Mnemonics
Angle addition and
Wigner-Eckart
We write the natural notation:
j1 ⊗ j2 ≡ (j1 + j2 ) ⊕ (j1 + j2 − 1) ⊕ ... ⊕ |j1 − j2 |
and we have
| j1 j2 , jm % =
coefficients
• Clebsch-Gordan
Spin-orbit
coupling
• Thomas precession
• Wigner-Eckart theorem
•
!
m1 ,m2
| j1 j2 , m1 m2 % & j1 j2 , m1 m2 | j1 j2 , jm %
with m = m1 + m2 and |j1 − j2 | ≤ j ≤ |j1 + j2 |
. – p.8/12
1
What we are doing is block diagonalizing
the rotation matrices:
2
Clebsch Gordon: spin half
so that
D
j1 +j2
.
| 1, 1 ! =
D(R) =
Dj1 +j2 −1

1 1
!
2 2

| 1 − 1 !
 2 2

 | −1 1 !

2 2
| − 12 − 12 !
|
X
X
X
= |
= |
1
2
1
2
1
2
1
2
!"
!

1 1
|
2 2

" 12 − 12 |


" − 12 12 | 

" − 12 − 12 |
"
1
2
1
2
| 1, 1 !
(3)
| 1, 1 !
Dj1 −j2
. – p.10/12
3
4
| 12 12 , 00 !
| 12 12 , 10 !
Example: the two spin 12 singlet:


X | 12 12 ! " 12 12 |


1
1 
 1 1
.  | 2 − 2 ! " 2 − 2 |
o 0! =  1 1
| 1,
1, 0 !
 |o
 |−
! " − 12 12 | 


2 2
X| − 12 − 12 ! " − 12 − 12 |
Expample: The state with Sz = 0 and Stot = 1 with two spin 12 .


| 12 12 ! " 12 12 |
X


1
1 
 1 1
.  | 2 − 2 ! " 2 − 2 |
| 1, 0 ! = 
(4)
 | 1, 0 !
 | −1 1 ! " −1 1 | 

2 2
2 2 
X| − 12 − 12 ! " − 12 − 12 |
(5)
= | −12 12 !" −12 12 | 0, 0 ! + | 12 −12 !" 12 −12 |o1, 0 !
1
= √ ( | 12 −12 ! − | −12 12 !)
2
= | − 12 12 !" − 12 12 | 1, 0 ! + | 12 − 12 !" 12 − 12 | 1, 0 !
1
= √ ( | −12 12 ! + | 12 −12 !)
2
Look CG coefficients up: we shall not derive them.
They are called “Clebsch Gordan” in Mathematica.
. – p.11/12
. – p.12/12
5
6
34. Clebsch-Gordan coefficients
34. Clebsch-Gordan coefficients
34. Clebsch-Gordan coefficients
010001-1
34. CLEBSCH-GORDAN COEFFICIENTS, SPHERICAL HARMONICS,
34. CLEBSCH-GORDAN COEFFICIENTS, SPHERICAL HARMONICS,
AND d FUNCTIONS
!
Note: A square-root sign is to be understood over every coefficient, e.g., for −8/15 read − 8/15.
1/2 "1/2
1
0
+1
1
0
0
+ 1/2 + 1/2 1
+ 1/2 ! 1/2 1/2 1/2 1
! 1/2 + 1/2 1/2 ! 1/2 ! 1
+ 1 + 1/2
3/2
+ 3/2 3/2 1/2
1 + 1/2 + 1/2
+ 1 ! 1/2
0 + 1/2
J
J
M
M
m1 m2
"
3
cos θ
m 1 m 2 Coefficien t s
2" 1/2 + 5/2
5/2 5/2 3/2
4π
.
.
"
+ 2 +1/2
1 + 3/2 + 3/2
.
.
3
.
.
