University Quantum Field Theory I · Natural Units, Relativistic States and the Feynman Propagator · 1.8
Complex propagator: particles forward, antiparticles back
Here the propagator starts keeping two books: which quantum travels, always forward in time, and which way the charge runs, always from φ† to φ. Learn to read ⟨0|Tφ(x)φ†(y)|0⟩ as one function serving a particle and its antiparticle, then evaluate it on both sides of the light cone.
Build the model
Connect the measurement to the mechanism.
A complex scalar field never does just one thing: φ annihilates a particle of charge q or creates its antiparticle, and φ† does the reverse, as UAQM-09.03's modes showed. So ⟨0|Tφ(x)φ†(y)|0⟩ holds two processes, and the time order picks one. With y earlier, φ†(y) makes a particle that φ(x) absorbs; with x earlier, φ(x) makes an antiparticle that φ†(y) absorbs.
Both quanta have positive energy, both runs move charge q from y to x, and the two amplitudes join into one function DF(x − y) that is even in x − y as well as Lorentz invariant, and so depends on the interval (x − y)² alone. This is the Stueckelberg-Feynman reading of negative frequencies, made exact by the field: the wave e(+iEx⁰) in φ is not a negative-energy state running backward from y to x but a positive-energy antiparticle running forward from x to y, and no sea is needed. Outside the light cone DF is a four-dimensional Euclidean Green's function, mK₁(ms)/(4π²s), falling as e(−ms) beyond a Compton length; inside it turns as e(−imλ) in proper time for either order; near the cone it tends to the mass-blind −1/(4π²(x² − iε)).
The limits are real: the fields are free, which story a spacelike leg tells depends on the observer, and the light-cone singularity is the one that loops will have to tame.
- Simple definition
- For a complex scalar field, DF(x − y) = ⟨0|Tφ(x)φ†(y)|0⟩ is the amplitude for one unit of charge q to pass from y to x: carried by a particle when y is earlier, by its antiparticle when x is earlier, with one invariant function for both.
- Example
- For a pion-mass field at equal times 1.0 fm apart, DF = mK₁(mr)/(4π²r) = 1.85 × 10⁻² fm⁻², or 7.2 × 10⁻⁴ GeV², 0.73 of the massless value. Observers boosted one way call it a π⁺ from y to x, the other way a π⁻ from x to y; all get this number.
First term: a particle made at y and absorbed at x. Second: an antiparticle made at x and absorbed at y. The T-product glues them into one function.
ħ = c = 1, metric (+, −, −, −); D(x − y) = ⟨0|φ(x)φ†(y)|0⟩ from UAQM-09.04; DF has dimension mass², in fm⁻² or GeV².
Only |t| appears: both orders turn as e(−iE|t|), positive energy. Integrated over r it is e(−im|t|)/(2m), the zero-momentum part.
t = x⁰ − y⁰ and r = x − y in fm; Eₚ = √(p² + m²), with p and m in fm⁻¹ through ħc = 197.327 MeV fm.
φφ lowers the charge by 2q and cannot return to the vacuum. A real field, its own antiparticle, has ⟨Tφφ⟩ = DF instead, and its lines carry no arrow.
[Q, φ] = −qφ and Q|0⟩ = 0; q is the particle's charge, +e for a π⁺ field.
From p⁰ = ip₄ and a Gaussian integral. At m = 0 it gives 1/(4π²r²) exactly; the exponent peaks at τ = r/(2m), where it equals −mr: the e(−mr) fall.
Spacelike, in the frame where x⁰ = y⁰; τ is a Schwinger parameter of dimension mass⁻², fm² here.
Real and positive. Tends to 1/(4π²s²) as ms → 0 and falls as e(−ms) beyond a Compton length: an amplitude, not a signal.
K₁ is the modified Bessel function; m = mc/ħ in fm⁻¹, s in fm, DF in fm⁻². Charged pion: ħ/mc = 1.414 fm.
The spacelike form continued by s → iλ. For mλ ≫ 1 its phase turns as e(−imλ): positive frequency in proper time, whichever event is first.
