University Quantum Field Theory I · Discrete Symmetries of Quantised Fields: C, P, T and CPT · 11.1

Parity on Fock space: scalar and Dirac quanta

Here the discrete symmetries stop being rules for wavefunctions and become operators on states of any number of particles. You define parity on ladder operators, find that asking the field to reflect locally fixes the antiparticle's phase for you, and see why a fermion pair answers differently from a pair of spin-0 quanta.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Advanced QM gave one Dirac wavefunction the parity law ψ → ηγ⁰ψ(t, −x) and read the positron's opposite sign off γ⁰v = −v at rest. Here parity becomes an operator on Fock space, fixed by what it does to quanta. It is unitary, because it keeps t and so must commute with e^(−iHt); it reverses momentum, a polar vector, and keeps spin, an axial one; so all it can do to an annihilator is reverse its momentum and multiply it by a phase, P a(p)P⁻¹ = η a(−p).

One demand remains: the field, the object every Lagrangian is built from, must go into a field at the mirror point. Relabelling p → −p under the invariant measure turns that demand into a condition linking particle and antiparticle. For a scalar it gives η′ = η*, so a particle-antiparticle pair carries no extra sign and has parity (−1)^L.

For the Dirac field the spinors intervene, γ⁰u(p̃, s) = u(p, s) but γ⁰v(p̃, s) = −v(p, s), so η_d = −η_b* and every fermion pair carries an extra −1: positronium's ground state is 0⁻. The phase η_b alone is a convention, rotated at will by P → Pe^(iαQ); the product η_bη_d = −1 is not. The construction stops where the dynamics do: P is a symmetry only if the interaction is P-even, as QED's is; the charged weak current, which couples to left-handed fermions alone, breaks it, and a theory whose fermion is ψ_L alone has no parity operator of this kind.

Simple definition: Parity on Fock space is the unitary operator P that reverses each quantum's momentum, keeps its spin and multiplies its ladder operators by an intrinsic phase, the phases chosen so that every field at (t, x) goes into a field at the mirror point (t, −x).
Example: With η_b = +1, P b†(p, s)|0⟩ = b†(−p, s)|0⟩ but P d†(p, s)|0⟩ = −d†(−p, s)|0⟩, so an e⁺e⁻ pair in an S wave has parity (+1)(−1) = −1: ground-state positronium is 0⁻, as the crossed polarisations of its two 511 keV photons confirm.

Parity on the Dirac ladder operatorsP b(p, s) P⁻¹ = η_b b(−p, s), P d(p, s) P⁻¹ = η_d d(−p, s)

The definition: momentum reversed, spin kept, one phase per species. P is unitary because it keeps t and so commutes with e^(−iHt).

b, d annihilate fermion and antifermion, mass dimension −1; p in GeV; s the spin label, kept; |η_b| = |η_d| = 1

Scalar field under parityP φ(t, x) P⁻¹ = η φ(t, −x) requires η′ = η*

Derived by relabelling p → −p. A pair carries ηη′ = +1; a real field, b = a, needs η = ±1, and the π⁰'s −1 is measured.

φ = ∫dp̃ [a e^(−ip⋅x) + b† e^(ip⋅x)], dimension 1; P a(p)P⁻¹ = η a(−p), P b(p)P⁻¹ = η′ b(−p); x in GeV⁻¹

Mirrored Dirac spinorsγ⁰u(p̃, s) = u(p, s), γ⁰v(p̃, s) = −v(p, s), with p̃ = (E, −p)

Derived from the Dirac-representation forms: γ⁰ flips the lower entries, which is what p → −p does to u and minus what it does to v.

u, v the spinors of UAQM-04, ūu = −v̄v = twice the mass, entries in GeV^½; the same spin label s on both sides

Dirac field under parityP ψ(t, x) P⁻¹ = η_b γ⁰ ψ(t, −x) requires η_d = −η_b*

Derived: after p → −p the d† term carries η_d* times the −1 of γ⁰v(p̃) = −v(p). No operator is reordered, so statistics plays no part.

ψ of mass dimension 3/2; γ⁰ is UAQM-05.06's S_P, quoted; η_b, η_d the ladder phases above

Rephasing by the conserved chargeP′ = P e^(iαQ): η_b → η_b e^(−iα), η_d → η_d e^(iα)

Shown. P′ is as good a parity as P, so η_b is a convention, +1 for the electron by agreement, while η_bη_d = −1 cannot move.

