University Quantum Field Theory I · Cross-Sections, Decay Rates and Unitarity · 5.6

The non-relativistic limit and the Yukawa potential

This is where field theory hands a result back to quantum mechanics and the two agree. One exchange diagram, taken to low speed, is the Born amplitude of a Yukawa potential, and the factors that make the match exact, 8π√s in one place and 2M at each vertex, show how relativistic bookkeeping differs from the kind you learned first.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Quantum Mechanics II began with a potential and computed scattering. Field theory begins with a Lagrangian and computes amplitudes, and nothing promises that a potential exists. At low speed one does, and this topic reads it off. Give the scalar Yukawa model a second complex scalar ψ′ of mass M′ with the same coupling g.

Then ψψ′ → ψψ′ has a single tree, φ exchange in the t channel, and in the centre-of-mass frame its amplitude g²/(|q|² + m²) depends on the momentum transfer alone, the mark of an instantaneous, static interaction. Dividing by 8π√s turns ℳ into the f whose square is dσ/dΩ; setting that f equal to the first Born amplitude −(μ/2π)Ṽ(q) gives Ṽ(q) = −ℳ/(4MM′), and so V(r) = −(g²/(4MM′)) e^(−mr)/(4πr): attractive, with range 1/m. The 4MM′ is not decoration. Relativistic states carry 2E in their norm, so a slow ψ acts as a source of strength g/(2M), and with that strength topic 1.09's energy of two static sources is the same function.

For identical ψ the second contraction supplies f(π − θ), the symmetrisation quantum mechanics imposes by hand. The match is first order and low speed by construction. It fails as the Born parameter 2μα_Y/m approaches 1, where the potential must be iterated and ψψ′ can bind; at relativistic speed; and at order g⁴, where unitarity demands an imaginary part no tree has.

Simple definition: The non-relativistic limit of an exchange amplitude names the potential whose first Born amplitude equals ℳ/(8π√s) at low speed; for scalar exchange between particles of masses M and M′ that potential has Ṽ(q) = −ℳ/(4MM′), a Yukawa potential.
Example: A φ of mass 140 MeV exchanged between a ψ and a ψ′ of 1.0 GeV each, with g = 1.0 GeV, gives range ħc/m = 1.41 fm and α_Y = g²/(16πMM′) = 0.0199, so V(1.0 fm) = −0.0199 × 197.3 MeV fm × e^(−0.709)/(1.0 fm) = −1.93 MeV.

The single tree for ψψ′ → ψψ′ℳ = g²/(m² − t) = g²/(|q|² + m²)

Topic 4.06's t-channel term, quoted. In the CM frame each particle keeps its energy, so t = −|q|² exactly: the static form needs no approximation yet.

g the coupling and m the φ mass, in GeV; q = p − p′ the CM momentum transfer, |q| = 2p sin(θ/2); ℳ dimensionless

The amplitude quantum mechanics calls ff(θ) = ℳ/(8π√s)

Derived from topic 5.04's dσ/dΩ = |ℳ|²/(64π²s), so that dσ/dΩ = |f|². At low speed √s → M + M′, the only place speed enters.

√s the CM energy in GeV; f in GeV⁻¹, times ħc = 0.1973 GeV fm for fm; elastic, so |p_f| = |pᵢ|

Reading the potential off the amplitudeṼ(q) = −ℳ/(4MM′)

Derived by setting f equal to the Born amplitude −(μ/2π)Ṽ, μ = MM′/(M + M′). The 4MM′ is the 2M and 2M′ that relativistic normalisation puts at the two vertices.

