3.1 Leptons. 3.1.1 Lepton multiplets and lepton numbers 3.1.2 Universal lepton interactions: the number of neutrinos 3.1.3 Neutrinos 3.1.4 Neutrino mixing and oscillations 3.1.5 Lepton numbers revisited 3.2 Quarks. 3.2.1 Evidence for quarks 3.2.2 Quark generation and quark numbers 3.3 Hadrons. 3.3.1 Flavour independence and charge multiplets 3.3.2 Quark model spectroscopy 3.3.3 Hadron magnetic moments and masses Problems. 05/03/14
F. Ould-Saada
1
¡
Hadron spectroscopy §  Study of hadron properties: masses, lifetimes, decay modes, spins, charges, and other QNs, led to Quark Model (Gell-­‐Mann, Zweig 1964) Baryons ≡ qqq
¡
Mesons ≡ qq
& p ≡ uud ; n ≡ udd
2
1
u(Q = + ) ; d(Q = − ) ⇒ ' +
−
3
3
(π ≡ ud ; π ≡ u d
Lepton scattering §  led to nucleon substructure and point-­‐like constituents €
€
¡
“Jet” production e+ e− → q + q → jet + jet
05/03/14
F. Ould-Saada
JADE@PETRA, DESY 2
¡
Jet production §  Production rate, angular distributions of jets match that of quarks §  Free quarks not observed, confined in hadrons e+ e− → q + q → jet + jet
¡
Modern theory of Strong Interactions: §  QCD – Quantum Chromo Dynamics – Ch.5 JADE@PETRA, DESY 05/03/14
F. Ould-Saada
3
05/03/14
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4
¡
6 flavours of spin-­‐1/2 quarks occurring in 3 generations of ▪  Up-­‐type with charge +2/3 à u,c,t ▪  Down-­‐type with charge -­‐1/3 à d,s,b §  Corresponding anti-­‐quarks " u%
$ ',
# d&
¡
"c %
$ ',
#s&
"t %
$ '
#b&
(GeV)
⎛ d ⎞
⎜⎜ ⎟⎟ ,
⎝ u ⎠
⎛ s ⎞
⎜⎜ ⎟⎟ ,
⎝ c ⎠
⎛ b ⎞
⎜⎜ ⎟⎟
⎝ t ⎠
Constituent masses, except Top which does not form bound-­‐states (why?) €
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F. Ould-Saada
5
¡
Weak Interactions responsible for up-­‐down transitions ¡
In SI and EMI quarks can only be created and destroyed in pairs ¡
e+ e− → γ → cc
N c = +1 for c; N c = −1 for c; N c = 0 for others
e+e− → γ →
/ cu
p ≡ uud → N u = 2, N d = 1, N s,c,b,t = 0
Quark number conservation: §  SI, EMI ¡
n → pe − νe
N f ≡ N( f ) − N( f )
( f = u, d, s, c, b, t)
Total quark number conserved N q ≡ N(q) − N(q ) ; c(1) → s(1) + u(1) + d (−1)
§  In WI ¡
Baryon Number conserved §  Nucleon is baryon, not pion 05/03/14
F. Ould-Saada
Nq
N ( q) − N ( q )
B≡
=
3
3
6
¡
Flavour independence à force between 2 quarks at same distance is independent of quark flavours involved – u,d,s,c,b,t §  us=ds; u-­‐ubar = d-­‐dbar ¡
Charge multiplets: §  Particles ~same mass, same quantum numbers (J,C,P,B, S,C,B’,...) but different Q §  Ex: (π+, π-, π0) , (p,n) à m(d)-­‐m(u) =(3±1)MeV/c2 §  à p(938)=uud ; n(940)=udd ¡
Flavor independence of strong force §  charge independence of nuclear forces àpp=pn forces à Mirror nuclei ¡
Symmetry between u and d quarks §  Isospin symmetry §  Simplifies interpretation of hadron physics 05/03/14
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7
¡
Mirror nuclei: same value of A but values of N and Z interchanged B ( A = 11, Z = 5, N = 6 ) !
