
Advanced Quantum Mechanics course sequence · Interactive lesson
Parity for Dirac spinors: γ⁰, scalars and pseudoscalars
Parity is the first transformation in this unit that small steps cannot build, so its spinor matrix has to be solved for rather than exponentiated. The answer, γ⁰, is why a Dirac bound state carries two orbital quantum numbers, why a pion couples through γ⁵, and which densities a mirror turns inside out.
UAQM-05.06 · Lesson 6 of 9 in this unit. Position is not a completion score.Build the model
No generator reaches the mirror, so its spinor matrix is solved for, and γ⁰ appears.
The key idea
Parity acts on a Dirac field as ψ′(t, x) = ηPγ⁰ψ(t, −x): the argument is reflected and the spinor multiplied by γ⁰, which in the Dirac representation keeps the upper two components and reverses the lower two.
Example. An electron with 0.300 MeV of kinetic energy moving along +z with spin up has u ∝ (1, 0, 0.476, 0). γ⁰ gives (1, 0, −0.476, 0), the spinor of the same electron moving along −z with spin still up: its helicity goes from +1 to −1.
The complete physical model and assumptions
Parity is a Lorentz transformation, ΛP = diag(1, −1, −1, −1), but with det ΛP = −1 it lies on no path from the identity, so no exponential of σμν produces its spinor matrix. Topic 5.02's covariance condition does not need one: it demands a matrix that commutes with γ⁰ and anticommutes with γ¹, γ² and γ³, and the Clifford algebra supplies γ⁰, unique up to a phase. So ψ′(t, x) = ηPγ⁰ψ(t, −x), derived, and everything else follows from the minus sign γ⁰ puts on the lower two components in the Dirac representation. The two blocks carry opposite intrinsic parity, so a free electron is mirrored into one with reversed momentum and unchanged spin, and a bound state whose upper block has orbital l must have a lower block with l ± 1. Every bilinear picks up γ⁰Γγ⁰, which separates the pairs proper transformations could not: ψ̄ψ from iψ̄γ⁵ψ, the current from the spin density. The model has three edges. In the non-relativistic limit the lower block shrinks to v/2c of the upper and parity collapses to Schrödinger's x → −x, which is why γ⁰ was never needed before. The phase ηP cancels in every density, so a single fermion's parity is a convention and only relative parities, such as Unit 8's electron and positron, are measured. And P is a symmetry only of a P-even Hamiltonian: electromagnetism and the strong force qualify, while the weak V − A current does not.
1No small step reaches the mirror
Units throughout: SI with x⁰ = ct, metric diag(1, −1, −1, −1), the Dirac representation, γ⁵ = iγ⁰γ¹γ²γ³ and Unit 4's u†u = 2E/c. Spatial inversion is ΛP = diag(1, −1, −1, −1), and det ΛP = −1. Every Λ reached in topics 5.01 to 5.04 is a product of exponentials of the generators J and K, and det eG = e(tr G) = 1 because those 4 × 4 generators are traceless. No continuous path from the identity reaches a determinant of −1, so no exponential of σμν can serve as SP either: it has to be solved for. Topic 5.02's condition S⁻¹γμS = Λμνγν never mentioned generators, so put ΛP into it. The time row gives SP⁻¹γ⁰SP = γ⁰ and the space rows give SP⁻¹γiSP = −γi. SP must commute with γ⁰ and anticommute with each of γ¹, γ² and γ³, a question about 4 × 4 matrices that the Clifford algebra answers on its own.
2γ⁰ solves the condition, and only γ⁰ does
Try γ⁰. It commutes with itself, and (γ⁰, γi) = 2g(0i) = 0 makes it anticommute with each γi, so the condition holds, derived. γ⁵ fails, because it anticommutes with γ⁰ as well. If S′ also works, S′SP⁻¹ commutes with all four γμ and so with all their products; topic 5.05's sixteen products span every 4 × 4 matrix, so S′SP⁻¹ is a multiple of 1. Hence SP = ηPγ⁰, and keeping ∫ψ†ψ d³x fixed forces |ηP| = 1, which is topic 5.03's S⁻¹ = γ⁰S†γ⁰ once more. The Hamiltonian form is the quickest check: x → −x turns H = cα⋅p + βmc² + V(r) into −cα⋅p + βmc² + V(r), and β = γ⁰, which anticommutes with α, turns it back, so ηPβ with x → −x commutes with H whenever V is even. No density fixes the phase, since bilinears carry |ηP|² = 1. Asking two inversions to return the field, up to the sign a 2π rotation already allows, leaves ηP = ±1 or ±i, and the electron is given +1 by convention.
