University Quantum Field Theory I · Natural Units, Relativistic States and the Feynman Propagator · 1.9
A classical source: Poisson quanta and the Yukawa energy
This is an interacting field theory you can solve exactly, and it pays twice. A prescribed source shows what an emitted quantum is: an on-shell Fourier component, counted by Poisson statistics. Hold two sources still and the same calculation hands you the Yukawa potential, the first force this course derives from a field.
Build the model
Connect the measurement to the mechanism.
Couple a free scalar field linearly to a prescribed c-number source, ℒ = ½(∂φ)² − ½m²φ² + jφ, and you have the simplest interacting field theory, solvable exactly at any strength. The field equation stays linear, so DR solves it, as topic 1.07 found for ⟨φ⟩; read after the source has gone, it adds to every annihilation operator a number, i j̃(p), the source's Fourier transform on the mass shell. The vacuum therefore ends as a coherent state of every mode, UAQM-11.07's one-mode result made relativistic, and the quanta come out Poisson-distributed with mean N̄ = ∫ d³p |j̃|²/((2π)³ 2Eₚ).
Only on-shell components count, so a source that changes smoothly over times long compared with 1/m emits essentially nothing. The vacuum persistence amplitude says the same through DF: ⟨0|0⟩ⱼ = exp(−½∫∫ j DF j) exactly, its modulus squared is e(−N̄), the Poisson chance of no emission, and its phase carries what emission does not. For two static sources held a time T that phase is e(−iET) with E = −g² e(−mr)/(4πr): like sources attract, with range 1/m, 1.41 fm for the pion, through spacelike momentum exchange that emits nothing.
The model stops where the sources stop being classical: they never recoil, the infinitely heavy limit of topic 5.06's scattering; a point source's infinite self-energy is dropped; and for m = 0 a source that changes velocity emits infinitely many soft quanta.
- Simple definition
- A classical source is a prescribed c-number function j(x) added to the Lagrangian as jφ. It leaves the vacuum in a coherent state, so the field holds Poisson-distributed quanta, and the vacuum persistence amplitude ⟨0|0⟩ⱼ records both the chance of no emission and the energy of the sources.
- Example
- A source pulse that leaves N̄ = 2.0 quanta on average lets the vacuum survive with probability |⟨0|0⟩ⱼ|² = e(−2.0) = 0.135, and gives one or two quanta with 0.271 each; two static sources with g²/(4π) = 1, 1.5 fm apart in the pion field, have E = −45.5 MeV.
Once the source is off it adds a number to each annihilation operator, aₒᵤₜ(p) = aᵢₙ(p) + i j̃(p), so the in-vacuum is a coherent state of every mode.
j a c-number source of mass dimension 3 (MeV³); φ of dimension 1 (MeV); DR from topic 1.07, (□ + m²)DR = −iδ⁴(x − y)
Only on-shell Fourier components radiate. A source that changes smoothly over times long compared with 1/m has almost none above p⁰ = m, so it emits essentially nothing.
Eₚ = √(p² + m²) in MeV; j̃ of mass dimension −1 (MeV⁻¹, or fm through ħc); d³p/Eₚ in MeV²; N̄ a pure number
Each mode is coherent and the modes are independent, so their counts add as independent Poisson variables. N̄ = 4 means 4 ± 2 quanta.
n = 0, 1, 2, … quanta of any momenta; variance equal to N̄; exact at any source strength, not a weak-source result
Exact, not a leading order: commutators of jφI are numbers, so the logarithm of the amplitude is quadratic in j and equals the second-order Dyson term.
DF from topic 1.05, mass dimension 2 (MeV²); d⁴x of dimension −4 (MeV⁻⁴ or fm⁴); the exponent is a pure number
DF's iε reproduces the Poisson P(0) that DR gave. The real part of the exponent counts emitted quanta; the imaginary part carries energy as a phase.