#
10 pp,15nn pairs
#
" same np − pairs
11
6 C ( A = 11, Z = 6, N = 5)#
#
15pp,10 nn pairs
$
11
5
¡
¡
Charge symmetry approximately verified à If different np-­‐pairs, effect due to fact that np pair not subject to Pauli principle §  test of charge independence ¡
11
5
B , 116C , 114Be
But evidence that strong force stronger for S=1 than S=0 state 05/03/14
F. Ould-Saada
Low lying energy levels of mirror nuclei 8
Nucleon ≡ N ≡ (np)
1 1
,+
2 2
p ≡
≡ (10) ; n ≡
Ladder operators allow transformations between states p and n
1 1
,−
2 2
I + ≡ I1 + iI 2 = (00 10 ) ; I − ≡ I1 − iI 2 = (10 00 )
≡(10)
Commutation rules :
[I3 , I ± ] = ± I ± ; [I + , I − ] = 2I3
N ' = (n'p') = GN = G (np)
Casimir operator :
SU(2) : groupof unitary2x2matriceswithdet=1
combination of gen. which commutes with all generators of the group
!2
C =[I+ I− + I− I+ ] + I = I + I + I = I
2
3
†
G ⋅ G = E ; det(G) = 1
2
1
2
2
2
3
SU(2) acts on a 2 dimensional space. Transfromation or rotation matrices:
G(θ i ) = e−iθi Ii ; where the generator of the group are discribed by Pauli matrices
! $ I1 '
σi
I i = 2 ; I = & I2 ) ∈ Isospinspace
% I3 (
" 0 1 %"1 %
σ1 = (
) ; σ = ( ) ; σ = ( ) ; 1= ( )
0 1
1 0
2
0 −i
i 0
3
1 0
0 −1
1 0
0 1
Commutationrules:
Tr{I i } = 0 ; [I i , I j ] = iεijk I k ; {I i , I j } = δij
Proton and neutron are eigenstates of I 3
I 3 p = + 12 p ; I 3 n = − 12 n
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F. Ould-Saada
I+ p ≡ $
#
"
I+ n ≡ $
#
"
I− p ≡ $
#
"
I− n ≡ $
#
'$ ' = 0
0 0 &# 0 &
0 1 %" 0 % "1 %
'$ ' = $ ' ≡ p
0 0 &#1 & # 0 &
0 0 %"1 % " 0 %
'$ ' = $ ' ≡ n
1 0 &# 0 & #1 &
0 0 %" 0 %
'$ ' = 0
1 0 &#1 &
9
¡
2-­‐dimensional space representation of SU(2) §
§
¡
Nucleon: (p,n) or quarks (u,d) Generators are 2x2 Pauli matrices 3-­‐dimensional space representation of SU(2) §
§
Basis: (π+, π0, π-) with isospin I=1 and I3=+1,0,-­‐1 Generators are 3x3 matrices ¡
4-­‐dimensional space representation of SU(2) §
§
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Δ(1232): I=3/2 à I3=3/2,1/2,-­‐1/2,-­‐3/2 Basis (Δ++, Δ-, Δ0, Δ-) F. Ould-Saada
10
¡  Isospin operators §  Anology with spin §  Commutation relations §  Ladder operator " Iˆ %
! $ x'
I = $ Iˆy ' ;
$ Iˆ '
# z&
this appendix: “-­‐” sign 05/03/14
F. Ould-Saada
x
Iˆ± ≡ Iˆx ,±iIˆy ;
y
;
z
[Iˆ , Iˆ ] = iIˆ
[Iˆ , Iˆ ] = iIˆ
y
z
x
i
j
k
;
[Iˆ , Iˆ ] = iIˆ
z
x
y
Iˆ3 ≡ Iˆz
[Iˆ , Iˆ ] = 2Iˆ ; [Iˆ , Iˆ ] = ± Iˆ
+
−
(Iso)spin −
§  Notice different convention in [Iˆ , Iˆ ] = iIˆ
3
±
"
$
1/ 2 : $
$
#
3
±
"
%
u%' $ + 1 ' "$↑%'
' = $ 2' = $ '
d '& $$# − 12 ''& $#↓'&
1
1
1
1
Iˆ3 u = + 2 u; Iˆ3 d = − 2 d; Iˆ3 u = − 2 u; Iˆ3 d = + 2 d ;
Iˆ+ u = 0; Iˆ+ d = u; Iˆ− u = d; Iˆ− d = 0
Iˆ u = −d ; Iˆ d = 0; Iˆ u = 0; Iˆ d = −u
+
+
−
−
11
¡
⇒ Iˆα A = ∑ βi Iˆα Ai ; α = +,−,3
state : A = ∑ βi Ai
i
i
Iˆα (ab) = ( Iˆα a)b + a( Iˆα b)
Iˆ (abc) = ( Iˆ a)bc + a( Iˆ b)c + ab( Iˆ c)
α
α
α
α
1
Iˆ 2 ≡ Iˆx2 + Iˆy2 + Iˆz2 = 2 ( Iˆ+ Iˆ− + Iˆ− Iˆ+ ) + Iˆ32
Iˆ 2 I,I3 = I(I +1) I,I3
Iˆ3 I,I3 = I3 I,I3 ;
Isospin operators & quark states §  State |A> composed of quarks and antiquarks |Ai> §  Family of (2I+1) states |I,I3> simultaneously eigenstates of I3and I2 with eigenvalues I(I+1) and I3 I3 = −I,−I +1,...I −1,I
Iˆ± I,I3 = C± (I,I3 ) I,I3 ± 1 ; C± (I,I3 ) = + [(I ∓ I3 )(I ± I3 +1)]
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12
# a;I,I = I &
3
%
(
a;I,I
=
I
−1
3
%
(;
%...