Q = ∫dp̃ Σₛ (b†b − d†d), dimensionless and P-even; α any real angle

Parity of a fermion-antifermion pairP|f f̄L, S⟩ = η_b* η_d* (−1)^L |f f̄L, S⟩ = (−1)^(L+1) |f f̄L, S⟩

Derived. ¹S₀ is 0⁻ and ³S₁ is 1⁻: para-positronium, η_c, J/ψ. A spin-0 particle-antiparticle pair gets (−1)^L instead.

L the orbital angular momentum of the relative motion, S the total spin, which does not enter; eigenvalues dimensionless

A unitary P that reverses momentum and keeps spin

Work in ħ = c = 1 with the metric (+, −, −, −), and write p̃ = (E, −p) and x̃ = (t, −x). Wigner's theorem, quoted from Advanced QM's Unit 1, lets a symmetry be unitary or antiunitary. Parity keeps t, so it must commute with H and with e^(−iHt); an antiunitary operator commuting with H would turn e^(−iHt) into e^(+iHt), time reversal's signature, so P is unitary. Its action on one quantum follows from the observables: momentum is a polar vector and reverses, angular momentum is axial and does not, because P commutes with every rotation. So spin is kept and helicity flips. On Fock space that leaves one phase per species: P b(p, s)P⁻¹ = η_b b(−p, s), P d(p, s)P⁻¹ = η_d d(−p, s), and P a(p)P⁻¹ = η a(−p) for a scalar. Adjoints give P b†(p, s)P⁻¹ = η_b* b†(−p, s): a fermion at rest is a parity eigenstate with eigenvalue η_b*, and a moving one goes to its mirror image.

Locality fixes the scalar antiparticle's phase

Take UAQM-09.03's complex field in topic 1.03's normalisation, φ(x) = ∫dp̃ [a(p)e^(−ip⋅x) + b†(p)e^(ip⋅x)], and let P b(p)P⁻¹ = η′ b(−p). Then Pφ(t, x)P⁻¹ = ∫dp̃ [η a(−p)e^(−ip⋅x) + η′* b†(−p)e^(ip⋅x)]. Relabel p → −p: dp̃ and Eₚ are even, and e^(∓ip⋅x) becomes e^{(∓ip⋅x̃)}. The result is η φ₊(x̃) + η′* φ₋(x̃), the positive- and negative-frequency parts. Each part alone is a nonlocal functional of the field, so the image is a local field only if the coefficients agree: η′* = η, and Pφ(t, x)P⁻¹ = η φ(t, −x), derived. The pair carries ηη′ = |η|² = +1, so orbital L gives parity (−1)^L; ρ⁰ → π⁺π⁻, a 1⁻ meson into an L = 1 pair, fits. A real field has b = a, forcing η = η*, so η = ±1, scalar or pseudoscalar, a sign no charge can rephase. The π⁰ is the measured −1, read from the planes of the Dalitz pairs in π⁰ → e⁺e⁻e⁺e⁻ (quoted).

The Dirac field: γ⁰ and the sign on d

Quote UAQM-09.07's field, ψ(x) = ∫dp̃ Σₛ[b(p, s)u(p, s)e^(−ip⋅x) + d†(p, s)v(p, s)e^(ip⋅x)]. Applying P and relabelling p → −p leaves η_b b(p, s)u(p̃, s)e^{(−ip⋅x̃)} + η_d* d†(p, s)v(p̃, s)e^(ip⋅x̃): the operators are back at p, the spinors sit at p̃. In the Dirac representation u = (√(E+m) ξ, σ⋅p ξ/√(E+m)) and v = (σ⋅p χ/√(E+m), √(E+m) χ), so reversing p flips the lower entries of u and the upper entries of v, with ξ and χ untouched, which is why spin is kept. γ⁰ = diag(1, 1, −1, −1) flips the lower entries, so u(p̃, s) = γ⁰u(p, s) but v(p̃, s) = −γ⁰v(p, s). The field becomes γ⁰[η_b b u e^{(−ip⋅x̃)} − η_d* d† v e^(ip⋅x̃)], which is η_bγ⁰ψ(t, −x), UAQM-05.06's S_P = γ⁰ quoted, only if η_d = −η_b*. Derived, with no operator reordered: the sign is the v spinor's, set at rest, where v has only lower entries.

