University Quantum Field Theory I · Interacting Fields: the Interaction Picture, the Gell-Mann-Low Formula and Wick's Theorem · 2.2
The Heisenberg field, |Ω⟩ and correlation functions
Before drawing a single diagram you should know what the diagrams compute. Once λ is on, the field loses its plane-wave expansion and the vacuum stops being the Fock vacuum, so what defines the theory is a set of vacuum correlation functions. Here is why, with the first-order vacuum written out.
Build the model
Connect the measurement to the mechanism.
Take φ⁴, L = ½∂μφ∂μφ − ½m²φ² − (λ/4!)φ⁴, and let the operators evolve with the full Hamiltonian. Two things survive the interaction and two do not. The field obeys its Euler-Lagrange equation as an operator identity, (∂² + m²)φ = −(λ/6)φ³, and keeps [φ(t, x), π(t, y)] = iδ³(x − y) at every time, because both are the t = 0 operators conjugated by the same e(iHt).
But the cubic term ties every mode to every other, so no expansion with constant a(p) and plane waves solves the equation; and |0⟩ is no longer the ground state, because Hᵢₙₜ turns it into pairs |k, −k⟩ and quartets with p₁ + p₂ + p₃ + p₄ = 0, each held down by an energy denominator of at least 2m or 4m but multiplied in norm by the volume of space, so that ⟨0|Ω⟩ → 0 as V → ∞. With neither the modes nor the vacuum known, the theory is specified by its correlation functions ⟨Ω|T φ(x₁)…φ(xₙ)|Ω⟩, Heisenberg fields in the true vacuum: defined without knowing the spectrum, Poincaré invariant, and exactly what perturbation theory computes. The price is that nothing comes free any more: the m in L need not be the measured mass, φ need not create one particle with unit amplitude, and particles and the S-matrix must be extracted from these functions in Unit 4.
Everything here is first order in λ, and the vanishing overlap warns the next topics not to lean on |Ω⟩ ≈ |0⟩.
- Simple definition
- An n-point correlation function is the vacuum expectation value ⟨Ω|T φ(x₁)…φ(xₙ)|Ω⟩ of time-ordered Heisenberg fields, taken in the ground state |Ω⟩ of the full Hamiltonian rather than in the free vacuum |0⟩.
- Example
- For φ⁴ in one spacetime dimension at g = λ/m³ = 1.0, the first-order vacuum is |Ω⟩ = |0⟩ − 0.044|2⟩ − 0.013|4⟩, and the equal-time two-point function ⟨Ω|φ²|Ω⟩ is 0.875 of the free 1/(2m); diagonalising H numerically gives 0.906.
The cubic term couples every mode to every other, so no superposition of free plane waves solves it; at λ = 0 it is UAQM-09.02's Klein-Gordon equation.
φ the Heisenberg field, mass dimension 1 (MeV); m in MeV; λ dimensionless; every term is mass³ in ħ = c = 1.
Both fields are their t = 0 forms conjugated by one unitary e(iHt), which preserves the commutator; at unequal times it is an operator, not UAQM-09.04's c-number.
δ³ in fm⁻³, mass³ in ħ = c = 1, matching [φ] + [π] = 1 + 2. π = ∂ₜφ because Lᵢₙₜ has no derivatives.
Zero only at λ = 0. aₚ(t) keeps [aₚ, aq†] = (2π)³δ³(p − q) at each t but rotates into three-field products, so φ has no expansion with constant aₚ.
aₚ(t) = e(iEₚ t) ∫ d³x e(−ip⋅x)(Eₚφ + iπ)/√(2Eₚ); p⋅x a 3-vector product; Eₚ = √(p² + m²); aₚ is mass(−3/2).
Four, two or no quanta: 6 counts where an annihilation part can meet a creation part to its right, 3 the ways two pairs close. ∫ d³x then sets total momentum to zero.
φ⁻ is the creation part of the t = 0 field; D₀ = ∫ d³k/((2π)³ 2Eₖ), mass², growing as Λ²/(8π²) with a cutoff Λ.
Zero total momentum keeps |Ω⟩ translation invariant, but the norm carries (2π)³δ³(0) = V and so grows with the volume of space.
|p₁…p₄⟩ = a†…a†|0⟩; first order in λ; the energy denominator E₁ + E₂ + E₃ + E₄ is at least 4m; mass dimension −6, four times −3/2.
