florida state universityhadrons in the quark modela winston roberts wroberts@fsu.edu florida state...

Hadrons in the Quark Modela

Winston Roberts

wroberts@fsu.edu

Florida State University

a24th Annual Hampton University Graduate Studies Program, 05/31/09 to 06/20/09, Newport News, VA.

Hadrons in the Quark Modela – p.

Hadrons in the Quark Modela – p.

• Introduction, Classification Scheme, Simple Predictions

• Meson Spectrum

• Baryons Spectrum

• Meson Transitions

• Baryons Transitions & Sundries

Hadrons in the Quark Modela – p.

What is a quark?In Finnegan’s Wake (J. Joyce), ‘three quarks for Muster Mark’

Hadrons in the Quark Modela – p.

Six flavors of quarks,

(

u

d

)

,

(

c

s

)

,

(

t

b

)

,

(

23

−13

)

Hadrons in the Quark Modela – p.

Six flavors of quarks,

(

u

d

)

,

(

c

s

)

,

(

t

b

)

,

(

23

−13

)

Masses: u ≈ 4 MeV, d ≈ 7 MeV, s ≈ 150 MeV, c ≈ 1.5 GeV, b ≈ 5 GeV, t ≈ 180 GeV.

Hadrons in the Quark Modela – p.

Six flavors of quarks,

(

u

d

)

,

(

c

s

)

,

(

t

b

)

,

(

23

−13

)

Masses: u ≈ 4 MeV, d ≈ 7 MeV, s ≈ 150 MeV, c ≈ 1.5 GeV, b ≈ 5 GeV, t ≈ 180 GeV.Note u and d almost degenerate, and their masses are much smaller than typical energyscale (a few hundred MeV) of the strong interaction. Can treat as two states of sameparticle −→ isospin, described in terms of SU(2).

Hadrons in the Quark Modela – p.

Six flavors of quarks,

(

u

d

)

,

(

c

s

)

,

(

t

b

)

,

(

23

−13

)

Masses: u ≈ 4 MeV, d ≈ 7 MeV, s ≈ 150 MeV, c ≈ 1.5 GeV, b ≈ 5 GeV, t ≈ 180 GeV.Note u and d almost degenerate, and their masses are much smaller than typical energyscale (a few hundred MeV) of the strong interaction. Can treat as two states of sameparticle −→ isospin, described in terms of SU(2).Extend to include s quark, treat using SU(3): each quark is a member of a flavor triplet.

Hadrons in the Quark Modela – p.

Six flavors of quarks,

(

u

d

)

,

(

c

s

)

,

(

t

b

)

,

(

23

−13

)

Masses: u ≈ 4 MeV, d ≈ 7 MeV, s ≈ 150 MeV, c ≈ 1.5 GeV, b ≈ 5 GeV, t ≈ 180 GeV.Note u and d almost degenerate, and their masses are much smaller than typical energyscale (a few hundred MeV) of the strong interaction. Can treat as two states of sameparticle −→ isospin, described in terms of SU(2).Extend to include s quark, treat using SU(3): each quark is a member of a flavor triplet.The only hadrons we know (those that have been confirmed non-controversially) aremesons (quark and antiquark) and baryons (three quarks).

Hadrons in the Quark Modela – p.

A short aside on quantum numbers

Hadrons in the Quark Modela – p.

A short aside on quantum numbers

Quarks have spin 1/2, positive parity, antiquarks have negative parity.

Hadrons in the Quark Modela – p.

A short aside on quantum numbers

Quarks have spin 1/2, positive parity, antiquarks have negative parity.

Two quarks, or a quark-antiquark pair, can have total spin 0 or 1.

Hadrons in the Quark Modela – p.

A short aside on quantum numbers

Quarks have spin 1/2, positive parity, antiquarks have negative parity.

Two quarks, or a quark-antiquark pair, can have total spin 0 or 1.

A baryon with no orbital angular momentum (S-wave) can have total spin 1/2 or 3/2.

Hadrons in the Quark Modela – p.

A short aside on quantum numbers

Quarks have spin 1/2, positive parity, antiquarks have negative parity.

Two quarks, or a quark-antiquark pair, can have total spin 0 or 1.

A baryon with no orbital angular momentum (S-wave) can have total spin 1/2 or 3/2.

For a meson with orbital angular momentum L between the quark and antiquark, theparity is +1 × (−1) × (−1)L = (−1)L+1

Hadrons in the Quark Modela – p.

A short aside on quantum numbers

Quarks have spin 1/2, positive parity, antiquarks have negative parity.

Two quarks, or a quark-antiquark pair, can have total spin 0 or 1.

A baryon with no orbital angular momentum (S-wave) can have total spin 1/2 or 3/2.

For a meson with orbital angular momentum L between the quark and antiquark, theparity is +1 × (−1) × (−1)L = (−1)L+1

For ground state baryons, the parity is positive; more on quantum numbers later

Hadrons in the Quark Modela – p.

For mesons (pseudoscalar, L = 0, S = 0, JP = 0−):

3 ⊗ 3 = 1 ⊕ 8

Hadrons in the Quark Modela – p.

rgr r

r r

r r

π0[

1√2

(

uu − dd)

]

η8

[

1√6

(

uu + dd − 2ss)

]

π− [du] π+[

−ud]

K0 [ds] K+ [us]

K− [su] K0[

−sd]

η1

[

1√3

(

uu + dd + ss)

]

Hadrons in the Quark Modela – p.

All (light) mesons fall into these multiplets. Thus, for the vectormesons (L = 0, S = 1, JP = 1−)

Hadrons in the Quark Modela – p.

rgr r

r r

r r

ρ0[

1√2

(

uu − dd)

]

φ8

[

1√6

(

uu + dd − 2ss)

]

ρ− [du] ρ+[

−ud]

K0∗ [ds] K+∗ [us]

K−∗ [su] K0∗ [−sd

]

φ1

[

1√3

(

uu + dd + ss)

]

Hadrons in the Quark Modela – p.

