Physics 710
March 12, 2010
Problem Set 15
Problem 12.6.1: E = Aer/a0 .
(1) No (, )-dependence implies E Y00 , so we must have
(2) Therefore E = RE, =0 =
1
U
r E,0
= 0 and m = 0.
which satises eqn. (12.6.5) with
(rE ) +
2
2
= 0:
[E V (r)](rE ) = 0.
(1)
As
Physics 711
January 29, 2010
Problem Set 10
Problem 1: Exercise 9.4.2 .
Solution:
(1)
r2 dr =
| = 4
0
4
a3
0
e2r/a0 r2 dr =
0
4
2!
= 1.
a3 (2/a0 )3
0
(2) This state = (r) is clearly even under (x, y, z ) (x, y, z ), so X = Y =
Z = 0. Therefore (X )2 = X
Physics 711
January 22, 2010
Problem Set 9
Problem 1: Exercise 7.4.2
1/2
X
= n|X |n
1/2
=
n|(a + a)|n =
2m
1/2
=
n|
2m
n + 1 n|n + 1 + n n|n 1
2m
n + 1|n + 1 +
= 0,
since n|n 1 = n|n + 1 = 0 by orthogonality of the energy eigenstates. Similarly,
P
= n|P |
Physics 711
February 5, 2010
Problem Set 11
Problem 10.2.3:
2
(1) Since H = Hx + Hy + Hz where Hx = Px /(2m) + X 2 (m 2 /2), etc., the problem
separates into 3 copies of the 1-dimensional harmonic oscillator. So the eigenstates of H are the tensor product
Physics 710
February 19, 2010
Problem Set 12
Problem 11.2.2: Since I = T ( )T ( ) = [I +(i / )G + O( 2 )] [I (i / )G + O( 2 )] =
I + (i / )(G G) + O( 2 ), therefore
0=
i
(G G) + O( 2 ).
Since this is true for all , it implies that G G = 0.
Problem 11.4.1:
Physics 710
March 5, 2010
Problem Set 14
Problem 12.5.2: (2 points)
(1) From the text (eqns. 12.5.23 and 12.5.24) we read o
(1
Jx /2) =
2
01
,
10
(1/2)
where the last, Jz
Then
0 i
,
i0
2
(1
Jz /2) =
10
,
0 1
2
, follows immediately from the denition of th
Physics 710
February 26, 2010
Problem Set 13
Problem 12.2.3: x = cos and y = sin imply
x
y
x +
y = cos x + sin y ,
x
y
=
x +
y = sin x + cos y .
=
Invert this to get
1
sin ,
1
= sin + cos .
x = cos
y
Therefore, since Lz = XPy Y Px , acting on wavefuncti
Physics 711
January 15, 2010
Problem Set 8
Problem 1: Exercise 7.3.1
Plug the power series expansion =
to get
n
n=0 cn y
into the equation + (2 y 2 ) = 0
cn n(n 1)y n2 + (2 y 2 )y n = 0.
n=0
Shift n n + 2 in the rst term, and n n 2 in the third term to ge
Physics 710-711-712
December 4, 2009
Problem Set 7
Problem 1: Exercise 5.3.1
The Hamiltonian is
12
P + Vr (X ) iVi
2m
where Vr is a real function and Vi a real constant. Therefore
H=
H =
12
1
(P )2 + Vr (X ) (i) Vi =
P + Vr (X ) + iVi = H,
2m
2m
so H is n
Physics 710-712
due October 2, 2009
Problem Set 1 (real version)
Let v , w, . be vectors in a real vector space V, and a, b, c, . be real scalars. Also,
let A, B , C , . be linear operators mapping V V, with I denoting the identity
operator. Finally, let
Physics 710-712
due October 2, 2009
Problem Set 1 (complex version)
Let |v , |w , . be vectors in a complex vector space V, and a, b, c, . be complex
scalars. Also, let A, B , C , . be linear operators mapping V V, with I denoting
the identity operator. F
Physics 710-712
due October 14, 2009
Problem Set 2
Problem 1: Do exercise 1.3.4 of the text.
Solution:
|V + W |2
=
V + W |V + W
2
V |V + V |W + W |V + W |W
=
2
= |V | + |W | + V |W + V |W
= |V |2 + |W |2 + 2 Re V |W
|V |2 + |W |2 + 2| V |W |
2
(1)
2
2
|V
Physics 710-712
due October 26, 2009
Problem Set 3
Problem 1: Do exercise 1.10.1 of the text.
Solution:
dx (ax) f (x)
=
=
=
y
y
(y ) f
(change variables ax = y )
a
a
1
y
(change of variables for integrals)
dy (y ) f
|a|
a
1
1
y
1
=
dy (y ) f
f (0) =
dy (
Physics 710-711-712
November 25, 2009
Problem Set 6
Problem 1: Exercise 5.1.2
2
In the X basis, (5.1.3) becomes m E = EE , whose solutions are = Aeikx + Beikx
2
where k = 2mE . This is the solution given in the exercise. If E < 0, these solutions
are not
Physics 710-711-712
November 16, 2009
Problem Set 5
Problem 1: Do exercise 4.2.1 of the text.
(1)
The possible outcomes are Lz = cfw_1, 0, 1, which are the eigenvalues of Lz .
Solution:
(2)
Lz | = 1 | implies
Solution:
1
| = 0 .
0
(Note that I have normal
Physics 710-712
due October 30, 2009
Problem Set 4
Problem 1: Do exercise 2.5.3 of the text.
Solution: The problem asks us to get the equations of motion using the Hamiltonian method
for the system shown in Figure 1.5 (p. 46) of the text. Using x1 and x2