University Quantum Field Theory I · Interacting Fields: the Interaction Picture, the Gell-Mann-Low Formula and Wick's Theorem · 2.6
The Gell-Mann-Low formula for correlation functions
Every Feynman diagram in this course is a term of one ratio, and this is the lesson that earns it. We cannot write down the interacting vacuum or the Heisenberg field, so we swap both for free fields and the free vacuum, pay with a phase and an overlap nobody can compute, and let a single division cancel the bill.
Build the model
Connect the measurement to the mechanism.
An interacting field theory predicts correlation functions ⟨Ω|T φ(x₁) … φ(xₙ)|Ω⟩, but both ingredients are out of reach: the Heisenberg field obeys a nonlinear equation nobody can solve, and the true vacuum |Ω⟩ is an unknown superposition of free multiparticle states. The Gell-Mann-Low formula replaces each by something computable. Every Heisenberg field becomes U†φI U, a free field dressed by evolution operators, and neighbouring operators merge, so the product becomes one chain of evolutions running forward in time. At the two ends, topic 2.05's projection makes |Ω⟩ out of |0⟩ by evolving along T(1 − iε), at the price of a phase e(−iE₀T) and an overlap ⟨0|Ω⟩ at each end.
Those factors cannot be computed, and they need not be: the same chain with no fields inserted, the vacuum-to-vacuum amplitude, carries exactly the same factors, so they cancel in the ratio, and the chain folds into one time-ordered exponential. The formula is exact; perturbation theory begins only when that exponential is expanded. It rests on three things it does not prove. It needs ⟨0|Ω⟩ ≠ 0, true in a box but lost as the volume grows, which is Haag's theorem in miniature.
It needs a gap above E₀, which massless quanta close. And it returns correlation functions of the field as written in the Lagrangian, whose link to measured particles waits for LSZ in Unit 4.
- Simple definition
- The Gell-Mann-Low formula writes a time-ordered correlation function in the interacting vacuum as a ratio of free-vacuum expectation values: free fields times the time-ordered exponential of the interaction, divided by the same expression with no fields.
- Example
- A scalar of mass m = 200 MeV pushed by a uniform source ℒᵢₙₜ = jφ, with j = (60 MeV)³, has ⟨Ω|φ|Ω⟩ = j/m² = 5.40 MeV; the formula's first-order term gives exactly that, even in a 50 fm box where |⟨0|Ω⟩|² is only 2.5 × 10⁻²¹.
Quoted from topic 2.03, with the pictures agreeing at t = 0. Each field becomes a free field between operators that can be expanded in the coupling.
φ, φI in GeV (mass dimension 1); t in GeV⁻¹; U(t, 0) = e(iH₀t) e(−iHt), dimensionless and unitary
Between two fields, the U of one and the U† of the next collapse into one evolution across the gap, so n fields leave a single chain.
t₁, t₂ in GeV⁻¹; U(t₁, t₂) = e(iH₀t₁) e(−iH(t₁ − t₂)) e(−iH₀t₂); shown by multiplying out
Derived by running topic 2.05's ket argument on ⟨0|e(−iHT): the tilt damps every excited state and leaves the ground state with two unknown factors.
E₀ in GeV, measured from H₀|0⟩ = 0; ⟨0|Ω⟩ dimensionless; ε a small positive number
Derived. It carries exactly the phase and overlap that the numerator carries, which is why dividing by it removes them.
dimensionless; follows from ⟨Ω|Ω⟩ = 1 with both projections inserted; E₀ grows in proportion to the volume V
Derived, and exact before any expansion. Without derivative couplings the exponent is i∫ℒᵢₙₜ d⁴x; every Feynman diagram in the course is a term of this ratio.
time integrals from −T to T with T → ∞(1 − iε); HI in GeV; both sides have mass dimension n
Haag's theorem in miniature: the overlap dies exponentially with V, while the ratio still returns ⟨Ω|φ|Ω⟩ = j/m² in every box.
uniform static source ℒᵢₙₜ = jφ: j in GeV³, box volume V in GeV⁻³, mass m in GeV; derived for the zero-momentum mode
Rewrite each Heisenberg field as a dressed free field
Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Take two fields with x⁰ ≥ y⁰, so the time-ordered product is just φ(x)φ(y). Topic 2.03 supplies the rewrite, quoted: φ(x) = U†(x⁰, 0) φI(x) U(x⁰, 0), where φI is the free field Advanced QM quantised (UAQM-09.02) and U(t, 0) = e(iH₀t) e(−iHt). Substituting gives U†(x⁰, 0) φI(x) U(x⁰, 0) U†(y⁰, 0) φI(y) U(y⁰, 0), and the middle pair collapses: U(x⁰, 0) U†(y⁰, 0) = e(iH₀x⁰) e(−iH(x⁰ − y⁰)) e(−iH₀y⁰) = U(x⁰, y⁰), shown by multiplying out. With n fields in time order the same step repeats at every gap, leaving U(0, x₁⁰) φI(x₁) U(x₁⁰, x₂⁰) φI(x₂) … φI(xₙ) U(xₙ⁰, 0), in which every operator between two fields evolves forward from the field on its right to the field on its left; only the two end operators, which reach back to t = 0, can run backward, and the next step absorbs them into forward evolutions from −T and to T. Nothing has been approximated, and the mass dimension is unchanged: each φI carries one power, each U none.
