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Scattering Theory

Attempt these from memory before reading.

  • What is a cross-section, and what are its dimensions?
  • Write the partial wave expansion of the scattering amplitude.
  • What is a phase shift, and what does δl=0\delta_l = 0 mean physically?
  • State the Born approximation and the condition for it to be valid.
  • What is the difference between a bound state and a resonance?

Check your answers against the text. Prompt 1 is the one that connects this page to experiment: everything on it exists to predict a number someone can measure.

Motivation: scattering is how anything is known

Section titled “Motivation: scattering is how anything is known”

Nobody has ever seen an atomic nucleus. What Rutherford saw in 1911 was that alpha particles occasionally came backwards, and he inferred from the angular distribution that the positive charge was concentrated in a region 10510^5 times smaller than the atom.

That is the pattern for the whole subject, and it has not changed. The electron’s magnetic moment, the proton’s internal structure, the existence of the up and down quarks — every one of these was inferred from how something scattered off something else. Scattering theory is not one topic among many; it is the interface between theory and experiment at every scale.

The central object is the scattering amplitude f(θ)f(\theta). Far from the target the wave function is a plane wave plus an outgoing spherical wave:

ψ∼eikz+f(θ) eikrr,\psi \sim e^{ikz} + f(\theta)\,\frac{e^{ikr}}{r},

and f(θ)f(\theta) contains everything. The differential cross-section is ∣f(θ)∣2|f(\theta)|^2, the total cross-section is its integral, and resonances show up as rapid variation of its phase. Solving a scattering problem means finding one complex-valued function of one angle.

Why expand in partial waves. A central potential is spherically symmetric, so angular momentum is conserved and each ll evolves independently. That decomposes a problem in three dimensions into a stack of one-dimensional problems — one radial equation per ll — each with a single number to describe it, the phase shift δl\delta_l. The amplitude becomes

f(θ)=1k∑l(2l+1) eiδlsin⁡δl Pl(cos⁡θ),f(\theta) = \frac{1}{k}\sum_l (2l+1)\,e^{i\delta_l}\sin\delta_l\,P_l(\cos\theta),

and the physics of a complicated 3D collision is now a list of angles. This is the same move as diagonalising an operator: the basis in which the problem decouples is the one worth using.

For a spherically symmetric potential V(r)V(r) The scattering amplitude can be expanded in partial waves:

f(θ)=12ik∑l=0∞(2l+1)(e2iδl−1)Pl(cos⁡θ)=1k∑l=0∞(2l+1)eiδlsin⁡δl Pl(cos⁡θ)f(\theta) = \frac{1}{2ik}\sum_{l=0}^{\infty}(2l + 1)(e^{2i\delta_l} - 1)P_l(\cos\theta) = \frac{1}{k}\sum_{l=0}^{\infty}(2l + 1)e^{i\delta_l}\sin\delta_l\,P_l(\cos\theta)

Where δl\delta_l is the phase shift for partial wave ll.

Optical theorem:

σtotal=4πk Im f(0)=4πk2∑l=0∞(2l+1)sin⁡2δl\sigma_{\text{total} = \frac{4\pi}{k}\,\text{Im}\,f(0) = \frac{4\pi}{k^2}\sum_{l=0}^{\infty}(2l+1)\sin^2\delta_l}

Partial wave unitarity bound: sin⁡2δl≤1\sin^2\delta_l \leq 1 So the maximum contribution of partial wave ll to the cross section is:

σlmax⁡=4πk2(2l+1)\sigma_l^{\max} = \frac{4\pi}{k^2}(2l + 1)

For a weak potential, the scattering amplitude to first order is:

f(θ,ϕ)≈−m2πℏ2∫e−ik"⋅r′V(r′) d3r′f(\theta, \phi) \approx -\frac{m}{2\pi\hbar^2}\int e^{-i\mathbf{k}"\cdot\mathbf{r}'}V(\mathbf{r}')\,d^3r'

Where k′\mathbf{k}' is the scattered wave vector and q=k′−k\mathbf{q} = \mathbf{k}' - \mathbf{k} is the momentum transfer.