Hadrons in the Quark Modela – p.

For baryons:

3 ⊗ 3 ⊗ 3 = (6 ⊕ 3) ⊗ 3 = 10 ⊕ 8 ⊕ 8 ⊕ 1

Hadrons in the Quark Modela – p.

rgr r

r r

r r

Σ0 [uds]

Λ [uds]

Σ− [dds] Σ+ [uus]

n [udd] p [uud]

Ξ− [dss] Ξ0 [uss]

Hadrons in the Quark Modela – p.

rr r

r r rr

r r

r

∆− [ddd] ∆0 [udd]

Σ∗0 [uds]Σ∗− [dds] Σ∗+ [uus]

∆+ [uud] ∆++ [uuu]

Ξ∗− [dss] Ξ∗0 [uss]

Ω− [sss]

Hadrons in the Quark Modela – p.

Imagine a simplified strong-interaction Hamiltonian Hs. The mass eigenvalues for vectormesons can be worked out as

Hadrons in the Quark Modela – p.

Imagine a simplified strong-interaction Hamiltonian Hs. The mass eigenvalues for vectormesons can be worked out as

〈K∗|Hs|K∗〉 >= 〈ds|Hs|ds〉 ≡ m1 + d+ s ≡ mK∗ ,

where m1 is an eigenvalue assumed to be common to all members of the multiplet(SU(3) symmetry).

Hadrons in the Quark Modela – p.

Imagine a simplified strong-interaction Hamiltonian Hs. The mass eigenvalues for vectormesons can be worked out as

〈K∗|Hs|K∗〉 >= 〈ds|Hs|ds〉 ≡ m1 + d+ s ≡ mK∗ ,

where m1 is an eigenvalue assumed to be common to all members of the multiplet(SU(3) symmetry).

Similarly:

〈uu|Hs|uu〉 ≡ m1 + 2u = mρ

〈ss|Hs|ss〉 ≡ m1 + 2s = mφ

Hadrons in the Quark Modela – p.

Imagine a simplified strong-interaction Hamiltonian Hs. The mass eigenvalues for vectormesons can be worked out as

〈K∗|Hs|K∗〉 >= 〈ds|Hs|ds〉 ≡ m1 + d+ s ≡ mK∗ ,

where m1 is an eigenvalue assumed to be common to all members of the multiplet(SU(3) symmetry).

Similarly:

〈uu|Hs|uu〉 ≡ m1 + 2u = mρ

〈ss|Hs|ss〉 ≡ m1 + 2s = mφ

We’re ignoring the u− d mass difference.

Hadrons in the Quark Modela – p.

Imagine a simplified strong-interaction Hamiltonian Hs. The mass eigenvalues for vectormesons can be worked out as

〈K∗|Hs|K∗〉 >= 〈ds|Hs|ds〉 ≡ m1 + d+ s ≡ mK∗ ,

where m1 is an eigenvalue assumed to be common to all members of the multiplet(SU(3) symmetry).

Similarly:

〈uu|Hs|uu〉 ≡ m1 + 2u = mρ

〈ss|Hs|ss〉 ≡ m1 + 2s = mφ

We’re ignoring the u− d mass difference.

Comparing equations, we find mK∗ =mφ+mρ

2. mφ = 1.020 GeV; mρ = 0.77 GeV

=⇒ mK∗ = 1.792

= 0.895 GeV. Experimental masses are 0.892 GeV (charged) and0.895 GeV (neutral).

Hadrons in the Quark Modela – p.

In addition: mφ −mK∗ = mK∗ −mρ=120 MeV.

Hadrons in the Quark Modela – p.

In addition: mφ −mK∗ = mK∗ −mρ=120 MeV.We can do the same for baryons: find similar results.Note: since ω and ρ are roughly degenerate, ω must be predominantly u and d quarks.ωρ −→ ω contains a small component of ss.

Hadrons in the Quark Modela – p.

Let’s look at pseudoscalars. Masses are π(140), K(496), η(550), η′(960).

Hadrons in the Quark Modela – p.

Let’s look at pseudoscalars. Masses are π(140), K(496), η(550), η′(960).

The π is an isotriplet, and so has no strange quarks. The η and η′ are both isosinglets,are both relatively heavy, and so may both have some strange content.

Hadrons in the Quark Modela – p.

Let’s look at pseudoscalars. Masses are π(140), K(496), η(550), η′(960).

The π is an isotriplet, and so has no strange quarks. The η and η′ are both isosinglets,are both relatively heavy, and so may both have some strange content.

〈us|Hs|us〉 ≡ m0 + u+ s ≡ mK ,

where m0 is the pseudoscalar equivalent of m1.

Hadrons in the Quark Modela – p.

Let’s look at pseudoscalars. Masses are π(140), K(496), η(550), η′(960).

The π is an isotriplet, and so has no strange quarks. The η and η′ are both isosinglets,are both relatively heavy, and so may both have some strange content.

〈us|Hs|us〉 ≡ m0 + u+ s ≡ mK ,

where m0 is the pseudoscalar equivalent of m1.

〈ud|Hs|ud〉 ≡ m0 + u+ d = m0 + 2u ≡ mπ+

Hadrons in the Quark Modela – p.

Let’s look at pseudoscalars. Masses are π(140), K(496), η(550), η′(960).

The π is an isotriplet, and so has no strange quarks. The η and η′ are both isosinglets,are both relatively heavy, and so may both have some strange content.

〈us|Hs|us〉 ≡ m0 + u+ s ≡ mK ,

where m0 is the pseudoscalar equivalent of m1.

