University Quantum Mechanics II · Quantum Scattering Theory · 15.7
Resonances and the Ramsauer-Townsend effect
Two features shape low-energy cross-sections, both read off a phase shift plotted against energy. Where it climbs steeply through π/2, one partial wave fills its unitarity limit and the projectile lingers; where it passes through π, that wave vanishes. You will derive the peak, time it, and meet the atom slow electrons cannot see.
Build the model
Connect the measurement to the mechanism.
A resonance is a partial wave whose phase shift changes by nearly π within a narrow band of energy, and the cause is a quasi-bound state: a level inside an attractive well that leaks out through a barrier, the centrifugal term ħ²l(l + 1)/2mr² when l ≥ 1 or a Coulomb barrier between charged fragments. Near the level's energy Eᵣ the steep function is δₗ; the smooth one is cot δₗ, which crosses zero. One term of its Taylor series, cot δₗ ≈ −2(E − Eᵣ)/Γ, turns sin²δₗ into a Lorentzian of full width Γ and σₗ into the Breit-Wigner form, which touches the unitarity limit 4π(2l + 1)/k² at Eᵣ whatever the potential.
The width is also a clock: a wave packet is delayed by 2ħ dδₗ/dE, and a short pulse that excites the level leaves it decaying as e(−Γt/ħ), so it lives about ħ/Γ. The same bookkeeping explains the opposite extreme. A partial wave contributes nothing when its phase shift is a multiple of π, and at energies so low that only the s wave matters, a δ₀ crossing π leaves the target almost invisible: the Ramsauer-Townsend minimum of slow electrons in argon.
The model fails several widths from Eᵣ, where the threshold law δₗ ∝ k(2l+1) takes over; under a background phase, which skews the peak; when resonances overlap or inelastic channels open; and in argon itself, whose minimum needs a many-electron potential with a long polarisation tail that no square well reproduces.
- Simple definition
- A scattering resonance is an energy Eᵣ where one phase shift δₗ climbs steeply through π/2, so that partial wave reaches its unitarity limit and the projectile is held for about ħ/Γ; the Ramsauer-Townsend effect is δ₀ passing through π, which removes the s wave.
- Example
- A p-wave electron resonance at Eᵣ = 1.00 eV peaks at 12π/k² = 12π × 3.81 eV Ų ÷ 1.00 eV = 144 Ų, using ħ²/2mₑ = 3.81 eV Ų; if Γ = 0.20 eV, the electron is held for ħ/Γ = 0.658 eV fs ÷ 0.20 eV = 3.3 fs.
Derived: the first Taylor term of cot δₗ, which is smooth where δₗ passes π/2. Trust it within a few Γ of Eᵣ, and only when Γ ≪ Eᵣ.
E, Eᵣ and Γ in J (or all in eV); δₗ in rad; Γ is defined by dδₗ/dE = 2/Γ at E = Eᵣ
Derived from sin²δₗ = 1/(1 + cot²δₗ). At Eᵣ it equals the unitarity limit 4π(2l + 1)/k² for any potential, and Γ is the full width at half maximum.
σₗ in m²; k in m⁻¹ with E = ħ²k²/2m, m the reduced mass; E, Eᵣ, Γ in one energy unit
Derived by stationary phase on the outgoing wave e(i(kr + 2δₗ)). At Eᵣ it is 4ħ/Γ; a hard sphere, δ₀ = −kR, gives −2R/v, an advance.
Δt in s; ħ = 1.055 × 10⁻³⁴ J s; E in J; δₗ in rad
Argued, not derived in full: the Lorentzian amplitude's Fourier transform decays as e(−Γt/2ħ), so the probability of finding the particle still trapped falls as e(−t/τ).
τ in s, Γ in J; with ħ = 6.582 × 10⁻¹⁶ eV s, a width of 1.00 eV means τ = 0.658 fs
Derived. The peak moves to Eᵣ − (Γ/2) tan δbg and a zero opens at Eᵣ + (Γ/2) cot δbg: the skewed line shape that a symmetric fit misreads.
δbg a slowly varying background phase in rad; δᵣₑₛ runs from 0 to π; other symbols as above
Derived by matching at R. As kR → 0 it becomes tan KR = KR, first nonzero root KR = 4.493: for an electron and R = 1.00 Å, V₀ = 76.9 eV.