M and M′ the ψ and ψ′ masses in GeV; Ṽ(q) = ∫ d³r e^(−iq⋅r) V(r), in GeV⁻²; low speed only

The Yukawa potential between ψ and ψ′V(r) = −(g²/(4MM′)) e^(−mr)/(4πr)

Derived by Fourier transform. Attractive, and an energy, since g²/(4MM′) is a pure number. As m → 0 it becomes a 1/r tail whose Born cross-section is Rutherford's.

r in GeV⁻¹ (1 GeV⁻¹ = 0.1973 fm), V in GeV; range R = 1/m, or ħc/(mc²) = 1.41 fm for m = 139.6 MeV

Identical ψ: the u channel is f(π − θ)f_ψψ(θ) = f(θ) + f(π − θ)

Shown: UAQM-02.05's symmetrised amplitude, produced by two contractions with no symmetrisation imposed. At 90° the rate is 4|f(90°)|², twice the classical sum.

f(θ) from the t-channel term, f(π − θ) from the u-channel term, √s → 2M; σ integrates |f_ψψ|² over half the sphere

When one Born term is enoughb = 2μα_Y/m ≪ 1, with α_Y = g²/(16πMM′)

UQMII-15.09's low-energy condition, quoted, in this model's variables. At b = 1.680, quoted, the potential binds ψψ′, which no finite order of the Born series can show.

μ the reduced mass and m the φ mass, in GeV; b = f(0)/R, the forward amplitude over the range, is dimensionless

One diagram, and in the CM frame it is already static

Natural units, ħ = c = 1, with the metric (+, −, −, −). Topic 2.01's scalar Yukawa model, quoted, couples a complex ψ of mass M to a real φ of mass m through −gψ†ψφ, with g a mass. Add a second complex scalar ψ′ of mass M′ coupled the same way, −gψ′†ψ′φ. ψ number and ψ′ number are separately conserved, so ψψ′ → ψψ′ has no s channel, since the pair cannot annihilate into a φ, and no u channel, since no vertex turns a ψ into a ψ′. One tree is left, topic 4.06's t-channel exchange, quoted: ℳ = g²/(m² − t). Now work in the centre-of-mass frame. Elastic scattering keeps |p|, so each particle keeps its energy, the φ carries q = p − p′ with no time component, and t = −|q|² = −4p² sin²(θ/2) exactly. Hence ℳ = g²/(|q|² + m²) at any speed in this frame: a propagator with no energy dependence, the mark of an instantaneous interaction. Low speed has not been used yet.

From ℳ to f: the one place speed enters

Topic 5.04 gives dσ/dΩ = |ℳ|²/(64π²s) for elastic scattering in the CM frame, quoted. Define f(θ) = ℳ/(8π√s); then dσ/dΩ = |f|², which is how Quantum Mechanics II defined its scattering amplitude, and topic 5.07 shows that the phase matches too, since this f obeys σₜₒₜ = (4π/p) Im f(0). So f = g²/(8π√s (|q|² + m²)), derived. Now slow down. √s = √(M² + p²) + √(M′² + p²) = M + M′ + p²/(2μ) + …, with μ = MM′/(M + M′), so at low speed f → g²/(8π(M + M′)(|q|² + m²)). The correction is a relative p²/(2μ(M + M′)): for M = M′ = 1 GeV and p = 0.1 GeV, √s = 2.00998 GeV, a shift of 0.50%. That is the size of what the non-relativistic limit throws away, and dimensions are already right: g² over a squared mass is a pure number, so f is a length, GeV⁻¹, as a scattering amplitude must be.

Matching the Born amplitude gives the potential

Quoted from UQMII-15.08: two particles interacting through V(r) have, in the CM frame, the first Born amplitude f_B = −(μ/2π) Ṽ(q), with Ṽ(q) = ∫ d³r e^(−iq⋅r) V(r) and μ the reduced mass. Set f_B equal to the low-speed f: Ṽ(q) = −(2π/μ) g²/(8π(M + M′)(|q|² + m²)), and since μ(M + M′) = MM′, Ṽ(q) = −g²/(4MM′(|q|² + m²)), derived. The inverse transform, ∫ d³q e^(iq⋅r)/((2π)³(|q|² + m²)) = e^(−mr)/(4πr), is UQMII-15.09's, quoted. So V(r) = −(g²/(4MM′)) e^(−mr)/(4πr), derived. Four checks. g²/(4MM′) is dimensionless, so V has the dimension of 1/r, an energy. V is symmetric in the two particles, as an interaction between them must be. V is negative: scalar exchange attracts. And as m → 0, V becomes a 1/r potential whose Born cross-section is Rutherford's. Writing α_Y = g²/(16πMM′) gives the familiar V = −α_Y e^(−mr)/r.