Defined without knowing the spectrum: G⁽²⁾ → DF(x₁ − x₂) as λ → 0, and every odd G⁽ⁿ⁾ vanishes by φ → −φ while |Ω⟩ is unique.
Heisenberg fields; ⟨Ω|Ω⟩ = 1; vacuum energy subtracted so that H|Ω⟩ = 0; mass dimension n, since [φ] = 1.
The Heisenberg field obeys a nonlinear equation
Work in ħ = c = 1 with the metric (+, −, −, −) and UQFTI-02.01's model, [φ] = 1 and λ dimensionless. The interaction has no derivatives, so π = ∂L/∂(∂ₜφ) = ∂ₜφ as before, and H = ∫ d³x [½π² + ½(∇φ)² + ½m²φ² + (λ/4!)φ⁴] = H₀ + Hᵢₙₜ. Let the operators evolve with the full H, φ(t, x) = e(iHt)φ(0, x)e(−iHt). Heisenberg's iφ̇ = [φ, H] returns φ̇ = π. For iπ̇ = [π, H], the equal-time rule [π(x), φ(y)⁴] = −4iφ(y)³δ³(x − y) gives [π(x), Hᵢₙₜ] = −(iλ/6)φ(x)³, so π̇ = ∇²φ − m²φ − (λ/6)φ³. Together, (∂² + m²)φ = −(λ/6)φ³: the Euler-Lagrange equation as an operator identity, derived. Every term has mass dimension 3, and at λ = 0 it is the Klein-Gordon equation UAQM-09.02 solved.
The commutator survives; the free expansion does not
Both fields at time t are their t = 0 forms conjugated by one unitary, so [φ(t, x), π(t, y)] = e(iHt)[φ(0, x), π(0, y)]e(−iHt) = iδ³(x − y) at every t, derived. What fails is the mode expansion. Any free form with constant aₚ, of any mass, makes (∂² + m²)φ a multiple of φ, and φ³ is not. To watch it fail mode by mode, define aₚ(t) = e(iEₚ t)∫ d³x e(−ip⋅x)(Eₚφ + iπ)/√(2Eₚ), with p⋅x the three-vector product; for a free field this is UAQM-09.02's constant aₚ. It still obeys [aₚ(t), aq†(t)] = (2π)³δ³(p − q), but after using π̇ every term cancels except daₚ/dt = −(iλ/(6√(2Eₚ)))e(iEₚ t)∫ d³x e(−ip⋅x)φ³. The annihilation operators rotate into three-field products, so aₚ†(t)|Ω⟩ has no reason to be a state of energy Eₚ. Dimensions: aₚ is mass(−3/2), and the right side is mass(−1/2), one power higher, as a time derivative requires.
The free vacuum is not an eigenstate of H
H₀|0⟩ = 0, but H|0⟩ = Hᵢₙₜ|0⟩, which is not a multiple of |0⟩. Split the t = 0 field as φ⁺ + φ⁻, with φ⁺ holding the annihilation operators, and expand (φ⁺ + φ⁻)⁴|0⟩. A φ⁺ standing rightmost kills |0⟩; a φ⁺ with j factors of φ⁻ to its right can meet any one of them, each meeting leaving the c-number D₀ = [φ⁺(x), φ⁻(x)] = ⟨0|φ(x)²|0⟩. One φ⁺ in the first, second or third place gives 3 + 2 + 1 = 6 ways to leave a pair; two φ⁺ close in 3 ways, the 3D₀² of ⟨0|φ⁴|0⟩. So φ⁴|0⟩ = [(φ⁻)⁴ + 6D₀(φ⁻)² + 3D₀²]|0⟩, shown. Integrating over x gives each piece zero total momentum: Hᵢₙₜ|0⟩ holds quartets with p₁ + p₂ + p₃ + p₄ = 0, pairs with ⟨k₁k₂|Hᵢₙₜ|0⟩ = λD₀(2π)³δ³(k₁ + k₂)/(4Eₖ), and the constant (λ/8)D₀²V times |0⟩. No odd number appears, because H is even under φ → −φ. A normal-ordered Hᵢₙₜ keeps only the quartets, as UQFTI-02.09 shows.
