University Quantum Field Theory I · Feynman Diagrams for Scalar Field Theories · 3.9
Tree diagrams are classical; loops count powers of ħ
When a field theorist says tree level they mean classical, and this is where the word is earned. Iterate the φ⁴ field equation with a source and the tree diagrams appear one at a time, symmetry factors included; restore ħ and every loop costs exactly one more power of it.
Build the model
Connect the measurement to the mechanism.
Couple φ⁴ theory to a c-number source J and ask what field it makes. Classically the answer solves a nonlinear equation, (□ + m²)φ = J − (λ/6)φ³, which a Green's function turns into an integral equation and iteration turns into a series in J: each step feeds three copies of the previous answer into one more vertex. Draw the terms and every one is a tree, carrying exactly the weight 1/S that Wick's theorem gives the same tree in ⟨0|φ(x)|0⟩J/⟨0|0⟩J, because expanding φ³ does the same counting as the 1/4! and the relabelled vertices.
So the trees are the classical field. What the iteration never makes is a loop: it only multiplies fields that end on sources. Restoring ħ says why that matters.
Each propagator inherits one ħ from the canonical commutator and each vertex 1/ħ from the Dyson exponent, so a connected diagram is weighted ħ(I − V) = ħ(L − 1), and every loop is one more power of ħ: a quantum correction to a classical answer. The trees have limits of their own. With DF they give an in-out matrix element, complex in general, and only DR gives the causal field a detector records; for a uniform source the series converges only while λφ₀²/m² stays below 8/9, beyond which the classical field exists but the sum of trees, order by order, diverges; and the whole tree approximation holds only where the field's action is large compared with ħ.
- Simple definition
- Tree diagrams with source insertions sum to the solution of the classical field equation with that source, and with ħ restored a connected diagram with L loops carries ħ(L − 1), so the loops are the quantum corrections to it.
- Example
- A uniform source J = 0.30 m³ with λ = 2: trees with 0, 1, 2 and 3 vertices give φ = (0.30 − 0.0090 + 0.00081 − 0.00010) m = 0.2917 m, the root of m²φ + λφ³/6 = J. The one-loop tadpole on the same line carries one extra ħ.
Derived from L = ½(∂φ)² − ½m²φ² − (λ/4!)φ⁴ + Jφ. The λ/6 is 4 × λ/4!: differentiating φ⁴ has already chosen the leg that points toward x.
φ: field (mass dimension 1); J: c-number source (dimension 3); m: mass; λ: dimensionless coupling
Derived. Iterating it in powers of J generates the trees; G = iDF gives the tree part of ⟨0|φ|0⟩J/⟨0|0⟩J, G = iDR the causal classical field.
G: Green's function, (□ + m²)G = δ⁴(x − y), mass dimension 2; DF and DR from topics 1.06 and 1.07
Derived twice: the multinomial in expanding φ³ gives it classically, topic 3.02's orbit count gives it from Wick's theorem, so the two series agree term by term.
S: automorphisms with x fixed; kᵢ: copies of subtree Tᵢ on a vertex; a source leg has S = 1. One vertex: 6; two: 12
Derived. Every line carries zero momentum, so G = 1/m². λ → 0 leaves the free φ₀; at λ = 2 and φ₀ = 0.30 m the trees sum to 0.2917 m, the exact root of m²φ + λφ³/6 = J.
φ₀: free response (mass dimension 1); coefficients 1, 1, 3, 12, 55 times (λ/6)ⁿ; converges only while λφ₀²/m² is below 8/9
Derived. A tree in ⟨φ⟩J comes out ħ⁰ however many vertices it has; each loop adds one ħ, so loops are the quantum corrections.
I: internal lines, each DF ∝ ħ; V: vertices, each −iλ/ħ (a source iJ/ħ counts as one); L = I − V + 1 loops
Derived. Classically λ enters only through the field's size, λφ²/m²; each loop costs one ħλ, which is why topic 3.05 found one more λ per loop.
