Phys 400 Lecture 11
More Angular Momentum Addition
Isospin
Phys 400 Lecture 11
1
Addition of Angular Momentum
A state with angular momentum magnitude = j
and z component = m can be written j,!m .
It is an eigenstate of the j 2 operator with eigenvalue ! 2 j ( j + 1),
and an eigenvalue of the jz operator with eigenvalue ! m .
The product state j1 ,!m1 j2 ,!m2 is an eigenstate of the
operator J z = jz1 + jz 2 with eigenvalue ! ( m1 + m2 ) .
The jz1 operator only acts on the 1st factor, ignoring the 2d.
The jz 2 operator only acts on the 2d factor, ignoring the 1st.
Phys 400 Lecture 11
2
Addition of Angular Momentum 2
(j
( j )( j ,!m
(j
z1
1
z1
j1 ,!m1
z1
1
+ jz 2 ) ( j1 ,!m1 j2 ,!m2
)
j2 ,!m2 ) + ( jz 2 ) ( j1 ,!m1 j2 ,!m2
)
)
)( j ,!m ) + ( j ,!m )( j j ,!m
( !m j ,!m )( j ,!m ) + ( j ,!m )( !m j ,!m )
( !m )( j ,!m j ,!m ) + ( !m )( j ,!m j ,!m )
( !m + !m )( j ,!m j ,!m )
1
1
1
1
1
1
1
Phys 400 Lecture 11
2
2
1
1
2
2
1
1
2
2
2
2
1
1
z2
2
1
2
2
1
2
2
2
2
2
2
3
Addition of Angular Momentum 3
j2 ,!m2 is (usually) not an eigenstate
The product state j1 ,!m
1
! ! 2
! !
2
2
of the operator J = j1 + j2 = j1 + 2 j1 ! j2 + j22
(
)
The product state is an eigenstate of j12 or j22 , with eigenvalues
2
2
! j1 ( j1 + 1) or ! j2 ( j2 + 1) .
! !
j
!
j
But it is (usually)
not
an
eigenstate
of
1
2 , because the
!
output of each j operator can be a mix of different m values.
Phys 400 Lecture 11
4
Addition of Angular Momentum 4
But there are linear combinations of j1 ,!m1 j2 ,!m2 products,
j
,!m
,!
j
,!m
with carefully selected coefficients,
and
1
1
2
2 values,
!
!
2
2
that are eigenstates of J = j1 + j2 .
(
)
For given values of j1 and j2 , these special linear combinations
have a range of J 2 eigenvalues. The eigenvalues are
2
! J ( J + 1) with J = ( j1 ! j2 ) ,!...!,!( j1 + j2 ) in steps of 1
And there are several linear combinations of j1 ,!m1 j2 ,!m2
products for each J value, with different eigenvalues of
J z = jz1 + jz 2 in the range ! ( !J,!!J + 1,!...!+J ! 1,!+J ) .
Phys 400 Lecture 11
5
Clebsch-Gordon Coefficients
We use upper-case J and M for the state(s) of total angular
momentum, and lower case j and m (with subscripts) for the
sources. Both sources have fixed j values, varying m values.
Phys 400 Lecture 11
6
Reading Clebsch-Gordon Tables
For two spin-1/2 contributions, we can have J = 0 and J = 1.
The J = 1, M = +1 state is just m1 = +1 2,!m2 = +1 2.
The J = 1, M = 0 state is 1 2 times ( m1 = +1 2,!m2 = !1 2 )
plus 1 2 times ( m1 = !1 2,!m2 = +1 2 ) .
Phys 400 Lecture 11
7
Reading Clebsch-Gordon Tables (2)
The J = 0, M = 0 state is +1 2 times ( m1 = +1 2,!m2 = !1 2 )
plus !1 2 times ( m1 = !1 2,!m2 = +1 2 ) .
The J = 1, M = –1 state is just m1 = !1 2,!m2 = !1 2.
Phys 400 Lecture 11
8
Reading Clebsch-Gordon Tables (3)
You can also use it backwards.
The state m1 = !1 2,!m2 = +1 2 is +1
plus !1 2 times (J = 0, M = 0).
Phys 400 Lecture 11
2 times (J = 1, M = 0)
9
Adding More Than Two Sources
If you have three sources of j, first add two to get the resulting
J, M values, then add the third source to each. This likely
involves using several of the C-G blocks, and products of
coefficients.
A particle decay to 2 other particles with spin, plus orbital
angular momentum, has 3 contributions (only relative orbital
angular momentum counts). A decay to 3 particles with spin
would have 5 contributions.
If a particle spin is zero, there’s no spin to add.
Phys 400 Lecture 11
10
When to Use Clebsch-Gordon Tables
If we know a composite system has some total angular
momentum J and third component M, the probability that the
two components of the system have particular values for their
third components m1 ,!m2 is the square of the C-G coefficient
(remember that the coefficient in the table is a square-root!)
If a particle decays to two particles with spin, and if the orbital
angular momentum is zero (not guaranteed, but this tends to
dominate), the probabilities of the final state spin combinations
are the squares of the CG coefficients.
