University Quantum Mechanics II · Quantum Scattering Theory · 15.4
Phase shifts, partial cross-sections and the unitarity limit
Scattering in three dimensions looks like a whole angular distribution to predict. For a central potential it collapses to one real angle per angular momentum. This lesson finds those angles, builds f(θ) and σ from them, and shows why no partial wave can scatter more than 4π(2l + 1)/k², however strong the force.
Build the model
Connect the measurement to the mechanism.
A real potential of finite range a can do only one thing to the partial wave of angular momentum l: shift its phase. Beyond r = a the particle is free, so each radial function is a mix of the free solutions jₗ and nₗ, and the mix is fixed by the one interior solution that stays finite at the origin; far away it reads sin(kr − lπ/2 + δₗ)/kr. A central force cannot change l and a real one cannot lose probability, so the outgoing spherical wave returns with its full flux and nothing but a phase, Sₗ = e(2iδₗ).
Subtracting the plane wave's own partial waves leaves f(θ) = (1/k)Σ(2l + 1)e(iδₗ) sin δₗ Pₗ(cos θ), and Legendre orthogonality turns ∫|f|² dΩ into σ = (4π/k²)Σ(2l + 1) sin²δₗ. Scattering becomes a list of angles, and each term has a ceiling set by the wavelength alone. The method pays when ka is small, because only waves with l up to about ka reach the potential.
It stops at a Coulomb tail, which never lets the wave become free; at absorption, which makes |Sₗ| less than 1; at high energy, where many waves each carry a little and the Born approximation takes over; and at spin-dependent forces and identical particles, which need more labels than l.
- Simple definition
- The phase shift δₗ is the angle by which a central potential advances or delays the lth partial wave outside its range; that wave then adds 4π(2l + 1) sin²δₗ/k² to the cross-section, and never more than 4π(2l + 1)/k².
- Example
- At k = 1.0 fm⁻¹ an s-wave phase shift of 30° gives σ₀ = (4π/k²) sin²30° = 12.6 fm² × 0.25 = 3.1 fm², a quarter of the s-wave ceiling 4π/k² = 12.6 fm² = 0.126 b.
Derived from the exterior form A[cos δₗ jₗ(kr) − sin δₗ nₗ(kr)], valid beyond the range a. A central V's whole far-field effect is one angle per l.
k = √(2mE)/ħ in m⁻¹ with m the reduced mass in kg; δₗ in rad, real for a real V and fixed only mod π.
Derived: a central V keeps l and a real V keeps flux, so only a phase survives. With the incoming wave normalised as the free one, this factor is the whole change.
Dimensionless; it multiplies e(ikr)/r in the lth partial wave. |Sₗ| = 1 for a real V, less than 1 if V absorbs.
Derived by subtracting the plane wave's partial waves, which leaves fₗ = (Sₗ − 1)/(2ik). Cross terms between different l shape dσ/dΩ = |f|².
f in m, k in m⁻¹; Pₗ are Unit 1's dimensionless Legendre polynomials; the sum runs over l = 0, 1, 2, …
Derived from ∫|f|² dΩ with ∫Pₗ Pₗ′ dΩ = 4πδₗₗ′/(2l + 1). Interference moves flux between angles but cancels from σ, so each l owns a separate share.
σ in m²; 1 b = 10⁻²⁸ m² = 100 fm². At k = 1.0 fm⁻¹, 4π/k² = 12.6 fm².
Derived from sin²δₗ ≤ 1. Only the wavelength sets it, so no potential can push a wave past it, and a computed σₗ that exceeds it is wrong.
In m². Reached at δₗ = 90° (mod 180°), where Sₗ = −1; four times the classical annulus π(2l + 1)/k².
Derived from continuity of Rₗ and Rₗ′. V = 0 gives γₗ = k jₗ′/jₗ and δₗ = 0; a hard wall, γₗ → ∞, gives tan δₗ = jₗ(ka)/nₗ(ka).
γₗ = Rₗ′/Rₗ just inside r = a, in m⁻¹; jₗ′ = djₗ/dx at x = ka. The interior normalisation drops out.
Outside the range, all the potential leaves is a phase
Beyond r = a the radial equation is the free one, so Rₗ is a mix of the previous lesson's free solutions, quoted here with n₀(x) = −cos x/x: Rₗ = A[cos δₗ jₗ(kr) − sin δₗ nₗ(kr)]. The mix is not ours to choose. Inside, only one solution stays finite at r = 0, so the interior fixes the ratio of the two exterior coefficients, and that ratio is tan δₗ. With jₗ(x) → sin(x − lπ/2)/x and nₗ(x) → −cos(x − lπ/2)/x, the far field is A sin(kr − lπ/2 + δₗ)/kr: the free wave, displaced. A real V makes the radial equation real, so δₗ is real, and flipping the sign of A shows it is fixed only modulo π. The sign carries physics. Attraction raises the local wave number, so the wave reaches r = a further along in phase and every far-field node moves inward by δₗ/k: at k = 1.0 fm⁻¹, δ₀ = +30° pulls each node in by 0.52 fm. Repulsion makes δₗ negative, and the most repulsive potential of range a, an impenetrable sphere, gives δ₀ = −ka.