From Re[i/(p² − m² + iε)] = πδ(p² − m²), which keeps only on-shell p; N̄ as in the mean-number card; dimensionless
Read off the cross term, ⟨0|0⟩ⱼ ∝ e(−iET). Like sources attract with range 1/m, 1.41 fm for the pion; each source's infinite self-energy is r-independent and dropped.
g a pure number, j = gδ³(x − xᵢ); r in fm and m = mc²/ħc in fm⁻¹; multiply by ħc = 197.327 MeV fm for MeV
A linear source is solved exactly by DR
Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout, with ħc = 197.327 MeV fm restored for numbers. Add to the free real scalar of UAQM-09.02 a prescribed c-number function: ℒ = ½(∂φ)² − ½m²φ² + jφ. The action is dimensionless and φ has mass dimension 1, so j has mass dimension 3. The field equation (□ + m²)φ = j is linear, so its operator solution is the free field plus topic 1.07's classical response: φ(x) = φᵢₙ(x) + i∫d⁴y DR(x − y)j(y), derived, with φᵢₙ the field before j acts. Once j has switched off, DR(x − y) = D(x − y) − D(y − x), two mode sums. With topic 1.03's covariant modes, φ = ∫d̃p [a(p)e(−ip⋅x) + a†(p)e(ip⋅x)] and d̃p = d³p/((2π)³2Eₚ), matching the coefficients of e(−ip⋅x) gives aₒᵤₜ(p) = aᵢₙ(p) + i j̃(p), derived, with j̃(p) = ∫d⁴y e(ip⋅y)j(y) at p⁰ = Eₚ. The source shifts every annihilation operator by a number and does nothing else. Dimensions: j̃ has mass dimension −1, like a(p). Free limit: j → 0 gives aₒᵤₜ = aᵢₙ.
The in-vacuum is coherent in every mode: Poisson quanta
Since aₒᵤₜ(p)|0⟩ = i j̃(p)|0⟩ for every p, the in-vacuum is an eigenstate of every out-annihilator: UAQM-11.07's coherent state, there for one mode, here for all of them, derived. To count, put the field in a box of volume V, where aₖ = a(pₖ)/√(2EₖV) obeys [aₖ, aₖ†] = 1. Mode k then holds a coherent state of amplitude αₖ = i j̃(pₖ)/√(2EₖV), so its count is Poisson with mean |αₖ|², and different modes are independent. Independent Poisson counts add to a Poisson count, so P(n) = e(−N̄)N̄ⁿ/n!, derived, with N̄ = Σ|αₖ|² → ∫d³p |j̃(p)|²/((2π)³2Eₚ). The result is exact at any source strength. The physics sits in the words on shell: only Fourier components with p⁰ = Eₚ ≥ m count. A point source gδ³(x)f(t) has j̃ = g f̃(Eₚ), so if f changes smoothly and slowly compared with 1/m, f̃ is negligible above m and the source emits essentially nothing. The mean energy carried off is ∫d̃p Eₚ|j̃(p)|², the same count weighted by Eₚ.
Vacuum persistence: DF's iε gives the chance of no emission
Now ask for ⟨0|0⟩ⱼ, the amplitude for the vacuum to survive the source. In Quantum Mechanics II's interaction picture, ℋI = −jφI and U = T exp(i∫d⁴x jφI). Values of jφI at two times commute to a c-number, j(x)j(y)[D(x − y) − D(y − x)], so the quoted identity eX eY = e(X+Y) e([X, Y]/2) merges the time-ordered product into exp(i∫jφI) times a number, whose vacuum expectation is again a number. The logarithm of ⟨0|0⟩ⱼ is therefore exactly quadratic in j and equals its second-order Dyson term, (i²/2)∫∫j(x)j(y)⟨0|Tφ(x)φ(y)|0⟩: ⟨0|0⟩ⱼ = exp(−½∫d⁴x d⁴y j(x)DF(x − y)j(y)), derived. In momentum space the exponent is −½∫d⁴p/(2π)⁴ |j̃(p)|² i/(p² − m² + iε), and topic 1.06's iε splits i/(p² − m² + iε) into i times a principal value, which is imaginary, plus πδ(p² − m²), which is real. Only the delta function reaches |⟨0|0⟩ⱼ|, and it puts p on shell: |⟨0|0⟩ⱼ|² = e(−N̄), derived, the Poisson P(0). DR counted the quanta; DF's iε finds the same number.
