University Quantum Field Theory I · Discrete Symmetries of Quantised Fields: C, P, T and CPT · 11.5

CPT invariance of Hermitian, Lorentz-invariant Lagrangians

Any Lorentz-invariant interaction you write down may break C, P or T, and the weak force breaks all three; none can break their product. This lesson shows why using nothing but the bilinear tables: count Lorentz indices, let an antiunitary operator conjugate the couplings, and see exactly which assumptions the result spends.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Charge conjugation, parity and time reversal are each a choice a theory may or may not make: the weak interaction breaks C and P, and CP and T violation have both been measured. Their product is different. Composing the operators of topics 11.01 to 11.03 gives one antiunitary map, Θ = CPT, that sends every boson field, fermion bilinear and derivative carrying n Lorentz indices to (−1)ⁿ times its Hermitian conjugate at −x: it is the total reflection x^μ → −x^μ, carried on spinors by γ⁵.

A Lorentz scalar has an even number of indices, so the signs cancel; antiunitarity turns each coupling g into g*; and a Hermitian Lagrangian already holds g* in front of each term's conjugate. Every term therefore goes into its own Hermitian partner, ℒ(x) into ℒ(−x), and the action is unchanged whatever C, P and T do separately. The argument has costs, and they mark where it stops: a Lagrangian built locally from fields with no fixed background tensor, Hermiticity, and anticommuting fermions, whose reordering sign is all that keeps a fermion mass term CPT-even.

Lüders, Pauli and Jost proved the stronger result with no Lagrangian at all, from the axioms of local quantum field theory; that theorem, and the spin-statistics theorem for interacting fields, are stated here and quoted.

Simple definition: CPT invariance is the statement that the antiunitary product Θ = CPT maps the Lagrangian density at x onto the Lagrangian density at −x, so the action, and with it the dynamics, is unchanged even when C, P and T are separately broken.
Example: The electron EDM term −(i/2) d ψ̄σ^μνγ⁵ψ F_μν is odd under P and under T, yet its fields carry 2 + 2 = 4 Lorentz indices, so CPT multiplies it by (−1)⁴ = +1 and leaves the real d alone: the term is CPT-even.

CPT on the Dirac fieldΘ ψ(x) Θ⁻¹ = η γ⁵ ψ*(−x)

Derived by composing P's γ⁰, C's γ² and T's γ¹γ³: all four γ's multiply to a phase times γ⁵, the spinor form of x → −x, since γ⁵γ^μγ⁵ = −γ^μ.

Θ = CPT, antiunitary; η a phase, |η| = 1, common to all Dirac fields; ψ has mass dimension 3/2, x is in GeV⁻¹

CPT on a Dirac bilinearΘ (ψ̄Γψ)(x) Θ⁻¹ = (−1)ⁿ (ψ̄Γψ)†(−x)

Derived, with one of its two minus signs supplied by anticommutation. Scalar, pseudoscalar and tensor come out even; vector and axial vector odd.

Γ a product of gamma matrices with n free Lorentz indices, γ⁵ counting none; ψ̄Γψ has mass dimension 3

CPT on boson fields and derivativesΘ B^{(μ₁…μₙ)}(x) Θ⁻¹ = (−1)ⁿ B^{(μ₁…μₙ)}†(−x)

Derived from topics 11.01 to 11.04, with every boson's CPT phase set to +1, which is always allowed. P and T each treat A⁰ and Aⁱ differently; CPT gives all four −1.

B a boson field with n indices: φ (n = 0) and A^μ (n = 1) of mass dimension 1, F^μν (n = 2) of dimension 2; ∂_μ adds 1 to n

A Hermitian, local, Lorentz-invariant LagrangianΘ ℒ(x) Θ⁻¹ = ℒ(−x)

Shown: (−1)^even = +1 and antiunitarity sends gᵢ to gᵢ*, so each term becomes its partner. ∫d⁴x ℒ is then unchanged, whatever C, P and T do.

