Parity as a unitary symmetry and parity-odd terms

Units are SI with ħ kept; nothing here is relativistic, so the course metric diag(1, −1, −1, −1) plays no part. Define parity by its action on the canonical pair: P x P⁻¹ = −x and P p P⁻¹ = −p. Wigner's theorem (lesson 1.1) allows a unitary or an antiunitary P, and a discrete symmetry may be either, so an argument is needed. Conjugate the commutator: P[xᵢ, pⱼ]P⁻¹ = [−xᵢ, −pⱼ] = iħδᵢⱼ. The left side is also P(iħδᵢⱼ)P⁻¹, which stays iħδᵢⱼ if P is linear and becomes −iħδᵢⱼ if it is antilinear. Only the linear choice is consistent, so P is unitary: derived. The same test applied to time reversal, which keeps x and flips p, gives the opposite verdict in lesson 1.4. On wavefunctions, P|x⟩ = |−x⟩ gives ⟨x|P|ψ⟩ = ψ(−x). With Quantum Mechanics II's Yₗᵐ(−r̂) = (−1)ˡYₗᵐ(r̂), used and not re-derived, a state Rₙₗ(r)Yₗᵐ has parity (−1)ˡ: +1 for hydrogen's 1s, −1 for its 2p.

Inversion is the matrix −1, which commutes with every rotation matrix, so P U(R) P⁻¹ = U(R) for every rotation operator U(R) = exp(−iφ n̂⋅J/ħ); no stray phase can enter, because rotations have no non-trivial one-dimensional representation. Expand to first order in φ. Since P is linear, the i passes through unchanged and P J P⁻¹ = J: derived. An antiunitary operator would have had to flip J, which is what time reversal will do. Orbital angular momentum is even directly, since L = x × p becomes (−x) × (−p) = L, so S = J − L is even as well. Next, P² commutes with x, p and S, a complete set for one particle, so by Schur's lemma, quoted, P² = e^(iα). Redefining P as e^(−iα/2)P makes P² = 1, with eigenvalues ±1. For a fixed set of particles the intrinsic phase left over is a convention; relative intrinsic parities matter only once particles can be created or destroyed.

A vector operator either flips under P, a polar vector (r, p, the dipole d = qr, the field E), or does not, an axial vector (L, S, J, the moment μ, the field B). For fields, invert the sources with the system: E from inverted charges flips, B from inverted currents does not, consistent with ∇ × E = −∂B/∂t, since ∇ is polar. Signs multiply. Like dotted with like gives a scalar: p², L⋅S, S₁⋅S₂. Axial dotted with polar gives a pseudoscalar: S⋅p, S⋅r, S⋅E, J⋅p. Polar crossed with polar is axial, as L = x × p shows. Anticipating lessons 1.5 and 1.6, time reversal flips p, L, S and B and keeps r and E, so S⋅p is P-odd but T-even, while S⋅r and S⋅E are odd under both: the electric-dipole signature of lesson 1.8. Applied fields need care. With the apparatus held fixed, the Stark term eE⋅r is P-odd on the atom alone; that is a polar vector supplied by the laboratory, not a broken law. A term εS⋅p with a constant ε has no apparatus to blame.

If [H₀, P] = 0 and P|n⟩ = π|n⟩, any P-odd Q obeys ⟨n|Q|n⟩ = ⟨n|P⁻¹(PQP⁻¹)P|n⟩ = −π²⟨n|Q|n⟩, so ⟨n|Q|n⟩ = 0: derived. Quantum Mechanics II's selection rule, used here, adds that ⟨k|Q|n⟩ = 0 unless πₖ = −πₙ. Now add εS⋅p. Its diagonal element vanishes, so a non-degenerate level does not move at first order, but the state does: each opposite-parity |k⟩ is admixed with amplitude ε⟨k|S⋅p|n⟩/(Eₙ − Eₖ). An odd observable lives in the cross term between the two parities, so ⟨S⋅p⟩ = 2εΣₖ|⟨k|S⋅p|n⟩|²/(Eₙ − Eₖ) is first order, derived, while the probability of odd parity is second order. The energy moves at second order, by ε²Σₖ|⟨k|S⋅p|n⟩|²/(Eₙ − Eₖ), and dE/dε = ⟨S⋅p⟩ checks both. Since P H(ε) P⁻¹ = H(−ε), energies are even in ε and ⟨S⋅p⟩ is odd: the mirror-image system has the same spectrum and the opposite handedness. The expansion fails once |ε⟨k|S⋅p|n⟩| is a sizeable fraction of |Eₙ − Eₖ|: in the figure's two-level model, at a quarter of the gap the exact value already falls 11% short.

For H₀ = p²/2m + V(r), [H₀, S⋅r] = −(iħ/m)S⋅p, so ⟨k|S⋅p|n⟩ = (im/ħ)(Eₖ − Eₙ)⟨k|S⋅r|n⟩ and every energy denominator cancels. The first-order state is (1 − iεmS⋅r/ħ)|n⟩, the start of exp(−iεmS⋅r/ħ): each spin turned about the local radial direction by the angle εmr. Then ⟨S⋅p⟩ = (iεm/ħ)⟨n|[S⋅r, S⋅p]|n⟩, and [S⋅r, S⋅p] = iħ(S⋅L + S²), so ⟨S⋅p⟩ = −εm⟨S⋅L + S²⟩: derived, and −(3/4)εmħ² in any s state of a spin-½ particle, whatever V is. Dimensions: ε is kg⁻¹ m⁻¹, so εmħ² is J kg m, the unit of S⋅p. Hydrogen tests the sum: 2p₁/₂ alone, with |⟨2p₁/₂|S⋅p|1s⟩| = 0.242ħ²/a₀ across the 10.2 eV gap that Lyman-α measures, gives 41.6% of it; higher p₁/₂ levels and the continuum give the rest. S⋅r commutes with r, so ⟨r⟩ stays zero at first order: this T-even term makes no dipole. The expansion needs εmR ≪ 1 over the state's size R. For a free particle R is unbounded, and the ground states are helicity eigenstates at |p| = |ε|mħ/2 instead.

In 1957 Wu and collaborators polarised ⁶⁰Co nuclei, spin J = 5, at millikelvin temperatures in a magnetic field and counted β electrons. The rate is described by W(θ) = 1 + A f (v/c) cos θ, with f the nuclear polarisation and θ the angle between J and the electron momentum pₑ. More electrons left against the spin, and modern measurements give A ≈ −1, the largest size positivity of W allows. All of this is quoted, not derived; the weak interaction is responsible. The measured quantity is ⟨J⋅pₑ⟩, a pseudoscalar. Invert the whole apparatus: the coil current still circulates the same way, so B and J are unchanged, while every electron momentum reverses. A parity-invariant law would therefore demand W(θ) = W(π − θ), that is A = 0. The observed A ≠ 0 means the decay Hamiltonian contains a P-odd piece, playing the part εS⋅p plays here. J and pₑ both flip under time reversal, so the correlation is T-even: it shows that P fails and says nothing about T.