Y11 = −
sin θ eiφ
+ 2 ! 1/2 1/5 4/5 5/2 3/2
8π
+ 1 + 1/2 4/5 ! 1/5 + 1/2 + 1/2
" #
+ 1 ! 1/2 2/5 3/5 5/2 3/2
5 3
1$
0
2
Y2 =
cos θ −
0 + 1/2 3/5 ! 2/5 ! 1/2 ! 1/2
4π 2
2
"
0 ! 1/2 3/5 2/5 5/2 3/2
15
! 1 + 1/2 2/5 ! 3/5 ! 3/2 ! 3/2
Y21 = −
sin θ cos θ eiφ
2
! 1 ! 1/2 4/5 1/5 5/2
8π
3/2 "1/2 + 2 2 1
"
! 2 + 1/2 1/5 ! 4/5 ! 5/2
1 15
+ 3/2 +1/2 1 + 1 + 1
! 2 ! 1/2
1
Y22 =
sin2 θ e2iφ
4 2π
1
+ 3/2 ! 1/2 1/4 3/4 2
Note: A square-root sign is to be understood over every coefficient, e.g., for −8/15 read −
AND
d FUNCTIONS
"
1/3 2/3 3/2 1/2
2/3 ! 1/3 ! 1/2 ! 1/2
0 ! 1/2 2/3 1/3 3/2
! 1 + 1/2 1/3 ! 2/3 ! 3/2
0
+ 1/2 + 1/2 3/4 ! 1/4 0
3
! 1 ! 1/2
1
1
+ 1/2 ! 1/2 1/2 1/2
3/2 "1 + 5/2
2
+3 3
2
5/2 5/2 3/2
! 1/2 + 1/2 1/2 ! 1/2 ! 1 ! 1
1 +2 +2
+ 3/2 + 1
1 + 3/2 + 3/2
2
+ 2 0 1/3 2/3
1
3
! 1/2 ! 1/2 3/4 1/4 2
3/2 1/2
+ 3/2 0 2/5 3/5 5/2
+ 1 + 1 2/3 !1/3
+1
+1
+1
! 3/2 + 1/2 1/4 ! 3/4 ! 2
+ 1/2 + 1 3/5 ! 2/5 + 1/2 + 1/2 + 1/2
+ 2 !1 1/15 1/3 3/5
! 3/2 ! 1/2 1
2/5 1/2
+ 3/2 ! 1 1/10
+ 1 0 8/15 1/6 ! 3/10
2
1
3
+ 1/2 0 3/5 1/15 ! 1/3 5/2
3/2 1/2
1
2
0 + 1 2/5 ! 1/2 1/10
0
0
0
! 1/2 + 1 3/10 ! 8/15 1/6 ! 1/2 ! 1/2 ! 1/2
+1 +1
+ 1 ! 1 1/5 1/2 3/10
+ 1/2 ! 1 3/10 8/15 1/6
0
1/2 1/2
2
1
2
1
0 ! 2/5
0 0 3/5
3
! 1/2 0 3/5 ! 1/15 ! 1/3 5/2 3/2
0
0
1/2 ! 1/2
0
! 1 + 1 1/5 ! 1/2 3/10
!1 !1
!1
! 3/2 + 1 1/10 ! 2/5 1/2 ! 3/2 ! 3/2
! 1/2 ! 1 3/5 2/5 5/2
+ 1 ! 1 1/6 1/2 1/3
0 ! 1 2/5 1/2 1/10
! 3/2 0 2/5 ! 3/5 ! 5/2
2
! 1 0 8/15 ! 1/6 ! 3/10
3
0 0 2/3
0 ! 1/3 2
1
! 2 + 1 1/15 ! 1/3 3/5 ! 2 ! 2
! 1 + 1 1/6 ! 1/2 1/3 ! 1 ! 1
! 3/2 ! 1
1
1
0
+1
1
0
0
+ 1/2 + 1/2 1
+ 1/2 ! 1/2 1/2 1/2 1
! 1/2 + 1/2 1/2 ! 1/2 ! 1
1/2"1/2
+2 +1
+1 +1
2
+2
1
+1 0
0 +1
Y"−m = (−1)m Y"m∗
j
d m! ,m
m−m!
= (−1)
0 ! 1 1/2 1/2 2
! 1 0 1/2 ! 1/2 ! 2
!1 !1 1
j
d m,m!
2 " 3/2
=
d "m,0 =
+2 !1
+1 0
0 +1
! 1 +2
1 + cos θ
θ
cos
2
2
√ 1 + cos θ
θ
3/2
d 3/2,1/2 = − 3
sin
2
2
√ 1 − cos θ
θ
3/2
d 3/2,−1/2 = 3
cos
2
2
1 − cos θ
θ
3/2
d 3/2,−3/2 = −
sin
2
2
3 cos θ − 1
θ
3/2
d 1/2,1/2 =
cos
2
2
3 cos θ + 1
θ
3/2
d 1/2,−1/2 = −
sin
2
2
3/2
d 3/2,3/2 =
! 1 ! 1 2/3 1/3 3
! 2 0 1/3 ! 2/3 ! 3
!2 !1 1
4π
Y m e−imφ
2# + 1 "
! 1/2 ! 1/2 1
"j1 j2 m1 m2 |j1 j2 JM #
= (−1)J−j1 −j2 "j2 j1 m2 m1 |j2 j1 JM #
θ
1 + cos θ
1/2
1 = cos θ
2
d 0,0
d 1/2,1/2 = cos
d 11,1 =
2
2
+2
θ
sin θ
1/2 1/2
3
2
1
1/2
d 1/2,−1/2 = − sin
d 11,0 = − √
1/2 ! 1/2 + 1
+1
+1
2
2
+ 3/2 ! 1/2 1/5 1/2 3/10
1 − cos θ
0
3
2
1
+ 1/2 + 1/2 3/5
0 ! 2/5
d 11,−1 =
0
0
0
! 1/2 + 3/2 1/5 ! 1/2 3/10
0
2
2/5
+ 3/2 ! 3/2 1/20 1/4 9/20 1/4
5/2 3/2