ℒ = Σ (gᵢ Oᵢ + gᵢ* Oᵢ†), each Oᵢ a product of fields at x with an even index count; ℒ has mass dimension 4

The same bilinear if fermions commutedΘ (ψ̄Γψ)(x) Θ⁻¹ = −(−1)ⁿ (ψ̄Γψ)†(−x)

Derived. QED's mass, kinetic and vertex terms would flip sign while −¼F_μνF^μν did not, so the argument uses spin-statistics; it does not prove it.

the bilinear law with the anticommutation sign removed; n and Γ as before

A fixed background vector breaks CPTΘ (b_μ ψ̄γ^μγ⁵ψ)(x) Θ⁻¹ = −b_μ (ψ̄γ^μγ⁵ψ)(−x)

Derived. Every index is contracted, but only one sits on a field, so N = 1. The term breaks Lorentz invariance, the input the theorem needed.

b_μ a real constant four-vector of mass dimension 1 (GeV), not a field: it neither flips nor moves

Three operators make one reflection

Work in ħ = c = 1 with the metric (+, −, −, −), and compose the operators of topics 11.01 to 11.03 on the Dirac field. Parity supplies γ⁰ and sends x⃗ to −x⃗; time reversal supplies γ¹γ³, sends t to −t and is antiunitary; charge conjugation supplies γ² and turns ψ into ψ*. Up to phases the product of all four gamma matrices is γ⁰γ¹γ²γ³ = −iγ⁵, so Θψ(x)Θ⁻¹ = η γ⁵ ψ*(−x), with |η| = 1 and every coordinate reversed. The γ⁵ is no accident: γ⁵γ^μγ⁵ = −γ^μ, so γ⁵ is the spinor image of the total reflection x^μ → −x^μ. Bosons follow the same pattern. A complex scalar goes to φ†(−x) once its CPT phase is chosen as +1. For the photon, P and T each keep A⁰ and flip A^i, and C flips everything, so A⁰ collects (+1)(+1)(−1) and Aⁱ collects (−1)(−1)(−1): every component gives −A^μ(−x). Parity and time reversal each treat time and space differently; their product treats all four components alike, which is why one index-counting rule can cover everything.

The bilinear rule, and the sign anticommutation pays

Write ψ̄Γψ = ψ†Mψ with M = γ⁰Γ and apply Θ in three moves. Antiunitarity conjugates the c-number matrix, M → M*. The field law turns ψ†Mψ into ψ(γ⁵M*γ⁵)ψ*, operators now in the wrong order, and restoring ψ* to the left costs −1: the fields anticommute, and the c-number (ψ, ψ†) at one point is removed by normal ordering. It is the sign that made the current C-odd in topic 11.02. Pulling γ⁰ through γ⁵ costs a second −1 and leaves ψ̄γ⁵Γ̄γ⁵ψ, with Γ̄ = γ⁰Γ†γ⁰ and ψ̄Γ̄ψ = (ψ̄Γψ)†. Since γ⁵ anticommutes with each γ^μ, γ⁵Γ̄γ⁵ = (−1)ⁿΓ̄, which derives the rule. For ψ̄ψ, iψ̄γ⁵ψ, ψ̄γ^μψ, ψ̄γ^μγ⁵ψ and ψ̄σ^μνψ the signs are +, +, −, −, +, the product of topic 11.04's C and T columns with Advanced QM's P column: the axial current gives (+1) × (−1)^μ × [−(−1)^μ] = −1. Two different fields leave a factor η₁*η₂, set to 1 by a common phase.