Turn both ends of the chain into the free vacuum
The chain still sits between ⟨Ω| and |Ω⟩, which nobody can write down. Topic 2.05 replaces the ket, quoted: evolving |0⟩ along T(1 − iε) damps each excited state by e(−ε(Eₙ − E₀)T) relative to the ground state, so U(0, −T)|0⟩ → e(−iE₀T) ⟨Ω|0⟩ |Ω⟩. The bra needs the same argument run from the left. Because H₀|0⟩ = 0, ⟨0|U(T, 0) = ⟨0|e(−iHT) = Σₙ ⟨0|n⟩ e(−iEₙT) ⟨n|, and the same tilt leaves ⟨0|Ω⟩ e(−iE₀T) ⟨Ω|, derived. The outer operators now merge as well: ⟨0|U(T, 0) U(0, x₁⁰) = ⟨0|U(T, x₁⁰) and U(xₙ⁰, 0) U(0, −T)|0⟩ = U(xₙ⁰, −T)|0⟩. So the correlation function is the limit of ⟨0|U(T, x₁⁰) φI(x₁) … φI(xₙ) U(xₙ⁰, −T)|0⟩ divided by e(−2iE₀T) |⟨0|Ω⟩|². The numerator is computable. The divisor is not, because E₀ and ⟨0|Ω⟩ belong to the exact ground state.
Divide by the same chain with no fields
The unknown divisor is fixed by a normalisation we already have. Apply both projections to ⟨Ω|Ω⟩ = 1: the bra brings ⟨0|U(T, 0), the ket brings U(0, −T)|0⟩, the two merge into U(T, −T), and 1 = ⟨0|U(T, −T)|0⟩ / [e(−2iE₀T) |⟨0|Ω⟩|²], derived. The vacuum-to-vacuum amplitude is therefore exactly the factor we could not compute, and dividing the numerator by it cancels the phase and the overlap together. The order of operations matters. Neither numerator nor denominator has a limit on its own: e(−2iE₀T) keeps turning, and E₀, an energy summed over all space, is proportional to the volume. Only the ratio converges. In the static-source model below, a box 20 fm on a side has E₀ = −607 MeV, so by T = 10 fm/c the common phase 2|E₀|T has turned through 62 rad, and the ratio does not notice.
Fold the chain into one time-ordered exponential
Each U(t₁, t₂) is the time-ordered exponential of −i∫HI dt from t₂ to t₁, quoted from topic 2.03 (UQMII-12.02). In the numerator the intervals [x₁⁰, T], [x₂⁰, x₁⁰], …, [−T, xₙ⁰] tile [−T, T] in order, and each field sits on the boundary between its neighbours, so every operator already stands to the left of everything earlier. One time-ordering symbol acting on one exponential over the whole interval produces exactly that arrangement: the numerator is ⟨0|T φI(x₁)…φI(xₙ) exp(−i∫HI dt)|0⟩, derived. The starting time order was arbitrary and the time-ordering symbol reorders whatever it is given, so the result holds for every ordering, and the left side is the full T-product. Without derivative couplings HI = −∫ℒᵢₙₜ d³x, so the exponent is i∫ℒᵢₙₜ d⁴x over the whole slab of spacetime, the object topic 2.04 showed to be Lorentz invariant. Free limit: switch the coupling off and U = 1, the denominator is 1, and the formula returns ⟨0|T φI(x₁)…φI(xₙ)|0⟩, as it must.
Test the formula where the exact answer is known
Make topic 1.09's classical source uniform: ℒᵢₙₜ = jφ for a scalar of mass m, with [j] = 3 so that [ℒᵢₙₜ] = 4. The Heisenberg equation is (□ + m²)φ = j. In the translation-invariant vacuum ⟨Ω|φ|Ω⟩ is a constant, so □ removes it and ⟨Ω|φ|Ω⟩ = j/m², exact and derived. Now the formula. At zeroth order the numerator is ⟨0|φI|0⟩ = 0, and the denominator is 1 + O(j²), so to first order the ratio is ij∫d⁴z DF(x − z). Integrating over every z picks out the momentum-space propagator at p = 0, i/(0 − m²) = −i/m², and the formula returns j/m², the exact value; the tilted time contour is what supplies the iε. With j = (60 MeV)³ and m = 200 MeV that is 5.40 MeV. Checks: the mass dimension is 3 − 2 = 1; the answer vanishes as j → 0; and it diverges as m → 0, the limit in which the gap closes and topic 2.05's projection fails: the divergence is the formula flagging its own assumption.
