University Quantum Field Theory I · Interacting Fields: the Interaction Picture, the Gell-Mann-Low Formula and Wick's Theorem · 2.7
Normal ordering and the contraction of two fields
Every internal line you will draw in a Feynman diagram is the number this topic pulls out of a pair of free fields. Take φ(x)φ(y) apart into an operator that knows the state and a c-number that does not, watch time ordering turn that c-number into DF, and learn which fields of a complex field may pair.
Build the model
Connect the measurement to the mechanism.
A free field is linear in ladder operators, φ = φ⁺ + φ⁻, with every annihilator in φ⁺ and every creator in φ⁻, and that linearity is all this topic uses. Multiply two such fields and only one term stands in the wrong order; moving it costs the commutator [φ⁺(x), φ⁻(y)], which is a number because [a, a†] is one. So φ(x)φ(y) splits into its normal-ordered part, an operator with zero vacuum value that carries everything about the state, plus a c-number that is the same in every matrix element.
Time ordering only chooses the c-number, D(x − y) when x is later and D(y − x) when y is, and the two glue into the Feynman propagator: Tφ(x)φ(y) = :φ(x)φ(y): + DF(x − y). DF is now earned as the contraction of the pair, not merely defined as a vacuum value, and since the Gell-Mann-Low formula turns every correlator into time-ordered products of free fields, every propagator in a Feynman diagram will be one such contraction. For a complex field the bookkeeping becomes a charge rule: a and b† commute, so φ never contracts with φ, only with φ†.
The identity stops where its premise does. An interacting Heisenberg field is linear in no set of ladder operators and its commutator is an operator; at coincident points the c-number D(0) diverges; and anticommuting fields bring signs, deferred to topic 7.02.
- Simple definition
- The contraction of two free fields is the c-number by which their time-ordered product exceeds their normal-ordered product; for a real scalar field it is the Feynman propagator DF(x − y), the same in every matrix element.
- Example
- For a π⁺ field at two points 2.0 fm apart at one time, T φ(x)φ†(y) = :φ(x)φ†(y): + 2.81 × 10⁻³ fm⁻², topic 1.08's DF, while T φ(x)φ(y) = :φ(x)φ(y): exactly: φ never contracts with φ.
Definition. φ⁺|0⟩ = 0 and ⟨0|φ⁻ = 0, so any product with every φ⁻ left of every φ⁺, a normal-ordered one, has zero vacuum value.
dp̃ = d³p/((2π)³2Eₚ); aₚ has mass dimension −1 and φ has 1 (GeV); ħ = c = 1, metric (+, −, −, −)
Derived, at any times: only φ⁺(x)φ⁻(y) is out of order, and swapping it costs a commutator that is a number because [a, a†] is.
D(x − y) = ∫dp̃ e(−ip⋅(x − y)) = ⟨0|φ(x)φ(y)|0⟩, UAQM-09.04's vacuum amplitude, quoted; mass dimension 2 (GeV²)
Derived. This c-number is defined as the contraction of φ(x) with φ(y); between vacua the normal-ordered part dies and ⟨0|Tφφ|0⟩ = DF remains.
DF(x − y) = θ(x⁰ − y⁰)D(x − y) + θ(y⁰ − x⁰)D(y − x), GeV²; topic 1.06's i/(p² − m² + iε) in momentum space
Derived. :φφ: is Hermitian, so in every state Im⟨Tφ(x)φ(y)⟩ = Im DF(x − y): the imaginary part never depends on the state.
real field; {A, B} = AB + BA; Re D(x − y) = ½⟨0|(φ(x), φ(y))|0⟩, GeV²
Shown, as an operator identity. :φφ: is not zero, but it lowers the charge by 2q, so it has no diagonal elements in a state of definite charge.
φ = ∫dp̃ [aₚ e(−ip⋅x) + bₚ† e(ip⋅x)], UAQM-09.03, quoted; φ⁺ holds a, φ⁻ holds b†; likewise for φ†φ†
Shown. Topic 1.08's propagator, now as a contraction: a particle from y to x or an antiparticle from x to y, so every charged line joins a φ† to a φ.
