CHAPTER 2 1. We have
(x) dkA(k)e
ikx
dk
N k
2 2
e
ikx
dk
N k
2 2
coskx
because only the even part of eikx = coskx + i sinkx contributes to the integral. The integral can be looked up. It yields
(x) N e
x 
so that
 (x) 
2
N2
2
2
e
2 
SOLUTIONS MANUAL
CHAPTER 1
1. The energy contained in a volume dV is
U( ,T)dV
U( ,T)r drsin d d
2
when the geometry is that shown in the figure. The energy from this source that emerges through a hole of area dA is
dE( ,T ) U ( ,T )dV dAcos 4 r2
CHAPTER 3. 1. The linear operators are (a), (b), (f) 2.We have
x
dx' x' (x')
(x)
To solve this, we differentiate both sides with respect to x, and thus get
d (x) dx x (x)
A solution of this is obtained by writing d / immediately state that
(x) Ce
1
CHAPTER 13 1. (a) electronproton system mr
me (1 5.45 10 4 )me 1 me / M p me 4 (b) electrondeuteron system mr (1 2.722 10 )me 1 me / M d m (c) For two identical particles of mass m, we have mr 2
2. One way to see that P12 is hermitian, is to n
CHAPTER 5.
1. We are given
dx(A (x) * (x)
dx (x) * A (x)
Now let (x) (x) (x) , where is an arbitrary complex number. Substitution into the above equation yields, on the l.h.s.
dx(A (x) dx (A ) *
A (x) *( (x) (A )*
(x) * (A )*  2 (A )*
On th
CHAPTER 4.
1. The solution to the left side of the potential region is As shown in Problem 315, this corresponds to a flux
j(x) k 2 A m B 
2
(x)
Ae
ikx
Be
ikx
.
The solution on the right side of the potential is as above, the flux is
(
CHAPTER 19 1. We have
M fi 1 3 d re V
i .r
V (r)
If V(r) = V(r), that is, if the p9otential is central, we may work out the angular integration as follows:
M fi 1 V r V (r)dr
2 2 0
0
d
0
sin d e
i r cos
with the choice of the vector
2 0
as d
CHAPTER 8 1. The solutions are of the form where un (x)
E
n1 n 2 n 3
(x, y,z)
un1 (x)un2 (y)un3 (z)
2 n x ,and so on. The eigenvalues are sin a a
E n1 E n2 E n3 2 2 2 (n 2ma2 1
2 n2 2 n3 )
2. (a) The lowest energy state corresponds to the lowest
CHAPTER 7 1. (a) The system under consideration has rotational degrees of freedom, allowing it to rotate about two orthogonal axes perpendicular to the rigid rod connecting the two masses. If we define the z axis as represented by the rod, then the H
CHAPTER 12. 1. With a potential of the form
V(r) 1 m 2
2 2
r
the perturbation reduces to
H1
2 1 1 dV (r) 2 L2 2 2S L 2 (J 2m c r dr 4mc 2 ( ) j( j 1) l(l 1) s(s 1) 4mc 2
S2 )
where l is the orbital angular momentum, s is the spin of the parti
CHAPTER 11 1. The first order contribution is
E
(1) n
nx n
2
2m
2
n  (A
A )(A
A ) n
To calculate the matrix element n  A2
A n n 1 n 1 ; n  A
AA
2 A A (A )  n we note that
n 1 n 1  so that (1) the first and last terms give
ze
CHAPTER 10 1. We need to solve
0 2 i i u 0 v u 2 v
1 1 2 i The eigenstate can be obtained by noting that it must be orthogonal to the + state, and 1 1 this leads to . 2 i
For the + eigenvalue we have u = iv , s that the normalized eigenstate i
CHAPTER 6 19. (a) We have Aa> = aa> It follows that <aAa> = a<aa> = a if the eigenstate of A corresponding to the eigenvalue a is normalized to unity. The complex conjugate of this equation is <aAa>* = <aA+a> = a* If A+ = A, then it follows
1
CHAPTER 14 1. The spinpart of the wave function is the triplet
ms ms ms
1 0 1
(1)
(2)
1 ( 2
(1)
(1)
(2)
(1)
(2)
)
(2)
This implies that the spatial part of the wave function must be antisymetric under the interchange of the coordinate
CHAPTER 16. 1. The perturbation caused by the magnetic field changes the simple harmonic oscillator Hamiltonian H0 to the new Hamiltonian H
H H0 q B L 2m
If we choose B to define the direction of the z axis, then the additional term involves B Lz. W
1
CHAPTER 15 1. With the perturbing potential given, we get
C(1s 2 p) eE0 i
210
z 
100
0
dte e
i t
t
where = (E21 E10). The integral yields 1 / ( C(1s2p) is
i ) so that the absolute square of
P(1s
We may use 
2 p) e E
2
2 0

(

CHAPTER 17 1. We start with Eq. (1719) . We define k as the z axis. This means that the polarization vector, which is perpendicular to k has the general form
( )
^cos i
^ sin j
This leads to
B A i 2
0
V
^ kk (^ cos i
^ sin ) j
B0 ( ^ cos j