University Quantum Field Theory I · Fermions in Perturbation Theory: the Dirac Propagator, Fermion Signs and Yukawa Theory · 7.2
Wick's theorem for fermions: contractions with signs
Unit 2 built Wick's theorem for fields that commute. Feed it Dirac fields and one thing changes: every term now carries the sign of the permutation that pulls its contracted pairs together. Fix that rule once, here, and every fermion sign later in the course, loops and exchange diagrams included, follows from it.
Build the model
Connect the measurement to the mechanism.
Topic 2.08's theorem writes a time-ordered product of free fields as the normal-ordered product plus every way of contracting pairs, and its proof moved the latest field's annihilation part rightwards, collecting a commutator at each field. The Dirac field of UAQM-09.07 obeys anticommutators instead, and the fermion theorem is what that one substitution does to the proof. Each hop of ψ⁺ past another fermion field costs −1, so a contraction left behind after j hops carries (−1)ʲ, the sign of the permutation that makes that pair adjacent.
The bookkeeping closes only if time ordering and normal ordering carry the same permutation sign; then a fermion moves freely inside either symbol at one minus sign per swap, and T(ψψ̄) − :ψψ̄: is a c-number, topic 7.01's SF, whichever field is later. Fermion number does the rest: ψ never contracts with ψ, nor ψ̄ with ψ̄, so n ψ's and n ψ̄'s pair in n! ways, not (2n − 1)!!, and the signed sum is a determinant, det SF(xᵢ − yⱼ), which vanishes when two rows coincide, the Pauli principle as algebra. At equal times, with a Slater determinant playing the vacuum, it is UAQM-10.05's direct-minus-exchange rule.
The theorem is exact for free fields, all the interaction picture needs, and false for interacting Heisenberg fields; and its signs are relative, since a sign common to every term only records the order in which the fields were written.
- Simple definition
- Wick's theorem for fermions writes a time-ordered product of free anticommuting fields as its normal-ordered product plus every way of contracting ψ with ψ̄, each term weighted by the sign of the permutation that brings its contracted pairs together.
- Example
- ⟨0|T ψ(x₁)ψ̄(y₁)ψ(x₂)ψ̄(y₂)ψ(x₃)ψ̄(y₃)|0⟩ has 3! = 6 terms, three with +1 and three with −1, where six real scalars would give 15 unsigned ones; pairing x₁ with y₂, x₂ with y₁ and x₃ with y₃ is one swap from the identity, so it enters with −1.
Defined, not derived: it makes T antisymmetric, T(ψ(x)ψ̄(y)) = −T(ψ̄(y)ψ(x)) at all times. Topic 7.01 needed it for frame independence; here it makes every contraction a c-number.
each F is a component of ψ or ψ̄, mass dimension 3/2 (GeV³⁄²); P counts the swaps from the written order to latest-first
Defined, like T. With signs in both T and :⋯:, fermion fields anticommute freely inside either symbol, which is what lets the induction step carry one field through the rest.
ψ⁺ holds the electron annihilators b and ψ̄⁻ the creators b†; each swap that puts creators left costs −1; ⟨0|:⋯:|0⟩ = 0
Reversing the pair gives −SF(x − y), not SF(y − x): the propagator still runs from ψ̄(y) to ψ(x), and only the T-product's sign has changed.
C[AB] ≡ T(AB) − :AB:; α, β = 1 to 4 are spinor indices; SF(x) has mass dimension 3 (GeV³), quoted from topic 7.01
Shown from the modes: the anticommutators involved, [b, d†]₊ and [d, b†]₊, both vanish. So n ψ's and n ψ̄'s pair in n! ways, against (2n − 1)!! for 2n real scalars: 6 against 15 at n = 3.