J₁ and Y₁ are Bessel functions; one value for x⁰ − y⁰ of either sign; a term −iδ(x²)/(4π) sits on the cone itself.
Which operator acts first decides what is made
ħ = c = 1 and the metric (+, −, −, −) throughout. UAQM-09.03's field, in topic 1.03's invariant normalisation, is φ(x) = ∫ dp̃ [a(p)e(−ip⋅x) + b†(p)e(ip⋅x)], with dp̃ = d³p/((2π)³2Eₚ): a removes a particle of charge q, b† adds an antiparticle of charge −q. On the vacuum only creation survives, so φ†(y)|0⟩ = ∫ dp̃ e(ip⋅y)|p⟩ is a particle made at y, and φ(x)|0⟩ = ∫ dp̃ e(ip⋅x)|p̄⟩ an antiparticle made at x. The bosonic T-product puts the later operator on the left: ⟨0|Tφ(x)φ†(y)|0⟩ = θ(x⁰ − y⁰)⟨0|φ(x)φ†(y)|0⟩ + θ(y⁰ − x⁰)⟨0|φ†(y)φ(x)|0⟩. Since ⟨0|φ(x)|p⟩ = e(−ip⋅x), the first term is D(x − y), the particle absorbed at x; ⟨0|φ†(y)|p̄⟩ = e(−ip⋅y) makes the second D(y − x), the antiparticle absorbed at y. Derived: DF is the particle's amplitude when y is first and the antiparticle's when x is first.
Positive energy in both directions
Put both terms over one integral. With t = x⁰ − y⁰ and r = x − y the first is ∫ dp̃ e(−iEₚ t + ip⋅r); in the second, relabel p → −p and it becomes ∫ dp̃ e(+iEₚ t + ip⋅r). So DF(t, r) = ∫ dp̃ e(ip⋅r − iEₚ|t|), derived: whichever event comes first, the phase advances as e(−iEₚ|t|) with the elapsed time, the mark of positive energy. Integrating over r keeps only p = 0 and gives e(−im|t|)/(2m). The untimed D projects instead to e(−imt)/(2m), which for t negative turns the other way: read as one quantum running from y back to x, that is negative energy, and the T-product never uses it there. This is Stueckelberg's 1941 and Feynman's 1949 reading of negative frequencies, here a theorem of the field: the e(+iEx⁰) wave in φ rides on b†, so no sea is needed, and a boson field could not fill one anyway. For a pion-mass field the phase turns once per 2π/m = 8.88 fm of time, 2.96 × 10⁻²³ s.
The charge runs from y to x, whatever the order
The charge Q = q∫ dp̃ [a†a − b†b] obeys, from the modes in one line, [Q, φ] = −qφ and [Q, φ†] = +qφ†: φ always lowers the charge by q where it acts, φ† always raises it. With y first, the interval between the events holds a particle of charge +q moving from y to x; with x first, an antiparticle of charge −q moving from x to y, which carries the same charge the same way. So the arrow a Feynman diagram puts on a charged-scalar line, from the φ† end to the φ end, records charge flow, not time, and an arrow pointing into the past is an antiparticle travelling into the future. At spacelike separation the order itself depends on the frame: a boost along the line joining the events at a speed above |x⁰ − y⁰|/|x − y| reverses it, as topic 1.05 showed, so one observer's π⁺ from y to x is another's π⁻ from x to y, while DF, invariant, is one number for both. That works only because particle and antiparticle share one mass, and the measured π⁺ and π⁻ masses agree to a few parts in 10⁴.
