University Quantum Field Theory I · Discrete Symmetries of Quantised Fields: C, P, T and CPT · 11.4

The bilinears under C and T, and the photon under P, C and T

Here the discrete symmetries stop being definitions and become a table you can use. A transpose with its reordering sign gives every bilinear its C sign, a complex conjugation gives its T sign, one demand on the QED vertex gives the photon its laws, and the table then shows QED keeping C, P and T one at a time.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topics 11.01 to 11.03 defined P, C and T on the fields, but Lagrangians are built from products of fields, so the working tool is a table of signs for the Hermitian bilinears ψ̄Γψ and for the photon. The P column is quoted from UAQM-05.06; the other two are derived here. C turns ψ into a column of ψ† and ψ̄ into a row of ψ, so a bilinear comes back transposed and in the wrong order: restoring the order costs −1 because the fields anticommute, and the transpose reverses a product of k distinct γ matrices, leaving the sign (−1)^(k(k+1)/2), even for ψ̄ψ, iψ̄γ⁵ψ and the axial current, odd for the vector current and the tensor.

T keeps ψ̄ on the left and ψ on the right, so nothing is reordered, but it is antilinear: each γ^μ returns (−1)^μ and each explicit i returns −1. The photon gets no laws of its own. Demanding that e ψ̄γ^μψA_μ be invariant makes A^μ copy the current: C-odd, a polar vector under P, and under T keeping A⁰ while reversing A, as the classical potentials of a charge and its current do.

Then the mass, kinetic, Maxwell and gauge-fixing terms and the vertex are each even, so QED conserves C, P and T separately. That is a fact about these terms, not a theorem: an axial photon coupling or an electron dipole term would break them, and only the product CPT is protected in general.

Simple definition: The C and T columns of the bilinear table give the sign that each Hermitian Dirac bilinear, component by component, acquires under charge conjugation or time reversal of the quantised field; the photon field then takes whatever signs keep e ψ̄γ^μψA_μ invariant.
Example: Of the 16 bilinear components, C is +1 on six, ψ̄ψ, iψ̄γ⁵ψ and the four ψ̄γ^μγ⁵ψ, and −1 on the four ψ̄γ^μψ and six ψ̄σ^μνψ; T is +1 on six, ψ̄ψ, ψ̄γ⁰ψ, ψ̄γ⁰γ⁵ψ and the three ψ̄σ⁽⁰ⁱ⁾ψ, and −1 on the other ten.

C on any Dirac bilinearC :ψ̄Γψ:(x) C⁻¹ = −:ψ̄ (iγ²γ⁰)Γᵀ(iγ²γ⁰) ψ:(x)

Derived. With (iγ²γ⁰)γ^μᵀ(iγ²γ⁰) = γ^μ and (iγ²γ⁰)² = −1, the sandwich returns a k-fold product reversed, times (−1)^(k−1); the reorder sign makes it (−1)^k.

Γ a product of γ matrices, ᵀ its transpose; ψ̄Γψ has mass dimension 3; the leading − is the anticommutation sign of topic 11.02

The C columnη_C = (−1)^(k(k+1)/2): +1 for ψ̄ψ, iψ̄γ⁵ψ, ψ̄γ^μγ⁵ψ−1 for ψ̄γ^μψ, ψ̄σ^μνψ

Derived: (−1)^k from the rule above times (−1)^(k(k−1)/2) for reversing k anticommuting matrices. Quantise with commutators and every entry flips.

k = 0, 4, 3, 1, 2 distinct γ's in each antisymmetrised Γ; C is unitary, so a factor i in front changes nothing

T on any Dirac bilinearT :ψ̄Γψ:(t, x) T⁻¹ = :ψ̄ (γ¹γ³)⁻¹Γ*(γ¹γ³) ψ:(−t, x)

Derived from topic 11.03's T ψ(t, x) T⁻¹ = η γ¹γ³ψ(−t, x). γ¹γ³ commutes with γ⁰, γ², γ⁵ and anticommutes with γ¹, γ³, so (γ¹γ³)⁻¹γ^μ*(γ¹γ³) = (−1)^μγ^μ.