For the Yukawa potential V(r)=(V0/r)e−μrV(r) = (V_0/r)e^{-\mu r}:

f(θ)=−2mV0ℏ2(μ2+q2),q=2ksin⁡(θ/2)f(\theta) = -\frac{2m V_0}{\hbar^2(\mu^2 + q^2)}, \quad q = 2k\sin(\theta/2)

Setting μ=0\mu = 0 (Coulomb potential), this reproduces the Rutherford scattering formula.

When the scattering energy is near a quasi-bound state, the phase shift passes through π/2\pi/2 (resonance):

δl(E)≈δbg+arctan⁡ ⁣(Γ/2ER−E)\delta_l(E) \approx \delta_{\text{bg} + \arctan\!\left(\frac{\Gamma/2}{E_R - E}\right)}

Where ERE_R is the resonance energy and Γ\Gamma is the width. The cross section has the Breit—Wigner form:

σl(E)=4πk2(2l+1)(Γ/2)2(E−ER)2+(Γ/2)2\sigma_l(E) = \frac{4\pi}{k^2}(2l+1)\frac{(\Gamma/2)^2}{(E - E_R)^2 + (\Gamma/2)^2}

At resonance (E=ERE = E_R): σlmax⁡=4πk2(2l+1)\sigma_l^{\max} = \frac{4\pi}{k^2}(2l+1) (unitarity limit).

Worked Example 12.1: Low-Energy Scattering and Scattering Length

For ss-wave scattering (l=0l = 0) at low energy (ka≪1ka \ll 1), only the l=0l = 0 phase shift contributes:

σ≈4πk2sin⁡2δ0≈4πas2\sigma \approx \frac{4\pi}{k^2}\sin^2\delta_0 \approx 4\pi a_s^2

Where the scattering length asa_s is defined by kcot⁡δ0→−1/ask\cot\delta_0 \to -1/a_s as k→0k \to 0.

For a hard sphere of radius aa: δ0=−ka\delta_0 = -ka (exact), giving as=aa_s = a and σ=4πa2\sigma = 4\pi a^2 (four times the geometric cross section πa2\pi a^2 --- a purely quantum result).

For the 3^3 He—4^4 He system: as≈1.4a_s \approx 1.4 nm (positive, indicating a repulsive effective potential). For the neutron—proton system (triplet): as≈5.4a_s \approx 5.4 fm (positive, with a bound state --- the deuteron). For singlet: as≈−23.7a_s \approx -23.7 fm (negative, indicating a virtual state).

Worked Example 12.2: Born Approximation for a Gaussian Potential

Consider V(r)=V0 e−r2/(2a2)V(r) = V_0\,e^{-r^2/(2a^2)}.

f(θ)=−m2πℏ2∫e−iq⋅rV0e−r2/(2a2) d3rf(\theta) = -\frac{m}{2\pi\hbar^2}\int e^{-i\mathbf{q}\cdot\mathbf{r}}V_0 e^{-r^2/(2a^2)}\,d^3r =−mV02πℏ2(2πa2)3/2e−q2a2/2=−mV0ℏ2(2π)1/2a3 e−2k2a2sin⁡2(θ/2)= -\frac{m V_0}{2\pi\hbar^2}(2\pi a^2)^{3/2}e^{-q^2 a^2/2} = -\frac{m V_0}{\hbar^2}(2\pi)^{1/2}a^3\,e^{-2k^2a^2\sin^2(\theta/2)}

The total cross section:

σ=∫∣f∣2 dΩ=2π∫0π∣f∣2sin⁡θ dθ\sigma = \int |f|^2\,d\Omega = 2\pi\int_0^\pi |f|^2\sin\theta\,d\theta

At low energy (ka≪1ka \ll 1): f≈−mV0ℏ2(2π)1/2a3f \approx -\frac{mV_0}{\hbar^2}(2\pi)^{1/2}a^3 (independent of θ\theta), giving:

σ≈4π(mV0ℏ2)22π a6=8π2m2V02a6ℏ4\sigma \approx 4\pi\left(\frac{mV_0}{\hbar^2}\right)^2 2\pi\,a^6 = \frac{8\pi^2 m^2 V_0^2 a^6}{\hbar^4}

The Born approximation is valid when ∣V0∣≪ℏ2/(ma2)|V_0| \ll \hbar^2/(ma^2) I.e., the potential is weak compared to the kinetic energy associated with the length scale aa.

