University Quantum Field Theory I · Interacting Fields: the Interaction Picture, the Gell-Mann-Low Formula and Wick's Theorem · 2.5

Projecting onto the interacting vacuum with T(1 − iε)

Every diagram in this course is computed around the free vacuum |0⟩, yet every quantity we want lives in the true vacuum |Ω⟩. This is the bridge: evolve |0⟩ along a time axis tipped just below the real line and all but |Ω⟩ dies away. Learn what it costs, an overlap and a gap, and why the same tip is the iε in every propagator.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The free vacuum |0⟩ is not the ground state of H = H₀ + Hᵢₙₜ, but it is a superposition of the true eigenstates, |0⟩ = Σₙ |n⟩⟨n|0⟩, with the interacting vacuum |Ω⟩ among them. Evolution with the full H only attaches a phase e^(−iEₙT) to each piece, and on the real time axis those phases turn forever: in a finite box nothing settles and T → ∞ has no limit. Tip the axis to T(1 − iε) and each phase acquires a modulus e^(−εEₙT); measured against the |Ω⟩ term, every excited piece shrinks as exp(−ε(Eₙ − E₀)T), so |Ω⟩, the lowest, outlasts the rest.

Dividing by its own factor e^(−iE₀T)⟨Ω|0⟩, and recognising e^(−iHT)|0⟩ as U(0, −T)|0⟩, writes the unknown vacuum through the free one and the very operator the Dyson series computes. The method rests on two facts about the spectrum. |0⟩ must overlap |Ω⟩, true in a box of volume V though the overlap falls as e^(−aV/2); and a gap must separate E₀ from everything |0⟩ reaches. Massless quanta close the gap in infinite volume, and there Gell-Mann and Low's slow switching of the coupling takes over.

Nor is the tilt a trick to forget once it has done its job: carried through the vertex times of the expansion, it is exactly the iε that makes every propagator Feynman's.

Simple definition: The interacting vacuum |Ω⟩ is projected out of the free vacuum |0⟩ by evolving |0⟩ with the full Hamiltonian for a time T sent to infinity along T(1 − iε), on which every excited component decays faster than the |Ω⟩ component does.
Example: In φ⁴ theory with m = 0.50 GeV the slowest-dying admixture is two quanta at rest, Δ = 2m = 1.0 GeV. With ε = 0.010 it shrinks, relative to the |Ω⟩ term, by e⁻¹ every 1/(εΔ) = 100 GeV⁻¹, which is 6.6 × 10⁻²³ s.

The evolved free vacuum in the eigenbasis of He^(−iHT)|0⟩ = Σₙ e^(−iEₙT) |n⟩⟨n|0⟩

Derived from completeness. Real-time evolution never changes |⟨n|0⟩|, it only turns phases, so on the real axis no term ever wins.

H full Hamiltonian; Eₙ its eigenvalues in GeV (J in SI); T in GeV⁻¹ (s, via ħ = 6.582 × 10⁻²⁵ GeV s); |n⟩ complete

Admixture of an excited state on the tilted axis|cₙ/c₀| = |⟨n|0⟩/⟨Ω|0⟩| exp(−ε(Eₙ − E₀)T) ≤ |⟨n|0⟩/⟨Ω|0⟩| e^(−εΔT)

Derived. The smallest gap among the states |0⟩ overlaps is the bottleneck: εΔT must diverge, so T → ∞ is taken before ε → 0⁺.

cₙ: coefficient of |n⟩ in e^(−iHT)|0⟩ with T → T(1 − iε); ε dimensionless; gap Δ in GeV (J in SI)

Projection onto the interacting vacuum|Ω⟩ = lim U(0, −T)|0⟩ / (e^(−iE₀T)⟨Ω|0⟩), T → ∞(1 − iε)

Derived. Writes the unknown vacuum through the free one and the operator the Dyson series computes; the unknown denominator cancels in topic 2.06.

