University Quantum Field Theory I · Discrete Symmetries of Quantised Fields: C, P, T and CPT · 11.2

Charge conjugation on quantised fields

Charge conjugation is where a field's statistics stops being bookkeeping. Define C as the swap of particle and antiparticle, push it through the mode expansions, and find exactly what pays for the antiparticle's opposite charge: the derivative for a scalar, the anticommutator for a Dirac field.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Advanced QM built ψ^c = iγ²ψ* from a wavefunction and showed that it obeys the Dirac equation with the opposite charge, but it could not make the current come out reversed: the conjugate wavefunction carries the same positive density, ψ^c†ψ^c = ψ†ψ, so the sign of the positron's current had to be stated. On Fock space the question becomes sharp. C is defined as the unitary operator that exchanges particle and antiparticle with momentum and spin kept, a with b for the complex scalar and b with d for the Dirac field, and leaves the vacuum alone; carried through the mode expansions it sends φ to φ† and ψ to iγ²(ψ†)ᵀ, the field form of ψ^c.

Applied to the normal-ordered current, the gamma matrices hand back +ψ̄γ^μψ, and the minus sign appears only when ψ is moved back past ψ̄, which costs −1 because the fields anticommute. The scalar current flips for a different reason, its antisymmetric derivative, with commuting fields. Either way the charge is C-odd exactly when the field has the statistics that also keeps its energy positive; with the other statistics Q would count quanta and H would change sign.

A Hermitian field has nothing to swap with and keeps only C = ±1. The argument stops at free fields: whether an interaction commutes with C is a property of its Lagrangian, which QED has and the weak interaction lacks.

Simple definition: Charge conjugation C is the unitary operator on Fock space that turns every particle into its antiparticle with the same momentum and spin and leaves the vacuum alone, so it carries a complex field into its Hermitian conjugate, times a spinor matrix for a Dirac field, and a real field into itself times a sign η_C = ±1.
Example: C turns an electron with |p| = 2.00 MeV along z and spin up into a positron with the same momentum, spin and energy E = 2.064 MeV; the electron number Q goes from +1 to −1, and the electric charge from −e to +e.

C on the Dirac ladder operatorsC bₛ(p) C⁻¹ = dₛ(p), C dₛ(p) C⁻¹ = bₛ(p), C|0⟩ = |0⟩

Defined, not derived. A phase in C b C⁻¹ = η d forces C d C⁻¹ = η* b and is absorbed into d, so C² = 1; C commutes with the free H and P^μ.

bₛ, dₛ remove an electron, a positron of momentum p and spin s; mass dimension −1, from (b, b†) = (2π)³ 2Eₚ δ³(p − q)

Complex scalar fieldC φ(x) C⁻¹ = φ†(x)

Derived in one line: a e^(−ip⋅x) + b† e^(ip⋅x) becomes b e^(−ip⋅x) + a† e^(ip⋅x), the expansion of φ†. Same point x: C is internal and moves nothing in spacetime.

φ has mass dimension 1 (GeV), x is in GeV⁻¹; the swap is C a(p) C⁻¹ = b(p), C b(p) C⁻¹ = a(p)

Dirac fieldC ψ(x) C⁻¹ = iγ²(ψ†)ᵀ(x) = iγ²γ⁰ψ̄ᵀ(x)

Derived from the swap and v(p, s) = iγ²u(p, s)*, quoted from UAQM-08.05 and inverted to u = iγ²v*. It is the operator form of ψ^c = iγ²ψ*, and C² = 1 follows.

ψ has mass dimension 3/2; (ψ†)ᵀ is the column of the ψₐ†; iγ² is real with (iγ²)² = 1 in the Dirac and Weyl bases

The Dirac currentC :ψ̄γ^μψ:(x) C⁻¹ = −:ψ̄γ^μψ:(x)

Derived. The matrices alone return +j^μ, since (iγ²γ⁰)γ^μᵀ(iγ²γ⁰) = γ^μ; the whole minus sign is the cost of moving ψ back past ψ̄ inside the normal ordering.

j^μ = :ψ̄γ^μψ: has mass dimension 3 and counts electrons minus positrons; the electric current is −e j^μ, e = 0.3028

Charge with the reordering sign εQ = Σₛ ∫ d³p/((2π)³ 2Eₚ) [bₛ†bₛ + ε dₛ†dₛ]

Derived from :ψ†ψ:. Only ε = −1 gives C Q C⁻¹ = −Q. With ε = +1, Q counts quanta and is C-even, while H = Σ E(b†b − d†d) would change sign under C.

