University Quantum Field Theory I · Discrete Symmetries of Quantised Fields: C, P, T and CPT · 11.8
Positronium's C = (−1)(L+S) derived
Advanced QM had to quote positronium's C = (−1)(L+S); with C now acting on Fock space you can earn it. Apply C to the bound pair, pay one sign for reordering fermion operators and two for relabelling momentum and spin, and every level's JPC follows, ready for the photon counting of topic 11.09.
Build the model
Connect the measurement to the mechanism.
Positronium shows C acting on a state rather than a field. At leading non-relativistic order the pair at rest is a superposition of b†(p, s) d†(−p, s′)|0⟩ weighted by a momentum wavefunction ψ̃(p) of orbital angular momentum L and spin coefficients χ(s, s′) of total spin S. Topic 11.02's C, swapping b† and d† with momentum and spin kept, hands back d†(p, s) b†(−p, s′)|0⟩: the positron now carries what the electron did.
Restoring the order costs −1, as the operators anticommute; renaming the dummy momentum p as −p gives ψ̃(−p) = (−1)L ψ̃(p); renaming the spins returns χ(s′, s) = (−1)(S+1) χ(s, s′), a Clebsch-Gordan symmetry that applies only because UAQM-08.05's v = iγ²u* makes the positron's spin labels a standard doublet. The product is C = (−1)(L+S), derived, and topic 11.01's parity on the same state, with no reordering but opposite intrinsic parities, gives P = (−1)(L+1). Only the symmetry of ψ̃ and χ was used, never their shape, and C and P commute with the QED Hamiltonian, so the eigenvalues survive every correction the leading form omits: para-positronium is 0⁻⁺, ortho-positronium 1⁻⁻, and the 2P levels 1⁺⁻ and 0⁺⁺, 1⁺⁺, 2⁺⁺.
The argument stops where its inputs stop. It needs a particle and its own antiparticle, so muonium has no C; L labels only the leading component, while S survives as the exact CP = (−1)(S+1); and the weak interaction, named only, breaks C.
- Simple definition
- Positronium's charge-conjugation parity is the eigenvalue C = (−1)(L+S) that the operator exchanging electron and positron returns on a level of orbital angular momentum L and total spin S; with P = (−1)(L+1) it fixes the level's JPC.
- Example
- Ortho-positronium, 1³S₁, has L = 0 and S = 1, so C = (−1)¹ = −1 and P = −1: JPC = 1⁻⁻, the photon's own quantum numbers, which is why virtual annihilation lifts it 87.6 GHz above where magnetism alone would put it.
Leading non-relativistic order, Eₚ ≈ m. Only ψ̃(−p) = (−1)L ψ̃(p) and the exchange symmetry of χ will be used, never the shape of either.
N = √(M/2)/m for ⟨Ps|Ps⟩ = 2M(2π)³δ³(0); ψ̃ in GeV(−3/2), ∫|ψ̃|² d³p/(2π)³ = 1; χ dimensionless
On a b†d† pair the phases multiply to |λ|² = 1, so the pair's C is free of convention, while neither the electron nor the positron alone is a C eigenstate.
Momentum and spin kept; no phase is needed with v = iγ²u* (UAQM-08.05); a phase λ* on b† forces λ on d†
Clebsch-Gordan symmetry from Quantum Mechanics II. It applies to the positron's label only because d†(p, s) rotates exactly as b†(p, s) does, derived here.
S = 0 or 1; s labels b† and s′ labels d†, both standard spin-½ doublets because iγ²S(R)* = S(R)iγ²
Derived. It squares to 1, as C² = 1 requires, and the radial quantum number n never enters. With commuting operators the first factor is +1 and every C reverses.
−1 from d†b† = −b†d†; (−1)L from renaming p as −p in ψ̃; (−1)(S+1) from renaming s ↔ s′ in χ; dimensionless
Derived. P's −1 is the relative intrinsic parity, C's the Pauli sign: alike in look, different in origin. A charged-scalar pair has neither, so its C = P = (−1)L.