R the well radius in m, V₀ its depth in J, k and K in m⁻¹; the condition for δ₀ = nπ
Expand cot δₗ, not δₗ
At a resonance δₗ climbs through π/2 so steeply that its Taylor series about Eᵣ converges only within Γ/2 of it, and the cross-section needs sin²δₗ anyway. The smooth function is cot δₗ, which simply crosses zero at Eᵣ. Keep its first term: cot δₗ ≈ −(E − Eᵣ)(dδₗ/dE)/sin²δₗ, evaluated at Eᵣ, which is −2(E − Eᵣ)/Γ once Γ is defined by dδₗ/dE = 2/Γ there. Then sin²δₗ = 1/(1 + cot²δₗ) = (Γ²/4)/[(E − Eᵣ)² + Γ²/4], and multiplying by the (4π/k²)(2l + 1) of the partial-wave sum gives the Breit-Wigner form. Check it three ways. The fraction is dimensionless, so σₗ is an area. At E = Eᵣ ± Γ/2, cot δₗ = ∓1, δₗ = 45° or 135°, and σₗ is half its peak, so Γ is the full width at half maximum. At Eᵣ, σₗ = 4π(2l + 1)/k², the unitarity limit: the strength of the potential decides where the peak sits and how wide it is, never how tall. The amplitude says the same thing compactly: aₗ = 1/(cot δₗ − i) = (Γ/2)/(Eᵣ − E − iΓ/2), a pole at the complex energy Eᵣ − iΓ/2.
A resonance needs something to hold the particle
The steep rise happens when the well holds a level at positive energy that can only leak out through a barrier. For l ≥ 1 the barrier is centrifugal. Take an electron in a square well of radius 1.00 Å, solve the p-wave matching numerically, and deepen the well. At V₀ = 34.3 eV, δ₁ passes 90° at Eᵣ = 1.33 eV with Γ = 1.23 eV, hardly a resonance. At 36.6 eV the level sits at 0.352 eV with Γ = 0.149 eV; at 37.1 eV, 0.172 eV with Γ = 0.050 eV. At 37.6 eV, where k₀R = π, it binds. The lower the level, the more barrier it tunnels through, and Γ/(kᵣR)³ settles near 1.33ħ²/2mR²: this is the threshold law Γ ∝ k(2l+1), quoted here from the barrier penetrability, with l = 1. The s wave has no centrifugal barrier, so a bare well gives no sharp s-wave peak; a level near zero energy shows up instead as a large scattering length. Sharp s-wave resonances need another barrier, such as the Coulomb repulsion between two charged fragments.
The width is a clock
A wave packet in partial wave l leaves with phase kr + 2δₗ − Et/ħ. Its peak sits where that phase is stationary in k, which gives r = v(t − Δt) with Δt = 2ħ dδₗ/dE, the Wigner delay. At Eᵣ, dδₗ/dE = 2/Γ, so a packet narrow in energy is held back by 4ħ/Γ. A hard sphere, δ₀ = −kR, gives Δt = −2R/v: the wave turns round at the surface, not the centre. A pulse short compared with ħ/Γ instead excites the whole line, and the Fourier transform of aₗ ∝ 1/(E − Eᵣ + iΓ/2) is a wave ∝ e(−iEᵣ t/ħ)e(−Γt/2ħ) for t > 0, whose probability decays with lifetime τ = ħ/Γ. This is an argument, not a full derivation: it ignores how k and Γ vary across the line. Nuclear tables use this conversion. The ground state of ⁸Be is an s-wave resonance 91.84 keV above two alpha particles, held by their Coulomb barrier, 4 × 1.44 MeV fm ÷ r ≈ 1.4 MeV at r = 4 fm. Its measured width, 5.57 eV, gives τ = 6.582 × 10⁻¹⁶ eV s ÷ 5.57 eV = 1.18 × 10⁻¹⁶ s, the half-life τ ln 2 = 8.2 × 10⁻¹⁷ s that the tables list, and Γ/Eᵣ = 6 × 10⁻⁵: a line that narrow needs a barrier over ten times higher than the level.
A background phase skews the peak
Real phase shifts rarely start from zero at a resonance. Write δₗ = δbg + δᵣₑₛ, where δᵣₑₛ rises from 0 to π across the line with tan δᵣₑₛ = (Γ/2)/(Eᵣ − E) and δbg varies slowly enough to be treated as constant. The cross-section is still (4π/k²)(2l + 1) sin²δₗ, but now the maximum comes where δbg + δᵣₑₛ = π/2, at E = Eᵣ − (Γ/2) tan δbg, and there is also a zero where δbg + δᵣₑₛ = π, at E = Eᵣ + (Γ/2) cot δbg. With δbg = 30° and Γ = 0.20 eV, the peak sits 0.058 eV below Eᵣ and the zero 0.173 eV above it, the two separated by Γ/sin 2δbg = 0.231 eV; at Eᵣ itself the cross-section is only cos²30° = 0.75 of the limit. At δbg = 90° the peak turns into a pure dip. This skewed shape is the Fano profile, named here and not developed; its practical lesson is that fitting a symmetric Breit-Wigner to a skewed line misplaces Eᵣ and misreads Γ.