(
%
(
$ a;I,I3 = −I '
¡
Hadron states exist in multiplets §  u,d : isospin doublets §  s,c,b,t: isospin singlets ¡
Isospin 1 §  Pion §  |0,0> identified as eta meson §  Notice different convention in this appendix: “-­‐” sign
§  I have used in the rest of my slides (for example for the pion) 1
1
# u& # d &
% (;% (
$ d ' $ −u '
1
1
u = q; 2 , − 2 ; d = q; 2 , − 2
1
1
1
1
u = − q; 2 , − 2 ; d = q; 2 , + 2
# us & # sd & #K + & # K 0 &
( →% 0 ( ; % − (
% (;%
ds
$ ' $ −su ' $K ' $ −K '
I = 1,I3 = +1 = −ud ⇒ Iˆ− I = 1,I3 = +1 = 2 I = 1,I3 = 0
1
Iˆ− (−ud ) = −( Iˆu)d − u( Iˆd ) = −dd + uu ⇒ I = 1,I3 = 0 =
(uu − dd )
2
Iˆ ⇒ I = 1,I = −1 = du
−
3
I = 0,I3 = 0 = αuu + βdd
1
Iˆ+ I = 0,I3 = 0 = 0 ; α 2 + β 2 = 1 ⇒ I = 0,I3 = 0 =
(uu + dd )
2
# −π + &
% 0 (
1
+
0
(uu − dd ) , π − = du
%π ( ; π = ud , π =
2
%π − (
$
'
$π + '
& 0)
1
+
0
(uu + dd ) , π − = du
&π ) ; π = ud , π =
2
&π − )
% (
05/03/14
F. Ould-Saada
13
Similar to spin ! BasicConcepts1c (s. 8-10)
1
2
1
× : 2 ⊗ 2 = 3⊕ 1
2
1
1× : 3 ⊗ 2 = 4 ⊕ 2
2
1 1
1
2
2
× × : 2 ⊗ 2 ⊗ 2 = 4S ⊕ 2 MS ⊕2 MA
2
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14
Notation for coupling of 2 isospin
1
particles
2
1 1
× : 2 ⊗ 2 = 3⊕ 1
2 2
2 = 2 ⋅ i + 1 → Isospin multiplicity
Result is 3 symmetric components (3) and 1 antisymmetric ( 1 )
1
1× : 3 ⊗ 2 = 4 ⊕ 2
2
1 1 1
× × : 2 ⊗ 2 ⊗ 2 = 4S ⊕2 MS ⊕2 MA
2 2 2
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15
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16
1
1(π ) × (N ) : 3 ⊗ 2 = 4 ⊕ 2
2
π ≡ (π +, π 0 , π − ) ; N ≡ ( p, n)
π N States → π N; I, I 3
Δ
++
≡
3
3
,+
2
2
=π p
+
Δ+ ≡
3
1
,+
2
2
=+
Δ0 ≡
3
1
,−
2
2
=
Δ− ≡
3
3
,−
2
2
= π −n
N* ≡
1
1
,+
2
2
=+
N* ≡
1
1
,−
2
2
=
05/03/14
1
3
1
3
π +n
π−p
2
3
2
3
π +n
π−p
+
+
−
−
%
'
2
0p '
π
'
3
&
2
0n
π
'
3
'
'(
1
3
1
3
Γ (Δ+ → π +n)
Γ ( Δ + → π 0 p)
=
1
2
π0p
π 0n
F. Ould-Saada
17
Δ0 ≡
3
1
,−
2
2
=
1
3
π−p
+
N* ≡
1
1
,−
2
2
=
2
3
π−p
−
π−p
π 0n $&
%⇒
1
0n
π
π 0n
&'
3
2
3
=
1
1 3
,−
2
3 2
−
1
2 1
,−
2
3 2
=
1
2 3
,−
2
3 2
+
1
1 1
,−
2
3 2
⇒ M (π − p → π − p) = 13 M 3/2 + 23 M1/2 $&
%
−
0
2
2
M (π p → π n ) = 3 M 3/2 − 3 M1/2 &'
I(Δ) = 3 / 2 ⇒ M 3/2 >> M1/2 @ Δ(1232)
2
σ total (π − p) = σ (π − p → π − p) + σ (π − p → π 0 n ) ∝ 13 M 3/2 &
+
(σ
π
p)
(
(
total
3
3
++
+
Δ ≡ 2,+ 2 = π p
=3
'
−
σ total (π p)
(
2
