University Quantum Field Theory I · Discrete Symmetries of Quantised Fields: C, P, T and CPT · 11.6

What CPT predicts, and a coupling that breaks P and CP

CPT is the one discrete symmetry a local relativistic field theory cannot break, so its predictions test the whole framework at once. Learn exactly what it promises, which tempting promises it does not make, and how one scalar with two couplings breaks P and CP while an electron's dipole moment keeps watch.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topic 11.05 showed that the antiunitary Θ = CPT leaves every Hermitian, Lorentz-invariant local Lagrangian unchanged, so it commutes with the Hamiltonian. This topic cashes that in. P and T each reverse momentum while only T reverses spin, so Θ keeps a particle's momentum, flips its spin and turns it into its antiparticle: at rest, spin up becomes the antiparticle with spin down. Θ then carries one propagator onto the other pole for pole, so masses and total widths agree to all orders, CP violation included; and because the moment operator is CPT-even while spin is CPT-odd, the moments are equal and opposite and the g-factors equal.

Partial widths are another matter. CPT turns an outgoing state into an incoming one, so it equates a → f with ā → f̄ only when the products do not rescatter; a measured difference needs a CP-violating phase and a final-state phase together, and is no evidence against CPT. The second half builds a theory that breaks P and CP yet keeps CPT: a scalar coupled through both ψ̄ψ and iψ̄γ⁵ψ, which no intrinsic parity can rescue.

Its fingerprint is an electron dipole moment proportional to gₛ gₚ mₑ, and the HfF⁺ bound tests it. The predictions stop where the theorem's assumptions stop: a mass or moment mismatch would mean nonlocality, broken Lorentz invariance or an energy unbounded below, and no local relativistic field theory could absorb it.

Simple definition: CPT's predictions are the equalities every local relativistic field theory must obey: particle and antiparticle share one mass and one total width and carry equal and opposite magnetic moments; equal partial widths are not among them.
Example: The electron's moment is 1.001 159 652 μ_B pointing against its spin, so CPT predicts 1.001 159 652 μ_B along the positron's spin; trapped e⁻ and e⁺ show g-factors equal to within 2.1 × 10⁻¹².

CPT on a one-particle stateΘ|ap, s_z⟩ = η|āp, −s_z⟩, since ΘPΘ⁻¹ = P, ΘJΘ⁻¹ = −J and ΘQΘ⁻¹ = −Q

Derived: P and T each reverse p, so CPT keeps it; only T reverses spin, and C exchanges particle and antiparticle. A particle at rest, spin up, goes to its antiparticle at rest, spin down.

Θ = CPT, antiunitary; η a phase; p the three-momentum in GeV; s_z in units of ħ; P, J, Q momentum, angular momentum, charge

One pole for particle and antiparticle⟨a|(z − H)⁻¹|a⟩ = ⟨ā|(z − H)⁻¹|ā⟩ for every complex z

Derived from ⟨α|A|β⟩ = ⟨Θβ|ΘA†Θ⁻¹|Θα⟩ with ΘHΘ⁻¹ = H. Masses and total widths agree to all orders, CP-violating interactions included.

z a complex energy in GeV; states at rest; the shared pole at z = M − iΓ/2 fixes the mass M and total width Γ, both in GeV

Opposite moments, equal g-factorsμ_ā = −μₐ and g_ā = gₐ, because ΘμΘ⁻¹ = +μ for μ = ½∫ x × j d³x

Derived. The x and the current each flip under Θ, so μ is CPT-even while spin is CPT-odd. Quoted test: g(e⁻)/g(e⁺) − 1 = (0.5 ± 2.1) × 10⁻¹².

μ the moment along the spin, in e GeV⁻¹ (J T⁻¹ in SI); g dimensionless; uses q_ā = −qₐ and m_ā = mₐ

A partial-width difference needs two phases|A|² − |Ā|² = −4A₁A₂ sin(φ₁ − φ₂) sin(δ₁ − δ₂)

Shown for two amplitudes, with Watson's final-state phases quoted. Either phase difference zero and Γ(a → f) = Γ(ā → f̄); summed over states that rescatter, the differences cancel.