U(0, −T) = e^(−iHT) e^(iH₀T), topic 2.03's evolution operator; H₀|0⟩ = 0; every factor dimensionless

Overlap with the free vacuum in a box|⟨Ω|0⟩|² = e^(−aV)

Non-zero in any finite box, so the projection works there; zero as V → ∞, so ⟨Ω|0⟩ may only appear where it cancels.

V box volume in GeV⁻³ (m³ in SI); a, a positive coupling-dependent density, in GeV³ (m⁻³); stated, not derived

Gap left by a photon pair in a periodic boxΔ = 4πħc/L

Derived from kinematics. The gap, and with it the damping, vanishes as L → ∞, so the time and volume limits fight and adiabatic switching replaces the tilt.

L box side in m; two photons of momenta ±2πħ/L, the lowest-energy zero-momentum state |0⟩ reaches in QED; 2.48 eV at L = 1 μm

The tilt seen in momentum space(p⁰)² − Eₚ²(1 − iε)² ≈ p² − m² + iε′, with ε′ = 2εEₚ²

Shown below: tilted vertex times turn each contraction's e^{(−iEₚ|t|)} into a decaying one, putting the poles at ±Eₚ(1 − iε), Feynman's quadrants.

p four-momentum in GeV, p² = (p⁰)² − |p|²; Eₚ = √(|p|² + m²) in GeV; ε′ in GeV²; the ε² term dropped

The free vacuum is a superposition of true eigenstates

Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Let H = H₀ + Hᵢₙₜ have eigenstates |n⟩ with energies Eₙ, the lowest being the interacting vacuum |Ω⟩ with energy E₀, and normal-order H₀ so that H₀|0⟩ = 0. Completeness gives e^(−iHT)|0⟩ = Σₙ e^(−iEₙT)|n⟩⟨n|0⟩, derived. Only some |n⟩ appear. In φ⁴, H commutes with the total momentum and with φ → −φ, while |0⟩ has zero momentum and an even number of quanta, so ⟨n|0⟩ vanishes unless |n⟩ shares both; topic 2.02's first-order admixture of two and four quanta at zero total momentum is the example. On the real time axis this evolution cannot isolate |Ω⟩. Each |⟨n|0⟩| stays fixed and only the relative phases exp(−i(Eₙ − E₀)T) turn. In a box the spectrum is discrete, the sum is almost periodic, and T → ∞ has no limit.

Tip the time axis and the ground state outlasts the rest

Replace T by T(1 − iε) with ε small and positive. Each term gains a modulus, |exp(−iEₙT(1 − iε))| = e^(−εEₙT), so measured against the |Ω⟩ term an excited term carries exp(−ε(Eₙ − E₀)T) ≤ e^(−εΔT), where Δ is the smallest gap among the states |0⟩ overlaps; in φ⁴ that is two quanta at rest, Δ = 2m to leading order. The order of limits follows: T → ∞ at fixed ε first, ε → 0⁺ last, because what must diverge is εΔT. A small tilt suffices. With m = 0.14 GeV, ε = 0.010 is a turn of only 0.57°, yet εΔ = 2.8 × 10⁻³ GeV and an admixture falls to 10⁻⁶ of its start by T = 13.8/(εΔ) = 4.9 × 10³ GeV⁻¹, or 3.2 × 10⁻²¹ s. The projected state does not depend on ε; ε sets only the speed. Turned the full 90°, T becomes imaginary time and e^(−iHT) becomes e^(−Hτ): lattice QCD extracts hadron masses exactly this way and matches measured masses to a few percent, quoted. Here ε stays small because the same contour must return real-time correlation functions.