1/2
! 2/5
7/2
+ 1/2 ! 1/2 9/20 1/4 ! 1/20 ! 1/4
1/5 + 1/2 + 1/2 + 1/2 + 1/2
1
3
2
! 1/2 + 1/2 9/20 ! 1/4 ! 1/20 1/4
!1
! 3/2 + 3/2 1/20 ! 1/4 9/20 ! 1/4 ! 1 ! 1
! 3/2 1/35 6/35 2/5
2/5
0 ! 3/10
! 1/2 12/35 5/14
+ 1/2 ! 3/2 1/5 1/2 3/10
5/2 3/2 1/2
+1/2 18/35 ! 3/35 ! 1/5
1/5
7/2
0 ! 2/5
! 1/2 ! 1/2 3/5
2
3
+3/2 4/35 ! 27/70 2/5 ! 1/10 ! 1/2 ! 1/2 ! 1/2 ! 1/2
! 3/2 + 1/2 1/5 ! 1/2 3/10 ! 2 ! 2
+ 1 ! 3/2 4/35 27/70 2/5 1/10
! 1/2 ! 3/2 1/2 1/2 3
4
3
2
1
0 ! 1/2 18/35 3/35 ! 1/5 ! 1/5
! 3/2 ! 1/2 1/2 ! 1/2 ! 3
+1
+1
+1
+1
! 1 +1/2 12/35 ! 5/14
5/2 3/2
0 3/10 7/2
! 3/2 ! 3/2 1
! 2 +3/2 1/35 ! 6/35 2/5 ! 2/5 ! 3/2 ! 3/2 ! 3/2
1/14 3/10 3/7 1/5
3/7 1/5 ! 1/14 ! 3/10
0 ! 3/2 2/7 18/35 1/5
3/7 ! 1/5 ! 1/14 3/10
1
0
4
3
2
! 1 ! 1/2 4/7 ! 1/35 ! 2/5 7/2
5/2
0
0
0
1/14 ! 3/10 3/7 ! 1/5
0
0
! 2 + 1/2 1/7! 16/35 2/5 ! 5/2 ! 5/2
+ 2 ! 2 1/70 1/10 2/7 2/5 1/5
! 1 ! 3/2 4/7
3/7 7/2
+ 1 ! 1 8/35 2/5 1/14 ! 1/10 ! 1/5
! 2 ! 1/2 3/7 ! 4/7 ! 7/2
0 ! 2/7
0 1/5
0 0 18/35
!
2
!
3/2
1
8/35 ! 2/5 1/14 1/10 ! 1/5
4
! 1 +1
2
1
3
! 2 +2
1/70 ! 1/10 2/7 ! 2/5 1/5
!1
!1
!1
!1
7/2
+ 7/2 7/2 5/2
+ 2 + 3/2
1 + 5/2 + 5/2
5/2
+ 2 + 1/2 3/7 4/7 7/2
+ 1 + 3/2 4/7 ! 3/7 + 3/2 + 3/2
+ 2 ! 1/2 1/7 16/35
+ 1 +1/2 4/7 1/35
0 +3/2 2/7 ! 18/35
2 "2 +44 4 3
+2
+2 +2 1 +3 +3
+1
3
2
+ 2 + 1 1/2 1/2
4
0
+ 1 + 2 1/2 ! 1/2 + 2
+2 +2
!1
+ 2 0 3/14 1/2 2/7
+ 1 +1 4/7
0 ! 3/7
0 +2 3/14 ! 1/2 2/7
"
3/2 " 3/2
j
d −m,−m!
d 22,2 =
3
+3 3
1 +2
+ 3/2 + 3/2
+ 3/2 + 1/2
+ 1/2 + 3/2
3/2
+ 3/2
3/7 1/5
+ 1 ! 2 1/14 3/10
1/5 ! 1/14 ! 3/10
0 ! 1 3/7
! 1 0 3/7 ! 1/5 ! 1/14 3/10
! 2 +1 1/14 ! 3/10
3/7 ! 1/5
# 1 + cos θ $2
2
1 + cos θ
d 22,1 = −
sin θ
2
√
6
2 =
d 2,0
sin2 θ
4
1 − cos θ
d 22,−1 = −
sin θ
2
# 1 − cos θ $2
d 22,−2 =
2
1 + cos θ
(2 cos θ − 1)
2
"
3
d 21,0 = −
sin θ cos θ
2
1 − cos θ
d 21,−1 =
(2 cos θ + 1)
2
d 21,1 =
4
!2
3
!2
2
!2
0 ! 2 3/14 1/2 2/7
! 1 ! 1 4/7
0 ! 3/7
4
3
! 2 0 3/14 ! 1/2 2/7 ! 3 ! 3
! 1 ! 2 1/2 1/2 4
! 2 ! 1 1/2 ! 1/2 ! 4
!2
d 20,0 =
#3
2
cos2 θ −
1$
2
!2
1
Figure 34.1: The sign convention is that of Wigner (Group Theory, Academic Press, New York, 1959), also used by Condon and Shortley (The
Theory of Atomic Spectra, Cambridge Univ. Press, New York, 1953), Rose (Elementary Theory of Angular Momentum, Wiley, New York, 1957),
and Cohen (Tables of the Clebsch-Gordan Coefficients, North American Rockwell Science Center, Thousand Oaks, Calif., 1974). The coefficients
here have been calculated using computer programs written independently by Cohen and at LBNL.