The vacuum and the Hamiltonian are parity-even

Since b(p, s) = η_b* P b(−p, s)P⁻¹, b(p, s)P|0⟩ = η_b* P b(−p, s)|0⟩ = 0, and likewise for d and a: every annihilator kills P|0⟩, so the unique vacuum gives P|0⟩ = |0⟩, a phase set to 1. For a real scalar P can be written out: P = exp(iG), G = (π/2)∫dp̃ [a†(p)a(p) − η a†(−p)a(p)], η = ±1. G is Hermitian and dimensionless, dp̃ carrying mass dimension 2 and a†a −2; [G, a(q)] = (π/2)(η a(−q) − a(q)) exponentiates to P a(q)P⁻¹ = η a(−q), and G|0⟩ = 0, shown. The free H = ∫dp̃ Σₛ Eₚ(b†b + d†d) returns at −p with |η_b|² = |η_d|² = 1, and E is even in p, so PHP⁻¹ = H; momentum reverses and Q is kept. In this free-field limit any phases work, so parity is a statement about interactions: with the P column of the bilinears quoted from UAQM-05.06, QED's −eψ̄γ^μψA_μ is P-even once A^μ reflects as a vector, topic 11.04's job.

Which phases are conventions, and which sign is physics

Q commutes with H and with P, so P′ = Pe^(iαQ) is an equally good parity. Since b lowers Q by one and d raises it, P′ carries η_b e^(−iα) and η_d e^(iα): η_d = −η_b* survives, η_bη_d = −1 does not move, and η_b alone is a convention, +1 for the electron by agreement. The same rotation can set a charged scalar's η to +1 but leaves ηη′ = +1. Limits check the rule. A self-conjugate field has no charge to rephase: a real scalar needs η = ±1, and a Majorana fermion, d = b, needs η_b = −η_b*, so ±i. At η_b = i the Dirac rule gives η_d = i too, the convention with P² = −1 on fermions. The relative sign is measured: para-positronium, ¹S₀, is 0⁻, its photons' polarisations crossed; η_c at 2984 MeV is 0⁻ and decays to p p̄ in ¹S₀, while π⁺π⁻ and K⁺K⁻, 0⁺ in an S wave, are forbidden and only upper limits exist; J/ψ at 3096.9 MeV is the 1⁻ ³S₁ state (all quoted).

Where parity stops: chirality and the massless limit

In UAQM-05.08's Weyl representation, quoted, ψ = (ψ_L, ψ_R) and γ⁰ is off-diagonal, so the field law reads Pψ_L(t, x)P⁻¹ = η_b ψ_R(t, −x) and Pψ_R(t, x)P⁻¹ = η_b ψ_L(t, −x): parity swaps the chiral halves. A massive electron needs both. For helicity −½ the right-chiral share of u†u is (E − |p|)/(2E), 0.065% at |p| = 10 MeV, and P, flipping the helicity, swaps the shares. As m → 0 the share falls as m²/(4E²), helicity becomes chirality, and P maps a purely left-handed quantum to a purely right-handed one. A theory whose fermion is ψ_L alone offers P nothing to map it into, so no unitary P obeys the local law; the charged weak current, coupling to ψ_L only, breaks parity there, named here and not developed. The non-relativistic limit closes the loop: at rest γ⁰ is +1 on u and −1 on v, the sign UAQM-08.07 found for wavefunctions, now carried by the operator d.

Change one variable at a time

Make the relationship visible.

Interactive model

Particle phase arg η30 °

Field: 0 scalar, 1 Dirac1

Pair orbital L0

With k = 1, sweep arg η: η and η′ turn opposite ways while the dot stays at −1, because P → Pe^(iαQ) moves conventions, not the product. At 90° both sit at +i. With k = 0 the mirror is the real axis; a real scalar, its own antiparticle, needs the two to meet, at 0° or 180° only.

ANTIPARTICLE arg η′150 °

PRODUCT η η′-1

PAIR PARITY (−1)^(L+k)-1

Live interpretationANTIPARTICLE arg η′: 150 °. PRODUCT η η′: −1. PAIR PARITY (−1)^(L+k): −1

Catch the common trap

Explain before calculating.