Count indices, conjugate couplings

A Hermitian, local, Lorentz-invariant Lagrangian is a sum of terms g O + g* O†, where O is a product, at one point x, of boson fields, fermion bilinears and derivatives with every index contracted by g_μν or ε^μνρσ. Each contraction pairs two indices and ε carries four, so the total count N on the fields is even. Each factor goes to its own sign times its conjugate at −x, and each ∂_μ adds −1, because ∂/∂x^μ of a function of −x changes sign. The factors are bosonic, so inside normal ordering their order reverses at no cost, and O(x) → (−1)^N O†(−x) = O†(−x). Antiunitarity turns g into g*, so g O(x) goes to g* O†(−x), the partner term, and the partner returns g O(−x): Θℒ(x)Θ⁻¹ = ℒ(−x), shown. The QED vertex −eψ̄γ^μψA_μ collects (−1)(−1) = +1, and ∫d⁴x ℒ is unchanged because d⁴x is. Limits agree: with every coupling off, the free terms are CPT-even by the same count; with every term C-, P- and T-even, CPT is just their product.

C, P and T may fail one at a time; CPT holds

The theorem earns its keep on terms that fail the separate tests. The electron electric-dipole term −(i/2) d ψ̄σ^μνγ⁵ψ F_μν flips sign under P and under T, so a measured electron EDM would break both; but iψ̄σ^μνγ⁵ψ and F_μν carry two indices each, N = 4, d is real, and Θ returns the term unchanged at −x. A chiral coupling ψ̄γ^μ(1 − γ⁵)ψ V_μ, with V_μ a vector, breaks P maximally: P turns it into ψ̄γ^μ(1 + γ⁵)ψ V_μ, and C, with either C-phase for V, changes it too, yet both pieces have N = 2 and CPT keeps it. A complex Yukawa g φ ψ̄₁ψ₂ + g* φ† ψ̄₂ψ₁ can carry CP violation when field rephasings cannot remove the phase of g, and it still survives, because g → g* is exactly the exchange of the two terms. The consequence, developed in topic 11.06, is a bookkeeping rule: in a CPT-invariant theory, T violation and CP violation come in equal amounts.

Commuting fermions would break it

One minus sign in the bilinear rule came from anticommutation, so take it away. Quantise ψ with commutators and every fermion bilinear picks up an extra −1: Θ(ψ̄Γψ)(x)Θ⁻¹ = −(−1)ⁿ(ψ̄Γψ)†(−x). In QED the Dirac kinetic term, the mass term and the vertex −eψ̄γ^μψA_μ each contain one bilinear and all change sign, while −¼F_μνF^μν contains none and does not, so ℒ goes to ℒ_Maxwell − ℒ_Dirac − ℒᵢₙₜ at −x. No phase convention rescues it, because a bilinear of a single field carries |η|² = 1. The bilinear argument therefore uses the spin-statistics connection rather than proving it. The sign is an old acquaintance: reordering d d† in the free Dirac Hamiltonian is what UAQM-09.06 and 09.07 found keeps the energy bounded below, with every norm positive, only with anticommutators, and the same reordering of ψ and ψ† is what keeps a fermion mass term CPT-even here.

The general theorem, quoted, and where CPT can fail

The bilinear argument needs a Lagrangian. The Lüders–Pauli–Jost theorem does not, and is quoted here. Its axioms: a unitary representation of the proper orthochronous Poincaré group with energy bounded below and a unique invariant vacuum; local fields in finite-dimensional Lorentz representations; and microcausality, fields commuting or anticommuting at spacelike separation. Jost's route continues vacuum expectation values into the complex Lorentz group, where x → −x is connected to the identity, and antiunitarity comes from reversing operator order. The spin-statistics theorem for interacting fields, extending UAQM-09.09, rests on the same axioms and is also quoted. Drop an input and CPT can fail: a fixed background vector in b_μψ̄γ^μγ⁵ψ neither flips nor moves, leaving N = 1, so the term is CPT-odd and Lorentz-violating; nonlocal interactions escape too. Experiment agrees so far: the K⁰ and K̄⁰ masses differ by less than 6 × 10⁻¹⁹ of their value (quoted).

Change one variable at a time

Make the relationship visible.