The first-order vacuum, and why it outgrows |0⟩
First-order perturbation theory, quoted, gives |Ω⟩ = |0⟩ − Σ |n⟩⟨n|Hᵢₙₜ|0⟩/Eₙ over n ≠ 0, each piece divided by its free energy, at least 2m for a pair and 4m for a quartet: ⟨k₁k₂|Ω⟩ = −λD₀(2π)³δ³(k₁ + k₂)/(8Eₖ²), and the quartet card above, derived. Both carry zero momentum, so |Ω⟩ stays translation invariant. But the norm of each piece contains (2π)³δ³(0) = V: the pairs alone have probability Vλ²D₀²/(1024πm), using ∫ d³k/((2π)³Eₖ⁴) = 1/(8πm). However small λ is, the admixture grows without limit with the volume, and higher orders, organised by Unit 3's vacuum diagrams, turn 1 − aV into e(−aV), quoted: in infinite volume ⟨0|Ω⟩ = 0. That is why UQFTI-02.05 projects |Ω⟩ out instead of expanding about |0⟩. Limiting check: with one mode, V → 1 and D₀ → 1/(2m), these give the quartic oscillator's ⟨2|Ω⟩ = −√2g/32 and ⟨4|Ω⟩ = −√6g/192, with g = λ/m³: the figure's model.
Correlation functions are what the theory is
With neither modes nor vacuum known in closed form, the interacting theory is specified by G⁽ⁿ⁾(x₁, …, xₙ) = ⟨Ω|T φ(x₁)…φ(xₙ)|Ω⟩: Heisenberg fields, time-ordered, in the normalised true vacuum, with the vacuum energy, (λ/8)D₀²V at first order, subtracted from H so that H|Ω⟩ = 0. This is a definition, and every later unit computes it. Each property rests on something stated. Translation invariance of |Ω⟩ makes G⁽ⁿ⁾ depend only on the differences xᵢ − xⱼ; Lorentz invariance of |Ω⟩, with microcausality making T frame-independent for spacelike pairs (UQFTI-02.04), makes them invariant functions; φ → −φ with a unique |Ω⟩ makes every odd G⁽ⁿ⁾ vanish. G⁽ⁿ⁾ has mass dimension n. Free limit: at λ = 0, |Ω⟩ → |0⟩, G⁽²⁾ → DF(x₁ − x₂) of Unit 1, and G⁽⁴⁾ becomes three products of DF, counted in UQFTI-02.09. In the figure's one-mode theory, G⁽²⁾ at equal times is ⟨φ²⟩, which falls to 1 − λ/(8m³) of its free value.
What the interacting field no longer promises
Three things the free field guaranteed are now open, stated here and settled in Unit 4. The mass: at first order the pairs in |Ω⟩ are exactly the vacuum of a free field with m² raised by λD₀/2 (the Medium example), a shift growing as Λ², so the m in L is not the measured mass, 139.570 MeV for a charged pion; the measured mass must be read from G⁽²⁾. The normalisation: in the free theory ⟨0|φ(0)|p⟩ = 1 for |p⟩ = √(2Eₚ)aₚ†|0⟩, but the interacting φ also makes multi-quantum states from |Ω⟩, so its amplitude for the physical one-particle state need not be 1; its square is the Z of UQFTI-04.03. The particles: aₚ†(t) no longer creates them, so asymptotic states and the S-matrix must be extracted from G⁽ⁿ⁾ by LSZ, not assumed. Without φ → −φ, even ⟨Ω|φ|Ω⟩ can be nonzero, as the Hard example finds. At λ → 0 every promise returns.
Change one variable at a time
Make the relationship visible.
Raise g from 0 to 1: the solid curve climbs at φ = 0 and pulls in its shoulders, and the n = 2 bar grows 3.46 times faster than n = 4. At g = 2 first order puts ⟨φ²⟩ at 0.75 of free; numerical diagonalisation gives 0.84, so trust first order only at small g.
