Skip to main content

Nuclear, particle & quantum-information sequence · Interactive lesson

Quark flavours and additive quantum numbers

Before you write an amplitude you have to say what the state is. This topic hands you one counting operator per flavour, shows how they fold into baryon number, isospin and hypercharge, and turns the whole set into a single charge relation that no hadron in the tables violates.

UPV-17.04 · Lesson 4 of 9 in this unit. Position is not a completion score.
Starting model: Quark flavours and additive quantum numbersThe (I₃, Y) weight plane, with three dashed lines of constant integer charge Q = I₃ + Y/2 running at slope −2. A content of 2 up, 1 down and 0 strange quarks gives B = 1.00 and S = 0, hence Y = 1.00 and I₃ = 0.5, placing the dot at Q = 1.00. The open ring appears only when that charge is an integer.Q = +1Q = 0Q = −1I₃Ycontent u² d¹ s⁰B = 1.00 S = 0Y = B + S = 1.00I₃ = 0.5Q = I₃ + Y/2 = 1.00ring drawn only for integer Q
Start with a picture. Change the model in the lab below.

Nuclear, particle & quantum-information sequence. These units continue into nuclear and particle physics and quantum information. Follow your university’s prerequisites, mathematical depth and assessment; this independent pathway is not a universal programme specification.

1

Build the model

Every flavour label is a diagonal operator you can count.

The key idea

An additive quantum number of a hadron is the eigenvalue of a diagonal, Hermitian flavour-counting operator: add one unit for each quark of that flavour, subtract one for each antiquark, and the total labels the state.

Example. The Ω⁻ is sss, so S = −3 and B = +1, giving Y = B + S = −2; with I₃ = 0 the relation Q = I₃ + Y/2 returns −1, exactly the charge measured.

The complete physical model and assumptions

The six flavours span five orders of magnitude in mass, so no single non-abelian symmetry usefully relates all of them. What survives is abelian. On the six-dimensional flavour space spanned by |u⟩ … |t⟩, each flavour carries a number operator Nf = nf − nf̄ that is diagonal, Hermitian and commutes with every other one, so a hadron can be a simultaneous eigenstate of all six, and when it is, each eigenvalue is the plain sum over its constituents — additive, unlike parity, which multiplies, or spin, which needs Clebsch-Gordan coefficients. Tradition dresses the counts in names: S = −Nₛ, C = +Nc, B′ = −Nb, T = +Nₜ, each label taking the sign of its quark's electric charge, alongside B = (1/3)Σf Nf. The payoff is that electric charge itself belongs to this commuting set, Q = (2/3)(Nᵤ + Nc + Nₜ) − (1/3)(Nd + Nₛ + Nb), so the generalised Gell-Mann–Nishijima relation Q = I₃ + Y/2 with Y = B + S + C + B′ + T is a linear identity among diagonal operators, not an empirical law awaiting a test. The cost is that nothing here is derived. The assignments are read off a spectrum, the symmetry that would relate multiplet partners is broken by mass, the labels are conserved only by the strong and electromagnetic Hamiltonians, and the masses themselves are scheme- and scale-dependent parameters rather than observables.

1Six flavours, six commuting counting operators

Work in the six-dimensional flavour space spanned by |u⟩, |d⟩, |s⟩, |c⟩, |b⟩ and |t⟩, together with its conjugate spanned by the six antiquarks. For each flavour define the number operator Nf = nf − nf̄, diagonal in that basis with eigenvalue +1 on its own quark, −1 on its own antiquark and 0 on the other ten single-particle states. Six diagonal matrices commute pairwise, so a state can be a simultaneous eigenstate of all of them, and for a composite state each eigenvalue is the plain sum over its constituents. That is what additive means, and it is the exception rather than the rule: parity multiplies, C multiplies, and angular momenta need Clebsch-Gordan coefficients. Electric charge follows the same additive pattern, +2/3 for u, c and t against −1/3 for d, s and b, and so does baryon number at +1/3 per quark. The proton uud therefore carries B = 1 and Q = 2/3 + 2/3 − 1/3 = +1 by counting alone, with no dynamics anywhere in the argument.

