∣ ∣ ∣ ∣ ∣ ∣
−b 1 0
1 −b 1
0 1 −b∣ ∣ ∣ ∣ ∣ ∣
= 0
wherea=√ ̄h 2 b.
−b(b^2 −1)−1(−b−0) = 0
b(b^2 −2) = 0There are three solutions to this equation:b= 0,b= +
√
2, andb=−√
2 ora= 0,a= + ̄h, and
a=− ̄h. These are the eigenvalues we expected forℓ= 1. For each of these three eigenvalues, we
should go back and find the corresponding eigenvector by using thematrix equation.
0 1 0
1 0 1
0 1 0
ψ 1
ψ 2
ψ 3
=b
ψ 1
ψ 2
ψ 3
ψ 2
ψ 1 +ψ 3
ψ 2
=b
ψ 1
ψ 2
ψ 3
Up to a normalization constant, the solutions are:
ψ+ ̄h=c
√^1
2
1
√^1
2
ψ0 ̄h=c
1
0
− 1
ψ− ̄h=c
−√ 1
2
1
−√ 1
2
.
We should normalize these eigenvectors to represent one particle.For example:
〈ψ+ ̄h|ψ+ ̄h〉 = 1|c|^2( 1
√
2∗
1 ∗ √^12∗)
√^1
2
1
√^1
2
= 2|c|^2 = 1c =1
√
2
Try calculating the eigenvectors ofLy.
You already know what the eigenvalues are.
18.10.5A 90 degree rotation about the z axis.
If we rotate our coordinate system by 90 degrees about the z axis, the old x axis becomes the new
-y axis. So we would expect that the state with angular momentum + ̄hin the x direction,ψ(+x), will
rotate intoψ(−y)within a phase factor. Lets do the rotation.
Rz(θz) =
eiθz 0 0
0 1 0
0 0 e−iθz
.
Rz(θz= 90) =
i 0 0
0 1 0
0 0 −i