1
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
;
χ

=
0
1
;
S
z
=
¯
h
2
1
0
0

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

i
i
0
The spin is placed in a uniform magnetic field,
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
±
=
∓
γB
0
¯
h/
2.
A small magnetic field
b
=
b
ˆ
y
is applied in addition to
B
, making the total Hamiltonian
H
=
H
(0)
+
H
with
H
=

γbS
y
. Use perturbation theory to find:
1.) (4 pts)
The first 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

χ
±
.
E
(1)
+
=
χ
+
 
γbS
y

χ
+
=

γb
¯
h
2
( 1
0 )
0

i
i
0
1
0
=

γb
¯
h
2
( 1
0 )
0
i
= 0
E
(1)

=
χ

 
γbS
y

χ

=

γb
¯
h
2
( 0
1 )
0

i
i
0
0
1
=

γb
¯
h
2
( 0
1 )

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

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

χ


H

χ
+

2
E
+

E

=
γ
2
b
2
¯
h
2
4
( 0
1 )
0

i
i
0
1
0
2

γB
0
¯
h/
2

γB
0
¯
h/
2
=
γ
2
b
2
¯
h
2
4
( 0
1 )
0
i
2

γB
0
¯
h
=

γb
2
¯
h
4
B
0
E
(2)

=

χ
+

H

χ


2
E


E
+
=
γ
2
b
2
¯
h
2
4
( 1
0 )
0

i
i
0
0
1
2
γB
0
¯
h/
2

(

γB
0
¯
h/
2)
=
γ
2
b
2
¯
h
2
4
( 1
0 )

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

.
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 Spring '08
 Staff
 pts, 1 L, 0 W, order energy corrections

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