21.7 Examples
21.7.1 Counting states forℓ= 3 Plus spin
Forℓ= 3 there are 2ℓ+ 1 = 7 different eigenstates ofLz. There are two different eigenstates of
Szfor spin^12. We can have any combination of these states, implying 2×7 = 14 possible product
states likeY 31 χ+.
We will argue based on adding vectors... that there will be two total angular momentum states that
can be made up from the 14 product states,j=ℓ±^12 , in this casej=^52 andj=^72. Each of these
has 2j+ 1 states, that is 6 and 8 states respectively. Since 6 + 8 = 14 this gives us the right number
of states.
21.7.2 Counting states for ArbitraryℓPlus spin
For angular momentum quantum numberℓ, there are (2ℓ+ 1) different m states, while for spin we
have 2 statesχ±. Hence the composite system has 2(2ℓ+ 1) states total.
Maxjz=ℓ+^12 so we have a state withj=ℓ+^12. This makes up (2j+ 1) = (2ℓ+ 2) states, leaving
(4ℓ+ 2)−(2ℓ+ 2) = 2ℓ=
(
2
(
ℓ−
1
2
)
+ 1
)
Thus we have a state withj=ℓ−^12 and that’s all.
21.7.3 Addingℓ= 4 toℓ=
As an example, we count the states for each value of total m (z component quantum number) if we
addℓ 1 = 4 toℓ 2 = 2.
audio
Totalm (m 1 ,m 2 )
6 (4,2)
5 (3,2) (4,1)
4 (2,2) (3,1) (4,0)
3 (1,2) (2,1) (3,0) (4,-1)
2 (0,2) (1,1) (2,0) (3,-1) (4,-2)
1 (-1,2) (0,1) (1,0) (2,-1) (3,-2)
0 (-2,2) (-1,1) (0,0) (1,-1) (2,-2)
-1 (1,-2) (0,-1) (-1,0) (-2,1) (-3,2)
-2 (0,-2) (-1,-1) (-2,0) (-3,1) (-4,2)
-3 (-1,-2) (-2,-1) (-3,0) (-4,1)
-4 (-2,-2) (-3,-1) (-4,0)
-5 (-3,-2) (-4,-1)
-6 (-4,-2)