Flux conservation makes the change a factor e(2iδₗ)
Write the far field as waves travelling in and out. Multiplied by the constant e(iδₗ), e(iδₗ) sin(kr − lπ/2 + δₗ)/kr = [e(2iδₗ) e(i(kr − lπ/2)) − e(−i(kr − lπ/2))]/(2ikr). Set δₗ = 0 and this is the free partial wave, so with this normalisation the incoming wave is exactly the free one and the potential's entire effect is Sₗ = e(2iδₗ) on the outgoing wave. Two symmetries force that form. A central V commutes with L² and with every component of L, so a wave that comes in with angular momentum l leaves with l: each partial wave is its own one-channel problem. A real V conserves probability, so the radial flux carried in by the incoming l wave must all leave in the outgoing l wave, and |Sₗ| = 1. An absorbing potential, modelled by giving V a negative imaginary part, breaks the second condition: the outgoing wave returns weakened, |Sₗ| is less than 1, and the missing flux is the absorption cross-section. This lesson names that case and does not develop it.
Subtract the plane wave and f(θ) is what is left
Far away the state must read e(ikz) + f(θ)e(ikr)/r. The plane wave's own l component is quoted from the previous lesson: (2l + 1) iˡ jₗ(kr) Pₗ(cos θ), which tends to (2l + 1) iˡ [e(i(kr − lπ/2)) − e(−i(kr − lπ/2))] Pₗ/(2ikr). Give the scattering solution, in the normalisation above, the same coefficient (2l + 1) iˡ, so the incoming waves agree, and subtract. The difference is (2l + 1) iˡ (Sₗ − 1) e(i(kr − lπ/2)) Pₗ/(2ikr), and since iˡ e(−ilπ/2) = 1 it equals (2l + 1)(Sₗ − 1)/(2ik) × e(ikr)/r × Pₗ(cos θ): purely outgoing, as a scattered wave must be. So f(θ) = Σ(2l + 1) fₗ Pₗ(cos θ) with fₗ = (Sₗ − 1)/(2ik) = e(iδₗ) sin δₗ/k, derived here, with the units of 1/k, a length. Because each fₗ is complex, dσ/dΩ = |f|² holds cross terms between waves: s and p together add (6/k²) sin δ₀ sin δ₁ cos(δ₀ − δ₁) cos θ, which tips the distribution forward or backward according to its sign.
Orthogonality strips the cross terms from σ
Integrate |f|² over all directions. With the beam along z nothing depends on the azimuth, so dΩ = 2π d(cos θ), and Unit 1's Legendre norm gives ∫Pₗ Pₗ′ dΩ = 4πδₗₗ′/(2l + 1). Every cross term between different l integrates to zero, and each diagonal term leaves (2l + 1)²|fₗ|² × 4π/(2l + 1). So σ = 4πΣ(2l + 1)|fₗ|² = (4π/k²)Σ(2l + 1) sin²δₗ, derived here, and each partial wave owns a separate share σₗ = 4π(2l + 1) sin²δₗ/k². Interference moves flux between angles without creating or destroying any, which is why the forward–backward tilt above never shows up in σ. Dimensions: 1/k² is an area, and a useful scale is 4π/k² = 12.6 fm² = 0.126 b at k = 1.0 fm⁻¹. The factor 2l + 1 is not a count of m states, since only m = 0 occurs for a beam along z; it is the weight the plane-wave expansion gives the lth wave, the quantum counterpart of the classical annulus between impact parameters l/k and (l + 1)/k.
Each partial wave has a ceiling fixed by the wavelength
Since sin²δₗ ≤ 1, σₗ ≤ 4π(2l + 1)/k², reached when δₗ = 90° (mod 180°). There Sₗ = −1: the outgoing wave returns with its sign reversed, the largest change a phase can make, and |Sₗ − 1|² = 4. That 4 is why the ceiling is four times the classical annulus π(2l + 1)/k²: the scattered wave must both cancel the unscattered outgoing wave and supply the reversed one. Nothing about V enters the bound, only k, so every computed σₗ must pass it, and it survives absorption because |1 − Sₗ| cannot exceed 2 while |Sₗ| ≤ 1. With l running up to about ka, it also caps the total near 4π(a + 1/k)². Real forces come close. In the neutron–proton spin-singlet channel, where the force is central, phase-shift analyses of measured data give δ₀ ≈ 63.6° at 5 MeV laboratory energy (quoted from a table). There k = 0.245 fm⁻¹, the ceiling 4π/k² is 2.09 b, and σ₀ reaches 80% of it, 1.67 b.