Why ⟨0|Tφφ|0⟩ vanishes, and what a real field does
Since Q|0⟩ = 0, ⟨0|[Q, φ(x)φ(y)]|0⟩ = 0, while the commutation rule makes it −2q⟨0|φ(x)φ(y)|0⟩. So ⟨0|φφ|0⟩ = 0 for either order, and the T-product with it: φφ lowers the charge by 2q and cannot return to the vacuum. In modes, the only term that could survive is ⟨0|a(p)b†(k)|0⟩, zero because a and b commute. Now write φ = (φ₁ + iφ₂)/√2, with φ₁ and φ₂ real fields of one mass, each with ⟨Tφᵢφⱼ⟩ = δᵢⱼDF. Then ⟨Tφφ†⟩ = ½(DF + DF) = DF and ⟨Tφφ⟩ = ½(DF − DF) = 0, derived. The cancellation needs the two masses equal, which is the phase symmetry that conserves Q. A real field, the π⁰ say, is its own antiparticle: ⟨Tφφ⟩ = DF and its lines carry no arrow. A charged line always runs from a φ† to a φ, so at every vertex the arrows coming in must match those going out, which is how Unit 2's rules will conserve charge.
A function of the interval alone, Euclidean outside the cone
DF is invariant (topic 1.05) and, from its two terms, even: swapping x − y for y − x swaps them. Inside the cone invariance alone would allow a dependence on the sign of x⁰ − y⁰, and D has one, since D(−x) = D(x)* differs from D(x) there; evenness removes it, so DF = F(x²), derived. Outside the cone take x⁰ = 0 in topic 1.06's integral and turn p⁰ = ip₄, which crosses no Feynman pole: DF = ∫ d⁴pE e(ip⋅r)/((2π)⁴(pE² + m²)), a Green's function in four Euclidean dimensions. Write 1/(pE² + m²) = ∫₀^∞ dτ e(−τ(pE² + m²)); the Gaussian gives e(−r²/(4τ))/(16π²τ²), so DF = (1/(16π²))∫₀^∞ dτ τ⁻² e(−m²τ − r²/(4τ)). At m = 0, u = 1/τ gives 1/(4π²r²) exactly. For m nonzero the quoted integral ∫₀^∞ dτ τ⁻² e(−aτ − b/τ) = 2√(a/b) K₁(2√(ab)) gives mK₁(mr)/(4π²r); the exponent peaks at τ = r/(2m), where it equals −mr. For a pion-mass field 2.0 fm apart that is 2.81 × 10⁻³ fm⁻², 0.44 of the massless 6.33 × 10⁻³ fm⁻²: the mass has begun to bite.
Inside the cone, quoted: proper time and a mass-blind core
Continuing s → iλ, with λ = √(x²) and x² → x² − iε, turns K₁ into a Hankel function, quoted: DF = m[Y₁(mλ) + iJ₁(mλ)]/(8πλ) for either sign of x⁰ − y⁰, plus −iδ(x²)/(4π) on the cone. For mλ ≫ 1 it goes as e(−imλ), a quantum at rest for proper time λ: positive frequency again, for both orders. For a pion-mass field DF = (−6.60 + 7.66i) × 10⁻³ fm⁻² at λ = 2.0 fm, and its phase falls by 0.350 rad between λ = 8.0 and 8.5 fm, close to m × 0.5 fm = 0.354 rad. Towards the cone the mass drops out: DF → −1/(4π²(x² − iε)) on both sides, and at equal times 0.20 fm apart the pion value is 0.974 of the massless one. That universal singularity is what loops will meet in Unit 6. Integrated over time, ∫ dt e(−iE|t|) = −2i/E turns DF into −i e(−mr)/(4πr), the three-dimensional Yukawa function that topic 1.09 uses.
Change one variable at a time
Make the relationship visible.
Set x⁰ − y⁰ to +1.9 fm, then to −1.9 fm: the solid arrow turns from a particle y → x into an antiparticle x → y, the dashed charge arrow stays y → x, and Re DF and Im DF do not move. Then drag |x − y| past 1.9 fm, where the curve turns spacelike and Im DF vanishes.
INTERVAL x²2.61 fm²
√|x²| ÷ PION ħ/mc1.143
x⁰ − y⁰ AS A TIME-6.34 10⁻²⁴ s
Re DF IF MASSLESS-0.0097 fm⁻²
Live interpretationINTERVAL x²: 2.61 fm². √|x²| ÷ PION ħ/mc: 1.143. x⁰ − y⁰ AS A TIME: −6.34 10⁻²⁴ s. Re DF IF MASSLESS: −0.0097 fm⁻²
Catch the common trap
Explain before calculating.
