florida state universityhadrons in the quark modela winston roberts [email protected] florida state...

Hadrons in the Quark Modela

Winston Roberts

[email protected]

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.

• Introduction, Classification Scheme, Simple Predictions

• Meson Spectrum

• Baryons Spectrum

• Meson Transitions

• Baryons Transitions & Sundries


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


Six flavors of quarks,

(

u

d

)

,

(

c

s

)

,

(

t

b

)

,

(

23

−13

)



(

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.



(

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).



(

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.



(

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).


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 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


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

3 ⊗ 3 = 1 ⊕ 8


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)

]


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


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)

]


For baryons:

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


rgr r

r r

r r

Σ0 [uds]

Λ [uds]

Σ− [dds] Σ+ [uus]

n [udd] p [uud]

Ξ− [dss] Ξ0 [uss]


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]


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





Similarly:



We’re ignoring the u− d mass difference.





Similarly:



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).


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


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.


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.




〈us|Hs|us〉 ≡ m0 + u+ s ≡ mK ,


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




〈us|Hs|us〉 ≡ m0 + u+ s ≡ mK ,


〈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)




〈us|Hs|us〉 ≡ m0 + u+ s ≡ mK ,


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


`

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.




〈us|Hs|us〉 ≡ m0 + u+ s ≡ mK ,


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


`

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


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)



√8

3(u− s)

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



√8

3(u− s)



|η′〉 = −|η8〉 sin θ + |η1〉 cos θ

−→ |η8〉 = |η〉 cos θ − |η′〉 sin θ

−→ η8 = η cos2 θ + η′ sin2 θ



√8

3(u− s)



|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´



√8

3(u− s)



|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


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.


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.


In positronium, the hyperfine splitting is 2πα3

~σ1·~σ2

m1m2|ψ(0)|2


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


For the meson case, ~S = 12

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


ρ = M + u+ d+ aud

; π = M + u+ d− 3aud

; π − ρ = 4aud


Similarly, K∗ −K = 4aus

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

cu, B∗ − B = 4a

bu, etc.


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.


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

m1m2+ ~σ1·~σ3

m1m3+ ~σ2·~σ3

m2m3


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

u2 +(~σ1+~σ2)·~σ3

um3


Black/whiteboard: color and baryon wave functions


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)


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]


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


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


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

)


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


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

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

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

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

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

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

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

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

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

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

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

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

Ω++ccc

Ξ+cc

Ξ++cc

Ω+cc

Ω0c ∆+∆0 ∆++∆−

Σ++c

Σ+c

Σ0c

Ω−Ξ0Ξ−

Ξ+c

Ξ0c

Σ− Σ+Σ0


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


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


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





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.





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


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)



2→ JP = 1

2

−, 3

2

−(∆, Σ, no Λ)

48 : 1 + 32→ JP = 1

2

−, 3

2

−, 5

2


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−.



2→ JP = 1

2

−, 3

2

−(∆, Σ, no Λ)

48 : 1 + 32→ JP = 1

2

−, 3

2

−, 5

2


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)


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〉




|J,M〉 =X

mS

ΨL,mL

“

~ρ, ~λ”


ΨL,mL

“

~ρ, ~λ”

=X

mρ


“

~λ”


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.




|J,M〉 =X

mS

ΨL,mL

“

~ρ, ~λ”


ΨL,mL

“

~ρ, ~λ”

=X

mρ


“

~λ”



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




|J,M〉 =X

mS

ΨL,mL

“

~ρ, ~λ”


ΨL,mL

“

~ρ, ~λ”

=X

mρ


“

~λ”



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.


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



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


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

#


Consider the Coulomb term: 1r12

+ 1r13

+ 1r23



+ 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)



+ 1r13

+ 1r23


〉, or 〈f(r12)〉 is easy: 〈f(r12)〉 =R



2ρ)ψ(ρ)R


~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.



+ 1r13

+ 1r23


〉, or 〈f(r12)〉 is easy: 〈f(r12)〉 =R



2ρ)ψ(ρ)R


~r13 =√

2m2

m1+m2~ρ+

q

32~λ =⇒ r13 =

r

2m22

(m1+m2)2ρ2 + 3

2λ2 + 2m2√

3(m1+m2)~ρ · ~λ.



Example: expand

1

|~ρ′ + ~λ′|≈X

ℓ

(−1)ℓ ρ′ℓ

λ′(ℓ+1)Yℓ(ρ′)Y

∗ℓ (λ′), ρ′ ≤ λ′



+ 1r13

+ 1r23


〉, or 〈f(r12)〉 is easy: 〈f(r12)〉 =R



2ρ)ψ(ρ)R


~r13 =√

2m2

m1+m2~ρ+

q

32~λ =⇒ r13 =

r

2m22

(m1+m2)2ρ2 + 3

2λ2 + 2m2√

3(m1+m2)~ρ · ~λ.



Example: expand

1

|~ρ′ + ~λ′|≈X

ℓ

(−1)ℓ ρ′ℓ


∗ℓ (λ′), ρ′ ≤ λ′

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



+ 1r13

+ 1r23


〉, or 〈f(r12)〉 is easy: 〈f(r12)〉 =R



2ρ)ψ(ρ)R


~r13 =√

2m2

m1+m2~ρ+

q

32~λ =⇒ r13 =

r

2m22

(m1+m2)2ρ2 + 3

2λ2 + 2m2√

3(m1+m2)~ρ · ~λ.



Example: expand

1

|~ρ′ + ~λ′|≈X

ℓ

(−1)ℓ ρ′ℓ


∗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


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.





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)


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


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−)


florida state universityhadrons in the quark modela winston roberts [email protected] florida state...

Documents