Through π, a partial wave switches off
Since σₗ ∝ sin²δₗ, a partial wave contributes nothing whenever δₗ is a multiple of π. At low energy every wave with l ≥ 1 is already small, because δₗ ∝ k(2l+1) for a short-range potential, so if δ₀ passes through π there, the whole target nearly vanishes. The square well shows the mechanism. Matching sin Kr inside to sin(kr + δ₀) outside gives K cot KR = k cot(kR + δ₀), and setting δ₀ = nπ leaves K cot KR = k cot kR: the interior wave reaches R with exactly the logarithmic slope a free wave would have there, so outside nothing records that a well was present. As kR → 0 the right side tends to 1/R and the condition becomes tan KR = KR, with first nonzero root KR = 4.493. For an electron and R = 1.00 Å that is V₀ = 3.81 eV Ų × (4.493/1.00 Å)² = 76.9 eV, a deep well whose interior wave has gained a node, not a weak one. It is the three-dimensional cousin of the perfect transmission over a square well met in one-dimensional scattering, where a whole number of half-wavelengths fits across the well.
Argon: a minimum a short-range well cannot make
The measured momentum-transfer cross-section of slow electrons in argon falls by more than an order of magnitude to a minimum at about 0.24 eV, then climbs again; krypton and xenon behave alike. Its zero-energy value follows from the measured scattering length A = −1.45 a₀: 4πA² = 7.4 × 10⁻²⁰ m². A short-range potential cannot pair so large a cross-section with so low a zero. Its tan δ₀ = −ak + O(k³) holds only odd powers of k, so a zero at kR ≪ 1 needs the k³ term to cancel the first, which forces a down to order R(kR)²: a target nearly invisible at every low energy. An electron near argon feels the induced-dipole tail V = −αd e²/(8πε₀r⁴), with αd = 11.07 a₀³, and the modified effective-range theory of O'Malley, Spruch and Rosenberg, quoted, not derived, adds an even term: tan δ₀ = −Ak − (παd/3a₀)k² + …. With A negative the two terms cancel at k = 3|A|a₀/(παd) = 0.125 a₀⁻¹, that is E = 13.6 eV × 0.125² = 0.21 eV, within about 11% of the measured 0.24 eV. The rest is a many-electron problem, exchange with eighteen electrons included, that this unit only quotes.
Change one variable at a time
Make the relationship visible.
Start at δbg = 0°: the peak sits on Eᵣ at the limit. Shrink Γ and the lifetime grows. Raise δbg to 30°: the peak slides below Eᵣ and a zero opens above it, where δₗ crosses 180°; a negative δbg puts the zero below, at 0°. That sin δ = 0 is what empties argon's s wave.
LIFETIME τ = ħ/Γ4.1 fs
DELAY AT Eᵣ, 4ħ/Γ16.5 fs
PEAK, WHERE δ = 90°1.000 eV
σₗ(Eᵣ) ÷ LIMIT1.00
Live interpretationLIFETIME τ = ħ/Γ: 4.1 fs. DELAY AT Eᵣ, 4ħ/Γ: 16.5 fs. PEAK, WHERE δ = 90°: 1.000 eV. σₗ(Eᵣ) ÷ LIMIT: 1.00
Catch the common trap
Explain before calculating.
For electrons, spin ignored, a p-wave (l = 1) resonance sits at Eᵣ = 0.50 eV with Γ = 20 meV. Deepening the well moves it to Eᵣ = 0.25 eV, and near threshold Γ ∝ k³. With ħ²/2mₑ = 3.81 eV Ų and ħ = 0.658 eV fs, what are the new peak σ₁ and the lifetime of the quasi-bound state?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyElectrons, spin ignored, show a p-wave (l = 1) resonance at Eᵣ = 1.20 eV with Γ = 0.080 eV. Find the peak partial cross-section, the phase shift and σ₁ at E = 1.24 eV, and the lifetime of the quasi-bound state. Use ħ²/2mₑ = 3.81 eV Ų.
- Check the model applies: Γ/Eᵣ = 0.080/1.20 = 0.067, so the line is narrow and the linear cot δ₁ is safe within a few widths.
- At Eᵣ the p wave reaches its unitarity limit: σ₁ = 12π/k² = 12π(ħ²/2mₑ)/Eᵣ = 12π × 3.81 eV Ų ÷ 1.20 eV = 120 Ų = 1.20 × 10⁻¹⁸ m².
