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Nuclear, particle & frontier orientation · Interactive lesson

Conserved numbers and discrete symmetries

Most reactions you can write down never happen. This is the audit that tells you which: balance charge, baryon number and each lepton family first, then ask which of the three interactions is on duty — because the weak one is allowed to change flavour and to tell left from right, and the other two are not.

UPIV-14.05 · Lesson 5 of 9 in this unit. Position is not a completion score.
Starting model: Conserved numbers and discrete symmetriesPolar plot of the beta electrons leaving polarised nuclei: W(θ) = 1 − (2h − 1) P β cos θ, with θ measured from the nuclear spin, which points up. The faint circle is the isotropic pattern parity would demand. At P = 0.60, β = 0.60 and left-handed fraction h = 1.00, the asymmetry is 0.36, so W(0°) = 0.64 against W(180°) = 1.36.⟨J⟩ nuclear spinW(θ) = 1 − (2h − 1) P β cos θ(2h − 1) P β = 0.36W(0°) = 0.64 W(180°) = 1.36θ = 0°θ = 180°
Start with a picture. Change the model in the lab below.

Nuclear, particle & frontier orientation. These numbered units provide nuclear, particle and frontier orientation, not a substitute for a complete specialist course. Check your university’s required depth and prerequisites.

1

Build the model

Every ledger must balance — and the weak interaction keeps its own books.

The key idea

A conserved quantum number is an additive label whose total is unchanged across a reaction — but only for the interactions that respect it, which is almost never all three.

Example. Σ⁻ → n + π⁻ takes strangeness from −1 to 0. The strong and electromagnetic interactions conserve strangeness exactly, so only the weak one is left, and the lifetime is 1.5 × 10⁻¹⁰ s rather than the 10⁻²³ s a strong decay takes.

The complete physical model and assumptions

The conservation rules of particle physics are two different kinds of thing, and the skill is telling them apart. A few are theorems: charge, energy-momentum and angular momentum follow from continuous symmetries by Noether's theorem, and CPT invariance follows from Lorentz invariance and locality, which is why a particle and its antiparticle are required to share a mass and a lifetime exactly. The rest is bookkeeping invented to record an absence. Baryon number was assigned because the proton does not decay; separate electron, muon and tau numbers because μ⁻ → e⁻γ has never been seen; strangeness because strange particles are made in pairs by the strong interaction and then live some 10¹⁴ times longer than it took to make them. Each was read off a missing process rather than derived, and each has a limit: the weak interaction changes flavour freely, isospin is only as good as the up–down mass difference is small, and neutrino oscillation has already broken the individual lepton-family numbers. The discrete symmetries cost most. Parity is exact at the strong and electromagnetic vertices, which are pure vector, and maximally violated by the charged weak current, which is V−A and couples to the left-handed chiral projection of a fermion field and to nothing else. So the ledger's honest output is a veto, not a rate: it says what cannot happen, and which interaction must be the one that does.

1Theorems, and rules invented to record an absence

Two different kinds of rule share the word conservation. Charge, energy-momentum and angular momentum are conserved because of Noether's theorem: each follows from a continuous symmetry of the action, and no experiment is expected to break them. CPT invariance is a theorem too, forced by Lorentz invariance, locality and a Hermitian Hamiltonian in any quantum field theory. Everything else on the ledger was invented to explain a missing process. Baryon number exists because the proton is stable — the partial lifetime for p → e⁺π⁰ now exceeds 10³⁴ years, some twenty-four orders of magnitude beyond the age of the universe. Separate electron, muon and tau numbers exist because μ⁻ → e⁻γ has never been seen, down to a branching ratio near 10⁻¹³. Strangeness exists because kaons and hyperons are produced copiously and then decay absurdly slowly. Assigning a quantum number records an absence in arithmetic; it does not explain it.

2The audit, in the order that decides fastest

Given a proposed reaction, work down a fixed list and stop at the first failure. Charge is exact and takes one line. Baryon number is exact at every Standard Model vertex: count quarks minus antiquarks and divide by three. Then run the three lepton-family ledgers separately — this is where μ⁻ → e⁻γ dies, with ΔLμ = −1 and ΔLₑ = +1 even though the total lepton number is untouched. Then energy: a decay needs a positive Q value, a collision needs invariant mass above threshold. Only after all that do you look at flavour, and flavour asks a different question. A change of strangeness, charm or beauty does not veto a process; it routes it, because the strong and electromagnetic interactions conserve flavour and the charged weak current does not. The first four tests answer whether the process can happen at all. The fifth answers who has to do it, and therefore how slowly.

