PHY 5645 Homework Solutions - D. Kimel
HW Set 5, Problem 1
Since the energy of the state is determined, but not the momentum, then clearly the momentum can
be p = k 0 . Thus we can write
*
|f = c 1 |k 0 + c 2 | ? k 0
(1)
The problem states, that in the x
PHY 5645 Homework Solutions - D. Kimel
HW Set 5, Problem 2
Working in a frame where 3 = 3 = 0, we have Ar 2 = 32 , Ap 2 = 3 2 . The resulting nine
p
r
r
p
Cartesian products separate into two groups as follows:
Ar 2 Ap 2 = AxAp x 2 + . + AxAp y 2 + AyAp x
PHY 5645 Homework Solutions - D. Kimel
HW Set 6, Problem 2: 2.11
%
For a one-dimensional SHO potential H = p 2 /2m + mg 2 x 2 /2; hence x = 1/i x, H = p/m, and
%
6
p = 1/i p, H = 1/i mg 2 /2 p, x 2 = mg 2 /2i ?2ix = ?mg 2 x. Hence x + g 2 x = 0, and
%
the
PHY 5645 Homework Solutions - D. Kimel
HW Set 6, Problem 3: 2.12
a. The initial state |J = e ?
x|J = x|e ?
ipa
ipa
|0 can be written in the position representation as
|0 = 0|e +
ipa
|x D = 0|x ? a D = x ? a|0
(1)
where Ive used the fact that the wave func
PHY 5645 Homework Solutions - D. Kimel
HW Set 6, Problem 3: 2.13
a. We can write x and p in terms of the creation and annihilation operators:
a ! + a , p = i mg a ! ? a
2
2mg
x=
(1)
Thus
x|n =
m|x =
2mg
n + 1 |n + 1 + n |n ? 1
2mg
m + 1 m + 1| + m m ? 1
PHY 5645 Homework Solutions - D. Kimel
HW Set 6, Problem 3: 2.14
a. Consider p v |x|J
p v |x|J =
X dx v p v |x v |x v x v |J =
1
2^
X dx v x v e ?
= i / v p v |J,
/p
i
pv xv
x v |J = i / v
/p
QED
1
2^
X dx v e ?
i
pv xv
x v |J
(2)
b. Using our substitutio
PHY 5645 Homework Solutions - D. Kimel
HW Set 7, Problem 1
!
First, let us choose a new coordinate system for spin in which z R n so that
33
W 6 B = gS z
(1)
and thus,
p2
p2
33
?W6B =
+ gS z
(2)
H=
2m
2m
and
S z |m = mi|m miX m
(3)
1
where
1
Xm
1
=
0
0
0
PHY 5645 Homework Solutions - D. Kimel
HW Set 7, Problem 2
The Hamiltonian is given by
H = AS x + BS y
(1)
Let us define
3
*
C A + Bj
(2)
3
We now choose a new z ? axis parallel to C, see the figure.
z
A
y
C
B
x
3
Then with the new coordinate system along
PHY 5645 Homework Solutions - D. Kimel
HW Set 8, Problem 1
3.1 Find the eigenvalues and eigenvectors of a y =
0 ?i
i
0
. First form the characteristic
equation
?V ?i
= 0 = V 2 ? 1, with solutions V = 1
?V
i
(1)
We then have to solve the equation
0 ?i
c1
i
PHY 5645 Homework Solutions - D. Kimel
HW Set 8, Problem 2
3.2 a. Let us write
3a
a 0 + ia 6 3
3a
a 0 ? ia 6 3
33
U = a 0 + ia 6 a =
3a
a 0 ? ia 6 3
3a
a 0 + ia 6 3
3a
a 0 + ia 6 3
a2 ? 3
a
0
=
2
3a
+ 2ia 0 a 6 3
(1)
2
a2 + 3
a
0
3a
where Ive used a 6 3 a
PHY 5645 Homework Solutions - D. Kimel
HW Set 8, Problem 3
3.3
I will be using direct product states, required for two independent particles by the interpretation of the
state vectors in terms of probabilities. Let the electron states form the first vecto
PHY 5645 Homework Solutions - D. Kimel
HW Set 8, Problem 4
3.4
For a spin-1 particle, the eigenstates are given by
S z |z, m = m|z, m, with m = 1, 0, ?1
(1)
Thus the matrix elements of the operator in question are
z, m v |S z S z + S z ? |x, m = 0
(2)
whe
PHY 5645 Homework Solutions - D. Kimel
HW Set 9, Problem 1
1.
a. The matrix representation of H is
b|H|b b|H|r
01
= V0
r|H|b r|H|r
(1)
10
by inspection.
b. From eq.(1), we see H = V 0 a x , or C = a x so therefore, remembering the eigenvalues of a x
1
, t
PHY 5645 Homework Solutions - D. Kimel
HW Set 9, Problem 2
We know from the class notes that for j =
1
2
,
K
D 2 R i K = e ? i S i K = e ?ia i 2 = cos
1
i
K
K
? ia i sin
2
2
(1)
a. Thus, for rotation about the y axis
cos
K
K
D R y K = cos ? ia y sin
=
2
2
PHY 5645 Homework Solutions - D. Kimel
HW Set 10, Problem 2
In terms of state vectors, the initial state of the system is
|z, l = 1, m = 1 + |z, 1, 0
|J =
2
!
And the probability we want is P = |n, 1, 0|J |2 . Now we know
i
i
? i Jz ^ ? i Jy ^
!