[a, a†] gives D(x − y) when x is later; [b, b†] gives D(y − x) when y is later; DF as for a real field of the same mass
Split the field by frequency, then fix an order
Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout. The free real field of UAQM-09.02, in topic 1.03's invariant normalisation, is quoted: φ(x) = ∫dp̃ [aₚ e(−ip⋅x) + aₚ† e(ip⋅x)], with dp̃ = d³p/((2π)³2Eₚ) and [aₚ, aq†] = (2π)³2Eₚ δ³(p − q). Call the first term φ⁺(x), the positive-frequency part holding every annihilator, and the second φ⁻(x) = φ⁺(x)†, holding every creator, so φ⁺|0⟩ = 0 and ⟨0|φ⁻ = 0. Normal ordering, written :⋯:, rewrites each term of a product with all creators left of all annihilators and generates no commutators: :φ(x)φ(y): = φ⁻(x)φ⁻(y) + φ⁻(x)φ⁺(y) + φ⁻(y)φ⁺(x) + φ⁺(x)φ⁺(y). For bosons it is symmetric in x and y, and every term ends in φ⁺ or begins with φ⁻, so ⟨0|:φ(x)φ(y):|0⟩ = 0. It is a rule for how a product is written, not a function of the operator: term by term, :φ⁺(x)φ⁻(y) − φ⁻(y)φ⁺(x): = 0, although that difference is the non-zero number met next.
One commutator separates the product from its normal order
Multiply out φ(x)φ(y) = (φ⁺(x) + φ⁻(x))(φ⁺(y) + φ⁻(y)). Three of the four terms already stand in normal order; only φ⁺(x)φ⁻(y) is out of order, and swapping it costs a commutator: φ⁺(x)φ⁻(y) = φ⁻(y)φ⁺(x) + [φ⁺(x), φ⁻(y)]. So φ(x)φ(y) = :φ(x)φ(y): + [φ⁺(x), φ⁻(y)], derived. The commutator takes one line, [φ⁺(x), φ⁻(y)] = ∫dp̃ dq̃ e(−ip⋅x + iq⋅y)[aₚ, aq†] = ∫dp̃ e(−ip⋅(x − y)), and it is UAQM-09.04's vacuum amplitude D(x − y) = ⟨0|φ(x)φ(y)|0⟩, quoted. It is a number, not an operator, because [a, a†] is one, and that is what makes the split useful: the operator content of the product sits in the normal-ordered part, and the remainder appears identically in every matrix element. Nothing here asked which time is later, so the result holds for any x and y. Dimensions: φ has mass dimension 1 and dp̃ has 2, so D has 2, as φφ must.
Time ordering picks D or its mirror: the contraction
T puts the later field on the left. If x is later, Tφ(x)φ(y) = φ(x)φ(y) = :φ(x)φ(y): + D(x − y). If y is later, Tφ(x)φ(y) = φ(y)φ(x) = :φ(y)φ(x): + D(y − x), and the normal-ordered part is the same operator as before. Together, Tφ(x)φ(y) = :φ(x)φ(y): + θ(x⁰ − y⁰)D(x − y) + θ(y⁰ − x⁰)D(y − x) = :φ(x)φ(y): + DF(x − y), derived, with DF topic 1.05's propagator. At equal times the points are spacelike and microcausality gives D(x − y) = D(y − x), so no convention is needed there. This c-number is defined as the contraction of φ(x) with φ(y), drawn in print as a bracket over the pair; between vacua the identity returns ⟨0|Tφφ|0⟩ = DF, as it must. Subtracting the plain product, Tφ(x)φ(y) − φ(x)φ(y) = θ(y⁰ − x⁰)[φ(y), φ(x)], a c-number that vanishes outside the light cone and, for a massless field, everywhere off it.