Dirac or Weyl basis: γ² is the only imaginary γ and γ⁵ is real; Γ* is the complex conjugate; nothing is reordered

The T column by countingη_T = (−1)^{(nₛ + nᵢ)}

Derived. iψ̄γ⁵ψ: −1. ψ̄γ^μψ, ψ̄γ^μγ⁵ψ: (−1)^μ. σ⁽⁰ⁱ⁾: +1, σ^(ij): −1. ∂₀: −1, ∂ᵢ: +1. With boson T phases +1, a term contracted with g_μν alone is T-odd exactly when nᵢ is odd.

nₛ: spatial indices; nᵢ: explicit factors of i, with iγ⁵, σ^μν = (i/2)[γ^μ, γ^ν] and ∂_μ = −i(i∂_μ) one each, γ⁵ none

The photon field under CC A^μ(x) C⁻¹ = −A^μ(x)

Derived from the C-odd current. Every mode then obeys C a(k, λ) C⁻¹ = −a(k, λ), so a state of n photons carries C = (−1)ⁿ, the fact topics 11.07 to 11.09 use.

A^μ has mass dimension 1; fixed by demanding that e ψ̄γ^μψA_μ, with e = 0.3028 dimensionless, be C-invariant

The photon field under TT A^μ(t, x) T⁻¹ = (−1)^μ A^μ(−t, x)

Derived from the vertex. A⁰ is kept and A reversed, as the classical potential of a charge and the vector potential of its current are, so E is T-even and B T-odd.

(−1)^μ = +1 for μ = 0, −1 for μ = 1, 2, 3; P, by the same argument, gives (−1)^μ A^μ(t, −x), a polar vector

C transposes the bilinear, then pays to reorder it

Work in ħ = c = 1 with the metric (+, −, −, −), and write (−1)^μ for +1 at μ = 0 and −1 at μ = 1, 2, 3. Topic 11.02 gave C ψ C⁻¹ = iγ²γ⁰ψ̄ᵀ and C ψ̄ C⁻¹ = ψᵀiγ²γ⁰ at the same point. Put both into :ψ̄Γψ: and the result is Σ ψᵢMᵢⱼψ̄ⱼ with M = (iγ²γ⁰)Γ(iγ²γ⁰), a ψ now standing left of a ψ̄. Moving it back costs −1 inside the normal ordering, and Mᵀ = (iγ²γ⁰)Γᵀ(iγ²γ⁰) because iγ²γ⁰ is antisymmetric and appears twice. For a single γ^μ the sandwich returns γ^μ, so the reorder sign alone makes the current C-odd, as in 11.02. For a product of k distinct γ's, inserting 1 = −(iγ²γ⁰)² between neighbours returns the factors reversed, times (−1)^(k−1); with the reorder sign that is (−1)^k, and undoing the reversal of k anticommuting matrices costs (−1)^(k(k−1)/2). So η_C = (−1)^(k(k+1)/2): +1 for ψ̄ψ (k = 0), −1 for ψ̄γ^μψ (1), −1 for ψ̄σ^μνψ (2), +1 for ψ̄γ^μγ⁵ψ (3) and +1 for iψ̄γ⁵ψ (4). C is unitary, so the factors of i that make bilinears Hermitian play no part; quantise with commutators and every entry flips.

T conjugates the bilinear and reorders nothing

Topic 11.03's T ψ(t, x) T⁻¹ = η γ¹γ³ψ(−t, x) sends ψ to ψ and, through its adjoint, ψ̄ to η*ψ̄(γ¹γ³)⁻¹, since γ⁰ is real and commutes with γ¹γ³. The bilinear keeps its order, so no anticommutation sign appears, and the phase cancels. What T adds is antilinearity: it conjugates every c-number it passes, so T :ψ̄Γψ:(t, x) T⁻¹ = :ψ̄ (γ¹γ³)⁻¹Γ*(γ¹γ³) ψ:(−t, x). In the Dirac and Weyl bases γ⁰, γ¹, γ³ and γ⁵ are real and γ² is imaginary. γ¹γ³ commutes with γ⁰, γ² and γ⁵ and anticommutes with γ¹ and γ³, and conjugation supplies γ²'s minus sign, so (γ¹γ³)⁻¹γ^μ*(γ¹γ³) = (−1)^μγ^μ for all four while γ⁵ is untouched; each explicit i becomes −i. Hence ψ̄ψ is T-even and iψ̄γ⁵ψ T-odd; ψ̄γ^μψ and ψ̄γ^μγ⁵ψ both take (−1)^μ; and ψ̄σ^μνψ, carrying an i, takes −(−1)^μ(−1)^ν, so σ⁽⁰ⁱ⁾ is even and σ^(ij) odd. The derivative changes only through its argument: ∂₀ → −∂₀, ∂ᵢ → +∂ᵢ. Writing ∂_μ = −i(i∂_μ), with the momentum i∂_μ taking (−1)^μ, all of it is one rule, η_T = (−1)^{(nₛ + nᵢ)}.

