Solution_Manual_Quantum_Mechanics_for_Sc

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Chapter 2 problem solutions 
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2.3.1 
2.3.1 
(i) From Eq. (2.17) in the book and the following text, we have the expression ds=λzo/s for the 
separation of the fringes. Here zo = 10cm and s = 5nm. We can calculate λ using the de Broglie 
formula λ = h/p. Remembering also that E = p2/2m for a free particle, we have 
 p = 2mE = 31 192 (9.11 10 kg) (1eV) (1.6 10 J/eV)− −⋅ × ⋅ ⋅ × = 5.4x10-25 kg./s 
 34 1 9 25/ / [(6.626 10 J s) (1 10 m)]/[(5 10 m) (5.4 10 )] 25mms o od z s hz spλ − − − −= = = × ⋅ × × × × × = 
(ii) The mass of the proton is 1836 times larger than the mass of the electron. Since p is proportional 
to the square root of m and d is inversely proportional to p, then the answer should be 
25mm / 1836 583 mμ= . 
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2.3.2 
2.3.2 
(i) The slit is of width d and the screen is zo away. Each point in the slit is the source of a spherically 
expanding wave. The wavefunction at a point (x, zo) on the screen is the sum of the waves from all 
the points in the slit. Since it is a continuous set of points, our summation limits to an integral. 
t
-d/2
d/2
zo
t
-d/2
d/2
zo
 
Taking the center of the slit as the origin, consider a small section of the slit at a distance t from the 
center having a width dt. To find the wave function at x on the screen, due to all the point sources 
within this section, we must propagate the expanding waves through the distance between (0, t) and 
(x, zo). This distance r is 
 ( )(1/ 2)2 2( )or z x t= + − 
The wave function at (x, zo) due to the section dt is 
 1( , ) exp( )od x z ikr dtr
ψ = 
The paraxial approximation that x << zo implies that r ~ zo. For the 1/r term, we can safely assume 1/r 
~ 1/ zo, and neglect that as a constant factor; i.e., 
 1( , ) exp( ) exp( )o
o
d x z ikr dt ikr dt
z
ψ = ∝ 
However, an exponential is very sensitive to small changes in its argument so we cannot assume r ~ 
zo in the phase term. The paraxial approximation x << zo allows us to expand the square root in a 
Taylor series keeping only first order terms in (x - t)2. Also using t < x << zo 
 
( )
2 2(1/ 2)2 2
2
2 2
2
1 2( ) 1
2
1 21
2 2
o o
o
o o
o o o
x t xtr z x t z
z
x xt x xtz z
z z z
⎡ ⎤+ −⎛ ⎞= + − ≈ +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
⎡ ⎤−⎛ ⎞ ⎡ ⎤≈ + = + −⎢ ⎥⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎣ ⎦
 
Hence, 
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2.3.2 
 
2
( , ) exp( ) exp exp
2o o o o
x xtd x z ikr dt ik z ik dt
z z
ψ
⎛ ⎞⎛ ⎞ ⎛ ⎞∝ = + −⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠
 
Integrating this over t we get the contribution from the entire slit 
 ( )
/ 2
/ 2
( , ) exp exp
d
o
d o
xtx z ik ik dt
z
ψ φ
−
⎛ ⎞= −∫ ⎜ ⎟
⎝ ⎠
 
where 
2
2o o
xz
z
φ ⎛ ⎞= +⎜ ⎟
⎝ ⎠
 
 ( )
sin
2 2( , ) exp sin exp( )
2
2
o o
o
o
o
kxd
z kxd zx z ik d i
kxdkx z
z
ψ φ φ
⎛ ⎞
⎜ ⎟⎛ ⎞ ⎝ ⎠= =⎜ ⎟ ⎛ ⎞⎝ ⎠
⎜ ⎟
⎝ ⎠
 
We have a function of the form ( )sin /x x , also known as the sinc function. Below is a plot of 
( , )ox zψ . 
7 .104 0 7 .104
0.27
0.15
0.58
1
distance on screen (microns)
 
