University Quantum Field Theory I · Discrete Symmetries of Quantised Fields: C, P, T and CPT · 11.3
Time reversal on quantised fields: an antiunitary T
Parity and charge conjugation are linear: pushed through a field, they change operators and labels but leave every number alone. Time reversal also conjugates every number it passes, and that one feature turns a(p) → a(−p) into t → −t, puts Kramers' minus sign on every fermion, and keeps T from ever being a quantum number to conserve.
Build the model
Connect the measurement to the mechanism.
Time reversal must leave every energy alone while reversing momentum and spin, and on Fock space that forces it to be antiunitary: a linear operator that kept H could never turn e(−iHt) into e(iHt), and one that flipped H would need energies below the vacuum's. So T is fixed by three rules: T a(p) T⁻¹ = a(−p) for bosons, momentum and spin reversed with UAQM-01.06's alternating sign for fermions, and complex conjugation of every c-number it passes. The third rule does the work.
Conjugating e(−ip⋅x) and relabelling p → −p gives a positive-frequency wave at (−t, x), so the scalar field goes to φ(−t, x), where the same ladder rule in a linear operator gives φ(t, −x), parity. For the Dirac field one spinor identity, γ¹γ³u(p, s) = (−1)(½ + s)u*(p̃, −s), gives T ψ(t, x) T⁻¹ = −γ¹γ³ψ(−t, x): Advanced QM's matrix, a conventional phase, and no star on the operator. Because T conjugates its own phase on the second pass, T² = (−1)F in every convention, Kramers' sign on each fermion.
The construction exists for any free field; whether it is a symmetry depends on interactions. QED commutes with T and so allows no electron electric dipole moment; the weak interaction's complex couplings do not, and T violation is measured in neutral kaons and B mesons. Nor does T supply a conserved quantum number, as P and C do: its invariance buys reciprocity between a process and its reverse.
- Simple definition
- Time reversal on quantum fields is the antiunitary operator T that leaves the vacuum and every energy unchanged, reverses every momentum and spin, and complex-conjugates every c-number it passes, so that T φ(t, x) T⁻¹ = φ(−t, x).
- Example
- With ηT = −1, an electron with p = 1.00 MeV along +z and spin +½ goes to −|−p, −½⟩: it now moves along −z with spin −½, so its helicity, +½, and its energy, 1.123 MeV, are unchanged, and a second T returns the original state times −1.
Derived: T conjugates e(−ip⋅x) to e(ip⋅x), and relabelling p → −p makes it the positive-frequency wave at (−t, x). A linear operator with the same rule gives φ(t, −x).
φ of mass dimension 1; T is defined by T a(p) T⁻¹ = a(−p), a of dimension −1, and T|0⟩ = |0⟩; t and x of dimension −1
A definition: UAQM-01.06's spin rule on the label. Its adjoint gives T|p, s⟩ = ηT*(−1)(½ − s)|−p, −s⟩, so momentum and spin reverse together and helicity survives.
b annihilates an electron, dimension −1; s = ±½ the rest-frame spin along z; ηT a phase, −1 here; d obeys the rule with ηd
Derived from iσ²ξₛ = (−1)(½ + s)ξ₋ₛ and iσ²σ = −σ*iσ². It holds for v because v = iγ²u* and the real matrix iγ² commutes with γ¹γ³.
p̃ = (Eₚ, −p); u, v of dimension ½; γ¹γ³ = diag(iσ², iσ²), real in the Dirac and chiral representations, (γ¹γ³)² = −1
Derived. No star on ψ: T keeps electrons as electrons. On a wavefunction it reads χ → γ¹γ³χ*(−t, x), UAQM-08.07's iγ¹γ³K times −i.
ψ of mass dimension 3/2; ηT = −1, and locality forces ηd = ηT* = −1; a general ηT gives ηTγ¹γ³
Derived: ηT meets its own conjugate on the second pass. One electron carries Kramers' −1; positronium and every state of bosons carry +1.
F the number of fermions plus antifermions; T²bT⁻² = −b, T²dT⁻² = −d and T²aT⁻² = a, whatever ηT is
Derived. Γ* flips each explicit i: ψ̄γ⁵ψ is T-even and the Hermitian iψ̄γ⁵ψ is T-odd. Topic 11.04 tabulates all sixteen bilinears.