Read the columns as the densities they are

In the non-relativistic reading every entry is a sign you already know. ψ̄γ⁰ψ = ψ†ψ is the number density, charge-weighted into the charge density: C-odd, T-even. ψ̄γ^iψ = ψ†α^iψ is the current: C-odd, and T-odd because velocities reverse. ψ̄γ^iγ⁵ψ = ψ†Σ^iψ is the spin density: C-even, because an electron and a positron with the same spin add rather than cancel, and T-odd, because spin reverses. Its time component ψ†γ⁵ψ, of order σ⃗⋅p⃗/m, is a helicity density and T-even, since T reverses spin and momentum together. With σ^(ij) = ε^(ijk)Σ^k, ψ̄σ^(ij)ψ is a magnetisation, T-odd like B, and ψ̄σ⁽⁰ⁱ⁾ψ an electric polarisation, T-even like E: the tensor bilinear transforms exactly as F^μν will. Every ψ̄Γψ has mass dimension 3, as a density in natural units must. One more check: the product of the C, P and T entries is +1, +1, −1, −1, +1 for ψ̄ψ, iψ̄γ⁵ψ, ψ̄γ^μψ, ψ̄γ^μγ⁵ψ and ψ̄σ^μνψ, that is (−1)ⁿ with n the number of Lorentz indices, the rule topic 11.05 turns into CPT invariance.

The photon's laws are read off the vertex

The photon has no spinor structure to compute with, so its laws come from a requirement: whatever C, P and T do to A^μ must leave e ψ̄γ^μψA_μ invariant, the coupling e = 0.3028 being a real number that no operator changes. The current is C-odd, so C A^μ C⁻¹ = −A^μ. Under P the current takes (−1)^μ at (t, −x), so A^μ goes to (−1)^μA^μ(t, −x), a polar vector. Under T it takes (−1)^μ at (−t, x), so A^μ goes to (−1)^μA^μ(−t, x): A⁰ kept, A reversed. That is the classical limit exactly. The scalar potential follows the charge density, which T keeps, and the vector potential follows the current, which T reverses, so E = −∇A⁰ − ∂ₜA is T-even and B = ∇ × A is T-odd, while under P, E is odd and B even. The assignment is universal: the scalar current i(φ†∂^μφ − (∂^μφ†)φ) of topic 11.02 has the same three signs, so one photon law serves every charged field. In Unit 8's covariant quantisation all four modes obey C a(k, λ) C⁻¹ = −a(k, λ), and ∂_μA^μ goes into ±∂_μA^μ, so the Gupta–Bleuler condition is carried into itself.

QED, one term at a time

Now test ℒ = ψ̄(iγ^μ∂_μ − m)ψ − ¼F_μνF^μν − (1/2ξ)(∂_μA^μ)² − eψ̄γ^μψA_μ. The mass term is even under all three, as ψ̄ψ is, and the vertex is even by construction. F^μν = ∂^μA^ν − ∂^νA^μ inherits the tensor's signs, C −1, P (−1)^μ(−1)^ν and T −(−1)^μ(−1)^ν, so F_μνF^μν = 2(B² − E²) is even. ∂_μA^μ = ∂ₜA⁰ + ∇⋅A goes to −∂_μA^μ under C and T and to +∂_μA^μ under P, so its square is even. The kinetic term needs care. Under T the time part collects −1 from i → −i and −1 from ∂₀, each space part −1 from i and −1 from γ^k, so the term returns itself at (−t, x). Under C the derivative ends on ψ̄, giving −i(∂_μψ̄)γ^μψ, which is iψ̄γ^μ∂_μψ minus the total derivative i∂_μ(ψ̄γ^μψ), dropped from the action. P is quoted from UAQM-05.06. Every term is even, so QED conserves C, P and T separately, derived at the level of the action; the regulator respects all three, so renormalised QED does too (quoted). Measured: |dₑ| is below 4.1 × 10⁻³⁰ e cm (quoted), 2 × 10⁻¹⁹ of eħ/(2mₑc) = 1.93 × 10⁻¹¹ e cm, where QED alone predicts exactly zero.

