University Quantum Field Theory I · Natural Units, Relativistic States and the Feynman Propagator · 1.2
What Advanced QM supplies, quoted and restated
This is the course's only review, and it is written as a ledger rather than a lesson: every result Advanced QM proved and this course leans on, stated once in ħ = c = 1 with its code and the unit that will use it, then never derived again. Know which lines are quoted, because every later derivation is only as sound as its inputs.
Build the model
Connect the measurement to the mechanism.
Advanced Quantum Mechanics built quantum fields that never interact, and that is exactly the raw material perturbation theory needs: in Unit 2 the interaction-picture field is the free field, so every Feynman diagram is assembled from operators, amplitudes and spin sums Advanced QM already proved. This topic therefore quotes rather than teaches. Each result enters once, with its UAQM code, rewritten in ħ = c = 1 with the metric (+, −, −, −) and Heaviside-Lorentz charge, and is labelled quoted wherever it returns: Noether's currents and the energy-momentum tensor, and the real, complex and Dirac mode expansions with normal-ordered H, P and Q; the vacuum amplitude D(x − y), whose Compton-length tail outside the light cone is harmless because the commutator D(x − y) − D(y − x) vanishes there; the spin sums and traces; spin-statistics for free fields; the Coulomb-gauge photon; and Bethe's one mass subtraction.
The model is exact for free quanta and stops in three places, which become the course's agenda. It never time-orders two fields, so it has no Feynman propagator. It has no interactions, so its |0⟩ is not the interacting vacuum and its masses and field normalisations will shift.
And its photon obeys ∇⋅A = 0 in one frame only, so nothing about it is manifestly covariant. Bethe's subtraction worked, but with a cutoff left in the answer; the systematic version of it is renormalisation.
- Simple definition
- Advanced QM's free-field toolkit is the set of exact results about non-interacting quantum fields, their modes, vacuum amplitude, commutator, spin sums, statistics and Coulomb-gauge photons, that this course quotes by topic code and builds interactions on without re-deriving.
- Example
- Quoted from UAQM-09.04 and restated in fm: for the charged pion ħ/mc = 197.327 MeV fm ÷ 139.57 MeV = 1.414 fm, so at equal times and r = 2.0 fm the amplitude D is still 2.8 × 10⁻³ fm⁻², 44% of its massless value, while [φ(x), φ†(y)] is exactly zero.
Normal-ordered, H = Σ ωₖ aₖ†aₖ and P = Σ k aₖ†aₖ. Unit 2's interaction-picture field is this operator, and topic 1.03 trades the box for a(p) = √(2EₚV) aₖ.
ωₖ = √(k² + m²) in MeV; V the box volume, in fm³ via ħc; [aₖ, aₖ′†] = δₖₖ′; φ has mass dimension 1.
At spacelike separation σ = √(−(x − y)²) it equals mK₁(mσ)/(4π²σ), quoted: a tail falling as e(−mσ) over 1/m, 1.414 fm for the charged pion, that never reaches zero.
Eₚ = √(p² + m²); D has mass dimension 2, read in fm⁻² through ħc = 197.327 MeV fm; the real field gives the same D.
A proper orthochronous transformation carries a spacelike x − y into y − x, so particle and antiparticle terms cancel; inside the cone they do not, and topic 1.07 builds DR there.
(x − y)² = (x⁰ − y⁰)² − r², r the spatial distance, metric (+, −, −, −); spacelike means (x − y)² is negative.
With UAQM-04.03's traces it turns a sum over unobserved spins into one trace; its partner Σ v v̄ = p̸ − m does the same for antifermions in Units 7, 9 and 10.
p̸ = γμ pμ, in MeV; ū = u†γ⁰; normalisation u†u = 2E, so ū u = 2m (UAQM-04.06 in ħ = c = 1).
Two helicities ±1 (UAQM-11.04) and causal E and B, but A is not an observable and ∇⋅A = 0 holds in one frame only: Unit 8 replaces it by a covariant four-component field.
ω = |k|, λ = 1, 2; Heaviside-Lorentz units, so SI's √(ħ/(2ε₀ωV)) becomes (2ωV)(−1/2); A has mass dimension 1.
What survives Bethe's one mass subtraction (UAQM-13.04). For 2s at K = m it gives 1048 MHz against the measured 1057.8 MHz, and each doubling of K adds 94 MHz.
m the electron mass, K the photon-energy cutoff, K₀ Bethe's mean excitation energy, ln(K₀/Ry) = 2.81 for 2s.