ηb, ηd the intrinsic parities of topic 11.01, |ηb| = 1; P b†(p, s) P⁻¹ = ηb* b†(−p, s) keeps spin and order
Derived. A meson seen with one of these, such as the 1⁻⁺ π₁(1600) candidate (quoted), cannot be a quark-antiquark pair; positronium has no such level.
J runs from |L − S| to L + S; P = −C needs S = 0 and so J = L; P = C needs S = 1, which cannot give 0⁻⁻
The pair on Fock space, and why its leading term suffices
Natural units, ħ = c = 1, and the metric (+, −, −, −) throughout. Positronium at rest is, at leading order, N ∫ d³p/(2π)³ ψ̃(p) χ(s, s′) b†(p, s) d†(−p, s′)|0⟩ summed over spins, with [b(p, s), b†(q, s′)]₊ = (2π)³ 2Eₚ δ³(p − q) δₛₛ′ and N = √(M/2)/m giving ⟨Ps|Ps⟩ = 2M(2π)³δ³(0). Dimensions check: N carries mass(−1/2), ψ̃ mass(−3/2), d³p mass³ and b†d† mass⁻², so the state has mass⁻¹, like any one-particle state. Setting Eₚ = m is safe: the typical momentum is mα/2 = 1.86 keV, the inverse of the Bohr radius 2/(mα) = 0.106 nm, so each particle's kinetic energy is mα²/8 = 3.40 eV, 7 parts per million of m. The exact eigenstate also holds relativistic corrections, extra pairs and photons, but C and P commute with the QED Hamiltonian (topic 11.04, quoted), so it keeps the eigenvalues of the leading term it grows from. Only two symmetries are used: ψ̃(−p) = (−1)L ψ̃(p), from YLM(−p̂) = (−1)L YLM(p̂), and the exchange symmetry of χ.
Three signs from one exchange
Topic 11.02, quoted: C b†(p, s) C⁻¹ = d†(p, s), C d†(p, s) C⁻¹ = b†(p, s) and C|0⟩ = |0⟩. Insert C⁻¹C between the operators and C|Ps⟩ = N ∫ ψ̃(p) χ(s, s′) d†(p, s) b†(−p, s′)|0⟩: the positron now carries p and s, the electron −p and s′. Three moves restore the original form. First, b† and d† create different fermions and anticommute, so d†(p, s) b†(−p, s′) = −b†(−p, s′) d†(p, s): a factor −1. Second, p is integrated over every direction, so rename it −p; d³p is unchanged and the coefficient becomes ψ̃(−p) = (−1)L ψ̃(p). Third, s and s′ are summed, so swap their names; the coefficient becomes χ(s′, s) = (−1)(S+1) χ(s, s′). Collecting, C|Ps⟩ = (−1)(1+L+S+1)|Ps⟩ = (−1)(L+S)|Ps⟩, derived: the result UAQM-08.07 could only quote. It is dimensionless and squares to 1, as C² = 1 demands. Notice what C did not do: it never touched a position or a momentum. The (−1)L appears only because electron and positron traded labels and a dummy variable was renamed.
Why the spin swap costs (−1)(S+1): d† as a standard doublet
The third move used the Clebsch-Gordan symmetry of two spin-½ labels: the singlet (↑↓ − ↓↑)/√2 is odd under s ↔ s′ and every triplet member even. That symmetry belongs to two labels that rotate the same way. b†(p, s) rotates with the standard spin-½ matrices D(R); does d†(p, s)? The field's d† term is the sum of d†(p, s) v(p, s) e(ip⋅x), with v = iγ²u* from UAQM-08.05. Because iγ² S(R)* = S(R) iγ² for every rotation, S(R) carries v(p, s) into v(Rp, s′) weighted by D(R)*, where u was weighted by D(R), and covariance of the field then forces U(R) d†(p, s) U(R)⁻¹ to be d†(Rp, s′) weighted by D(R) itself: b†'s own rule, derived. At rest, in the Dirac representation, v(0, ↑) is proportional to (0, 0, 0, 1), a spin-down negative-energy spinor whose absence is a spin-up positron, and v(0, ↓) to −(0, 0, 1, 0), with the same positive factor: that minus sign is the one standard phases need. A convention that labels v by the negative-energy spinor's own spin changes χ's coefficients and the swap sign with them; the physical C stays (−1)(L+S).
