Physics 472 Solutions Exam 1 Spring 2007
1. (a) (2 pt) The eigenvectors ξ±corresponding to ±1
2¯hsatisfy
1
2¯hÃ∓1 1
1∓1!Ã a
b!= 0 or ∓a+b= 0 and ξ±=aÃ1
±1!.
Normalizing using ξ†
±ξ±= 1 gives
ξ±=q1
2Ã1
±1!.
(b) (2 pt) ψ(0) can be expanded as ψ(0) = C+ξ++C−ξ−where
C±=ξ†
±ψ(0) = q1
2(1 ±1) Ãcos(α)
sin(α)!=q1
2(cos(α)±sin(α)) .
(c) (1 pt) P(1
2¯h) = (cos(α) + sin(α))2/2 = (1 + sin(2α))/2.
(d) (2 pt)
hSzi=³cos(α) sin(α)´Ã1
2¯h0
0−1
2¯h!Ãcos(α)
sin(α)!=1
2¯h(cos2(α)−sin2(α)) = 1
2¯hcos(2α).
2. Scandium (Sc) has one delectron outside a core consisting of argon and a closed 4s shell,
i.e. the electron configuration is (Ar)(4s)2(3d)1.
(a) (2 pt) ψ(~r) is an eigenstate of Jzwith eigenvalue −1
2¯h, so only −1
2¯hwill be observed
with probability 1.
(b) (3 pt) The product states are not eigenstates of ~
J2, but the only possible values of
jare j=3
2and j=5
2. The results of a measurement of ~
J2are therefore 15
4¯h2and
35
4¯h2. From the 2 ×1
2table,
s2
5Y−1
2(θ, φ)χ1
2
=q2
5³q2
5|5
2−1
2i − q3
5|3
2−1
2i´
−s3
5Y0
2(θ, φ)χ−1
2
=−q3
5³q3
5|5
2−1
2i+q2
5|3
2−1
2i´
so
ψ(~r) = −R31(r)³1
5|5
2−1
2i+2√6
5|3
2−1
2i´.
The results are 35¯h2/4 with probability 1/25 and 15¯h2/4 with probability 24/25.
(c) (2 pt) The |3
2
1
2istate is
|3
2
1
2i=q3
5Y1
2(θ, φ)χ−1
2
−q2
5Y0
2(θ, φ)χ1
2
.