Γ any 4 × 4 matrix of numbers; (γ¹γ³)⁻¹ = −γ¹γ³; ηT cancels and no operators are reordered
Why T has to be antiunitary on Fock space
Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Time reversal must turn forward evolution into backward evolution, T e(−iHt) T⁻¹ = e(iHt). A linear T could do that only with THT⁻¹ = −H, sending a one-particle state of energy Eₚ ≥ m to −Eₚ, and Fock space has no such state: the free H = ∫ₚ Eₚ a†(p)a(p) is never negative. An antilinear T conjugates the i in the exponent and needs THT⁻¹ = +H instead. That is the Fock-space face of Wigner's dichotomy, quoted from UAQM-01.04: T(c|α⟩) = c*T|α⟩ and ⟨Tα|Tβ⟩ = ⟨β|α⟩. T must also reverse the momentum P and the angular momentum J and keep the charge Q, since a film run backwards shows the same particles moving the other way. So define it on quanta, T a(p) T⁻¹ = a(−p) with T|0⟩ = |0⟩, and let antilinearity extend it to every state. The rule keeps [a(p), a†(q)] = (2π)³2Eₚ δ³(p − q), whose right side is real, and relabelling p → −p, which leaves ∫ₚ = ∫d³p/((2π)³2Eₚ) and Eₚ alone, returns H unchanged and P reversed.
Conjugating the mode function turns p → −p into t → −t
Push T through the real field φ(x) = ∫ₚ [a(p)e(−ip⋅x) + a†(p)e(ip⋅x)]. The operators follow the rule and the c-number exponentials are conjugated: T φ(t, x) T⁻¹ = ∫ₚ [a(−p)e(ip⋅x) + a†(−p)e(−ip⋅x)]. Relabel p → −p. The exponent i(Eₚ t − p⋅x) becomes i(Eₚ t + p⋅x) = −ip⋅x′ with x′ = (−t, x), so the integrand is the original at x′ and T φ(t, x) T⁻¹ = φ(−t, x), derived. A linear U with the same ladder rule keeps −i(Eₚ t − p⋅x), the relabelling makes it −i(Eₚ t + p⋅x), and U φ(t, x) U⁻¹ = φ(t, −x): parity. Reversing momenta is common to both; the conjugation moves the reflection from space into time. The commutator confirms it. With π = ∂ₜφ, T π(t, x) T⁻¹ = −π(−t, x), so T[φ(t, x), π(t, y)]T⁻¹ = −iδ³(x − y), which matches T(iδ³)T⁻¹ only because T flips the i. For a complex field, a → a(−p) and c → c(−p) give φ(−t, x), not φ†, and Q = ∫ₚ (a†a − c†c) is unchanged.
Dirac quanta: the spin rule's sign and one spinor identity
T must now reverse spin as well, and UAQM-01.06's rule Θ|s⟩ = (−1)(½ − s)|−s⟩, for Θ = −iσyK, supplies the phases: T b(p, s) T⁻¹ = ηT(−1)(½ − s)b(−p, −s) and T d(p, s) T⁻¹ = ηd(−1)(½ − s)d(−p, −s), s = ±½ the rest-frame spin along z, d labelled by the positron's spin through v(p, s) = iγ²u*(p, s). Push T through ψ = ∫ₚ Σₛ [b u e(−ip⋅x) + d†v e(ip⋅x)] and relabel p → −p, s → −s: the b term becomes ηT Σₛ (−1)(½ + s) b(p, s) u*(p̃, −s) e(−ip⋅x′), with p̃ = (Eₚ, −p). In the Dirac representation γ¹γ³ = diag(iσ², iσ²) is real, and iσ² sends ξ₊ = (1, 0) to −ξ₋ and ξ₋ = (0, 1) to ξ₊, so iσ²ξₛ = (−1)(½ + s)ξ₋ₛ; with iσ²σ = −σ*iσ² this gives γ¹γ³u(p, s) = (−1)(½ + s)u*(p̃, −s), derived. So the b term is ηTγ¹γ³ times its original at x′, and the d† term, by the same identity for v, is ηd*γ¹γ³ times its original. A local field needs one matrix for both: ηd = ηT*. With this course's ηT = −1, T ψ(t, x) T⁻¹ = −γ¹γ³ψ(−t, x).
