130_notes.dvi

Before rotation the state is

ψ

(x)

+ ̄h=





1

(^21)
√
2
1
2





The rotated state is.

ψ′=





i 0 0

0 1 0

0 0 −i









1

(^21)
√
2
1
2



=





i

(^21)
√
2
−i
2





Now, what remains is to check whether this state is the one we expect. What isψ(−y)? We find that
state by solving the eigenvalue problem.

Lyψ(−y)=− ̄hψ(−y)

̄h

√

2 i





0 1 0

−1 0 1

0 −1 0









a

b

c



=− ̄h





a

b

c











√ib

2

i(c√−a)

2

−√ib

2





=





a

b

c





Settingb= 1, we get the unnormalized result.

ψ−(y)=C





√i

2

1

√−i

2





Normalizing, we get.

ψ(−y)=





i

(^21)
√ 2
−i
2





This is exactly the same as the rotated state. A 90 degree rotationabout the z axis changesψ+(x)

intoψ(−y).

18.10.6Energy Eigenstates of anℓ= 1 System in a B-field

Recall that the Hamiltonian for a magnetic moment in an external B-field is

H=

μBB

̄h

Lz.

As usual, we find the eigenstates (eigenvectors) and eigenvalues of a system by solving the time-
independent Schr ̈odinger equationHψ=Eψ. We see that everything in the Hamiltonian above is
a (scalar) constant except the operatorLz, so that

Hψ=

μBB

̄h

Lzψ=constant∗(Lzψ).