3Upper and lower components carry opposite parity
With γ⁰ = diag(1, 1, −1, −1) and ψ = (φ, χ), parity sends φ(t, x) to ηPφ(t, −x) and χ(t, x) to −ηPχ(t, −x): the two blocks carry opposite intrinsic parity. Unit 4's u(p, s) has lower block cσ⋅p χₛ/(E + mc²), odd in p, so γ⁰u(p, s) = u(−p, s), derived, while x → −x turns e(ip⋅x/ħ) into e(−ip⋅x/ħ) to match. The mirrored electron moves the other way with its spin unchanged, so its helicity reverses, as an axial spin and a polar momentum require. In v(p, s) the large block is the lower one, and γ⁰v(p, s) = −v(−p, s): an extra minus sign, which Unit 8 turns into the opposite intrinsic parity of electron and positron, seen in the perpendicular photon polarisations of para-positronium decay that Wu and Shaknov measured in 1950 (quoted). Limit: at 1 keV the lower block of u is 3.1% of the upper and γ⁰ barely acts, which is why Schrödinger theory never needed it.
4A bound state carries two orbital quantum numbers
In a central potential P commutes with H and with J, so stationary states can carry a parity as well as j and m. Give the upper block orbital l and the lower block l′, in spinor harmonics with Ω(−r̂) = (−1)l Ω(r̂). The upper block contributes ηP(−1)l and the lower −ηP(−1)(l′), and a single eigenvalue needs (−1)(l′) = −(−1)l: l′ = l ± 1, derived. Both couple with spin ½ to the same j, so l and l′ are j − ½ and j + ½, and l′ = 2j − l. Thus 1s₁/₂ has a p-wave lower block, 2p₁/₂ an s-wave, 2p₃/₂ a d-wave and 3d₃/₂ a p-wave: the spectroscopic letter names the upper block, and ηP(−1)l is Schrödinger's parity, the non-relativistic limit. 2s₁/₂ and 2p₁/₂ share n, j and Dirac's energy but not parity, so no P-even term mixes them; the Lamb shift moves them 1057.8 MHz apart (measured) without mixing them. Unit 7's exact 1s state (quoted) sets the size: lower ÷ upper = (1 − ε)/(Zα) with ε = √(1 − Z²α²), a probability share (1 − ε)/2 of 1.3 × 10⁻⁵ in hydrogen and 12.9% in U⁹¹⁺.
5One γ⁰ on each side sorts the bilinears
From ψ′ = ηPγ⁰ψ(t, −x), ψ′† = ηP*ψ†(t, −x)γ⁰, so ψ̄′ = ηP*ψ̄(t, −x)γ⁰ and every bilinear obeys ψ̄′Γψ′(t, x) = ψ̄γ⁰Γγ⁰ψ(t, −x), derived, with the phase gone. Γ = 1 gives +1: ψ̄ψ is a true scalar. Γ = γ⁵ gives −1, because γ⁵ anticommutes with γ⁰: a pseudoscalar. Since (ψ̄γ⁵ψ)* = −ψ̄γ⁵ψ, the real density is iψ̄γ⁵ψ. Blocks show the rule: ψ̄ψ = φ†φ − χ†χ pairs each block with itself, so the γ⁰ signs square away, while iψ̄γ⁵ψ = −2 Im(φ†χ) pairs upper with lower and keeps one minus. The vector ψ̄γμψ keeps its time part and reverses its space part, a polar current. The axial vector ψ̄γμγ⁵ψ does the opposite: its space part survives, as a spin density must, and its time part ψ†γ⁵ψ flips. The tensor splits like the field strength, σ(0i) reversing like E and σij not, like B. Topic 5.05 could not tell S from P or V from A under proper transformations; parity can, and these are Unit 1's classes carried to spinors.