〈ud|Hs|ud〉 ≡ m0 + u+ d = m0 + 2u ≡ mπ+

For η8, wave function is 1√6

`

uu+ dd− 2ss´

, and 〈η8|Hs|η8〉 = m0 + 16(4u+ 8s)

Hadrons in the Quark Modela – p.

Let’s look at pseudoscalars. Masses are π(140), K(496), η(550), η′(960).

The π is an isotriplet, and so has no strange quarks. The η and η′ are both isosinglets,are both relatively heavy, and so may both have some strange content.

〈us|Hs|us〉 ≡ m0 + u+ s ≡ mK ,

where m0 is the pseudoscalar equivalent of m1.

〈ud|Hs|ud〉 ≡ m0 + u+ d = m0 + 2u ≡ mπ+

For η8, wave function is 1√6

`

uu+ dd− 2ss´

, and 〈η8|Hs|η8〉 = m0 + 16(4u+ 8s)

Combining: 4K − π = 3η8 (Gell-Mann-Okubo). Putting in known masses gives η8 =

0.613 GeV, to be compared with 0.550 and 0.960 GeV.

Hadrons in the Quark Modela – p.

Let’s look at pseudoscalars. Masses are π(140), K(496), η(550), η′(960).

The π is an isotriplet, and so has no strange quarks. The η and η′ are both isosinglets,are both relatively heavy, and so may both have some strange content.

〈us|Hs|us〉 ≡ m0 + u+ s ≡ mK ,

where m0 is the pseudoscalar equivalent of m1.

〈ud|Hs|ud〉 ≡ m0 + u+ d = m0 + 2u ≡ mπ+

For η8, wave function is 1√6

`

uu+ dd− 2ss´

, and 〈η8|Hs|η8〉 = m0 + 16(4u+ 8s)

Combining: 4K − π = 3η8 (Gell-Mann-Okubo). Putting in known masses gives η8 =

0.613 GeV, to be compared with 0.550 and 0.960 GeV.

Note: this relation works better if the squares of the masses are (justified by chiralsymmetry?) substituted:4K2 − π2 = 3η2

8 , giving η8 = 0.562 GeV

Hadrons in the Quark Modela – p.

There is a non-vanishing matrix element of the Hamiltonian between the SU(3) singletand isoscalar member of the octet:〈η8|Hs|η1〉 =

√8

3(u− s)

Hadrons in the Quark Modela – p.

There is a non-vanishing matrix element of the Hamiltonian between the SU(3) singletand isoscalar member of the octet:〈η8|Hs|η1〉 =

√8

3(u− s)

This means that the singlet and octet states are not mass eigenstates of theHamiltonian: the physical states will be mixtures:

Hadrons in the Quark Modela – p.

There is a non-vanishing matrix element of the Hamiltonian between the SU(3) singletand isoscalar member of the octet:〈η8|Hs|η1〉 =

√8

3(u− s)

This means that the singlet and octet states are not mass eigenstates of theHamiltonian: the physical states will be mixtures:

Define |η〉 = |η8〉 cos θ + |η1〉 sin θ

|η′〉 = −|η8〉 sin θ + |η1〉 cos θ

Hadrons in the Quark Modela – p.

There is a non-vanishing matrix element of the Hamiltonian between the SU(3) singletand isoscalar member of the octet:〈η8|Hs|η1〉 =

√8

3(u− s)

This means that the singlet and octet states are not mass eigenstates of theHamiltonian: the physical states will be mixtures:

Define |η〉 = |η8〉 cos θ + |η1〉 sin θ

|η′〉 = −|η8〉 sin θ + |η1〉 cos θ

−→ |η8〉 = |η〉 cos θ − |η′〉 sin θ

−→ η8 = η cos2 θ + η′ sin2 θ

Hadrons in the Quark Modela – p.

There is a non-vanishing matrix element of the Hamiltonian between the SU(3) singletand isoscalar member of the octet:〈η8|Hs|η1〉 =

√8

3(u− s)

This means that the singlet and octet states are not mass eigenstates of theHamiltonian: the physical states will be mixtures:

Define |η〉 = |η8〉 cos θ + |η1〉 sin θ

|η′〉 = −|η8〉 sin θ + |η1〉 cos θ

−→ |η8〉 = |η〉 cos θ − |η′〉 sin θ

−→ η8 = η cos2 θ + η′ sin2 θ

In GMO relation: 4K − π = 3`

η cos2 θ + η′ sin2 θ´

Hadrons in the Quark Modela – p.

There is a non-vanishing matrix element of the Hamiltonian between the SU(3) singletand isoscalar member of the octet:〈η8|Hs|η1〉 =

√8

3(u− s)

This means that the singlet and octet states are not mass eigenstates of theHamiltonian: the physical states will be mixtures:

Define |η〉 = |η8〉 cos θ + |η1〉 sin θ

|η′〉 = −|η8〉 sin θ + |η1〉 cos θ

−→ |η8〉 = |η〉 cos θ − |η′〉 sin θ

−→ η8 = η cos2 θ + η′ sin2 θ

In GMO relation: 4K − π = 3`

η cos2 θ + η′ sin2 θ´

Solution can be found: tan2 θ = 4K−π−3η3η′−4K+π

≈ 0.18

Note that if the squares of the masses are used, this value is much smaller, ≈ 0.03

Hadrons in the Quark Modela – p.

Recall that π+ = m0 + u+ d ≡ m0 + 2u and ρ+ = m1 + 2u

Hadrons in the Quark Modela – p. 10

What’s the difference between m0 and m1? It’s not isospin, since both ρ and π areisovectors.