Write the projector through U(0, −T)

On the tilted axis e^(−iHT)|0⟩ → e^(−iE₀T)⟨Ω|0⟩|Ω⟩, so |Ω⟩ = lim e^(−iHT)|0⟩/(e^(−iE₀T)⟨Ω|0⟩) as T → ∞(1 − iε). Topic 2.03's evolution operator with reference time 0, U(t, t′) = e^(iH₀t) exp(−iH(t − t′)) e^{(−iH₀t′)}, quoted, gives U(0, −T) = e^(−iHT)e^(iH₀T), and e^(iH₀T)|0⟩ = |0⟩ because H₀|0⟩ = 0. Hence |Ω⟩ = lim U(0, −T)|0⟩/(e^(−iE₀T)⟨Ω|0⟩), derived, and the same steps at the far end give the bra, ⟨Ω| = lim ⟨0|U(T, 0)/(e^(−iE₀T)⟨0|Ω⟩). Check it twice. Every factor is dimensionless, as a relation between unit-normalised states must be. In the free limit Hᵢₙₜ → 0, E₀ = 0, ⟨Ω|0⟩ = 1 and U = 1, and the formula returns |Ω⟩ = |0⟩. The denominator is unknown, but it is the same number in every correlation function of the theory, which is why topic 2.06 can cancel it without ever computing it.

First assumption: an overlap that survives only in a box

If ⟨Ω|0⟩ were zero, the tilted evolution would converge on the lowest state |0⟩ does reach, and the formula would divide by zero. At weak coupling in a finite box the overlap is non-zero, but it need not be close to 1, stated: its logarithm is extensive, |⟨Ω|0⟩|² = e^(−aV). A toy picture shows why. Vacuum fluctuations in regions further apart than about 1/m are independent, so |0⟩ and |Ω⟩ differ cell by cell, and the overlaps multiply. If each cell of volume (1/m)³ contributes an amplitude 0.999, a box 10/m on a side holds 10³ cells and an overlap of 0.37, while a box 100/m on a side holds 10⁶ cells and an overlap of about 3 × 10⁻⁴³⁵. As V → ∞ the free and interacting vacua become orthogonal, the seed of the Haag theorem stated in topic 2.06. The working rule follows: keep V finite until the end of every calculation, and let ⟨Ω|0⟩ appear only where it cancels.

Second assumption: a gap, and what replaces it

The damping is e^(−εΔT), so everything hangs on a non-zero gap. In φ⁴ with a non-zero mass m the gap is 2m. In QED an electron-positron pair sits about 2mₑc² = 1.022 MeV above the vacuum, from the measured electron mass, but two photons with momenta ±2πħ/L in a periodic box of side L need only Δ = 4πħc/L, derived: 2.48 eV for L = 1 μm, 4.1 × 10⁵ times less. To bring that admixture to 10⁻⁶, εT must reach 13.8ħ/Δ = 3.7 × 10⁻¹⁵ s, against 8.9 × 10⁻²¹ s for the pair, and the demand grows in proportion to L, so the limits V → ∞ and T → ∞ no longer commute. The named substitute is Gell-Mann and Low's adiabatic switching: multiply Hᵢₙₜ by e^(−η|t|), evolve along the real axis from t = −∞, and divide by ⟨0|U_η(0, −∞)|0⟩. Their theorem, quoted, says that if this ratio has a limit as η → 0⁺ it is an eigenstate of H; the switching factor makes the time integrals converge as the tilt did.

The same tilt is the iε in every propagator

In the expansion every vertex time runs along the tilted contour, t = t′(1 − iε) with t′ real. Topic 1.05's contraction at spatial momentum p depends on time as e^{(−iEₚ|t|)}/2Eₚ, quoted. For positive t′ the tilt gives e^{(−iEₚt′)}e^{(−εEₚt′)}; for negative t′ the other time order gives e^(iEₚt′)e^(εEₚt′); both equal exp(−iEₚ(1 − iε)|t′|), so the tilt has replaced Eₚ by Eₚ(1 − iε) and the contraction decays both ways. The Fourier integral over t′ now converges at both ends and gives i(1 − iε)/((p⁰)² − Eₚ²(1 − iε)²), shown in the Hard example. Dropping ε², the denominator is p² − m² + 2iεEₚ²: topic 1.06's prescription with ε′ = 2εEₚ², positive for every p, and poles at ±Eₚ(1 − iε), in the fourth and second quadrants. Tilting the other way would make each contraction grow and give the anti-Feynman poles, and the same reversal would amplify the excited states instead of damping them. The sign of the iε and the existence of the projection are one fact. Check: ε′ is a mass squared, as p² − m² is.