Q is dimensionless; ε is the sign in d d† = ε d†d + c-number: −1 for anticommutators, +1 for commutators

A Hermitian fieldC φ(x) C⁻¹ = η_C φ(x), η_C = ±1

Derived: asking for a multiple of φ makes η_C real, and unitarity makes |η_C| = 1. No rephasing removes it: π⁰ → γγ is seen, π⁰ → 3γ is not (BR below 3.1 × 10⁻⁸).

φ = φ† of mass dimension 1; η_C is the intrinsic C-parity: +1 for the π⁰, −1 for the photon field (topic 11.04)

Define C by what it does to quanta

Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Charge conjugation is defined, not derived, and the definition is kept as small as possible: exchange each particle with its antiparticle, keep momentum and spin, leave the vacuum alone. For the complex scalar that is C a(p) C⁻¹ = b(p) and C b(p) C⁻¹ = a(p); for the Dirac field C bₛ(p) C⁻¹ = dₛ(p) and back, with C|0⟩ = |0⟩. On Fock states this is a relabelling, so it preserves every inner product and is unitary. It can be written down: with dμ = d³p/((2π)³ 2Eₚ), C = exp[i(π/2) Σₛ ∫dμ (bₛ† − dₛ†)(bₛ − dₛ)] reverses b − d, keeps b + d, and so swaps b and d for either statistics. A phase, C b C⁻¹ = η d, forces C d C⁻¹ = η* b if ψ is to go into a single field, and redefining d absorbs it, so C² = 1. The free H = Σₛ ∫dμ Eₚ(b†b + d†d) and the momentum treat b and d alike, so C commutes with both: the positron it makes has the electron's p, s and E.

The scalar field, and a current flipped by its derivative

Put the swap into φ(x) = ∫dμ [a(p) e^(−ip⋅x) + b†(p) e^(ip⋅x)]: a becomes b and b† becomes a†, which is the expansion of φ†, so C φ(x) C⁻¹ = φ†(x) at the same x. The Noether current j^μ = i:(φ†∂^μφ − (∂^μφ†)φ): then goes to i:(φ∂^μφ† − (∂^μφ)φ†):, and because the operators commute inside the normal ordering this is −j^μ. The sign comes from the antisymmetry of the derivative, not from statistics. In modes the derivative brings −iEₚ from e^(−iEₚt) and +iEₚ from e^(iEₚt), which is why, with commutators, Q = ∫dμ (a†a − b†b) and why ⟨p|j^μ(0)|p⟩ = +2p^μ for a particle and −2p^μ for its antiparticle, an energy, as mass dimension 3 − 2 = 1 says. A K⁺ and a K⁻ with |p| = 1.000 GeV and E = 1.115 GeV give j⁰ = ±2.230 GeV, equal and opposite. The free terms ∂_μφ†∂^μφ − m²φ†φ are symmetric under φ ↔ φ†, so C is a symmetry of the free scalar theory.

The Dirac field: ψ^c derived, not posited

For ψ(x) = Σₛ ∫dμ [bₛ(p) u(p, s) e^(−ip⋅x) + dₛ†(p) v(p, s) e^(ip⋅x)] the swap gives Σₛ ∫dμ [dₛ u e^(−ip⋅x) + bₛ† v e^(ip⋅x)]: operators exchanged, spinors untouched. Now quote UAQM-08.05's convention v(p, s) = iγ²u(p, s)*. In the Dirac and Weyl bases γ² is imaginary and symmetric, so iγ² is real with (iγ²)² = 1, and conjugating the convention gives u = iγ²v*. The transformed field is therefore iγ² Σₛ ∫dμ [dₛ v* e^(−ip⋅x) + bₛ† u* e^(ip⋅x)], and the sum is exactly (ψ†)ᵀ, the column of the four components ψₐ†. Hence C ψ C⁻¹ = iγ²(ψ†)ᵀ = iγ²γ⁰ψ̄ᵀ, the operator form of ψ^c = iγ²ψ*. Three checks: both sides have mass dimension 3/2; applying C twice returns (iγ²)(iγ²)ψ = ψ; and the adjoint, taken component by component, gives C ψ̄ C⁻¹ = ψᵀ iγ²γ⁰, which the current needs next.