S̃: the action at λ = 1 with source √λ J; with ħ restored λ has dimension 1/ħ, and ħλ is the dimensionless coupling
Turn the field equation into a recursion
Add a c-number source to φ⁴ theory, L = ½(∂φ)² − ½m²φ² − (λ/4!)φ⁴ + Jφ, in natural units with the metric (+, −, −, −). The Euler-Lagrange equation is (□ + m²)φ = J − (λ/6)φ³, and λ/6 = 4 × λ/4! because differentiating φ⁴ picks one of its four legs. Let G = iDF, so (□ + m²)G = δ⁴(x − y) by topic 1.06's (□ + m²)DF = −iδ⁴, and the equation becomes φ = ∫G J − (λ/6)∫G φ³. Iterate in powers of J. The free response is φ⁽⁰⁾(x) = ∫d⁴y G(x − y)J(y); the next term is φ⁽¹⁾ = −(λ/6)∫d⁴z G(x − z)φ⁽⁰⁾(z)³, cubic in J; and expanding (φ⁽⁰⁾ + φ⁽¹⁾)³ to first order in φ⁽¹⁾ gives φ⁽²⁾ = (λ²/12)∫d⁴z G(x − z)φ⁽⁰⁾(z)²∫d⁴w G(z − w)φ⁽⁰⁾(w)³, quintic. Dimensions check: G has mass dimension 2 and J dimension 3, so ∫d⁴y G J has −4 + 2 + 3 = 1, that of φ, and λ is dimensionless, so every later term matches. Draw each G as a line and each J as a cross: every term is a tree rooted at x.
The same trees from Wick's theorem
Now quantise. Topic 1.09 drove a free field with a source; switch φ⁴ on and ask for ⟨0|φ(x)|0⟩J/⟨0|0⟩J. The Gell-Mann-Low exponent now holds i∫d⁴x (Jφ − λφ⁴/4!), so each source insertion brings iJ and an integral, each vertex −iλ, each contraction DF, and dividing by ⟨0|0⟩J removes every vacuum bubble (topic 3.03), topic 1.09's −½∫J DF J among them. One source: ∫d⁴y DF(x − y) iJ(y) = ∫G J = φ⁽⁰⁾. One vertex and three sources, with S = 6 for the three interchangeable source legs (topic 3.02): (1/6)∫d⁴z DF(x − z)(−iλ)[∫DF iJ]³ = −(λ/6)∫G(x − z)φ⁽⁰⁾(z)³ = φ⁽¹⁾. The factors of i always cancel like this. A tree rooted at x has one line for each vertex and one for each source, and DF(−iλ) = −λG while DF(iJ) = G J. So the iteration with G = iDF reproduces the trees of the quantum one-point function factor for factor, and that one-point function is those trees plus diagrams with loops.
Why every symmetry factor agrees
Cut a rooted tree at its root vertex and three subtrees hang below it, falling into types Tᵢ that appear kᵢ times, with k₁ + k₂ + … = 3. Classically φ³ = (sum of trees)³ produces that unordered triple 3!/∏kᵢ! times, and the λ/6 in front leaves λ/∏kᵢ!. So if each subtree carries 1/S(Tᵢ), the new tree carries 1/S with S = ∏kᵢ! S(Tᵢ)(kᵢ). That recursion is exactly the count of automorphisms of a tree whose root x is held fixed: swap identical branches at any vertex, permute inside each branch. Topic 3.02 showed that the quantum weight is 1/S with the same S, so the two series agree at every order. Check at order λ²: one tree, two source legs on the root and a three-leg branch, S = 2 × 3! = 12, against the iteration's λ²/12. What the iteration never does is join two legs of one vertex to each other, or one branch to another: it only multiplies fields that end on sources. Those contractions, such as the tadpole −(iλ/2)DF(0) on the line from x, exist only in the quantum expansion, and every one of them closes a loop.
DF or DR: one equation, two boundary conditions
Nothing in the iteration chose DF. Use G = iDR instead; (□ + m²)DR = −iδ⁴ as well (topic 1.07), so the same trees with the same weights now build a real field that vanishes before the source acts: the causal classical solution. That it is also the tree-level expectation value in the initial vacuum, the field a detector records, is quoted here, not derived. The DF series is the in-out matrix element ⟨0ₒᵤₜ|φ|0ᵢₙ⟩/⟨0ₒᵤₜ|0ᵢₙ⟩ instead, complex in general: once the source has acted, DF keeps only positive frequencies, e(−iEt). The two differ only in boundary conditions, because DF − DR = D(y − x) (topic 1.07) solves the free equation. At first order φF − φR = i∫d⁴y D(y − x)J(y), which samples the source only on the mass shell p² = m², so wherever no line of a tree can go on shell, as for a static source, the two choices give identical trees. For a static source, integrating either propagator over time gives −i e(−mr)/(4πr), so G is topic 1.09's Yukawa profile, of range 1/m: ħc/(139.57 MeV) = 1.41 fm for the pion.
