University Quantum Field Theory I · Feynman Diagrams for Scalar Field Theories · 3.3

Vacuum bubbles exponentiate and cancel

This is the step that makes the Gell-Mann-Low formula usable. Every numerator is crowded with vacuum bubbles, each proportional to the volume of spacetime; here you prove they sum to an exponential the denominator divides out exactly, and find out what that exponential was: the energy of the vacuum.

Open the lab ↓Practice & worked examples ↓

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Connect the measurement to the mechanism.

Work in natural units ħ = c = 1 with the metric (+, −, −, −). Expand a Gell-Mann-Low numerator with Wick's theorem and, beside the diagrams you want, come pieces that touch no external point: vacuum bubbles. They are not small. A connected bubble is unchanged when all its vertices slide together, so it carries one factor of the volume of spacetime, (2π)⁴δ⁴(0) = 2TL³, infinite as T and L grow.

The way out is combinatorial. No line joins a bubble to the rest, so a diagram's value factorises; the 1/n! of the Dyson series is spent relabelling vertices, except for a 1/nᵢ! when nᵢ identical copies of a bubble Vᵢ can be interchanged; and because every order is summed, the number of copies of each bubble is a free choice. So all diagrams sum to the bubble-free ones times exp(Σᵢ Vᵢ), and the denominator is that exponential with nothing attached: it divides out exactly, whatever the divergent Vᵢ are. The exponent has a meaning.

Topic 2.05's denominator turns as e^(−2iE₀T), so the part of Σᵢ Vᵢ that grows with T is −2iE₀T, and E₀ is a constant energy density times L³. It stops where its assumptions do: it needs ⟨0|Ω⟩ ≠ 0, so a finite box; reading the exponent as an energy needs a static setting, for with a time-dependent source the bubbles sum to topic 1.09's vacuum-persistence amplitude instead; and gravity, which couples to energy itself, sees the density that particle physics drops.

Simple definition: A vacuum bubble is a piece of a diagram joined to none of the external points; summed over every number of copies, the bubbles multiply each diagram by the same factor exp(Σᵢ Vᵢ), which is exactly the Gell-Mann-Low denominator, so they cancel.
Example: Of the 105 contractions in the order-λ² denominator of φ⁴ theory, 9 give two separate figure-eights, weight 9/1152 = 1/128 = (1/2!)(1/8)²: the V₈²/2! term of exp(V₈ + …), the 1/2! surviving because the two bubbles are identical.

A diagram that carries bubblesdiagram = E × Πᵢ Vᵢ^(nᵢ)/nᵢ!

Derived from topic 3.02's 1/S: a disjoint union has S = S_E × Πᵢ Sᵢ^(nᵢ) nᵢ!, since swapping identical bubbles is an automorphism. That nᵢ! is all that survives of the Dyson 1/n!.

E: bubble-free part, in GeV² for a two-point function; Vᵢ: connected bubble, dimensionless, 1/Sᵢ included; nᵢ: its copies

Bubbles exponentiateΣ all diagrams = (Σ bubble-free diagrams) × exp(Σᵢ Vᵢ)

Derived. For each bubble alone, the sum over nᵢ of Vᵢ^(nᵢ)/nᵢ! is exp(Vᵢ); the bubbles are independent choices, so the product of the exponentials is the exponential of the sum.

the sum runs over every external part E and every list n₁, n₂, …, each met once because every order n is summed

The Gell-Mann-Low ratio is bubble-free⟨Ω|T φ(x₁)…φ(xₙ)|Ω⟩ = Σ diagrams with no vacuum bubble

Derived, to all orders and order by order, without knowing any Vᵢ. Disconnected pieces that each touch an external point stay. At λ = 0 there are no bubbles and the denominator is 1.

denominator ⟨0|T exp(−i∫H_I dt)|0⟩ = exp(Σᵢ Vᵢ): its only external part is the empty diagram, worth 1

Each connected bubble carries spacetime's volumeVᵢ ∝ (2π)⁴δ⁴(0) = ∫d⁴z = 2TL³

Derived. Slide every vertex of a bubble together and nothing changes, so one ∫d⁴z is free. A term with n bubbles grows as (2TL³)ⁿ; the logarithm of the sum is extensive.

