Hadrons in the Quark Modela
Winston Roberts
[email protected]
Florida State University
a
24th 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
c
t
,
,
,
d
s
b
2
3
− 31
!
Hadrons in the Quark Modela – p.
Six flavors of quarks,
u
c
t
,
,
,
d
s
b
2
3
− 31
!
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
c
t
,
,
,
d
s
b
2
3
− 31
!
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 energy
scale (a few hundred MeV) of the strong interaction. Can treat as two states of same
particle −→ isospin, described in terms of SU(2).
Hadrons in the Quark Modela – p.
Six flavors of quarks,
u
c
t
,
,
,
d
s
b
2
3
− 31
!
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 energy
scale (a few hundred MeV) of the strong interaction. Can treat as two states of same
particle −→ 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
c
t
,
,
,
d
s
b
2
3
− 31
!
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 energy
scale (a few hundred MeV) of the strong interaction. Can treat as two states of same
particle −→ 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) are
mesons (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, the
parity 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, the
parity 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, J P = 0− ):
3 ⊗ 3̄ = 1 ⊕ 8
Hadrons in the Quark Modela – p.
r
r
K 0 [ds̄]
r
h
√1
h 2
η8 √16
π − [dū]
π0
r
η1
√1
3
rg
i + r
uū − dd¯ π −ud¯
i
uū + dd¯ − 2ss̄
r
¯
−sd
i
uū + dd¯ + ss̄
K − [sū]
h
K + [us̄]
K̄ 0
Hadrons in the Quark Modela – p.
All (light) mesons fall into these multiplets. Thus, for the vector
mesons (L = 0, S = 1, J P = 1− )
Hadrons in the Quark Modela – p.
r
r
K 0∗ [ds̄]
r
h
√1
h 2
φ8 √16
ρ− [dū]
ρ0
r
K +∗ [us̄]
rg
φ1
√1
3
r 0∗
¯
K̄ −sd
i
uū + dd¯ + ss̄
K −∗ [sū]
h
i + r
uū − dd¯ ρ −ud¯
i
uū + dd¯ − 2ss̄
Hadrons in the Quark Modela – p.
Hadrons in the Quark Modela – p.
For baryons:
3 ⊗ 3 ⊗ 3 = (6 ⊕ 3̄) ⊗ 3 = 10 ⊕ 8 ⊕ 8 ⊕ 1
Hadrons in the Quark Modela – p.
r
r
n [udd]
p [uud]
r
rg
Σ− [dds]
r
Σ0 [uds]
Σ+ [uus]
Λ [uds]
r
Ξ− [dss]
r
Ξ0 [uss]
Hadrons in the Quark Modela – p.
r
r
∆− [ddd] ∆0 [udd]
r
r
∆+ [uud] ∆++ [uuu]
r
Σ∗− [dds]
r
Σ∗0 [uds]
r
r
Σ∗+ [uus]
r
Ξ∗− [dss] Ξ∗0 [uss]
r
Ω− [sss]
Hadrons in the Quark Modela – p.
Imagine a simplified strong-interaction Hamiltonian Hs . The mass eigenvalues for vector
mesons can be worked out as
Hadrons in the Quark Modela – p.
Imagine a simplified strong-interaction Hamiltonian Hs . The mass eigenvalues for vector
mesons can be worked out as
hK ∗ |Hs |K ∗ i >= hds|Hs |dsi ≡ 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 vector
mesons can be worked out as
hK ∗ |Hs |K ∗ i >= hds|Hs |dsi ≡ m1 + d + s ≡ mK ∗ ,
where m1 is an eigenvalue assumed to be common to all members of the multiplet
(SU(3) symmetry).
Similarly:
huu|Hs |uui ≡ m1 + 2u = mρ
hss|Hs |ssi ≡ m1 + 2s = mφ
Hadrons in the Quark Modela – p.
Imagine a simplified strong-interaction Hamiltonian Hs . The mass eigenvalues for vector
mesons can be worked out as
hK ∗ |Hs |K ∗ i >= hds|Hs |dsi ≡ m1 + d + s ≡ mK ∗ ,
where m1 is an eigenvalue assumed to be common to all members of the multiplet
(SU(3) symmetry).
Similarly:
huu|Hs |uui ≡ m1 + 2u = mρ
hss|Hs |ssi ≡ 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 vector
mesons can be worked out as
hK ∗ |Hs |K ∗ i >= hds|Hs |dsi ≡ m1 + d + s ≡ mK ∗ ,
where m1 is an eigenvalue assumed to be common to all members of the multiplet
(SU(3) symmetry).