Misconception check

Because η_b can be chosen at will, +1, −1 or i, it is tempting to think η_d can be too, so that the opposite parity of electron and positron is a bookkeeping choice, undone by setting both to +1. It cannot be. η_b is free only because P can be multiplied by e^(iαQ), a symmetry of every charge-conserving theory, and that rephasing turns η_b and η_d in opposite senses, leaving η_bη_d = −1 fixed. η_d on its own is not free once P must send ψ(t, x) into a multiple of γ⁰ψ(t, −x): any other choice leaves PψP⁻¹ a sum of positive- and negative-frequency parts with different coefficients, which is no local field, and no P-invariant interaction could be built from it. Nor is the sign a statistics effect, as it is for C in topic 11.02: no operator is reordered, and the sign would survive commutators. It lives in γ⁰v = −v at rest, and para-positronium's 0⁻ is its measured face.

Apply P to ψ(x) = ∫dp̃ Σₛ[b(p, s)u(p, s)e^(−ip⋅x) + d†(p, s)v(p, s)e^(ip⋅x)], with P b(p, s)P⁻¹ = η_b b(−p, s) and P d(p, s)P⁻¹ = η_d d(−p, s), and require Pψ(t, x)P⁻¹ = η_bγ⁰ψ(t, −x). Which statement is right?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron of mass 0.511 MeV moves along +z with |p| = 1.20 MeV and spin up along z. In the Dirac representation, write u(p, ↑) and v(p, ↑) = iγ²u(p, ↑)* numerically, reverse the momentum, and check γ⁰u(p̃, ↑) = u(p, ↑) and γ⁰v(p̃, ↑) = −v(p, ↑) entry by entry. What share of u†u sits in the lower components?

E = √(1.20² + 0.511²) MeV = 1.30427 MeV, so √(E + m) = √1.81527 MeV^½ = 1.3473 MeV^½ and |p|/√(E + m) = 0.8907 MeV^½. With ξ = (1, 0) and σ_zξ = ξ, u(p, ↑) = (1.3473, 0, 0.8907, 0) MeV^½.
UAQM-08.05's v = iγ²u*, with iγ² = ((0, iσ₂), (−iσ₂, 0)) in 2 × 2 blocks and iσ₂ = ((0, 1), (−1, 0)), gives v(p, ↑) = (0, −0.8907, 0, 1.3473) MeV^½: its two-spinor is χ = (0, 1), and σ_zχ = −χ puts the minus sign on the upper entry.
Reverse p: only the entries proportional to |p| change sign, so u(p̃, ↑) = (1.3473, 0, −0.8907, 0) and v(p̃, ↑) = (0, 0.8907, 0, 1.3473) MeV^½.
With γ⁰ = diag(1, 1, −1, −1): γ⁰u(p̃, ↑) = (1.3473, 0, 0.8907, 0) = u(p, ↑), and γ⁰v(p̃, ↑) = (0, 0.8907, 0, −1.3473) = −v(p, ↑). Both identities hold entry by entry.
Checks: u†u = (E + m) + (E − m) = 1.81527 + 0.79327 = 2.60854 MeV, which is 2E, and ūu = 1.81527 − 0.79327 = 1.02200 MeV, twice the mass. The lower entries carry 0.79327/2.60854 = 30.4% of u†u, (E − m)/(2E): a moving u is no γ⁰ eigenspinor, yet the identity holds because γ⁰ and p → −p do the same thing to it.

Answeru(p, ↑) = (1.347, 0, 0.891, 0) MeV^½ and v(p, ↑) = (0, −0.891, 0, 1.347) MeV^½; γ⁰u(p̃) = u(p) and γ⁰v(p̃) = −v(p) hold entry by entry, and 30.4% of u†u = 2E sits in the lower components.

MediumThe φ(1020) meson, spin 1 and mass 1019.46 MeV, decays most often to K⁺K⁻, each kaon of mass 493.68 MeV, and rarely to e⁺e⁻. Treat K⁺ and K⁻ as particle and antiparticle of one complex spin-0 field, and the leptons as Dirac quanta. Assuming the decays conserve parity, find the φ's parity from the kaon channel, the partial waves open to the e⁺e⁻ pair, and the momentum and speed of each kaon.