Phys 400 Lecture 11
11
The Big PDG Clebsch-Gordon Table
Phys 400 Lecture 11
12
Example 1
If an atomic electron has L = 1, it could have total J = 1/2.
If it has total MJ = –1/2, what is the probability that the
electron spin is up?
Phys 400 Lecture 11
13
Answer 1
If an atomic electron has L = 1, it could have total J = 1/2.
If it has total MJ = –1/2, what is the probability that the
electron spin is up?
Answer: 2/3
Phys 400 Lecture 11
14
Example 2
A particle with spin-1 decays into two spin-1 particles, with
zero orbital angular momentum. What is the probability that
one is m1 = +1 and the other is m2 = –1?
Phys 400 Lecture 11
15
Answer 2
If the initial M value is not specified, it’s equally likely to be
+1, 0, or –1. The M = +1 and –1 cases don’t give the specified
final state. Of the 1/3 of the decays from M = 0, half give the
specified state, so the answer is 1/6. (If we can’t distinguish
particle 1 from 2, we’d add the other row, giving 1/3).
No
Yes
Maybe
No
Phys 400 Lecture 11
16
Isospin and Strong Interactions
Isospin symmetry allows some predictions for pion-nucleon
interactions (at energies to low for many pions to be created).
Isospin also allows some predictions for decays.
We describe the initial and final states using isospin and CG
coefficients. Some states are linear combinations of different
isospins, some are pure.
Then we assume that the interaction links isospin components
in the initial state to the same components in the final state.
Phys 400 Lecture 11
21
Isospin
Isospin is a property of hadrons that obeys the same rules as
angular momentum. We use the symbol I (analogous to J) for
the magnitude, and there is a “z-component” we’ll call Iz.
The proton is I = 1/2, with Iz = +1/2.
The neutron is I = 1/2, with Iz = –1/2.
The ! + is I = 1, with Iz = +1.
The ! 0 is I = 1, with Iz = 0.
The ! – is I = 1, with Iz = –1.
Phys 400 Lecture 11
22
Isospin 2
The ! - baryons have isospin I = 3/2. There are 4 of them.
The !++ has Iz = +3/2.
The !+ has Iz = +1/2.
The !0 has Iz = –1/2.
The !– has Iz = –3/2.
The ! - baryons also have real spin J = 3/2 equal to their
isospin, and the proton and neutrons have real spin J = 1/2
equal to their isospin, but that’s mostly coincidence.
Pions have I = 1 but J = 0 so they aren’t equal
(but their partners the rho mesons do have I = J = 1).
Phys 400 Lecture 11
23
Conservation of Isospin
Isospin is conserved in strong interactions.
So when a ! decays, the final state particles (a nucleon and a
pion) must have the same isospin magnitude and z-component
as the initial state.
And the Clebsch-Gordon coefficients of the linear
combinations determine (relative) rates.
Phys 400 Lecture 11
24
Isospin Example 1
Compare the decay rates for
! ++ " # + p
+
+
+
0
! "# n
! "# p
!0 " # 0n
!0 " # $ p
$
$
! "# n
Phys 400 Lecture 11
25
Isospin Answer 1a
The ! is isospin I = 3/2. The different charge states are
I3 = 3/2, 1/2, –1/2, –3/2.
The pion is isospin I=1 with I3 = +1, 0, –1.
The proton and neutron are I=1/2 with I3 = +1/2, –1/2.
Phys 400 Lecture 11
26
Isospin Answer 1b
! + p = 3 2, + 3 2
+
! n=
3 2, +1 2 + 2 1 2, +1 2
3
2 3 2, +1 2 " 1 2, +1 2
! p=
0
3
2 3 2, "1 2 + 1 2, "1 2
! n=
0
"
! p=
3
3 2, "1 2 " 2 1 2, "1 2
Phys 400 Lecture 11
3
! " n = 3 2, " 3 2
27
Isospin Answer 1c
So the relative decay amplitudes are
!
++
+
+
" # p :! 3 2, + 3 2 3 2, + 3 2 = 1
+
! " # n :! 3 2, +1 2
+
! " # p :! 3 2, +1 2
0
! " # n :! 3 2, $1 2
0
0
$
! " # p :! 3 2, $1 2
0
3 2, +1 2 + 2 1 2, +1 2
3
2 3 2, +1 2 $ 1 2, +1 2
3
2 3 2, $1 2 + 1 2, $1 2
3
3 2, $1 2 $ 2 1 2, $1 2
3
! $ " # $ n :! 3 2, $ 3 2 3 2, $ 3 2 = 1
Phys 400 Lecture 11
=
1
3
=
2
3
=
2
3
=
1
3
28
Isospin Answer 1d
Square to get the relative decay rates:
!
++
+
$
$
" # p and ! " # n should be equal
! + " # 0 p and ! 0 " # 0 n should be 2/3 as much
! + " # + n and ! 0 " # $ p should be 1/3 as much
++
+
$
So ! ,!! ,!! ,!! should have the same total decay rates,
0
! + " # 0 p : ! + " # + n should be 2:1
! 0 " # 0 n : ! 0 " # $ p should be 2:1
Phys 400 Lecture 11
29
For Next Time
Read Chapter 4 (Symmetries), section 4 (discrete symmetries)
Phys 400 Lecture 11
35