Change one variable at a time

Make the relationship visible.

Interactive model

Tilt ε in T(1 − iε)0.10

Gap Δ = E₁ − E₀2.0 GeV

Set ε = 0: both curves oscillate forever at full height, so the real-axis limit does not exist. Raise ε and the 2Δ term dies first, then the Δ term; drag Δ down toward 0.2 GeV, a nearly closed gap, and even ε = 0.5 leaves the slowest term alive at T = 30 GeV⁻¹.

TILT BELOW THE REAL AXIS5.7 °

SLOWEST DAMPING RATE εΔ0.200 GeV

Δ TERM LEFT AT T = 30 GeV⁻¹0.0025

εΔT REACHED AT T = 30 GeV⁻¹6.00

Live interpretationTILT BELOW THE REAL AXIS: 5.7 °. SLOWEST DAMPING RATE εΔ: 0.200 GeV. Δ TERM LEFT AT T = 30 GeV⁻¹: 0.0025. εΔT REACHED AT T = 30 GeV⁻¹: 6.00

Catch the common trap

Explain before calculating.

In φ⁴ theory with m = 0.40 GeV, the free vacuum is evolved as e^(−iHT)|0⟩ with T → ∞(1 − iε) and ε = 0.010. Which admixture, measured against the |Ω⟩ term, dies most slowly and so sets how long T must be, and at what rate?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyIn φ⁴ theory with m = 0.30 GeV, the free vacuum is evolved along T(1 − iε) with ε = 0.020. How large must T be for every excited admixture, relative to the |Ω⟩ term, to have fallen to 10⁻⁴ of its starting size? Give T in GeV⁻¹, in seconds and as a light-travel distance.

Relative to the |Ω⟩ term an admixture of |n⟩ carries exp(−ε(Eₙ − E₀)T), so the slowest is the state with the smallest gap that |0⟩ overlaps. Momentum conservation and φ → −φ allow only zero total momentum and an even number of quanta: two quanta at rest, Δ = 2m = 0.60 GeV to leading order.
Require e^(−εΔT) ≤ 10⁻⁴, that is εΔT ≥ ln 10⁴ = 9.21, with εΔ = 0.020 × 0.60 GeV = 0.012 GeV.
T ≥ 9.21/(0.012 GeV) = 768 GeV⁻¹. Restore ħ: 768 GeV⁻¹ × 6.582 × 10⁻²⁵ GeV s = 5.05 × 10⁻²² s; with ħc = 0.1973 GeV fm the light-travel distance is 151 fm.
Checks: εΔT is dimensionless, as an exponent must be; as ε → 0 the required T diverges, the real-axis failure; and four quanta at rest, gap 4m, are by then down to (10⁻⁴)² = 10⁻⁸, so two quanta really are the bottleneck.

AnswerT ≈ 7.7 × 10² GeV⁻¹ = 5.1 × 10⁻²² s, a light-travel distance of about 1.5 × 10² fm, set by two quanta at rest (Δ = 0.60 GeV); four-quanta admixtures are then below 10⁻⁸.

MediumTruncate the zero-momentum, even sector to two states: the free vacuum |0⟩ (energy 0) and a two-quanta state |χ⟩ at rest (energy ω = 1.00 GeV), coupled by ⟨χ|Hᵢₙₜ|0⟩ = g = 0.150 GeV. Find E₀, |⟨Ω|0⟩| and the gap exactly, compare with perturbation theory, and find the excited admixture left after dividing e^(−iHT)|0⟩ by e^(−iE₀T)⟨Ω|0⟩ at T = 1000 GeV⁻¹, ε = 0.010.