Count the signs in the Dirac current

C sends annihilators to annihilators, so it passes through normal ordering, and both rules give C :ψ̄γ^μψ: C⁻¹ = :ψᵀ M ψ̄ᵀ: = Σᵢⱼ :ψᵢ Mᵢⱼ ψ̄ⱼ:, with M = (iγ²γ⁰)γ^μ(iγ²γ⁰). To read this as a current ψ̄(…)ψ, move each ψᵢ to the right of ψ̄ⱼ. Inside the normal ordering that move costs exactly −1 for anticommuting fields, leaving −ψ̄Mᵀψ. The matrices cost nothing on balance: iγ²γ⁰ is antisymmetric, so its two transposes give (−1)², and the identity (iγ²γ⁰)γ^μᵀ(iγ²γ⁰) = γ^μ, checked one μ at a time, gives Mᵀ = γ^μ. Hence C j^μ C⁻¹ = −j^μ, and the minus sign is the anticommutator and nothing else: quantise ψ with commutators and the same lines return +j^μ. This is what UAQM-08.05 could only state. For c-number spinors ψ^c†ψ^c = ψᵀ(iγ²)(iγ²)ψ* = ψ†ψ, the same positive density, because a wavefunction has no operator order to pay for.

Charge and energy choose the statistics together

The mode sums say the same without gamma matrices. Let ε be the sign in d d† = ε d†d + (c-number): −1 for anticommutators, +1 for commutators. Normal ordering ψ†ψ and ψ†(−iα⋅∇ + βm)ψ gives Q = Σₛ ∫dμ (b†b + ε d†d) and H = Σₛ ∫dμ Eₚ(b†b − ε d†d). With ε = −1, Q counts electrons minus positrons and is C-odd, while H is positive and C-even. With ε = +1 both fail at once: Q counts quanta and is C-even, and H = Σ E(b†b − d†d) is unbounded below and C-odd, so the swap is not even a symmetry of the free theory. The scalar is the mirror image, Q = ∫dμ (a†a − ε b†b) and H = ∫dμ Eₚ(a†a + ε b†b), and needs ε = +1. That the statistics which keep H positive are the ones which reverse Q is the free-field face of spin-statistics, whose general theorem topic 11.05 quotes. Measured, |q(e⁺) + q(e⁻)| is below 4 × 10⁻⁸ e (quoted), as one field with ε = −1 requires.

A real field keeps only a sign

A Hermitian field φ = φ† has one set of quanta and nothing to swap them with, so the most C can do is C a(p) C⁻¹ = η_C a(p). Asking that C φ C⁻¹ = η_C φ, a local field into a multiple of itself, matches a's coefficient to η_C and a†'s to η_C*, so η_C is real; unitarity gives |η_C| = 1, hence η_C = ±1. Rephasing a would change φ itself, so the sign is physical. A complex field is two such fields: with φ = (φ₁ + iφ₂)/√2, C φ C⁻¹ = φ† means φ₁ has C = +1 and φ₂ has C = −1. Since C Q C⁻¹ = −Q, only states with Q = 0 can be C eigenstates: an electron cannot, a π⁰ or a photon can. A C-invariant QED coupling proportional to j^μA_μ needs C A_μ C⁻¹ = −A_μ, so n photons carry (−1)ⁿ, as topic 11.04 builds. The measured π⁰ → γγ branching ratio, 98.8%, makes its η_C = +1 and forbids 3γ, and the measured bound is BR(3γ) below 3.1 × 10⁻⁸ (both quoted). The weak interaction breaks C and is only named here.

Change one variable at a time

Make the relationship visible.

Interactive model

Field κ: −1 scalar, +1 Dirac1

Statistics ε: −1 anti-, +1 commutators−1

Antiparticles in right packet2

Start at κ = +1, ε = −1: the dashed curve mirrors the solid one and Q reverses. Flip ε to +1 and the curves coincide, so C leaves Q alone. Then set κ = −1: now the scalar needs ε = +1, commutators, to get the mirror back.

