21.8.2 Using the Lowering Operator to Find Total Spin States
The total spin lowering operator is
S−=S−(1)+S(2)−.First lets remind ourselves of what the individual lowering operatorsdo.
S(1)−χ(1)+ = ̄h√
1
2
(
3
2
)
−
(
1
2
)(
− 1
2
)
χ(1)− = ̄hχ(1)−Now we want to identifyχ 11 =χ(1)+χ(2)+. Lets operate on this equation withS−. First the RHS gives
S−χ(1)+χ(2)+ =(
S(1)−χ(1)+)
χ(2)+ +χ(1)+(
S−(2)χ(2)+)
= ̄h(
χ(1)−χ(2)+ +χ(1)+χ(2)−)
.
Operating on the LHS gives
S−χ 11 = ̄h√
(1)(2)−(1)(0)χ 10 =√
2 ̄hχ 10.So equating the two we have
√
2 ̄hχ 10 = ̄h(
χ(1)−χ(2)+ +χ(1)+χ(2)−)
.
χ 10 =1
√
2
(
χ
(1)
−χ(2)
+ +χ(1)
+χ(2)
−)
.
Now we can lower this state. Lowering the LHS, we get
S−χ 10 = ̄h√
(1)(2)−(0)(−1)χ1(−1)=√
2 ̄hχ 1 ,− 1.Lowering the RHS, gives
S−
1
√
2
(
χ(1)+χ(2)− +χ(1)−χ(2)+)
= ̄h1
√
2
(
χ(1)−χ(2)− +χ(1)−χ(2)−)
=
√
2 ̄hχ(1)−χ(2)−⇒ χ 1 ,− 1 =χ(1)−χ(2)−Therefore we have found 3 s=1 states that work together. Theyare all symmetric under interchange
of the two particles.
There is one state left over which is orthogonal to the three states we identified. Orthogonal state:
χ 00 =1
√
2
(
χ(1)+χ(2)− −χ(1)−χ(2)+)
We have guessed that this is ans= 0 state since there is only one state and it has m=0. We could
verify this by using theS^2 operator.