Write the kaon pair as ∫dp̃ ψ̃(p) a†(p)b†(−p)|0⟩. P a†(p)P⁻¹ = η* a†(−p) and, since η′ = η*, P b†(p)P⁻¹ = η b†(−p), so P gives |η|² ∫dp̃ ψ̃(p) a†(−p)b†(p)|0⟩. Relabelling p → −p with ψ̃(−p) = (−1)^L ψ̃(p) leaves the eigenvalue (−1)^L, whatever η is.
Spin-0 daughters carry no spin, so J = 1 needs L = 1 and the kaon pair is P-odd. Parity conservation makes the φ(1020) a 1⁻ state, the assignment the tables quote.
For e⁺e⁻ the eigenvalue is η_b*η_d*(−1)^L = (−1)^(L+1), so P = −1 needs L even. With J = 1 that leaves L = 0 with S = 1, ³S₁, and L = 2 with S = 1, ³D₁; ¹P₁ and ³P₁ are P-even and excluded.
Kinematics: each kaon has E = M/2 = 509.73 MeV, so |p| = √(509.73² − 493.68²) MeV = 126.9 MeV and β = |p|/E = 0.249.
Contrast: had the leptons shared a parity, as the kaons do, a 1⁻ lepton pair would have to be a P wave. The extra −1 of η_d = −η_b* puts it in ³S₁ or ³D₁ instead, the waves through which a single virtual photon couples to e⁺e⁻, which is how e⁺e⁻ → φ(1020) makes the meson at colliders (quoted).

AnswerThe φ(1020) is 1⁻; its e⁺e⁻ pair can only be ³S₁ or ³D₁; each kaon has |p| = 127 MeV and β = 0.249.

HardAn electron of mass 0.511 MeV moves along +z with |p| = 2.00 MeV and helicity −½. In the Weyl representation, find the shares of u†u in the left- and right-chiral halves, then the shares for its parity image P|p, s⟩, and compare the smaller share with the massless estimate m²/(4E²). Then explain why a theory whose fermion is ψ_L alone admits no parity operator of this kind.

In the Weyl representation γ⁰ = ((0, 1), (1, 0)) in 2 × 2 blocks and ψ = (ψ_L, ψ_R), so Pψ(t, x)P⁻¹ = η_bγ⁰ψ(t, −x) splits into Pψ_L(t, x)P⁻¹ = η_bψ_R(t, −x) and Pψ_R(t, x)P⁻¹ = η_bψ_L(t, −x).
u(p, s) = (√(p⋅σ) ξ, √(p⋅σ̄) ξ). For helicity h, (σ⋅p̂)ξ = 2hξ, so the halves are √(E − 2h|p|) ξ and √(E + 2h|p|) ξ. Here E = √(2.00² + 0.511²) MeV = 2.0642 MeV and h = −½: left √(E + |p|) = 2.016 MeV^½, right √(E − |p|) = 0.2535 MeV^½.
Shares of u†u = 2E = 4.1285 MeV: left 4.0642/4.1285 = 98.44%, right 0.0642/4.1285 = 1.56%.
P b†(p, s)P⁻¹ = η_b* b†(−p, s) keeps the spin and reverses p, so the image has helicity +½. Its spinor u(p̃, s) = γ⁰u(p, s), since p̃⋅σ = p⋅σ̄, has the halves exchanged: left 1.56%, right 98.44%.
Massless estimate: (E − |p|)/(2E) = m²/(2E(E + |p|)) ≈ m²/(4E²) = 0.2611/17.04 = 1.53%, against the exact 1.56%. It vanishes as m/E → 0, where helicity is chirality and P turns a purely left-handed quantum into a purely right-handed one.
A Lagrangian built from ψ_L alone, ψ_L† iσ̄^μ∂_μψ_L plus couplings to ψ_L, contains no right-handed field for P to produce, so no unitary P obeys the local law; a Dirac mass term −m(ψ_L†ψ_R + ψ_R†ψ_L) would need ψ_R anyway. The charged weak current, coupling to ψ_L only, is where parity fails, named here only.

AnswerBefore P: helicity −½, 98.4% left-chiral and 1.56% right-chiral. After P: helicity +½, 1.56% left and 98.4% right. The estimate m²/(4E²) = 1.53% vanishes as m/E → 0. With ψ_L alone, P has no field to map into.