ANTIPARTICLE WEIGHT w = κε-1

Q OF THE STATE-1

Q AFTER C1

C Q C⁻¹ + Q (ZERO IF C-ODD)0

Live interpretationANTIPARTICLE WEIGHT w = κε: −1. Q OF THE STATE: −1. Q AFTER C: 1. C Q C⁻¹ + Q (ZERO IF C-ODD): 0

Catch the common trap

Explain before calculating.

Misconception check

The tempting account is that the matrix iγ² does the work: ψ^c = iγ²ψ* complex-conjugates the spinor, so the current must come out reversed. It does not. Put c-number spinors through the conjugation and ψ^c†ψ^c = ψ†ψ, the same positive density, which is why Advanced QM could state, but not derive, that the positron's current is reversed. In the operator calculation the gamma matrices return +ψ̄γ^μψ; the minus sign appears only when ψᵀ(…)ψ̄ᵀ is reordered into ψ̄(…)ψ, a step that costs −1 for anticommuting fields and nothing for commuting ones. Quantise the Dirac field with commutators and the same C leaves the current unchanged. The twin slip runs the other way for the scalar: its current is C-odd because the derivative is antisymmetric between φ and φ†, with commuting fields, and anticommutators would spoil it.

Suppose the Dirac field were quantised with commutators, [b, b†] = [d, d†] = (2π)³ 2Eₚ δ³(p − q), and C were still defined as the swap b ↔ d with C|0⟩ = |0⟩. What would C do to the normal-ordered charge Q and Hamiltonian H?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA free-field state holds two electrons of energy 1.20 MeV each and one positron of energy 0.80 MeV. Find the electron number Q and the energy H before and after C, first with the anticommutators the Dirac field obeys, then as if it were quantised with commutators.

Write both operators with the reordering sign ε of d d† = ε d†d + c-number: Q = Σ(b†b + ε d†d) and H = Σ Eₚ(b†b − ε d†d), normal ordered. C swaps b and d with p and s kept, so after C the state holds two positrons of 1.20 MeV and one electron of 0.80 MeV.
Anticommutators, ε = −1: before C, Q = 2 − 1 = +1 and H = 2 × 1.20 + 0.80 = 3.20 MeV. After C, Q = 1 − 2 = −1 and H = 0.80 + 2 × 1.20 = 3.20 MeV. The electric charge −eQ goes from −e to +e.
Commutators, ε = +1: before C, Q = 2 + 1 = 3 and H = 2.40 − 0.80 = 1.60 MeV. After C, Q = 1 + 2 = 3 and H = 0.80 − 2.40 = −1.60 MeV.
Checks: Q is a pure number and H carries MeV; every energy exceeds m = 0.511 MeV; and for a state with no positrons ε drops out of both, so the two quantisations agree on the electron-only states, as they must.

AnswerAnticommutators: Q goes from +1 to −1 while H stays 3.20 MeV. Commutators: Q stays 3 while H goes from 1.60 MeV to −1.60 MeV, so the swap would not even be a symmetry.

MediumA positron moves along +z with |p| = 2.00 MeV, spin up. In the Dirac basis build u(p, ↑), then v(p, ↑) = iγ²u(p, ↑)*, and use them to find ⟨e⁺|:j^μ(0):|e⁺⟩ for μ = 0 and 3 in invariant normalisation. What fixes the sign, and what would commutators give?