The state lives in the normal-ordered part
Because the contraction is a c-number, ⟨ψ|Tφ(x)φ(y)|ψ⟩ = ⟨ψ|:φ(x)φ(y):|ψ⟩ + DF(x − y) for every normalised |ψ⟩: all dependence on the state sits in the first term. Adding the plain product to its reverse gives (φ(x), φ(y)) = 2:φ(x)φ(y): + D(x − y) + D(y − x), and D(y − x) = D(x − y)* for a real field, so :φ(x)φ(y): = ½(φ(x), φ(y)) − Re D(x − y), derived. That operator is Hermitian, its expectation values are real, and Im⟨ψ|Tφ(x)φ(y)|ψ⟩ = Im DF(x − y) in every state, empty or full. The real part is where the state shows. In a coherent state, aₚ|α⟩ = αₚ|α⟩, the annihilators of a normal-ordered product act on |α⟩ and the creators on ⟨α|, so ⟨α|:φ(x)φ(y):|α⟩ = φcl(x)φcl(y), the product of the classical field values ⟨α|φ|α⟩. Normal ordering is the classical part; the contraction is the quantum remainder, there even where the classical field is zero.
Complex fields: only φ with φ† contracts
UAQM-09.03's complex field is quoted: φ = ∫dp̃ [aₚ e(−ip⋅x) + bₚ† e(ip⋅x)], so its φ⁺ holds a and its φ⁻ holds b†, while φ† has b in (φ†)⁺ and a† in (φ†)⁻. Repeat the argument. In φ(x)φ(y) the one out-of-order pair is φ⁺(x)φ⁻(y), and [aₚ, bq†] = 0, so Tφ(x)φ(y) = :φ(x)φ(y): exactly, with no c-number, and likewise for φ†φ†: the contraction of φ with φ is zero, shown. That is an operator statement, stronger than topic 1.08's vanishing vacuum value, and :φφ: itself is not zero: each of its terms absorbs particles or emits antiparticles, two in all, lowering the charge by 2q. In φ(x)φ†(y) the crossing pairs are a with a†, giving D(x − y) when x is later, and b with b†, giving D(y − x) when y is later, so Tφ(x)φ†(y) = :φ(x)φ†(y): + DF(x − y), topic 1.08's propagator, shown. A check: with φ = (φ₁ + iφ₂)/√2 and real fields contracting as δᵢⱼDF, φ with φ gives ½(DF − DF) = 0. Dirac fields anticommute, so T carries a sign; their contraction waits for topic 7.02.
Three checks: one mode, many quanta, a measurement
Average the field over a periodic box of volume V and only the zero-momentum mode survives: φ̄(t) = (a e(−imt) + a† e(imt))/√(2mV), with [a, a†] = 1 and contraction e(−im|t|)/(2mV). Then ⟨n|φ̄²|n⟩ = (2n + 1)/(2mV), the oscillator's ⟨n|x²|n⟩ = (2n + 1)/(2Mω) with Mω → mV, and for positive t, ⟨n|Tφ̄(t)φ̄(0)|n⟩ = [(n + 1)e(−imt) + n e(imt)]/(2mV): in the familiar n + 1 of stimulated plus spontaneous emission, n comes from :⋯: and 1 from the contraction. In a coherent state with α = √n̄ e(iθ), holding n̄ quanta on average, the normal-ordered part and the contraction stand at t = 0 in the ratio 4n̄cos²θ to 1, so at n̄ = 10⁶ and θ = 0 the contraction is a 2.5 × 10⁻⁷ correction: the classical limit. The normal-ordered part need not be positive. Squeezed light has been measured 15 dB below the vacuum noise, quoted: a quadrature X with variance 0.032 of the vacuum's, so ⟨:X²:⟩ = −0.968 of the contraction. Coincident points need care: φ(x)² = :φ(x)²: + D(0), with D(0) divergent, taken up in topic 2.09.
Change one variable at a time
Make the relationship visible.
Set n̄ = 0: the dashed curve lies flat on zero, as a normal-ordered product has no vacuum value. Raise n̄ or turn θ: solid and dashed move together, the gap between them staying Re DF = cos mt in units of 1/(2mV) and shrinking only against the scale. At negative mt the lower curves part: T swaps D(x − y) for D(y − x).