Why each vertex carries 1/(2M)

A second route to the same Ṽ explains the factor. Topic 1.03 normalised states as ⟨p|k⟩ = 2Eₚ(2π)³δ³(p − k), quoted; non-relativistic quantum mechanics uses (2π)³δ³(p − k). At low speed each relativistic state is √(2M) times a non-relativistic one, and the first-order non-relativistic element is −iṼ(q) times (2π)⁴δ⁴(P_f − Pᵢ), the delta function of total energy and momentum. So iℳ = √(2M ⋅ 2M ⋅ 2M′ ⋅ 2M′) × (−iṼ) = (2M)(2M′)(−iṼ), the 2M belonging to the vertex where ψ enters and leaves and the 2M′ to the ψ′ vertex: Ṽ = −ℳ/(4MM′) again, derived. Equivalently, the slow field ψ ≈ e^(−iMt)χ/√(2M) turns gψ†ψφ into (g/(2M))χ†χφ, so a slow ψ is a static source of dimensionless strength g/(2M). Topic 1.09 found two static point sources of strength g to have energy −g² e^(−mr)/(4πr); for strengths g₁ and g₂ the cross term is −g₁g₂ e^(−mr)/(4πr), and with g₁ = g/(2M) and g₂ = g/(2M′) that is this V(r) exactly, now reached through scattering. It also says why heavy particles couple weakly at fixed g, the fading of a dimensionful coupling that topic 2.01 predicted.

Identical ψ: the u channel returns as f(π − θ)

For ψψ → ψψ topic 4.06 gave ℳ = g²[1/(m² − t) + 1/(m² − u)], quoted. In the CM frame u = −|p + p′|² = −4p² cos²(θ/2), which is t with θ replaced by π − θ. Divide by 8π√s with √s → 2M: f_ψψ(θ) = f(θ) + f(π − θ), with f(θ) = g²/(16πM(4p² sin²(θ/2) + m²)), shown. That is UAQM-02.05's amplitude for identical spin-0 bosons, which the field theory produces from its two contractions without being told to symmetrise. A detector cannot tell the two ψs apart, so σ integrates |f(θ) + f(π − θ)|² over half the sphere, topic 5.02's 1/2!. At 90° the two terms are equal and the rate is 4|f(90°)|², twice the 2|f(90°)|² that distinguishable particles would give when either is counted. As p → 0 both terms tend to f(0) and σ → 8π f(0)², twice the distinguishable 4π f(0)², the factor 2 of UAQM-02.05. Once pR ≫ 1 the two peaks, at 0° and 180°, no longer overlap and the interference dies away.

Where the match stops

The match is first order in g² and leading order in p/M, and each has an edge. Strength: one Born term suffices only while the scattered wave is small within the range, b = 2μα_Y/m = f(0)/R ≪ 1, UQMII-15.09's low-energy condition in this model's variables, quoted. As b grows, the higher Born terms, which in field theory are box and ladder diagrams exchanging two, three or more φs, catch up with the first; at b = 1.680, a numerical result quoted for the Yukawa potential, ψψ′ acquires an s-wave bound state, a pole that no finite sum of diagrams contains, so the potential must be iterated: named here, not done. Unitarity: the tree f is real, but the optical theorem demands Im f(0) = pσ/(4π), of order g⁴, which topic 5.07 shows the next order supplies. Speed: corrections of relative order p²/M² come from √s, from the 2E in each norm and from q⁰ in other frames, and with them the idea of an instantaneous potential goes. The measured check is Yukawa's own, run backwards: a nuclear force of range about 1.4 fm asks for ħc/R ≈ 141 MeV, and the charged pion's mass is 139.57 MeV, quoted.

Change one variable at a time

Make the relationship visible.