T² = (−1)F, Kramers pairs, and no T quantum number
Apply the rule twice. The second T meets the phase ηT and conjugates it, so T²b(p, s)T⁻² = ηT*ηT(−1)(½ − s)(−1)(½ + s)b(p, s) = −b(p, s) for every ηT; the same holds for d, while T²a(p)T⁻² = a(p). With T²|0⟩ = |0⟩ this is T² = (−1)F, F the number of fermions plus antifermions, derived; on the field T²ψT⁻² = (γ¹γ³)²ψ = −ψ, UAQM-08.07's UU* = −1 again. Rephasing T by e(iθN), N the fermion number, shifts ηT to ηTe(−iθ) and ηd to ηde(iθ), so the overall phase is a convention and the sign is not. In a T-invariant theory every state with odd F has an orthogonal partner of the same energy, since ⟨α|Tα⟩ = ⟨T²α|Tα⟩ = −⟨α|Tα⟩: an electron at rest with spin +½ pairs with its own image, −1 times the electron at rest with spin −½. What T cannot supply is a conserved eigenvalue. If T|α⟩ = λ|α⟩, the rephased e(iθ)|α⟩ has eigenvalue e(−2iθ)λ, so a T-parity means nothing. Invariance instead links a process to its reverse: with THT⁻¹ = H, conjugating the −i in S's time-ordered exponential and reversing the time order give T S T⁻¹ = S†, so ⟨f|S|i⟩ = ⟨Ti|S|Tf⟩, shown.
Every factor of i flips: bilinears and couplings
Antilinearity reaches every number inside a product of fields. From T ψ T⁻¹ = −γ¹γ³ψ(−t, x), its adjoint and the real γ⁰, a bilinear ψ̄Γψ, with Γ any 4 × 4 matrix of numbers, goes to ψ̄(γ¹γ³)⁻¹Γ*γ¹γ³ψ at (−t, x), derived. Three features matter. ηT cancels against its conjugate, as any phase does in a bilinear; and, unlike charge conjugation, no operators are reordered, so no anticommutation sign enters, and Γ is conjugated rather than transposed, so an i written into Γ flips. In the Dirac representation γ⁵ is real and commutes with γ¹γ³, so ψ̄γ⁵ψ is T-even while the Hermitian pseudoscalar iψ̄γ⁵ψ is T-odd: one matrix, two signs, decided by a single i. The current is the physical check. γ⁰ is real and commutes with γ¹γ³, so the charge density ψ̄γ⁰ψ stays; γ¹ and γ³ anticommute with γ¹γ³ and γ² is imaginary, so all three components of ψ̄γψ reverse, as moving charge filmed backwards must. Couplings are c-numbers too, T(gO)T⁻¹ = g*TOT⁻¹, so a phase that no field redefinition removes breaks T. Topic 11.04 builds the full table.
Checks: dimensions, two limits and a measurement
Dimensions: T and ηT are pure numbers and touch no mass dimension, so φ(−t, x) balances φ at dimension 1 and −γ¹γ³ψ(−t, x) balances ψ at 3/2. Non-relativistic limit: for |p| ≪ m only the upper block of u survives, and the one-electron wavefunction ⟨0|ψ(x)T|χ⟩ = −ηT*γ¹γ³χ*(−t, x) = γ¹γ³χ*(−t, x) acts on it as iσ²K, which is −1 times UAQM-01.06's Θ = −iσyK; on four components it is UAQM-08.07's iγ¹γ³K times −i. High-energy limit: γ¹γ³ commutes with γ⁵, so T sends ½(1 − γ⁵)ψ to −γ¹γ³ times ½(1 − γ⁵)ψ(−t, x), keeping chirality, and it keeps helicity because spin and momentum reverse together; parity flips both, so a theory of left-handed fields can still conserve T. Measurement: QED is T-invariant, topic 11.04 supplying the photon, while an electron electric dipole moment, −dₑ σ⋅E in the non-relativistic limit, pairs a T-odd spin with a T-even field, so QED predicts dₑ = 0. The 2023 limit puts |dₑ| below 4.1 × 10⁻³⁰ e cm (quoted), 1.1 × 10⁻¹⁹ of e times the reduced Compton wavelength, 3.86 × 10⁻¹¹ cm.
Change one variable at a time
Make the relationship visible.
Set t = 0: the T image matches the original exactly for every p and δ, while the parity image is its mirror. Set δ = 0: the T and P images coincide at every t. Turn δ again and they split, because only T conjugates the phase of the state. Then drag t and watch both images run towards −x.
ENERGY E = √(p² + m²)244.1 MeV
WAVELENGTH 2πħc/p6.20 fm
ORIGINAL AT x = 0, ⟨φ⟩/A-0.89
T IMAGE AT x = 0, ⟨φ⟩/A0.05
Live interpretationENERGY E = √(p² + m²): 244.1 MeV. WAVELENGTH 2πħc/p: 6.20 fm. ORIGINAL AT x = 0, ⟨φ⟩/A: −0.89. T IMAGE AT x = 0, ⟨φ⟩/A: 0.05
Catch the common trap
Explain before calculating.