Hadrons in the Quark Modela – p. 10

ρ has S = 1, π has S = 0. In this simple framework, this is the only feature that can giverise to the difference between m0 and m1, so we might guess that the mass difference isdue to a spin-spin interaction, ~s1 · ~s2. This is usually called the contact hyperfineinteraction.

Hadrons in the Quark Modela – p. 10

In positronium, the hyperfine splitting is 2πα3

~σ1·~σ2

m1m2|ψ(0)|2

Hadrons in the Quark Modela – p. 10

Let’s assume that something similar will work here:

m(q1q2) = M +m1 +m2 + a~σ1 · ~σ2

m1m2

m(q1q2q3) = M ′ +m1 +m2 +m2 + a′3X

i>j

~σi · ~σj

mimj

Hadrons in the Quark Modela – p. 10

For the meson case, ~S = 12

(~σ1 + ~σ2) =⇒ 〈~σ1 · ~σ2〉 = 2S(S + 1) − 3

Hadrons in the Quark Modela – p. 10

ρ = M + u+ d+ aud

; π = M + u+ d− 3aud

; π − ρ = 4aud

Hadrons in the Quark Modela – p. 10

Similarly, K∗ −K = 4aus

, and assuming that this can be applied to all mesons,D∗ −D = 4a

cu, B∗ − B = 4a

bu, etc.

Hadrons in the Quark Modela – p. 10

Does it work? Choosing u ≈ 330 Mev, s ≈ 550 MeV, c ≈ 1.5 GeV, b ≈ 5 GeV andρ− π = 630 MeV =⇒K∗ −K ≈ 378 (396) MeV;D∗ −D ≈ 139 (141) MeV;B∗ − B ≈ 42 (46) MeV.However, note D∗

s −Ds ≈ 83 (144) MeV.

Hadrons in the Quark Modela – p. 10

For baryons, the hyperfine Hamiltonian becomes Hhyp = ~σ1·~σ2

m1m2+ ~σ1·~σ3

m1m3+ ~σ2·~σ3

m2m3

Hadrons in the Quark Modela – p. 11

Let’s choose m1 = m2 = u, say. Then Hhyp = ~σ1·~σ2

u2 +(~σ1+~σ2)·~σ3

um3

Hadrons in the Quark Modela – p. 11

Black/whiteboard: color and baryon wave functions

Hadrons in the Quark Modela – p. 11

Full set of symmetric flavor wave functions are: ∆++ = uuu, ∆−− = ddd, Ω− = sss

∆+ = 1√3(uud+ udu+ duu) ∆0 = 1√

3(ddu+ dud+ udd),

Ξ∗0 = 1√3(ssu+ sus+ uss) Ξ∗− = 1√

3(ssd+ sds+ dss)

Σ∗+ = 1√3(uus+ usu+ suu) Σ∗− = 1√

3(dds+ dsd+ sdd)

Σ∗0 = 1√6(uds+ dus+ usd+ sud+ dsu+ sdu)

Hadrons in the Quark Modela – p. 11

For the two mixed symmetric representations:

p − 1√6(udu+ duu− 2uud) 1√

2(udu− duu)

n 1√6(udd+ dud− 2ddu) 1√

2(udd− dud)

Σ+ 1√6(usu+ suu− 2uus) 1√

2(usu− suu)

Σ− 1√6(dsd+ sdd− 2dds) 1√

2(dsd− sdd)

Ξ0 1√6(uss+ sus− 2ssu) 1√

2(uss− sus)

Ξ− 1√6(dss+ sds− 2ssd) 1√

2(dss− sds)

Σ0 1√6

h

sud+du√2

+ usd+dsu√2

1√2

h

−sud+du√2

+ usd+dsu√2

i

−2 ud+du√2

si

Λ 1√2

h

dsu−usd√2

+ s du−ud√2

i

1√6

h

s du−ud√2

+ usd−dsu√2

− 2 du−ud√2

si

Λ1(antisymmetric) =1√6

[(s(ud− du) + (usd− dsu) + (du− ud)s]

Hadrons in the Quark Modela – p. 11

What about other quarks? Would an SU(4) classification scheme work? What aboutSU(5)

Hadrons in the Quark Modela – p. 12

SU(4) multiplets would have to have the SU(3) multiplets as ‘submultiplets’

Hadrons in the Quark Modela – p. 12

For mesons: 4N

4 = 1L

15 (NN

N = 1L

N2 − 1)

For baryons: 4N

4N

4 = 20L

20L

20L

4

(NN

NN

N =N(N+1)(N+2)

6

L N(N2−1)3

L N(N2−1)3

L N(N−1)(N−2)6

)

Hadrons in the Quark Modela – p. 12

sD

0D

sD

–D

0K

–π π +K

– K

(a)

sD

DD

sD

−ρ +ρ K

(b)

*0

K*−

*+K*0

D 0*D*−

*−

*+

−

*+

−cdcu−

cs−

us−ds−

su− sd− ud

−

uc−sc−

dc−

0ρ ωφψJ/

uc−sc−

dc−

−cdcu−

cs−

+

D+

+

K0

us−ds−

su− sd−

du−

du−

0D

ηη′ ηc

π 0

ud−

K 0*

C

I

Y

Hadrons in the Quark Modela – p. 12

............................................................................................................................................................................................................................................

......................................................................................................................

........................................................... ...........................................................

.................................................................................................................................................................................

...........................................................

........................................................... .......................................................... .......................................................... ........................................................................................................................................................................................................................................... .....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..............................................................................................................................................................................................................................................................................................................................................................

..............................................................................................................................................................................................................................................................................................................................................................

.....................................................................................................................................................

