222 3 Quantum Mechanics – II
Clearly the statesψ( 3
2 ,
3
2)
andψ( 3
2 ,−
3
2)
can be formed in only one wayψ(
3
2
,
3
2
)
=φ(1,1)φ(
1
2
,
1
2
)
(1)
ψ(
3
2
,−
3
2
)
=φ(1,−1)φ(
1
2
,−
1
2
)
(2)
We now use the ladder operatorsJ+andJ−to generate the second and the
third states.
J+φ(j,m)=[(j−m)(j+m+1)](^12)
φ(j,m+1) (3)
J−φ(j,m)=[(j+m)(j−m+1)]
(^12)
φ(j,m−1) (4)
Applying (4) to (1) on both sides
J−ψ
(
3
2
,
3
2
)
=
[(
3
2
+
3
2
)(
3
2
−
3
2
+ 1
)] 1 / 2
=
√
3 ψ(
3
2
,
1
2
)
=J−φ(1,1)φ(
1
2
,
1
2
)
=φ(1,1)J−(
1
2
,
1
2
)
+φ(
1
2
,
1
2
)
J−φ(1,1)=φ(1,1)φ(
1
2
,−
1
2
)
+φ(
1
2
,
1
2
)√
2 φ(1,0)Thusψ(
3
2
,
1
2
)
=
√
2
3
φ(1,0)φ(
1
2
,
1
2
)
+
√
1
3
φ(1,1)φ(
1
2
,−
1
2
)
(5)
Similarly, applyingJ+operator given by (3) to the stateψ( 3
2 ,−
3
2)
we
obtainψ(
3
2
,−
1
2
)
=
√
2
3
φ(1,0)φ(
1
2
,−
1
2
)
+
√
1
3
φ(1,−1)φ(
1
2
,
1
2
)
(6)
TheJ= 1 /2 state can be obtained by making it as a linear combinationψ(
1
2
,
1
2
)
=aφ(1,1)φ(
1
2
,−
1
2
)
+bφ(1,0)φ(
1
2
,
1
2
)
(7)
For normalization reason,
a^2 +b^2 =1(8)
We can obtain one other relation by making (7) orthogonal to (5)a√
1
3
+b√
2
3
= 0
Ora=−√
2 b (9)
Same result is obtained by applyingJ+operator to (7).J+ψ(1/ 2 , 1 /2)= 0Solving (8) and (9),=√
2
3
,b=−