Two static sources: the cross term is an energy
Hold two point sources, j = g[δ³(x − x₁) + δ³(x − x₂)], for a time T long compared with r = |x₁ − x₂| and 1/m, switched on and off slowly so that nothing is emitted; g is a pure number, because δ³ carries mass dimension 3. The cross term of ½∫∫jDFj is g²∫dx⁰∫dy⁰ DF(x⁰ − y⁰, r) ≈ g²T∫dt DF(t, r). The time integral sets p⁰ = 0 and leaves ∫d³p e(ip⋅r) i/(−|p|² − m²)/(2π)³ = −i e(−mr)/(4πr), derived by closing the |p| contour on the pole at |p| = im. So ⟨0|0⟩ⱼ carries the factor exp(ig²T e(−mr)/(4πr)), and writing it as e(−iET) gives E(r) = −g²e(−mr)/(4πr), derived. A check: topic 1.07 found the static field g e(−mr)/(4πr) from DR, and the classical energy of the second source in it, −g times that field, is the same E, because D̃R and D̃F agree at p⁰ = 0. Each source's own term is the same integral at r = 0: infinite, but independent of r, so it shifts the energy by a constant and is dropped.
Checks against limits and the measured pion
Dimensions: g is a pure number and mr is dimensionless, so E = −g²e(−mr)/(4πr) is an inverse length, an energy once multiplied by ħc. Massless limit: E → −g²/(4πr), Coulomb's form with the opposite sign. Scalar exchange attracts like sources, as topic 7.05 finds for every pair, while topic 9.03's photon makes like charges repel. Free limit: g → 0 gives E → 0, N̄ → 0 and ⟨0|0⟩ⱼ → 1. Numbers: the range is λ = ħc/(mc²) = 197.327 MeV fm ÷ 139.57 MeV = 1.41 fm for the charged pion and 1.46 fm for the neutral one, and at r = 3.0 fm with g²/(4π) = 1 the energy is −(65.78 MeV) e(−2.122) = −7.88 MeV, about an eighth of the massless value. Measured: in 1935 Yukawa ran this argument backwards, from a nuclear-force range near 2 fm to a quantum of about 100 MeV, and the charged pion, found in 1947, has mc² = 139.57 MeV; the long-range tail of the measured nucleon–nucleon force falls off on this 1.4 fm scale, quoted. Real pions are pseudoscalar, so their exchange is spin-dependent (topic 7.09); the scalar model keeps the range and drops the spin.
Where a classical source stops
Three idealisations limit the model, each in a known place. The sources never recoil: j is prescribed, so this is the limit of infinitely heavy sources, and nucleons of 939 MeV, under seven times the pion's mass, are only approximately static; topic 5.06 rebuilds the same potential from the scattering of recoiling particles. A point source has an infinite self-energy, −g²/(8πr) as r → 0, which a finite source size or, later in the course, renormalisation tames. And the massless field is delicate. A source whose velocity jumps, like a scattered charge, has |j̃|² ∝ 1/E² at low E, so dN̄ ∝ dE/E: the mean number diverges logarithmically at the soft end while the radiated energy stays finite there, and the chance e(−N̄) of emitting nothing is zero. That infrared catastrophe returns with photons in topics 14.07 to 14.09, where this same coherent state resolves it.
Change one variable at a time
Make the relationship visible.