The rules of a quotation: a code, ħ = c = 1, and a use
Each Advanced QM result enters this course once, in one line, with three tags: the topic code that proved it, its form in ħ = c = 1 with the metric (+, −, −, −) fixed in topic 1.01, and the unit that will use it. Restating is mechanical, but it has to be done once and carefully. Drop every ħ and c and set ε₀ = 1, so the Heaviside-Lorentz charge is e = √(4πα) = 0.3028. Advanced QM's ħω(a†a + ½) becomes ω a†a once normal ordered; the photon's mode amplitude √(ħ/(2ε₀ωV)) becomes (2ωV)(−1/2); the Compton length ħ/mc becomes 1/m, read back as 1.414 fm for the charged pion and 386.2 fm for the electron by multiplying by ħc = 197.327 MeV fm. A one-line dimension check guards each restatement: V has mass dimension −3 and ω has 1, so (2ωV)(−1/2) has mass dimension 1, as a scalar or photon field must. Whatever is not on this page is derived in the course, and saying which is which is part of every answer.
Free fields are the raw material, not a warm-up
UAQM-09.01 supplies the field Euler-Lagrange equation and Noether's theorem: each continuous symmetry of L gives a current with ∂μ jμ = 0, a phase symmetry gives the charge density, and translations give the energy-momentum tensor Tμν, whose charges are H = ∫ d³x T⁰⁰ and Pi = ∫ d³x T⁰ⁱ. UAQM-09.02 expanded the real field in a box with [aₖ, aₖ′†] = δₖₖ′, H = Σ ωₖ aₖ†aₖ and P = Σ k aₖ†aₖ. UAQM-09.03 added the complex field, aₖ for particles and bₖ† for antiparticles, with charge Q = q Σ (aₖ†aₖ − bₖ†bₖ). UAQM-09.05 and 09.07 gave the Dirac field ψ = Σ (2EₚV)(−1/2)(b u e(−ip⋅x) + d† v e(ip⋅x)) with (b, b†) = (d, d†) = 1 and H = Σ Eₚ (b†b + d†d): positrons of positive energy and no sea. All of it is quoted. It matters because in Unit 2 the interaction-picture field is precisely this free operator, and interactions enter only through a time-ordered exponential built from it, so every vertex in Units 3 to 15 is a product of these modes. Units 1 to 6 use the scalar fields, and Unit 7 onwards the Dirac field. Topic 1.03 lets the box grow, with a(p) = √(2EₚV) aₖ, so that one-particle states acquire a Lorentz-invariant normalisation the box could not give them, and builds the unitary action of the Poincaré group on Fock space from these same Noether charges.
D leaks outside the light cone; the commutator does not
UAQM-09.04 supplies the two functions topics 1.05 to 1.07 are built on. The vacuum amplitude D(x − y) = ⟨0|φ(x)φ†(y)|0⟩ depends only on the invariant interval and, inside the cone, on the sign of x⁰ − y⁰. At spacelike separation σ = √(−(x − y)²) it is mK₁(mσ)/(4π²σ), quoted, and never vanishes: for the charged pion at σ = 2.0 fm it is 2.8 × 10⁻³ fm⁻², 44% of the massless 1/(4π²σ²). The commutator [φ(x), φ†(y)] = D(x − y) − D(y − x) is the particle amplitude minus the antiparticle amplitude for the reverse trip. At spacelike separation a proper orthochronous transformation carries x − y into y − x, so the two terms are equal and the commutator is exactly zero; inside the cone no such transformation exists and the Pauli-Jordan function survives. Keep the roles apart: D, and from topic 1.05 DF, are amplitudes, while signals travel only through commutators. Topic 1.05 builds DF from the same two orderings, and topic 1.07 keeps the commutator only where x⁰ exceeds y⁰, multiplying it by θ(x⁰ − y⁰), to make DR.
Spin: two quoted identities and one conditional theorem
Units 7, 9 and 10 never sum over spins by hand. UAQM-04.08's spin sums, Σₛ u ū = p̸ + m and Σₛ v v̄ = p̸ − m, turn a sum over unobserved spins into a trace, and UAQM-04.03's traces finish it: Tr 1 = 4, Tr(γμγν) = 4g(μν), any odd number of γ's has zero trace, and Tr(γμγνγργσ) = 4(g(μν)g(ρσ) − g(μρ)g(νσ) + g(μσ)g(νρ)). Together they give, in a line derived here, Σ|ū(p′)u(p)|² = 4(p⋅p′ + m²), which at p′ = p is 8m², two spin states each with ū u = 2m: a one-line check on the normalisation. UAQM-09.09's spin-statistics connection is quoted with its conditions, because Unit 7's fermion signs rest on it: for free fields in three space dimensions, with Lorentz invariance, energy bounded below, positive norm and microcausality, integer spin requires commutators and half-integer spin anticommutators. The theorems for interacting fields, by Lüders and Zumino and by Burgoyne, are quoted again in Unit 11, not proved.