1 "1/2
+ 1 + 1/2
3/2
+ 3/2 3/2 1/2
1 + 1/2 + 1/2
+ 1 ! 1/2
0 + 1/2
1/3 2/3 3/2 1/2
2/3 ! 1/3 ! 1/2 ! 1/2
0 ! 1/2
! 1 + 1/2
3
+3 3
2
+2 +1 1 +2 +2
2" 1
2/3 1/3
1/3 ! 2/3
! 1 ! 1/2
2
+ 2 0 1/3 2/3
1
3
+ 1 + 1 2/3 !1/3
+1
+1
+1
+ 2 !1 1/15 1/3 3/5
2
1/6 ! 3/10
3
1" 1 + 22 2 1 + 10 + 01 8/15
2/5 ! 1/2 1/10
0
0
+1 +1 1 +1 +1
+ 1 ! 1 1/5 1/2
0
+ 1 0 1/2 1/2
2
1
0
0 0 3/5
0 + 1 1/2 ! 1/2
0
0
0
! 1 + 1 1/5 ! 1/2
7
+ 1 ! 1 1/6 1/2 1/3
0 0 2/3
0 ! 1/3 2
1
! 1 + 1 1/6 ! 1/2 1/3 ! 1 ! 1
Y"−m = (−1)m Y"m∗
!
8/15.
N ot a t ion :
J
J
M
M
...
...
m1 m2
34. Clebsch-Gordan
coefficients 010001-1
J
J
...
3
N
otθa t ion : 2" 1/2 5/2
cos
m 1 m 2 Coefficien t s
M
M + 5/2. . .5/2 3/2
4π
.
.
"
+ 2 +1/2
1 + 3/2 + 3/2
+ 1/2 ! 1/2 1/2 1/2 1
. HARMONICS,
.
3m
34.
COEFFICIENTS,
SPHERICAL
"
1 θm
! 1/2 ! 1
1/2CLEBSCH-GORDAN
! 1/2 + 1/2
.
.
Y11 = −
sin
eiφ2
+ 2 ! 1/2 1/5 4/5 5/2 3/2
3
8π
+ 1 + 1/2 4/5 ! 1/5 + 1/2 + 1/2
! 1/2 ! 1/2
5/21
Y10 =
cos θ
"
m
m
2" 1/2 + 5/2 5/2 3/2
Coefficien t s + 1 ! 1/2 2/5 3/5 5/2 3/2
#AND
$
1
2 d1 FUNCTIONS
5
3
4π
J
J
...
Y20 =
. cos2 θ. −
0 + 1/2 !3/5 ! 2/5 ! 1/2
"
N ot!a1/2
t ion :
+ 3/2
+ 2 3/2
+1/2 sign
1 +is3/2
Note:
A square-root
to be
understood over 4π
every2. coefficient,
M
M
...
. 2 e.g., for −8/15 read − 8/15.
1"1/2
3
"
0 ! 1/2 3/5 2/5 5/2 3/2
3/2 1/2
.
.
Y11 = −
sin θ eiφ + 1 + 1/2 + 3/2
! 1/2
1/5
4/5
5/2
3/2
15
m
m
2/5
!
1
+
1/2
!
3/5
!
3/2
!
3/2
"
11++21/2
1
2
+ 1/2
Y+211/2
= −3
sin θ cos θ eiφ
1/2 "1/2 + 1 + 11 + 1/2
8π
!
1/5
+
1/2
4/5
2
! 1 ! 1/2 4/5 1/5 5/2
5/2
+ 1 ! 1/2 1/3 02/3 3/2 1/2
" #
Y10 = " 8π
cos θ
m 2 Coefficien t s
"1/2 + 2 2 1
2" 1/23/2
!m
2 1+ 1/2
1/5 ! 4/5 ! 5/2
+ 5/2 5/2 3/2
02/3 !01/3 ! 1/2
+ 1/21+$
1/2
1
0 + 1/2
+ 1 !1/2
1/2 2/5 3/51 4π
5/2
5 3
15 3/2
.
.
+1/2+ 3/2
1 +1 +1
2 θ e2iφ + 2 +1/2 + 3/2
1 + 3/2
1
Y20 =
cos2 θ − + 1/2 ! 1/2 1/2 01/2
Y!222/5
= "
sin
.
. ! 2 ! 1/2
3/5
0
+
1/2
!
1/2
!
1/2
1/3
3/2
! 1/21 2/3
3
4 2π sin θ eiφ
! 1 1/3 ! 2/3 ! 3/2 Y 1 = −
!2
1/2 + 1/2 1/2 !!11/2
4π 2
.