  1. Symmetrisation applies to the full wavefunction: For fermions, the overall wavefunction (spatial ⊗\otimes spin ⊗\otimes any other degrees of freedom) must be antisymmetric. A symmetric spatial part requires an antisymmetric spin part (singlet), and vice versa. Do not apply (anti)symmetrisation to spatial and spin parts separately without ensuring the correct combined symmetry.

  2. The variational principle gives an upper bound: Etrial≥E0E_{\text{trial} \geq E_0} always. If you obtain a variational energy lower than the known exact ground state energy, you have made an error in the calculation (wrong normalisation, incorrect matrix element, or the trial function is not in the correct Hilbert space).

  3. Fermi’s Golden Rule applies to transitions to a continuum: For transitions to discrete states, use the Rabi formula instead. The density of states ρ(Ef)\rho(E_f) is essential --- if it is zero, the transition rate is zero regardless of the matrix element.

  4. The Born approximation assumes a weak potential: The condition is m∣V0∣a/ℏ2≪1m|V_0|a/\hbar^2 \ll 1 where aa is the range of the potential. For strong potentials (like the nuclear potential or hard spheres), the Born approximation gives qualitatively wrong results. Use partial wave analysis instead.

  5. Resonances require careful treatment: Near a resonance, perturbation theory breaks down. The Breit—Wigner formula is non-perturbative in the width Γ\Gamma. The scattering length can be much larger than the range of the potential near a resonance (the unitarity limit).

Problem 19: Exchange Energy in Lithium

Lithium (Z=3Z = 3) has the electron configuration 1s22s11s^2 2s^1. Using the variational method with ZeffZ_{\text{eff}} for the 1s1s electrons:

(a) Calculate ZeffZ_{\text{eff}} for the 1s1s electrons, treating the 2s2s electron as a perturbation.

(b) Calculate the ionisation energy (removing the 2s2s electron) and compare with the experimental value of 5.39 eV.

(c) Explain why the 2s2s electron is effectively screened by Zeff≈1.26Z_{\text{eff} \approx 1.26}.

Solution:

(a) For the 1s1s electrons, the effective charge is reduced from Z=3Z = 3 by screening from the other 1s1s electron and partial penetration of the 2s2s electron. The 1s1s electrons screen each other partially: using the helium result, Zeff(1s)≈Z−5/16=3−0.3125=2.69Z_{\text{eff}(1s) \approx Z - 5/16 = 3 - 0.3125 = 2.69}.

(b) The 2s2s electron sees an effective nuclear charge of Zeff(2s)≈3−2×0.85=1.3Z_{\text{eff}(2s) \approx 3 - 2 \times 0.85 = 1.3} (Slater’s rules). The energy:

E_{2s} = -\frac{Z_{\text{eff}^2}{n^2}\times 13.6\ \text{eV} = -\frac{1.3^2}{4}\times 13.6 = -\frac{1.69}{4}\times 13.6 = -5.75\ \text{eV}}

The ionisation energy is ∣E2s∣=5.75|E_{2s}| = 5.75 eV, close to the experimental 5.39 eV. The discrepancy reflects the crudeness of the Slater screening constants.

(c) The 2s2s electron has significant radial extent beyond the 1s1s core, so it sees a nearly bare nuclear charge at small rr but is screened by both 1s1s electrons at large rr. The effective charge Zeff≈1.26Z_{\text{eff} \approx 1.26} (using Hartree—Fock) represents this average screening.