Parity: no reordering, but opposite intrinsic parities
Topic 11.01, quoted: P b(p, s) P⁻¹ = ηb b(−p, s) and P d(p, s) P⁻¹ = ηd d(−p, s), with ηd = −ηb* forced by γ⁰v(p̃, s) = −v(p, s). Taking adjoints, P b†(p, s) P⁻¹ = ηb* b†(−p, s) and P d†(p, s) P⁻¹ = −ηb d†(−p, s), so P|Ps⟩ = −|ηb|² N ∫ ψ̃(p) χ(s, s′) b†(−p, s) d†(p, s′)|0⟩. The operators keep their order and their spins, so one move finishes the job: rename p as −p, for (−1)L. Hence P = (−1)(L+1), derived on Fock space, agreeing with UAQM-08.07's wavefunction argument. Set the two results side by side. C's extra −1 is statistics, the price of reordering fermion operators; P's extra −1 is the relative intrinsic parity of fermion and antifermion; neither depends on a phase convention. A charged-scalar pair has neither: commuting operators, no spin, and an antiparticle with its particle's parity, so π⁺π⁻ with orbital L has C = P = (−1)L. The ρ⁰(770), a 1⁻⁻ meson, decays to π⁺π⁻ in a P wave, with C = P = −1 as that rule requires (quoted).
The JPC of the n = 1 and n = 2 levels, and what C forbids
With Eₙ = −mα²/(4n²) = −6.80 eV/n², the two rules label every level: 1¹S₀ is 0⁻⁺ (para), 1³S₁ is 1⁻⁻ (ortho), 2¹S₀ and 2³S₁ repeat them, 2¹P₁ is 1⁺⁻, and 2³P₀, 2³P₁, 2³P₂ are 0⁺⁺, 1⁺⁺, 2⁺⁺, derived. Two consequences follow at once. 2¹P₁ and 2³P₁ share JP = 1⁺ but have opposite C, so nothing C-conserving mixes them; in muonium the spin-orbit term in (s₁ − s₂)⋅L, whose coefficient goes as 1/m₁² − 1/m₂², does mix them, and in positronium that coefficient vanishes, as C requires. And a single photon has C = −1 (topic 11.04, quoted), so every one-photon transition changes C: a P-to-S line joins ¹P₁ to ¹S₀ or ³PJ to ³S₁, never singlet to triplet, at any multipole order. Ortho-positronium shares the photon's 1⁻⁻, and the hyperfine splitting shows it: virtual one-photon annihilation acts on 1³S₁ alone, supplying (1/4)mα⁴/h = 87.6 GHz of the leading (7/12)mα⁴/h = 204.4 GHz, against 203.39 GHz measured (all quoted). Topic 11.09 turns these C values into photon counts, tested by the measured 142 ns and 125 ps lifetimes.
