Chapter 11 Total Angular Momentum

Chapter 11 Total Angular Momentum
With possibilities of both orbital L and spin S angular momentum in a system we must consider
that total angular momentum J is conserved J = L + S .
!
S and
!
L have constant projections on
!
J
J Z
!
Li ! J = L 2 + ! Li ! S = L 2 + 1 2 J 2 ! L 2 ! S 2
( ) = 1 2 J 2 + L 2 ! S 2
( )
S
J
L
= 1 ( j( j + 1) + l(l + 1) ! s(s + 1) )" 2 = cons tant
2
!
Si ! J = S 2 + ! Li ! S = S 2 + 1 2 J 2 ! L 2 ! S 2
( ) = 1 2 J 2 + L 2 ! S 2
( )
= 1 ( j( j + 1) ! l(l + 1) + s(s + 1) )" 2 = cons tant
2
In a system in which the Hamiltonian commutes with J and J 2 , [ H , J I ]=0 and [ H , J 2 ]=0 ,
we need to establish how J is related to L and S.
!
1) Given L x L ! = i" L ! and
!
J x J ! = L ! + S
!
!
S x ! S = i" ! S then
!
J x ! J = i" ! J ,
( ) x ( L ! + S
! ) = L ! x L ! + S ! x S ! + Sx ! L ! + Lx ! S
! #% $ &
!
J x ! J = i" ! J QED quod erat demonstrantum
cancel to zero
% = i" ! J
!
J follows the angular momentum algebra.
2) !
"
J 2 ,L 2 #
$ = 0 J2 commutes with L 2
!L
"
2 + S 2 + 2LiS , L 2 #
$ = !
"
L2 ,L 2 #
#$ % &% $
+ !
"
S2 ,S 2 #
#% $ &% $
+ 2 S ! L X " X ,L2 #
#$ % &% $
+ 2 S ! L Y " Y ,L2 #
#$ % &% $
+ 2 S ! L
Z " Z
,L 2 #
#$ % &% $
= 0
=0
=0
[L 2 ,L X ]=0
[L 2 ,L Y ]=0
[L 2 ,L Z ]=0
3) !
"
J 2 ,S 2 #
$ = 0 J2 commutes with S 2
!L
"
2 + S 2 + 2LiS , L 2 #
$ = !
"
L2 ,L 2 #
#$ % &% $
+ !
"
S2 ,S 2 #
#% $ &% $
+ 2 L ! S X " X ,S2 #
#% $ &% $
+ 2 L ! S Y " Y ,S2 #
#% $ &% $
+ 2 L ! S Z " Z ,S2 #
#% $ &% $
= 0
=0
=0
[S 2 ,S X ]=0
[S 2 ,L Y ]=0
[S 2 ,L Z ]=0
4) !
"
J 2 ,J Z
#
$ = 0 or !
"
J 2 ,J I
#
$ = 0 I = 1,2,3 J2 commutes with J x
J Y
J Z
!L
"
2 + S 2 + 2LiS , L Z
+ S Z
#
$ = !
"
L2 ,L Z
#
#$ % &% $
+ !
"
L2 ,S Z
#
#$ % &% $
+ !
"
S2 ,L Z
#
#% $ &% $
+ !
"
S2 ,S Z
#
#% $ &% $
=0
=0
+ 2S X
!" L X
,L Z
#
#% $ &% $ + 2S L Y
!" ,L #
Y Z
#$ % &% $ + 2S L
Z
!" z
,L Z
#
#$ % &% $
=%i"L Y
=+i"L X
+ 2L X
!" S X
,S Z
#
#% $ &% $ + 2L S Y
!" ,S #
Y Z
#% $ &% $ + 2L LS
Z
!" z
,S Z
#
#% $ &% $
=%i"S Y
=+i"S X
= %2i"S X
L Y
+ 2i"S Y
L X
% 2i"L X
S Y
+ 2i"L Y
S X
= 0
=0
=0
=0
=0

Set of Commuting Operators for the System- Symmetries of the Hamiltonian
The complete set of commuting operators is then H, J 2 , J Z
, L 2 , S 2 . Each operator represents
a constant of motion for the system, and each represented by a quantum number n, j, m j
,l , s
The most general wave function can be written
Clesbch"Gordan Coef
! = n, j, m n , j, mj
,l , s = % !
L,S
C
,l , s j mL
n
,m S " l,m l
" #$%
s,m S
n,m l ,m s
m j =m l +m s
( )
R(r ) Y lm #$
& S
H n, j, m j
,l , s = E n
n, j, m j
,l , s
J 2 n, j, m j
,l , s = j( j + 1)& 2 n, j, m j
,l , s
J Z
n, j, m j
,l , s = m j
& n, j, m j
,l , s
L 2 n, j, m j
,l , s = l(l + 1)& 2 n, j, m j
,l , s
S 2 n, j, m j
,l , s = s(s + 1)& 2 n, j, m j
,l , s
Addition of Angular Momentum
J = L + S
j = | L + S | ........ | L ! S |
m j
= + j, + ( j ! 1),+ j ! 2
( ) ....... ! j = 2 j + 1 states
Classification of J States for the Hydrogen Atom J=L+1/2
J = L + 1/ 2 :
J = 0 + 1/ 2 j = 1/ 2 m j
= ±1/ 2
2s+1 L J
= 2 S 1/2
e-
J = 1+ 1/ 2 j = 3 / 2 m j
= ±3 / 2, ±1/ 2
j = 1/ 2 m j
= ±1/ 2
2s+1 L J
= 2 P 3/2, 1/2
r
P+
J = 2 + 1/ 2 j = 5 / 2 m j
= ±5 / 2 = ±3 / 2, ±1/ 2
2s+1 L J
= 2 D 5/2, 3/2, 1/2
j = 3 / 2 m j
= ±3 / 2, ±1/ 2
j = 1/ 2 m j
= ±1/ 2