+ 2 ! 1/2+ 3/2
1/5! 1/2
4/5 1/4
5/2 3/4
3/2 2 . 1
+ 1/2
1
8π 2/5 5/2 3/2
"
0
0 ! 1/2 3/5
+ 1 + 1/2+ 1/2
1/5 3/4
+ 1/2! 1/4
+ 1/2 0
4/5+!1/2
!31/2 ! 1/2 1
" # 5/2
! 1 ! 1/2
1
2"
1
$
15
2/5
!
1
+
1/2
!
3/5
!
3/2
!
3/2
1/2 5/2
1/2 3/22
1
! 1/2
3/2 "15 +35/2 25/2 13/2
+ 1 ! 1/2+ 1/2
2/5
3/5
Y21 = −
sin θ cos θ +e2iφ+ 1 + 31 + 23 + 22
Y20 =
cos θ −
3/5
! 2/5
+ 1/2
1/2!!1/2
1/2! 1/2
!1 !1
2
! 12 !11/2
+ 3/2 4/5
8π
+ 3/24π
+1
+23/2 1/5 5/2 0 + 1/2! 1/2
1"1/2+ 2 + 3/2
3/2
"
0 ! 1/2
3/5
5/2 23/2
"
2+ 2 1 2
0 1/3
2/3"1/2
1
! 1/2 !
1/2 2/5
3/4 1/4
1/2 3
3/2
3/2
2 +01/22/5 1/5
5/2
5/2 !3/2
1/2
+!
3/2
3/5 ! 4/5
15
2/5
! 1 + 1/2
3/2 !!23/2
!1/3
+1 +1 1
+ 1 1 + 1+ 1 Y+11= − + 1/2
2/3
1 15
+ 1 + 1/2
+ 1/2
1/2+ 1
! 3/2 +
1/2! 3/5
1/4 !!3/4
+sin
1 θ3/5
1/2 + 1/2 + 1/2
+1/2
++3/2
cos!θ!2/5
e2iφ!+1/2
2
2
1
2
sin2 θ e2iφ + 1 ! 1/2 +1/3
! 1 ! 1/2 4/5 1/5 5/2
8π
3/2 Y2 =
1/153/21/31/2 3/5
2 !12/3
! 3/2 ! 1/2 1
2/5"1/2
+ 3/2 ! 1 1/10 3/2
1/2 + 2
4 2π
2
1
! 2 + 1/2 1/5 ! 4/5 ! 5/2
+ 3/2
!!1/2
1/4 33/4 2 2"1 1
! 1/21/6
1/2
8/15
+2/3
1 !
01/3
! 3/10
! 3/2
+ 1/2 0 3/5 1/15 ! 1/3 5/2
3/2 1/2
1
15
1" 1 + 22 2 0 + 1/2
+ 3/2 +1/2 1 + 1 + 1
2 0 0 0 0
2iφ + 1 3/10 ! 8/15
1
3/4
+2/5
1/2
+ 1/2
! 1/4
00+!
11/2
! 1/2
! 2 ! 1/2
1
sin2 θ!e1/2
1/6 ! 1/2 ! 1/2 ! 1/2
2/3
1/31/10
3/2 0Y2 =
43/102π
+1 +1 1 +1 +1
1
1
+ 3/2 ! 1/2 1/4 3/4 2
! 1 + 1/2 1/3 ! 2/3
1/21/2
+ 1+!1/2
!3/2
1 1/5
+ 1/2 ! 1 3/10 8/15 1/6
1/2
1
! 1/2
3/2 "1 + 5/2
2
3/4 ! 1/4
0
0
+ 1/20 + 1/2
0
+ 1 0 3/2
1/2 1/2 3 2
1
5/2 5/2
2
1
0 ! 2/5
0 3/5
3
! 1/2
3/5 ! 1/15 ! 1/3 5/2 3/2
! 1 ! 1/20! 1/2
1
2"
1
5/2
+ 1/2
0 + 1+ 3/2
1/2 ! 1/2
3/10
! 1 + 1 1/5
!3/2
1/21/2
1
! 3/2 + 1 1/10
3/2 ! 3/2
+ 1/2! 2/5
! 1/2 1/2 !1/2
"1! 1/2! 1 ! 1! 1 ! !11
2
+ 3 03 02 0
+ 3/2 + 1
1 + 3/2
1/2"1/2 + 11 1 0
! 0
Y1 =
Note: A square-root sign is to be understood over every coefficient,
e.g.,1 for
read − 8/15.
0
0 −8/15
+ 1/2 + 1/2
2" 1
1"1
AND
d FUNCTIONS
34. CLEBSCH-GORDAN COEFFICIENTS, SPHERICAL
HARMONICS,
...
...