3Flavour and isospin are approximate by construction

Strangeness is conserved exactly by the strong and electromagnetic interactions and broken by the weak one, by at most one unit for each W exchanged, so a weak decay has |ΔS| ≤ 1 and |ΔS| = 2 processes, which need two W's, are suppressed far below that. One rule settles the whole strange-particle puzzle of the 1950s. In π⁻ + p → K⁰ + Λ⁰ the strangeness ledger reads 0 → (+1) + (−1) = 0, so the strong interaction is free to make the pair, and it does in about 10⁻²³ s. Neither product can then decay strongly, because that would change S on its own, so the Λ⁰ lives 2.63 × 10⁻¹⁰ s — some 10¹⁴ times its own production time. Isospin is approximate for a different reason: it treats u and d as one doublet, which works only because md − mᵤ and α are both small beside the QCD scale. The error is set by those ratios themselves — md − mᵤ ≈ 2.5 MeV against a QCD scale near 200 MeV, and α ≈ 1/137, each about one per cent — so isospin predictions are worth about one per cent and no more. The proton–neutron splitting, 1.3 MeV on 939 MeV, is smaller still because the quark-mass and electromagnetic shifts partly cancel.

4Parity, and what makes an observable pseudoscalar

Parity inverts every spatial coordinate, r → −r. Polar vectors — position, momentum, electric field — change sign under it. Axial vectors, built from a cross product, do not: L = r × p picks up two sign changes, and spin transforms the same way. So a spin dotted into a momentum, ⟨J⟩⋅p̂, changes sign under P where an ordinary scalar does not. It is a pseudoscalar, and if the interaction responsible commuted with P its average would have to vanish exactly. That is what makes the Wu experiment a proof rather than a hint. Cobalt-60 polarised near 0.01 K emits its beta electrons preferentially opposite to the nuclear spin, following 1 − P (v/c) cos θ. With a polarisation near 0.6 and v/c near 0.6 the front-to-back modulation is about 36 per cent, and it reverses when the polarising field reverses, which is what rules out an apparatus artefact.

5V−A: why the violation is maximal, and only there

Write the projector PL = (1 − γ⁵)/2. The charged weak vertex is ū γμ (1 − γ⁵) u, which is 2 ū γμ PL u: the W couples to the left-handed chiral field and has no coupling whatever to the right-handed one. In the massless limit chirality equals helicity, so the neutrino from a weak decay is purely left-handed and its mirror image, a right-handed neutrino, has zero amplitude. A mirror-reflected weak process does not merely run at a different rate — the leading term does not run at all. That is what maximal means, and it is the same statement as the vector and axial couplings being equal in size, gA/gV = −1 at the quark level, dressed by QCD to −1.27 for the nucleon. The electromagnetic and strong vertices are pure γμ, which commutes with parity, so they cannot generate a cos θ term at all. The neutral weak current is the intermediate case: the Z couples to both chiralities of every charged fermion, with unequal strengths fixed by the fermion's charge, its weak isospin and sin²θW ≈ 0.231, so it violates parity by an amount that varies from fermion to fermion — the electron's left–right coupling asymmetry is only about 0.15 — rather than maximally across the board.

6C, T, CPT, and why CP violation is the prize

Charge conjugation fails as badly as parity: apply C to a left-handed neutrino and you get a left-handed antineutrino, which the weak current does not couple to and which has never been seen. The combination CP maps a left-handed neutrino onto a right-handed antineutrino, which does exist, so CP very nearly survives — and for the seven years after Wu it was taken to be exact. It is not. The long-lived neutral kaon, nominally CP-odd, decays to π⁺π⁻ with a branching ratio of 1.97 × 10⁻³. CPT then forces a T violation of exactly the same size, and the K⁰ ↔ K̄⁰ rate difference has been measured directly. In B mesons the effect is not small at all: sin 2β ≈ 0.70. This matters beyond spectroscopy, because Sakharov showed that a universe ending up with more matter than antimatter needs CP violation. The CKM phase supplies some, and falls short of the observed baryon asymmetry by roughly ten orders of magnitude — one of the clearest places the Standard Model is known to be incomplete.

2

Key equations

Read the units, assumptions and limits before substituting.

Baryon number

B = (nq − nq̄)/3

The proton is the lightest B = 1 state, so B conservation is what makes ordinary matter stable.

Every quark +1/3, every antiquark −1/3; leptons and gauge bosons 0. Additive over the whole final state.

Lepton-family numbers

ΔLₑ = ΔLμ = ΔLτ = 0

Vetoes μ⁻ → e⁻γ. Oscillation breaks it, so strictly only the total L survives.

+1 for ℓ⁻ and ν_ℓ, −1 for ℓ⁺ and ν̄_ℓ, 0 for everything else; one ledger per generation.

Parity of a two-body state

P = P₁ P₂ (−1)L

Turns parity into an arithmetic test on a reaction, not just a label attached to one state.

Intrinsic parity: quark +1, antiquark −1, photon −1, so P(π) = −1. L is the relative orbital angular momentum.

The V−A charged current

jμ = ū γμ (1 − γ⁵) u = 2 ū γμ PL u

The W couples to no right-handed field at all, so parity violation sits in the vertex, not in a fitted constant.

PL = (1 − γ⁵)/2 projects the left-handed chiral field; PL + PR = 1 and PL² = PL.