The function crosses zero each time the argument of the sine is a non-zero multiple of π , so the 
widths of each of the side fringes is oz
d
λ . More importantly, the width of the bright central lobe is 
2 oz
d
λ , which says the electrons diffract more as 
(a) the slit gets narrower for given λ 
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2.3.2 
(b) the wavelength gets larger for a given slit width 
The zo term just says the farther the screen, the larger the spot. 
(ii) The intensity of light at a point on the phosphorescent screen is proportional to the probability 
density of the electron at that point 
 2 2( , ) ( , ) sinc
2o o o
kxdI x z x z
z
ψ ⎛ ⎞= = ⎜ ⎟
⎝ ⎠
 
-d/2
d/2
zo
w
-d/2
d/2
zo
w
 
(iii) Two finite slits 
For two slits we can essentially use the technique in part (i). We integrate over each slit separately 
and add the results. Let the slit separation (center to center) be w and the variable of integration t 
measured from the center of each slit. The limits of integration are t = –d/2 to t = d/2. 
For the top slit this results in 
 
2 2(1/ 2)
2 2
2
2 2
2
( ) 2( )1 2 2( ) 1
2 2
1 ( / 2) 2( / 2) ( / 2) ( / 2)1
2 2
o o
o
o o
o o o
w wx t x twr z x t z
z
x w x w t x w x w tz z
z z z
⎡ ⎤⎛ ⎞− + − −⎢ ⎥⎜ ⎟⎛ ⎞= + − − ≈ +⎢ ⎥⎜ ⎟⎜ ⎟
⎝ ⎠ ⎢ ⎥⎜ ⎟
⎢ ⎥⎝ ⎠⎣ ⎦
⎡ ⎤− − − − −⎛ ⎞ ⎡ ⎤≈ + = + −⎢ ⎥⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎣ ⎦
 
Note that this is the same form as we had for the single slit just with x shifted to / 2x w− . For the 
bottom slit the shift is / 2x x w→ + i.e., 
 
2( / 2) ( / 2)( , ) exp exp
2o o o o
x w x w td x z ik z ik dt
z z
ψ
⎛ ⎞− −⎛ ⎞ ⎛ ⎞= + −⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠
 from the top slit 
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2.3.2 
 
2( / 2) ( / 2)( , ) exp exp
2o o o o
x w x w td x z ik z ik dt
z z
ψ
⎛ ⎞+ +⎛ ⎞ ⎛ ⎞= + −⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠
 from the bottom slit 
Integrating each and adding gives 
2 2
( / 2) ( / 2)sin sin
( / 2) ( / 2)2 2exp exp
( / 2) ( / 2)2 2
2 2
o o
o o
o o
o o
k x w d k x w d
x w x wz zd ik z d ik z
k x w d k x w dz z
z z
− +⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞− +⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎝ ⎠+ + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− +⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 
Ignoring constant phase factors and constants we have 
 
2 2
( / 2) ( / 2)sin sin
( / 2) ( / 2)2 2exp exp
( / 2) ( / 2)2 2
2 2
o o
o o
o o
k x w d k x w d
x w x wz zik ik
k x w d k x w dz z
z z
− +⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟− +⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎝ ⎠+⎜ ⎟ ⎜ ⎟− +⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎝ ⎠
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 
Plotting the modulus squared of this we get 
2 .104 1 .104 0 1 .104 2 .104
1
2
3
4
distance along screen (microns)
In
te
ns
ity
 
Note: Given the numbers in the problem 22 0/w z was very small so neglecting this will still give 
essentially the correct plot. However, in general it may not be the case that the slit spacing is 
negligible. 
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2.4.1 
2.4.1 
We will use the following test for linearity for a non-zero solution ψ(x). 
 If ψ(x) is a solution then aψ(x) is also a solution where a is an arbitrary constant. 
If we substitute aψ(x) into each of the equations we get the following. 
i) ( ) ( ) 0daz ag z z
dz
ψ ψ+ = 
We see that the a's will cancel, meaning that therefore aψ(x) is also a solution, so this equation is 
LINEAR. 
ii) 2 ( )( ) ( ) 0d za z a z
dz
ψψ ψ+ = 
The constant a cannot be canceled in this equation, so it is NOT LINEAR. (In other words, given 
that ψ(x) is a solution of ( )( ) ( ) 0d zz z
dz
ψψ ψ+ = , the only value of a for which aψ(x) is also a solution 
is a = 1.) 
iii) 
2
2
( ) ( ) ( )d z d za ab ac z
dz dz
ψ ψ ψ+ = 
We see that the a's will cancel, so this is LINEAR. 
iv) 
3
3
( ) 1d za
dz
ψ
= 
The constant a cannot be canceled in this equation, so it is NOT LINEAR. 
v) 
2
2 2
2
( ) ( )(1 ( ) ) ( )d z d za a a z ag z
dz dz
ψ ψψ ψ+ + = 
The constant a cannot be canceled in this equation, so it is NOT LINEAR. 
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2.6.1 
2.6.1 
The normalized wavefunctions for the various different levels in the potential well are 
 ( ) 2 sinn
z z
n zz
L L
πψ ⎛ ⎞= ⎜ ⎟
⎝ ⎠
 