What the table forbids, and what it leaves to the Lagrangian

Separate C, P and T conservation is a property of QED's particular terms, and the table shows which additions would end it. Add an axial coupling ψ̄γ^μγ⁵ψA_μ: the axial current is C-even against a C-odd photon, and P-axial against a polar A, so the term breaks C and P, while T, which treats both currents alike, survives, and so does CP. The electron dipole term −(i/2)d ψ̄σ^μνγ⁵ψF_μν has nᵢ = 3 and one γ⁵: T-odd and P-odd, but C-even. F_μνF̃^μν, proportional to E⋅B and contracted with ε^μνρσ, leaves three spatial indices: P-odd and T-odd, but for an abelian field it is the total derivative 2∂_μ(ε^μνρσA_ν∂_ρA_σ) and changes no amplitude. Lorentz invariance and Hermiticity exclude none of these terms; what keeps the other two out of QED is more specific. The axial current of a massive electron is not conserved, ∂_μ(ψ̄γ^μγ⁵ψ) = 2imψ̄γ⁵ψ, so the axial coupling is not gauge invariant, and the dipole term has mass dimension 5, so [d] = −1 and it is not renormalisable. Only the product CPT is guaranteed in general, by topic 11.05. Two cautions keep the table honest. For bilinears of two different fields, ψ̄₁Γψ₂, the intrinsic phases enter as η₁*η₂ and C also exchanges the two fields, so the entries hold only once those phases are fixed. And the signs belong to operators; whether an interaction respects them is a question about its Lagrangian, which the weak interaction answers by breaking C and P, here only named.

Change one variable at a time

Make the relationship visible.

Interactive model

k = 0 to 4: Γ = 1, γ^μ, σ^μν, γ^μγ⁵, iγ⁵1

Spatial indices on the component1

Start at k = 1, the QED vertex: the photon copies the current and every term bar points right. At k = 3 the axial current's C and P bars flip and the photon's cannot, so ψ̄γ^μγ⁵ψA_μ breaks C and P, not T. Change the spatial indices: P and T bars flip in the first two columns, never in the term.

LORENTZ INDICES n1

SPATIAL INDICES nₛ1

CPT SIGN OF ψ̄Γψ-1

SYMMETRIES THE TERM BREAKS0

Live interpretationLORENTZ INDICES n: 1. SPATIAL INDICES nₛ: 1. CPT SIGN OF ψ̄Γψ: −1. SYMMETRIES THE TERM BREAKS: 0

Catch the common trap

Explain before calculating.

Misconception check

The tempting shortcut is to treat T as parity in time: reverse the time component of every four-vector, keep the space components, leave the c-numbers alone. It gets every bilinear except ψ̄ψ wrong. The current ψ̄γ^μψ keeps its time component under T, because charge density does not care which way the film runs, and reverses its space components, because velocities do; the objects whose time components flip are the coordinate x⁰ and the derivative ∂₀. And T is antilinear, so it conjugates every i: iψ̄γ⁵ψ is T-odd although γ⁵ is untouched, and the tensor's σ⁽⁰ⁱ⁾ parts are even while σ^(ij) are odd, the reverse of the shortcut. The working rule is the derived one: (−1)^μ for each index and −1 for each explicit i, with ∂_μ counted as −i times the momentum i∂_μ.

QED's vertex has already fixed the photon's C, P and T laws. Suppose a small axial coupling g_A ψ̄γ^μγ⁵ψA_μ, with g_A real, is added for the electron. Which statement is right?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyUsing the photon's C, P and T laws, find how ∂_μA^μ and A_μA^μ transform, and decide whether QED's gauge-fixing term −(1/2ξ)(∂_μA^μ)² and a photon mass term ½M²A_μA^μ are C-, P- and T-even. What, then, keeps the photon massless? Convert the measured bound, M below 1 × 10⁻¹⁸ eV (quoted), into the range ħ/(Mc) of a static field.