Interactive model

CM momentum p100 MeV

Exchanged φ mass m140 MeV

Set p = 10 MeV: both curves go flat, the solid at 1 and the dashed at 4, an s wave. Double p and m together and nothing moves, since only pR matters. Push pR past 1 and the forward peak narrows to about 1/(pR) radians: the scattering has begun to resolve the range.

RANGE R = ħc/m1.41 fm

pR = p/m0.71

σ(ψψ′) ÷ 4π|f(0)|²0.329

σ(ψψ) ÷ σ(ψψ′)1.820

Live interpretationRANGE R = ħc/m: 1.41 fm. pR = p/m: 0.71. σ(ψψ′) ÷ 4π|f(0)|²: 0.329. σ(ψψ) ÷ σ(ψψ′): 1.820

Catch the common trap

Explain before calculating.

Misconception check

The tempting shortcut is to read the potential straight off the amplitude, Ṽ(q) = −ℳ, as if ℳ were a non-relativistic matrix element. It gives V = −g² e^(−mr)/(4πr), and dimensions kill it at once: in the scalar Yukawa model g is a mass, so that V has mass dimension 3, not the 1 of an energy. What is missing is the normalisation. Relativistic states carry 2E each, so at low speed ℳ = −(2M)(2M′)Ṽ and the potential carries g²/(4MM′). The second slip is the sign. Knowing that like charges repel, students read a positive ℳ between two ψs as repulsion. But a positive ℳ means a positive f, and the Born amplitude −(μ/2π)Ṽ is positive for an attractive well. Scalar exchange attracts ψψ, ψψ′ and ψψ̄ alike; the repulsion of like charges belongs to the photon's vector exchange, met with QED.

In the scalar Yukawa model with a second scalar ψ′ of mass M′, the tree amplitude for ψψ′ → ψψ′ at low speed is ℳ = g²/(|q|² + m²), where g has mass dimension 1. Which potential has a first Born amplitude equal to ℳ/(8π√s)?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyIn the scalar Yukawa model with a second scalar ψ′, take g = 1.5 GeV, M = 0.94 GeV, M′ = 1.88 GeV and a φ of mass m = 0.20 GeV. Find the range of the ψψ′ potential in fm, its strength α_Y = g²/(16πMM′), and V at r = 1.0 fm in MeV, and check the strength against topic 1.09's static sources.

Range: R = 1/m, restored as ħc/(mc²) = 197.327 MeV fm / 200 MeV = 0.987 fm.
Strength: g²/(4MM′) = 2.25 GeV²/(4 × 0.94 × 1.88 GeV²) = 2.25/7.069 = 0.3183, a pure number as it must be; α_Y = 0.3183/(4π) = 0.02533.
Restore ħc: V(r) = −α_Y ħc e^(−r/R)/r. At r = 1.0 fm, r/R = 1.0135 and e^(−1.0135) = 0.3629, while α_Y ħc = 0.02533 × 197.327 MeV fm = 4.998 MeV fm, so V = −4.998 × 0.3629 MeV = −1.81 MeV.
Topic 1.09's form, −g₁g₂ e^(−mr)/(4πr), with source strengths g₁ = g/(2M) = 0.798 and g₂ = g/(2M′) = 0.399: g₁g₂/(4π) = 0.3183/(4π) = 0.02533, the same α_Y. The heavier ψ′ is the weaker source, at half the strength, because the coupling g is a mass.

AnswerR = 0.987 fm, α_Y = 0.0253 and V(1.0 fm) = −1.81 MeV, attractive; the static-source picture gives the same α_Y.

MediumFor ψψ′ → ψψ′ with M = 1.0 GeV, M′ = 1.5 GeV, m = 0.25 GeV and g = 1.2 GeV, at CM momentum p = 0.15 GeV, check that one Born term is enough, then find dσ/dΩ at θ = 90° in μb sr⁻¹ and the total cross-section in mb. How large is the correction from √s?