Similarly:
huu|Hs |uui ≡ m1 + 2u = mρ
hss|Hs |ssi ≡ m1 + 2s = mφ
We’re ignoring the u − d mass difference.
m +m
Comparing equations, we find mK ∗ = φ 2 ρ . mφ = 1.020 GeV; mρ = 0.77 GeV
= 0.895 GeV. Experimental masses are 0.892 GeV (charged) and
=⇒ mK ∗ = 1.79
2
0.895 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.
hus|Hs |usi ≡ 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.
hus|Hs |usi ≡ m0 + u + s ≡ mK ,
where m0 is the pseudoscalar equivalent of m1 .
hud|Hs |udi ≡ 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.
hus|Hs |usi ≡ m0 + u + s ≡ mK ,
where m0 is the pseudoscalar equivalent of m1 .
hud|Hs |udi ≡ m0 + u + d = m0 + 2u ≡ mπ+
For η8 , wave function is
√1
6
`
´
uū + dd¯ − 2ss̄ , and hη8 |Hs |η8 i = 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.
hus|Hs |usi ≡ m0 + u + s ≡ mK ,
where m0 is the pseudoscalar equivalent of m1 .
hud|Hs |udi ≡ m0 + u + d = m0 + 2u ≡ mπ+
For η8 , wave function is
√1
6
`
´
uū + dd¯ − 2ss̄ , and hη8 |Hs |η8 i = 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.
hus|Hs |usi ≡ m0 + u + s ≡ mK ,
where m0 is the pseudoscalar equivalent of m1 .
hud|Hs |udi ≡ m0 + u + d = m0 + 2u ≡ mπ+
For η8 , wave function is
√1
6
`
´
uū + dd¯ − 2ss̄ , and hη8 |Hs |η8 i = 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 chiral
symmetry?) substituted:4K 2 − π 2 = 3η82 , 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) singlet
and isoscalar member
of the octet:
√
hη8 |Hs |η1 i = 38 (u − s)
Hadrons in the Quark Modela – p.
There is a non-vanishing matrix element of the Hamiltonian between the SU(3) singlet
and isoscalar member
of the octet:
√
hη8 |Hs |η1 i = 38 (u − s)
This means that the singlet and octet states are not mass eigenstates of the
Hamiltonian: 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) singlet
and isoscalar member
of the octet:
√
hη8 |Hs |η1 i = 38 (u − s)
This means that the singlet and octet states are not mass eigenstates of the
Hamiltonian: the physical states will be mixtures:
Define |ηi = |η8 i cos θ + |η1 i sin θ
|η ′ i = −|η8 i sin θ + |η1 i cos θ
Hadrons in the Quark Modela – p.
There is a non-vanishing matrix element of the Hamiltonian between the SU(3) singlet
and isoscalar member
of the octet:
√
hη8 |Hs |η1 i = 38 (u − s)
This means that the singlet and octet states are not mass eigenstates of the
Hamiltonian: the physical states will be mixtures:
Define |ηi = |η8 i cos θ + |η1 i sin θ
|η ′ i = −|η8 i sin θ + |η1 i cos θ
−→ |η8 i = |ηi cos θ − |η ′ i 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) singlet
and isoscalar member
of the octet:
√
hη8 |Hs |η1 i = 38 (u − s)
This means that the singlet and octet states are not mass eigenstates of the
Hamiltonian: the physical states will be mixtures:
Define |ηi = |η8 i cos θ + |η1 i sin θ
|η ′ i = −|η8 i sin θ + |η1 i cos θ
−→ |η8 i = |ηi cos θ − |η ′ i 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) singlet
and isoscalar member
of the octet:
√
hη8 |Hs |η1 i = 38 (u − s)
This means that the singlet and octet states are not mass eigenstates of the
Hamiltonian: the physical states will be mixtures:
Define |ηi = |η8 i cos θ + |η1 i sin θ
|η ′ i = −|η8 i sin θ + |η1 i cos θ
−→ |η8 i = |ηi cos θ − |η ′ i sin θ
−→ η8 = η cos2 θ + η ′ sin2 θ
`
´
In GMO relation: 4K − π = 3 η cos2 θ + η ′ sin2 θ
4K−π−3η
Solution can be found: tan2 θ = 3η
′ −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. 1
What’s the difference between m0 and m1 ? It’s not isospin, since both ρ and π are
isovectors.
Hadrons in the Quark Modela – p. 1
ρ has S = 1, π has S = 0. In this simple framework, this is the only feature that can give
rise to the difference between m0 and m1 , so we might guess that the mass difference is
due to a spin-spin interaction, ~s1 · ~s2 . This is usually called the contact hyperfine
interaction.