Restoring ħ: one per line, one inverse per vertex
Put ħ back with c = 1 and hold the classical Lagrangian fixed, so m is an inverse length. Only two places in the quantum theory hold ħ. The canonical commutator [φ(t, x), π(t, y)] = iħδ³(x − y) of UAQM-09.02 is met by φ = √ħ φ̂, with φ̂ the ħ = 1 field, so the free two-point function, and with it every DF, carries one ħ. The Dyson exponent is −(i/ħ)∫HI dt = (i/ħ)∫d⁴x Lᵢₙₜ, so each vertex is −iλ/ħ and each source iJ/ħ. A connected diagram with I internal lines and V vertices therefore carries ħ(I − V), and topic 3.05's L = I − V + 1 turns that into ħ(L − 1). For ⟨φ⟩J, count each source as a vertex and add the line to x: a tree has one line per vertex and per source, so it is ħ⁰ however many vertices it has, while the tadpole correction to the same line has three lines against one vertex and one source, so ħ¹. Each loop is one power of ħ: loops are the quantum corrections to the classical field. In path-integral language this is UQMII-11.03's stationary phase, named here only.
ħ and λ travel together, and where the trees stop
If loops count ħ, why did topic 3.05 find one more λ per loop? Rescale φ = ψ/√λ and J = K/√λ: the action becomes S̃[ψ, K]/λ with S̃ free of λ, so the weight e(iS/ħ) sees ħ and λ only as the product ħλ, the number natural units call λ. Classically λ survives only through the field's size, as λφ²/m², so tree level does not mean weak coupling. For a uniform source every line carries zero momentum, G = 1/m², and the trees sum to the root of m²φ + λφ³/6 = J: φ = φ₀ − (λ/6)φ₀³/m² + (λ²/12)φ₀⁵/m⁴ − (λ³/18)φ₀⁷/m⁶ + …, with φ₀ = J/m² and every term at ħ⁰. The series converges only while λφ₀²/m² stays below 8/9, the value set by the cubic's branch points φ = ±im√(2/λ); beyond it the classical field still exists, but the sum of trees, taken order by order, diverges. And the classical answer is good only while the action the field carries is large compared with ħ, which is exactly when the loops, each worth one more ħλ, are small.
Change one variable at a time
Make the relationship visible.
Step the vertices kept from 0 to 4 at J/m³ = {j|2} and the open dot closes on the filled one from alternate sides. Slide J past the vertical dashes, where λJ²/m⁶ passes 8/9, and the trees swing ever wider while the solid curve stays smooth. Raise λ and that radius moves in.
CLASSICAL φ/m AT THE MARKER0.4662
SUM OF THE TREES KEPT0.4688
TREE SERIES RADIUS J/m³0.667
λJ²/m⁶, CONVERGES BELOW 8/90.500
Live interpretationCLASSICAL φ/m AT THE MARKER: 0.4662. SUM OF THE TREES KEPT: 0.4688. TREE SERIES RADIUS J/m³: 0.667. λJ²/m⁶, CONVERGES BELOW 8/9: 0.500
Catch the common trap
Explain before calculating.
In φ⁴ theory with a source J, restore ħ. Compare two contributions to ⟨0|φ(x)|0⟩J/⟨0|0⟩J: a tree with three vertices and seven source insertions, and the one-loop tadpole correction to the line from x to a single source. Which powers of ħ do they carry?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA connected, amputated φ⁴ diagram contributing to the six-point function has 5 internal lines. Restore ħ: find its number of vertices and loops, its power of ħ, and how it compares with the tree-level six-point diagram.
- Every φ⁴ vertex has four legs and every internal line uses two of them, so 4V = 2I + E = 10 + 6 = 16, and V = 4.
- Loops: L = I − V + 1 = 5 − 4 + 1 = 2 (topic 3.05), consistent with V = L + E/2 − 1 = 2 + 3 − 1 = 4.
- Powers of ħ: each internal DF carries ħ and each vertex −iλ/ħ, so the diagram carries ħ(5 − 4) = ħ¹, which is ħ(L − 1) as it must be.