2e
3 |z,
PHY 5645 Homework Solutions - D. Kimel
HW Set 11, Problem 2.
We were given the wave function,
x 2 ? y 2 + 2ixy
+ 4z + 3 e ?Jr
f=N
r
r2
which we recognize as
(1)
32^ Y 2 + 4 4^ Y 0 + 3 4^ Y 0 e ?Jr
0
31
15 2
As a state vector, this can be written
f=N
|f =
PHY 5645 Homework Solutions - D. Kimel
HW Set 11, Problem 3.
3.16 (a)
In this problem, I will use symmetry to a great extent.
First, by symmetry we see L y = L x and L 2 = L 2 . Thus we only need to calculate L x and
y
x
L 2 . It is clear that
x
(1)
lm|L
PHY 5645 Homework Solutions - D. Kimel
HW Set 12, Problem 1.
3.21
a) Note that
> d jmm K
2
v
m=
>jm|e ?
m
=
>jm v |e +
i
iJ K
y
|jm v jm|e ? J y K |jm v D m
i
m
Jy K
m
i
i
i
Jz
J
|jmjm|e ? J y K |jm v = jm v |e + J y K z e ? J y K |jm v
(1)
We remember t
PHY 5645 Homework Solutions - D. Kimel
HW Set 12, Problem 2.
3.26
a) We will be using
TL =
M
> L 1 M 1 L 2 M 2 |L 1 L 2 LX L Z L
M
M
1
1
2
2
(1
M1M2
We identify U = X and V = Z in Eq. (1). Thus we can write
(2)
T 1 = 1 U 1 V1 ? 1 U 1 V1
10
01
1
2
2
(3)
T
PHY 5645 Homework Solutions - D. Kimel
HW Set 12, Problem 3.
3.28
a) In this problem, we will be using
X L 11 Z L 22 =
M
M
>L 1 M 1 L 2 M 2 |L 1 L 2 LMT L
M
(1)
LM
Remembering r 1 = z, r 1 1 = @x iy / 2 , we see
0
xy = i r 1 1 r 1 1 ? r 1 r 1 = i T 2 2 ?
PHY 5645 Homework Solutions - D. Kimel
HW Set 13, Problem 1.
3.29
In the expression for H int , we recognize that S 2 = 1 S 2 + S 2 + S + , S ? and
+
?
x
4
1
2
2
2
S y = ? 4 S + + S ? ? S + , S ? j, with S = S x iS y and S + , S ? = 2 3 2 ? S 2 . Thus
S
z
PHY 5645 Homework Solutions - D. Kimel
HW Set 13, Problem 2.
The coupled states diagonalize 3 1 6 3 2 =
SS
we can write the coupled states as
|Sm =
>
S 2 ?S 2 ?S 2
1
2
2
. Let the single particle states be |n|sm. Thus
s 1 m 1 s 2 m 2 |Sm|s 1 m 1 |s 2 m 2
PHY 5645 Homework Solutions - D. Kimel
HW Set 2, Problem 2
We consider the matrix
131
(1)
020
014
with characteristic equation V ? 1 V ? 2 V ? 4 = 0, giving eigenvalues V = 1, 2, 4 and
eigenvectors,
?5
1
0
0
1,
1
30
?2
1
1
2,
1
10
0
4
(2)
3
The matrix o
PHY 5645 Homework Solutions - D. Kimel
HW Set 2, Problem 3
We consider the matrix
cos S
sin S
(1)
? sin S cos S
with characteristic equation V ? e iS V ? e ?iS = 0, giving eigenvalues V = e iS , e ?iS , and eigenvectors
1
2
?i
1
e iS , 1
2
i
1
e ? iS
(2
PHY 5645 Homework Solutions - D. Kimel
HW Set 3, Problem 1
a
Let us write the arbitrary normalize state
f
as e iJ
b
1 ? f2
, with f real. Then we can
write
R 2 =
?2
6
1 ? f2
f
?2
6
?2
9
?2
f
9
1?f
= ?45f 2 ? 60f 1 ? f 2 + 85
2
(1)
Next consider the eigenv
PHY 5645 Homework Solutions - D. Kimel
HW Set 3, Problem 2
After a measurement of A which give the result a 1 , the system is in the state f = d 1 . If B is now
measured, the probabilities of obtaining the values b 1 and b 2 are given by expanding d 1 in
PHY 5645 Homework Solutions - D. Kimel
HW Set 3, Problem 3
The wave function is given by
fx =
1 |x | a, fx = 0, otherwise
2a
(1)
Thus, since the average momentum is obviously = 0, the uncertainty in momentum is given by
Ap =
f|p 2 |f
(2)
So
f|p 2 |f =
Now
PHY 5645 Homework Solutions - D. Kimel
HW Set 4, Problem 1: 1.20
Remember,
AA 2 = A 2 ? A 2
(1)
Our arbitrary state with which we evaluate the expectation value can be written
cos J
| = cos J| + + sin Je iK | ?
, in the | +, | ? representation.
sin Je iK
PHY 5645
HW Set 4, Problem 2, Solutions - D. Kimel
2. 1.21 For the infinite square well of width a
2 sin n^x
a
a
f n x =
, n = 1, 2, 3, .
(1)
By symmetry,
f n |x|f n x n = a , p n = 0
2
(2)
Thus
Ax 2 n = x 2 n ?
2
a
2
and, using p 2 f n = ? 2
=2
a
X0
d2
d