By a widely used abuse of notation, one usually does not write the symbolD(j), and this
sometimes leads to some confusion as to what a representation really is.
It is a good exercise to compute the representation matrices in thelowest dimensional
representations. This is done with the help of the matrix elements,
(D(j)(Ja)
)m′,m=〈j,m′|Ja|j,m〉 (8.16)
which in turn may be carried out by using the formulas^6
J 3 |j,m〉 = m ̄h|j,m〉
J±|j,m〉 =
√(j∓m)(j±m+ 1) ̄h|j,m± 1 〉 (8.17)
This allows us to compute
(D(j)(J 3 )
)m′,m= m ̄hδm′,m
(D(j)(J+)
)m′,m=
√(j−m)(j+m+ 1) ̄hδm′,m+1
(D(j)(J−)
)m′,m=
√(j+m)(j−m+ 1) ̄hδm′,m− 1 (8.18)
From these formula, it is manifest that each of these representations is irreducible. We
obviously haveD(0)= 0, the trivial representation. Next, we find, forj= 1/2,
(D(
(^12) )
(J 3 )
)=
̄h
2
(1 0
0 − 1
)(D(
(^12) )
(J+)
)= ̄h
(0 1
0 0
) (D(
(^12) )
(J 1 )
)=
̄h
2
(0 1
1 0
)(D(
(^12) )
(J−)
)= ̄h
(0 0
1 0
) (D(
(^12) )
(J 2 )
)=
̄h
2
(0 −i
i 0
)(8.19)
which are just the Pauli matrices for spin 1/2. Next, forj= 1, we find,
(D(1)(J 3 )
)= ̄h
1 0 0
0 0 0
0 0 − 1
(D(1)(J+)
)=
√
2 ̄h
0 1 0
0 0 1
0 0 0
(D(1)(J
1 )
)=
̄h
√
2
0 1 0
1 0 1
0 1 0
(D(1)(J−)
)=
√
2 ̄h
0 0 0
1 0 0
0 1 0
(D(1)(J
2 )
)=
̄h
√
2
0 −i 0
i 0 −i
0 i 0
(8.20)
(^6) Recall our notations ofJ±=J 1 ±iJ 2.