Matching logarithmic derivatives fixes each δₗ
Everything above took δₗ as known; here is how to get it. Integrate the radial equation outward from the regular solution, which starts as rˡ, and stop at r = a. Its normalisation is arbitrary, but its logarithmic derivative γₗ = Rₗ′/Rₗ at r = a is not, and it is all the outside needs. Rₗ and Rₗ′ are continuous at r = a, so γₗ is too, and equating it with the log derivative of A[cos δₗ jₗ(kr) − sin δₗ nₗ(kr)] gives tan δₗ = [k jₗ′(ka) − γₗ jₗ(ka)]/[k nₗ′(ka) − γₗ nₗ(ka)], derived here. For l = 0 the form in u = rR is cleaner, k cot(ka + δ₀) = u′/u at r = a, and a square well of depth V₀, where u = sin Kr with K² = k² + 2mV₀/ħ², gives tan(ka + δ₀) = (k/K) tan Ka. The figure's default, ka = 1.0 and 2mV₀a²/ħ² = 2, has Ka = √3 = 1.732 and tan(ka + δ₀) = −3.55; u is positive and u′ negative at r = a, so ka + δ₀ = π − 1.296 and δ₀ = 48.4°. Two checks close the loop: V₀ → 0 makes K → k and δ₀ → 0, and a hard wall, u(a) = 0, gives δ₀ = −ka.
Change one variable at a time
Make the relationship visible.
At s = 0 the solid curve lies on the dashed one. At ka = 1.0 drag s to −30: the node moves out and δ₀ = −46.8°, near the hard-sphere −57.3°. At ka = 0.2, s = 2.5 gives δ₀ = 88.9° and a full bar; s = 20, a well eight times deeper, gives δ₀ = 179.1° and an empty bar.
PHASE SHIFT δ₀0.845 rad
σ₀ ÷ (4π/k²) = sin²δ₀0.560
σ₀ ÷ πa²2.24
LIMIT 4π/k² ÷ πa²4.00
Live interpretationPHASE SHIFT δ₀: 0.845 rad. σ₀ ÷ (4π/k²) = sin²δ₀: 0.560. σ₀ ÷ πa²: 2.24. LIMIT 4π/k² ÷ πa²: 4.00
Catch the common trap
Explain before calculating.
A spinless particle with k = 0.50 fm⁻¹ scatters from a short-range central potential. A phase-shift analysis gives δ₀ = 60° and δ₁ = 30°, with every higher phase shift negligible. What is the total cross-section?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA numerical solution of the s-wave radial equation for a short-range attractive potential of range a = 2.0 fm, at k = 0.80 fm⁻¹, shows u = rR with its first node beyond the range at r = 3.10 fm. Find δ₀, the s-wave cross-section σ₀, and the fraction of the s-wave unitarity limit it reaches.
- Outside the range u ∝ sin(kr + δ₀), so its nodes sit where kr + δ₀ = nπ. This node has kr = 0.80 fm⁻¹ × 3.10 fm = 2.480.
- Take n = 1: δ₀ = π − 2.480 = 0.662 rad = 37.9°. Any other n adds a multiple of π, which changes nothing that can be measured.
- Check the sign: the free wave's node would sit at r = π/k = 3.93 fm, so the potential has pulled it in by 0.83 fm = δ₀/k, as attraction should.
- 4π/k² = 4π/(0.80 fm⁻¹)² = 19.63 fm², so σ₀ = 19.63 fm² × sin²37.9° = 19.63 fm² × 0.3775 = 7.41 fm² = 0.0741 b.
- The fraction of the s-wave ceiling is sin²δ₀ = 0.3775. With ka = 1.6 the p wave also reaches the potential, so σ₀ is not yet the total: the p wave alone could add up to 3 × 19.63 = 58.9 fm².
Answerδ₀ = 37.9° (mod 180°); σ₀ = 7.41 fm² = 0.0741 b, which is 37.7% of the s-wave limit 4π/k² = 19.6 fm².
MediumIn the spin-singlet state the neutron–proton force is central, so their relative motion is this lesson's spinless problem, with reduced mass μc² = 469.46 MeV. Model the force as a square well of depth V₀ = 15.1 MeV and range a = 2.50 fm. A 1.00 MeV neutron strikes a proton at rest, so E = 0.500 MeV in the centre-of-mass frame. Find δ₀ by matching at r = a, then σ₀, and compare with the measured phase shift.
- With ħc = 197.33 MeV fm: k = √(2μc²E)/ħc = √(2 × 469.46 × 0.500)/197.33 = 0.1098 fm⁻¹ and K = √(2μc²(E + V₀))/ħc = √(2 × 469.46 × 15.600)/197.33 = 0.6133 fm⁻¹, so ka = 0.2745 and Ka = 1.5333.