The lowest energy state is n = 1, and we are given zL = 1 nm. 
The probability of finding the electron between 0.1 and 0.2 nm from one side of the well is, using 
nanometer units for distance, 
 
( )
[ ]
[ ]
0.2 0.22 2
1
0.1 0.1
0.2
0.10.2
0.1
2sin ( )
1 cos(2 )
0.1 cos(2 )
10.1 sin(2 0.2) sin(2 0.1)
2
0.042
P z dz z dz
z dz
z dz
ψ π
π
π
π π
π
= =∫ ∫
= −∫
= − ∫
= − × − ×
=
 
(Note: For computation purposes, remember that the argument of the sine is in radians and not 
degrees. For example, when we say sin(π ) = 0, it is implicit here that we mean π radians.) 
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2.6.2 
2.6.2 
(i) Odd, since 
 
( ) ( )
( ) ( ) ( ) ( )
sin
sin sin
f x x
f x x x f x
=
− = − = − = −
 
(ii) Neither even nor odd, since 
 
( ) ( )
( ) ( ) ( )
exp cos( ) sin( )
exp cos( ) sin
f x ix x i x
f x ix x i x
= = +
− = − = −
 
(iii) Even, since 
 
( )
( )
2 2
2 2
( )( )
( )
f x x a x a x a
f x x a f x
= − + = −
− = − =
 
(iv) Even, since 
 
( ) ( )
( )
exp exp( ) 2cos( )
2cos( ) 2cos( ) ( )
f x ix ix x
f x x x f x
= + − =
− = − = =
 
(v) Odd, since 
 
( )
( )
2
2
( 1)
( 1) ( )
f x x x
f x x x f x
= −
− = − − = −
 
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2.6.3 
2.6.3 
(i) ( )sin 7 / zz Lπ 
Yes (this is the solution for 7n = for such a simple well) 
(ii) ( )cos 2 / zz Lπ 
No (this does not fit the boundary conditions at the walls of the well, not being zero amplitude at the 
walls) 
(iii) ( ) ( )0.5sin 3 / 0.2sin /z zz L z Lπ π+ 
No (this is a superposition of two eigenfunctions, but that is not a solution of the time-independent 
Schrödinger equation) 
(iv) ( )exp( 0.4 )sin 2 / zi z Lπ− 
Yes (This is the solution for 2n = , with a complex factor that makes no difference in the time-
independent Schrödinger equation.) 
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2.6.4 
2.6.4 
We have an infinite potential well of width zL in each of the three dimensions. 
(i) Given the note at the end of the problem, we can write the solution as a product of three solutions 
 
 ( ) 2 2 2, , sin sin sinx y z yx zn n n
z z z z z z
n yn x n zx y z
L L L L L L
ππ πψ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= × ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
 
With the allowed energies of the three dimensional system being 
 2 2 21 ( )x y zn n n n x y zE E E E E n n n∞= + + = + + 
where 
 
22
1 2 z
E
m L
π∞ ⎛ ⎞= ⎜ ⎟
⎝ ⎠
 
is the lowest allowed energy of a particle in a one-dimensional infinite potential well. 
 (ii) The lowest allowed state corresponds to ( , , ) (1,1,1)x y zn n n = and has an energy 1 13E E∞= . In this 
state the wave function in each of the three directions looks like 
 
