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C4 Option 2
What we will cover
• The Breit-Wigner resonance curve• Addition of angular momentum• Clebsch-Gordon coefficients
– Their derivation with raising and lowering operators
• Examples– Combining spin ½– Nucleon wave function– Decay of the �– The �(1232) resonance
C4 Option 3
•Broad states which can be formed by collisions between the particles into which the decay are referred to as resonances•They are described by the Breit-Wignerformula
( ) ( )
resonance theforming particles
two theof spins theare ,resonance theofspin theis
)121212
resonance theof mass is E
FWHM width total theis
f channeldecay of width partial is
i channelformation of width partial is
:where
4
4/4
21
21
R
22
2
ss
J
s)(s()J(
g
ph
EE
gE
j j
f
i
R
ji
+++=
=Γ=Γ
ΓΓ
=
Γ+−ΓΓ
=
�
λ
λπσ
Non-relativistic Briet-Wigner
�
C4 Option 4
The form of the Breit Wigner• The form derives from the exponential time dependence of the decay
– The energy dependence can then be found by taking the Fourier transform of the time domain
• The wave function of a non-stationary decaying state of central angular frequency �R = ER/� and lifetime �=�/� is:
• The intensity ��* follows the exponential decay law I∝e-�t
• The Fourier transform of the decay is:
• The amplitude as a function of E is:
1 where,)0()0()( )2/(2/ ==== Γ+−−− ceeet RR ittti�
ωτω ψψψ
�/ where)()(0
Edtetg ti == �∞
ωψω ω
� � Γ−−∝== −+Γ−
2/)(1
)0()()( )]()2/[(
iEEdtedtetE
R
EEitiEt Rψψχ
C4 Option 5
Form of the Breit Wigner• The cross section �(E) is proportional to *:
• The cross section falls to half the maximum at E-ER=±�/2• Optical theorem arguments lead to see Perkins 4th edition, pp57
for more details.• Spin multiplicity factors for the initial and final state are (2s1+1)(2s2+1) and
(2J+1), respectively. The cross section is scaled by final/initial.• The width scale if you are looking at formation in i and decay in j is �i�j/�2
• The Lorentz invariant relativistic form is:
( ) 4/4/
)(22
2
max Γ+−Γ=REE
E σσ
2max 4 �πσ =
( ) REMMsMs
ME =
Γ+−Γ= 02222
0
20
max where)(
)(0
σσ
C4 Option 6
Example: �++(1232)
Asymmetric width due to phase space variation over
resonance
Formed in elastic p scattering (�=�i =�f)
C4 Option 7
Addition of angular momentum
• Let J1 and J2 be 2 angular momentum operators i.e.– The orbital angular momentum (J1) and the spin (J2) of the
same particle– The angular momentum of spinless particle 1 (J1) and
spinless particle 2 (J2)
• These operators squared and their projection operators follow these relations
),(),( and ),()1(),(
),(),( and ),()1(),(
22222222222222
1111111111112
1
mjYmmjYJmjYjjmjYJ
mjYmmjYJmjYjjmjYJ
z
z
=+=
=+=
C4 Option 8
Addition of angular momentum• Consider the operators:
– Jz=J1z+J2z and– J2=(J1x+J2x)2+ (J1y+J2y)2 +(J1z+J2z)2
acting on the wave function – Y(j1,m1; j2,m2) = Y(j1,m1)Y(j2,m2)
• Y(j1,m1; j2,m2) is an eigenfunction of Jz with eigenvalueM=m1+m 2
• Y(j1,m1; j2,m2) is not in general an eigenfunction of J2
• However, linear combinations of Y(j1,m1; j2,m2) can produce eigenfunctions Y(J,M,j1,j2) of J2 with eigenvalues J(J+1)
��−=−=
=2
22
21
1
11
22112121 ),,,(),,,(j
jmjj
j
jm
) ,m; j,mY(jmmMJCjjMJY
Clebsch Gordoncoefficients
C4 Option 9
Raising and lowering operators• To derive Clebsh Gordon (CG) coefficients we need to make
use of raising and lowering operators• Angular momentum operators obey the commutation relation:
– [Ji,Jj]=i�ijkJk where �ijk=+1 or -1 if i, j, k are even (ie yzx) or odd permutations (ie zyx), respectively, and 0 otherwise (ie zzx)
• The only operator that commutes with Jx, Jy and Jz is:
• We define states as eigenfunctions of J2 and Jz with eigenvalues:
0],[ and 22222 =++= izyx JJJJJJ
jjjjm
mjmmjJmjjjmjJ z
,1,.....,1, where
,, and ,)1(,2
−+−−==+=
C4 Option 10
Raising and lowering operators
• Now we define raising and lowering operators:
which satisfy: [Jz,J±]=±J±
• Using this commutation relation:
yx iJJJ ±=±
),)(1(),(
,)1(
,)1(
,)(,
mjJmmjJJ
mjJm
mjJJ
mjJJJmjJJ
z
z
zz
−−
−
−
−−−
−=
−=
−=
−= In an analogous fashion it is possible to show that J+|j,m› is an
eigenfunction of Jz with eigenvalue m+1
1,, ���
mjCmjJ =
Generic constants that are dependent on j and m
C4 Option 11
Raising and lowering operators• Noting that J+ is the complex conjugate of J-
)1()1(),( so )1()1(
therefore,
],[
)(
but
,,
so, phasearbitrary an ignoring )()1( therefore
)(1,1,)(1,,
)1(,,)1(1,,
22
22
2222
22
2
***
±−+=−−+=
−−=
+−++=
−++=
=
==+=++=+
+=+=+
±
+−
+−
+−
+−
+−
+++
−−−
mmjjmjCmmjjC
JJJJJ
JJiJJJJJJ
JJJJiJJJJ
mjCmjJJ
CmCmC
mCmjmjmCmjJmj
mCmjmjmCmjJmj
zz
yxzzyx
xyyxyx
C4 Option 12
Clebsch Gordon coefficients• We can now use the expression for C2 to derive the C-G
coefficients• Consider 2 particles |j1,m1› and |j2,m2› forming a combined
state |j,m› where j1 and j2 are 1 and 1/2, respectively. Therefore, j is 1/2 or 3/2.