ψ lowers the fermion number N = Σ(b†b − d†d) by one and ψ̄ raises it by one; a vacuum value needs a net change of zero
Derived by extending topic 2.08's induction: each hop of the latest field's ψ⁺ costs −1, so a contraction left after j hops carries (−1)ʲ, the permutation's sign.
each pair taken with i written before j; P counts the swaps that make every pair adjacent, the rest keeping their written order
Derived: n! signed terms. Two equal rows, x₂ = x₁ with one spinor index, make it vanish, the Pauli principle; a complex scalar sums the same products unsigned, a permanent, which does not.
n × n, the (i, j) entry being SF(xᵢ − yⱼ) at the spinor indices of ψ(xᵢ) and ψ̄(yⱼ); mass dimension 3n
Why both orderings must carry signs
Natural units, ħ = c = 1, and the metric (+, −, −, −) throughout. Split the free Dirac field of UAQM-09.07 as ψ = ψ⁺ + ψ⁻, with the electron annihilators b in ψ⁺ and the positron creators d† in ψ⁻; in ψ̄, ψ̄⁺ holds d and ψ̄⁻ holds b†. Write [A, B]₊ = AB + BA: every anticommutator of these pieces is a c-number. A contraction is T(ψψ̄) − :ψψ̄:, and it is useful only if it is a number. When x is the later point, T(ψ(x)ψ̄(y)) − :ψ(x)ψ̄(y): = [ψ⁺(x), ψ̄⁻(y)]₊, a number. When y is later, the signed T = −ψ̄(y)ψ(x) gives −[ψ̄⁺(y), ψ⁻(x)]₊, a number again; an unsigned T = +ψ̄(y)ψ(x) would give 2:ψ̄(y)ψ(x): plus a number, an operator, and no expansion in contractions could close. An unsigned normal ordering fails sooner still: ψ⁺ψ̄⁻ − ψ̄⁻ψ⁺ is built from bₚ b†q − b†q bₚ = δpq − 2b†q bₚ, an operator. So T and :⋯: both carry (−1)ᴾ, and a fermion field can then be moved anywhere inside either symbol for one minus sign per swap. Topic 7.01 needed the sign in T for Lorentz invariance; the algebra needs it too.
Only ψ meets ψ̄
Try to contract two ψ's. With x later, C[ψα(x)ψβ(y)] = [ψ⁺α(x), ψ⁻β(y)]₊, every term of which is a multiple of [b, d†]₊ = 0; with y later it is −[ψ⁺β(y), ψ⁻α(x)]₊ = 0 for the same reason, and C[ψ̄ψ̄] vanishes through [d, b†]₊ = 0. The reason outlives free fields: with N = Σ(b†b − d†d), [N, ψ] = −ψ, so ψψ lowers N by two and has no vacuum value in any theory that conserves N. The one surviving contraction is topic 7.01's, quoted: C[ψ(x)ψ̄(y)] = SF(z) = ∫ d⁴p/(2π)⁴ i(p̸ + m) e(−ip⋅z)/(p² − m² + iε), with z = x − y, and C[ψ̄(y)ψ(x)] = −SF(z). Dimensions check: d⁴p brings 4 and i(p̸ + m)/p² brings −1, so SF carries 3, as ψψ̄ does. The counting changes: 2n real scalars have (2n − 1)!! complete pairings, but n ψ's and n ψ̄'s have n!, so 3 against 2 at n = 2, 15 against 6 at n = 3 and 105 against 24 at n = 4. That is the count of topic 2.07's complex scalar, now with signs. A Majorana field, its own antiparticle, would let ψ meet ψ; it is named, not used.
The induction step, with one new sign per hop
Topic 2.08's statement and induction are quoted, not repeated: with x₁ the latest time, φ₁⁻ already stands on the left, and moving φ₁⁺ rightwards through the (n − 1)-field result leaves one commutator, a contraction, at each field it passes. Two things change for fermions. Both sides of the theorem are now antisymmetric under a swap of two fields, so relabelling the latest field as F₁ costs the same sign on each side and changes nothing. And each hop is now F₁⁺Fₖ = −FₖF₁⁺ + [F₁⁺, Fₖ]₊, whose anticommutator is C[F₁Fₖ] from Fₖ's creation part and zero from its annihilation part. So after hopping over j uncontracted fields, F₁⁺ leaves C[F₁Fₖ] multiplied by (−1)ʲ, and j is exactly the number of swaps that would carry Fₖ leftwards to stand beside F₁. The piece of F₁⁺ that reaches the right-hand end has paid −1 per hop, which is the sign :⋯: assigns to an annihilator moved there. Every term of the n-field theorem appears once, with its permutation sign. Contractions are c-numbers, so once their sign is paid they can stand anywhere.