Where the derivation applies, and where it stops
Every input was generic: a fermion, its own antifermion and a non-relativistic leading term. So the rule holds for true muonium, μ⁺μ⁻, and, with a colour-singlet factor that is symmetric and adds no sign (stated), for quarkonium: J/ψ at 3096.9 MeV is ³S₁ and 1⁻⁻, formed directly in e⁺e⁻ collisions through one virtual photon, while the 0⁻⁺ ηc is not (quoted). It fails where its inputs fail. Muonium, μ⁺e⁻, and hydrogen are not C eigenstates at all, since C turns them into μ⁻e⁺ and antihydrogen. L labels only the leading component: spin-dependent forces mix ³S₁ with ³D₁, both 1⁻⁻. S survives, because CP = (−1)(S+1) is exact, so an exact level carries JPC and S, nothing finer. Statistics enters exactly once, through d†b† = −b†d†; with commuting operators the same steps give (−1)(L+S+1), which is why the figure lets you flip that sign. And C is exact only within QED: the weak interaction, named only, breaks it, at a level far below what positronium measurements resolve.
Change one variable at a time
Make the relationship visible.
Hold L = 0 and step S from 0 to 1: the spin bar flips and C goes from +1, para-positronium's 0⁻⁺, to −1, ortho's 1⁻⁻. Raise L to 1 and the curves, C and P all flip. Set the reordering sign to +1, the bookkeeping of commuting operators, and every C reverses while P does not.
C EIGENVALUE-1
P EIGENVALUE-1
CP1
Live interpretationC EIGENVALUE: −1. P EIGENVALUE: −1. CP: 1
Catch the common trap
Explain before calculating.
Positronium has an n = 3 level with L = 2, S = 0 and J = 2. Working on Fock space with this lesson's conventions, what is its JPC?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyGround-state positronium's MS = +1 state has χ(↑, ↑) = 1, and its two MS = 0 states have χ(↑, ↓) = 1/√2 with χ(↓, ↑) = +1/√2 or −1/√2. Apply C to each, writing every sign, and give C and JPC. Then check that setting Eₚ = m in the normalisation N is justified.
- MS = +1: C|Ps⟩ = N ∫ ψ̃(p) d†(p, ↑) b†(−p, ↑)|0⟩ = −N ∫ ψ̃(p) b†(−p, ↑) d†(p, ↑)|0⟩, the minus sign from anticommuting d† past b†.
- Rename p as −p. A 1S wavefunction depends on |p| only, so ψ̃(−p) = ψ̃(p) and the state returns with the factor −1: C = −1.
- MS = 0: after C and reordering, −(1/√2) N ∫ ψ̃(p)[b†(−p, ↓) d†(p, ↑) ± b†(−p, ↑) d†(p, ↓)]|0⟩. Renaming p gives the bracket [b†(p, ↓) d†(−p, ↑) ± b†(p, ↑) d†(−p, ↓)], which is +1 times the original bracket for the upper sign and −1 times it for the lower, so the states return with −1 and +1.
- So the triplet, MS = 0 and +1 alike, has C = −1, and the singlet C = +1. The two triplet members agreeing is the check that d† is a standard doublet: C commutes with rotations, so all three MS must share one C. With P = (−1)(0+1) = −1 the levels are 0⁻⁺ and 1⁻⁻.
- Normalisation: momenta in ψ̃ are about mα/2 = 1.86 keV, so Eₚ − m ≈ ½p²/m = mα²/8 = 3.40 eV, which is 6.7 × 10⁻⁶ of m = 511 keV.
AnswerSinglet: C = +1, JPC = 0⁻⁺. All three triplet states: C = −1, JPC = 1⁻⁻. Eₚ exceeds m by only 3.40 eV, 7 parts per million, so Eₚ = m is safe.
MediumPositronium's 2P levels decay to 1S by emitting one photon, whose C = −1 (topic 11.04). Decide which of 2³PJ → 1³S₁, 2³PJ → 1¹S₀, 2¹P₁ → 1¹S₀ and 2¹P₁ → 1³S₁ conserve C, show that angular momentum and parity alone would allow 2³P₁ → 1¹S₀ and 2¹P₁ → 1³S₁, and find the wavelength of the allowed lines at leading order.
- JPC from this lesson: 2³PJ are J⁺⁺ with J = 0, 1, 2; 2¹P₁ is 1⁺⁻; 1³S₁ is 1⁻⁻; 1¹S₀ is 0⁻⁺.