Beta asymmetry from a polarised source

W(θ) ∝ 1 + A P (v/c) cos θ, A = −1

The cos θ term is the pseudoscalar ⟨J⟩⋅p̂ in disguise, so a non-zero A is parity violation measured directly.

θ from the nuclear spin, P the polarisation, v/c the electron speed; A = −1 for ⁶⁰Co, a pure Gamow–Teller J → J − 1 decay, is maximal violation.

CPT theorem

m(X) = m(X̄) and τ(X) = τ(X̄)

So a measured CP violation forces an equal T violation; the K⁰ mass equality tests it below 10⁻¹⁸.

Follows from Lorentz invariance, locality and a Hermitian Hamiltonian — it is derived, not fitted.
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

1.00
0.5 to 1; steps of 0.05.
0.60
0 to 1; steps of 0.05.
0.60 c
0.1 to 0.9 c; steps of 0.05.

Try this. Set h to 0.5 and every ray lands back on the circle: equal left- and right-handed coupling conserves parity whatever the polarisation. Push h to 1 for pure V−A, then drop β to 0.15 and the pattern rounds out again — a slow electron cannot show its chirality.

Live results

ASYMMETRY (2h−1) P β
0.36
W AT θ = 0°
0.64
W AT θ = 180°
1.36
UP / DOWN COUNTS
0.471

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics IV modelPolar plot of the beta electrons leaving polarised nuclei: W(θ) = 1 − (2h − 1) P β cos θ, with θ measured from the nuclear spin, which points up. The faint circle is the isotropic pattern parity would demand. At P = 0.60, β = 0.60 and left-handed fraction h = 1.00, the asymmetry is 0.36, so W(0°) = 0.64 against W(180°) = 1.36.⟨J⟩ nuclear spinW(θ) = 1 − (2h − 1) P β cos θ(2h − 1) P β = 0.36W(0°) = 0.64 W(180°) = 1.36θ = 0°θ = 180°
Polar plot of the beta electrons leaving polarised nuclei: W(θ) = 1 − (2h − 1) P β cos θ, with θ measured from the nuclear spin, which points up. The faint circle is the isotropic pattern parity would demand. At P = 0.60, β = 0.60 and left-handed fraction h = 1.00, the asymmetry is 0.36, so W(0°) = 0.64 against W(180°) = 1.36.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Left-handed current fraction h: 1.00; Nuclear polarisation P: 0.60; Electron speed β = v/c: 0.60 c

Baseline 1.00; choose 0.5 to 1, in steps of 0.05.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The error here is treating helicity and chirality as the same word. Textbooks say the neutrino is left-handed, students carry that straight over to the electron, and then cannot explain why an electron emitted with the wrong helicity is suppressed rather than forbidden. Helicity is spin⋅p̂, and for anything with mass it is frame-dependent: overtake a 1 MeV electron and its helicity flips, so helicity cannot be what a Lorentz-invariant vertex couples to. Chirality is the invariant label the projector (1 − γ⁵)/2 selects, and that is what the W actually sees. The two agree only in the massless limit; a fermion of mass m and energy E carries a wrong-chirality admixture of order m/E. Everything quantitative in this topic lives in that gap — the beta asymmetry carries a factor v/c instead of 1, and π⁺ → e⁺ν is suppressed by (mₑ/mμ)² rather than being forbidden outright.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

Audit p → e⁺ + π⁰ and ν̄ₑ + p → n + e⁺ against charge, baryon number and electron-family number. Say which is forbidden, and for the survivor name the interaction that can mediate it and find its threshold. Use mₚ = 938.272, mₙ = 939.565, mₑ = 0.511 MeV/c².

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.

MediumExample 2

The Λ⁰ (uds, B = +1, S = −1) is produced in π⁻ + p → K⁰ + Λ⁰ and decays as Λ⁰ → p + π⁻ with τ = 2.63 × 10⁻¹⁰ s. Show that the production conserves strangeness while the decay does not, and use the ratio of the lifetime to the strong timescale to name the interaction behind each.

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.

HardExample 3

The π⁺ has spin 0 and decays through the V−A current. Explain why R = Γ(π⁺ → e⁺νₑ) / Γ(π⁺ → μ⁺νμ) equals (mₑ/mμ)² [(mπ² − mₑ²)/(mπ² − mμ²)]², evaluate it with mπ = 139.57, mμ = 105.66 and mₑ = 0.511 MeV/c², and say why a phase-space argument gets the answer backwards. Measured value: 1.233 × 10⁻⁴.

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.

6

Quick check

Select an answer and explain it before checking.

Cobalt-60 nuclei are polarised at about 0.01 K, and the beta electrons come out preferentially opposite to the nuclear spin, with a modulation proportional to v/c that reverses when the polarising field is reversed. What has the measurement established?
7

Review & continue

A conserved quantum number is an additive label whose total is unchanged across a reaction — but only for the interactions that respect it, which is almost never all three.

  • 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.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.