6A pseudoscalar density is a divergence of spin
For a slow electron the lower row of the Dirac equation gives χ ≈ −(iħ/2mc)σ⋅∇φ. Then φ†χ − χ†φ = −(iħ/2mc)∇⋅(φ†σφ), so iψ̄γ⁵ψ ≈ (ħ/2mc)∇⋅(φ†σφ), derived: the divergence of the spin density, scaled by ħ/2mc = 1.93 × 10⁻¹³ m. Dimensions: a length times m⁻¹ times m⁻³ is m⁻³, a density like ψ̄ψ. A mirror keeps every spin and reverses ∇, carrying a hedgehog of outward spins into an inward one, so the density flips. For a packet of size L it is about ħ/(2mcL) of ψ†ψ: relativistic, and odd in any parity eigenstate, so it integrates to zero. Couplings follow the classes. gψ̄ψΦ conserves parity with a scalar field Φ and gψ̄iγ⁵ψπ with a pseudoscalar π, and the pion's parity of −1 was measured in 1951 from π⁻d → nn at rest, whose final state can only be ³P₁ (quoted). The model stops where the Hamiltonian is not P-even. The weak V − A current broke parity in ⁶⁰Co decay in 1957, and in caesium a contact term in ψ†γ⁵ψ joins 6s₁/₂ to the s-wave lower blocks of the p₁/₂ states, which gives the 6s → 7s transition a parity-violating electric-dipole amplitude, measured to 0.35% in 1997.
Key equations
Read the units, assumptions and limits before substituting.
Covariance condition for the inversion
SP⁻¹ γ⁰ SP = γ⁰ and SP⁻¹ γi SP = −γi, from ΛP = diag(1, −1, −1, −1)
Turns covariance into pure algebra: find the matrix that commutes with γ⁰ and anticommutes with each γi. The two equations are the time and space rows of one condition.
SP a dimensionless 4 × 4 matrix; i = 1, 2, 3; det ΛP = −1, so no exponential of the generators J and K reaches ΛP.Parity on the Dirac field
ψ′(t, x) = ηP γ⁰ ψ(t, −x), with |ηP| = 1
Derived, and unique up to ηP, because only multiples of 1 commute with all four γμ. The upper pair of components keeps its sign and the lower pair reverses.
ψ in m(−3/2); ηP a dimensionless phase, +1 by convention for the electron; γ⁰ = diag(1, 1, −1, −1) in the Dirac representation.Mirror image of a free spinor
γ⁰ u(p, s) = u(−p, s)
Derived from the lower block cσ⋅p χₛ/(E + mc²), which is odd in p: momentum reverses, spin does not, and helicity Σ⋅p̂ changes sign.
u in (kg m s⁻¹)¹⁄², normalised to u†u = 2E/c; p in kg m s⁻¹; s labels the rest-frame spin, which is unchanged.Orbital pairing in a parity eigenstate
ηP(−1)l = −ηP(−1)(l′), so l′ = l ± 1 = 2j − l
1s₁/₂ hides a p-wave and 2p₁/₂ an s-wave. The parity ηP(−1)l is read from the upper block, which is Schrödinger's label in the non-relativistic limit.
l and l′ the orbital quantum numbers of the upper and lower blocks, j the total angular momentum; all dimensionless.How any bilinear transforms
ψ̄′Γψ′(t, x) = ψ̄ γ⁰Γγ⁰ ψ(t, −x)
γ⁰Γγ⁰ = +Γ for 1, γ⁰, γiγ⁵ and σij, and −Γ for γ⁵, γi, γ⁰γ⁵ and σ(0i): scalar against pseudoscalar, polar vector against axial vector.
Γ any of topic 5.05's sixteen matrices; each density in m⁻³; the phase enters as |ηP|² = 1 and drops out.Pseudoscalar density of a slow electron
iψ̄γ⁵ψ ≈ (ħ/2mc) ∇⋅(φ†σφ)
Derived from χ ≈ −(iħ/2mc)σ⋅∇φ. A pseudoscalar density is the divergence of an axial field: the mirror keeps every spin but reverses ∇, so the density changes sign.
φ the upper block, in m(−3/2); φ†σφ the spin density, in m⁻³; ħ/2mc = 1.93 × 10⁻¹³ m for the electron.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at Z = 1: the dashed lower curve is flat, about α/2 of the upper, so a bare x → −x passes for parity. Raise Z to 92 and step the operation to 1: the p-wave flips and the overlap drops to 0.741. Step to 2 and γ⁰ flips it back.