A complex scalar field φ = ∫ₚ [a(p)e(−ip⋅x) + c†(p)e(ip⋅x)] is given the antiunitary T with T a(p) T⁻¹ = a(−p) and T c(p) T⁻¹ = c(−p). What is T φ(t, x) T⁻¹, and what happens to the charge Q = ∫ₚ (a†a − c†c)?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron is in the state |χ⟩ = (3|p, +½⟩ + 4i|p, −½⟩)/5, with p = 2.40 MeV along +z, m = 0.511 MeV and |p, s⟩ = b†(p, s)|0⟩. Using T b(p, s) T⁻¹ = ηT(−1)(½ − s)b(−p, −s) with ηT = −1, find T|χ⟩, the rest-frame spin expectation ⟨σ⟩ before and after, the helicity and the energy, and T²|χ⟩.
- Take the adjoint of the rule: T b†(p, s) T⁻¹ = ηT*(−1)(½ − s)b†(−p, −s). With T|0⟩ = |0⟩ and ηT = −1, T|p, +½⟩ = −|−p, −½⟩ and T|p, −½⟩ = +|−p, +½⟩.
- T is antilinear, so the coefficient 4i becomes −4i: T|χ⟩ = (3(−|−p, −½⟩) − 4i|−p, +½⟩)/5 = −(4i|−p, +½⟩ + 3|−p, −½⟩)/5.
- Spin: for amplitudes (a, b) on (+½, −½), ⟨σ⟩ = (2 Re a*b, 2 Im a*b, |a|² − |b|²). Before, (0.6, 0.8i) gives a*b = 0.48i and ⟨σ⟩ = (0, 0.96, −0.28). After, (−0.8i, −0.6) gives a*b = −0.48i and ⟨σ⟩ = (0, −0.96, +0.28): every component reversed.
- Helicity: before, p̂ = +ẑ and ⟨S⋅p̂⟩ = ½(−0.28) = −0.14; after, p̂ = −ẑ and ⟨S⋅p̂⟩ = −½(+0.28) = −0.14, unchanged. Energy: E = √(2.40² + 0.511²) MeV = √6.0211 MeV = 2.454 MeV for both, as THT⁻¹ = H requires.
- Apply T again: T|−p, +½⟩ = −|p, −½⟩ and T|−p, −½⟩ = +|p, +½⟩, while antilinearity turns the amplitudes −4i/5 and −3/5 into +4i/5 and −3/5. So T²|χ⟩ = (4i/5)(−|p, −½⟩) − (3/5)|p, +½⟩ = −|χ⟩.
AnswerT|χ⟩ = −(4i|−p, +½⟩ + 3|−p, −½⟩)/5; ⟨σ⟩ goes from (0, 0.96, −0.28) to (0, −0.96, 0.28); the helicity, −0.14 in units of ħ, and E = 2.454 MeV are unchanged; T²|χ⟩ = −|χ⟩ for any ηT.
MediumAn electron has p = 1.20 MeV along +x, m = 0.511 MeV. In the Dirac representation, with ξ₊ = (1, 0) and ξ₋ = (0, 1), build u(p, ±½), apply γ¹γ³ = diag(iσ², iσ²) and compare with u*(p̃, ∓½), p̃ = (E, −p). Then show that with ηT = −1 the b part of T ψ T⁻¹ is −γ¹γ³ times the b part of ψ(−t, x), and find what the d† part requires of ηd.
- E = √(1.44 + 0.2611) MeV = 1.3043 MeV, so √(E + m) = √1.8153 MeV = 1.3473 √MeV and p/√(E + m) = 0.8907 √MeV. Along x, σ⋅p = pσₓ, which swaps ξ₊ and ξ₋.
- u(p, s) = (√(E + m)ξₛ, σ⋅p ξₛ/√(E + m)) gives u(p, +½) = (1.3473, 0, 0, 0.8907) and u(p, −½) = (0, 1.3473, 0.8907, 0), in √MeV. Check: u†u = 1.8153 + 0.7933 = 2.6085 MeV = 2E.
- iσ² sends (a, b) to (b, −a) in each block, so γ¹γ³u(p, +½) = (0, −1.3473, 0.8907, 0) and γ¹γ³u(p, −½) = (1.3473, 0, 0, −0.8907).
- Reversing p flips the sign of each lower block: u(p̃, −½) = (0, 1.3473, −0.8907, 0) and u(p̃, +½) = (1.3473, 0, 0, −0.8907), both real. Hence γ¹γ³u(p, +½) = −u*(p̃, −½) and γ¹γ³u(p, −½) = +u*(p̃, +½): the factor (−1)(½ + s).