CLASSICAL PART, NORMAL-ORDERED-2.82 (2mV)⁻¹
RE DF = RE D-0.42 (2mV)⁻¹
IM DF, TIME-ORDERED-0.91 (2mV)⁻¹
IM D, PLAIN PRODUCT0.91 (2mV)⁻¹
Live interpretationCLASSICAL PART, NORMAL-ORDERED: −2.82 (2mV)⁻¹. RE DF = RE D: −0.42 (2mV)⁻¹. IM DF, TIME-ORDERED: −0.91 (2mV)⁻¹. IM D, PLAIN PRODUCT: 0.91 (2mV)⁻¹
Catch the common trap
Explain before calculating.
For a free complex scalar field φ, take x inside the forward light cone of y, so x⁰ is later than y⁰ and the separation is timelike. Which pair of operator identities is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe π⁰ field (m = 135.0 MeV) is averaged over a periodic cube of side 6.0 fm, leaving one mode: φ̄(t) = (a e(−imt) + a† e(imt))/√(2mV). Write φ̄(0)² as a normal-ordered operator plus a c-number, evaluate the c-number in fm⁻², and use it to find ⟨n|φ̄(0)²|n⟩ for n = 0 and for n = 3 pions at rest.
- At t = 0 the split is φ̄⁺ = a/√(2mV) and φ̄⁻ = a†/√(2mV). In (φ̄⁺ + φ̄⁻)² only φ̄⁺φ̄⁻ is out of order, so φ̄² = :φ̄²: + [φ̄⁺, φ̄⁻] = :φ̄²: + [a, a†]/(2mV) = :φ̄²: + 1/(2mV).
- Numbers: m = 135.0 MeV ÷ 197.327 MeV fm = 0.6841 fm⁻¹ and V = (6.0 fm)³ = 216 fm³, so 2mV = 295.6 fm² and the c-number is 1/(2mV) = 3.38 × 10⁻³ fm⁻², or 1.32 × 10⁻⁴ GeV².
- The operator part is :φ̄²: = (a² + 2a†a + a†²)/(2mV). In |n⟩ the terms a² and a†² change the number of quanta by two and have no diagonal element, while 2a†a gives 2n. So ⟨n|φ̄²|n⟩ = (2n + 1)/(2mV).
- Vacuum: the normal-ordered part gives 0, leaving 3.38 × 10⁻³ fm⁻², the contraction alone. Three pions: 7/(2mV) = 2.37 × 10⁻² fm⁻², of which 6/(2mV) is normal-ordered and 1/(2mV) is the same c-number as in the vacuum.
- Checks: fm⁻² is mass², the dimension of φ̄². The oscillator x = (a + a†)/√(2Mω) has ⟨n|x²|n⟩ = (2n + 1)/(2Mω), recovered with Mω → mV. The c-number falls as 1/L³: the field averaged over a larger box fluctuates less.
Answerφ̄(0)² = :φ̄(0)²: + 1/(2mV), with 1/(2mV) = 3.38 × 10⁻³ fm⁻²; ⟨0|φ̄²|0⟩ = 3.38 × 10⁻³ fm⁻² and ⟨3|φ̄²|3⟩ = 2.37 × 10⁻² fm⁻².
MediumA periodic cube of side 5.0 fm holds three π⁺ and one π⁻ at rest, the state |3, 1⟩. The box average of the charged-pion field (m = 139.57 MeV) is φ̄(t) = (a e(−imt) + b† e(imt))/√(2mV). Find ⟨3, 1|Tφ̄(t)φ̄†(0)|3, 1⟩ at t = +2.0 fm/c and t = −2.0 fm/c, separating the contraction from the normal-ordered part, and find ⟨3, 1|Tφ̄(t)φ̄(0)|3, 1⟩.
- By this topic's result, Tφ̄(t)φ̄†(0) = :φ̄(t)φ̄†(0): + e(−im|t|)/(2mV): the a, a† pair supplies e(−imt) for positive t, the b, b† pair e(imt) for negative t, together the zero-mode part of DF.
- With φ̄†(0) = (b + a†)/√(2mV), :φ̄(t)φ̄†(0): = (ab e(−imt) + a†a e(−imt) + b†b e(imt) + b†a† e(imt))/(2mV). In |3, 1⟩ only a†a and b†b have diagonal elements, so the normal-ordered part is (3e(−imt) + e(imt))/(2mV).