Hadrons in the Quark Modela – p. 1
In positronium, the hyperfine splitting is
σ1 ·~
σ2
2πα ~
3 m1 m2
|ψ(0)|2
Hadrons in the Quark Modela – p. 1
Let’s assume that something similar will work here:
m(q1 q 2 ) = M + m1 + m2 + a
~
σ1 · ~
σ2
m1 m2
3
X
~
σi · ~
σj
m(q1 q2 q3 ) = M ′ + m1 + m2 + m2 + a′
mi mj
i>j
Hadrons in the Quark Modela – p. 1
~=
For the meson case, S
1
2
(~
σ1 + ~
σ2 ) =⇒ h~
σ1 · ~
σ2 i = 2S(S + 1) − 3
Hadrons in the Quark Modela – p. 1
ρ=M +u+d+
a
;
ud
π =M +u+d−
3a
;
ud
π−ρ=
4a
ud
Hadrons in the Quark Modela – p. 1
4a
, and assuming that this can be applied to all mesons,
Similarly, K ∗ − K = us
∗ − B = 4a , etc.
,
B
D∗ − D = 4a
cu
bu
Hadrons in the Quark Modela – p. 1
Does it work? Choosing u ≈ 330 Mev, s ≈ 550 MeV, c ≈ 1.5 GeV, b ≈ 5 GeV and
ρ − π = 630 MeV =⇒
K ∗ − K ≈ 378 (396) MeV;
D∗ − D ≈ 139 (141) MeV;
B ∗ − B ≈ 42 (46) MeV.
However, note Ds∗ − Ds ≈ 83 (144) MeV.
Hadrons in the Quark Modela – p. 1
For baryons, the hyperfine Hamiltonian becomes Hhyp =
~
σ1 ·~
σ2
m1 m2
+
~
σ1 ·~
σ3
m1 m3
+
~
σ2 ·~
σ3
m2 m3
Hadrons in the Quark Modela – p. 1
Let’s choose m1 = m2 = u, say. Then Hhyp =
~
σ1 ·~
σ2
2
u
+
(~
σ1 +~
σ2 )·~
σ3
um3
Hadrons in the Quark Modela – p. 1
Black/whiteboard: color and baryon wave functions
Hadrons in the Quark Modela – p. 1
Full set of symmetric flavor wave functions are: ∆++ = uuu, ∆−− = ddd, Ω− = sss
∆+ =
√1 (uud
3
+ udu + duu) ∆0 =
√1 (ddu
3
+ dud + udd),
1
√1 (ssu + sus + uss) Ξ∗− = √
(ssd + sds + dss)
3
3
Σ∗+ = √1 (uus + usu + suu) Σ∗− = √1 (dds + dsd + sdd)
3
3
Ξ∗0 =
Σ∗0 =
√1 (uds
6
+ dus + usd + sud + dsu + sdu)
Hadrons in the Quark Modela – p. 1
For the two mixed symmetric representations:
− √1 (udu + duu − 2uud)
p
n
Σ+
Σ−
Ξ0
Ξ−
Σ0
Λ
6
1
√ (udd + dud − 2ddu)
6
√1 (usu + suu − 2uus)
6
1
√ (dsd + sdd − 2dds)
6
1
√ (uss + sus − 2ssu)
6
√1 (dss + sds − 2ssd)
6 h
1
√
√
√
s ud+du
+ usd+dsu
6
2 i
2
√
s
−2 ud+du
h 2
i
dsu−usd
du−ud
1
√
√
+s √
2
2
2
1
√
(udu − duu)
2
1
√
(udd − dud)
2
1
√
(usu − suu)
2
1
√
(dsd − sdd)
2
1
√
(uss − sus)
2
1
√
(dss − sds)
2 h
i
ud+du
usd+dsu
1
√
√
−s √
+
2
2
2
1
√
6
h
√
+
s du−ud
2
usd−dsu
√
2
−2
du−ud
√
s
2
i
1
Λ1 (antisymmetric) = √ [(s(ud − du) + (usd − dsu) + (du − ud)s]
6
Hadrons in the Quark Modela – p. 1
What about other quarks? Would an SU(4) classification scheme work? What about
SU(5)
Hadrons in the Quark Modela – p. 1
SU(4) multiplets would have to have the SU(3) multiplets as ‘submultiplets’
Hadrons in the Quark Modela – p. 1
For mesons: 4
N
4=1
L
15 (N
N
N =1
L
N 2 − 1)
N N
L L L
For baryons: 4
4
4 = 20
20
20
4
2
N N
L
N (N +1)(N +2)
N (N −1) L
(N
N
N =
6
3
N (N 2 −1)
3
L
N (N −1)(N −2)
)
6
Hadrons in the Quark Modela – p. 1
Ds+
cs− −
−
cu
cd
D0
K0
π
–
ds−
−
du
K
−
su
–
I
cs− −
−
cu
cd
D*0
K *0
ρ
K0
uc−
D0
Ds−
Ds* +
Y
−
D+
+
K
π 0 η us− −
ηc η ′ sd− ud π +
– dc− −
sc
D
C
(a)
−
du
ds−
(b)
D* +
K *+
ρ 0 ω us− − +
− J/
ψ φ sd− ud ρ
su
K *−
− dc− −
*
D
sc
K *0
uc−
D*0
Ds* −
Hadrons in the Quark Modela – p. 1
++
Ω
......... ccc
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+ .................................................................. ++
cc.................. .... .................... cc
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.