2Why the s quark carries S = −1

The flavour labels predate the quarks. Strangeness was assigned in the early 1950s so that the K⁺ came out with S = +1, and when the quark model arrived the K⁺ turned out to be us̄ — which forces S = −1 on the s quark itself. The surviving rule is worth one line of memory: each flavour label takes the sign of that quark's electric charge. The down-type quarks s and b, at charge −1/3, carry S = −1 and B′ = −1; the up-type quarks c and t, at +2/3, carry C = +1 and T = +1. Antiquarks reverse every one of them. So D⁰ = cū has C = +1, B⁻ = bū has B′ = −1, and Λ⁰ = uds has S = −1. The convention looks perverse until you see what it buys: with these signs one formula covers every hadron in the tables, and with the signs flipped that same formula fails for every hadron carrying net strangeness, charm or bottomness.

3Hypercharge collects everything but isospin

Define Y = B + S + C + B′ + T, the sum of every additive label except the isospin component, and the claim is Q = I₃ + Y/2. This is not an experimental law but a linear identity among diagonal operators, and you can check it in three lines. Put B = (1/3)Σf Nf, I₃ = (1/2)(Nᵤ − Nd), S = −Nₛ, C = Nc, B′ = −Nb and T = Nₜ into the definition. Hypercharge becomes Y = (1/3)Nᵤ + (1/3)Nd − (2/3)Nₛ + (4/3)Nc − (2/3)Nb + (4/3)Nₜ. Adding I₃ to Y/2 gives (2/3)Nᵤ − (1/3)Nd − (1/3)Nₛ + (2/3)Nc − (1/3)Nb + (2/3)Nₜ, which is the charge operator itself. Test it on states: K⁺ = us̄ has B = 0 and S = +1, so Y = 1, and I₃ = +1/2, giving Q = +1. D⁰ = cū has B = 0, C = +1, so Y = 1, while the ū supplies I₃ = −1/2, giving Q = 0.

4Isospin is the only non-abelian piece left

Why does the u–d pair get a doublet while every heavier flavour gets its own U(1)? Mass. At μ = 2 GeV in MS-bar the up and down masses are about 2.16 and 4.67 MeV, a difference far below ΛQCD ≈ 200 MeV, so the strong Hamiltonian barely distinguishes them and SU(2) isospin is a good approximate symmetry. The strange quark near 93 MeV is comparable to ΛQCD, so SU(3) flavour is broken but still useful: each unit of strangeness shifts the baryon decuplet by a nearly constant step, 1232 → 1385 → 1532 → 1672 MeV, spacings of 153, 147 and 140 MeV. The charm, bottom and top masses — 1.27, 4.18 and about 172.5 GeV — are hopeless, which is why nobody draws SU(4) multiplets and means it. What is left is the Cartan pair (I₃, Y): two commuting charges, hence the two-dimensional weight diagrams on which every SU(3) multiplet is plotted.

5Which Hamiltonian conserves which label

Every additive flavour number commutes with the strong and electromagnetic Hamiltonians, so those interactions conserve each one separately. The charged weak current does not: it turns one flavour into another, and at tree level it changes any one flavour number by at most one unit, so a first-order weak decay has |ΔS| ≤ 1 and |ΔS| = 2 requires second order — which is what makes K⁰–K̄⁰ mixing so slow. Reading a lifetime then reduces to checking the ledger. Δ⁺⁺ → pπ⁺ changes no flavour number and goes strong, in 5.6 × 10⁻²⁴ s. Σ⁰ → Λ⁰γ conserves strangeness but needs a photon, so it goes electromagnetic, in 7.4 × 10⁻²⁰ s. Λ⁰ → pπ⁻ has ΔS = +1 and must go weak, in 2.6 × 10⁻¹⁰ s: nearly fourteen orders of magnitude of lifetime read off one conservation law. Charge, baryon number and colour survive all three interactions; the individual flavour numbers do not.