A₁, A₂ real magnitudes of two amplitudes to one final state; φ weak phases, odd for ā; δ strong phases, shared; all in rad

A scalar with both couplingsLᵢₙₜ = −φψ̄(gₛ + i gₚ γ⁵)ψ

Derived from the bilinear table: the gₛ term keeps P only if φ's parity is +1, the gₚ term only if it is −1, so P, T and CP fail together while C and CPT survive.

φ a real scalar of mass M in GeV; gₛ, gₚ real and dimensionless; Lᵢₙₜ in GeV⁴; the i makes iψ̄γ⁵ψ Hermitian

The electron dipole moment it induces|dₑ| = e|gₛ gₚ| mₑ [ln(M²/mₑ²) − 3/2]/(8π²M²), for M ≫ mₑ

The factors gₛ gₚ, mₑ and 1/M² are derived from symmetry and dimensions; the loop coefficient is quoted. For M ≪ mₑ it tends to e|gₛ gₚ|/(8π²mₑ).

dₑ in e GeV⁻¹, 1 GeV⁻¹ = 1.973 × 10⁻¹⁴ cm; mₑ = 5.11 × 10⁻⁴ GeV and M in GeV; the bracket is a loop integral's large-M limit

What CPT does to energy, momentum, spin and charge

Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Topic 11.05 showed, and this topic uses, that Θ = CPT sends each Hermitian field product to itself at −x times (−1)ⁿ, where n counts its Lorentz indices. Apply that to the conserved charges. Energy and momentum are integrals of T^(0ν), two indices, so ΘHΘ⁻¹ = H and ΘPΘ⁻¹ = P. The charge integrates the current j⁰, one index, so ΘQΘ⁻¹ = −Q. Angular momentum integrates xⁱ T^(0j) − x^j T⁽⁰ⁱ⁾, and the explicit x supplies one more flip, so ΘJΘ⁻¹ = −J. The ladder operators of topics 11.01 to 11.03 say the same state by state: P and T each reverse p, so CPT keeps it; T alone reverses the spin; C swaps b for d. Hence Θ|a; p, s_z⟩ = η|ā; p, −s_z⟩: a particle at rest with spin up becomes its antiparticle at rest with spin down. Compare CP, which reverses p and keeps the spin.

One pole: equal masses and equal total widths

For an antiunitary Θ, ⟨α|A|β⟩ = ⟨Θβ|ΘA†Θ⁻¹|Θα⟩. Take α = β = a at rest and A = (z − H)⁻¹ with z a complex energy. Then A† = (z* − H)⁻¹, and Θ conjugates z* back to z while leaving H alone, so ⟨a|(z − H)⁻¹|a⟩ = ⟨ā|(z − H)⁻¹|ā⟩: one analytic function. A stable particle's pole sits at its mass, an unstable one's at M − iΓ/2, and particle and antiparticle share it, so m_ā = mₐ and Γ_ā = Γₐ to every order, derived. In the free-field limit this is trivial, since one Dirac field carries one m; the content is that loop corrections, CP-violating ones included, cannot split the pair. CP would force the same, but CP fails for kaons, so their masses test CPT: interference against the K_L − K_S splitting of 3.48 × 10⁻¹² MeV gives |m_K⁰ − m_K̄⁰|/m_K below 6 × 10⁻¹⁹, about 3 × 10⁻¹⁰ eV. The μ⁺ and μ⁻ lifetimes agree to (2 ± 8) × 10⁻⁵. Both quoted.

Opposite magnetic moments, equal g-factors

The moment operator μ = ½∫ x × j d³x contains an explicit x and a vector current, each flipped by Θ, so ΘμΘ⁻¹ = +μ while spin is CPT-odd. For a Hermitian operator the antiunitary rule gives equal real expectation values, ⟨a, ↑|μ_z|a, ↑⟩ = ⟨ā, ↓|μ_z|ā, ↓⟩: the antiparticle's moment along its own spin is the opposite, μ_ā = −μₐ, derived. Write μ = (g/2)(q/m)S for each; with q_ā = −qₐ and m_ā = mₐ the two signs cancel and g_ā = gₐ. So the electron's moment points against its spin and the positron's along it, both 1.001 159 652 μ_B. Limiting check: at tree level both have Dirac's g = 2, and Schwinger's α/(2π) correction depends on the charge only through e², so it is shared. Quoted tests: e⁻ and e⁺ g-factors agree to (0.5 ± 2.1) × 10⁻¹²; the antiproton's moment is −1 times the proton's 2.792 847 μ_N to 1.5 parts per billion; its charge-to-mass ratio is −1 times the proton's to 16 parts per trillion.