Set m to 0 and drag τ: nothing moves, because a point source with one time scale emits 2α/π quanta however fast it acts. Then restore m = 1 mπ and lengthen τ past 1/m = 1.41 fm/c: the bars collapse onto n = 0 and only the phase of ⟨0|0⟩ is left.
MEAN NUMBER N̄0.677
P(0) = |⟨0|0⟩|²0.508
P(n ≥ 1) = 1 − |⟨0|0⟩|²0.492
ADIABATIC RATIO mτ0.71
Live interpretationMEAN NUMBER N̄: 0.677. P(0) = |⟨0|0⟩|²: 0.508. P(n ≥ 1) = 1 − |⟨0|0⟩|²: 0.492. ADIABATIC RATIO mτ: 0.71
Catch the common trap
Explain before calculating.
A source pulse acting on a scalar field gives ⟨0|0⟩ⱼ = exp(−0.35 + 0.80i). What is the probability that the field is left holding exactly one quantum?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyIn a scalar model of the pion field (mc² = 139.57 MeV), two static point sources with g²/(4π) = 0.80 sit 2.0 fm apart. Find the range of the interaction, the energy E in MeV, its ratio to the massless value at the same separation, and the force between the sources in MeV fm⁻¹ and in newtons.
- Range: λ = ħc/(mc²) = 197.327 MeV fm ÷ 139.57 MeV = 1.4138 fm, so mr = 2.0/1.4138 = 1.4146 and e(−mr) = 0.24302.
- Restore ħc in E = −g²e(−mr)/(4πr): E = −(g²/(4π)) ħc e(−mr)/r = −0.80 × 197.327 MeV fm × 0.24302 ÷ 2.0 fm = −19.2 MeV. Dimensions: g² is a pure number and ħc/r is an energy.
- Massless comparison: −0.80 × 197.327 MeV fm ÷ 2.0 fm = −78.9 MeV, so at this separation the mass cuts the energy by e(−mr) = 0.243; at r much less than 1.41 fm the two would agree.
- Force: F = −dE/dr = −(g²/(4π)) ħc e(−mr)(1/r² + m/r) = −0.80 × 197.327 × 0.24302 × (0.2500 + 0.3537) MeV fm⁻¹ = −23.16 MeV fm⁻¹, where m = 0.7073 fm⁻¹ and the minus sign means attraction.
- In SI: 1 MeV fm⁻¹ = 1.602 × 10⁻¹³ J ÷ 10⁻¹⁵ m = 160.2 N, so |F| = 23.16 × 160.2 N = 3.71 × 10³ N, the weight of about 380 kg acting between two particles.
Answerλ = 1.41 fm; E = −19.2 MeV, 0.243 of the massless −78.9 MeV; F = −23.2 MeV fm⁻¹ = −3.71 × 10³ N, attractive.
MediumA point source j(x) = g δ³(x) e(−|t|/τ) with g²/(4π) = 1.0 and τ = 0.50 fm/c acts on a massless scalar field. Find the mean number of quanta it leaves and show that it does not depend on τ, then find the vacuum persistence probability, the chances of exactly one and of three or more quanta, and the mean energy radiated, in total and per quantum.
- Fourier transform on shell: j̃(p) = g∫dt e(iEₚ t) e(−|t|/τ) = 2gτ/(1 + Eₚ²τ²). Dimensions: g is a pure number and τ a length, so j̃ has mass dimension −1, as N̄ requires.
- With m = 0 the angular integral gives d³p/((2π)³2Eₚ) = E dE/(4π²), so N̄ = (1/(4π²))∫₀^∞ E|j̃|² dE = (g²τ²/π²)∫₀^∞ E dE/(1 + E²τ²)² = (g²τ²/π²) × 1/(2τ²) = g²/(2π²). τ cancels: with m = 0 the only scale is τ, and a pure number cannot depend on it.
- With g² = 4π × 1.0, N̄ = 4π/(2π²) = 2/π = 0.637.