- Numbers: m = 0.7073 fm⁻¹, V = 125 fm³, 2mV = 176.8 fm², 1/(2mV) = 5.655 × 10⁻³ fm⁻², and m|t| = 1.4146 rad, with cos = 0.1556 and sin = 0.9878.
- t = +2.0 fm/c: the total is (4e(−imt) + e(imt))/(2mV) = (5cos mt − 3i sin mt)/(2mV) = (4.40 − 16.8i) × 10⁻³ fm⁻², of which the contraction is (0.88 − 5.59i) × 10⁻³ fm⁻². The coefficient 4 = 3 + 1 is three stimulated by the π⁺ present plus one from the contraction.
- t = −2.0 fm/c: the total is (3e(−imt) + 2e(imt))/(2mV) = (4.40 + 5.59i) × 10⁻³ fm⁻², now with n₋ + 1 = 2 on the antiparticle wave. With no pions both results fall to e(−im|t|)/(2mV), the vacuum limit.
- Tφ̄(t)φ̄(0) has no c-number, since [a, b†] = 0, so it equals :φ̄(t)φ̄(0):, whose terms a², b†a and b†² all change n₊ or n₋: the expectation value is 0, though the operator is not, removing two π⁺ with amplitude √6 e(−imt)/(2mV).
AnswerAt t = +2.0 fm/c, (4.40 − 16.8i) × 10⁻³ fm⁻², the contraction giving (0.88 − 5.59i) × 10⁻³ fm⁻²; at t = −2.0 fm/c, (4.40 + 5.59i) × 10⁻³ fm⁻²; ⟨3, 1|Tφ̄(t)φ̄(0)|3, 1⟩ = 0.
HardTwo events share a spatial point, with y⁰ − x⁰ = 3.0 fm/c, and φ is the free π⁰ field, m = 135.0 MeV. Show that Tφ(x)φ(y) − φ(x)φ(y) is a c-number, evaluate it using topic 1.08's interior form Im DF = mJ₁(mλ)/(8πλ), quoted, where λ is the proper time between the events, and find it again for events 3.0 fm apart at one time, and for a massless field.
- Both products differ from :φ(x)φ(y): by c-numbers: the plain product by D(x − y) at any times, the T-product, with y later, by D(y − x). The normal-ordered parts cancel, so Tφ(x)φ(y) − φ(x)φ(y) = D(y − x) − D(x − y) = [φ(y), φ(x)], the same number in every state.
- For a real field D(x − y) = ⟨0|φ(x)φ(y)|0⟩ = D(y − x)*, so the difference is 2i Im D(y − x). With y later, D(y − x) = DF(y − x), whose imaginary part inside the cone is mJ₁(mλ)/(8πλ), quoted, with λ = 3.0 fm. So the difference is i mJ₁(mλ)/(4πλ).
- Numbers: m = 135.0/197.327 fm⁻¹ = 0.6841 fm⁻¹, mλ = 2.052 and J₁(2.052) = 0.5728, so mJ₁(mλ)/(4πλ) = 0.6841 × 0.5728/(37.70 fm) = 1.04 × 10⁻² fm⁻², or 4.05 × 10⁻⁴ GeV². It is purely imaginary, since Re DF = Re D.
- Spacelike: 3.0 fm apart at one time, microcausality (UAQM-09.04) gives D(x − y) = D(y − x), so the difference is 0 and T needs no decision. This is why equal-time tests cannot tell D from DF.
- Massless: mJ₁(mλ)/λ tends to m²/2 and then to 0 as m → 0, so off the light cone T and the plain product coincide; a massless field's commutator lives on the cone. Towards the cone, λ → 0, the massive value tends to m²/(8π) = 1.86 × 10⁻² fm⁻², and ours is 2J₁(mλ)/(mλ) = 0.558 of it. Dimensions: fm⁻² is mass², as for φφ.
AnswerTφ(x)φ(y) − φ(x)φ(y) = [φ(y), φ(x)] = i mJ₁(mλ)/(4πλ) = 1.04 × 10⁻² i fm⁻² (4.05 × 10⁻⁴ i GeV²); it is zero for the spacelike pair and, away from the light cone, for a massless field.