.
.
.
.
.
.
− ......... ... .........
0
....... .
.
.
.
.
........
.......... −
Ξ ...
Σ ...
Ξ
∆...
Σ ...
Ξ
... Ξ
...
Ω
...Σ
... Ξ
...
...
∆
... Ω ... ∆
...Σ
...
...
Ξ
...
Ω
...Σ
...
++
∆
...
Σ
Hadrons in the Quark Modela – p. 1
Hadrons in the Quark Modela – p. 1
Hadrons in the Quark Modela – p. 1
ij
= brij −
Hconf
ij
=
Hhyp
"
4αs
3rij
+c
4αcon 8π
4αten 1
Si · Sj δ 3 (rij ) +
3
3mi mj 3
3mi mj rij
3Si · rij Sj · rij
− Si · Sj
2
rij
!#
,
ij
ij
ij
= HSO(cm)
HSO
+ HSO(TP)
ij
HSO(cm)
=
4αs
3r3
ij
HSO(TP)
=
»
2
3r
#
–" ~
~j
S
Si
1
1
~
+
+
·L
mi
mj
mi
mj
ij
∂Hconf
∂r
"
~j
~i
S
S
+ 2
m2i
mj
#
~
·L
Hadrons in the Quark Modela – p. 1
Hadrons in the Quark Modela – p. 1
Hadrons in the Quark Modela – p. 1
Hadrons in the Quark Modela – p. 1
Hadrons in the Quark Modela – p. 1
Hadrons in the Quark Modela – p. 1
Hadrons in the Quark Modela – p. 1
(1)
λ-type: φs χλ =2 10; Φλ
(2)
φλ χs =4 8; Φλ
√1
2
(3)
(φρ χρ − φλ χλ ) =2 8; Φλ
(4)
φA χρ =2 1; Φλ
(1)
ρ-type: φs χρ =2 10; Φρ
(2)
φρ χs =4 8; Φρ
√1
2
(3)
(φλ χρ + φρ χλ ) =2 8; Φρ
(4)
φA χλ =2 1; Φρ
(1)
A-type: φA χs =4 1; ΦA
√1
2
(2)
(φλ χρ − φρ χλ ) =2 8; ΦA
Hadrons in the Quark Modela – p. 1
I can write the ground state wave function in a notation (1S)3 . With this notation, the first
orbital excitation can be written (1S)2 (1P ).
Hadrons in the Quark Modela – p. 1
I can write the ground state wave function in a notation (1S)3 . With this notation, the first
orbital excitation can be written (1S)2 (1P ).
There are 3 possibilities for this: ~
r1 ψ, ~
r2 ψ, ~
r3 ψ, depending on which quark is excited (ψ
is a (1S)3 wave function).
Hadrons in the Quark Modela – p. 1
I can write the ground state wave function in a notation (1S)3 . With this notation, the first
orbital excitation can be written (1S)2 (1P ).
There are 3 possibilities for this: ~
r1 ψ, ~
r2 ψ, ~
r3 ψ, depending on which quark is excited (ψ
is a (1S)3 wave function).
I can create states with particular symmetries by taking linear combinations: assuming
all masses equal,
ψρ = √1 (~
r1 − ~
r2 ) ψ, ψλ = √1 (~
r1 + ~
r2 − 2~
r3 ) ψ, ψS = √1 (~
r1 + ~
r2 + ~
r3 ) ψ
2
6
3
Hadrons in the Quark Modela – p. 1
I can write the ground state wave function in a notation (1S)3 . With this notation, the first
orbital excitation can be written (1S)2 (1P ).
There are 3 possibilities for this: ~
r1 ψ, ~
r2 ψ, ~
r3 ψ, depending on which quark is excited (ψ
is a (1S)3 wave function).