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

Ω++ccc

Ξ+cc

Ξ++cc

Ω+cc

Ω0c ∆+∆0 ∆++∆−

Σ++c

Σ+c

Σ0c

Ω−Ξ0Ξ−

Ξ+c

Ξ0c

Σ− Σ+Σ0

Hadrons in the Quark Modela – p. 12

Hadrons in the Quark Modela – p. 12

Hadrons in the Quark Modela – p. 12

Hijconf = brij − 4αs

3rij+ c

Hijhyp =

"

4αcon

3mimj

8π

3Si · Sjδ

3(rij) +4αten

3mimj

1

r3ij

3Si · rijSj · rij

r2ij− Si · Sj

!#

,

HijSO = H

ij

SO(cm)+H

ij

SO(TP)

Hij

SO(cm)=

4αs

3r3

»

1

mi

+1

mj

–

"

~Si

mi

+~Sj

mj

#

· ~L

Hij

SO(TP)=

2

3r

∂Hijconf

∂r

"

~Si

m2i

+~Sj

m2j

#

· ~L

Hadrons in the Quark Modela – p. 13

Hadrons in the Quark Modela – p. 14

Hadrons in the Quark Modela – p. 14

Hadrons in the Quark Modela – p. 14

Hadrons in the Quark Modela – p. 14

Hadrons in the Quark Modela – p. 14

Hadrons in the Quark Modela – p. 14

λ-type: φsχλ =2 10; Φ(1)λ

φλχs =4 8; Φ(2)λ

1√2

(φρχρ − φλχλ) =2 8; Φ(3)λ

φAχρ =2 1; Φ(4)λ

ρ-type: φsχρ =2 10; Φ(1)ρ

φρχs =4 8; Φ(2)ρ

1√2

(φλχρ + φρχλ) =2 8; Φ(3)ρ

φAχλ =2 1; Φ(4)ρ

A-type: φAχs =4 1; Φ(1)A

1√2

(φλχρ − φρχλ) =2 8; Φ(2)A

Hadrons in the Quark Modela – p. 15

I can write the ground state wave function in a notation (1S)3. With this notation, the firstorbital excitation can be written (1S)2(1P ).

Hadrons in the Quark Modela – p. 16

I can write the ground state wave function in a notation (1S)3. With this notation, the firstorbital excitation can be written (1S)2(1P ).

There are 3 possibilities for this: ~r1ψ, ~r2ψ, ~r3ψ, depending on which quark is excited (ψis a (1S)3 wave function).

Hadrons in the Quark Modela – p. 16

I can write the ground state wave function in a notation (1S)3. With this notation, the firstorbital excitation can be written (1S)2(1P ).

There are 3 possibilities for this: ~r1ψ, ~r2ψ, ~r3ψ, depending on which quark is excited (ψis a (1S)3 wave function).

I can create states with particular symmetries by taking linear combinations: assumingall masses equal,ψρ = 1√

2(~r1 − ~r2)ψ, ψλ = 1√

6(~r1 + ~r2 − 2~r3)ψ, ψS = 1√

3(~r1 + ~r2 + ~r3)ψ

Hadrons in the Quark Modela – p. 16

I can write the ground state wave function in a notation (1S)3. With this notation, the firstorbital excitation can be written (1S)2(1P ).

There are 3 possibilities for this: ~r1ψ, ~r2ψ, ~r3ψ, depending on which quark is excited (ψis a (1S)3 wave function).

I can create states with particular symmetries by taking linear combinations: assumingall masses equal,ψρ = 1√

2(~r1 − ~r2)ψ, ψλ = 1√

6(~r1 + ~r2 − 2~r3)ψ, ψS = 1√

3(~r1 + ~r2 + ~r3)ψ

We can set the center of mass to be at the origin: ~r1 + ~r2 + ~r3 = 0, and ψS becomestrivial: only two, mixed-symmetric excitations possible with L = 1. If Rcm 6= 0, the ψS

state corresponds to the 3 quarks in the (1S)3, all moving with one unit of anguarmomentum relative to some origin.

Hadrons in the Quark Modela – p. 16

I can write the ground state wave function in a notation (1S)3. With this notation, the firstorbital excitation can be written (1S)2(1P ).

There are 3 possibilities for this: ~r1ψ, ~r2ψ, ~r3ψ, depending on which quark is excited (ψis a (1S)3 wave function).

I can create states with particular symmetries by taking linear combinations: assumingall masses equal,ψρ = 1√

2(~r1 − ~r2)ψ, ψλ = 1√

6(~r1 + ~r2 − 2~r3)ψ, ψS = 1√

3(~r1 + ~r2 + ~r3)ψ

We can set the center of mass to be at the origin: ~r1 + ~r2 + ~r3 = 0, and ψS becomestrivial: only two, mixed-symmetric excitations possible with L = 1. If Rcm 6= 0, the ψS

state corresponds to the 3 quarks in the (1S)3, all moving with one unit of anguarmomentum relative to some origin.

The fully symmetric wave functions possible are: 1√2

“

Φ(1)λψλ + Φ

(1)ρ + ψρ

”

: 210

1√2

“

Φ(2)λψλ + Φ

(2)ρ + ψρ

”

: 48

1√2

“

Φ(3)λψλ + Φ

(3)ρ + ψρ

”

: 28

1√2

“

Φ(4)λψλ + Φ

(4)ρ + ψρ

”

: 21

Hadrons in the Quark Modela – p. 16

Possible angular momentum values:210 : 1 + 1

2→ JP = 1

2

−, 3

2

−(∆, Σ, no Λ)

48 : 1 + 32→ JP = 1

2

−, 3

2

−, 5

2

−(nucleon, Λ, Σ, no ∆)

28 : 1 + 12→ JP = 1

2

−, 3

2

−(same as 48)

21 : 1 + 12→ JP = 1

2

−, 3

2

−(Λ only)

Hadrons in the Quark Modela – p. 17

Possible angular momentum values:210 : 1 + 1

2→ JP = 1

2

−, 3

2

−(∆, Σ, no Λ)

48 : 1 + 32→ JP = 1

2

−, 3

2

−, 5

2

−(nucleon, Λ, Σ, no ∆)

28 : 1 + 12→ JP = 1

2

−, 3

2

−(same as 48)

21 : 1 + 12→ JP = 1

2

−, 3

2

−(Λ only)

Number of states in the multiplet is 2 x 10 + 4 x 8 + 2 x 8 + 2 x 1=70. Multiplet is oftenreferred to as 70, 1−.

