0.5, −0.125, −0.4375 and −0.2891 for l = 0 to 4, so the terms (2l + 1)jₗPₗ are 0.841471, 0.451753, −0.038772, −0.027583 and −0.002630.
- Real part, with i⁰ = 1, i² = −1 and i⁴ = 1: 0.841471 + 0.038772 − 0.002630 = 0.877613, against 0.877583.
- Imaginary part, with i¹ = i and i³ = −i: 0.451753 + 0.027583 = 0.479336, against 0.479426.
- The terms shrink because jₗ(1) ≈ 1/(2l + 1)!!, so (2l + 1)jₗ(1) ≈ 1/(2l − 1)!!: 1, 0.33, 0.067 and 0.0095 for l = 1 to 4, against exact values 0.90, 0.31, 0.063 and 0.0091. At kr = 1 only l ≲ 1 are large, as the impact-parameter rule predicts.
AnswerThrough l = 4 the sum is 0.877613 + 0.479336i against e(0.5i) = 0.877583 + 0.479426i; the errors, 3 × 10⁻⁵ and 9 × 10⁻⁵, are the first omitted terms, l = 6 and l = 5.
HardSquare Rayleigh's formula and average it over a sphere of radius r to show that Σ (2l + 1)jₗ(kr)² = 1. Then take 14.1 MeV neutrons on ⁵⁸Ni with the empirical nuclear radius R = 1.25 A¹⁄³ fm: find kR, the share of the incident beam each partial wave carries at r = R, and how many partial waves a calculation must keep.
- |e(ikz)|² = 1 everywhere. Its average over directions at fixed r is (1/2)∫|Σ iˡ(2l + 1)jₗPₗ|² d(cos θ). Cross terms between different l vanish by Legendre orthogonality and |iˡ|² = 1, so the average is Σ (2l + 1)² jₗ² × (1/2) × 2/(2l + 1) = Σ (2l + 1)jₗ(kr)² = 1.
- Kinematics, nonrelativistic: plab c = √(2 × 939.565 × 14.1) MeV = 162.8 MeV, and in the centre-of-mass frame p = plab × 58/59 = 160.0 MeV/c, so k = 160.0/197.33 fm⁻¹ = 0.811 fm⁻¹. Relativity would raise p by 0.4%, since T/mc² = 0.015.
- R = 1.25 × 58¹⁄³ fm = 1.25 × 3.871 fm = 4.84 fm, so kR = 0.811 × 4.84 = 3.92, or 3.924 carried unrounded, the value used below.
- Shares at x = 3.924 for l = 0 to 6: 0.032, 0.055, 0.400, 0.356, 0.128, 0.026 and 0.003. Cumulatively, l ≤ 3 holds 0.843, l ≤ 4 holds 0.971 and l ≤ 5 holds 0.996, and adding l = 6 brings the total to 0.9997, with l ≥ 7 supplying the last 0.0003, as the sum rule demands: a check on the arithmetic.
- Impact parameters bₗ = (l + ½)/k: b₃ = 4.32 fm lies inside R, b₄ = 5.55 fm just outside, b₅ = 6.78 fm well outside. Keep l = 0 to 4, five partial waves, and add l = 5, with 2.6% of the beam at the edge, as the convergence test.
- Checks: the count matches lₘₐₓ ≈ kR = 3.9, and each share is a pure number, as a probability must be. Measured elastic angular distributions of 14 MeV neutrons on medium nuclei are strongly forward-peaked with diffraction minima, the mark of several interfering partial waves; one s wave would be isotropic.
AnswerkR = 3.9. Waves l ≤ 4 carry 97% of the incident beam at r = R, so keep l = 0 to 4 and test convergence with l = 5, which carries 2.6%.