Reading signs off a contraction diagram
Counting swaps one at a time invites slips, so draw instead. Write the fields in a row and join each contracted pair by an arc above them. Bringing one end of an arc to the other passes every field in between: those belonging to arcs nested inside it come in pairs, and the rest are uncontracted or belong to arcs that cross it, so the permutation sign is −1 per crossing and −1 per uncontracted field beneath an arc; a full contraction has none of the latter. Each arc whose ψ̄ stands left of its ψ adds another −1, because C[ψ̄ψ] = −SF. Take ⟨0|T ψ(x₁)ψ̄(y₁)ψ(x₂)ψ̄(y₂)ψ(x₃)ψ̄(y₃)|0⟩, the figure's order 0. Pairing x₁y₂, x₂y₃, x₃y₁ draws three mutually crossing arcs, one of them ψ̄-first: (−1)⁴ = +1. Pairing x₁y₂, x₂y₁, x₃y₃ nests one arc inside another, the inner one ψ̄-first: −1. In general the term pairing xᵢ with yσ(i) carries the sign of σ, so the six terms are det[SF(xᵢ − yⱼ)], and n pairs give n! terms, half of each sign once n ≥ 2. Checks: n = 1 returns SF, the result has mass dimension 3n, and two equal rows, x₂ = x₁ with one spinor index, give zero where a complex scalar's unsigned sum would not.
The equal-time case is UAQM-10.05's determinant
Put every operator at one time and T has nothing left to reorder: the written order stands. Now let a Slater determinant |Φ⟩ play the vacuum. For an empty orbital c annihilates |Φ⟩ and c† creates; for an occupied one the roles swap, c† annihilating |Φ⟩ and c creating a hole, just as b and d† share ψ. Normal ordering relative to |Φ⟩ carries the proof through unchanged, with contractions ⟨Φ|cᵢ†cₖ|Φ⟩ = δᵢₖnᵢ, and ⟨cᵢ†cⱼ†⟩ = ⟨cₗcₖ⟩ = 0 by particle number, the analogue of C[ψψ] = 0. In ⟨Φ|cᵢ†cⱼ†cₗcₖ|Φ⟩ the pairing i with k, j with l nests, sign +, and i with l, j with k crosses, sign −: ⟨cᵢ†cₖ⟩⟨cⱼ†cₗ⟩ − ⟨cᵢ†cₗ⟩⟨cⱼ†cₖ⟩, the direct-minus-exchange rule UAQM-10.05 proved by pushing annihilators to the vacuum, and for p creators and p annihilators the p × p determinant it stated. That relative minus sign is the one that puts helium's 2³S triplet below its 2¹S singlet, measured 0.796 eV apart. The path integral reaches the same determinants by treating classical fermion fields as Grassmann numbers, θθ′ = −θ′θ; that route is named here, not taken.
Bilinears move freely, and where the theorem stops
Interaction densities hold fermion fields in pairs, gφψ̄ψ or igφψ̄γ⁵ψ, and carrying a pair past any other field costs (−1)² = +1. Whole vertices therefore commute inside T: the n! orderings of n identical vertices in the Dyson series stay identical and the 1/n! still cancels, exactly as in Unit 3. Signs enter only through the way contractions thread the individual fields, and that threading is what the next three topics exploit: contractions with external states in topic 7.03, the closed loop in topic 7.04 and the exchange of identical fermions in topic 7.05. The theorem has a clear boundary. It is an identity among free fields, linear in one set of ladder operators, so it is exact for the interaction-picture fields inside the Dyson series and false for interacting Heisenberg fields, whose vacuum correlators reach it only through the Gell-Mann-Low formula. And it decides nothing about which order the external fields are written in: that choice multiplies every term by one common sign.
Change one variable at a time
Make the relationship visible.
In order 0, step k from 1 to 6: pairings 1 to 3 shift the ψ̄ cyclically and keep +1, pairings 4 to 6 swap two and take −1, so the six terms build det SF. Order 1 flips every sign, being three swaps from order 0; order 2 flips them back. Crossings plus ψ̄-first arcs always match the sign.