- At 1.24 eV: cot δ₁ = −2(1.24 − 1.20)/0.080 = −1, so δ₁ = 135° and sin²δ₁ = 0.50.
- σ₁ = 0.50 × 12π × 3.81 eV Ų ÷ 1.24 eV = 57.9 Ų, slightly under half the peak because the limit itself falls as 1/E.
- Lifetime: τ = ħ/Γ = 0.658 eV fs ÷ 0.080 eV = 8.2 fs.
Answerσ₁ = 120 Ų (1.20 × 10⁻¹⁸ m²) at 1.20 eV; δ₁ = 135° and σ₁ = 57.9 Ų at 1.24 eV; τ = 8.2 fs.
MediumA phase-shift analysis of electron scattering finds δ₁ = 40° at 0.950 eV and δ₁ = 110° at 1.010 eV, with no background phase. Find Eᵣ and Γ, the lifetime, the Wigner delay at Eᵣ, and how far an unscattered electron of energy Eᵣ would travel in that delay.
- Write cot δ₁ = 2(Eᵣ − E)/Γ at both energies: cot 40° = 1.1918 = 2(Eᵣ − 0.950 eV)/Γ and cot 110° = −0.3640 = 2(Eᵣ − 1.010 eV)/Γ.
- Subtract: 1.1918 + 0.3640 = 2(0.060 eV)/Γ, so Γ = 0.120 eV ÷ 1.5557 = 0.0771 eV.
- Back-substitute: Eᵣ = 0.950 eV + 1.1918 × 0.0771 eV/2 = 0.996 eV. Check with the second point: 1.010 − 0.3640 × 0.0386 = 0.996 eV. Γ/Eᵣ = 0.077, so the linear expansion is fair.
- Lifetime τ = ħ/Γ = 0.658 eV fs ÷ 0.0771 eV = 8.5 fs; the delay of a narrow packet at Eᵣ is 2ħ(2/Γ) = 4ħ/Γ = 34 fs.
- Speed at Eᵣ: v = c√(2Eᵣ/mₑc²) = 3.00 × 10⁸ m s⁻¹ × √(2 × 0.996/511 000) = 5.92 × 10⁵ m s⁻¹, so the delay is worth v × 34.1 fs = 2.0 × 10⁻⁸ m.
AnswerEᵣ = 0.996 eV, Γ = 0.077 eV, τ = 8.5 fs, Wigner delay 34 fs, equivalent to about 20 nm of free flight.
HardModel a slow electron on an atom as an attractive square well of radius R = 1.00 Å (ħ²/2mₑ = 3.81 eV Ų). Find the depth that puts the s-wave Ramsauer-Townsend zero at zero energy, then the depth that moves it to 0.240 eV, and that well's scattering length. Compare with argon, whose scattering length is A = −1.45 a₀.
- With δ₀ = nπ the matching condition is K cot KR = k cot kR. At kR → 0 the right side is 1/R, so tan KR = KR: KR = 4.49341, and V₀ = 3.81 eV Ų × (4.49341/1.00 Å)² = 76.9 eV.
- Near that root, tan x/x ≈ 1 + x₀(x − x₀) with x₀ = 4.49341, and tan kR/kR ≈ 1 + (kR)²/3. Equating gives KR = x₀ + (kR)²/(3x₀); with KR ≈ k₀R + (kR)²/(2x₀), where k₀² = 2mₑV₀/ħ², this means k₀R = x₀ − (kR)²/(6x₀).
- At 0.240 eV, (kR)² = 0.240/3.81 = 0.0630, so k₀R = 4.49341 − 0.0630/26.96 = 4.49107 and V₀ = 3.81 eV × 4.49107² = 76.847 eV, 0.080 eV shallower, just the E/3 that this expansion predicts. The zero is touchy: 0.01 eV of depth moves it by 0.03 eV, so the depth must be carried to the meV. The exact condition, solved numerically at V₀ = 76.847 eV, puts the zero at 0.241 eV.
- Scattering length a = R − tan(k₀R)/k₀ = +0.0104 Å, close to R(kR)²/6 = 0.0105 Å, so the zero-energy cross-section is 4πa² = 1.4 × 10⁻³ Ų.
- Argon has 4πA² = 4π(0.767 Å)² = 7.4 Ų, more than 5000 times larger, and a negative A. A short-range well makes a low-energy zero only by being almost invisible; argon's deep minimum between large values needs its long polarisation tail.
AnswerV₀ = 76.9 eV for a zero at zero energy; V₀ = 76.847 eV puts it at 0.24 eV (0.241 eV from the exact condition), with a = +0.0104 Å and 4πa² = 1.4 × 10⁻³ Ų, more than 5000 times below argon's 7.4 Ų.