What Haag's theorem says, and why a box is the working answer
The same model gives the denominator exactly. A uniform source moves only the zero-momentum mode, so the ground state is that mode displaced to φ = j/m², with E₀ = −j²V/(2m²) and |⟨0|Ω⟩|² = exp(−j²V/(2m³)), derived in the Hard example. With the numbers above, the overlap is 0.954 in a 5 fm box, 0.048 at 20 fm and 2.5 × 10⁻²¹ at 50 fm: in infinite volume the two vacua do not overlap at all. Haag's theorem, stated, not proved, makes this general: in infinite volume, a field joined to a free field at one time by a unitary operator, with a Poincaré-invariant vacuum, is itself free, so the U(t, 0) used above cannot exist there. The working answer is topic 2.05's: work in a finite box, where every step is legitimate, form the ratio, in which V cancels, then let V → ∞. The path-integral route, UQMII-11.02 applied to fields, reaches the same ratio without an interaction picture; it is named, not used. No measurement tests the formula alone; what is built on it is tested, up to the electron's g − 2 in Unit 14, where theory and the measured aₑ agree to better than a part in 10⁹, quoted.
Change one variable at a time
Make the relationship visible.
Drag L from 5 to 20 fm: the dashed and thin curves spin 64 times faster, because E₀ grows as L³, yet the solid ratio does not move at all. Then lower ε and watch the ratio ring for longer before it settles: that is why T must run to infinity along the tilted ray.
OVERLAP |⟨0|Ω⟩|²0.519
VACUUM ENERGY E₀-131.2 MeV
PHASE 2|E₀|T AT mT = 1215.7 rad
RATIO TRANSIENT AT mT = 120.050
Live interpretationOVERLAP |⟨0|Ω⟩|²: 0.519. VACUUM ENERGY E₀: −131.2 MeV. PHASE 2|E₀|T AT mT = 12: 15.7 rad. RATIO TRANSIENT AT mT = 12: 0.050
Catch the common trap
Explain before calculating.
In the static-source model, the numerator and denominator of the Gell-Mann-Low ratio for ⟨Ω|φ|Ω⟩ are evaluated at large T in a box of volume V. The volume is then doubled with j and m held fixed. What happens?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA real scalar of mass m = 200 MeV is coupled to a uniform static source, ℒᵢₙₜ = jφ with j = (60.0 MeV)³. Use the Gell-Mann-Low formula to first order in j to find ⟨Ω|φ|Ω⟩, and check the result against the Heisenberg equation of motion.
- Dimensions first. [ℒᵢₙₜ] = 4 and [φ] = 1, so [j] = 3, and j = 2.16 × 10⁵ MeV³. A one-point function has mass dimension 1, so the answer must be j/m² times a pure number.
- Expand the ratio. At zeroth order the numerator is ⟨0|φI(x)|0⟩ = 0. At first order it is ⟨0|T φI(x) i∫d⁴z jφI(z)|0⟩ = ij∫d⁴z DF(x − z), using ⟨0|T φI φI|0⟩ = DF from topic 1.05. The denominator's first-order term is ⟨0|φI|0⟩ = 0 too, so it is 1 + O(j²).
- Integrate. With topic 1.06's DF(x − z) = ∫d⁴p/(2π)⁴ i e(−ip⋅(x − z))/(p² − m² + iε), the z integral gives (2π)⁴δ⁴(p), leaving i/(−m²) = −i/m². So ⟨Ω|φ|Ω⟩ = ij(−i/m²) = j/m².
- Numbers: j/m² = 2.16 × 10⁵ MeV³ / (4.00 × 10⁴ MeV²) = 5.40 MeV.
- Exact check. The Heisenberg field obeys (□ + m²)φ = j. In the translation-invariant vacuum ⟨Ω|φ|Ω⟩ is a constant, so □ removes it and m²⟨Ω|φ|Ω⟩ = j: the same j/m². The equation is linear in j, so every higher order of the ratio must cancel.
Answer⟨Ω|φ|Ω⟩ = j/m² = 5.40 MeV, exact; it has mass dimension 1, vanishes as j → 0, and diverges as m → 0, where the gap closes.
MediumFor the same source, j = (60.0 MeV)³ and m = 200 MeV, in a periodic box of side L = 5.00 fm, evaluate the denominator ⟨0|T exp(i∫jφI d⁴x)|0⟩ to order j² over −T ≤ t ≤ T, take T → ∞(1 − iε), and read off E₀ and the overlap. Check E₀ against second-order perturbation theory.