A_μA^μ = (A⁰)² − A⋅A. Under C every factor gives −1; under P and T, A⁰ keeps its sign and A reverses. Each product pairs two equal signs, so A_μA^μ is even under all three, evaluated at the transformed point.
∂_μA^μ = ∂ₜA⁰ + ∇⋅A. C gives −∂_μA^μ. Under P, ∂ₜ and A⁰ are kept while ∇ and A both reverse: +∂_μA^μ at (t, −x). Under T, ∂ₜ reverses and A⁰ does not, while ∇ is kept and A reverses: −∂_μA^μ at (−t, x). The counting rule agrees: no spatial index survives and the one ∂ gives nᵢ = 1, so T-odd.
Squares are even, so both terms are C-, P- and T-even. Dimensions: [A] = 1, so [M²A_μA^μ] = 2 + 2 = 4 and [(∂_μA^μ)²] = 4 with ξ dimensionless, as a Lagrangian density needs. Because ∂_μA^μ only changes sign, the Gupta–Bleuler condition is carried into itself.
So C, P and T allow a photon mass. Gauge invariance is what forbids it: under A_μ → A_μ + ∂_μλ, F_μν is unchanged but A_μA^μ gains 2A^μ∂_μλ + ∂_μλ∂^μλ. The gauge-fixing term breaks it too, but only among unphysical modes, which the Ward identity keeps decoupled (quoted).
Range: ħ/(Mc) = ħc/(Mc²) = 197.327 MeV fm ÷ (1 × 10⁻²⁴ MeV) = 1.97 × 10²⁶ fm = 1.97 × 10¹¹ m, about 1.3 times the Earth–Sun distance of 1.496 × 10¹¹ m.

Answer∂_μA^μ is C-odd, P-even and T-odd; A_μA^μ is even. Both terms are C-, P- and T-even, so only gauge invariance excludes the mass. The bound puts the range above 2.0 × 10¹¹ m, about 1.3 AU.

MediumAn electron has momentum 0.750 MeV along +z and spin along +z. With invariant normalisation its forward matrix elements are ⟨:ψ̄γ^μψ:(0)⟩ = 2p^μ and ⟨:ψ̄γ^μγ⁵ψ:(0)⟩ = 2m s^μ, quoted from Advanced QM. Evaluate both, then use only the C and T columns to predict them for the positron with the same momentum and spin and for the time-reversed electron, and check the T prediction directly.

Kinematics: E = √(0.750² + 0.511²) MeV = 0.9075 MeV, so 2p^μ = (1.815, 0, 0, 1.500) MeV. Spin along the motion gives s^μ = (|p|/m, 0, 0, E/m), with s⋅p = 0 and s² = −1, so 2m s^μ = (2|p|, 0, 0, 2E) = (1.500, 0, 0, 1.815) MeV. Dimensions: each operator has mass dimension 3 and each state −1, leaving MeV.
C: topic 11.02's C sends |e⁻(p, s)⟩ to |e⁺(p, s)⟩ and equals its inverse, so ⟨e⁺|O|e⁺⟩ = ⟨e⁻|C O C⁻¹|e⁻⟩ = η_C ⟨e⁻|O|e⁻⟩. The vector element becomes (−1.815, 0, 0, −1.500) MeV: the positron counts negatively, and the electric current −e j^μ flips with it. The axial element stays (1.500, 0, 0, 1.815) MeV: spin density counts both alike.
T: topic 11.03's T sends |e⁻(p, s)⟩ to a phase times |e⁻(−p, −s)⟩. Being antiunitary, it gives ⟨Te|O|Te⟩ = ⟨e|T⁻¹OT|e⟩*, which for a Hermitian bilinear with η_T fixed is η_T times the real ⟨e|O|e⟩. Both columns give (−1)^μ: vector (1.815, 0, 0, −1.500) MeV, axial (1.500, 0, 0, −1.815) MeV.
Direct check: the reversed electron has p′ = −p and spin along −z, so 2p′^μ = (1.815, 0, 0, −1.500) MeV. Both vectors flipped, so its helicity is unchanged and s′^μ = (|p|/m, 0, 0, −E/m), giving 2m s′^μ = (1.500, 0, 0, −1.815) MeV. Both agree; the axial time component, 2|p|, is a helicity density and so T-even.
Contrast P, quoted: |e⁻(−p, s)⟩ reverses the momentum but not the spin, so the vector element again takes (−1)^μ while the axial one takes −(−1)^μ, giving (−1.500, 0, 0, 1.815) MeV with the helicity reversed. P and T agree on the vector current and disagree on the axial one.