H = [[0, g], [g, ω]] has eigenvalues (ω ∓ √(ω² + 4g²))/2. With √(1.00 + 0.090) GeV = 1.04403 GeV: E₀ = −0.02202 GeV, E₁ = 1.02202 GeV, gap Δ = 1.04403 GeV.
The ground eigenvector obeys −E₀v₀ + gv₁ = 0, so v₁/v₀ = E₀/g = −0.14677 and |⟨Ω|0⟩| = 1/√(1 + 0.14677²) = 0.98940; the orthogonal excited state has |⟨1|0⟩| = 0.14521.
Perturbation theory gives E₀ ≈ −g²/ω = −0.0225 GeV and v₁/v₀ ≈ −g/ω = −0.150, both within 2.2% of the exact values, the error being the next order. As g → 0, E₀ → 0 and ⟨Ω|0⟩ → 1, the free limit.
On the tilted axis the |Ω⟩ component of e^(−iHT)|0⟩ has modulus |⟨Ω|0⟩| e^(−εE₀T) = 0.98940 × e^(0.2202) = 1.233, more than 1: evolution in complex time is not unitary, which is why one divides by e^(−iE₀T)⟨Ω|0⟩ instead of renormalising by hand.
After that division the |1⟩ coefficient has modulus 0.14677 × e^(−εΔT) = 0.14677 × e^(−10.440) = 0.14677 × 2.92 × 10⁻⁵ = 4.3 × 10⁻⁶.

AnswerE₀ = −0.0220 GeV, |⟨Ω|0⟩| = 0.9894, Δ = 1.044 GeV; at T = 1000 GeV⁻¹ (6.6 × 10⁻²² s) the projected state is |Ω⟩ plus a 4.3 × 10⁻⁶ admixture of the excited state.

HardA contraction carries spatial momentum |p| = 0.40 GeV for a quantum of mass m = 0.30 GeV, and the vertex times run along t′(1 − iε) with ε = 0.0050. Find how fast the contraction now decays in real time, Fourier transform it to find the momentum-space denominator, ε′ and the poles, evaluate |D̃_F| on shell, and say what the opposite tilt would do.

Eₚ = √(0.40² + 0.30²) GeV = 0.50 GeV. On the tilted contour the contraction e^{(−iEₚ|t|)}/2Eₚ becomes e^{(−iEₚ|t′|)} e^{(−εEₚ|t′|)}/2Eₚ for either time order, decaying with time constant 1/(εEₚ) = 400 GeV⁻¹ = 2.63 × 10⁻²² s, or 78.9 fm of light travel.
Fourier transform. The positive-t′ half gives i/(p⁰ − Eₚ(1 − iε)), convergent because the exponent has real part −εEₚt′; the negative-t′ half gives −i/(p⁰ + Eₚ(1 − iε)). Their sum, divided by 2Eₚ, is i(1 − iε)/((p⁰)² − Eₚ²(1 − iε)²).
Expand: (p⁰)² − Eₚ²(1 − 2iε − ε²) = p² − m² + 2iεEₚ² + ε²Eₚ². So ε′ = 2εEₚ² = 2 × 0.0050 × 0.25 GeV² = 2.5 × 10⁻³ GeV², while the real shift ε²Eₚ² = 6.3 × 10⁻⁶ GeV² vanishes faster as ε → 0. Poles: p⁰ = ±(0.5000 − 0.0025i) GeV, in the fourth and second quadrants, the Feynman placement.
On shell, p⁰ = Eₚ, the denominator is close to iε′, so |D̃_F| ≈ 1/ε′ = 400 GeV⁻²: finite only because ε is non-zero, and divergent as ε → 0, which is the one-particle pole. Dimensions: mass⁻², as i/(p² − m²) must be.
Opposite tilt, t′(1 + iε): the contraction grows as e^{(+εEₚ|t′|)}, the time integrals diverge, and formally the poles move to ±Eₚ(1 + iε), the anti-Feynman i/(p² − m² − iε′). The same reversal would amplify, not damp, the excited states in the vacuum projection.

AnswerEₚ = 0.50 GeV; decay time 400 GeV⁻¹ (2.6 × 10⁻²² s); denominator p² − m² + iε′ with ε′ = 2.5 × 10⁻³ GeV² and poles at ±(0.5000 − 0.0025i) GeV; |D̃_F| ≈ 400 GeV⁻² on shell. The opposite tilt gives anti-Feynman poles and no projection.