σ total (π + p) ∝ M 3/2
()
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18
σ total (π + p )
=3
−
σ total (π p )
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19
¡
Hadrons B ≡ qqq ; B ≡ q q q
§  Baryons and anti-­‐baryons: half-­‐integral spin §  Mesons and anti-­‐mesons: integral spin ¡
Strange particles M ≡ qq
p = uud ; n = udd
π + = ud ;π 0 = uu , dd ; π − = u d
§  First discovered in 1947 in cloud chamber photographs of cosmic rays. §  K-­‐mesons or Kaons first called V-­‐particles §  Hyperons involve protons in their decays ▪  Produced in SIs (t~10-­‐23s) but decay through WIs(t~10-­‐10s) K 0 → π + + π −
π − + p →Λ 0 + K 0
Λ0 → π −+p
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F. Ould-Saada
20
¡
Associated production §  Pais, 1952: strange particles produced in pairs §  New quantum number, strangeness S, was introduced in 1953 by Gell-­‐
Mann and Nishijima. S ≡ −N s = −( N(s) − N(s ))
§  S is conserved in strong & EM interactions but not in weak interactions. π − + p → Λ0 + K 0
€
S: 0
0
−1
+1 → ΔS = 0
Λ0 → π − + p
S:
−1
0
0
→ ΔS =1
K 0 →π + +π −
S:
05/03/14
F. Ould-Saada
+1
0
0
→ ΔS =1
21
¡
Kaons §  K+ and K-­‐ (m~494MeV) and K0 (m~498MeV) §  Spin 0 and odd parity,close to pion but Isospin ½ §  K0 : Particle (S=1) distinct from antiparticle (S=-­‐1) ▪  Leads to important effects ¡
Hyperons §  Spin ½ like proton: Λ0 (M~1115MeV, I=0) §  Σ+,-­‐,0(M~1190MeV, I=1): Ξ0,-­‐ ,
(M~1190MeV, I=½, S=-­‐2 ) §  more massive hyperons with spin 3/2, 5/2,... 05/03/14
F. Ould-Saada
1
2
⎧ I 3 = + 12 K +
⎨
1
0
⎩ I 3 = − 2 K
S = −1, I =
⎧ I 3 = + 12 K 0
⎨
1
I
=
−
K−
⎩ 3
2
S = 1, I =
1
2
K − p →Λπ 0
π − p→Σ0K 0 ; Σ0→Λγ
π ± p→Σ±K +
Σ±→nπ ± ; Σ+→pπ 0
K − p→Ξ−π +K 0 ; Ξ−→Λπ −
Ξ0→Λπ 0 ; Λ→pπ −
22
¡
GNN relation ¡  Electric charge Q, Baryon number B, Strangeness S, ¡
3rd component of isospin I3 , Hypercharge Y=B+S With the hand of this relation 2 hyperons predicted in 1953 Σ0, Ξ- discovered in 1958 and 1959 P
J =0
05/03/14
F. Ould-Saada
−
Q = I3 +
B+S
Y
= I3 +
2
2
P
J =
1+
2
23
3 ⊗ 3 = 1⊕ 8
¡
¡
Meson= quark-­‐antiquark combination 3 ⊗ 3 = 1 ⊕ 8
Total spin: §  S=0 (antisymmetric spin wave function) and S=1 (symmetric)
¡
2S+1l
J
àPossible
P=(–1)l+1 C=(–1)l+S
; JPC
1S
–+
0 ; 0 3S
––
1;1
¡
1P
+–
1;1
3P
(allowed) states:
++
0;0
3P
++
1;1
3P
++
2;2
……..