AnswerPositron: vector (−1.815, 0, 0, −1.500) MeV, axial (1.500, 0, 0, 1.815) MeV. Time-reversed electron: vector (1.815, 0, 0, −1.500) MeV, axial (1.500, 0, 0, −1.815) MeV, matching 2p′^μ and 2m s′^μ.

HardA neutral spin-0 meson φ of mass m couples to photons through ℒ = (κ/4)φF_μνF^μν + (κ̃/4)φF_μνF̃^μν, with F̃^μν = ½ε^μνρσF_ρσ and κ, κ̃ real. Find the C, P and T signs φ needs for each term to be invariant, and what both together break. Then set κ = 0 for the π⁰ (m = 134.977 MeV, τ = 8.43 × 10⁻¹⁷ s, BR(γγ) = 98.82%, quoted), extract |κ̃|, and say what an admixture κ = 0.10κ̃ would do to the photon polarisations.

From A^μ's laws, F^μν = ∂^μA^ν − ∂^νA^μ has C −1, P (−1)^μ(−1)^ν and T −(−1)^μ(−1)^ν. So F_μνF^μν = 2(B² − E²) is even under all three, while F_μνF̃^μν, proportional to E⋅B, is C-even, P-odd (E odd, B even) and T-odd (E even, B odd). The counting rule agrees: ε leaves three spatial indices.
The κ term needs φ C-, P- and T-even, a 0⁺⁺ state. The κ̃ term needs φ C-even, P-odd and T-odd, a 0⁻⁺ state, and that T-odd sign is exactly the table's entry for iψ̄γ⁵ψ, carried by a pseudoscalar made of a fermion pair. With both non-zero, η_C = +1 keeps C, but no single η_P or η_T serves both terms: P, T and CP are broken, CPT is not.
Rate: [φFF] = 5, so [κ] = [κ̃] = −1. Both amplitudes vanish when either ε is replaced by its photon's k, the Ward identity, so the polarisation sum Σ ε_μ ε_ν* → −g_μν is allowed and gives Σ|M|² = 2(κ² + κ̃²)(k₁⋅k₂)² = (κ² + κ̃²)m⁴/2, the cross term cancelling. Then Γ = ½ × (1/2m) × (1/8π) × Σ|M|² = (κ² + κ̃²)m³/(64π), the first ½ for identical photons; GeV⁻² × GeV³ leaves a width in GeV.
π⁰ numbers: Γ_γγ = 0.9882 × ħ/τ = 0.9882 × (6.5821 × 10⁻¹⁶ eV s ÷ 8.43 × 10⁻¹⁷ s) = 7.716 eV. With κ = 0, κ̃² = 64πΓ/m³ = 201.06 × 7.716 × 10⁻⁹ GeV ÷ 2.4591 × 10⁻³ GeV³ = 6.309 × 10⁻⁴ GeV⁻², so |κ̃| = 0.02512 GeV⁻¹, or 0.02512 × 0.197327 fm = 4.956 × 10⁻³ fm with ħc restored.
Polarisation: in the φ rest frame, with linear polarisations at relative angle χ about the photon axis, k⋅ε = 0 leaves M ∝ κ cos χ + κ̃ sin χ, the sign of the second term set by the orientation of ε^μνρσ. For κ = 0 the rate goes as sin²χ, perpendicular planes, as the π⁰'s double Dalitz pairs show (quoted). κ = 0.10κ̃ moves the maximum to χ = atan(κ̃/κ) = 84.3°, a 5.7° tilt that reverses under P and under T.
Limits: κ̃ → 0 gives cos²χ, parallel planes, the 0⁺⁺ case. The total rate, proportional to κ² + κ̃², rises by only 1.0% for κ = 0.10κ̃, so a lifetime cannot see the admixture; only the P- and T-odd tilt can.

AnswerC(φ) = +1 for both; P = T = +1 for κ, P = T = −1 for κ̃; together they break P, T and CP, not C. For the π⁰, |κ̃| = 2.51 × 10⁻² GeV⁻¹ (4.96 × 10⁻³ fm); κ = 0.10κ̃ tilts the polarisation maximum from 90° to 84.3°.