- In T ψ T⁻¹ the coefficient of b(p, s)e(−ip⋅x′) is ηT(−1)(½ + s)u*(p̃, −s). With ηT = −1, s = +½ gives +u*(p̃, −½) = −γ¹γ³u(p, +½), and s = −½ gives −u*(p̃, +½) = −γ¹γ³u(p, −½): both are −γ¹γ³u(p, s).
- The d† term carries ηd*(−1)(½ + s)v*(p̃, −s) = ηd*γ¹γ³v(p, s), because v = iγ²u* and the real iγ² commutes with γ¹γ³. Matching −γ¹γ³ needs ηd* = −1, so ηd = −1 = ηT*.
Answerγ¹γ³u(p, ±½) = ∓u*(p̃, ∓½), with entries ±1.3473 and ±0.8907 √MeV; the b part of T ψ T⁻¹ is −γ¹γ³ times that of ψ(−t, x) for both spins, and locality fixes ηd = −1.
HardQED is T-invariant. Use T S T⁻¹ = S† to relate e⁺e⁻ → μ⁺μ⁻ to its reverse, μ⁺μ⁻ → e⁺e⁻, at the same √s = 0.300 GeV, spins averaged and the electron mass neglected (mμ = 0.10566 GeV, α = 1/137.036). Find the ratio of the two cross-sections, evaluate both in nb from topic 9.06's tree-level σ(e⁺e⁻ → μ⁺μ⁻) = (4π/3)(α²/s)β(3 − β²)/2 (quoted), and check the high-energy and threshold limits.
- The QED interaction ψ̄γμψAμ is T-even: the current's (j⁰, −j) at (−t, x) meets the photon's (A⁰, −A), quoted from topic 11.04. Conjugating the −i in S's time-ordered exponential and reversing the time order then gives T S T⁻¹ = S†, so ⟨f|S|i⟩ = ⟨Ti|S|Tf⟩, and |M(i → f)| = |M(Tf → Ti)| because T's phases have modulus one.
- Tf has every momentum and spin reversed. Summed over spins the flips are a relabelling, and in the centre-of-mass frame a rotation by π about the normal to the scattering plane turns every momentum back, so Σ|M|² is the same for e⁺e⁻ → μ⁺μ⁻ and μ⁺μ⁻ → e⁺e⁻ at equal √s and angle. With dσ/dΩ = (pf/pᵢ)(1/4)Σ|M|²/(64π²s) and four spin states averaged on both sides, σ(μμ → ee)/σ(ee → μμ) = (pₑ/pμ)².
- Kinematics: pₑ = √s/2 = 0.1500 GeV and pμ = √(s/4 − mμ²) = √(0.022500 − 0.011164) GeV = 0.10647 GeV, so β = pμ/pₑ = 0.7098, β² = 0.5038, and the ratio is 1/β² = 1.985.
- Numbers: (4π/3)(α²/s) = 4π(1/137.036)²/(0.2700 GeV²) = 2.478 × 10⁻³ GeV⁻², or 965.1 nb with 1 GeV⁻² = 0.3894 mb. So σ(e⁺e⁻ → μ⁺μ⁻) = 965.1 nb × 0.7098 × (3 − 0.5038)/2 = 855.0 nb, and σ(μ⁺μ⁻ → e⁺e⁻) = 855.0 nb × 1.985 = 1697 nb.
- Limits: as β → 1 both tend to (4π/3)(α²/s), as they must once every mass is negligible and the two processes look alike. At threshold, β → 0, σ(e⁺e⁻ → μ⁺μ⁻) vanishes like β for want of phase space, while σ(μ⁺μ⁻ → e⁺e⁻) grows like 1/β, the 1/v law of slow incoming particles; this is tree level, and Coulomb attraction modifies both near threshold.
- Checks: σ ∝ α²/s has mass dimension −2, an area. At tree level the reversed amplitude is the same s-channel diagram, and Casimir's trick makes Σ|M|² the product of the same electron and muon traces, so it is unchanged, as reciprocity says; T invariance adds that the equality survives every QED loop.
Answerσ(μ⁺μ⁻ → e⁺e⁻)/σ(e⁺e⁻ → μ⁺μ⁻) = 1/β² = 1.985 at √s = 0.300 GeV: 855 nb for e⁺e⁻ → μ⁺μ⁻ and 1.70 × 10³ nb for the reverse.