Totals must agree; single channels need not

CPT reverses time, so it turns an outgoing state into an incoming one: ⟨f, out|a⟩ = ⟨ā|f̄, in⟩ up to a phase, with f̄ the CPT image of f. The decay of ā needs ⟨f̄, out|ā⟩ instead. At lowest order in the decay interaction H_w, with no rescattering, in equals out, and Hermiticity closes the gap: |⟨f|H_w|a⟩| = |⟨f̄|H_w|ā⟩|, equal partial widths. Beyond that they can differ. Write A(a → f) = A₁e^{(i(φ₁+δ₁))} + A₂e^{(i(φ₂+δ₂))}, where the weak phases φ come from complex couplings and flip sign for ā, while the strong phases δ come from final-state rescattering (Watson's theorem, quoted) and do not. Then |A|² − |Ā|² = −4A₁A₂ sin(φ₁ − φ₂) sin(δ₁ − δ₂): a difference needs CP violation and a final-state phase together. Measured: B⁰ → K⁺π⁻ against B̄⁰ → K⁻π⁺ shows an asymmetry of −0.083, over twenty standard deviations from zero. Summed over every state the products rescatter into, the differences cancel, which is how the totals stay equal.

A scalar with both couplings breaks P, T and CP

Add a real scalar φ of mass M with Lᵢₙₜ = −φψ̄(gₛ + i gₚγ⁵)ψ, gₛ and gₚ real. Topic 11.04's table gives ψ̄ψ the signs (+, +, +) under (P, C, T) and iψ̄γ⁵ψ the signs (−, +, −), its T sign coming entirely from T's antilinearity acting on the i that Hermiticity demands. A real φ may carry intrinsic signs η_P, η_C, η_T = ±1, but only one set. The gₛ term is P-invariant only if η_P = +1, the gₚ term only if η_P = −1, so with both present P fails whatever φ's parity, and T fails the same way. Both bilinears are C-even, so η_C = +1 keeps C, which leaves CP acting like P: broken. CPT asks each term for η_Pη_Cη_T = +1, and both give it, as 11.05 promised. One caveat sharpens the claim: the chiral rotation ψ → e^(iαγ⁵)ψ can turn gₛ + i gₚγ⁵ into a pure scalar, but it moves the phase into the electron mass. Only the angle between the Yukawa phase and the mass is physical, so at mₑ = 0 the violation could be rotated away.

Why an electron dipole moment needs both couplings

UAQM-01.08 showed that a permanent electric dipole moment dₑ violates both P and T. Here that becomes a rule on couplings. With gₚ = 0 the theory keeps P with η_P = +1, so dₑ = 0; with gₛ = 0 it keeps P with η_P = −1, so again dₑ = 0. Sharper: P maps the theory with (gₛ, gₚ) onto the one with (gₛ, −gₚ) and reverses dₑ, so dₑ is odd in gₚ; relabelling φ → −φ flips both couplings and changes nothing, so dₑ is odd in gₛ too. At one loop dₑ ∝ gₛ gₚ, derived. It also needs a chirality flip, hence a factor mₑ, and dimensions then demand 1/M² for a heavy scalar. The loop coefficient is quoted: |dₑ| = e|gₛ gₚ|mₑ[ln(M²/mₑ²) − 3/2]/(8π²M²). At M = 1 GeV that is |gₛ gₚ| × 1.74 × 10⁻¹⁸ e cm, and the HfF⁺ bound of 4.1 × 10⁻³⁰ e cm gives |gₛ gₚ| < 2.4 × 10⁻¹². For gₛ = gₚ that is 1.5 × 10⁻⁶, below the electron's own Higgs coupling mₑ/v = 2.1 × 10⁻⁶.