Coulomb-gauge photons: two helicities, one preferred frame
UAQM-11.03 quantised only the two transverse components of A, with ελ perpendicular to k and A⁰ fixed instantaneously by the charges through Poisson's equation (UAQM-11.01); UAQM-11.04 showed that each quantum carries momentum k and helicity ±1. Two features make this photon awkward for a covariant theory. The equal-time commutator of A with −E is i times the transverse delta function, not iδ³, and the commutators of A do not vanish outside the light cone, so A is not an observable, although E and B, whose commutators are derivatives of the massless Pauli-Jordan function, are causal. And ∇⋅A = 0 holds in one frame only: after a boost a gauge transformation is needed to restore it, so no intermediate step is manifestly Lorentz covariant. That is the comparison Unit 8 makes. It quantises all four components of Aμ in Feynman gauge, pays for covariance with photons of negative norm, removes them with the Gupta-Bleuler condition, and recovers Advanced QM's two transverse photons as the physical states.
Bethe's one subtraction, and the agenda Advanced QM left
UAQM-13.03 found that the photon cloud adds δm/m = 4αK/(3πm) to a free electron, 0.31% at K = m and growing linearly with the cutoff K. UAQM-13.04 wrote the Hamiltonian in the measured mass, and that single subtraction left an ns shift (4α(Zα)⁴m/(3πn³)) ln(K/K₀) (UAQM-13.05): 1048 MHz for 2s at K = m against the measured 1057.8 MHz, all quoted. Unit 6 re-reads the subtraction as an on-shell mass counterterm, and Unit 15 removes the cutoff that remains, which moves the 2s result by 94 MHz for every doubling of K. What Advanced QM did not do is this course's agenda. It never time-ordered two fields, so it has no Feynman propagator: topics 1.05 to 1.08. Its fields never interact: Unit 2 brings the interaction picture, the Dyson series and Wick's theorem, and Units 3 to 6 the diagrams, LSZ, rates and loops. Its photon is not covariant: Unit 8. Everything else on this page is used without proof.
Change one variable at a time
Make the relationship visible.
Start at t = 0.8 fm, r = 2.0 fm: one solid curve joins dot and ring, so D(x − y) = D(y − x) and the commutator cancels. Drag r below t and the curve splits into two sheets with dot and ring apart: no boost reverses the order. Then drag m to move the tail scale.
INTERVAL x²-3.36 fm²
MIN BOOST β TO FLIP ORDER (1 = NONE)0.40
COMPTON LENGTH ħ/mc1.409 fm
TAIL e(−m√(−x²)), SPACELIKE ONLY0.272
Live interpretationINTERVAL x²: −3.36 fm². MIN BOOST β TO FLIP ORDER (1 = NONE): 0.40. COMPTON LENGTH ħ/mc: 1.409 fm. TAIL e(−m√(−x²)), SPACELIKE ONLY: 0.272
Catch the common trap
Explain before calculating.
For the charged pion, UAQM-09.04 gives D(x − y) ≈ 2.8 × 10⁻³ fm⁻² for two events at equal times 2.0 fm apart, 44% of its massless value. A student concludes that measuring φ at y can disturb a measurement at x, though no light signal could join them. Which reply is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAdvanced QM's equal-time vacuum amplitude (UAQM-09.04, quoted) carries a factor e(−mr) outside the light cone. Restate it in fm: at what distance has that factor fallen to 1.0 × 10⁻² for the charged pion (139.570 MeV) and for the electron (0.511 MeV)? How long does light take to cross the pion's distance, and what is [φ(x), φ†(y)] for two events that far apart at equal times?
- In ħ = c = 1 the factor is e(−mr) with m an inverse length. Restore ħc to read 1/m in fm: 1/m = ħc/(mc²) = 197.327/139.570 = 1.4138 fm for the pion and 197.327/0.511 = 386.16 fm for the electron.
- Set e(−mr) = 1.0 × 10⁻², so r = ln(100)/m = 4.6052/m: 4.6052 × 1.4138 fm = 6.511 fm for the pion, and 4.6052 × 386.16 fm = 1778 fm = 1.778 pm for the electron.
- In natural units a light signal needs t = r = 6.51 fm. Restoring c, t = 6.511 × 10⁻¹⁵ m ÷ 2.998 × 10⁸ m s⁻¹ = 2.17 × 10⁻²³ s.
- Two events at equal times a distance r apart have (x − y)² = −r², which is negative, so they are spacelike, and UAQM-09.04's commutator vanishes exactly there whatever the size of D.
- Checks: mr/ħc is MeV × fm ÷ MeV fm, a pure number, as an exponent must be; and the range ħ/mc shrinks to zero as c → ∞, which is why non-relativistic quantum mechanics never meets the tail.
Answerr = 6.51 fm for the pion and 1.78 × 10³ fm (1.78 pm) for the electron; light crosses the pion distance in 2.17 × 10⁻²³ s; the commutator is exactly zero at both separations.