2T: elapsed time, in GeV⁻¹ (1 GeV⁻¹ = 6.58 × 10⁻²⁵ s); L³: box volume, in GeV⁻³ (1 fm³ = 130.1 GeV⁻³)

The exponent is the vacuum energyΣᵢ Vᵢ → −2iE₀T + ln|⟨0|Ω⟩|² as T → ∞(1 − iε)

Derived by matching topic 2.05's limit of ⟨0|U(T, −T)|0⟩. Each Vᵢ carries one L³, so E₀ is a constant density times L³, and ln|⟨0|Ω⟩|² is proportional to L³ as well.

E₀: energy of |Ω⟩ above the normal-ordered free vacuum, in GeV (1 GeV = 1.602 × 10⁻¹⁰ J); ⟨0|Ω⟩: a pure number

First order: the figure-eightE₀/L³ = (λ/8) D_F(0)²

Derived, and equal to ⟨0|H_I|0⟩/L³ = (λ/4!) × 3D_F(0)², first-order perturbation theory. λ = 1 and Λ = 1 GeV give 2.0 × 10⁻⁵ GeV⁴ = 4.2 × 10³² J m⁻³.

λ dimensionless; D_F(0) = ⟨0|φ²|0⟩ in GeV², Λ²/(8π²) for massless φ cut off at |k| = Λ; 1 GeV⁴ = 2.085 × 10³⁷ J m⁻³

Sort each contraction by what reaches the external points

Wick's theorem turns the Gell-Mann-Low numerator into a sum of full contractions, each drawn as a diagram. Cut every diagram into its connected components and sort them: a component holding at least one external point belongs to the external part E, and a component holding none is a vacuum bubble. No line joins E to a bubble, so the integrand is a product, and so is the value. At order λ the two-point numerator has 5!! = 15 contractions. Twelve put x and y on the vertex and make the tadpole, weight 12/4! = 1/2. Three join x straight to y and close the vertex on itself: D_F(x − y) times the figure-eight V₈ = (−iλ/8)∫d⁴z D_F(0)², the term topic 3.01 left for here. The external part need not be connected. In the four-point function D_F(x₁ − x₂)D_F(x₃ − x₄) has two pieces, but each touches an external point, so it is not a bubble.

Identical bubbles keep a 1/nᵢ!

Go to order λ², with vertices z and w and weight (1/2!)(1/4!)² = 1/1152 per contraction. With the tadpole on one vertex and a figure-eight on the other there are 2 × 12 × 3 = 72 contractions, weight 1/16 = (1/2)(1/8): the factor 2 for choosing which vertex carries the tadpole has eaten the 1/2!. With x joined to y and both vertices closed into figure-eights there are only 3 × 3 = 9, weight 1/128 = (1/2!)(1/8)². Swapping z and w now returns the same contractions rather than new ones, so nothing eats the 1/2!. Topic 3.02's rule says this in general. A diagram is weighted 1/S, and the automorphisms of a disjoint union are those of each piece together with every permutation of identical pieces, so S = S_E × Πᵢ Sᵢ^(nᵢ) nᵢ! and the diagram equals E × Πᵢ Vᵢ^(nᵢ)/nᵢ!.

Summing every order frees the choices

A diagram is now fixed by two independent choices: its external part E, and the number of copies nᵢ of each connected bubble Vᵢ. At a fixed order the choices are tied, because the vertices must add up to n. The Dyson series sums every n, though, so each combination of E with n₁, n₂, … appears exactly once, at whatever order its vertices total. The sum therefore factorises into the sum over E times, for each bubble, the sum over nᵢ of Vᵢ^(nᵢ)/nᵢ!, which is exp(Vᵢ): all diagrams add up to (Σ_E E) × exp(Σᵢ Vᵢ). The denominator ⟨0|T exp(−i∫H_I dt)|0⟩ is the same sum with no external points, whose only external part is the empty diagram, worth 1; it is exp(Σᵢ Vᵢ) itself. Dividing, every correlation function is the sum of its bubble-free diagrams. Nothing here used the value of any Vᵢ, which is fortunate, since each is infinite. At λ = 0 every Vᵢ vanishes and the free-field limit returns.