Hadrons in the Quark Modela – p. 17

Possible angular momentum values:210 : 1 + 1

2→ JP = 1

2

−, 3

2

−(∆, Σ, no Λ)

48 : 1 + 32→ JP = 1

2

−, 3

2

−, 5

2

−(nucleon, Λ, Σ, no ∆)

28 : 1 + 12→ JP = 1

2

−, 3

2

−(same as 48)

21 : 1 + 12→ JP = 1

2

−, 3

2

−(Λ only)

Number of states in the multiplet is 2 x 10 + 4 x 8 + 2 x 8 + 2 x 1=70. Multiplet is oftenreferred to as 70, 1−.

Examples of states in this mutliplet areN(1520), N(1535), ∆(1620), ∆(1700),Λ(1405), Λ(1520)

Hadrons in the Quark Modela – p. 17

What about higher excitations? Easier to illustrate for the general case (not SU(3)),separately for Λ-type and Σ-type states.

Hadrons in the Quark Modela – p. 18

What about higher excitations? Easier to illustrate for the general case (not SU(3)),separately for Λ-type and Σ-type states.

We build components of the wave function as Clebsch-Gordan sums:

|J,M〉 =X

mS

ΨL,mL

“

~ρ, ~λ”

χ(S,mS)〈L,mL, S,mS |J,M〉

ΨL,mL

“

~ρ, ~λ”

=X

mρ

ψnρ,ℓρ,mρ(~ρ)ψnλ,ℓλ,mλ

“

~λ”

〈ℓρ,mρ, ℓλ,mλ|L,mL〉

Hadrons in the Quark Modela – p. 18

What about higher excitations? Easier to illustrate for the general case (not SU(3)),separately for Λ-type and Σ-type states.

We build components of the wave function as Clebsch-Gordan sums:

|J,M〉 =X

mS

ΨL,mL

“

~ρ, ~λ”

χ(S,mS)〈L,mL, S,mS |J,M〉

ΨL,mL

“

~ρ, ~λ”

=X

mρ

ψnρ,ℓρ,mρ(~ρ)ψnλ,ℓλ,mλ

“

~λ”

〈ℓρ,mρ, ℓλ,mλ|L,mL〉

Wave functions must still be fully symmetric in identical quarks. Excitations in λ are (12)symmetry (λ symmetry). Radial excitations in ρ also have λ symmetry. Orbitalexcitations in ρ have ρ symmetry.

Hadrons in the Quark Modela – p. 18

What about higher excitations? Easier to illustrate for the general case (not SU(3)),separately for Λ-type and Σ-type states.

We build components of the wave function as Clebsch-Gordan sums:

|J,M〉 =X

mS

ΨL,mL

“

~ρ, ~λ”

χ(S,mS)〈L,mL, S,mS |J,M〉

ΨL,mL

“

~ρ, ~λ”

=X

mρ

ψnρ,ℓρ,mρ(~ρ)ψnλ,ℓλ,mλ

“

~λ”

〈ℓρ,mρ, ℓλ,mλ|L,mL〉

Wave functions must still be fully symmetric in identical quarks. Excitations in λ are (12)symmetry (λ symmetry). Radial excitations in ρ also have λ symmetry. Orbitalexcitations in ρ have ρ symmetry.

Simplified flavor wave functions: ΣQ = uuq, 1√2(ud+ du)Q, ddQ

ΛQ = 1√2(ud− du)Q, ΩQ = ssQ

ΞQ = 1√2(us− su)Q, 1√

2(ds− sd)Q, Ξ′

Q = 1√2(us+ su)Q, 1√

2(ds+ sd)Q

Hadrons in the Quark Modela – p. 18

What about higher excitations? Easier to illustrate for the general case (not SU(3)),separately for Λ-type and Σ-type states.

We build components of the wave function as Clebsch-Gordan sums:

|J,M〉 =X

mS

ΨL,mL

“

~ρ, ~λ”

χ(S,mS)〈L,mL, S,mS |J,M〉

ΨL,mL

“

~ρ, ~λ”

=X

mρ

ψnρ,ℓρ,mρ(~ρ)ψnλ,ℓλ,mλ

“

~λ”

〈ℓρ,mρ, ℓλ,mλ|L,mL〉

Wave functions must still be fully symmetric in identical quarks. Excitations in λ are (12)symmetry (λ symmetry). Radial excitations in ρ also have λ symmetry. Orbitalexcitations in ρ have ρ symmetry.

Simplified flavor wave functions: ΣQ = uuq, 1√2(ud+ du)Q, ddQ

ΛQ = 1√2(ud− du)Q, ΩQ = ssQ

ΞQ = 1√2(us− su)Q, 1√

2(ds− sd)Q, Ξ′

Q = 1√2(us+ su)Q, 1√

2(ds+ sd)Q

Space-spin wave function must have the same symmetry as flavor wave function.