Interactive model

Coupling phase δ60 °

Field indices N (odd: background b_μ)2

Commuting fermions q (0 no, 1 yes)0

Sweep δ with N = 2 and q = 0: the arrow lands in the small ring at every phase, so a complex coupling never breaks CPT, while the last readout shows a linear map would match only at δ = 0° or 180°. Set N = 1, a background vector, or q = 1, commuting fermions, and the arrow points away.

CPT SIGN s1

MISMATCH WITH PARTNER0 |g|

IF CPT WERE LINEAR1.73 |g|

Live interpretationCPT SIGN s: 1. MISMATCH WITH PARTNER: 0 |g|. IF CPT WERE LINEAR: 1.73 |g|

Catch the common trap

Explain before calculating.

A Lagrangian holds the free kinetic and mass terms of two Dirac fields ψ₁ and ψ₂ and of a complex vector field W_μ, plus g ψ̄₁γ^μψ₂W_μ + g* ψ̄₂γ^μψ₁W_μ†, with g = |g|e^(iδ) and δ = 0.4. Which statement about Θ = CPT is right?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThree candidate additions to QED, with ψ the electron field, φ a real scalar, and κ and g real: (a) the Pauli term −κ ψ̄σ^μνψ F_μν, (b) g φ iψ̄γ⁵ψ, (c) g φ ψ̄γ⁵ψ. Find each one's CPT sign from index counting and Hermiticity, then evaluate κ = aₑ e/(4 mₑ), quoted from the vertex correction, for aₑ = 0.00115965, in MeV⁻¹ and in fm.

(a) ψ̄σ^μνψ is Hermitian with n = 2, and F_μν = ∂_μA_ν − ∂_νA_μ carries 2 more (one from the derivative, one from A). N = 4, (−1)⁴ = +1 and κ is real, so the term returns unchanged at −x: CPT-even.
(b) iψ̄γ⁵ψ is Hermitian with n = 0 and φ carries no index, so N = 0, g is real and the term is CPT-even although the bilinear is a pseudoscalar: CPT never asks about intrinsic parity.
(c) ψ̄γ⁵ψ is anti-Hermitian: (ψ̄γ⁵ψ)† = −ψ̄γ⁵ψ. The rule gives g φ ψ̄γ⁵ψ → g φ (−ψ̄γ⁵ψ)(−x), so the term is CPT-odd. It is also not Hermitian, and its Hermitian part, half of the term plus its conjugate, is zero: it can never belong to a Lagrangian.
Dimensions: [ψ̄σ^μνψ] = 3 and [F_μν] = 2, so [κ] = 4 − 5 = −1. With e = √(4πα) = 0.30282 and mₑ = 0.51100 MeV, κ = 0.00115965 × 0.30282 / (4 × 0.51100 MeV) = 1.7180 × 10⁻⁴ MeV⁻¹.
Restore ħc = 197.327 MeV fm: κ = 1.7180 × 10⁻⁴ MeV⁻¹ × 197.327 MeV fm = 0.033902 fm, a length, as mass dimension −1 requires.

Answer(a) and (b) are CPT-even; (c) is CPT-odd because it is not Hermitian. κ = 1.718 × 10⁻⁴ MeV⁻¹ = 3.390 × 10⁻² fm.

MediumTake the Dirac kinetic term iψ̄γ^μ∂_μψ and the pseudovector coupling (f/m_π) ψ̄γ^μγ⁵ψ ∂_μπ⁰, with ψ the proton field of mass mₚ = 938.27 MeV, f²/4π = 0.075 (quoted) and m_π = 139.57 MeV. Track every CPT sign in both, then rewrite the coupling as an equivalent pseudoscalar one on shell and find its strength.