I can create states with particular symmetries by taking linear combinations: assuming
all masses equal,
ψρ = √1 (~
r1 − ~
r2 ) ψ, ψλ = √1 (~
r1 + ~
r2 − 2~
r3 ) ψ, ψS = √1 (~
r1 + ~
r2 + ~
r3 ) ψ
2
6
3
We can set the center of mass to be at the origin: ~
r1 + ~
r2 + ~
r3 = 0, and ψS becomes
trivial: 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 anguar
momentum relative to some origin.
Hadrons in the Quark Modela – p. 1
I can write the ground state wave function in a notation (1S)3 . With this notation, the first
orbital excitation can be written (1S)2 (1P ).
There are 3 possibilities for this: ~
r1 ψ, ~
r2 ψ, ~
r3 ψ, depending on which quark is excited (ψ
is a (1S)3 wave function).
I can create states with particular symmetries by taking linear combinations: assuming
all masses equal,
ψρ = √1 (~
r1 − ~
r2 ) ψ, ψλ = √1 (~
r1 + ~
r2 − 2~
r3 ) ψ, ψS = √1 (~
r1 + ~
r2 + ~
r3 ) ψ
2
6
3
We can set the center of mass to be at the origin: ~
r1 + ~
r2 + ~
r3 = 0, and ψS becomes
trivial: 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 anguar
momentum relative to some origin.
”
“
(1)
(1)
1
The fully symmetric wave functions possible are: √ Φλ ψλ + Φρ + ψρ : 2 10
2
”
“
(2)
(2)
1
√
Φλ ψλ + Φρ + ψρ : 4 8
2 “
”
(3)
(3)
1
√
Φλ ψλ + Φρ + ψρ : 2 8
2 “
”
(4)
(4)
1
√
Φλ ψλ + Φρ + ψρ : 2 1
2
Hadrons in the Quark Modela – p. 1
Possible angular momentum values:
2 10 : 1 + 1 → J P = 1 − , 3 − (∆, Σ, no Λ)
2
2
2
48
:1+
28
:1+
21
:1+
3
2
1
2
1
2
→ JP =
→ JP =
→ JP =
1−
,
2
1−
,
2
1−
,
2
3− 5−
, 2 (nucleon,
2
3−
(same as 48 )
2
3−
(Λ only)
2
Λ, Σ, no ∆)
Hadrons in the Quark Modela – p. 1
Possible angular momentum values:
2 10 : 1 + 1 → J P = 1 − , 3 − (∆, Σ, no Λ)
2
2
2
48
:1+
28
:1+
21
:1+
3
2
1
2
1
2
→ JP =
→ JP =
→ JP =
1−
,
2
1−
,
2
1−
,
2
3− 5−
, 2 (nucleon,
2
3−
(same as 48 )
2
3−
(Λ only)
2
Λ, Σ, no ∆)
Number of states in the multiplet is 2 x 10 + 4 x 8 + 2 x 8 + 2 x 1=70. Multiplet is often
referred to as 70, 1− .
Hadrons in the Quark Modela – p. 1
Possible angular momentum values:
2 10 : 1 + 1 → J P = 1 − , 3 − (∆, Σ, no Λ)
2
2
2
48
:1+
28
:1+
21
:1+
3
2
1
2
1
2
→ JP =
→ JP =
→ JP =
1−
,
2
1−
,
2
1−
,
2
3− 5−
, 2 (nucleon,
2
3−
(same as 48 )
2
3−
(Λ only)
2
Λ, Σ, no ∆)
Number of states in the multiplet is 2 x 10 + 4 x 8 + 2 x 8 + 2 x 1=70. Multiplet is often
referred to as 70, 1− .
Examples of states in this mutliplet are
N (1520), N (1535), ∆(1620), ∆(1700), Λ(1405), Λ(1520)
Hadrons in the Quark Modela – p. 1
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. 1
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 i =
ΨL,mL
“
X
mS
ΨL,mL
“
”
ρ
~, ~
λ χ(S, mS )hL, mL , S, mS |J, M i
” X
“ ”
~
ρ
~, λ =
ψnρ ,ℓρ ,mρ (~
ρ) ψnλ ,ℓλ ,mλ ~λ hℓρ , mρ , ℓλ , mλ |L, mL i
mρ
Hadrons in the Quark Modela – p. 1
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 i =
ΨL,mL
“
X
mS
ΨL,mL
“
”
ρ
~, ~
λ χ(S, mS )hL, mL , S, mS |J, M i
” X
“ ”
~
ρ
~, λ =
ψnρ ,ℓρ ,mρ (~
ρ) ψnλ ,ℓλ ,mλ ~λ hℓρ , mρ , ℓλ , mλ |L, mL i
mρ
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. Orbital
excitations in ρ have ρ symmetry.