Y10 =
! 1/2 ! 1/2 1
1 "1/2
N ot a t ion :
010001-1
010001-1
0 ! 1 1/2 1/2 2
! 1 0 1/2 ! 1/2 ! 2
!1 !1 1
"
+ 5/2
5/2
3/2
! 1/2 + !
1/2
1 !1
+ 2 + 1 1 11/6
+2
+ 2 1/2
3/5 !2/5
1/21/2
! 1 ! 1/2
5/2
2/513/4
1/10
0 !!
+1/2
3/2 1/4
3/2
+111/2
+ 3/2 2
!+1/2
3/2 1/31/2
+ 3/2 0 2/5 3/5+ 1 !5/2
2/5 !
! 3/2
0 ! 1/2
3/5 1/4
! 5/22
8/150 ! 1/6
23/2 1/2
! 1 +03/2
! 3/10
3
+02 0 02/3
1
1/3 2/3
0 ! 1/3 3 2 2 1
! 1/2
3/4
2/5
3/5 ! 25/2
! 3/2
+ 1/2
1/4
! 3/4
+ 1/2 + 1 3/5 ! 2/5!+11++1/2
1/2 1/3
++1/2
1/15
!
1/3
!
2
!
2
3/5
! 2 ++11/2
+ 11/6
1! 1 + 1! 1 + 1
2/3!+!1/3
1
1/2
! 3/2 + 1/2
1/4!!13/4 !
! 3/2
12
+ 1 3/5 ! 2/5 + 1/2 + 1/2 + 1/2
1 1/32/53 1/2
! 3/2+!3/2
!101/15
+ 3/2 ! 1 1/10 + 22/5
! 11/2
!1/2
1 ! 1 2/3
1/21/3
1/2 3/5
2
! 3/2 ! 1/2 1
1/10
−m 2 0 m3/5
1
1 !
018/15
! 3/10
21/2
3
! 2 0 1/3
! 3 ! 1/3 5/2 "j3/2
!01/3
5/2
+ 1/2
3/5! 2/3
1/15
! 1/2
! 2 3/2
1/21/6
= (−1) Y"m∗ +1/15
" 1
1 j2 m1/2
1 m2 |j1 j2 JM #
1Y"" +11/2
1
+2
2
2/5
0
+
1
!
1/2
1/10
0
0
0
1/6
!
1/2
+
1
!
8/15
!
1/2
!
1/2
!
1/2
3/10
0
! 1/2 + 1 3/10 ! 8/15 1/6 ! 1! 1/2
! 1/2 4π Y m e−imφ
!2 !1 1
! 1 1 ! 1/2
"
+1 +1 1 +1 +1
= (−1)J−j1 −j2 "j2 j1 m2 m1 |j2 j1 JM #
= 3/10
m,0 1/2
+ 1 ! 1 d1/5
+ 1/2 ! 1 3/10 8/15 1/6
2# + 1 "
3/10
+ 1 0 1/2 1/2
2 +11/2 0! 1 3/10
2
1
01/6
! 2/5
0 0 8/15
3/5
3
! 1/2 0 3/5 ! 1/15 ! 1/3 5/2 3/2
1 m−m
! 2/5
3j 2
0 + 1 1/2
! 1/2 ! 0j !01/2 0j 0 3/5
! 1/15
!3/2
1/3
5/2
3/2
3/10
!1 +1
1/5 ! 1/2
!
1
!
1
!
1
!
3/2
+
1
! 2/5 1/2 ! 3/2 ! 3/2θ
1/10
3
" 3/2 + 3
1 + cos θ
1/2
d m,m! = d −m,−m!
3
2
d 10,0 = cos θ ! 1/2
d 1/2,1/2
= cos
d 11,1 =
3/10
! 1d m! ,m
! 1 = (−1)
! 1+ 1 ! 1 1/6
! 3/2
! 2/5 1/2
! 1 3/5
2/5 5/2
13/2
1/2 1/10
1/3+ 1 1/10
2/5 ! 3/2
0 !!
1/2
2
+ 3/2 + 3/2
1 +2
+2
83/52 ! 5/2
! 3/2 0 2/5 !
8/15
2
!
1
!
1/6
!
3/10
0
3
0
0
0
2
1
2/3
!