- C of the final state is C(atom) × C(photon) = −C(atom), so C conservation needs the atom's C to change sign. 2³PJ (+) → 1³S₁ (−) and 2¹P₁ (−) → 1¹S₀ (+) pass; 2³PJ (+) → 1¹S₀ (+) and 2¹P₁ (−) → 1³S₁ (−) fail.
- Without C, 2³P₁ → 1¹S₀ (J = 1 → 0) and 2¹P₁ → 1³S₁ (J = 1 → 1) would be ordinary E1 lines, parity going from + to − with an odd-parity photon. Only C removes them, and as every photon has C = −1 whatever its multipole, they stay forbidden at all orders.
- In general one photon flips (−1)(L+S); an E1 line changes L by 1, so S cannot change. For E1 lines ΔS = 0 is exact in positronium, since an E1 photon has CP = (−1)(−1) = +1 and the atom's CP = (−1)(S+1) is exact, not the approximate rule it is in atoms with unequal masses; an M1 photon, with CP = −1, must instead change S, as in 1³S₁ → 1¹S₀ + γ.
- Energy: Eₙ = −mα²/(4n²) with mα²/4 = 6.803 eV, so E(2P) − E(1S) = (3/4)(6.803 eV) = 5.102 eV, fine structure neglected. Restoring ħ and c, λ = hc/ΔE = 1239.84 eV nm ÷ 5.102 eV = 243.0 nm.
AnswerAllowed: 2³PJ → 1³S₁ + γ and 2¹P₁ → 1¹S₀ + γ. The two singlet-triplet lines are forbidden by C at every multipole order. λ = 243.0 nm, twice the 121.5 nm of hydrogen with an infinitely heavy nucleus.
HardAt leading order the 1S hyperfine splitting of positronium is quoted as a magnetic spin-spin term (1/3)mα⁴ plus a contact term from virtual one-photon annihilation, (1/4)mα⁴ at n = 1 and proportional to |ψ(0)|². Decide which of 1¹S₀, 1³S₁, 2³S₁, 2³P₁ and 3³D₁ that term can shift, evaluate the n = 1 splitting in GHz against the measured 203.39 GHz, and find the shift of 2³S₁.
- A single virtual photon is made by the vector field Aμ: J = 1, P = −1 and C = −1 (topic 11.04), so it carries JPC = 1⁻⁻, and an annihilation conserving J, P and C can only reach levels that are 1⁻⁻ too.
- JPC from this lesson: 1¹S₀ is 0⁻⁺ and 2³P₁ is 1⁺⁺, so both are excluded; 1³S₁, 2³S₁ and 3³D₁ are 1⁻⁻ and pass.
- The contact term needs the pair at one point, so it goes as |ψ(0)|², zero whenever L ≥ 1. 3³D₁ is 1⁻⁻ and still unshifted at this order: the right JPC is necessary, not sufficient.
- n = 1, restoring h: mα⁴ = 0.511 MeV × (1/137.036)⁴ = 1.4490 × 10⁻³ eV, which is 350.38 GHz. Magnetic part 350.38/3 = 116.79 GHz, annihilation part 350.38/4 = 87.59 GHz, sum (7/12)(350.38) = 204.39 GHz.
- Measured 203.39 GHz is 0.49% lower, the size of the order-α correction that closes most of the gap (quoted). Without the 1⁻⁻ annihilation term the prediction would be 116.79 GHz, 43% short.
- 2³S₁: |ψₙ(0)|² ∝ 1/n³ for S states, so its annihilation shift is 87.59 GHz ÷ 8 = 10.95 GHz.
AnswerOnly 1³S₁ and 2³S₁ are shifted (3³D₁ has ψ(0) = 0). n = 1 splitting 204.4 GHz at leading order against 203.39 GHz measured, 87.6 GHz of it from annihilation; 2³S₁ is raised by 10.9 GHz.