Hadrons in the Quark Modela – p. 1
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 i =
ΨL,mL
“
X
mS
ΨL,mL
“
”
ρ
~, ~
λ χ(S, mS )hL, mL , S, mS |J, M i
” X
“ ”
~
ρ
~, λ =
ψnρ ,ℓρ ,mρ (~
ρ) ψnλ ,ℓλ ,mλ ~λ hℓρ , mρ , ℓλ , mλ |L, mL i
mρ
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. Orbital
excitations in ρ have ρ symmetry.
Simplified flavor wave functions: ΣQ = uuq,
ΛQ =
ΞQ =
√1 (ud − du)Q, ΩQ = ssQ
2
1
√ (us − su)Q, √1 (ds − sd)Q,
2
2
√1 (ud
2
Ξ′Q =
+ du)Q, ddQ
1
√
(us
2
+ su)Q,
√1 (ds
2
+ sd)Q
Hadrons in the Quark Modela – p. 1
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 i =
ΨL,mL
“
X
mS
ΨL,mL
“
”
ρ
~, ~
λ χ(S, mS )hL, mL , S, mS |J, M i
” X
“ ”
~
ρ
~, λ =
ψnρ ,ℓρ ,mρ (~
ρ) ψnλ ,ℓλ ,mλ ~λ hℓρ , mρ , ℓλ , mλ |L, mL i
mρ
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. Orbital
excitations in ρ have ρ symmetry.
Simplified flavor wave functions: ΣQ = uuq,
ΛQ =
ΞQ =
√1 (ud − du)Q, ΩQ = ssQ
2
1
√ (us − su)Q, √1 (ds − sd)Q,
2
2
√1 (ud
2
Ξ′Q =
+ du)Q, ddQ
1
√
(us
2
+ su)Q,
√1 (ds
2
+ sd)Q
Space-spin wave function must have the same symmetry as flavor wave function.
Hadrons in the Quark Modela – p. 1
Let’s denote wave function components by S, L, nρ , ℓρ , nλ , ℓλ
Hadrons in the Quark Modela – p. 1
Let’s denote wave function components by S, L, nρ , ℓρ , nλ , ℓλ
For ΣQ with J P =
1+
,
2
1
, 0, 0, 0, 0, 0.
2λ
1
1
,
0,
1,
0,
0,
0
, 0, 0, 0, 1, 0. Other components
2λ
2λ
3
, 2, 0, 0, 0, 2; 12 ρ , 0, 0, 1, 0, 1 and 21 ρ , 1, 0, 1, 0, 1
2
components to see are
3
, 2, 0, 2, 0, 0;
2
first component of wave function is
Next easiest
are
Hadrons in the Quark Modela – p. 1
Let’s denote wave function components by S, L, nρ , ℓρ , nλ , ℓλ
For ΣQ with J P =
1+
,
2
first component of wave function is
1
, 0, 0, 0, 0, 0.
2λ
1
1
,
0,
1,
0,
0,
0
, 0, 0, 0, 1, 0. Other components
2λ
2λ
3
, 2, 0, 0, 0, 2; 12 ρ , 0, 0, 1, 0, 1 and 21 ρ , 1, 0, 1, 0, 1
2
components to see are
3
, 2, 0, 2, 0, 0;
2
JP
ΣQ
ΛQ
1+
21
1+
22
1+
23
1+
24
1+
25
1+
26
1+
27
1−
21
1−
22
1−
23
1
, 0, 0, 0, 0, 0
2λ
1
, 0, 1, 0, 0, 0
2λ
1
, 0, 0, 0, 1, 0
2λ
3
, 2, 0, 2, 0, 0
2
3
, 2, 0, 0, 0, 2
2
1
, 0, 0, 1, 0, 1
2ρ
1
, 1, 0, 1, 0, 1
2ρ
1
, 1, 0, 1, 0, 0
2ρ
1
, 1, 0, 0, 0, 1
2λ
3
, 1, 0, 0, 0, 1
2
1
, 0, 0, 0, 0, 0
2ρ
1
, 0, 1, 0, 0, 0
2ρ
1
, 0, 0, 0, 1, 0
2ρ
1
, 0, 0, 1, 0, 1
2λ
1
, 1, 0, 1, 0, 1
2λ
3
, 1, 0, 1, 0, 1
2
3
, 2, 0, 1, 0, 1
2
1
, 1, 0, 1, 0, 0
2λ
3
, 1, 0, 1, 0, 0
2
1
, 1, 0, 0, 0, 1
2ρ
Next easiest
are
Hadrons in the Quark Modela – p. 1
ij
Hconf
ij
=
Hhyp
3
X
i<j=1
"
„
«
3
X
brij
2αCoul
=
−
.