The next three states all have the same energy 2,3,4 16E E∞= and correspond to ( , , ) (2,1,1)x y zn n n = , 
( , , ) (1, 2,1)x y zn n n = , ( , , ) (1,1,2)x y zn n n = . In these three cases, the wave function along two of the 
dimensions looks as above, but along the third dimension it is the next higher state of the infinite well 
 
(iii) These last three states have equal energy and hence are degenerate (i.e., they have a degeneracy 
of 3) because the width of the well is the same in all three dimensions. 
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2.7.1 
2.7.1 
To check orthogonality on [-1,1] we take the inner product of the two functions. If they are 
orthogonal on the interval, then the inner product (orthogonality integral) is zero, i.e., 
 
1
*
1
( ) ( ) 0f x g x dx
−
=∫ 
Since the interval is symmetric about 0x = , we can use the parity of the integrand to determine by 
inspection whether the integral is zero or not. But we should be careful with periodic functions, as in 
(v). 
(i) orthogonal 
 
1
3
1
0x
−
=∫ (since 3x is odd) 
(ii) not orthogonal 
 
1
4
1
0x
−
≠∫ (since 4x is even) 
(iii) not orthogonal 
 
1
1
sin 0x xdx
−
≠∫ (since sinx x is even) 
(iv) not orthogonal 
 
1 1
1 1
exp cos sin
2 2 2
i x x xx dx x ix dxπ π π
− −
−⎛ ⎞ ⎛ ⎞= −∫ ∫⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 
note that cos( / 2)xπ and sin( / 2)xπ are periodic with period 4, so the interval is less than a period. 
 
1
1
0 sin 0
2
xi x dxπ
−
⎛ ⎞− ≠∫ ⎜ ⎟
⎝ ⎠
 
(since cos
2
xx π is odd and sin
2
xx π is even) 
(v) orthogonal 
 
1 1
1 1
exp( 4 ) [cos(4 ) sin(4 )] 0ix dx x i x dxπ π π
− −
− = − =∫ ∫ 
because the period of both cos(4 )xπ and sin(4 )xπ is 2, so the interval is an integral number of 
periods (in this case one full period). The integral of a sinusoid over one period is zero. 
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2.7.2 
2.7.2 
(i) 20 1 2( ) 1, ( ) , ( ) ... ( ) nnf x f x x f x x f x x= = = = 
For a counter example to show these functions are not all orthogonal, we can choose the x and x3 
functions, leading to an orthogonality integral 
 
1 1
3 4
1 1
0x x dx x dx
− −
× = ≠∫ ∫ (since 4x is even) 
(ii) Let the unnormalized functions be labeled as ( )ih x . So we have, for our first member of this set 
of functions 
 0 0( ) ( ) 1h x f x= = 
To normalize a function we divide it by the appropriate normalization factor, which is the square root 
of the integral of its modulus squared over the interval of interest. For this function, the integral of its 
modulus squared over the interval of interest is 
 
1 12
0
1 1
( ) 1 2f x dx dx
− −
= =∫ ∫ 
Thus the normalized version is 0
1( )
2
g x = 
We can check that 0 ( )g x is normalized; i.e., 
 
1 2
0
1
( ) 1g x
−
=∫ (1) 
Now we try to construct a function that is orthogonal to 0 ( )g x by constructing a combination of 
0 ( )g x and 1( )f x that is orthogonal to 0 ( )g x . One appropriate way of writing such a combination is 
1 1 10 0( ) ( ) ( )h x f x a g x= + . We want this to be orthogonal to 0 ( )g x , so we require 
 ( )
1
1 10 0 0
-1
( ) ( ) ( ) 0f x a g x g x dx+ =∫ 
so, using Eq. (1) 
1
1 0 10
-1
( ) ( )f x g x dx a= −∫ 
1
10 1 1
-1
Hence 0 . Thus, in this particular case, ( ) ( )
2
xa dx h x f x x− = = = =∫ . 
 Normalizing 
1 12 2
1
1 1
( ) 2 / 3h x dx x dx
− −
= =∫ ∫ 
Hence 1
3( )
2
g x x= 
So now we have constructed a function 1 0( ) that is normalized and orthogonal to ( )g x g x . 
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2.7.2 
c) Similarly, we now try to construct a function 2 ( )h x that is orthogonal to 0 ( )g x and 1( )g x by 
adding amounts of these two functions and some of the next linearly independent function in our set, 
i.e., 
 2 2 20 0 21 1( ) ( ) ( ) ( ) ( )h x f x a g x a x g x= + + 
Orthogonalizing to g0 and g1 gives 
 