Ground state mesons L=0: 9 pseudoscalars
JPC = 0– +
05/03/14
2 ⊗ 2 = 1⊕ 3
F. Ould-Saada
9 vectors
(JPC = 1– –)
24
¡
From the total angular momentum ! ! ! 2 ! !
! !
J = L + S = L2 + S 2 + 2 L ⋅ S
(
)
§  Account for spin-­‐space coupling §  With ¡
¡
¡
! ! J(J +1) − L(L +1) − S(S +1)
L⋅S =
2
L−S ≤ J ≤ L+S
There are no q-­‐ qbar mesons with exotic quantum numbers such as JPC=0+-­‐ , 1-­‐+ ,… On the other hand exotic states (non-­‐q-­‐
qbar) with normal quantum numbers are not forbidden: gluballs, hybrids, pentaquarks … It is the duty of hadrons spectroscopy to confirm the existing hadrons and to find the missing ones. 05/03/14
F. Ould-Saada
25
u,d,s
!
S = 1 2 + 1 2 + 1 2 →S = 1 2 , 3 2
! !
L = 0 →J = S
Hyperons JP=3/2+ Baryons JP=1/2+
*
*
*
05/03/14
F. Ould-Saada
*
Not symmetric enough à something is missing! *
26
The Ω- (S=-3) predicted to have a mass of 1680 MeV, discovered in 1964 at
mΩ = 1674 ±3 MeV, τΩ=82 ps
05/03/14
F. Ould-Saada
27
Top does not form
bound-states,
simply too heavy
¡  Charmed and bottom hadrons C ≡ N c ≡ (N (c) − N (c ) )
~
B ≡ − N b ≡ −(N (b) − N (b ) )
P
J =
3+
2
baryons
¡  Hypercharge B + S + C + B! + (T )
Y
Q = I3 +
= I3 +
2
2
C: Charm; B! : Bottom; (T :Top )
Y = B + S + C + B! + (T )
05/03/14
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29
¡  Charmed mesons 05/03/14
F. Ould-Saada
30
05/03/14
S
C
S
C
F. Ould-Saada
31
Symmetry again and again ¡
¡
After the discovery of the t lepton in 1975, no doubt for a 3rd quark family Anomaly-­‐free theory Q(l − ) + Q(ν l ) + 3× [Q(qu ) + Q(qd )] = 0
N c = 3; l = e, µ, τ ; qu = u, c, t; qd = d, s, b
§  Sum of charges within all families is 0 §  Existence of 6 quark flavours required within the CKM quark mixing matrix even before the discovery of charm quark in 1974! ¡
b-­‐quark discovered in 1976 §  JPC=0-­‐+ B-­‐mesons §  2 isospin doublets and 4 isospin singlets 05/03/14
F. Ould-Saada
! +
# B
# Bo
"
$ !
$ ! Bo
bu
&≡#
& #
& # bd & ; # B −
% "
% "
$ !
$
bd
& ≡ ##
&&
&
% " bu %
Bso ≡ ( bs ) ;
Bso ≡ ( bs ) Bc+ ≡ ( bc) ;
Bc− ≡ ( bc ) 32
Braibant et al.
05/03/14
F. Ould-Saada
33
K *+ → K 0 + π +
us → ds +ud τ =1.3×10−23 s
π 0 (uu, dd) → γγ
τ = 0.8 × 10
−16
s
π + (ud) → µ +ν µ
τ = 2.6 × 10
−8
s
Λ(1116) → pπ −, nπ 0
sud → uud+du
05/03/14
τ =2.6×10−10 s
F. Ould-Saada
34
¡
Dirac theory §  Point-­‐like fermion with electric charge eqe, mass m, spin ½ has magnetic dipole moment §  Magnetic moments eq e! " eq M P %
1
1
'' µ N
µ q ≡ q, Sz = µ̂ z q, Sz = =
= $$
2
2
2mq c # mq &
e!