A π⁺ field φ has quanta of charge +e. Event x is 2.0 fm/c earlier than event y and 1.0 fm away from it. What does ⟨0|Tφ(x)φ†(y)|0⟩ describe, and which function gives its value?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA charged-kaon field, measured mass 493.677 MeV, has K⁺ quanta and K⁻ antiquanta. In the lab, event x is 0.60 fm/c later than event y and 1.00 fm away. Say which process ⟨0|Tφ(x)φ†(y)|0⟩ describes in the lab and in a frame moving at 0.80c along x − y, then evaluate DF in fm⁻² and GeV² and compare it with a massless field.
- Interval: x² = (0.60 fm)² − (1.00 fm)² = −0.64 fm², spacelike, with s = √(0.64 fm²) = 0.80 fm. In the lab x is later, so φ†(y) makes a K⁺ at y and φ(x) absorbs it: charge +e from y to x.
- In the frame at β = 0.80, γ = 5/3: ct′ = γ(0.60 − 0.80 × 1.00) fm = −0.33 fm and r′ = γ(1.00 − 0.80 × 0.60) fm = 0.87 fm, with x² still −0.64 fm². Now x is earlier, so φ(x) makes a K⁻ that φ†(y) absorbs: charge −e from x to y, the same charge flow.
- DF depends on x² alone, so evaluate it at equal times with r = s. ħ/mc = 197.327/493.677 fm = 0.3997 fm, so m = 2.502 fm⁻¹, ms = 2.001 and K₁(2.001) = 0.1396. Then DF = mK₁(ms)/(4π²s) = 2.502 × 0.1396/(39.48 × 0.80) fm⁻² = 1.106 × 10⁻² fm⁻².
- Units: 1 fm⁻² = (0.197327 GeV)² = 0.03894 GeV², so DF = 4.31 × 10⁻⁴ GeV², dimension mass² as it must be. Massless check: 1/(4π²s²) = 3.96 × 10⁻² fm⁻², so the kaon amplitude is msK₁(ms) = 0.279 of it; two Compton lengths of spacelike distance cost a factor of 3.6.
AnswerSpacelike, s = 0.80 fm: a K⁺ from y to x in the lab, a K⁻ from x to y at 0.80c. Either way DF = 1.11 × 10⁻² fm⁻² = 4.3 × 10⁻⁴ GeV², 0.28 of the massless 3.96 × 10⁻² fm⁻².
MediumFor a π⁺ field, measured mass 139.570 MeV, integrate ⟨0|Tφ(t, x)φ†(0)|0⟩ over all x to project onto zero momentum. Evaluate the result at t = +1.0 × 10⁻²³ s and t = −1.0 × 10⁻²³ s, compare it with the untimed ⟨0|φ(t, x)φ†(0)|0⟩ projected the same way, and check that the projection solves the zero-momentum Klein-Gordon equation with a −iδ(t) source.
- In the mode form, ∫ d³x e(ip⋅x) = (2π)³δ³(p) leaves only p = 0: ∫ d³x DF = e(−im|t|)/(2m), of dimension mass⁻¹. Here 1/(2m) = 197.327/(2 × 139.570) fm = 0.7069 fm, or 3.582 GeV⁻¹.
- Phase: m|t|/ħ = 139.570 MeV × 1.0 × 10⁻²³ s / (6.582 × 10⁻²² MeV s) = 2.120 rad. So at t = +1.0 × 10⁻²³ s and at t = −1.0 × 10⁻²³ s alike the projection is 0.7069 fm × e(−2.120i) = (−0.369 − 0.603i) fm: the π⁺ running forward and the π⁻ running forward turn the same way.
- The untimed D projects to e(−imt)/(2m) for every t. It equals DF at t = +1.0 × 10⁻²³ s, but at t = −1.0 × 10⁻²³ s it is (−0.369 + 0.603i) fm, turning as e(+im|t|): read as one quantum from 0 back to x, energy −139.57 MeV. The T-product uses ⟨0|φ†(0)φ(x)|0⟩ there, which projects to e(+imt)/(2m) = e(−im|t|)/(2m), energy +139.57 MeV.