- Order j vanishes, since ⟨0|φI|0⟩ = 0. Order j² is ((ij)²/2) ∫dt dt′ ∫d³z d³w DF(z − w), and the spatial integrals keep only the zero-momentum mode: ∫d³z d³w DF(z − w) = V e(−im|t − t′|)/(2m).
- Over the square −T ≤ t, t′ ≤ T, ∫∫e(−im|t − t′|) dt dt′ = 4T/(im) + 2(1 − e(−2imT))/m². Multiplying by −j²V/(4m) gives the order-j² term i j²VT/m² − (j²V/(2m³))(1 − e(−2imT)).
- Tilt: along T(1 − iε) the term e(−2imT) dies, leaving 1 + i j²VT/m² − j²V/(2m³). At this order topic 2.05 predicts e(−2iE₀T) |⟨0|Ω⟩|² ≈ 1 − 2iE₀T − (1 − |⟨0|Ω⟩|²). The term growing with T is the phase, so E₀ = −j²V/(2m²); the constant is the overlap deficit, 1 − |⟨0|Ω⟩|² = j²V/(2m³).
- Numbers: L = 5.00 fm / 197.327 MeV fm = 0.02534 MeV⁻¹, so V = 1.627 × 10⁻⁵ MeV⁻³. Then E₀ = −(2.16 × 10⁵)² × 1.627 × 10⁻⁵ / (2 × 200²) MeV = −9.49 MeV, and j²V/(2m³) = 0.0474.
- Second-order Rayleigh-Schrödinger theory: Hᵢₙₜ = −j∫φ d³x links |0⟩ only to one zero-momentum quantum of energy m, with |⟨1|Hᵢₙₜ|0⟩|² = j²V/(2m), so ΔE = −j²V/(2m²) = −9.49 MeV, the same. Per unit volume that is −j²/(2m²) = −0.0759 MeV fm⁻³, the classical minimum of ½m²φ² − jφ at φ = j/m².
AnswerE₀ = −j²V/(2m²) = −9.49 MeV and 1 − |⟨0|Ω⟩|² ≈ j²V/(2m³) = 0.047 for the 5.00 fm box; both grow in proportion to V, and both cancel in the ratio.
HardSolve the source model exactly in a periodic box of side L. Find |⟨0|Ω⟩|² for L = 5.00, 20.0 and 50.0 fm with j = (60.0 MeV)³ and m = 200 MeV; show that the Gell-Mann-Low ratio for ⟨Ω|φ|Ω⟩ at finite T does not depend on L; and find how large T must be, with ε = 0.05, for the ratio to lie within 0.1% of its limit.
- Only the zero-momentum mode couples. With φ₀ = (a + a†)/√(2mV) it has H = m a†a − g(a + a†), g = j√(V/(2m)). Completing the square, H = m(a† − c)(a − c) − mc² with c = g/m, so |Ω⟩ is the coherent state a|Ω⟩ = c|Ω⟩, E₀ = −mc², and |⟨0|Ω⟩|² = e(−c²), derived: topic 1.09's Poisson chance of no quanta, with mean c² = j²V/(2m³).
- Numbers: c² = 0.0474 × (L/5.00 fm)³ = 0.0474, 3.04 and 47.4, so |⟨0|Ω⟩|² = 0.954, 0.048 and 2.5 × 10⁻²¹. The free vacuum is a fine approximation in the small box and useless in the large one.
- The ratio at finite T: the first-order numerator is ij∫dt∫d³z DF(x − z), and ∫d³z DF(x − z) = e(−im|t − x⁰|)/(2m), with no V in it. Integrating t along the tilted contour from −T(1 − iε) to T(1 − iε), at x⁰ = 0, gives (j/m²)[1 − e(−imT(1 − iε))]. Higher orders cancel between numerator and denominator, as Wick's theorem in topic 2.08 will show.
- So the ratio is the same in every box, although numerator and denominator each carry e(−c²), which ranges over 21 orders of magnitude here, and the phase e(−2iE₀T), whose rate grows as L³.
- The leftover transient has modulus e(−εmT). Requiring e(−εmT) ≤ 10⁻³ gives mT ≥ ln(1000)/0.05 = 138.2, so T ≥ 138.2 × 197.327 MeV fm / 200 MeV = 136 fm/c.
Answer|⟨0|Ω⟩|² = 0.954, 0.048 and 2.5 × 10⁻²¹ at L = 5.00, 20.0 and 50.0 fm; the ratio (j/m²)[1 − e(−imT(1 − iε))] is the same in every box and lies within 0.1% of j/m² = 5.40 MeV once T ≥ 136 fm/c.