Change one variable at a time

Make the relationship visible.

Interactive model

Weak-phase difference Δφ60 °

Strong-phase difference Δδ45 °

Amplitude ratio r = A₂/A₁0.5

Set Δφ or Δδ to zero and each channel's solid and open bars match. With both on, what f₁ gains f₂ loses, so the two total bars never part: CPT fixes the sum, not the split. At r = 0 there is no second amplitude to interfere with.

Γ(a → f₁)0.496 Γ₀

Γ(ā → f̄₁)1.108 Γ₀

CP ASYMMETRY IN f₁0.382

CP ASYMMETRY IN f₂-0.683

Live interpretationΓ(a → f₁): 0.496 Γ₀. Γ(ā → f̄₁): 1.108 Γ₀. CP ASYMMETRY IN f₁: 0.382. CP ASYMMETRY IN f₂: −0.683

Catch the common trap

Explain before calculating.

A real scalar φ is added to QED with Lᵢₙₜ = −φψ̄(gₛ + i gₚ γ⁵)ψ, both couplings non-zero. You may give φ any intrinsic signs η_P, η_C, η_T = ±1. Which statement is right?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron in a Penning trap's 3.00 T field has g = 2.002 319 304 36, its moment pointing against its spin. Use CPT to predict the positron's g-factor, which way its moment points, which spin state lies lower, and its spin-flip frequency. What frequency difference would a fractional g difference of 2.1 × 10⁻¹² produce?

CPT maps e⁻ with spin up to e⁺ with spin down, and μ is CPT-even, so ⟨μ_z⟩ is equal in the two states: the positron's moment along its own spin is opposite to the electron's. It points along the positron's spin, with the same size (g/2)μ_B = 1.001 159 652 μ_B.
With q(e⁺) = +e and equal masses, μ = (g/2)(q/m)S returns g(e⁺) = g(e⁻) = 2.002 319 304 36. The energy is −μ⋅B, so the positron lies lower with its spin along B, the electron with its spin against B.
A spin flip changes the energy by gμ_B B for either particle, so ν = g(μ_B/h)B = 2.002 319 304 36 × 13.996 245 GHz T⁻¹ × 3.00 T = 84.07 GHz. Dimensions: (J T⁻¹)(T)/(J s) = s⁻¹.
A fractional difference of 2.1 × 10⁻¹² in g shifts ν by the same fraction: 2.1 × 10⁻¹² × 84.07 × 10⁹ Hz = 0.18 Hz.

Answerg(e⁺) = g(e⁻) = 2.002 319 304 36; the positron's moment lies along its spin, so spin along B is lower; ν = 84.07 GHz for both; a 2.1 × 10⁻¹² difference is 0.18 Hz.

MediumA spin-0 particle a decays into f₁ and f₂, which rescatter into each other; the strong eigenchannels are η₁ and η₂, with f₁ = (η₁ + η₂)/√2 and f₂ = (η₂ − η₁)/√2. a reaches η₁ and η₂ with amplitudes 1 and r = 0.40, weak phases differing by Δφ = 50° and strong phases by Δδ = 30°. In units of Γ₀, a's width into η₁ alone, find the four partial widths of a and ā, each channel's CP asymmetry, and both totals.

For ā the weak phases flip sign and the strong ones do not. Projecting onto f₁ gives Γ(a → f₁) = ½[1 + r² + 2r cos(Δφ + Δδ)] and Γ(ā → f̄₁) = ½[1 + r² + 2r cos(Δδ − Δφ)]; in f₂ the interference term changes sign.
With 1 + r² = 1.160, 2r = 0.80, cos 80° = 0.1736 and cos(−20°) = 0.9397: Γ(a → f₁) = ½(1.160 + 0.139) = 0.649, Γ(ā → f̄₁) = ½(1.160 + 0.752) = 0.956, Γ(a → f₂) = 0.511 and Γ(ā → f̄₂) = 0.204, all in Γ₀.
Check with the two-phase formula: Γ(a → f₁) − Γ(ā → f̄₁) = −2r sin Δφ sin Δδ = −0.80 × 0.7660 × 0.500 = −0.306, and the f₂ difference is +0.306.
Asymmetries [Γ(ā) − Γ(a)]/[Γ(ā) + Γ(a)]: f₁ gives 0.306/1.605 = 0.191, f₂ gives −0.306/0.715 = −0.429. They differ in size because the channels differ in rate; what cancels is the rate differences.
Totals: 0.649 + 0.511 = 1.160 for a and 0.956 + 0.204 = 1.160 for ā, equal as CPT demands. Limiting case: at Δδ = 0 both widths into f₁ become 0.837, and at Δφ = 0 both become 0.926; one phase alone produces nothing.