⟨2|Ω⟩, FIRST ORDER-0.044
⟨4|Ω⟩, FIRST ORDER-0.0128
|⟨0|Ω⟩|²0.9979
⟨φ²⟩ IN |Ω⟩ ÷ IN |0⟩0.875
Live interpretation⟨2|Ω⟩, FIRST ORDER: −0.044. ⟨4|Ω⟩, FIRST ORDER: −0.0128. |⟨0|Ω⟩|²: 0.9979. ⟨φ²⟩ IN |Ω⟩ ÷ IN |0⟩: 0.875
Catch the common trap
Explain before calculating.
In φ⁴ theory with Hᵢₙₜ = (λ/4!)∫ d³x φ⁴, not normal ordered, which free-field states does the true vacuum |Ω⟩ contain at first order in λ, besides |0⟩ itself?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyPut a φ⁴ field with the charged-pion mass, m = 139.570 MeV, and λ = 1.0 in a periodic cube of side L = 4.0 fm. At first order, what is the amplitude in |Ω⟩ of the state with all four quanta in the k = 0 mode? Check its dimension and its behaviour as L grows.
- In the box the k = 0 part of the t = 0 field is (a₀ + a₀†)/√(2mV), with [a₀, a₀†] = 1, and the normalised state is |4₀⟩ = (a₀†)⁴|0⟩/√24. Only the (φ⁻)⁴ term of φ⁴ reaches it, through ⟨4₀|(a₀†)⁴|0⟩ = √24.
- Integrating over the box gives a factor V: ⟨4₀|Hᵢₙₜ|0⟩ = (λ/24) × V × √24/(2mV)² = λ√24/(96m²V). With m = 139.570/197.327 fm⁻¹ = 0.7073 fm⁻¹ and V = 64 fm³, this is 1.594 × 10⁻³ fm⁻¹ = 0.3145 MeV.
- The energy denominator is 4m = 558.3 MeV, so ⟨4₀|Ω⟩ = −0.3145/558.3 = −5.63 × 10⁻⁴. Equivalently −(√6/192)λ/(m³V), with m³V = 22.65: the box holds 22.65 cubic Compton lengths.
- Checks: λ/(m³V) is dimensionless, as an amplitude must be. The zero mode taken alone is the quartic oscillator with g = λ/(m³V), so this is the figure's ⟨4|Ω⟩ at g = 0.0442. Doubling L divides it by 8, to −7.0 × 10⁻⁵.
- Small per state is not small in total: the number of quartets with zero total momentum below a fixed energy grows as V³, while each probability falls as V⁻², so their summed probability grows as V, the volume factor of the explanation.
Answer⟨4₀|Ω⟩ = −5.6 × 10⁻⁴, a probability of 3.2 × 10⁻⁷; it falls as 1/L³ per state while the number of such states grows as L⁹.
MediumShow that the two-quantum part of the first-order φ⁴ vacuum is exactly the vacuum of a free field with a shifted mass, find the shift, and evaluate it for m = 139.570 MeV, λ = 0.50 and a momentum cutoff Λ = 1.0 GeV on D₀.
- A free field of mass m′, m′² = m² + δm², has H′ = H₀ + ½δm²∫ d³x φ², with the shift treated as a perturbation of the mass-m theory. The counting that gave φ⁴|0⟩ gives φ²|0⟩ = [(φ⁻)² + D₀]|0⟩, and ⟨k₁k₂|∫ d³x (φ⁻)²|0⟩ = (2π)³δ³(k₁ + k₂)/Eₖ.
- So ⟨k₁k₂|H′ − H₀|0⟩ = ½δm²(2π)³δ³(k₁ + k₂)/Eₖ, and dividing by −2Eₖ, the vacuum of mass m′ contains ⟨k₁k₂|0′⟩ = −δm²(2π)³δ³(k₁ + k₂)/(4Eₖ²): pairs at zero total momentum, nothing else.
- φ⁴'s pairs are ⟨k₁k₂|Ω⟩ = −λD₀(2π)³δ³(k₁ + k₂)/(8Eₖ²). The two agree for every k when δm² = λD₀/2. At first order the pairs in |Ω⟩ are the vacuum of a field of mass² m² + λD₀/2; only the quartets are new.