Hadrons in the Quark Modela – p. 18

Let’s denote wave function components by S,L, nρ, ℓρ, nλ, ℓλ

Hadrons in the Quark Modela – p. 19

Let’s denote wave function components by S,L, nρ, ℓρ, nλ, ℓλ

For ΣQ with JP = 12

+, first component of wave function is 1

2 λ, 0, 0, 0, 0, 0. Next easiest

components to see are 12 λ, 0, 1, 0, 0, 0 1

2 λ, 0, 0, 0, 1, 0. Other components are

32, 2, 0, 2, 0, 0; 3

2, 2, 0, 0, 0, 2; 1

2 ρ, 0, 0, 1, 0, 1 and 1

2 ρ, 1, 0, 1, 0, 1

Hadrons in the Quark Modela – p. 19

Let’s denote wave function components by S,L, nρ, ℓρ, nλ, ℓλ

For ΣQ with JP = 12

+, first component of wave function is 1

2 λ, 0, 0, 0, 0, 0. Next easiest

components to see are 12 λ, 0, 1, 0, 0, 0 1

2 λ, 0, 0, 0, 1, 0. Other components are

32, 2, 0, 2, 0, 0; 3

2, 2, 0, 0, 0, 2; 1

2 ρ, 0, 0, 1, 0, 1 and 1

2 ρ, 1, 0, 1, 0, 1

JP ΣQ ΛQ

12

+

112 λ, 0, 0, 0, 0, 0 1

2 ρ, 0, 0, 0, 0, 0

12

+

212 λ, 0, 1, 0, 0, 0 1

2 ρ, 0, 1, 0, 0, 0

12

+

312 λ, 0, 0, 0, 1, 0 1

2 ρ, 0, 0, 0, 1, 0

12

+

432, 2, 0, 2, 0, 0 1

2 λ, 0, 0, 1, 0, 1

12

+

532, 2, 0, 0, 0, 2 1

2 λ, 1, 0, 1, 0, 1

12

+

612 ρ, 0, 0, 1, 0, 1 3

2, 1, 0, 1, 0, 1

12

+

712 ρ, 1, 0, 1, 0, 1 3

2, 2, 0, 1, 0, 1

12

−1

12 ρ, 1, 0, 1, 0, 0 1

2 λ, 1, 0, 1, 0, 0

12

−2

12 λ, 1, 0, 0, 0, 1 3

2, 1, 0, 1, 0, 0

12

−3

32, 1, 0, 0, 0, 1 1

2 ρ, 1, 0, 0, 0, 1

Hadrons in the Quark Modela – p. 19

Hijconf =

3X

i<j=1

„

brij

2− 2αCoul

3rij

«

.

Hijhyp =

3X

i<j=1

"

2αcon

3mimj

8π

3Si · Sjδ

3(rij) +2αten

3mimj

1

r3ij

3Si · rijSj · rij

r2ij− Si · Sj

!#

,

HijSO = H

ij

SO(cm)+H

ij

SO(TP)

Hij

SO(cm)=

2αs

3r3ij

"

~rij × ~pi · ~Si

m2i

− ~rij × ~pj · ~Sj

m2j

− ~rij × ~pj · ~Si − ~rij × ~pi · ~Sj

mimj

#

Hij

SO(TP)= − 1

2rij

∂Hijconf

∂rij

"

~rij × ~pi · ~Si

m2i

− ~rij × ~pj · ~Sj

m2j

#

Hadrons in the Quark Modela – p. 20

Consider the Coulomb term: 1r12

+ 1r13

+ 1r23

Hadrons in the Quark Modela – p. 21

Consider the Coulomb term: 1r12

+ 1r13

+ 1r23

Since r12 ∝ ρ, calculating 〈 1r12

〉, or 〈f(r12)〉 is easy: 〈f(r12)〉 =R

d3ρd3λΨ′∗(ρ, λ)f(r12)Ψ(ρ, λ) =R

d3ρψ′∗(ρ)f(√

2ρ)ψ(ρ)R

d3λψ′∗(λ)ψ(λ)

Hadrons in the Quark Modela – p. 21

Consider the Coulomb term: 1r12

+ 1r13

+ 1r23

Since r12 ∝ ρ, calculating 〈 1r12

〉, or 〈f(r12)〉 is easy: 〈f(r12)〉 =R

d3ρd3λΨ′∗(ρ, λ)f(r12)Ψ(ρ, λ) =R

d3ρψ′∗(ρ)f(√

2ρ)ψ(ρ)R

d3λψ′∗(λ)ψ(λ)

~r13 =√

2m2

m1+m2~ρ+

q

32~λ =⇒ r13 =

r

2m22

(m1+m2)2ρ2 + 3

2λ2 + 2m2√

3(m1+m2)~ρ · ~λ.

How do we evaluate 〈 1r13

〉, or 〈f(r13)〉?One of two ways.

Hadrons in the Quark Modela – p. 21

Consider the Coulomb term: 1r12

+ 1r13

+ 1r23

Since r12 ∝ ρ, calculating 〈 1r12

〉, or 〈f(r12)〉 is easy: 〈f(r12)〉 =R

d3ρd3λΨ′∗(ρ, λ)f(r12)Ψ(ρ, λ) =R

d3ρψ′∗(ρ)f(√

2ρ)ψ(ρ)R

d3λψ′∗(λ)ψ(λ)

~r13 =√

2m2

m1+m2~ρ+

q

32~λ =⇒ r13 =

r

2m22

(m1+m2)2ρ2 + 3

2λ2 + 2m2√

3(m1+m2)~ρ · ~λ.

How do we evaluate 〈 1r13

〉, or 〈f(r13)〉?One of two ways.