Each connected bubble grows with the volume of spacetime

A connected bubble is a product of propagators D_F between its vertices, integrated over every vertex position, and shifting all the vertices by one four-vector changes nothing. Change variables to one vertex and the separations from it. The separations give a factor that stops growing once the box is much larger than the range of D_F, apart from ultraviolet divergences where vertices meet, and the last ∫d⁴z gives the volume of spacetime, 2TL³ for a box of volume L³ held for an elapsed time 2T. In momentum space the same factor is (2π)⁴δ⁴(0): every vertex conserves momentum, and with no external momentum one delta function is left with zero argument. For the figure-eight, V₈ = (−iλ/8)D_F(0)² × 2TL³. A term with n bubbles therefore grows as (2TL³)ⁿ, so the bubble sum is not extensive, but its logarithm is: one power of the volume, the mark of an energy.

The exponent is the energy of the vacuum

Topic 2.05 projected the interacting vacuum out of the free one: as T runs to infinity along T(1 − iε), ⟨0|U(T, −T)|0⟩ → |⟨0|Ω⟩|² e^(−2iE₀T), E₀ being the energy of |Ω⟩ above the normal-ordered free vacuum. Matching the part that grows with T gives −2iE₀T; the constant left over is ln|⟨0|Ω⟩|², also proportional to L³, which is why the overlap falls exponentially with the volume. At first order only the figure-eight contributes, so E₀/L³ = (λ/8)D_F(0)². That is first-order perturbation theory: ⟨0|H_I|0⟩/L³ = (λ/4!)⟨0|φ⁴|0⟩ = (λ/24) × 3D_F(0)², the 3 counting the pairings of four fields. It has dimension M⁴, as an energy density must, is positive, as the mean of the positive operator (λ/4!)φ⁴ must be, and vanishes with λ. At order λ² the melon and dumbbell bubbles hold the four- and two-quantum states topic 2.02 found mixed into |Ω⟩, and they lower E₀, as any second-order shift of a ground state does: stated, not derived.

A constant density that only gravity can see

Cut a massless field off at |k| = Λ and D_F(0) = Λ²/(8π²), so the figure-eight gives E₀/L³ = λΛ⁴/(512π⁴): 2.0 × 10⁻⁵ GeV⁴ for λ = 1 and Λ = 1 GeV, or 4.2 × 10³² J m⁻³ with ħ and c restored, growing as Λ⁴ without limit. That is λ/(32π²) of the free zero-point density Λ⁴/(16π²), which normal ordering H₀ already removed. A constant added to H shifts every energy alike, and correlation functions, amplitudes and cross-sections depend only on differences; the cancellation of the bubbles is that statement, so the density is dropped. The Casimir force is no exception: it is how the zero-point energy changes with the spacing of two plates, finite and measured, and the constant density cancels from it. Gravity couples to energy itself, and there a constant density acts as a cosmological constant. The measured dark-energy density, quoted from the Planck 2018 fit, is 5.2 × 10⁻¹⁰ J m⁻³, about (2.2 meV)⁴, some 8 × 10⁴¹ times below the 1 GeV estimate: the cosmological-constant problem, named and not solved.

Change one variable at a time

Make the relationship visible.

Interactive model

Cutoff Λ on ⟨φ²⟩1.00 GeV

Box side L3.00 fm

Bubbles kept N2

Keep N = 2 and widen the box from 2 to 4 fm: E₀ grows as L³, the phase turns eight times faster, and the dashed curves leave the ±1 band ever sooner. Each extra bubble kept buys only about 0.4 rad of agreement; the summed bubbles never leave the band.