Problem 20: Partial Wave Analysis for Square Well

Consider scattering from the attractive square well V(r)=−V0V(r) = -V_0 for r<ar < a and V(r)=0V(r) = 0 for r>ar > a.

(a) Show that the ss-wave phase shift satisfies:

δ0=−ka+arctan⁡ ⁣(kκtan⁡(κa))\delta_0 = -ka + \arctan\!\left(\frac{k}{\kappa}\tan(\kappa a)\right)

Where κ=2m(V0+E)/ℏ\kappa = \sqrt{2m(V_0 + E)}/\hbar and k=2mE/ℏk = \sqrt{2mE}/\hbar.

(b) Show that a bound state exists at energy E=−∣E∣E = -|E| when κ0a=π/2\kappa_0 a = \pi/2 where κ0=2m(V0−∣E∣)/ℏ\kappa_0 = \sqrt{2m(V_0 - |E|)}/\hbar.

(c) Show that the scattering length diverges as a new bound state appears.

Solution:

(a) Inside the well (r<ar < a), the radial wavefunction for l=0l = 0 is u(r)=Asin⁡(κr)u(r) = A\sin(\kappa r). Outside (r>ar > a), u(r)=Bsin⁡(kr+δ0)u(r) = B\sin(kr + \delta_0).

Matching uu and u′u' at r=ar = a:

κcos⁡(κa)=kcos⁡(ka+δ0)/sin⁡(ka+δ0)⋅k\kappa\cos(\kappa a) = k\cos(ka + \delta_0)/\sin(ka + \delta_0)\cdot k

Wait: κcot⁡(κa)=kcot⁡(ka+δ0)\kappa\cot(\kappa a) = k\cot(ka + \delta_0).

cot⁡(ka+δ0)=κkcot⁡(κa)\cot(ka + \delta_0) = \frac{\kappa}{k}\cot(\kappa a) ka+δ0=\arccot ⁣(κkcot⁡(κa))=arctan⁡ ⁣(kκtan⁡(κa))ka + \delta_0 = \arccot\!\left(\frac{\kappa}{k}\cot(\kappa a)\right) = \arctan\!\left(\frac{k}{\kappa}\tan(\kappa a)\right) δ0=−ka+arctan⁡ ⁣(kκtan⁡(κa))\delta_0 = -ka + \arctan\!\left(\frac{k}{\kappa}\tan(\kappa a)\right)

(b) A bound state has E<0E < 0 So k=iκ′k = i\kappa' where κ′=2m∣E∣/ℏ\kappa' = \sqrt{2m|E|}/\hbar. The bound state condition is that the exterior solution decays exponentially: u(r)=Be−κ′ru(r) = Be^{-\kappa' r}. Matching:

κcot⁡(κa)=−κ′\kappa\cot(\kappa a) = -\kappa'

As ∣E∣→0|E| \to 0: κ′→0\kappa' \to 0 So κcot⁡(κa)→0\kappa\cot(\kappa a) \to 0 Giving κa=π/2\kappa a = \pi/2 (the threshold for the first bound state).

(c) The scattering length as=−lim⁡k→0(δ0/k)a_s = -\lim_{k \to 0}(\delta_0/k). As κa→π/2\kappa a \to \pi/2, tan⁡(κa)→∞\tan(\kappa a) \to \infty So:

δ0≈−ka+arctan⁡(∞)=−ka+π/2\delta_0 \approx -ka + \arctan(\infty) = -ka + \pi/2 as=−1k(−ka+π2)=a−π2ka_s = -\frac{1}{k}\left(-ka + \frac{\pi}{2}\right) = a - \frac{\pi}{2k}

As k→0k \to 0: as→±∞a_s \to \pm\infty (diverges), changing sign as the bound state appears.

Scattering theory extracts information about forces by firing particles at targets and analyzing how they deflect. The differential cross-section tells you the probability of scattering into a given solid angle, and it is directly measurable. For a central potential, the scattering amplitude is decomposed into partial waves, each corresponding to a specific angular momentum. The scattering length characterizes low-energy scattering and diverges when a new bound state appears, signaling a resonance. The optical theorem connects the total cross-section to the forward scattering amplitude, reflecting the fact that particles removed from the forward beam must go somewhere. Partial wave analysis turns a complex three-dimensional scattering problem into a sum of one-dimensional problems.

