Lecture 7: the Quark Model and SU(3)Flavor

Lecture 7: The Quark Model and SU(3)flavor

September 13, 2018 Review: Isospin

• Can classify hadrons with similar mass (and same spin & P) but diﬀerent charge into multiplets • Examples:

 π+  p N ≡ Π ≡ π0 n   π− p = 1 1 2 2 π+ = |1, 1i π0 = |1, 0i n = 1 − 1 2 2 π− = |1, −1i

• Isospin has the same algebra as spin: SU(2)

I Can confirm this by comparing decay or scattering rates for different members of the same isomultiplet I Rates related by normal Clebsh-Gordon coefficients Review: Strangeness

• In 1950’s a new class of hadrons seen

I Produced in πp interaction via Strong Interaction I But travel measureable distance before decay, so decay is weak ⇒ ∃ conserved quantum number preventing the strong decay Putting Strangeness and Isospin together

• Strange hadrons tend to be heavier than non-strange ones with the same spin and parity J P Name Mass (MeV) 0− π± 140 K± 494 1− ρ± 775 K∗± 892

• Associate strange particles with the isospin multiplets Adding Particles to the Axes: Pseudoscalar Mesons

• Pions have S = 0 • From π−p → Λ0K0 deﬁne K0 • Three charge states ⇒ I = 1 to have S = 1 • • Draw the isotriplet: If strangeness an additive quantum number, ∃ anti-K0 with S = −1 • Also, K+ and K− must be particle-antiparticle pair: (eg from φ → K+K−)

But this is not the whole story There are 9 pseudoscalar mesons (not 7)! The Pseudoscalar Mesons

• Will try to explain this using group theory Introduction to Group Theory (via SU(2) Isospin)

• Fundamental SU(2) representation: a doublet

u 1 0 χ = so u = d = d 0 1

• Inﬁnitesmal generators of isospin rotations:

0 1 0 −i 1 0 τ = , τ = , τ = 1 1 0 2 i 0 3 0 −1

• τ matrices satisfy commutation relations 1 1 1 τi, τj = iijkτk 2 2 2 Commutation relations deﬁne the SU(2) algebra • ∃ higher representations: N × N matrices with N = 2I + 1 • Also, there is an operator that commutes with all the τ’s: 2 2 1 1 2 I = ~τ = Σi τ 2 4 i and there are raising and lowering operators 1 τ± = (τ1 ± iτ2) 2 Raising and Lowering Operators

• In Quantum Mechanics, can start with state |J, Jz = Ji and construct all other states of same J using lowering operator J− + • Similarly with Isopsin, if start with π = |I = 1,Iz = 1i and 0 − construct π and π using lowering operator τ− • If we introduce strangeness, can we navigate among all the mesons in the same way?

I Need second lowering operator to navigate in the S direction

Extend group from SU(2) to SU(3) Group Theory Interpretation of Meson Spectrum

• Raising and lowering operators to navigate around the multiplet • Gell Man and Zweig: Patterns of multiplets explained if all hadrons were made of quarks

I Mesons: qq 3 ⊗ 3 = 1 ⊕ 8 I Baryons: qqq 3⊗3⊗3 = 1⊕8⊕8⊕10 • Particles with same spin, parity • In those days, 3 flavors and charge conjugation (extension to more flavors symmetry described as multiplet discussed later) • Are the quarks real or just I Different Iz and S (or constructs? More on that next Y = B + S) week Defining SU(3)

• SU(3): All unitary transformations on 3 component complex vectors without the overall phase rotation (U(1))

U†U = UU† = 1 det U = 1  8  X U = exp i λaθa/2 a=1

• The fundamental representation of SU(3) are 3 × 3 matrices

 0 1 0   0 −i 0   1 0 0  λ1 =  1 0 0  λ2 =  i 0 0  λ3 =  0 −1 0  0 0 0 0 0 0 0 0 0  0 0 1   0 0 −i   0 0 0  λ4 =  0 0 0  λ5 =  0 0 0  λ6 =  0 0 1  1 0 0 i 0 0 0 1 0  0 0 0   1 0 0  1 λ7 =  0 0 −i  λ8 = √  0 1 0  0 i 0 3 0 0 −2

• Commutation relations: λa λb λc , = ifabc 2 2 2 1 1 where f123 = 1, f147 = f246 = f257 = f345 = 2 , f156 = f367 = − 2 and √ f458 = f678 = 3/2. SU(3) Raising and Lowering Operators

• SU(3) contains 3 SU(2) subgroups embedded in it

Isospin : λ1 λ2 √ λ3 U − spin : λ6 λ7 √3λ8 − λ3 V − spin : λ λ 3λ + λ 4 5 8 3 • Fundamental representation: A triplet • For each subgroup, can form raising and lowering operators • Deﬁne group structure starting at one corner and using raising and • Any two subgroups enough to lowering operators navigate through multiplet • Deﬁne “highest weight state” ψ as state where both I+ψ = 0 and V+ψ = 0 • Quarks (u, d, s) have p = 1, q = 0 while antiquarks (u, d, s) have p = 0, q = 1

p+1 (V−) φmax = 0 q+1 (I−) φmax = 0 structure : (p, q) More on Quarks and SU(3)

