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 x
If we look at 
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 defining the z
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
^ sin ) i
We a
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. When H
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
zero, and the second
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 coordinates of the two parti
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 note that
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

(
2
210 2
z 
2
100 2

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 particle in the we
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
x2 /2
(1
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 is
2, We no
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
(x) Ce
ikx
De
ik
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 Hamilt
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 values o
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 the r.h.s. we g
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 that
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
The total
Metalografa y Tratamientos Trmicos
X
1
X TEMPLE Y REVENIDO
X.1 Temple y Revenido ordinario
Es una operacin que se realiza calentando a una temperatura por encima del punto de
transformacin Ac3 o Ac1, enfriando con tal velocidad que se produzca un cons