1 1
2 2
21 20
1 1
3 1 20 and - 
2 2 3
a x xdx a x dx
− −
= − = = =∫ ∫ 
So 22
1( ) 
3
h x x= − 
Normalizing, we have 
1 12 4 2
2
1 1
1 2 8( ) ( ) 
9 3 45
h x dx x x dx
− −
= + − =∫ ∫ 
and hence 2 22
1 45 1 5( ) (3 1) (3 1)
3 8 2 2
g x x x= − = − 
d) We write 
 0 0 1 1 , 1 1( ) ( ) ( ) ( ) ..... ( ) i i i i i i ih x f x a g x a g x a g x− −= + + + + 
Orthogonalizing to ( )jg x , we have 
 
1 1 1
1 1 1
1
1
( ) ( ) 0 ( ) ( ) ( ) ( ) 0 
 ( ) ( ) 
i j i j ij j j
ij i j
h x g x dx f x g x dx a g x g x dx
a f x g x dx
− − −
−
= ⇒ + =∫ ∫ ∫
⇒ = − ∫
 
e) For the function 3( )g x , we start with 3 3 30 0 31 1 32 2( ) ( ) ( ) ( ) ( )h x f x a g x a g x a g x= + + + 
Orthogonalizing to 0 ( )g x , 1( )g x , and 2 ( )g x leads to 
 