6Masses are parameters, and topness never labels a hadron

Everything above is bookkeeping on labels that are integers or simple fractions. Mass is not like that. A quark is never isolated, so there is no rest frame to weigh it in; what appears in the Lagrangian is a renormalised parameter depending on both scheme and scale. The MS-bar values mᵤ ≈ 2.16 MeV, md ≈ 4.67 MeV and mₛ ≈ 93 MeV are quoted at μ = 2 GeV, and one-loop running carries mₛ to roughly 117 MeV at μ = 1 GeV — the same quark, a different number. Nor do they add up to anything you weigh: 2mᵤ + md = 8.99 MeV is under 1% of the proton's 938.3 MeV, the rest being the kinetic and field energy of the confined quarks and gluons, while the constituent masses near 336 MeV used in magnetic-moment fits are a different object again. The top makes the point sharpest: its width near 1.48 GeV gives τ ≈ 4.4 × 10⁻²⁵ s, some seven times shorter than the ħ/ΛQCD ≈ 3.3 × 10⁻²⁴ s the strong interaction needs to dress it, so topness is a perfectly good quantum number that no hadron ever carries.

2

Key equations

Read the units, assumptions and limits before substituting.

Generalised Gell-Mann–Nishijima relation

Q = I₃ + Y/2, Y = B + S + C + B′ + T

Fixes the charge from flavour counting alone, so an assignment that fails it is wrong before any dynamics is written.

All dimensionless, Q in units of e. Y is strong hypercharge, B baryon number, I₃ the third isospin component.

Flavour counting operators

Nf = nf − nf̄S = −Nₛ, C = +Nc, B′ = −Nb, T = +Nₜ

Turns a quark content into every additive label by counting, with antiquarks subtracting rather than adding.

Integers. Each label takes the sign of that quark's electric charge, so the s quark itself carries S = −1.

Electric charge as a diagonal operator

Q = (2/3)(Nᵤ + Nc + Nₜ) − (1/3)(Nd + Nₛ + Nb)

Substituting the definitions of I₃, B and Y turns this line into Q = I₃ + Y/2, which is why that relation cannot fail.

Diagonal in the |u⟩ … |t⟩ basis, so [Q, Nf] = 0 and every flavour eigenstate is already a charge eigenstate.

Baryon number and the isospin doublet

B = (1/3) Σf Nf, I₃ = (1/2)(Nᵤ − Nd)

B is the U(1) trace part of the flavour charges; I₃ is the only one that generates a good approximate non-abelian symmetry, SU(2) isospin.

B is +1/3 per quark, −1/3 per antiquark. Only u and d form a doublet; s, c, b and t are singlets with I = I₃ = 0.

What conserves flavour, and what does not

[Hstrong, Nf] = [Hₑₘ, Nf] = 0, [Hweak, Nf] ≠ 0

Lets you read the interaction off a lifetime: a strangeness-changing decay is weak, so 10⁻¹⁰ s or longer, not 10⁻²³ s.

The charged weak current changes a flavour number by at most one unit at tree level, so |ΔS| ≤ 1; ΔQ = ΔB = 0 for all three interactions.

Quark mass runs with the scale

m̄(μ) = m̄(μ₀)[αₛ(μ)/αₛ(μ₀)]κκ = 12/(33 − 2nf)

Shows the mass is a Lagrangian parameter tied to a scheme and a scale, never a rest energy you could weigh.

MS-bar masses in MeV or GeV, κ dimensionless, one loop, nf active flavours. With αₛ ≈ 0.30 at 2 GeV, 0.50 at 1 GeV and nf = 3, mₛ(2 GeV) = 93 MeV becomes about 117 MeV at 1 GeV.
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

2
0 to 3; steps of 1.
1
0 to 3; steps of 1.
0
0 to 3; steps of 1.