Live results
- ENERGY ε = E/mc²
- 0.8012
- LOWER ÷ UPPER (1 − ε)/Zα
- 0.3322
- LOWER-BLOCK SHARE
- 9.939 %
- OVERLAP WITH THE STATE
- 1.0000
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Nuclear charge Z: 82; Operation: 0 none, 1 x → −x, 2 with γ⁰: 0
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The tempting error is to carry Schrödinger's rule over unchanged: parity sends ψ(t, x) to ψ(t, −x), and a state's parity is (−1)l for a single l. For the Dirac equation x → −x reverses cα⋅p but not βmc², so the reflected spinor satisfies a different equation; the lower block, which carries σ⋅p, comes out with the wrong sign. γ⁰ is the repair, not a convention. The same slip gives all four components of 1s₁/₂ the label l = 0, but the lower block is a radial function times (cos θ, sin θ e(iφ)), an l = 1 function, and only the γ⁰ sign lets an s-wave and a p-wave share one parity. Hydrogen hides the error, because its lower block holds 1.3 × 10⁻⁵ of the probability. In hydrogen-like uranium it holds 12.9%, and the bare mirror image overlaps the true state by only 0.741.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
An electron with kinetic energy 0.300 MeV moves along +z with its spin up along z. Write Unit 4's u(p, s) with numbers, apply the parity matrix γ⁰, and identify the electron the result describes: its momentum, spin and helicity. Then check what happens to u†u and ūu.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- E = T + mc² = 0.300 + 0.511 = 0.811 MeV, so pc = √(E² − m²c⁴) = √(0.8110² − 0.5110²) MeV = 0.6298 MeV, and the lower-to-upper ratio is a = pc/(E + mc²) = 0.6298/1.3220 = 0.4764.
- With χₛ = (1, 0) and σzχₛ = χₛ, u(pẑ, ↑) = √(E/c + mc) (1, 0, 0.4764, 0). The upper block carries the spin; the lower block, cσ⋅p χₛ/(E + mc²), carries the sign of the momentum.
- γ⁰ = diag(1, 1, −1, −1) gives √(E/c + mc) (1, 0, −0.4764, 0). For momentum −pẑ the lower block is cσ⋅(−pẑ)χₛ/(E + mc²) = −0.4764 χₛ, so the result is exactly u(−pẑ, ↑). Under z → −z the plane wave e(ipz/ħ) becomes e(−ipz/ħ), consistent with it.
- The spin is still up and the momentum is reversed, so the helicity Σ⋅p̂ goes from +1 to −1, as an axial spin and a polar momentum require.
- Norms, in units of mc: E/c + mc = 1.3220/0.5110 = 2.587, so u†u = 2.587 × (1 + 0.4764²) = 3.174 = 2E/mc², and ūu = 2.587 × (1 − 0.4764²) = 2.000. γ⁰ is unitary and preserves both. Limit: as T → 0, a → 0 and γ⁰ acts on u as the identity, Schrödinger's bare x → −x.
Answer. γ⁰u(pẑ, ↑) = √(E/c + mc) (1, 0, −0.476, 0) = u(−pẑ, ↑): momentum −0.630 MeV/c along z, spin still up, helicity −1. u†u = 3.17 mc and ūu = 2.00 mc, both unchanged.
MediumExample 2
Hydrogen-like lead, Z = 82, is in its 1s₁/₂ state with spin up. Unit 7's exact solution, quoted, has upper block g(r)χ↑/√(4π) and lower block ik g(r)(σ⋅r̂)χ↑/√(4π), with k = (1 − ε)/(Zα) and ε = √(1 − Z²α²). Show that the lower block is a p-wave and the state has parity +1. Find the share of probability in the lower block and the overlap of the state with its bare mirror image ψ(t, −x).
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- σ⋅r̂ χ↑ = (cos θ, sin θ e(iφ)) = √(4π/3)(Y₁₀χ↑ − √2 Y₁₁χ↓): only l = 1 harmonics, so l′ = 1 = 2j − l with j = ½ and l = 0. Under r̂ → −r̂ the lower block changes sign and the upper does not.
- With ηP = 1 the upper block gives (−1)⁰ = +1 and the lower block −(−1)¹ = +1, the γ⁰ sign cancelling the p-wave's. The two agree, so the state is a parity eigenstate with eigenvalue +1.