1/3
! 1/2 ! 1 3/5
2/5 1/25/2
7/2
0 ! 1 2/5 1/2 1/10
θ
sin θ
2"
3/2
3
2
1
+
3/2
+
1/2
1/2
1/2
1
1/15 ! 1/3 3/5 ! 2 ! 2
1/3 ! 1 ! 1
! 1 + 1 1/6 ! 1/2 7/2
d 1/2,−1/2! 3/2
d 1,0 = − √
= −! 1sin 1
! 3/2 ! 20 + 12/5
3/5 1/2
! 5/2
+ 1/2 +!3/2
! 1/2 + 1 + 1
+1
2 1 + 5/2 + 5/2
! 1 0 8/15 ! 1/6 ! 3/10 + 2 +33/2 + 7/2
2
2
5/2 1/2 2
! 1 ! 1 2/3 1/3 3
0 ! 1 1/2
!1
1/2 1/5 1/2 3/10
2 ! 13/70 1/2
−m 3/5 m! 2m∗
! 2 + 1 1/15 !Y1/3
! 3/2 ! 1++!3/2
4/7! 1/2
7/2! 2 5/2 3/2"
01/2
2 +
1/3 !3/5
2/3 ! 30 ! 2/5
= (−1) Y" + 2 +!1/2
1 − cos θ
2 JM #
3"j1 j22m1 m21|j1 j0
1/2
"
d 11,−1 =
+ 3/2 4/7 ! 3/7! 1+ 3/2
+ 3/2
4π
2 ! 1! 1/2
1 3/10
0 J−j01 −j20
! 1/2 + 3/2 !1/5
0
! 1 1 + 3/2
"
m
−imφ
3
! 1 ! 1 2/3 + 11/3
2
= (−1)
"j2 j1 m2 m1 |j2 j1 JM #
d
=
Y" e
1/2 1/7 16/35
+ 3/2 ! 3/2 1/20 1/4 9/20 1/4
! 2 0 1/3 ! 2/3 ! 3 ++ 21 !+1/2
"j1m,0
j!22/5
m m2#2+
|j11j5/2
# 1/2
"
2 JM3/2
1/35
2/51 7/2
4/7
+ 1/2 ! 1/2 9/20 1/4 ! 1/20 ! 1/4
4
4π 2"2
!
+"
1/2
2/7 ! 18/35 1/5
j
! 2 !j 1 1 0 +3/2
1
1/20 1/4
2
J−j
−j+21/2 +31/2 + 1/2
! 1/2 + 1/2 9/20 ! 1/4 !1/2
3/2
mj −imφ m−m
θ3
1 + cos θ
13/2
Spin-orbit interaction
Spin-orbit coupling and Thomas
precession
Now consider a spin circling a nucleus. In the rest-frame of the
electron, due to the Lorenz-transformation of the electric £eld,
there is a B-£eld
Bef f =
B
v
q
× E = 3v × r
c
cr
so that assuming the electron is rotating as in the £gure
HLS = µBef f · S
µZe
= 3 (mv × r) · S
r mc
µZe
= − 3 L·S
r mc
Ze2
= − 3 2 2L · S
r mc
E
. – p.1/5
9
10
Thomas precession
Spin-orbit coupling and Thomas
precession
Now the tricky part is that in the comoving reference frame of
the electron with S rotates about L
dS
= ω0 × S
dt
1
=
L×S
mr2
B
However, the frequency is now shifted between the reference
frame in the lab and the electron, so the Larmor-freqency is
E
shifted by a factor γ = √
≈1+
1
1−v 2 /c2
which can be expanded as
v2
2c2
. – p.2/5
11
12
Thomas precssion, continued
Thomas precession, continued
In the lab reference frame
This gives a contribution to the spin precession in the lab frame
of
dS
1 v2
=
L×S
dt
mr2 2c2
2
mv
1
=
L×S
r 2m2 r
2
Ze
1
=
L×S
r2 2m2 r
2
Ze
=
L×S
2r3 m2 c2
dS
v2
=
ω0 × S
dt
2c2
v2
=
L×S
2mc2 r2
. – p.4/5
13
14
. – p.3/5
Example
Spin orbit result
Spin coupling example: n = 3 levels of an atom. S = 1/2 and
This gives a net result 1/2 of the previous value, giving a net
spin-orbit interaction
HLS
H = γL2 + 2α L · S
Ze2
= − 3 2 2L · S
2r m c
= γL2 + α(J 2 − L2 − S 2 )
. – p.5/5
15
16
. – p.6/7
Energy levels n=3
Wigner-Eckart theorem
This has Lmax = 2 with spin 1/2 multiplicity 2(5 + 3 + 1) = 18
deg = 2 L = 0 J = 1/2 E = 0γ + α( 32 12 − 0 − 34 )
deg = 4 L = 1 J = 3/2 E = 2γ + α( 52 32 − 2 − 34 )
deg = 2
J = 1/2 E = 2γ + α( 32 12 − 2 − 34 )
deg = 6 L = 2 J = 5/2 E = 6γ + α( 72 52 − 6 − 34 )
deg = 4
J = 3/2 E = 6γ + α( 52 32 − 6 − 34 )
sum=18
. – p.7/7
17
18
Operators form representations too!
Definition of representation
Example: x,y,z
Recall we defined the rotation operators:
rY10
| a ! ⇒ D(R) | a !
=r
!
3
4π
cos θ =
rY1+ = r √!2 L+ Yl0
R
rY1− = r √!2 L− Yl0
By a representation | j, m ! we we mean exactly
j
!
| j, m ! ⇒ Dm,m
! (R) | j, m !
!
3
z
4π
!
!
3 iθ
3
= −r 8π
e sin θ = − 8π
(x + iy)
!
!
3 −iθ
3
= r 8π
e
sin θ = 8π
(x − iy)
Yll (θ, φ)) = cl (x + iy)l
R
(−1)l
cl = l
2 l!