E = √(p² + m²) = √(4.000 + 0.261) MeV = 2.064 MeV, and E + m = 2.575 MeV. With ξ = (1, 0), u = (√(E + m) ξ, σ_z|p| ξ/√(E + m)) = (1.6048, 0, 1.2463, 0) √MeV.
In this basis iγ² is real and sends (u₁, u₂, u₃, u₄) to (u₄, −u₃, −u₂, u₁), so v = (0, −1.2463, 0, 1.6048) √MeV. Checks: v†v = 1.553 + 2.575 = 4.128 MeV = 2E, and v̄v = −1.022 MeV, minus twice the mass, as a v spinor must give.
The positron is created by d†v e^(ip⋅x) in ψ and removed by d v̄ e^(−ip⋅x) in ψ̄, so the current contains v̄γ^μv d d†. Inside the normal ordering d d† becomes −d†d, hence ⟨e⁺|:j^μ(0):|e⁺⟩ = −v̄γ^μv = −2p^μ.
Numbers: v̄γ⁰v = v†v = 4.128 MeV and v̄γ³v = 4.000 MeV = 2|p|, so ⟨j⁰⟩ = −4.128 MeV and ⟨j³⟩ = −4.000 MeV. Dimensions: j^μ has mass dimension 3 and the two invariantly normalised states −2 together, leaving 1, an energy.
Dividing by the norm 2E, the positron carries electron number −1, so electric charge −e × (−1) = +e, and j³/j⁰ = |p|/E = 0.969 is its velocity, which falls to 0 at rest and rises to 1 at high energy. With commutators d d† gives +d†d and the same spinors give +2p^μ: the positron would count as an electron.

Answer⟨e⁺|:j^μ(0):|e⁺⟩ = −2p^μ: j⁰ = −4.13 MeV and j³ = −4.00 MeV, ratio 0.969. The sign is the anticommutator's, since the c-number v̄γ^μv = +2p^μ is positive.

HardKeep one electron mode and one positron mode with the same p and s, box-normalised so (b, b†) = (d, d†) = 1, spanning |0⟩, |e⁻⟩ = b†|0⟩, |e⁺⟩ = d†|0⟩ and |e⁻e⁺⟩ = b†d†|0⟩. Show that U = exp[i(π/2)(b† − d†)(b − d)] acts as C, find its matrix and eigenvalues, and for |χ⟩ = 0.6|e⁻⟩ + 0.8|e⁺⟩ find ⟨χ|U|χ⟩ and ⟨Q⟩ before and after C.

Set c = (b − d)/√2 and c′ = (b + d)/√2. The cross anticommutators of b and d vanish, so (c, c†) = (c′, c′†) = 1 and (c, c′†) = 0, and (b† − d†)(b − d) = 2c†c. As c†c has eigenvalues 0 and 1, U = exp(iπc†c) = 1 − 2c†c.
[c†c, c] = −c gives U c U⁻¹ = e^(−iπ)c = −c, while c′ anticommutes with c and c†, so it commutes with c†c and U c′ U⁻¹ = c′. Then b = (c′ + c)/√2 goes to (c′ − c)/√2 = d, and d goes to b. Since c|0⟩ = 0, U|0⟩ = |0⟩: U is C on this pair.
On the basis: U|0⟩ = |0⟩, U|e⁻⟩ = |e⁺⟩, U|e⁺⟩ = |e⁻⟩, and U b†d†|0⟩ = d†b†|0⟩ = −b†d†|0⟩. In the order (|0⟩, |e⁻⟩, |e⁺⟩, |e⁻e⁺⟩) the rows are (1, 0, 0, 0), (0, 0, 1, 0), (0, 1, 0, 0) and (0, 0, 0, −1); U² = 1 and the trace is 0.
Eigenvalues: +1 for |0⟩ and for (|e⁻⟩ + |e⁺⟩)/√2, −1 for (|e⁻⟩ − |e⁺⟩)/√2 and for the pair |e⁻e⁺⟩. The pair's −1 is the reordering sign again: for bosons a†b†|0⟩ goes to b†a†|0⟩ = +a†b†|0⟩, so a boson pair in one mode has C = +1. Topic 11.08 adds orbital and spin exchange.
U|χ⟩ = 0.6|e⁺⟩ + 0.8|e⁻⟩, so ⟨χ|U|χ⟩ = 2 × 0.6 × 0.8 = 0.96. With Q = b†b − d†d, ⟨Q⟩ = 0.36 − 0.64 = −0.28 before C and +0.28 after. Q maps (|e⁻⟩ + |e⁺⟩)/√2 onto (|e⁻⟩ − |e⁺⟩)/√2: a C-odd operator links C = +1 to C = −1, so every C eigenstate has ⟨Q⟩ = 0.

AnswerU swaps b and d and fixes |0⟩; its eigenvalues are +1, +1, −1, −1 (trace 0), the e⁻e⁺ pair having C = −1 where a boson pair has +1; ⟨χ|U|χ⟩ = 0.96, and ⟨Q⟩ goes from −0.28 to +0.28.