wang
(Wang)
#1
in the second factor. For the second spin operator, the roles arereversed, and it acts as the
identity. One may summarize this by
S 1 on H 1 → S 1 ⊗I 2 on H
S 2 on H 2 → I 1 ⊗S 2 on H (8.27)
In this way, both operators are now defined on the same Hilbert spaceH, and may be added
to obtain total spin,
S=S 1 ⊗I 2 +I 1 ⊗S 2 or simply S=S 1 +S 2 (8.28)
but the latter, although commonly used, is a bit of an abuse of notation.
It is straightforward to check thatSsatisfies the angular momentum algebra as well. By
definition,Sgives a 4-dimensional representation ofSO(3). The question is now whether
this representation is reducible or not, and if it is reducible what its irreducible components
are.
To shed light on this question, we first calculate the totalS^3 =Szof each basis state.
Fortunately, the states have been arranged so that they are eigenstates ofS^3 , with eigenvalues
m, and we get
m= +1 | 1 ,+〉⊗| 2 ,+〉
m= 0 | 1 ,+〉⊗| 2 ,−〉, | 1 ,−〉⊗| 2 ,+〉
m=− 1 | 1 ,−〉⊗| 2 ,−〉 (8.29)
There is a single state withm=±1, which indicates that this 4-dimensional representation
contains thej= 1 irreducible representation ofSO(3). Thus, we identify
|j= 1,m= +1〉 = | 1 ,+〉⊗| 2 ,+〉
|j= 1,m=− 1 〉 = | 1 ,−〉⊗| 2 ,−〉 (8.30)
The way we can get all the states of this representation is by starting with these states, and
applyingS−successively. Since
Si−|i,+〉= ̄h|i,−〉 Si−|i,−〉= 0
Si+|i,−〉= ̄h|i,+〉 Si+|i,+〉= 0 (8.31)
we have
S−|j= 1,m= +1〉 =
√
2 ̄h|j= 0,m= 0〉
= (S 1 −⊗I 2 +I 1 ⊗S 2 −)| 1 ,+1〉⊗| 2 ,+〉
= ̄h| 1 ,−〉⊗| 2 ,+〉+ ̄h| 1 ,+〉⊗| 2 ,−〉 (8.32)