MediumYukawa theory in Unit 7 needs S, the sum over both spins s and s′ of |ū(p′, s′) u(p, s)|². Using UAQM-04.08's spin sums and UAQM-04.03's traces, both quoted, evaluate S for an electron of momentum 1.00 MeV scattered elastically through 60° by a static target, then check the result at 0° and in the massless limit.
- The bilinear's conjugate is (ū(p′)u(p))* = ū(p)u(p′), so |ū(p′)u(p)|² = ū(p′)u(p) ū(p)u(p′). Summing over s replaces u(p)ū(p) by p̸ + m, and over s′ replaces u(p′)ū(p′) by p̸′ + m, closing the chain into S = Tr[(p̸′ + m)(p̸ + m)].
- UAQM-04.03's traces: Tr(p̸′p̸) = 4p⋅p′, the two terms with a single γ vanish, and the identity has trace 4, giving 4m², so S = 4(p⋅p′ + m²), derived here from the quoted identities and of mass dimension 2, since ū u has dimension 1.
- Kinematics: E = √(1.00² + 0.511²) MeV = 1.1230 MeV. A static target keeps E′ = E and |p′| = |p|, so p⋅p′ = E² − |p|² cos 60° = 1.2611 − 0.5000 = 0.7611 MeV².
- S = 4(0.7611 + 0.2611) MeV² = 4 × 1.0222 MeV² = 4.09 MeV².
- At 0°, p⋅p′ = m² and S = 8m² = 2.09 MeV²: two spin states, each with ū u = 2m and no flip, give 2 × (2m)², confirming the normalisation u†u = 2E.
- As m → 0, S → 4p⋅p′ = 8E² sin²(θ/2), 2.00 MeV² at 60° for E = 1.00 MeV, and zero forward: the scalar bilinear joins opposite chiralities (UAQM-05.08), so a massless fermion must flip its helicity, which a forward scattering cannot do and conserve angular momentum along the beam.
AnswerS = 4(p⋅p′ + m²) = 4.09 MeV² at 60°; it falls to 8m² = 2.09 MeV² at 0° and tends to 8E² sin²(θ/2) as m → 0.
HardUAQM-13.03 gave a free electron's non-relativistic electromagnetic mass as δm/m = 4αK/(3πm) for photons up to K. Weisskopf's relativistic self-mass grows only as (3α/(2π)) ln(K/m), a leading logarithm quoted here, whose dimensionally regularised form Unit 12 derives. Compare the two at K = m, at K = mₚ = 938.272 MeV and at K = 1.22 × 10¹⁹ GeV; then find how far Bethe's 2s shift, (α⁵m/(6π)) ln(K/K₀), moves when K doubles.
- In ħ = c = 1 both forms are dimensionless, K/m being a ratio of energies. Their coefficients are 4α/(3π) = 4/(3π × 137.036) = 3.097 × 10⁻³ and 3α/(2π) = 3/(2π × 137.036) = 3.484 × 10⁻³.
- At K = m the linear form gives 0.31%, while the logarithm vanishes: the leading-log form holds only for K ≫ m, the constant beside it being set by the regulator.
- At K = mₚ, K/m = 938.272/0.511 = 1836.2. Linear: 3.097 × 10⁻³ × 1836.2 = 5.69, a shift of 569% that exceeds the mass itself, so the non-relativistic treatment has failed. Logarithmic: 3.484 × 10⁻³ × ln 1836.2 = 3.484 × 10⁻³ × 7.515 = 0.0262, or 2.6%.
- At K = 1.22 × 10¹⁹ GeV, K/m = 1.22 × 10²² MeV ÷ 0.511 MeV = 2.39 × 10²². Linear: 7.4 × 10¹⁹. Logarithmic: 3.484 × 10⁻³ × 51.53 = 0.180, or 18%. Covariance turns a hopeless divergence into a mild one.
- Bethe's 2s shift has n = 2 and Z = 1, so its prefactor is 4α⁵m/(3π × 8) = α⁵m/(6π) = 5.610 × 10⁻⁷ eV, or 135.6 MHz after dividing by h = 4.1357 × 10⁻¹⁵ eV s.
- Doubling K adds 135.6 MHz × ln 2 = 94.0 MHz, 8.9% of the measured 1057.8 MHz: the residual cutoff dependence that Unit 15 removes by matching Bethe's low-energy part to the relativistic vertex.
Answerδm/m = 0.31% (linear) against 0 (leading log) at K = m; 569% against 2.6% at K = mₚ; 7.4 × 10¹⁹ against 18% at 1.22 × 10¹⁹ GeV. Doubling K moves Bethe's 2s shift by 94.0 MHz, 8.9% of the measured splitting.