ARC CROSSINGS3
ARCS WITH ψ̄ FIRST1
SIGN OF THIS TERM1
PERMUTATION SIGN sgn σ1
Live interpretationARC CROSSINGS: 3. ARCS WITH ψ̄ FIRST: 1. SIGN OF THIS TERM: 1. PERMUTATION SIGN sgn σ: 1
Catch the common trap
Explain before calculating.
Using Wick's theorem for free Dirac fields, which expression equals ⟨0|T ψ̄(y₁)ψ(x₁)ψ̄(y₂)ψ(x₂)|0⟩? Spinor indices are suppressed: each SF carries the indices of its own ψ and ψ̄.
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyExpand T(ψ(x₁)ψ(x₂)ψ̄(y₁)ψ̄(y₂)) for free Dirac fields by Wick's theorem with every sign, writing Sᵢⱼ = SF(xᵢ − yⱼ) and keeping uncontracted fields in their written order. Count the terms and compare with four real scalar fields.
- Only ψ–ψ̄ pairs contract, so of the six pairs just x₁y₁, x₁y₂, x₂y₁ and x₂y₂ can appear; ψ(x₁)ψ(x₂) and ψ̄(y₁)ψ̄(y₂) give zero.
- Single contractions cost −1 for each field the ψ̄ hops to reach its ψ. y₁ hops ψ(x₂) to reach x₁: −S₁₁ :ψ(x₂)ψ̄(y₂):. y₂ hops ψ̄(y₁) and ψ(x₂): +S₁₂ :ψ(x₂)ψ̄(y₁):. y₁ already stands beside x₂: +S₂₁ :ψ(x₁)ψ̄(y₂):. y₂ hops ψ̄(y₁): −S₂₂ :ψ(x₁)ψ̄(y₁):.
- Full contractions: x₁y₁ with x₂y₂ draws two crossing arcs, sign −; x₁y₂ with x₂y₁ nests, sign +. So ⟨0|T ψ(x₁)ψ(x₂)ψ̄(y₁)ψ̄(y₂)|0⟩ = −S₁₁S₂₂ + S₁₂S₂₁.
- Checks: swapping x₁ and x₂ turns this into −S₂₁S₁₂ + S₂₂S₁₁, minus itself, as antisymmetry demands; at x₂ = x₁ with equal spinor indices it vanishes; its mass dimension is 4 × 3/2 = 6 = 3 + 3.
- Count: 1 normal-ordered term, 4 single contractions and 2 double ones, 7 in all. Four real scalars allow all 6 pairs and 3 complete pairings: 1 + 6 + 3 = 10 terms.
AnswerT = :ψ(x₁)ψ(x₂)ψ̄(y₁)ψ̄(y₂): − S₁₁:ψ(x₂)ψ̄(y₂): + S₁₂:ψ(x₂)ψ̄(y₁): + S₂₁:ψ(x₁)ψ̄(y₂): − S₂₂:ψ(x₁)ψ̄(y₁): − S₁₁S₂₂ + S₁₂S₂₁: 7 terms, against 10 for four real scalars.
MediumIn Yukawa theory HI = gφψ̄ψ, with the vertex normal ordered, and the second-order Dyson term is ((−i)²/2!) ∫d⁴x d⁴y T[HI(x)HI(y)]. Find the sign of each term in which exactly one fermion contraction joins the two vertices, show that these terms add rather than cancel, and give the coefficient that survives.
- Write the fermion fields as ψ̄α(x)ψα(x)ψ̄β(y)ψβ(y), indices summed. ψψ and ψ̄ψ̄ pairs vanish and same-vertex pairs are excluded by normal ordering, so of the six pairs only ψ(x) with ψ̄(y) and ψ̄(x) with ψ(y) remain.
- ψα(x) and ψ̄β(y) stand adjacent in ψ-first order: +SF(x − y)αβ :ψ̄α(x)ψβ(y): = +:ψ̄(x)SF(x − y)ψ(y):.