μ = MM′/(M + M′) = 0.600 GeV and α_Y = g²/(16πMM′) = 1.44/(16π × 1.5) = 0.01910. The amplitude's numerator is A = 2μα_Y = g²/(8π(M + M′)) = 0.02292 GeV, so f(θ) = A/(|q|² + m²).
Born check: b = A/m = 0.0917 ≪ 1. Equivalently f(0) = A/m² = 0.3667 GeV⁻¹ = 0.0724 fm, small against the range R = 0.789 fm.
At 90°, |q|² = 2p² = 0.0450 GeV², so f = 0.02292/(0.0450 + 0.0625) GeV⁻¹ = 0.2132 GeV⁻¹ and dσ/dΩ = 0.04545 GeV⁻² × 0.3894 mb GeV² = 0.01770 mb sr⁻¹ = 17.7 μb sr⁻¹.
Total: σ = 2π∫ A² d(cos θ)/(2p²(1 − cos θ) + m²)² = 4πA²/(m²(m² + 4p²)) = 4π f(0)²/(1 + 4p²/m²) = 4π × 0.1345/2.44 GeV⁻² = 0.6925 GeV⁻² = 0.270 mb. Check: as p → 0 it tends to 4π f(0)², isotropic s-wave scattering.
Relativistic √s = √1.0225 + √2.2725 = 1.0112 + 1.5075 = 2.5187 GeV against M + M′ = 2.5 GeV, so the tree f with the exact √s is smaller by a factor 0.9926 and the cross-sections by 0.985.

Answerb = 0.092, so one Born term suffices; dσ/dΩ(90°) = 17.7 μb sr⁻¹ and σ = 0.270 mb, both lowered 1.5% by the exact √s.

HardIdentical ψ of mass M = 1.0 GeV exchange a φ of mass m = 0.30 GeV with g = 1.2 GeV. At CM momentum p = 0.20 GeV find dσ/dΩ at 90° and the total cross-section in mb, compare both with distinguishable particles of the same mass and coupling, and check the p → 0 limit.

ℳ = g²[1/(m² − t) + 1/(m² − u)] with t = −2p²(1 − cos θ) and u = −2p²(1 + cos θ). Dividing by 8π√s with √s → 2M gives f_ψψ = f(θ) + f(π − θ), f(θ) = A/(a − d cos θ), where A = g²/(16πM) = 0.02865 GeV, a = 2p² + m² = 0.170 GeV² and d = 2p² = 0.080 GeV². Born check: here A = 2μα_Y, so b = A/m = 0.095 ≪ 1.
At 90°: f = A/a = 0.1685 GeV⁻¹, and dσ/dΩ = |2f|² = 0.1136 GeV⁻² = 0.0442 mb sr⁻¹, twice the 0.0221 mb sr⁻¹ of distinguishable particles counted either way.
σ = ½∫|f(θ) + f(π − θ)|² dΩ = ∫f² dΩ + ∫f(θ)f(π − θ) dΩ. The direct term is 4πA²/(m²(m² + 4p²)) = 0.010313/(0.09 × 0.25) GeV⁻² = 0.4584 GeV⁻², the distinguishable cross-section.
Interference: 2πA²∫d(cos θ)/(a² − d² cos²θ) = (2πA²/(ad)) ln((a + d)/(a − d)) = (0.0051566/0.0136) × ln 2.778 GeV⁻² = 0.3792 × 1.0217 GeV⁻² = 0.3874 GeV⁻².
σ = 0.8457 GeV⁻² × 0.3894 mb GeV² = 0.329 mb, against 0.4584 × 0.3894 = 0.178 mb for distinguishable particles, a ratio of 1.85. As p → 0 both terms tend to 4πA²/m⁴ and σ → 8πA²/m⁴ = 0.992 mb, exactly twice the distinguishable 0.496 mb, UAQM-02.05's factor 2; at pR = 0.67 the interference is already down to 85% of the direct term.

Answerdσ/dΩ(90°) = 0.0442 mb sr⁻¹ and σ = 0.329 mb, 1.85 times the distinguishable 0.178 mb; the ratio rises to 2 as p → 0.