Try this. Start from the proton, two up and one down, sitting on the Q = +1 line. Trade one up for a strange quark: I₃ falls by a half, Y by a whole unit, and the dot lands on Q = 0 as the Λ⁰. Push the strange count to three with no light quarks and it drops to Y = −2 on Q = −1, the Ω⁻.

Live results

BARYON NUMBER B
1.00
STRANGENESS S
0
HYPERCHARGE Y
1.00
CHARGE Q
1.00

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

Interactive Physics V modelThe (I₃, Y) weight plane, with three dashed lines of constant integer charge Q = I₃ + Y/2 running at slope −2. A content of 2 up, 1 down and 0 strange quarks gives B = 1.00 and S = 0, hence Y = 1.00 and I₃ = 0.5, placing the dot at Q = 1.00. The open ring appears only when that charge is an integer.Q = +1Q = 0Q = −1I₃Ycontent u² d¹ s⁰B = 1.00 S = 0Y = B + S = 1.00I₃ = 0.5Q = I₃ + Y/2 = 1.00ring drawn only for integer Q
The (I₃, Y) weight plane, with three dashed lines of constant integer charge Q = I₃ + Y/2 running at slope −2. A content of 2 up, 1 down and 0 strange quarks gives B = 1.00 and S = 0, hence Y = 1.00 and I₃ = 0.5, placing the dot at Q = 1.00. The open ring appears only when that charge is an integer.

Predict → change one variable → check

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

Saved baseline. Up quarks nᵤ: 2; Down quarks nd: 1; Strange quarks nₛ: 0

Baseline 2; choose 0 to 3, in steps of 1.
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 commonest error is to hand the s quark S = +1, on the grounds that it is the strange one. The convention runs the other way: each flavour label takes the sign of that quark's electric charge, so the s quark carries S = −1 and the s̄ carries S = +1 — which is exactly what K⁺ = us̄ needs to keep the S = +1 that Gell-Mann assigned it a decade before the quark model. Flip that sign and everything downstream fails quietly rather than loudly. The Λ⁰ = uds comes out with Y = B + S = +2 instead of 0, so Q = I₃ + Y/2 returns +1 for a neutral baryon; the Ω⁻ moves from Y = −2 to Y = +4 and is assigned charge +2; and the relation looks broken. It is not. Q = I₃ + Y/2 is an identity among diagonal operators and cannot fail, so when it appears to, the sign convention is what to check.

5

Practice & worked examples

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

EasyExample 1

The Dₛ⁺ meson has quark content c s̄. Assign its baryon number, strangeness, charm and third isospin component, then use Q = I₃ + Y/2 to confirm its charge.

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 Ω⁻ is sss, of mass 1672.45 MeV. (a) Give S, B, Y, I₃ and Q. (b) Its lightest strangeness-conserving hadronic channel is Ξ⁰K⁻, with masses 1314.86 and 493.68 MeV. Decide which interaction must drive the decay, and estimate the lifetime scale.

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.

HardExample 3

The Standard-Model top width is Γₜ = GF mₜ³ (1 − x)² (1 + 2x) |Vtb|² / (8π√2), with x = MW²/mₜ². Take mₜ = 172.5 GeV, MW = 80.38 GeV, GF = 1.1664 × 10⁻⁵ GeV⁻² and |Vtb| = 1. Find Γₜ and τₜ, compare with the hadronisation time ħ/ΛQCD at ΛQCD ≈ 0.2 GeV, and say what this does to topness as a hadron label.

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.

6

Quick check

Select an answer and explain it before checking.

A baryon is built from one up quark and two strange quarks. Which line gives its third isospin component, hypercharge and charge correctly?
7

Review & continue

An additive quantum number of a hadron is the eigenvalue of a diagonal, Hermitian flavour-counting operator: add one unit for each quark of that flavour, subtract one for each antiquark, and the total labels the state.

  • 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.