- Poisson: |⟨0|0⟩ⱼ|² = P(0) = e(−0.637) = 0.529, P(1) = 0.637 × 0.529 = 0.337 and P(2) = 0.107, so P(n ≥ 3) = 1 − 0.529 − 0.337 − 0.107 = 0.027.
- Energy: Ē = (1/(4π²))∫₀^∞ E²|j̃|² dE = (g²τ²/π²) × π/(4τ³) = g²/(4πτ) = 1.0/(0.50 fm) = 2.0 fm⁻¹, which is 395 MeV after multiplying by ħc. Per quantum, Ē/N̄ = π/(2τ) = 620 MeV: halving τ keeps N̄ and doubles the energy of each quantum.
AnswerN̄ = g²/(2π²) = 0.637 for any τ; |⟨0|0⟩ⱼ|² = 0.529, P(1) = 0.337, P(n ≥ 3) = 0.027; Ē = 395 MeV in total, 620 MeV per quantum.
HardTake the same pulse, j(x) = g δ³(x) e(−|t|/τ), now with g²/(4π) = 2.0 in the pion field, mc² = 139.57 MeV. Reduce N̄ to a one-dimensional integral, evaluate it for τ = 1.0 fm/c and τ = 5.0 fm/c, give the vacuum persistence probability in each case, and compare with a massless field, for which N̄ = g²/(2π²) at any τ.
- On shell Eₚ ≥ m, and after the angles d³p/((2π)³2Eₚ) = √(E² − m²) dE/(4π²). With j̃ = 2gτ/(1 + E²τ²) and x = Eτ, N̄ = (g²/π²) I(μ), where I(μ) = ∫μ^∞ √(x² − μ²) dx/(1 + x²)² and μ = mτ = τ/λ with λ = ħc/(mc²) = 1.414 fm.
- The integral is elementary: write (1 + x²)⁻² as −∂/∂b of 1/(b + x²) at b = 1 and integrate before differentiating. The result, I(μ) = ½[1 − μ² asinh(1/μ)/√(1 + μ²)], passes three checks: numerical quadrature agrees to six figures, I(0) = ½ returns the massless g²/(2π²), and I → 1/(3μ²) for μ ≫ 1.
- τ = 1.0 fm/c: μ = 0.7073, μ² = 0.5003, asinh(1/μ) = 1.1460 and √(1 + μ²) = 1.2249, so I = ½(1 − 0.5003 × 1.1460/1.2249) = 0.2660. With g²/π² = 8π/π² = 2.546, N̄ = 0.677 and |⟨0|0⟩ⱼ|² = e(−0.677) = 0.508.
- τ = 5.0 fm/c: μ = 3.5365, μ² = 12.507, asinh(1/μ) = 0.27913 and √(1 + μ²) = 3.6752, and because the bracket nearly cancels the product is carried unrounded, μ² asinh(1/μ)/√(1 + μ²) = 0.949888, so I = ½(1 − 0.949888) = 0.02506, N̄ = 2.546 × 0.02506 = 0.0638 and |⟨0|0⟩ⱼ|² = 0.938.
- Massless field: N̄ = g²/(2π²) = 4/π = 1.27 at both durations, with |⟨0|0⟩ⱼ|² = 0.280. The pion's mass cuts emission to 53% of that at τ = 1.0 fm/c and to 5.0% at τ = 5.0 fm/c: once the pulse is slower than 1/m = 1.41 fm/c its on-shell Fourier components fade, and the source carries its Yukawa field along instead of radiating. The cusp at t = 0 keeps the fall a power, 1/μ², rather than an exponential.
AnswerN̄ = (g²/π²) I(mτ) = 0.677 with |⟨0|0⟩ⱼ|² = 0.508 at τ = 1.0 fm/c, and 0.0638 with |⟨0|0⟩ⱼ|² = 0.938 at τ = 5.0 fm/c, against 1.27 and 0.280 for a massless field.