ClaimTrue?Counterexample
The quantum cross-section of a hard sphere matches the classical onenolow-energy ss-wave scattering gives σ=4πa2\sigma = 4\pi a^2, four times the classical πa2\pi a^2, because the wave diffracts around the sphere
A resonance is a bound statenoit is a quasi-bound state: metastable, with a finite lifetime, appearing as a pole of the SS-matrix on the unphysical sheet rather than on the real axis
The Born approximation needs a weak potentialincompleteit needs ∣V∣\lvert V\rvert small compared with the kinetic energy and the potential to vary slowly on the scale of the wavelength
Higher partial waves matter at low energynothe centrifugal barrier suppresses them, and only the l=0l = 0 wave contributes; that is why low-energy cross-sections are isotropic
Phase shifts are small for weak scatteringnot necessarilya resonance makes δl\delta_l swing through π\pi rapidly even for a weak potential
Cross-sections cannot exceed the geometric sizenothe hard-sphere result above is larger, and a resonance can exceed it by orders of magnitude
Scattering conserves probability, so SS is unitaryyesand this is the constraint that makes the phase shifts real for real potentials

The first row is the one that most strongly marks the difference between classical and quantum intuition. A classical ball bearing thrown at a basketball is shadowed by it; a long-wavelength quantum particle is diffracted by it, and the forward diffraction adds exactly the extra 3πa23\pi a^2.

The third row is the one that catches people applying the Born approximation to a Coulomb field, where it fails not because the potential is strong but because it is long-ranged, and the amplitude has to be treated separately.

  • The inverse scattering problem. Given the scattering data, can the potential be reconstructed? In one dimension the answer is yes, exactly, via the Gel’fand-Levitan equation — and that result is the mathematical foundation of soliton theory.
  • Few-body scattering. Three bodies do not reduce to pairwise problems, and the Faddeev equations are the framework; Efimov states, where three bodies bind even though no pair does, are the celebrated prediction.
  • Scattering in quantum field theory. The SS-matrix is the central object and its analyticity properties are what make dispersion relations possible; the modern amplitudes programme tries to compute it without a Hamiltonian at all.
  • Cold atoms and Feshbach resonances. A magnetic field tunes a resonance through threshold, letting experimentalists dial the interaction strength between two atoms from zero to infinity. That control is what makes precision many-body physics possible.
  • Scattering and quantum chaos. How chaotic classical dynamics shows up in scattering data — through the statistics of resonance spacings — connects to random matrix theory.

Mix these with one-dimensional problems, angular momentum, approximation methods, Symplectic Structure and KAM Theory and Ordinary Differential Equations.

  1. (Angular momentum.) Derive the partial wave expansion by expanding a plane wave in spherical harmonics and identifying each ll term as an incoming and an outgoing wave. Show that only the phase of each can change, and that this is why one number per ll suffices.
  2. (Operators.) Prove that the SS-matrix is unitary for a real potential, and explain why unitarity is what makes the phase shifts real. What goes wrong with absorption?
  3. (One-dimensional problems.) Show that the 1D transmission and reflection coefficients of the barrier problem are the l=0l = 0 partial-wave result in disguise, and identify what the higher ll waves add in three dimensions.
  4. (Approximation methods.) Derive the Born approximation as first-order perturbation theory in V^\hat{V}, and show that it is the Fourier transform of the potential evaluated at the momentum transfer. Then explain why that makes electron diffraction a way of photographing potentials.
  5. (Classical mechanics.) Derive the Rutherford cross-section classically from the Coulomb force, and show that quantum mechanics gives the same answer. That agreement is unusual — identify what makes it work.
  6. (Research.) State the Gel’fand-Levitan equation and explain in what sense it solves the inverse problem exactly. Why does the reconstruction need the bound state data as well as the scattering data?