• Clearly the maximum (3/2)and minimum (-3/2) m states are:
01,1 and 1,120,1 ,0,121,1
0, and ,1,
: of definition theand operatorslower and raising theusing
,1,1,
,1,1,
21
21
21
21
21
21
21
21
23
23
21
21
23
23
=−−==
=−−=
−−=−
=
−−−
−−
−
JJJ
JJ
C
C4 Option 13
CG coefficients
• Now operating on the combined state
• In analogous fashion one finds:
• So we have all the j=3/2 states and the CG coefficients
21
21
31
21
21
32
21
23
21
21
21
21
21
21
21
23
23
23
,1,1,0,1,
,1,1,0,12
,1,1,3,
−+=
−+=
== −− JJ
21
21
31
21
21
32
21
23 ,1,1,0,1, −+−=−
C4 Option 14
Clebsch Gordon coefficients• To find the j=1/2 states we note that:
• Similarly with J-:31
32
21
21
21
21
21
21
22
21
21
21
21
21
21
and
toleading 02
therefore
,1,12,1,10,
apply Now .1with
,0,1,1,1,
−==
=+
+==
=+
+−=
+
+
ba
ba
baJ
Jba
ba
21
21
32
21
21
31
21
21 ,1,1,0,1, −−=−
operators applyingby computed beCan
,,
:case General
21
21 2211
±
�=
J
mjmjCj,mmm
jmmm
C4 Option 15
Tabulations of CG Coefficients from the PDG
C4 Option 16
Table of CG coefficients: 1×1/2 and 1/2×1/2
113/23/11
3/13/203/13/20
3/23/1111
212121212121
23
21
21
21
21
23
21
23
21
23
21
23
23
−−−+−
−−+
−+++
−−−+++Mmm
J
1
11001
1011
21
21
21
21
21
21
21
21
21
21
21
21
21
−−−−
−
−Mmm
J
C4 Option 17
Example: 2 spin ½ particles • Two spin ½ particles can combine to form 4 possible spin
states:– ��, � , �, (particle 1 and particle 2)
• The total spin can be s=1 a triplet state or s=0 a singlet with respect to sz
• Using CG coefficients we can write the irreducible representation:
)(),,,,(0,1
,,1,1
)(),,,,(0,1
,,1,1
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
↓↑−↑↓=−−−=
↓↓=−−=−
↓↑+↑↓=−+−=
↑↑== } Symmetric triplet
Antisymmetric singlet
C4 Option 18
Example: the nucleon wavefunction• Three quarks of spin ½ produce a baryon with spin ½
– Therefore, one quark’s spin antiparallel to the other two: ��
• The spin state is a mixture of symmetric and antisymmetricstates– The flavour wavefunction also has mixed symmetric, but total spin-
flavour wavefunction is symmetric
• Flavour asymmetry impossible with uuu, ddd and sss states– No spin a ½ states of this type observed
• The spin wavefunction of a proton with spin +1/2 written as the product of one quark’s spin wavefunction and that of the remaining pair
– Arbitrary choice of u-quark pair– Using our previously derived spin 1/2×1 CG coefficients
),()0,1(),()1,1(),( 21
21
32
21
21
32
21
21
duuduup χχχχχ +−=
C4 Option 19
Example: nucleon wave functions • Now substituting the wave functions for 2 spin ½ particles:
– Symmetric under interchange of the 2 u quarks• To make the other symmetric pairings: swap 1st with 3rd and 2nd with 3rd quark and
add the new terms. Normalised to 1.
• Other examples:– Make neutron with a u�d– Use two s quarks and a u or d to make �0 and �-, respectively– Wave functions for � triplet and �
( ) ( )↑↑↓↑↓↑↓↑↑↑ −−=
↑↓+↑↓==↑↑
duuduuduu
uuuu
61
61
32
21
p
: toleads )(0,1 and 1,1 χχ
)2
2
2(181
↓↑↑↑↓↑↑↑↓
↑↑↓↓↑↑↑↓↑
↑↑↓↑↓↑↓↑↑↑
−−+
−−+
−−=
uuduuduud
uduuduudu
duuduuduup
Swap 1st and 3rd quarks
Swap 2nd and 3rd quarks
C4 Option 20
Example: decay of the �• The � is an isospin singlet (I=0) and it decays almost solely to pπ- and n π0
• To a first approximation the decay is a s�u quark transition which changes the isospin by ½
• The pion-nucleon state is in a state I=1/2 with a third component I3=-1/2• The CG coefficients for spin 1 (pion) and ½ (nucleon) give
Therefore
• The measured ratio is 1.8 isospin not an exact symmetry especially in weak decay
pn −−=
−−−=−
ππ 320
31
21
21
32
21
21
31
21
21 ,1,1,0,1,
2,
,
)()(
2
31
32
2
21
210
2
21
21
0 =��
�
�
��
�
� −=
−
−=
→Λ→Λ
−−
π
ππσπσ
n
p
np
Assuming phase space the same due to the mass of n and p being approximately equal and the masses of the π being approximately equal
C4 Option 21
Summary
• Breit Wigner resonance form• Derivation of Clebsch Gordon coefficients• Some examples of their use