1 1 1
3 3 3 2
30 31 32
1 1 1
1 3 2 3 1 50 - (3 1) 0
2 2 5 2 2 2
a x dx a x xdx a x x dx
− − −
= − = = = = − =∫ ∫ ∫ 
So 3 33
2 3( ) ( ) 5 3
5 2
h x x x x x= − = − 
Normalizing, we have 
1 12 6 2 4
3
1 1
8( ) (25 9 30 )
7
h x dx x x x dx
− −
= + − =∫ ∫ 
Hence 33
1 7( ) (5 3 )
2 2
g x x x= − 
So finally we have 
 2 30 12 3
1 3 1 5 1 7( ) ( ) ( ) (3 1) ( ) (5 3 )
2 2 2 2 2 2
g x g x x g x x g x x x= = = − = − 
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2.7.2 
f) The above is not the only set of orthogonal normalized functions for this interval in powers of x. If 
we start with a different function in the series we will in general get a new set. For instance, we could 
make the following choice for our first function 
 ( )0 3h f x x= = 
Normalizing this choice gives ( )0 32g x x= 
Then for our second function, we would have 
 ( ) ( ) ( )1 2 10 0h x f x a g x= + 
Orthogonalizing to ( )0g x leads to 
1
2
10
1
3 0
2
a x xdx
−
= − =∫ 
Hence ( ) 21h x x= 
Normalizing this function gives ( ) 21 52g x x= 
and so on. Note that these functions g are a different and orthonormal set; for example, we have no 
function that is simply proportional to x2 in the set of functions in part (e) above. 
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	Introduction
	Problems
	Chapter 2 Problems
	Section 2.3
	Section 2.4
	Section 2.6
	Section 2.7
	Section 2.8
	Section 2.9
	Section 2.10
	Section 2.11
	Chapter 3 Problems
	Section 3.1
	Section 3.3
	Section 3.6
	Section 3.7
	Section 3.8
	Section 3.10
	Section 3.11
	Section 3.12
	Section 3.13
	Section 3.14
	Chapter 4 Problems
	Section 4.1
	Section 4.2
	Section 4.6
	Section 4.8
	Section 4.10
	Section 4.11
	Section 4.12
	Chapter 5 Problems
	Section 5.1
	Section 5.2
	Section 5.4
	Chapter 6 Problems
	Section 6.2
	Section 6.3
	Section 6.4
	Section 6.5
	Section 6.6
	Chapter 7 Problems
	Section 7.1
	Section 7.2
	Section 7.4
	Chapter 8 Problems
	Section 8.3
	Section 8.4
	Section 8.5
	Section 8.6
	Section 8.8
	Section 8.10
	Chapter 9 Problems
	Section 9.1
	Section 9.2
	Chapter 10 Problems
	Section 10.5
	Chapter 11 Problems
	Section 11.1
	Section 11.2
	Section 11.4
	Chapter 12 Problems
	Section 12.4
	Chapter 13 Problems
	Section 13.1
	Section 13.4
	Section 13.6
	Chapter 14 Problems
	Section 14.1
	Section 14.3
	Section 14.5
	Chapter 15 Problems
	Section 15.1
	Section 15.5
	Section 15.6
	Section 15.7
	Chapter 16 Problems
	Section 16.1
	Section 16.2
	Section 16.3
	Chapter 17 Problems
	Section 17.2
	Section 17.3
	Section 17.4
	Chapter 18 Problems
	Section 18.3
	Chapter 19 Problems
	Section 19.1
	Section 19.3
	Problem Solutions
	Chapter 2 Solutions
	2.3.1
	2.3.2
	2.4.1
	2.6.1
	2.6.2
	2.6.3
	2.6.4
	2.7.1
	2.7.2
	2.8.1
	2.8.2
	2.8.3
	2.8.4
	2.8.5
	2.8.6
	2.8.7
	2.9.1
	2.9.2
	2.9.3
	2.9.4
	2.10.1
	2.11.1
	2.11.2
	2.11.3
	Chapter 3 Solutions
	3.1.1
	3.1.2
	3.1.3
	3.3.1
	3.6.1
	3.6.2
	3.6.3
	3.6.4
	3.7.1
	3.7.2
	3.7.3
	3.8.1
	3.10.1
	3.11.1
	3.12.1
	3.12.2
	3.12.3
	3.12.4
	3.12.5
	3.13.1
	3.14.1
	3.14.2
	Chapter 4 Solutions
	4.1.1
	4.1.2
	4.2.1
	4.6.1
	4.8.1
	4.10.1
	4.10.2
	4.10.3
	4.10.4
	4.10.5
	4.10.6
	4.11.1
	4.11.2
	4.11.3
	4.11.4
	4.11.5
	4.11.6
	4.11.7
	4.11.8
	4.12.1
	Chapter 5 Solutions
	5.1.1
	5.1.2
	5.1.3
	5.2.1
	5.2.2
	5.2.3
	5.4.1
	5.4.2
	5.4.3
	Chapter 6 Solutions
	6.2.1
	6.2.2
	6.3.1
	6.3.2
	6.3.3
	6.4.1
	6.5.1
	6.5.2
	6.6.1
	6.6.2
	Chapter 7 Solutions
	7.1.1
	7.1.2
	7.2.1
	7.2.2
	7.2.3
	7.4.1
	7.4.2
	7.4.3
	Chapter 8 Solutions
	8.3.1
	8.4.1
	8.5.1
	8.6.1
	8.6.2
	8.8.1
	8.8.2
	8.10.1
	8.10.2
	8.10.3
	Chapter 9 Solutions
	9.1.1
	9.1.2
	9.2.1
	9.2.2
	Chapter 10 Solutions
	10.5.1
	10.5.2
	10.5.3
	10.5.4
	10.5.5
	10.5.6
	10.5.7
	10.5.8
	10.5.9
	Chapter 11 Solutions
	11.1.1
	11.2.1
	11.2.2
	11.2.3
	11.2.4
	11.4.1
	11.4.2
	11.4.3
	Chapter 12 Solutions
	12.4.1
	12.4.2
	12.4.3
	Chapter 13 Solutions
	13.1.1
	13.4.1
	13.4.2
	13.4.3
	13.4.4
	13.6.1
	Chapter 14 Solutions
	14.1.1
	14.3.1
	14.3.2
	14.5.1
	Chapter 15 Solutions
	15.1.1
	15.1.2
	15.5.1
	15.6.1
	15.6.2
	15.6.3
	15.7.1
	Chapter 16 Solutions
	16.1.1
	16.1.2
	16.2.1
	16.3.1
	16.3.2
	16.3.3
	Chapter 17 Solutions
	17.2.1
	17.3.1
	17.4.1
	17.4.2
	Chapter 18 Solutions
	18.3.1
	18.3.2
	18.3.3
	18.3.4
	Chapter 19 Solutions
	19.1.1
	19.3.1