Two Spin ½ Particles
S 1
= 1/ 2 S 2
= 1/ 2 J = 1/ 2 + 1/ 2 = 1 , 0
j = 1 m j
= ±1,0 Spin Triplet State ! Symmetric
|1,1> = " +1/2 (1)" +1/2 (2) ##
( ) #$ + $#
|1,0 >= 1 2 " +1/2 (1)" !1/2 (2) + " !1/2 (1)" +1/2 (2)
|1,!1>= " !1/2 (1)" !1/2 (2) $$
j = 0 m j
= 0 Spin Singlet ! AntiSymmetric state
| 0,0 > = 1 (
2 " +1/2 (1)" !1/2 (2) ! " !1/2 (1)" +1/2 (2)) #$ ! $#
L=1+1/2 States and Clebsch=Gordan Coefficients
J = 1+ 1/ 2 = 3 / 2 ,1/ 2
j = 3 / 2 m l
= ±3 / 2, ±1/ 2 ! | 3 / 2,3 / 2 > | 3 / 2,+1/ 2 > | 3 / 2,"1/ 2 > | 3 / 2,"3 / 2 >
j = 1/ 2 m l
= ±1/ 2 ! |1/ 2,1/ 2 > |1/ 2,"1/ 2 >
| j,m j
> =
# C ml
| l,m
m s l
> | s,m s
>
m l +m s =m j
| 3 / 2,+3 / 2 > = |1, +1> |1/ 2,+1/ 2 >
| 3 / 2,!3 / 2 > = |1,!1> |1/ 2,!1/ 2 >
| 3 / 2,1/ 2 > =
|1/ 2,1/ 2 > =
| 3 / 2,!1/ 2 > =
|1/ 2,!1/ 2 > =
2
3 |1,0 >|1/ 2,+1/ 2 > + 1 3
1
3 |1,0 >|1/ 2,+1/ 2 > ! 2 3
|1,1>|1/ 2,!1/ 2 >
|1,1>|1/ 2,!1/ 2 >
1
3 |1,!1>|1/ 2,+1/ 2 > + 2
|1,0 >|1/ 2,!1/ 2 >
3
2
3 |1,!1>|1/ 2,!1/ 2 > ! 1 3
|1,0 >|1/ 2,!1/ 2 >

ParaHelium and OrthoHelium
H =
2
$ p 1
2m !
'
&
2e2
) +
% 4"# 0
r 1 (
2
$ p 2
2m !
'
&
2e2
) +
% 4"# 0
r 2 (
e 2
4"# 0
r
!# "#
12
$
H ee Interaction term
e-
r 12
e-
E = Z 2
n E + Z 2
2 0
n E + e 2
2 0
4"# 0
r
!#"
# 12
$
J = 1/ 2 + 1/ 2
j = 1 m j
= ±1 ,0
*E 12 ~30eV
2S+1 L J
r 2
r 1
+2e
j = 0 m j
= 0
+(1,2) = (, 100
(1) , 100
(2)) - A (1 ,2) , S (1,2) - A (1 ,2) E = 4E 0
+ 4E 0
+ 30eV = !79eV
( )
+(1,2) = 1 2 , (1) , (2) + , (1) , (2)
100 nlm nlm 100
!###### "###### $
- A (1,2) , S (1,2) - A (1 ,2) E = 4 1 E + 4 0
n E + *E 2 0 12
r 12 < r 12
e- e-
( )
+(1,2) = 1 2 , (1) , (2) ! , (1) , (2)
100 nlm nlm 100
!###### "###### $
-S (1,2) , A (1,2) - S (1 ,2) E = 4 1 E + 4 0
n E ! *E 2 0 12
r 12 > r 12
e- e-
para ! He
ortho ! He
HUND’s RULES (not covered)
1) Highest Spin States have the lowest energy. Consistent w Pauli Principle
2) For a given S state highest L lies lowest. Consistent w Pauli Principle
3) If Shell < half filled lowest J lies lowest, if >half filled highest J lies lowest.
Pauli Principle -The total wave function must me antisymmetric for electron states.
S states 0 A
L States (-1) L 0 S 1 A 2 S 3 A …….
1)
4 He 2
: 1s 2 S = 1/ 2 + 1/ 2 S = 1 S , 0 A
L = 0 + 0 L = 0 S
J = 0 + 0 J = 0
1 S 0
ground state
2)
12 C 6
: 1s 2 2s 2 2p 2 S = 1/ 2 + 1/ 2 S = 1 S , 0 A
L = 1+1 L = 2 S , 1 A , 0 S
J = 1+1 J = 2, 1, 0
3 S 1
ground state