Build hadrons by adding quark quantum numbers

By UK A Level Physics · independent study · myAudience

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Build hadrons by adding quark quantum numbers

Quark composition explains several quantum-number patterns in hadrons. This lesson assumes fractions and particle-family bookkeeping. Only up, down and strange quarks and their antiquarks are required here. In units of e, their charges are u: +2/3, d: −1/3 and s: −1/3. Each quark has baryon number +1/3; only the strange quark has strangeness −1. Antiquarks reverse all these additive quantum numbers: ū has charge −2/3, and s̄ has charge +1/3 and strangeness +1.

A proton has valence composition uud: its charge is 2/3 + 2/3 − 1/3 = +1 and B = 1. A neutron is udd: its charge is zero and B = 1. An antiproton is ūūd̄ and an antineutron is ūd̄d̄, giving B = −1 in each case and charges −1 and zero respectively. These valence labels account for additive quantum numbers; they are not literal diagrams of three stationary coloured balls or a complete description of the internal quantum state.

Charged pions have π⁺ = ud̄ and π⁻ = dū. A neutral pion involves a quantum combination of uū and dd̄ components; either component gives Q = B = S = 0, but calling it only one component misses the neutral-state caveat. For the required kaons, K⁺ = us̄, K⁰ = ds̄, K⁻ = sū and K̄⁰ = sd̄. Each has B = 0. The first two have S = +1, the latter two S = −1, even when the electric charge is zero.

Beta-minus changes a down quark into an up quark through the weak interaction: d → u + e⁻ + ν̄ₑ. In a neutron, one d in udd changes, leaving uud, a proton. Beta-plus changes u → d + e⁺ + νₑ in a suitable nucleus, taking a proton's uud to udd. The baryon number of the changing quark stays +1/3 while its charge changes by one unit; the emitted charged lepton balances that change.

When solving a composition problem, add Q, B and S separately, then compare all three with the stated particle. Charge alone cannot distinguish d from s, and neutrality alone cannot distinguish K⁰ from K̄⁰. The Particle Data Group's quark-model review also describes hadrons beyond this simple syllabus model. Those do not make this model useless, but they limit claims that every possible hadron must consist of exactly two or three valence constituents. Use the model for the named assessed particles and identify its limits.

The disks group constituents without depicting literal orbits. For uud, the charge sum is 2/3+2/3−1/3 = +1; for udd it is 2/3−1/3−1/3 = 0. Both contain three quarks with B = 1/3 each, giving B = 1. A meson contains a quark and an antiquark, so B = 1/3−1/3 = 0. The pion u anti-d and kaon u anti-s have identical charge +1 but different strangeness. Charge alone cannot identify a hadron. Anti-s carries S = +1, whereas s carries S = −1. An antiproton anti-u anti-u anti-d has Q/e = −1 and B = −1; neutral does not imply zero baryon number for an antineutron.

**Supplied unfamiliar data:** Λ⁰ = uds, Q/e = 0, B = 1, S = −1, rest energy 1116 MeV; K⁺ = u anti-s, Q/e = +1, B = 0, S = +1, rest energy 494 MeV; p = uud, Q/e = +1, B = 1, S = 0, rest energy 938 MeV; π⁻ = d anti-u, Q/e = −1, B = 0, S = 0, rest energy 140 MeV; π⁺ has Q/e = +1, B = 0, S = 0, rest energy 140 MeV. All leptonic family numbers are zero. Values are deliberately rounded supplied teaching data, not a new experimental mass measurement.

**Worked strong-production check.** Test π⁻ + p → K⁺ + Λ⁰. Initial (Q/e,B,S) = (0,1,0); final = (1,1,0). Strangeness balances, but charge fails, so reject. Now test π⁻ + p → K⁺ + Λ⁰ + π⁻. Both sides give (0,1,0). The initial rest-energy sum is 1078 MeV and final rest-energy sum 1750 MeV. At least 672 MeV of extra centre-of-momentum energy above the initial rest sum is required. This is not a fixed-target beam threshold: momentum conservation makes that threshold larger. Additive conservation is necessary, not proof that a reaction occurs or dominates.

**Worked weak-decay check.** Λ⁰ → p + π⁻ gives charge 0 = +1−1, B 1 = 1+0 and ΔS = +1. The final rest sum is 1078 MeV, leaving 38 MeV for final kinetic energy in the Λ rest frame. The two momenta must be equal and opposite; the energies are not necessarily equal because the masses differ. A strong decay would violate its strangeness conservation, whereas the specified weak change is permitted. Do not infer a lifetime from the ledger alone.

**Family trap.** Compare μ⁻ → e⁻ + anti-νₑ + νμ with μ⁻ → e⁻ + anti-νμ + νₑ. Both balance charge and total lepton number +1, but only the first balances electron and muon families separately in this curriculum model. For the second, electron-family final number is +2 and muon-family final number is −1, versus initial 0 and +1. Each family account must be written. Neutrino oscillations are beyond this assessed bookkeeping model; they should not be invoked to rescue the wrong elementary decay equation.

Worked example: Determine Q/e, B and S for a meson with composition ds̄ and identify the required kaon.

Q/e = −1/3 + 1/3 = 0.

B = +1/3 − 1/3 = 0.

S = 0 + 1 = +1, matching K⁰ rather than K̄⁰.

Answer: K⁰: Q = 0, B = 0, S = +1.

Check your understanding: In neutron beta-minus decay, which valence quark changes?

Answer: A down quark changes into an up quark.

Why: udd becomes uud; charge conservation also needs the electron, and lepton-number conservation needs the antineutrino.