VACUUM ENERGY DENSITY E₀/L³2.61 MeV fm⁻³

BOX VACUUM ENERGY E₀70.5 MeV

PHASE E₀t AT t = 20 fm/c7.14 rad

BUBBLES IN TOP TERM AT 20 fm/c7

Live interpretationVACUUM ENERGY DENSITY E₀/L³: 2.61 MeV fm⁻³. BOX VACUUM ENERGY E₀: 70.5 MeV. PHASE E₀t AT t = 20 fm/c: 7.14 rad. BUBBLES IN TOP TERM AT 20 fm/c: 7

Catch the common trap

Explain before calculating.

Misconception check

The tempting story is that vacuum bubbles are dropped because they are small, or higher order, or because disconnected diagrams are unphysical. None of it is true. Each connected bubble is proportional to the volume of spacetime: with λ = 1 and a 1 GeV cutoff the figure-eight alone has magnitude about 4 × 10⁶⁶ for one cubic metre held for one second, so the numerator on its own is dominated by bubbles at every order and means nothing. They leave because the denominator is exactly their exponential, a cancellation that holds whatever their divergent values. And the rule is specific: only components touching no external point cancel. In the four-point function D_F(x₁ − x₂)D_F(x₃ − x₄) is disconnected yet bubble-free, and it stays; it describes two particles passing untouched, which topic 3.04 sets aside by a different argument.

In a box much larger than the range of D_F, hold the elapsed time 2T fixed and double the volume. What happens to the Gell-Mann-Low denominator ⟨0|T exp(−i∫H_I dt)|0⟩?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyMassless φ⁴ theory with λ = 0.60 is regulated by a sharp cutoff Λ = 2.0 GeV on spatial momenta. Keeping only the figure-eight in the exponent of the Gell-Mann-Low denominator, find the vacuum energy density in GeV⁴ and MeV fm⁻³, the energy E₀ of a box of side 1.0 fm, and how long its denominator takes to turn through 1 rad.

The exponent is V₈ = (−iλ/8)D_F(0)² × 2TL³, and its growth with T must be −2iE₀T, so E₀/L³ = (λ/8)D_F(0)² at this order.
D_F(0) = ⟨0|φ²|0⟩ = ∫d³k/((2π)³ 2|k|) up to |k| = Λ, which is Λ²/(8π²) = 4.0/78.96 = 0.05066 GeV².
E₀/L³ = (0.60/8) × (0.05066 GeV²)² = 0.075 × 2.566 × 10⁻³ GeV⁴ = 1.925 × 10⁻⁴ GeV⁴. Dimension M⁴, an energy density; positive, as ⟨0|H_I|0⟩ for positive λ must be; and zero in the free limit λ → 0.
Restore ħc = 0.197327 GeV fm: 1 GeV⁴ = 1 GeV/(0.197327 fm)³ = 130.1 GeV fm⁻³, so E₀/L³ = 25.05 MeV fm⁻³, or 4.0 × 10³³ J m⁻³. A box of (1.0 fm)³ holds E₀ = 25.05 MeV.
The denominator is e^(−iE₀t), so one radian takes t = ħ/E₀ = (197.327 MeV fm ÷ 25.05 MeV)/c = 7.88 fm/c = 2.6 × 10⁻²³ s. Double the box side and E₀ grows 8 times, so the phase turns 8 times faster.
Against the free zero-point density, ∫d³k/(2π)³ |k|/2 = Λ⁴/(16π²) = 16/157.9 = 0.1013 GeV⁴, the ratio is 1.90 × 10⁻³ = λ/(32π²): independent of Λ, so both grow as Λ⁴ without limit.

AnswerE₀/L³ ≈ 1.9 × 10⁻⁴ GeV⁴ ≈ 25 MeV fm⁻³ (4.0 × 10³³ J m⁻³); a (1.0 fm)³ box has E₀ ≈ 25 MeV and its denominator turns 1 rad in 7.9 fm/c ≈ 2.6 × 10⁻²³ s; 0.19% of the free zero-point density.

MediumIn φ⁴ theory, expand the Gell-Mann-Low ratio for ⟨Ω|T φ(x)φ(y)|Ω⟩ to order λ². Sort the full contractions of the order-λ² numerator by whether they carry vacuum bubbles, give the weight of each bubble-carrying class, and show that the denominator removes exactly those classes.