Kinetic term: Γ = iγ^μ has one index and the derivative on ψ adds one, so the sign is (−1)² = +1, and the term goes to its own conjugate at −x: (iψ̄γ^μ∂_μψ)† = −i(∂_μψ̄)γ^μψ, the i conjugated by antiunitarity on the way.
Integrate by parts: −i(∂_μψ̄)γ^μψ = iψ̄γ^μ∂_μψ − i∂_μ(ψ̄γ^μψ). The term returns to itself at −x plus a total derivative that ∫d⁴x drops, so it is CPT-even in exactly the sense that it is Hermitian: up to a surface term.
Coupling: ψ̄γ^μγ⁵ψ is Hermitian with n = 1, ∂_μπ⁰ carries one index from the derivative, π⁰ is real and f/m_π is real. N = 2, so the coupling is CPT-even, pseudoscalar pion and all.
On shell the Dirac equation gives ∂_μ(ψ̄γ^μγ⁵ψ) = 2imₚψ̄γ⁵ψ, so integrating by parts turns the coupling into −(2 mₚ f/m_π) π⁰ iψ̄γ⁵ψ, with N = 0 and again CPT-even, as it must be: a surface term cannot change the CPT sign.
Numbers: f = √(4π × 0.075) = 0.97081, so f/m_π = 0.97081 / 139.57 MeV = 6.9557 × 10⁻³ MeV⁻¹, or 1.3726 fm after multiplying by ħc; mass dimension −1, as a dimension-5 operator needs. The pseudoscalar strength is g = 2 × 938.27 × 0.97081 / 139.57 = 13.053, dimensionless, with g²/4π = 13.56.

AnswerBoth are CPT-even, the kinetic term up to the surface term −i∂_μ(ψ̄γ^μψ). f/m_π = 6.956 × 10⁻³ MeV⁻¹ (1.373 fm), equivalent on shell to g = 13.05, g²/4π = 13.6.

HardAdd to QED the term −b_μ ψ̄γ^μγ⁵ψ, with b_μ a real constant four-vector whose spatial part has the illustrative size |b⃗| = 1.0 × 10⁻³¹ GeV. Decide whether the term is Hermitian, Lorentz invariant and CPT invariant, find the energy splitting it gives an electron at rest, and convert that splitting into a spin-precession frequency and period.

Hermiticity and dimension: ψ̄γ^μγ⁵ψ is Hermitian and b_μ is real, so the term is Hermitian. [ψ̄γ^μγ⁵ψ] = 3, so [b] = 4 − 3 = 1: b is an energy.
Lorentz: the index is contracted, but b_μ stays fixed while the fields transform, so boosting or rotating the electron and not b changes the term. b⃗ picks out a direction, and even rotation invariance is lost.
CPT: the bilinear has n = 1 and goes to −(ψ̄γ^μγ⁵ψ)(−x); b_μ is a real c-number, untouched by conjugation and not reflected. The term goes to minus itself, N = 1, CPT-odd. The theorem is not contradicted, because its Lorentz-invariance input is exactly what fails.
Energy: the Hamiltonian density is ℋ′ = b_μψ̄γ^μγ⁵ψ = b₀ψ†γ⁵ψ − b⃗⋅ψ†Σ⃗ψ, using ψ̄γ^kγ⁵ψ = ψ†Σ^kψ. At rest ψ†γ⁵ψ averages to zero, being of order σ⃗⋅p⃗/E, and spin along or against b⃗ shifts the energy by ∓|b⃗|: ΔE = 2|b⃗| = 2.0 × 10⁻³¹ GeV = 2.0 × 10⁻²² eV.
Restore ħ: ν = ΔE/h = 2.0 × 10⁻²² eV / (4.1357 × 10⁻¹⁵ eV s) = 4.8 × 10⁻⁸ Hz, a period of 1/ν = 2.1 × 10⁷ s, about 240 days. Against mₑc² = 5.11 × 10⁵ eV the splitting is 3.9 × 10⁻²⁸ of the rest energy.

AnswerHermitian, Lorentz-violating and CPT-odd (N = 1). ΔE = 2|b⃗| = 2.0 × 10⁻²² eV, ν = 4.8 × 10⁻⁸ Hz, a period of about 240 days (2.1 × 10⁷ s).