• Remember from last time: Y Q = I + 3 2 • All the quarks have Baryon number 1/3 (Baryons are qqq states) • The quark quantum numbers are:

Quark B Q I3 SY u 1/3 2/3 1/2 0 1/3 d 1/3 −1/3 −1/2 0 1/3 s 1/3 −1/3 0 -1 −2/3 u −1/3 −2/3 −1/2 0 −1/3 d −1/3 1/3 1/2 0 −1/3 s −1/3 1/3 0 1 2/3 Combining SU(3) states: 2 quarks

• Combining two SU(3) objects gives 3 × 3 = 9 possible states

uu √1 (ud + du) √1 (ud − du) 2 2 dd √1 (us + su) √1 (us − su) 2 2 ss √1 (ds + sd) √1 (ds − sd) 2 2 6 3 symmetric anti − symmetric

3 ⊗ 3 = 6 ⊕ 3

• Triplet is a 3 (not a 3): 1 I √ (ud − du) has I = 0, Iz = 0, and Y = 2/3 2 1 1 1 I √ (us − su) has I = , Iz = and Y = −1/3 2 2 2 1 1 1 I √ (ds − sd) has I = , Iz = − and Y = −1/3 2 2 2 Combining SU(3) states (adding a 3rd quark)

• 3 ⊗ 3 ⊗ 3 = 3 ⊗ (6 ⊕ 3) = 10s ⊕ 8M,S ⊕ 8M,A ⊕ 1 • Start with the fully symmetric combination with the 6:

uuu 3 such states √1 (ddu + udd + dud) 6 such states 3 √1 (dsu + uds + sud + sdu + dus + usd) 1 such state 6 • Now, the mixed symmetry combination with the 6:

√1 [(ud + du)u − 2uud] 8 such states 6 • Now the 3:

√1 [(ud − du)s + (usd − dsu) + (du − ud)s] 8 such states 6 • One totally antisymmetric state

ijkqiqjqk Combining SU(3) states (qq)

• Start with π+ = u d • Now add strange quarks: 4 combinations • Using: us ds us ds I |ui = − d − K+ K0 K− K0

I − d = + |ui • One missing combination we can construct using raising or lowering We ﬁnd: operators 1 √ I− ud = √ dd − |uui 0 2 (dd + uu − 2ss)/ 6 ≡ η = |I = 1 I = 0i 3 These 8 states are called an octet 0 1 π = √ dd − |uui • 2 One additional independent combination: the singlet state Doing this again: π− = d u √ (uu + dd + ss)/ 6 Pseudoscalar Mesons (J P = 0−)

Example II3 S Meson Combo Decay Mass (MeV) 1 1 0 π+ ud µ+ν 140 1 0 0 π0 √1 (dd − uu) γγ 135 2 1 −1 0 π− du µ−ν 140 1 1 + + 2 2 +1 K us µ ν 494 1 1 0 2 − 2 +1 K ds ππ 498 1 1 − − 2 2 −1 K us µ ν 494 1 1 0 2 − 2 −1 K ds ππ 498 0 0 0 η √1 (dd + uu − 2ss) see next page 8 6 0 0 0 η √1 (dd + uu + ss) see next page 0 3 Observations about the Pseudoscalar Mesons

• Mass of strange mesons larger than non-strange by about 170 MeV

I Strange quark has a larger mass than up and down I Leads to SU(3) breaking in H

• η8 and η0 would be degenerate if SU(3) were perfect symmetry

I Degenerate PT: The states can mix. Physical states are: • η: Mass=549 Decay: η → 2γ • η0: Mass=958 Decays:η0 → ηππ or 2γ Vector Mesons (J P = 1−)

II3 S Meson Combo Decay Mass (MeV) 1 1 0 ρ+ ud π+π0 776 1 0 0 ρ0 √1 (dd − uu) π+π− 776 2 1 −1 0 ρ− du π−π0 776 1 1 ∗+ 2 2 +1 K us Kπ 892 1 1 ∗0 2 − 2 +1 K ds Kπ 892 1 1 ∗− 2 2 −1 K us Kπ 892 1 1 ∗0 2 − 2 −1 K ds Kπ 892 0 0 0 ω √1 (uu + dd) 783 3π 2 0 0 0 φ ss 1019 KK

See comments on next page Comments on Vector Mesons (1−)

• Unlike the pseudoscalars which decay weakly, the vectors can decay strongly, eg

ρ0 → π+π− K∗0 → K+π−

• Octet-nonet mixing is maximal in the case of the vector mesons

I The φ is all ss while the ω is all uu and dd 3− Baryon Decouplet (J = 2 )

• Here is the ∆ that we talked about earlier • These are strongly decaying resonances 1− Baryon Octet (J = 2 )

• Lightest baryons • Decay weakly (except proton which is stable) Comments on Antiparticles

• For mesons, particle and antiparticle are in the same multiplet − + I π is antiparticle of π 0 I π is its own antiparticle I The multiplet is called “self-charge conjugate” • For baryons, the antiparticles are in diﬀerent multiplets

