(
Ly
̄h) 3
=
1
√
2 i1
2
0 2 0
−2 0 2
0 −2 0
=
(
Ly
̄h)
...
All the odd powers are the same. All the nonzero even powers are the same. The ̄hs all cancel out.
We now must look at the sums for each term in the matrix and identify the function it represents.
- Then= 0 term contributes 1 on the diagonals.
- Then= 1, 3 , 5 ,...terms sum to sin(θ)
(
iLy
̄h)
.
- Then= 2, 4 , 6 ,...terms (with a -1 in the matrix) are nearly the series for^12 cos(θ). Then= 0
term is is missing so subtract 1. The middle matrix element is twice the other even terms.
eiθLy/ ̄h=
1 0 0
0 1 0
0 0 1
+ sin(θ)√^1
2
0 1 0
−1 0 1
0 −1 0
+^1
2
(cos(θ)−1)
1 0 − 1
0 2 0
−1 0 1
Putting this all together, we get
Ry(θy) =
1
2 (1 + cos(θy))
√^1
2 sin(θy)1
2 (1−cos(θy))
−√^12 sin(θy) cos(θy) √^12 sin(θy)
1
2 (1−cos(θy)) −
√^1
2 sin(θy)1
2 (1 + cos(θy))
.
18.11.6Derive Spin^12 Operators
We will again use eigenstates ofSz, as the basis states.
χ+ =(
1
0
)
χ− =(
0
1
)
Szχ± = ±̄h
2
χ±Sz =̄h
2(
1 0
0 − 1
)
Its easy to see that this is the only matrix that works. It must be diagonal since the basis states are
eigenvectors of the matrix. The correct eigenvalues appear on the diagonal.
Now we do the raising and lowering operators.
S+χ+ = 0
S+χ− =√
s(s+ 1)−m(m+ 1) ̄hχ+= ̄hχ+