!
(2l + 1)(2l)!
4π
They transform exactly like wavefunctions.
. – p.2/??
19
. – p.4/??
20
Transformation of operators
Review of angle addition
The definition of the operators Tmj is that it transforms exactly
like | j, m ! under rotations. i.e.
We write the natural notation:
j1 ⊗ j2 ≡ (j1 + j2 ) ⊕ (j1 + j2 − 1) ⊕ ... ⊕ |j1 − j2 |
j
j
D† (R)Tmj D(R) = Dm,m
! (R)Tm!
and we have
| j1 j2 , jm % =
!
m1 ,m2
Thus
| j1 j2 , m1 m2 % & j1 j2 , m1 m2 | j1 j2 , jm %
j
!
| j, m ! ⇒ Dm,m
! (R) | j, m !
R
with m = m1 + m2 and |j1 − j2 | ≤ j ≤ |j1 + j2 |.
We rewrite this as
!
| j1 , m1 %⊗ | j2 , m2 % =
| j1 j2 j, m % & j1 j2 , jm | j1 j2 , m1 m2 %
and
j
j
Tmj ⇒ Dm,m
! (R)Tm!
R
j,m
. – p.5/??
. – p.6/??
21
Wigner-Eckart
Transformation of products
j1
j2
| j1 , m 1 ! ⊗ | j2 , m 2 ! ⇒ D m
! Dm ,m!
1 ,m
2
R
1
R
1
2
j1
j2
Tmj11 | j2 , m2 ! ⇒ Dm
! Dm ,m!
1 ,m
2
2
Angle addition formula
!
| j1 , m1 !⊗ | j2 , m2 ! =
| j1 j2 j, m ! # j1 j2 , jm | j1 j2 , m1 m2 !
| j1 , m!1 ! ⊗ | j2 , m!2 !
Tmj1!
1
22
j,m
| j2 , m!2 !
Wigner-Eckart generalization:
!
Tmj11 | j2 , m2 ! =
Nj1 ,j2 ,j | j, m ! # j1 j2 , jm | j1 j2 , m1 m2 !
Therefore up to normalization Tmj11 | j2 , m2 !, transforms exactly
like | j1 , m1 !⊗ | j2 , m2 ! since Tmj11 and | j1 , m1 ! transform iden-
j,m
tically.
(Note that the implicit j1 , j2 is dropped in the W.E. form) Taking
the scalar product with # jm | we get
# j, m |Tmj11 | j2 , m2 ! = Nj1 ,j2 ,j # j1 j2 , jm | j1 j2 , m1 m2 !
. – p.7/9
23
. – p.8/9
24
Confusing historical notation:
For historical reasons one uses Tqk instead of Tmj11 and
Nj1 ,j2 ,j ≡
What’s it good for?
Confusing historical notation:
For historical reasons one uses
Nj1 ,j2 ,j ≡
Tqk
instead of
Tmj11
" j ||T j1 | | j2 #
√
2j2 + 1
leaving us with formula 3.10.31 in the textbook.
" j, m |Tmj11 | j2 , m2 # = Nj1 ,j2 ,j " j1 j2 , jm | j1 j2 , m1 m2 #
and
From only a single number for each pair
of representations j and j2 , all matrix
elements are determined by Clebsch-Gordan
coeffecients, as well as a host of selection
rules where we now the CG coefficients vanish.
" j ||T j1 | | j2 #
√
2j2 + 1
leaving us with formula 3.10.31 in the textbook.
" j, m |Tmj11 | j2 , m2 # = Nj1 ,j2 ,j " j1 j2 , jm | j1 j2 , m1 m2 #
. – p.9/9
25
26
. – p.9/9
Electric quadrupole
Selection rules - electric dipole
As a second example, lets take electric quadrupoletransitions;
! j, m |x2 − y 2 | j2 , m2 #
As an example, lets take electric dipole transitions; when is
! j, m |!z | j2 , m2 "
nonzero. (I have not bothered working out exactly what value of
nonzero so that j1 = 1 and m1 = 0. The answer is equivalent
m1 , the operator x2 −y 2 corresponds to.) The answer is equivalent
to the question when can ! 2, j2 , j; m | 2, j2 , m1 , m2 # be nonzero.
to the question when can ! j1 , j2 , j; m | j1 , j2 , 0, m2 " be nonzero.
This means that |j − j2 | ≤ 2 and |m − m2 | ≤ 2. However
This means that |j − j2 | ≤ 1 and m = m2 . However, it turns out
! 2, j2 , j; m | 2, j2 , m1 , m2 # is zero unless j + j1 + 2 is even, so
that ! 1, j2 , j; m | 1, j2 , 0, m " must be zero by parity, so we have
in fact only j = j2 ± 2 or j = j2 is allowed and j = j2 ± 1 is
in fact only j = j1 ± 1.
forbidden.
. – p.11/??
. – p.10/??
27
28