AnswerΓ(a → f₁) = 0.649, Γ(ā → f̄₁) = 0.956, Γ(a → f₂) = 0.511, Γ(ā → f̄₂) = 0.204, in Γ₀; asymmetries +0.191 in f₁ and −0.429 in f₂; both totals 1.160 Γ₀.

HardA scalar φ of mass M = 20.0 MeV couples only to electrons, as −φψ̄(gₛ + i gₚγ⁵)ψ with gₛ = G cos θ and gₚ = G sin θ. (a) At fixed G, which θ maximises the one-loop electron dipole moment? (b) There, what largest G does the HfF⁺ bound |dₑ| < 4.1 × 10⁻³⁰ e cm allow? (c) For that G, find φ's lifetime and mean decay length at an energy of 1.00 GeV. (d) What does the bound say at θ = 0?

(a) dₑ ∝ gₛ gₚ = G² sin θ cos θ = ½G² sin 2θ, largest at θ = 45°, where gₛ gₚ = ½G². At θ = 0 or 90° the theory keeps P and T, so dₑ vanishes identically.
(b) M/mₑ = 39.1, so the large-M form applies: ln(M²/mₑ²) − 3/2 = 7.334 − 1.5 = 5.834, and |dₑ|/|gₛ gₚ| = e mₑ × 5.834/(8π²M²) = e × 5.11 × 10⁻⁴ × 5.834/(78.96 × 4.00 × 10⁻⁴) GeV⁻¹ = 0.0944 e GeV⁻¹. The exact Feynman-parameter integral, done numerically, gives 0.0947, 0.27% higher.
Convert with 1 GeV⁻¹ = 1.973 × 10⁻¹⁴ cm: 0.0947 × 1.973 × 10⁻¹⁴ = 1.87 × 10⁻¹⁵ e cm per unit gₛ gₚ. So gₛ gₚ < 4.1 × 10⁻³⁰/1.87 × 10⁻¹⁵ = 2.2 × 10⁻¹⁵, G² < 4.4 × 10⁻¹⁵ and G < 6.6 × 10⁻⁸.
(c) Topic 7.07's rates, quoted, add without interference once spins are summed: Γ = G²M(cos²θ β³ + sin²θ β)/(8π), with β = √(1 − 4mₑ²/M²) = 0.99869; at 45° the bracket is (β³ + β)/2 = 0.99739. So Γ = 4.39 × 10⁻¹⁵ × 0.0200 GeV × 0.99739/(8π) = 3.48 × 10⁻¹⁸ GeV.
τ = ħ/Γ = 6.582 × 10⁻²⁵ GeV s/3.48 × 10⁻¹⁸ GeV = 1.89 × 10⁻⁷ s and cτ = 56.6 m. At E = 1.00 GeV, γβ = p/M = 49.99, so the mean decay length is 2.83 km.
(d) At θ = 0 the coupling is a pure scalar, P and T hold, and dₑ = 0 at every order: the bound says nothing about G. Checks: [dₑ] = e GeV/GeV² = e GeV⁻¹, a length; for M ≪ mₑ the integral would give e gₛ gₚ/(8π²mₑ) instead.

Answer(a) θ = 45°. (b) G < 6.6 × 10⁻⁸, so gₛ = gₚ < 4.7 × 10⁻⁸. (c) τ = 1.9 × 10⁻⁷ s, cτ = 57 m, about 2.8 km at 1.00 GeV. (d) No bound: a pure scalar induces no dipole moment.