2
3r
ij
i<j=1
2αten 1
2αcon 8π
Si · Sj δ 3 (rij ) +
3
3mi mj 3
3mi mj rij
3Si · rij Sj · rij
− Si · Sj
2
rij
!#
,
ij
ij
ij
= HSO(cm)
HSO
+ HSO(TP)
ij
=
HSO(cm)
2αs
3
3rij
"
~
rij
~i
~j
~i − ~
~j
×p
~i · S
~
rij × p
~j · S
~
rij × p
~j · S
rij × p
~i · S
−
−
2
2
mi
mj
mi mj
ij
=−
HSO(TP)
1
2rij
ij
∂Hconf
∂rij
"
~
rij
~i
~j
×p
~i · S
~
rij × p
~j · S
−
m2i
m2j
#
#
Hadrons in the Quark Modela – p. 2
Consider the Coulomb term:
1
r12
+
1
r13
+
1
r23
Hadrons in the Quark Modela – p. 2
Consider the Coulomb term:
1
r12
+
1
r13
+
1
r23
Since r12 ∝ ρ, calculating h r1 i, or hf (r12 )i is easy: hf (r12 )i =
12
√
R
R 3 3
R
′∗
d ρd λΨ (ρ, λ)f (r12 )Ψ(ρ, λ) = d3 ρψ ′∗ (ρ)f ( 2ρ)ψ(ρ) d3 λψ ′∗ (λ)ψ(λ)
Hadrons in the Quark Modela – p. 2
Consider the Coulomb term:
1
r12
+
1
r13
+
1
r23
Since r12 ∝ ρ, calculating h r1 i, or hf (r12 )i is easy: hf (r12 )i =
12
√
R
R 3 3
R
′∗
d ρd λΨ (ρ, λ)f (r12 )Ψ(ρ, λ) = d3 ρψ ′∗ (ρ)f ( 2ρ)ψ(ρ) d3 λψ ′∗ (λ)ψ(λ)
r
q
√
2m2
3~
2
2
2 + 3 λ2 + √ 2m2
~
r13 = m 2m
ρ
~
+
λ
=⇒
r
=
ρ
ρ
~·~
λ.
13
2
+m
2
2
(m +m )
1
2
1
2
3(m1 +m2 )
How do we evaluate h r1 i, or hf (r13 )i?
13
One of two ways.
Hadrons in the Quark Modela – p. 2
Consider the Coulomb term:
1
r12
+
1
1
+
r13
r23
Since r12 ∝ ρ, calculating h r1 i, or hf (r12 )i is easy: hf (r12 )i =
12
√
R
R 3 3
R
′∗
d ρd λΨ (ρ, λ)f (r12 )Ψ(ρ, λ) = d3 ρψ ′∗ (ρ)f ( 2ρ)ψ(ρ) d3 λψ ′∗ (λ)ψ(λ)
r
q
√
2m2
3~
2
2
2 + 3 λ2 + √ 2m2
~
r13 = m 2m
ρ
~
+
λ
=⇒
r
=
ρ
ρ
~·~
λ.
13
2
+m
2
2
(m +m )
1
2
1
2
3(m1 +m2 )
How do we evaluate h r1 i, or hf (r13 )i?
13
One of two ways.
Example: expand
1
|~
ρ′ + ~
λ′ |
≈
X
ℓ
(−1)
ℓ
ρ′ℓ
ˆ′ )Y ∗ (λ̂′ ), ρ′ ≤ λ′
Y
(
ρ
ℓ
ℓ
λ′(ℓ+1)
Hadrons in the Quark Modela – p. 2
Consider the Coulomb term:
1
r12
+
1
1
+
r13
r23
Since r12 ∝ ρ, calculating h r1 i, or hf (r12 )i is easy: hf (r12 )i =
12
√
R
R 3 3
R
′∗
d ρd λΨ (ρ, λ)f (r12 )Ψ(ρ, λ) = d3 ρψ ′∗ (ρ)f ( 2ρ)ψ(ρ) d3 λψ ′∗ (λ)ψ(λ)
r
q
√
2m2
3~
2
2
2 + 3 λ2 + √ 2m2
~
r13 = m 2m
ρ
~
+
λ
=⇒
r
=
ρ
ρ
~·~
λ.
13
2
+m
2
2
(m +m )
1
2
1
2
3(m1 +m2 )
How do we evaluate h r1 i, or hf (r13 )i?
