Opposite intrinsic parity and time reversal for spinors

Units are SI with c and ħ kept, the metric is diag(1, −1, −1, −1), the Dirac representation is used throughout, and the electron's charge is −e. Quoted in one line from Unit 5: ψ′(t, x) = η_Pγ⁰ψ(t, −x) is the only linear map that keeps the Dirac equation covariant under x → −x, and nothing fixes the phase η_P beyond |η_P| = 1. At rest the solutions are u(0, s)e^{(−imc²t/ħ)}, with only upper components, and v(0, s)e^(+imc²t/ħ), with only lower ones. Since γ⁰ = diag(1, 1, −1, −1), γ⁰u = +u and γ⁰v = −v: a positive-energy electron at rest has parity +η_P, a negative-energy one −η_P. Derived, in one matrix product. Moving plane waves are not eigenstates, since P sends p to −p, but states of definite orbital angular momentum are. A bound state whose upper components have orbital l has parity η_P(−1)^l, and its lower components, with l ± 1, agree because γ⁰ adds the missing −1. For a negative-energy state the lower components are the large ones, so its parity is −η_P(−1)^l′, with l′ their orbital.

From topic 8.03, quoted: the vacuum is the filled sea, assigned parity +1, and a Slater determinant of parity eigenstates has the product of their eigenvalues. Empty a sea state at rest, eigenvalue −η_P, and that factor leaves the product: measured from the vacuum, the hole has (−η_P)⁻¹ = −η_P*, since |η_P| = 1. Topic 8.05's ψ^c = iγ²ψ* agrees by another route. If ψ′ = η_Pγ⁰ψ(t, −x), then (ψ′)^c = iγ²η_P*γ⁰ψ*(t, −x) = −η_P*γ⁰(iγ²ψ*)(t, −x), using γ⁰* = γ⁰ and γ²γ⁰ = −γ⁰γ²: the positron's wavefunction carries −η_P*. Choose η_P = 1 and the electron is even, the positron odd. Choose η_P = i, as some authors do so that P² = −1 on fermions, and both carry i. Either way η(e⁻)η(e⁺) = −|η_P|² = −1, derived, with no convention left in it. That product is the physics, and it holds for every particle the Dirac equation describes: muons, taus and quarks.

A fermion and its antifermion with orbital angular momentum L in their relative coordinate have parity η(e⁻)η(e⁺)(−1)^L = (−1)^(L+1), derived; the total spin S does not enter, since P acts on space only. In positronium 1¹S₀ is 0⁻, 1³S₁ is 1⁻, and 2³P₀, 2³P₁, 2³P₂ are 0⁺, 1⁺, 2⁺. Hydrogen's rule is (−1)^L because the electron and proton numbers never change, so their phases are set by convention and cancel from every amplitude. In positronium the factor −1 is common to every state, so an E1 transition still links S to P exactly as in hydrogen and spectroscopy cannot see the sign. It shows only when the pair becomes something without that factor. Quarkonium follows the same rule, quoted: J/ψ at 3097 MeV, a ³S₁ cc̄ pair, is 1⁻, and the P-wave χ_c0, χ_c1, χ_c2 at 3415, 3511 and 3556 MeV are 0⁺, 1⁺, 2⁺. So η_c, a ¹S₀ pair of parity −1, cannot decay to two pions, whose J = 0 state has parity +1, and none is seen.

Para-positronium annihilates into two photons in 125 ps, measured. In its rest frame they leave along ±k with polarisations ε₁, ε₂ transverse to k. A J = 0 amplitude is a rotational scalar linear in each ε, and transversality leaves two: ε₁⋅ε₂, even under parity, and k̂⋅(ε₁ × ε₂), odd because k reverses while ε₁ × ε₂ does not. This is Yang's 1950 argument, derived here in outline. A 0⁻ state decays with |M|² ∝ sin²χ, χ the angle between the linear polarisations: crossed. A 0⁺ state would give cos²χ: parallel. Compton scattering reads a 511 keV photon's polarisation, preferring planes perpendicular to ε. With the quoted Klein-Nishina cross-section, both photons scattered through θ in planes at angle Δ give N(Δ) ∝ (κ − sin²θ)² − sin⁴θ cos 2Δ for 0⁻, derived; N(90°)/N(0°) peaks at 2.84 near θ = 82° and tends to 1 as θ → 0. Wu and Shaknov (1950, quoted) measured 2.04 ± 0.08 where theory for their finite detectors gave 2.00. Para-positronium is odd, and the relative parity is measured.

Quoted from Unit 1: time reversal is antiunitary, and for spin ½ it is Θ = −iσ_y K. For the Dirac equation try T = UK, U a constant unitary 4 × 4 matrix, and ask that ψ_T(t, x) = Uψ*(−t, x) solve the same equation. Conjugating iħ∂ₜψ = Hψ gives iħ∂ₜψ_T = (UH*U⁻¹)ψ_T, so we need UH*U⁻¹ = H for H = cα⋅p + βmc² + qφ(x). Conjugation sends p = −iħ∇ to −p and leaves the real φ alone, so the conditions are Uα*U⁻¹ = −α and Uβ*U⁻¹ = β. In the Dirac representation α₁, α₃ and β are real and α₂ is imaginary, so U must anticommute with α₁ and α₃ and commute with α₂ and β. The product α₁α₃ does exactly that, and iγ¹γ³ = −iα₁α₃ = −diag(σ₂, σ₂), derived. On each block −σ₂K = −i(−iσ_y K), Unit 1's rule times the phase −i, and in the non-relativistic limit, where only the upper block survives, T is Θ up to that phase. Check: UΣ*U⁻¹ = −Σ, so spin reverses with momentum and helicity is kept. A magnetic field breaks the symmetry unless A is reversed too.

Apply T twice: T² = UKUK = UU* = (−diag(σ₂, σ₂))(−diag(σ₂*, σ₂*)) = diag(σ₂σ₂*, σ₂σ₂*) = −1, since σ₂* = −σ₂. A phase on U cancels against its conjugate, and a change of representation S gives SUU*S⁻¹ = −1 again: T² = −1 belongs to spin ½, not to a convention. Kramers' theorem follows, derived. If Hψ = Eψ with real φ, then Tψ has energy E, and antiunitarity gives ⟨ψ|Tψ⟩ = ⟨T²ψ|Tψ⟩ = −⟨ψ|Tψ⟩, so ⟨ψ|Tψ⟩ = 0. Every level of one electron in any electrostatic field, however lopsided, is at least a doublet; for N electrons T² = (−1)^N, so the guarantee holds for odd N. Only a magnetic field splits the pair, by gμ_B B: at g = 2.00 and B = 0.340 T that is 3.94 × 10⁻⁵ eV, or 9.52 GHz, the X-band where ESR spectrometers find such doublets. Two results are not derived here. C = (−1)^(L+S), which sends para-positronium to two photons and ortho-positronium to three, needs Unit 9 and is quoted; CPT, which makes the e⁺ and e⁻ masses equal, measured to 8 parts in 10⁹, is stated.