- The tree has L = 0: V = E/2 − 1 = 2 and I = (4 × 2 − 6)/2 = 1, so it carries ħ(1 − 2) = ħ⁻¹ and λ². The two-loop diagram is down by ħ² and by λ², that is by (ħλ)², one ħλ per loop.
AnswerV = 4 and L = 2; the diagram carries ħ¹ against the tree's ħ⁻¹, a relative factor of (ħλ)², one ħλ for each loop.
MediumA spatially uniform source J(t) = J₀ cos ωt drives φ⁴ theory with λ = 3.0, J₀ = 0.20 m³ and ω = 0.50 m. Find the free response, then the one-vertex tree including the new harmonic it creates, and say whether the choice of DF or DR matters here.
- A uniform field obeys d²φ/dt² + m²φ = J − (λ/6)φ³, and G acting on cos Ωt multiplies it by 1/(m² − Ω²), for either propagator, whenever Ω ≠ m. Free response: φ⁽⁰⁾ = A cos ωt with A = J₀/(m² − ω²) = (0.20/0.75) m = 0.2667 m.
- Cube it: φ⁽⁰⁾³ = A³[(3/4) cos ωt + (1/4) cos 3ωt], with A³ = 0.018963 m³. The vertex feeds both frequencies back through G.
- One-vertex tree: φ⁽¹⁾ = −(λ/6)A³[(3/4) cos ωt/(m² − ω²) + (1/4) cos 3ωt/(m² − 9ω²)] = −0.5 × (0.018963 m³) × (1.000 cos ωt − 0.200 cos 3ωt)/m² = −0.00948 m cos ωt + 0.00190 m cos 3ωt.
- The fundamental falls by 3.6%, to 0.2572 m. The harmonic is driven above m, where m² − 9ω² = −1.25 m² is negative, so it responds with the opposite sign to its drive, as a forced oscillator above resonance does.
- DF − DR is the free solution D(y − x), which samples the source only at frequency m. Neither 0.5 m nor 1.5 m is on shell, so both propagators give the same tree; at ω = m/3 the harmonic would be on shell and the iteration would fail. Check: the periodic solution found numerically has 0.2580 m and 0.0018 m, the gaps being the order-λ² trees.
Answerφ(t) ≈ 0.2572 m cos ωt + 0.0019 m cos 3ωt through one vertex, the same for DF and DR since neither frequency equals m; the full classical solution has 0.2580 m and 0.0018 m.
HardFor a uniform source in φ⁴ theory, find every order-λ³ tree in ⟨φ⟩J with its symmetry factor and check that their sum matches the classical iteration. Then evaluate the tree series through λ³ for λ = 1.2 and J = 0.60 m³, and compare it with the exact classical field.
- A rooted tree with three four-leg vertices has 2V + 1 = 7 source legs. Two shapes exist: the chain, whose root and second vertex each hold two sources and a branch, ending on a vertex with three sources; and the fork, whose root holds one source and two identical three-source branches.
- Automorphisms with x fixed: the chain swaps the source pair at each of its first two vertices and permutes the last three, S = 2 × 2 × 3! = 24; the fork swaps its branches and permutes inside each, S = 2 × 3! × 3! = 72. Total weight (1/24 + 1/72)λ³ = λ³/18.
- Iteration: φ⁽³⁾ = −(λ/6)G[3φ⁽⁰⁾²φ⁽²⁾ + 3φ⁽⁰⁾(φ⁽¹⁾)²] with G = 1/m². The first term is the chain, 9(λ/6)³ = λ³/24, and the second the fork, 3(λ/6)³ = λ³/72: −(λ³/18)φ₀⁷/m⁶ in all, one term per tree.
- Numbers: φ₀ = J/m² = 0.60 m and (λ/6)φ₀²/m² = 0.072, so φ = (0.6000 − 0.0432 + 0.00933 − 0.00269) m = 0.5634 m.
- Newton's method on m²φ + λφ³/6 = J gives the exact classical field 0.56410 m, so the λ³ truncation is low by 0.00066 m, 0.12%. The four order-λ⁴ trees, together +55(λ/6)⁴φ₀⁹/m⁸ = +0.00089 m, overshoot by less; λφ₀²/m² = 0.43 lies inside 8/9, so the alternating series converges.
AnswerThe chain (S = 24) and the fork (S = 72) give λ³/18, the iteration's 12(λ/6)³; through λ³ the trees give φ = 0.5634 m against the exact 0.5641 m.