- With the cutoff, D₀ = [Λ√(Λ² + m²) − m² ln((Λ + √(Λ² + m²))/m)]/(8π²) = 1.213 × 10⁴ MeV², close to the leading Λ²/(8π²) = 1.267 × 10⁴ MeV². Then δm² = 0.50 × 1.213 × 10⁴/2 = 3.03 × 10³ MeV², and m′ = √(1.948 × 10⁴ + 3.03 × 10³) MeV = 150.0 MeV.
- Checks: δm² is mass², since λ is dimensionless and D₀ is mass²; m′ → m as λ → 0. Doubling Λ to 2.0 GeV gives D₀ = 4.996 × 10⁴ MeV² and m′ = 178.8 MeV: the shift grows as Λ², so the m in L cannot be the measured 139.570 MeV unless it is tuned against Λ.
Answerδm² = λD₀/2 = 3.0 × 10³ MeV²: the pairs belong to the vacuum of a 150.0 MeV field, not 139.6 MeV, 7.5% heavier at Λ = 1.0 GeV and 28% at 2.0 GeV.
HardIn the scalar Yukawa model, Lᵢₙₜ = −gψ†ψφ, take ψ with M = 938.272 MeV, φ with m = 139.570 MeV, g = 100 MeV and a cutoff Λ = 1.0 GeV. Find ⟨Ω|φ|Ω⟩ at first order both from the Heisenberg equation and from the first-order vacuum, then find how |⟨0|Ω⟩|² depends on the volume.
- Heisenberg route: ∂L/∂φ gives (∂² + m²)φ = −gψ†ψ. In a translation-invariant |Ω⟩, ∂²⟨φ⟩ = 0, so m²⟨φ⟩ = −g⟨Ω|ψ†ψ|Ω⟩ = −gD₀(M) + O(g²), with D₀(M) = ⟨0|ψ†ψ|0⟩ = ∫ d³p/((2π)³ 2Eₚ), now at mass M.
- State route: ψ†ψφ|0⟩ = [ψ†⁻ψ⁻ + D₀(M)]φ⁻|0⟩, so Hᵢₙₜ|0⟩ holds ψψ̄φ triplets of zero charge and momentum, and a single φ quantum with ⟨k|Hᵢₙₜ|0⟩ = gD₀(M)(2π)³δ³(k)/√(2m). Charge conservation pairs ψ with ψ̄, but Lᵢₙₜ is odd in φ, so nothing forbids a single φ quantum, unlike in φ⁴.
- Dividing by −m: ⟨k|Ω⟩ = −gD₀(M)(2π)³δ³(k)/(m√(2m)). Only this piece survives in ⟨Ω|φ(x)|Ω⟩ = 2 Re ∫ d³k/(2π)³ ⟨0|φ(x)|k⟩⟨k|Ω⟩, with ⟨0|φ(x)|k⟩ = 1/√(2m) at k = 0, giving −gD₀(M)/m²: the two routes agree.
- Numbers: D₀(M) = [Λ√(Λ² + M²) − M² ln((Λ + √(Λ² + M²))/M)]/(8π²) = 7.03 × 10³ MeV², so ⟨φ⟩ = −100 × 7030/139.570² MeV = −36.1 MeV. Dimension: g is mass¹ and D₀/m² is mass⁰, matching [φ] = 1.
- Volume: in a box the zero-mode amplitude is α = −gD₀√(V/(2m))/m, with |α|² = g²D₀²V/(2m³) = 9.09 × 10⁴ MeV³ × V = 0.0118 V/fm³. Kept alone, this piece is UQFTI-01.09's constant classical source, whose vacuum is a coherent state with Poisson quanta, quoted: |⟨0|Ω⟩|² = e(−|α|²).
- So |⟨0|Ω⟩|² = 0.988 for V = 1 fm³, 0.23 for a 5 fm cube and 7 × 10⁻⁶ for a 10 fm cube, and zero in infinite volume. LSZ (Unit 4) needs ⟨Ω|φ|Ω⟩ = 0, which shifting φ by a constant arranges.
Answer⟨Ω|φ|Ω⟩ = −gD₀(M)/m² = −36 MeV by both routes; |⟨0|Ω⟩|² = e(−0.0118 V/fm³), 0.23 in a 5 fm cube and 7 × 10⁻⁶ in a 10 fm cube.