- Check: f = e(−im|t|)/(2m) has f′ = −i sgn(t) e(−im|t|)/2, which jumps by −i at t = 0, so f″ = −m²f − iδ(t). Thus (d²/dt² + m²)f = −iδ(t), the zero-momentum part of topic 1.06's (□ + m²)DF = −iδ⁴(x); the untimed e(−imt)/(2m) gives zero on the right instead.
Answer∫ d³x DF = e(−im|t|)/(2m) = (−0.369 − 0.603i) fm at both t = ±1.0 × 10⁻²³ s. The untimed D gives (−0.369 + 0.603i) fm at the negative time, a phase running backward that the T-product never uses.
HardThe B⁰ₛ meson field is complex, with B⁰ₛ quanta and B̄⁰ₛ antiquanta, but weak interactions give its real components φ₁ and φ₂ slightly different masses, with measured splitting ħΔmₛ, Δmₛ = 17.765 ps⁻¹. Ignoring widths and CP violation, find ⟨0|Tφ(x)φ†(0)|0⟩ and ⟨0|Tφ†(x)φ†(0)|0⟩ projected onto zero momentum, and the fraction of B⁰ₛ made at rest that is found as B̄⁰ₛ after 0.100 ps.
- With φ = (φ₁ + iφ₂)/√2 and independent real fields, ⟨Tφᵢφⱼ⟩ = δᵢⱼDF(mᵢ). Expanding φφ† and φ†φ†, the cross terms drop out: ⟨Tφφ†⟩ = ½[DF(m₁) + DF(m₂)] and ⟨Tφ†φ†⟩ = ½[DF(m₁) − DF(m₂)], zero only if m₁ = m₂, the condition for the charge to be conserved.
- φ†(0) makes a B⁰ₛ; at the later point φ(x) absorbs a B⁰ₛ and φ†(x) absorbs a B̄⁰ₛ. Projecting with ∫ d³x DF = e(−im|t|)/(2m), the two amplitudes are ½[e(−im₁t)/(2m₁) ± e(−im₂t)/(2m₂)] for t positive, with + for B⁰ₛ and − for B̄⁰ₛ.
- Write m₁ and m₂ as m̄ ± Δm/2. ħΔmₛ = 6.582 × 10⁻²² MeV s × 17.765 × 10¹² s⁻¹ = 1.169 × 10⁻⁸ MeV, and with the measured m̄ = 5366.9 MeV, 1/(2m₁) and 1/(2m₂) differ by a fraction Δm/m̄ = 2.2 × 10⁻¹², so a common 1/(2m̄) will do. The amplitudes become e(−im̄t)/(2m̄) times cos(Δm t/2) and −i sin(Δm t/2).
- The fraction found as B̄⁰ₛ is sin²(Δm t/2). At t = 0.100 ps, Δmₛ t/2 = 17.765 × 0.100/2 = 0.888 rad, so sin² = 0.602 and cos² = 0.398. The conversion is complete at t = π/Δmₛ = 0.177 ps.
- Checks: as Δm → 0 the B̄⁰ₛ amplitude vanishes and ⟨Tφ†φ†⟩ = 0 returns; the amplitude ratio |tan(Δm t/2)| = 1.23 is dimensionless. Against measurement: the B⁰ₛ lifetime, 1.52 ps, spans 27 rad of Δmₛ t, about 4.3 full oscillations, so the neglected decay damps them without hiding them, which is how Δmₛ is measured.
Answer⟨Tφφ†⟩ = ½[DF(m₁) + DF(m₂)] and ⟨Tφ†φ†⟩ = ½[DF(m₁) − DF(m₂)], projecting to cos(Δm t/2) and −i sin(Δm t/2); after 0.100 ps a fraction 0.60 is B̄⁰ₛ, with complete conversion at 0.177 ps.