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Physics 325
Solution to Second Midterm Exam
Winter 2004
PART I (19 pts)
Spin 1
/
2:
This problem is about a spin 1
/
2 particle. The eigenstates of
S
2
and
S
z
are
labelled
χ
±
with:
χ
+
=
±
1
0
²
;
χ

=
±
0
1
²
;
S
z
=
¯
h
2
±
10
0

1
²
;
S
x
=
¯
h
2
±
01
²
;
S
y
=
¯
h
2
±
0

i
i
0
²
The spin is placed in a uniform magnetic Feld,
±
B
=
B
0
ˆ
z
, giving a Hamiltonian
ˆ
H
=
H
(0)
=

γB
0
S
z
.
The eigenstates of
H
(0)
are then
χ
±
with eigenvalues
E
±
=
∓
0
¯
h/
2.
A small magnetic Feld
±
b
=
b
ˆ
y
is applied in addition to
±
B
, making the total Hamiltonian
H
=
H
(0)
+
H
²
with
H
²
=

γbS
y
. Use perturbation theory to Fnd:
1.) (4 pts)
The Frst order energy corrections to
E
+
and
E

.
Solution:
Because the two states,
χ
±
, have diﬀerent energies, we use nondegenerate perturbation theory:
E
(1)
±
=
±
χ
±

H
²

χ
±
²
.
E
(1)
+
=
±
χ
+

y

χ
+
²
=

γb
¯
h
2
(1 0)
±
0

i
i
0
²±
1
0
²
=

¯
h
2
±
0
i
²
=0
E
(1)

=
±
χ

y

χ

²
=

¯
h
2
(0 1)
±
0

i
i
0
0
1
²
=

¯
h
2
±

i
0
²
2.) (5 pts)
The second order energy corrections to
E
+
and
E

.
Solution:
Because there are only two states, the sum over states,
∑
m
³
=
n
, in the expression for
E
(2)
n
reduces
to a single term:
E
(2)
+
=
±
χ


H
²

χ
+
²
2
E
+

E

=
γ
2
b
2
¯
h
2
4
³
³
³
±
0

i
i
0
1
0
²
³
³
³
2

0
¯
h/
2

0
¯
h/
2
=
γ
2
b
2
¯
h
2
4
³
³
³
±
0
i
²
³
³
³
2

0
¯
h
=

2
¯
h
4
B
0
E
(2)

=
±
χ
+

H
²

χ

²
2
E


E
+
=
γ
2
b
2
¯
h
2
4
³
³
³
±
0

i
i
0
0
1
²
³
³
³
2
0
¯
h/
2

(

0
¯
h/
2)
=
γ
2
b
2
¯
h
2
4
³
³
³
±

i
0
²
³
³
³
2
0
¯
h
=
2
¯
h
4
B
0
3.) (4 pts)
The Frst order corrections to the wavefunctions
χ
+
and
χ

.
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