QuantumPhysics.dvi

Total angular momentum may be defined onHjust as we did when we added two spin 1/2

systems,

J=J 1 ⊗I 2 +I 1 ⊗S 2 (8.40)

The basis states ofHmay be organized according to eigenvaluesmofJ 3 =Jz,

m=j 1 +

1

2

|j 1 ,j 1 〉⊗| 2 ,+〉

m=j 1 −

1

2

|j 1 ,j 1 〉⊗| 2 ,−〉, |j 1 ,j 1 − 1 〉⊗| 2 ,+〉

m=j 1 −

3

2

|j 1 ,j 1 − 1 〉⊗| 2 ,−〉, |j 1 ,j 1 − 2 〉⊗| 2 ,+〉

··· ···

m=−j 1 +

1

2

|j 1 ,−j 1 + 1〉⊗| 2 ,−〉, |j 1 ,−j 1 〉⊗| 2 ,+〉

m=−j 1 −

1

2

|j 1 ,−j 1 〉⊗| 2 ,−〉 (8.41)

There is a unique highestJ 3 state withj=j 1 + 1/2, so the tensor product contains once

the representationj=j 1 + 1/2, and we identity

|j,+j〉 = |j 1 ,+j 1 〉⊗| 2 ,+〉

|j,−j〉 = |j 1 ,−j 1 〉⊗| 2 ,−〉 (8.42)

Acting withJ−=J 1 −+S 2 −, we obtain all the states of this representation exactly once. For

example, at the levelm=j 1 − 1 /2, we obtain,

J−|j,+j〉 = ̄h

√

2 j 1 |j,j− 1 〉

=

(

J 1 −⊗I 2 +I 1 ⊗S 2 −

)

|j 1 ,j 1 〉⊗| 2 ,+〉

= ̄h

√

2 j 1 |j 1 ,j 1 − 1 〉⊗| 2 ,+〉+ ̄h|j 1 ,j 1 〉⊗| 2 ,−〉 (8.43)

As a result, we obtain a formula for the state

|j,j− 1 〉=

1

√

2 j

(√

2 j 1 |j 1 ,j 1 − 1 〉⊗| 2 ,+〉+|j 1 ,j 1 〉⊗| 2 ,−〉

)

(8.44)

At each value ofm=−j,···,+j, there is exactly one state belonging to the representation

j. Atm=j−1, the linear combination orthogonal to|j,j− 1 〉is given by

1

√

2 j

(

|j 1 ,j 1 − 1 〉⊗| 2 ,+〉−

√

2 j 1 |j 1 ,j 1 〉⊗| 2 ,−〉

)

(8.45)