I 10 ⇒ 10 I 8 ⇒ 8 I S ⇒ −S; Iz ⇒ −Iz • Upper right corner moves to lower left

I Baryon number = 1 ⇒ Baryon number = −1 Fermi Statistics: Why Color

• Imposition of Fermi Statistics on Baryon States ++ I ∆ = uuu, spin=3/2, s-wave: These are all symmetric under interchange I Need another degree of freedom to antisymmetrize (must have at least 3 possible states, since we are antisymmetrizing 3 objects) • We’ll in 2 weeks, that the QCD Lagrangian is deﬁned by SU(3)color interaction

I Gluons are color octets I Observable hadrons are color singlets I The color singlet states are anti-symmetric under color exchange • This solves the Fermi statistics problem • All hadrons are color singlets • Thus quark wave function before adding color is symmetric SU(3) Breaking and Mass Relations

• SU(3) symmetry: all members of multiplet have same mass

I Mass depends on binding energy: cannot calculate, since perturbative calculations not possible for low energy QCD • Reasons why the physical masses in multiple are diﬀerent

I Diﬀerence in quark masses

• md > mu by a few MeV, ms heavier by ∼ 170 MeV

I Coulomb energy diﬀerence associated with the electrical energy between pairs of quarks 2 • Of order e /R0. With R0 ∼ 0.8 fm, this ∼ 2 MeV

I Magnetic energy differences associated with the magnetic moments of the quarks (hyperfine interaction) Vector-Pseudoscalar Meson Mass Differences

0 0 • mπ = 135 MeV, mρ = 775 MeV but SU(3) wave function is the same • Only diﬀerence is the spin of the particles • Both are ` = 0 states of the qq pair • Diﬀerence is spin: S = 0 or S = 1 • Consider magnetic dipole energy

αs U ∝ S~1 · S~2 mimj

• Meson mass is A m(q1q2) = m1 + m2 + S~1 · S~2 mimj

• Good agreement with data using

mu, md = 0.307 GeV ms = 0.490 GeV A = 0.06 GeV Some comments on quark masses

• Mass values from previous page not universal • Quarks are bound in hadrons: the quark “mass” needed in these calculations affected by binding energy • Baryons and mesons have different wave functions and hence different binding energies

I Thus slightly different effective masses • Above masses called “constituent mass” • Completely different from “current mass” that is used in QFT Lagrangian

I Current masses for quarks are a few MeV Baryon Masses

• Similar idea as for mesons D E D E D E S~1 · S~2 S~1 · S~3 S~2 · S~3 0 m(q1q2q3) = m1 +m2 +m3 +A   m1m2 m1m3 m2m3

• Good agreement with data using

mu, md = 0.365 GeV ms = 0.540 GeV A = 0.026 GeV

• Notice these values aren’t the same as for the mesons. Baryon Magnetic Moments (I)

• Baryons are not Dirac particles

I They have structure (quark bound states) • Magnetic moment diﬀers from that of pointlike fermions

I µproton = 2.792µN and µneutron = −1.913µN I where µN is the nuclear magneton e~ µN = 2mp • Quarks are Dirac particles so for a given spin-state we can calculate the magnetic moment. Thus for spin up quarks 2 e~ µu = hu ↑ |µˆz|u ↑i = 3 mu −1 e~ µd = hd ↑ |µˆz|d ↑i = 3 mu

• Hadron magnetic moments can be built from the quarks Baryon Magnetic Moments (II)

• Combine the 3 quarks by ﬁrst combining 2 and then adding the third • Since color state asymmetric quark wave function without color is symmetric • Let’s see how this works for the protom

I Isopsin of the first two quarks • I = 0: Antisymmetric state. Two quarks must be different 1 ψ0 = √ (ud − du) 2 1 • I = 1: Symmetric state ψ1: uu, √ (ud − du), dd 2 1 1 I Combine with 3rd quark and require I = 2 , Iz = 2 • I = 0: Third quark must be a u • I = 1: Use Clebsch-Gordon coeff. 3-quark state is p2/3d(uu) − p1/3u( √1 (ud − du)), 2 I Now add spin which must be antisymmetric for ψ0 and symmetric for ψ1 I Final result for spin up proton:

Φ = p1/8 [4/3uud ↑↑↓ −2/3uud ↑↓↑ −2/3uud ↓↑↑ −2/3udu ↑↑↓ +4/3udu ↑↓↑ −2/3udu ↓↑↑ −2/3duu ↑↑↓ −2/3duu ↑↓↑ +4/3duu ↓↑↑] Baryon Magnetic Moments (III)

• Combining terms from previous page: 1 |p ↑i = √ (2u ↑ u ↑ d ↓ −u ↑ u ↓ d ↑ −u ↓ u ↓ d ↑) 6

• This gived 4 1 1 µ = (µ + µ − µ ) + (µ − µ + µ ) + (−µ + µ + µ ) p 6 u u d 6 u u d 6 u u d 4 1 = µ − µ 3 u 3 d

• Similarly 4 1 µ = µ − µ n 3 d 3 u

• Predictions reasonably reproduce data values