: nuclear magneton
2M P
2M P
M
M
µu =
µ N , µd = − P µ N , µs = − P µ N
3mu
3md
3ms
µN ≡
§  Constituent quark masses mu ≈ md ≈ m = 0.344GeV ; ms = 0.539GeV
Λ (1116) = uds ⎫
⎬
s(ud ) = 0
⎭
MP
⇒ µΛ = µ s = −
µN
3ms
05/03/14
F. Ould-Saada
35
¡
½+ baryons B with quark configuration aab §  aa in symmetric spin-­‐1 state with parallel spins à magnetic moment 2µa §  ½ * 1 à ½ B; S = 12 ,Sz = 12 =
2
3
−
b; S = 12 ,Sz = − 12 aa; S = 1,Sz = 1
1
3
b; S = 12 ,Sz = 12 aa; S = 1,Sz = 0
1
1
1
1
µ B ≡ B;S = , Sz = µ̂ z B;S = , Sz =
2
2
2
2
µ B = 23 ( 2µ a − µ b ) + 13 µ b = 43 µ a − 13 µ b
µ p = 43 µu − 13 µ d =
05/03/14
F. Ould-Saada
MP
µN ;
m
mu ≈ md ≈ m
36
¡
Quark model is not bad at reproducing magnetic moments … §  The agreement is however not perfect 05/03/14
F. Ould-Saada
37
¡
Hadron masses §  1/2+ Baryon octet M Ξ − M Σ = M Ξ − M Λ = M Λ − M N = ms − mu,d
M Ω − M Ξ* = M Ξ* − M Σ* = M Σ* − M Δ = ms − mu,d
§  3/2+ Baryon decuplet §  à ms-­‐mu,d~120-­‐200MeV/c2 §  L=0 à differences in masses due to spin structure of states §  interaction energy between 2 spin ½ particles with magnetic moments 05/03/14
F. Ould-Saada
! !
µ ⋅µ
ΔE ∝ i 3 j
rij
! $ ei ' !
; if pointlike µi = & ) Si
% mi (
8π ei e j
2 ! !
ΔE =
ψ (0) Si ⋅ S j
3 mi m j
ψ (0) : wave function at the origin rij = 0
! !
Si ⋅ S j
Chromomagnetic interaction ⇒ ΔM ∝
mi m j
38
% 3
2
'
−
"
!2
! ! 2 !2 !2
! !
! ! ' 4
S ≡ S1 + S2 = S1 + S2 + 2 S1 ⋅ S2 ⇒ S1 ⋅ S2 = &
' + 1 "2
'( 4
(
)
M (meson ) = m1 + m2 + ΔM
! !
Si ⋅ S j
ΔM ∝
mi m j
M K = m + ms −
05/03/14
(S = 0)
(S = 1)
%
3a 1
' ΔM ( J P = 0 − ) = −
4 m1m2
'
&
' ΔM J P = 1− = a 1
(
) 4mm
'
1 2
(
a : constant
3a
4mms
F. Ould-Saada
39
! !
Si ⋅ S j
ΔM ∝ ∑
, i, j = 1, 3
m
m
i j
i< j
3+
2
! !
decuplet (↑↑↑) ⇒ S1 + S2
!2 !2
! !
! ! ! ! ! ! "2
2
= S1 + S2 + 2 S1 ⋅ S2 = 2" ⇒ S1 ⋅ S2 = S1 ⋅ S3 = S2 ⋅ S3 =
4
b% 1
2 (
M Σ* = 2m + ms + ' 2 +
*
4 & m mms )
1+
2
! ! !
octet (↑↑↓) ⇒ S1 + S2 + S3
(
)
2
!2 !2 !2
! ! ! ! ! !
3! 2
= S1 + S2 + S3 + 2 S1 ⋅ S2 + S1 ⋅ S3 + S2 ⋅ S3 =
4
" " " " " "
3! 2
⇒ S1 ⋅ S2 + S1 ⋅ S3 + S2 ⋅ S3 = −
4
(
)
2
(
(
)
)
!
!
! ! 2
! ! −3" 2
Λ = uds : S(ud) = 0 ⇒ Su + Sd = 0 ⇒ Su ⋅ Sd =
4
! !
! ! ! !
b % Su ⋅ Sd Su ⋅ Ss Sd ⋅ Ss (
M Λ = mu + md + ms + 2 '
+
+
*
" & mu md mu ms md ms )
! !
! ! ! ! ! ! ! !
%
b Su ⋅ Sd S1 ⋅ S2 + S1 ⋅ S3 + S2 ⋅ S3 − Su ⋅ Sd (
3b
=2m + ms + 2 ' 2 +
* ⇒ M Λ = 2m + ms −
" & m F. Ould-Saada
mms
4m 2
05/03/14
)
(
)
40
05/03/14
F. Ould-Saada
41
¡  Pages 107-­‐108 §  3.1 , 3.2, 3.3, 3.4 §  3.6 §  3.12, 3.13, 3.14, 3.15, 3.16 05/03/14
F. Ould-Saada
42
05/03/14
F. Ould-Saada
From Thomson
43