These sample solutions were prepared by Bess Fang.
Note: All comments within square brackets refer to the previous line.
Problem 1
We compute |1ih1|, |2ih2|, |3ih3|:
|1ih1| =
b
1
0
1 0
1, 0 =
,
00
1 3 1
3, 4 =
5 4 5
1 3 1
3, −4 =
|3ih3| =
b
5 −4 5
|2ih2| =
b
1 9 12 ,
25 12 16
1 9 −12 25 −12 16
.
Substituting these into
ρ = |1iw1 h1| + |2iw2 h2| + |3iw3 h3| =
b
results in the set of equations

9
9


w
+
w3 =
w
+
2
1


25
25



12
12
w2 − w3 =

25
25




16
16


w2 + w3 =
25
25

12 


20 



3
20 



8 


20
which yield
1 12 3 20 3 8

12
3


=
,
w
=
1


8
32



15
w2 =
,

32





 w3 = 5 .
32
As expected, the weights have unit sum, w1 + w2 + w3 = 1.
1
Problem 2
We compute
Z
hp|F |xi =
dx0 dp0 hp|x0 i hx0 |F |p0 i hp0 |xi
0
0
∞
e−ipx /~ −(a|x0 |+b|p0 |)/~ e−ip x/~
√
e
dx0 dp0 √
2π~
2π~
−∞
Z ∞
Z ∞
1
0
0
0
0
dx0 e−(ipx +a|x |)/~
dp0 e−(ip x+b|p |)/~ .
=
2π~ −∞
−∞
Z
=
Since the two integrals are of exactly the same form, we only need to evaluate
one of them:
Z ∞
0
0
dx0 e−(ipx +a|x |)/~
−∞
Z 0
Z ∞
0
0
0 (−ipx0 +ax0 )/~
=
dx e
+
dx0 e(−ipx −ax )/~
−∞
0
Z 0
Z ∞
0
0
0
(−ip(−x0 )+a(−x0 ))/~
=
d(−x ) e
+
dx0 e(−ipx −ax )/~
+∞
0
0
0
[replacing x by −x , note the changed limits]
Z ∞
Z ∞
0
0 (ip−a)x0 /~
=
dx e
+
dx0 e(−ip−a)x /~
0
0
~
~
2a~
=
.
+
= 2
a − ip a + ip
a + p2
Likewise, we have for the 2nd integral
Z ∞
0
0
dp0 e−(ip x+b|p |)/~ =
−∞
2b~
.
+ x2
b2
Therefore,
hp|F |xi =
2~
a
b
.
2
2
2
π a + p b + x2
2
Problem 3
For the position wave function hx|λi, the eigenket equation Λ|λi = |λiλ reads
hx|Λ|λi = hx|λiλ =
~ ∂ 1
p0 x + x0
hx|λi
~
i ∂x
or, after a simple rearrangement,
x
∂
x0
−λ .
loghx|λi = −i
∂x
x0
This differential equation is solved by
i
2
hx|λi = C(λ) e− 2 (x/x0 −λ) ,
where C(λ) is the multiplicative integration constant that could depend on λ,
but not on x.
Now, we take care of the normalization by considering hλ|λ0 i:
Z
0
hλ|λ i = dx hλ|xihx|λ0 i
Z
i
i
2
0 2
= dx C(λ)∗ e 2 (x/x0 −λ−) C(λ0 )e− 2 (x/x0 −λ )
Z
i
2 −λ02 )
i x (λ0 −λ)
∗
0
(λ
= C(λ) C(λ ) e 2
dx e x0
i
02
2
= C(λ)∗ C(λ0 ) e 2 (λ −λ ) 2πx0 δ(λ0 − λ)
= |C(λ)|2 2πx0 δ(λ0 − λ) .
[the exponential factor becomes 1 when λ = λ0 ,
and C(λ0 ) becomes C(λ)]
Since we require hλ|λ0 i = δ(λ0 − λ), we have
|C(λ)|2 2πx0 = 1, .
Taking C(λ) > 0,
C(λ) = C = √
1
.
2πx0
so that, finally,
hx|λi = √
i
1
2
e− 2 (x/x0 −λ) .
2πx0
3
Problem 4
(a) We have the Heisenberg equations of motion
∂H
dP
=−
=F
dt
∂X
and
dX
∂H
P
=
=
,
dt
∂P
M
which are solved by
P (t) = P (t0 ) + F T ,
T
FT2
X(t) = X(t0 ) + P (t0 ) +
,
M
2M
(1)
(2)
where T = t − t0 . Then
T
FT2
[X(t), X(t0 )] = X(t0 ) + P (t0 ) +
, X(t0 )
M
2M
T
=
P (t0 ), X(t0 )
M
[since X(t0 ) commutes with itself and the numerical term]
T
= −i~ .
M
(b) First we use Eqs. (1) and (2) to express P (t0 ) and P (t) in terms of X(t0 )
and X(t),
M
X(t) − X(t0 ) −
T
M
P (t) =
X(t) − X(t0 ) +
T
P (t0 ) =
1
FT ,
2
1
FT ,
2
Then we use these in
i~
∂
hx, t|x0 , t0 i = −hx, t|P (t)|x0 , t0 i
∂x
M
1
= − (x − x0 ) − F T hx, t|x0 , t0 i
T
2
and
i~
∂
hx, t|x0 , t0 i = hx, t|P (t0 )|x0 , t0 i
0
∂x
M
1
0
=
(x − x ) − F T hx, t|x0 , t0 i .
T
2
4
Now we divide by hx, t|x0 , t0 i and arrive at
M
1
∂
loghx, t|x0 , t0 i = − (x − x0 ) − F T ,
∂x
T
2
∂
M
1
i~ 0 loghx, t|x0 , t0 i =
(x − x0 ) − F T .
∂x
T
2
i~
Next, we need to express the Hamilton operator as a function of X(t) and X(t0 )
with X(t) to the left of X(t0 ) in products,
1
P (t)2 − F X(t)
2M
2
1
1 M
=
X(t) − X(t0 ) + F T − F X(t)
2M T
2
h
i
M
2
2
=
X(t) + X(t0 ) − X(t)X(t0 ) − X(t0 )X(t)
2T 2
F 2T 2
F
X(t) + X(t0 ) +
−
2
8M
M h
T i
2
2
=
X(t)
+
X(t
)
−
2X(t)X(t
)
−
i~
0
0
2T 2
M
2 2
F
F T
−
X(t) + X(t0 ) +
2
8M
[we are using the commutator found in (a)]
H =
which we now use in
i~
∂
hx, t|x0 , t0 i = hx, t|H|x0 , t0 i
∂t
M h
T i F
F 2T 2
0 2
0
=
(x − x ) − i~
− (x + x ) +
2T 2
M
2
8M
0
×hx, t|x , t0 i .
Thus, we have
i~
∂
M
F
F 2T 2
i~
0
0 2
loghx, t|x0 , t0 i =
(x
−
x
(x
+
x
)
+
−
.
)
−
2
∂t
2T
2
8M
2T
5