13
One of two ways.
Example: expand
1
|~
ρ′ + ~
λ′ |
≈
X
ℓ
(−1)
ℓ
ρ′ℓ
ˆ′ )Y ∗ (λ̂′ ), ρ′ ≤ λ′
Y
(
ρ
ℓ
ℓ
λ′(ℓ+1)
Evaluate hYℓ (ρˆ′ )i using dΩρ , hYℓ (λ̂′ )i using dΩλ , and remaining radial integrals involve
powers of ρ and λ
Hadrons in the Quark Modela – p. 2
Consider the Coulomb term:
1
r12
+
1
1
+
r13
r23
Since r12 ∝ ρ, calculating h r1 i, or hf (r12 )i is easy: hf (r12 )i =
12
√
R
R 3 3
R
′∗
d ρd λΨ (ρ, λ)f (r12 )Ψ(ρ, λ) = d3 ρψ ′∗ (ρ)f ( 2ρ)ψ(ρ) d3 λψ ′∗ (λ)ψ(λ)
r
q
√
2m2
3~
2
2
2 + 3 λ2 + √ 2m2
~
r13 = m 2m
ρ
~
+
λ
=⇒
r
=
ρ
ρ
~·~
λ.
13
2
+m
2
2
(m +m )
1
2
1
2
3(m1 +m2 )
How do we evaluate h r1 i, or hf (r13 )i?
13
One of two ways.
Example: expand
1
|~
ρ′ + ~
λ′ |
≈
X
ℓ
(−1)
ℓ
ρ′ℓ
ˆ′ )Y ∗ (λ̂′ ), ρ′ ≤ λ′
Y
(
ρ
ℓ
ℓ
λ′(ℓ+1)
Evaluate hYℓ (ρˆ′ )i using dΩρ , hYℓ (λ̂′ )i using dΩλ , and remaining radial integrals involve
powers of ρ and λ
Should work with any choice of functions used to expand wave function
Hadrons in the Quark Modela – p. 2
Hadrons in the Quark Modela – p. 2
Hadrons in the Quark Modela – p. 2
Hadrons in the Quark Modela – p. 2
Hadrons in the Quark Modela – p. 2
Hadrons in the Quark Modela – p. 2
Hadrons in the Quark Modela – p. 2
Hadrons in the Quark Modela – p. 2
Hadrons in the Quark Modela – p. 2
Hadrons in the Quark Modela – p. 2
So we have a spectrum, and wave functions. Are we done? Can we do anything else (do
we declare victory and go work on tans?)?
Hadrons in the Quark Modela – p. 2
So we have a spectrum, and wave functions. Are we done? Can we do anything else (do
we 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. 2
So we have a spectrum, and wave functions. Are we done? Can we do anything else (do
we 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; meson
semileptonic decays; meson rare decays; meson strong decays; baryon radiative
decays; baryon semileptonic decays; baryon rare decays; baryon strong decays...
Hadrons in the Quark Modela – p. 2
So we have a spectrum, and wave functions. Are we done? Can we do anything else (do
we 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; meson
semileptonic decays; meson rare decays; meson strong decays; baryon radiative
decays; baryon semileptonic decays; baryon rare decays; baryon strong decays...
Each of these represents a wealth of data. For instance, in the case of baryon
semileptonic decays: form factors, total and differential decay rates, polarization
asymmetries, decays to excited states, etc.
Hadrons in the Quark Modela – p. 2
So we have a spectrum, and wave functions. Are we done? Can we do anything else (do
we 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; meson
semileptonic decays; meson rare decays; meson strong decays; baryon radiative
decays; baryon semileptonic decays; baryon rare decays; baryon strong decays...
Each of these represents a wealth of data. For instance, in the case of baryon
semileptonic decays: form factors, total and differential decay rates, polarization
asymmetries, decays to excited states, etc.
Sketch of treatment of semileptonic decays (mesons), strong decays (baryons)
Hadrons in the Quark Modela – p. 2
Semileptonic decay: decay of a hadron to a final state containing a hadron and a pair of
leptons. Mediated by weak interaction: understanding these requires understanding the
interplay between the dynamics of the weak interaction leading to the decay, and the
strong interaction that provides the binding in the parent and daughter hadron
Hadrons in the Quark Modela – p. 2
Semileptonic decay: decay of a hadron to a final state containing a hadron and a pair of
leptons. Mediated by weak interaction: understanding these requires understanding the
interplay between the dynamics of the weak interaction leading to the decay, and the
strong 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-leptonic
decay (weak hadron decay to a final state that has no leptons: eg Λ → pπ − )
Hadrons in the Quark Modela – p. 2