- Numbers: Zα = 82 × 0.00729735 = 0.59838, ε = √(1 − 0.35806) = 0.80121, and k = 0.19879/0.59838 = 0.33221. The slow-electron estimate Zα/2 = 0.29919 is 10% low.
- Because |σ⋅r̂ χ↑|² = 1, the lower block holds k²/(1 + k²) of the probability. With k² = (1 − ε)/(1 + ε) this is (1 − ε)/2 = 0.0994.
- The bare mirror ψ(t, −x) keeps the upper block and reverses the lower, so its overlap with ψ is (1 − k²)/(1 + k²) = ε = 0.801; with γ⁰ included the overlap is exactly 1. For hydrogen, Z = 1, the same formulae give a share of 1.33 × 10⁻⁵ and an overlap of 0.99997: the non-relativistic limit, where x → −x alone passes for parity.
Answer. The lower block is l′ = 1 and the parity is +1. In Pb⁸¹⁺ the lower block holds 9.94% of the probability, and the bare mirror image overlaps the state by only 0.801, against exactly 1 with γ⁰. In hydrogen the figures are 1.33 × 10⁻⁵ and 0.99997.
HardExample 3
For the same 1s₁/₂ state in hydrogen-like uranium, Z = 92, find ψ̄ψ and iψ̄γ⁵ψ as functions of r and θ and say which is even and which odd under parity. Evaluate their ratio on the z axis exactly and from the slow-electron formula iψ̄γ⁵ψ ≈ (ħ/2mc)∇⋅(φ†σφ). Then integrate both over space and compare ∫ψ̄ψ d³x with the measured binding energy.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- With φ = gχ↑/√(4π) and χ = ik g(σ⋅r̂)χ↑/√(4π), ψ̄ψ = φ†φ − χ†χ = (1 − k²)g²/(4π), the same in every direction, so even under r̂ → −r̂.
- φ†χ = ik g² χ↑†(σ⋅r̂)χ↑/(4π) = ik g² cos θ/(4π), so iψ̄γ⁵ψ = −2 Im(φ†χ) = −2k g² cos θ/(4π): a dipole pattern, odd, because cos θ reverses under r̂ → −r̂.
- Ratio: iψ̄γ⁵ψ ÷ ψ̄ψ = −2k cos θ/(1 − k²), and with k² = (1 − ε)/(1 + ε) this is −(Zα/ε) cos θ. For Z = 92, Zα = 0.67136 and ε = 0.74113, so on the z axis the ratio is −0.906 for z above the nucleus and +0.906 below, at every r.
- Slow-electron formula with the Schrödinger g ∝ e(−Zr/a₀): φ†σφ = g²ẑ/(4π) and ∇⋅(φ†σφ) = −(2Z/a₀)g² cos θ/(4π). Times ħ/2mc, with ħ/(mca₀) = α, this is −Zα g² cos θ/(4π): a ratio of ∓0.671, low by the factor ε, or 26%. For hydrogen both give ∓0.00730.
- ∫cos θ dΩ = 0, so ∫iψ̄γ⁵ψ d³x = 0: a P-odd density has no net value in a parity eigenstate. And ∫ψ̄ψ d³x ÷ ∫ψ†ψ d³x = (1 − k²)/(1 + k²) = ε = 0.7411.
- Check: for a Coulomb field the relativistic virial theorem gives E = mc²∫ψ̄ψ d³x for a normalised state (quoted), so the binding is mc²(1 − ε) = 132.28 keV. The measurement at GSI is 0.46 keV lower, the 1s Lamb shift of QED and nuclear size: agreement to 0.35%.
Answer. ψ̄ψ = (1 − k²)g²/(4π), even; iψ̄γ⁵ψ = −2k g² cos θ/(4π), odd. On the z axis their ratio is ∓0.906 exactly and ∓0.671 from the slow-electron formula. ∫iψ̄γ⁵ψ d³x = 0, and ∫ψ̄ψ d³x = 0.741, giving a binding of 132.3 keV against 131.8 keV measured.
Quick check
Select an answer and explain it before checking.
Review & continue
Parity acts on a Dirac field as ψ′(t, x) = ηPγ⁰ψ(t, −x): the argument is reflected and the spinor multiplied by γ⁰, which in the Dirac representation keeps the upper two components and reverses the lower two.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.