Example: expand

1

|~ρ′ + ~λ′|≈X

ℓ

(−1)ℓ ρ′ℓ

λ′(ℓ+1)Yℓ(ρ′)Y

∗ℓ (λ′), ρ′ ≤ λ′

Hadrons in the Quark Modela – p. 21

Consider the Coulomb term: 1r12

+ 1r13

+ 1r23

Since r12 ∝ ρ, calculating 〈 1r12

〉, or 〈f(r12)〉 is easy: 〈f(r12)〉 =R

d3ρd3λΨ′∗(ρ, λ)f(r12)Ψ(ρ, λ) =R

d3ρψ′∗(ρ)f(√

2ρ)ψ(ρ)R

d3λψ′∗(λ)ψ(λ)

~r13 =√

2m2

m1+m2~ρ+

q

32~λ =⇒ r13 =

r

2m22

(m1+m2)2ρ2 + 3

2λ2 + 2m2√

3(m1+m2)~ρ · ~λ.

How do we evaluate 〈 1r13

〉, or 〈f(r13)〉?One of two ways.

Example: expand

1

|~ρ′ + ~λ′|≈X

ℓ

(−1)ℓ ρ′ℓ

λ′(ℓ+1)Yℓ(ρ′)Y

∗ℓ (λ′), ρ′ ≤ λ′

Evaluate 〈Yℓ(ρ′)〉 using dΩρ, 〈Yℓ(λ′)〉 using dΩλ, and remaining radial integrals involvepowers of ρ and λ

Hadrons in the Quark Modela – p. 21

Consider the Coulomb term: 1r12

+ 1r13

+ 1r23

Since r12 ∝ ρ, calculating 〈 1r12

〉, or 〈f(r12)〉 is easy: 〈f(r12)〉 =R

d3ρd3λΨ′∗(ρ, λ)f(r12)Ψ(ρ, λ) =R

d3ρψ′∗(ρ)f(√

2ρ)ψ(ρ)R

d3λψ′∗(λ)ψ(λ)

~r13 =√

2m2

m1+m2~ρ+

q

32~λ =⇒ r13 =

r

2m22

(m1+m2)2ρ2 + 3

2λ2 + 2m2√

3(m1+m2)~ρ · ~λ.

How do we evaluate 〈 1r13

〉, or 〈f(r13)〉?One of two ways.

Example: expand

1

|~ρ′ + ~λ′|≈X

ℓ

(−1)ℓ ρ′ℓ

λ′(ℓ+1)Yℓ(ρ′)Y

∗ℓ (λ′), ρ′ ≤ λ′

Evaluate 〈Yℓ(ρ′)〉 using dΩρ, 〈Yℓ(λ′)〉 using dΩλ, and remaining radial integrals involvepowers of ρ and λ

Should work with any choice of functions used to expand wave function

Hadrons in the Quark Modela – p. 21

Hadrons in the Quark Modela – p. 22

Hadrons in the Quark Modela – p. 22

Hadrons in the Quark Modela – p. 22

Hadrons in the Quark Modela – p. 22

Hadrons in the Quark Modela – p. 22

Hadrons in the Quark Modela – p. 23

Hadrons in the Quark Modela – p. 23

Hadrons in the Quark Modela – p. 23

Hadrons in the Quark Modela – p. 24

So we have a spectrum, and wave functions. Are we done? Can we do anything else (dowe declare victory and go work on tans?)?

Hadrons in the Quark Modela – p. 25

So we have a spectrum, and wave functions. Are we done? Can we do anything else (dowe declare victory and go work on tans?)?

Plethora of data that can be treated in such a model, once we have wave functions:

Hadrons in the Quark Modela – p. 25

So we have a spectrum, and wave functions. Are we done? Can we do anything else (dowe declare victory and go work on tans?)?

Plethora of data that can be treated in such a model, once we have wave functions:

Magnetic moments; meson radiative transitions; meson decay constants; mesonsemileptonic decays; meson rare decays; meson strong decays; baryon radiativedecays; baryon semileptonic decays; baryon rare decays; baryon strong decays...

Hadrons in the Quark Modela – p. 25

So we have a spectrum, and wave functions. Are we done? Can we do anything else (dowe declare victory and go work on tans?)?

Plethora of data that can be treated in such a model, once we have wave functions:

Magnetic moments; meson radiative transitions; meson decay constants; mesonsemileptonic decays; meson rare decays; meson strong decays; baryon radiativedecays; baryon semileptonic decays; baryon rare decays; baryon strong decays...

Each of these represents a wealth of data. For instance, in the case of baryonsemileptonic decays: form factors, total and differential decay rates, polarizationasymmetries, decays to excited states, etc.

Hadrons in the Quark Modela – p. 25

So we have a spectrum, and wave functions. Are we done? Can we do anything else (dowe declare victory and go work on tans?)?

Plethora of data that can be treated in such a model, once we have wave functions:

Magnetic moments; meson radiative transitions; meson decay constants; mesonsemileptonic decays; meson rare decays; meson strong decays; baryon radiativedecays; baryon semileptonic decays; baryon rare decays; baryon strong decays...

Each of these represents a wealth of data. For instance, in the case of baryonsemileptonic decays: form factors, total and differential decay rates, polarizationasymmetries, decays to excited states, etc.

Sketch of treatment of semileptonic decays (mesons), strong decays (baryons)

Hadrons in the Quark Modela – p. 25

Semileptonic decay: decay of a hadron to a final state containing a hadron and a pair ofleptons. Mediated by weak interaction: understanding these requires understanding theinterplay between the dynamics of the weak interaction leading to the decay, and thestrong interaction that provides the binding in the parent and daughter hadron

Hadrons in the Quark Modela – p. 26

Semileptonic decay: decay of a hadron to a final state containing a hadron and a pair ofleptons. Mediated by weak interaction: understanding these requires understanding theinterplay between the dynamics of the weak interaction leading to the decay, and thestrong interaction that provides the binding in the parent and daughter hadron

Contrasted with leptonic decays (meson decay to a pair of leptons), or weak non-leptonicdecay (weak hadron decay to a final state that has no leptons: eg Λ → pπ−)

Hadrons in the Quark Modela – p. 26