- Match u′/u at r = a: inside, u = sin Kr gives K cot Ka; outside, u = sin(kr + δ₀) gives k cot(ka + δ₀). So tan(ka + δ₀) = (k/K) tan Ka = 0.1790 × 26.65 = 4.771. Ka is below π/2, so u and u′ are both positive at r = a and ka + δ₀ = arctan 4.771 = 1.3642 rad.
- δ₀ = 1.3642 − 0.2745 = 1.0897 rad = 62.4°. Checks: V₀ → 0 makes K → k and δ₀ → 0. Ka sits just below π/2 because 15.1 MeV is just short of the 16.4 MeV at which this well would bind, which is why δ₀ is large and sensitive: V₀ = 15.0 or 15.2 MeV gives 61.0° or 63.9°.
- 4π/k² = 4π/(0.1098 fm⁻¹)² = 1042 fm² = 10.4 b, so σ₀ = 1042 fm² × sin²62.4° = 1042 fm² × 0.786 = 819 fm² = 8.19 b, 79% of the s-wave ceiling.
- Higher waves: l = 1 needs an impact parameter near 1/k = 9.1 fm, far outside a = 2.50 fm, and the same matching for l = 1 gives δ₁ = 0.08°. The singlet cross-section here is s wave alone.
- Compare: phase-shift analyses of measured neutron–proton data give about 62° at this energy (quoted from a table), within half a degree of the model; that measured phase gives σ₀ = 8.1 b.
Answerδ₀ = 62.4°, against about 62° from measured data; σ₀ = 8.19 b, 79% of the unitarity limit 4π/k² = 10.4 b.
HardA spinless particle meets an attractive square well of range a = 2.00 fm and strength K₀a = √(2mV₀) a/ħ = 2.00, arriving with k = 0.750 fm⁻¹. Find δ₀ and δ₁ by matching logarithmic derivatives, the total cross-section, and the forward and backward differential cross-sections, and justify dropping l ≥ 2.
- Inside, the wave number is K = √(k² + K₀²), so ka = 1.500 and Ka = √(1.500² + 2.00²) = 2.500. Since l ≈ kb and the well only reaches b ≤ a, waves up to l ≈ 1.5 feel it: keep s and p, and test d at the end.
- s wave: tan(ka + δ₀) = (ka/Ka) tan Ka = 0.600 × (−0.747) = −0.448. At r = a, u = sin 2.5 is positive and u′ ∝ cos 2.5 is negative, so ka + δ₀ lies in the second quadrant: π − 0.4214 = 2.7202 rad, and δ₀ = 1.2202 rad = 69.9°.
- p wave: inside R₁ = j₁(Kr), so γ₁a = Ka j₁′(Ka)/j₁(Ka) with j₁(2.5) = 0.4162 and j₁′(2.5) = j₀(2.5) − 2j₁(2.5)/2.5 = 0.2394 − 0.3330 = −0.0936, giving γ₁a = −0.562. At ka = 1.5: j₁ = 0.3962, n₁ = −0.6964, j₁′ = 0.1368, n₁′ = 0.8814.
- tan δ₁ = [1.5(0.1368) + 0.562(0.3962)]/[1.5(0.8814) − 0.562(0.6964)] = 0.4278/0.9307 = 0.460, so δ₁ = 24.7°. The same formula for l = 2 gives δ₂ = 1.1°, worth 0.04 fm², so dropping it costs 0.1%.
- With 4π/k² = 22.34 fm²: σ₀ = 22.34 × sin²69.9° = 19.7 fm² (ceiling 22.3 fm²) and σ₁ = 3 × 22.34 × sin²24.7° = 11.7 fm² (ceiling 67.0 fm²), so σ = 31.4 fm² = 0.314 b.
- Angles: k f(θ) = e(iδ₀) sin δ₀ + 3e(iδ₁) sin δ₁ cos θ, so k²|f|² = sin²δ₀ + 6 sin δ₀ sin δ₁ cos(δ₀ − δ₁) cos θ + 9 sin²δ₁ cos²θ. The three coefficients are 0.882, 6(0.939)(0.418)(0.704) = 1.658 and 9(0.1745) = 1.570, so dσ/dΩ is 4.110/k² = 7.31 fm² sr⁻¹ at θ = 0 and 0.794/k² = 1.41 fm² sr⁻¹ at θ = 180°.
Answerδ₀ = 69.9° (mod 180°) and δ₁ = 24.7°; σ = 31.4 fm² = 0.314 b; dσ/dΩ = 7.31 fm² sr⁻¹ forward and 1.41 fm² sr⁻¹ backward, a ratio of 5.2 made by s–p interference that cancels from σ.