The order-λ² numerator is (1/2!)(−iλ/4!)² ∫d⁴z d⁴w ⟨0|T φ(x)φ(y)φ(z)⁴φ(w)⁴|0⟩: ten fields give 9!! = 945 full contractions, each weighted 1/(2! × 24²) = 1/1152.
Contract φ(x) with φ(y). The other eight fields give 7!! = 105: 9 close both vertices into figure-eights, 24 join z to w by four lines (the melon V₄₈), 72 give a self-loop at each vertex and two lines between (the dumbbell V₁₆). Weights 9/1152 = (1/2!)(1/8)², 24/1152 = 1/48, 72/1152 = 1/16.
Put x and y on one vertex and close the other into a figure-eight: 2 × (4 × 3) × 3 = 72 contractions, weight 72/1152 = 1/16 = (1/2) × (1/8). This is the tadpole times V₈, with no 1/2!, because the two pieces differ.
The remaining 945 − 105 − 72 = 768 contractions tie both vertices to x or y: bubble-free connected diagrams, 288 + 288 + 192 for two tadpoles in series, the tadpole on a tadpole and the sunset.
Denominator through λ²: 1 + V₈ + V₈²/2! + V₄₈ + V₁₆. The order-λ² part of (D_F + tadpole + C₂)(1 + V₈ + V₈²/2! + V₄₈ + V₁₆), C₂ the bubble-free λ² sum, is C₂ + tadpole × V₈ + D_F(V₈²/2! + V₄₈ + V₁₆): exactly the classes of steps 2 to 4.

Answer177 of the 945 contractions carry bubbles, and they are exactly D_F(x − y)(V₈²/2! + V₄₈ + V₁₆) + tadpole × V₈. The denominator removes them, leaving D_F + tadpole + the three connected λ² diagrams built from the other 768.

HardWithout evaluating a single integral, test exponentiation of the φ⁴ vacuum bubbles at order λ³. Count the contractions in the order-λ³ denominator, subtract what exp(Σᵢ Vᵢ) builds from the connected bubbles of orders λ and λ², and so predict Σ 1/S over the connected three-vertex bubbles; then find those diagrams and check.

Write g for −iλ and set every D_F and every ∫d⁴z to 1, so each diagram is worth its weight 1/S times a power of g. Exponentiation is a counting identity, so it holds for these weights: Σ cₙgⁿ = exp(Σ kₙgⁿ), with cₙ the total and kₙ the connected weight at order n.
Order 3: twelve fields give 11!! = 10395 contractions, each weighted 1/(3! × 24³) = 1/82944, so c₃ = 10395/82944 = 385/3072 = 0.12533.
From orders 1 and 2, k₁ = 1/8 and k₂ = 1/48 + 1/16 = 1/12. The g³ term of the exponential is k₃ + k₁k₂ + k₁³/3!, so k₃ = 385/3072 − 32/3072 − 1/3072 = 352/3072 = 11/96 = 0.11458.
Connected bubbles with three vertices of degree four: a triangle with a self-loop at each vertex, S = 3! × 2³ = 48; a triangle with every side doubled, S = 48; two vertices joined by three lines and each by one line to a third that carries a self-loop, S = 2 × 3! × 2 = 24; a chain with double links and a self-loop at each end, S = 2 × 2² × 2² = 32.
Σ 1/S = 1/48 + 1/48 + 1/24 + 1/32 = (2 + 2 + 4 + 3)/96 = 11/96, as predicted. In contractions: 1728 + 1728 + 3456 + 2592 = 9504 connected, and 27 + 216 + 648 = 891 for V₈³/3!, V₈ × V₄₈ and V₈ × V₁₆, total 10395.

AnswerThe connected order-λ³ bubbles weigh Σ 1/S = 11/96 ≈ 0.115, so the logarithm of the counted denominator is g/8 + g²/12 + 11g³/96 + …, confirmed by the four diagrams with S = 48, 48, 24 and 32.