- For ψ̄α(x) with ψβ(y), carry ψβ(y) to second place past ψ̄β(y) and ψα(x): two swaps, +1. The pair is ψ̄-first, so its contraction is −SF(y − x)βα, leaving −SF(y − x)βα :ψα(x)ψ̄β(y):, and swapping inside :⋯: costs one more −1: +:ψ̄(y)SF(y − x)ψ(x):.
- Both terms carry +, and each is the other with x and y relabelled, as are the scalar fields φ(x)φ(y) beside them. They add to twice one of them, which cancels the 1/2!, leaving (−ig)² = −g² as the coefficient.
- Drop the reversal sign and the second term comes out −, the pair cancels, and fermion-scalar scattering would vanish at order g²: absurd. Dimensions: [φ] = 1 and [ψ̄ψ] = 3 make g dimensionless, and d⁴x d⁴y (−8) with ψ̄ SF ψ (3/2 + 3 + 3/2) and φφ (2) gives 0, as an S-matrix term must.
AnswerTwo terms survive, both with sign +, and they are equal after x ↔ y, so the 1/2! cancels: −g² ∫d⁴x d⁴y ψ̄(x)SF(x − y)ψ(y) φ(x)φ(y), with the remaining fields normal ordered or contracted further.
HardFor free Dirac fields let G = ⟨0|T ψ(x₁)ψ̄(y₁)ψ(x₂)ψ̄(y₂)ψ(x₃)ψ̄(y₃)|0⟩. Find the sign of each complete pairing from its contraction diagram, show that G vanishes when x₂ = x₁ with equal spinor indices, find the sign relating G to the same fields written ψ(x₁)ψ(x₂)ψ(x₃)ψ̄(y₁)ψ̄(y₂)ψ̄(y₃), and count every term of the full Wick expansion.
- Only ψ–ψ̄ arcs exist, so a complete pairing sends each xᵢ to yσ(i): 3! = 6 pairings, against 15 for six real scalars. With the fields in places 1 to 6, the arc from ψ(xᵢ) to ψ̄(yσ(i)) is ψ̄-first whenever σ(i) is smaller than i.
- Identity: three separate arcs, +1. Each single swap, such as x₁y₂, x₂y₁, x₃y₃, has no crossings and one ψ̄-first arc: −1. The cycle x₁y₂, x₂y₃, x₃y₁ has three crossings and one ψ̄-first arc: +1. The cycle x₁y₃, x₂y₁, x₃y₂ is one arc enclosing two ψ̄-first arcs: +1. These are the signs of σ, so G = det[SF(xᵢ − yⱼ)].
- Set x₂ = x₁ with one spinor index, so rows 1 and 2 coincide. Pair each σ with the σ′ that swaps the targets of x₁ and x₂: equal products, opposite signs, so the six terms cancel in three pairs and G = 0. A complex scalar, all signs +, gives 2(D₁₂D₁₃D₃₁ + D₁₁D₁₃D₃₂ + D₁₁D₁₂D₃₃), with Dᵢⱼ = DF(xᵢ − yⱼ): no exclusion.
- Regrouping ψ(x₁)ψ̄(y₁)ψ(x₂)ψ̄(y₂)ψ(x₃)ψ̄(y₃) into ψ(x₁)ψ(x₂)ψ(x₃)ψ̄(y₁)ψ̄(y₂)ψ̄(y₃), ψ(x₂) passes ψ̄(y₁) and ψ(x₃) passes ψ̄(y₂) and ψ̄(y₁): three swaps, so the grouped correlator is −G. For n pairs the swap count is n(n − 1)/2.
- Full expansion: a term with r contractions picks r of the three ψ's in 3!/(r!(3 − r)!) ways, r of the three ψ̄'s in as many, and matches them in r! ways, so r = 0 to 3 give 1 + 9 + 18 + 6 = 34 terms. Six real scalars allow any pairs: 1 + 15 + 45 + 15 = 76. Dimension: G carries 6 × 3/2 = 9, three propagators' worth.
Answer+1 for the identity and both cycles, −1 for the three swaps, so G = det[SF(xᵢ − yⱼ)], zero at x₂ = x₁ with equal indices; the grouped order gives −G; the full expansion has 34 terms, against 76 for six real scalars.