 Principles of Quantum Mechanics as Applied to Chemistry and Chemical Physics

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```two or more identical nuclei.
2.8 Particle in a three-dimensional box
A simple example of a three-dimensional system is a particle confined to a
rectangular container with sides of lengths a, b, and c. Within the box there is
no force acting on the particle, so that the potential V (r) is given by
V (r)  0, 0 < x < a, 0 < y < b, 0 < z < c
 1, x , 0, x . a; y , 0, y . b; z , 0, z . c
The wave function \u142(r) outside the box vanishes because the potential is
infinite there. Inside the box, the wave function obeys the Schro¨dinger equation
(2.70) with the potential energy set equal to zero
ÿ&quot;2
2m
@2\u142(r)
@x2
 @
2\u142(r)
@ y2
 @
2\u142(r)
@z2
!
 E\u142(r) (2:75)
The standard procedure for solving a partial differential equation of this type is
to assume that the function \u142(r) may be written as the product of three
functions, one for each of the three variables
\u142(r)  \u142(x, y, z)  X (x)Y (y)Z(z) (2:76)
Thus, X (x) is a function only of the variable x, Y (y) only of y, and Z(z) only of
z. Substitution of equation (2.76) into (2.75) and division by the product XYZ
give
ÿ&quot;2
2mX
d2 X
dx2
 ÿ&quot;
2
2mY
d2Y
dy2
 ÿ&quot;
2
2mZ
d2 Z
dz2
 E (2:77)
The first term on the left-hand side of equation (2.77) depends only on the
variable x, the second only on y, and the third only on z. No matter what the
values of x, or y, or z, the sum of these three terms is always equal to the same
constant E. The only way that this condition can be met is for each of the three
terms to equal some constant, say Ex, Ey, and Ez, respectively. The partial
differential equation (2.77) can then be separated into three equations, one for
each variable
d2 X
dx2
 2m
&quot;2
ExX  0, d
2Y
dy2
 2m
&quot;2
EyY  0, d
2 Z
dz2
 2m
&quot;2
EzZ  0
(2:78)
where
Ex  Ey  Ez  E (2:79)
2.8 Particle in a three-dimensional box 61
Thus, the three-dimensional problem has been reduced to three one-dimen-
sional problems.
The differential equations (2.78) are identical in form to equation (2.34) and
the boundary conditions are the same as before. Consequently, the solutions
inside the box are given by equation (2.40) as
X (x) 

2
a
r
sin
nxðx
a
, nx  1, 2, 3, . . .
Y (y) 

2
b
r
sin
n yðy
b
, ny  1, 2, 3, . . . (2:80)
Z(z) 

2
c
r
sin
nzðz
c
, nz  1, 2, 3, . . .
and the constants Ex, Ey, Ez are given by equation (2.39)
Ex  n
2
x h
2
8ma2
, nx  1, 2, 3, . . .
Ey 
n2y h
2
8mb2
, ny  1, 2, 3, . . . (2:81)
Ez  n
2
z h
2
8mc2
, nz  1, 2, 3, . . .
The quantum numbers nx, ny, nz take on positive integer values independently
of each other. Combining equations (2.76) and (2.80) gives the wave functions
inside the three-dimensional box
\u142nx,n y,nz(r) 

8
v
r
sin
nxðx
a
sin
nyðy
b
sin
nzðz
c
(2:82)
where v  abc is the volume of the box. The energy levels for the particle are
obtained by substitution of equations (2.81) into (2.79)
Enx,ny,nz 
h2
8m
n2x
a2
 n
2
y
b2
 n
2
z
c2
\ufffd \ufffd
(2:83)
Degeneracy of energy levels
If the box is cubic, we have a  b  c and the energy levels become
Enx,ny,nz 
h2
8ma2
(n2x  n2y  n2z) (2:84)
The lowest or zero-point energy is E1,1,1  3h2=8ma2, which is three times the
zero-point energy for a particle in a one-dimensional box of the same length.
The second or next-highest value for the energy is obtained by setting one of
62 Schro¨dinger wave mechanics
the integers nx, ny, nz equal to 2 and the remaining ones equal to unity. Thus,
there are three ways of obtaining the value 6h2=8ma2, namely, E2,1,1, E1,2,1,
and E1,1,2. Each of these three possibilities corresponds to a different wave
function, respectively, \u1422,1,1(r), \u1421,2,1(r), and \u1421,1,2(r). An energy level that
corresponds to more than one wave function is said to be degenerate. The
second energy level in this case is threefold or triply degenerate. The zero-
point energy level is non-degenerate. The energies and degeneracies for the
first six energy levels are listed in Table 2.1.
The degeneracies of the energy levels in this example are the result of
symmetry in the lengths of the sides of the box. If, instead of the box being
cubic, the lengths of b and c in terms of a were b  a=2, c  a=3, then the
values of the energy levels and their degeneracies are different, as shown in
Table 2.2 for the lowest eight levels.
Degeneracy is discussed in more detail in Chapter 3.
Table 2.1. Energy levels for a particle in a three-
dimensional box with a  b  c
Energy Degeneracy Values of nx, ny, nz
3(h2/8ma2) 1 1,1,1
6(h2/8ma2) 3 2,1,1 1,2,1 1,1,2
9(h2/8ma2) 3 2,2,1 2,1,2 1,2,2
11(h2/8ma2) 3 3,1,1 1,3,1 1,1,3
12(h2/8ma2) 1 2,2,2
14(h2/8ma2) 6 3,2,1 3,1,2 2,3,1 2,1,3 1,3,2 1,2,3
Table 2.2. Energy levels for a particle in a three-
dimensional box with b  a=2, c  a=3
Energy Degeneracy Values of nx, ny, nz
14(h2/8ma2) 1 1,1,1
17(h2/8ma2) 1 2,1,1
22(h2/8ma2) 1 3,1,1
26(h2/8ma2) 1 1,2,1
29(h2/8ma2) 2 2,2,1 4,1,1
34(h2/8ma2) 1 3,2,1
38(h2/8ma2) 1 5,1,1
41(h2/8ma2) 2 1,1,2 4,2,1
2.8 Particle in a three-dimensional box 63
Problems
2.1 Consider a particle in a one-dimensional box of length a and in quantum state n.
What is the probability that the particle is in the left quarter of the box
(0 < x < a=4)? For which state n is the probability a maximum? What is the
probability that the particle is in the left half of the box (0 < x < a=2)?
2.2 Consider a particle of mass m in a one-dimensional potential such that
V (x)  0, ÿa=2 < x < a=2
 1, x ,ÿa=2, x . a=2
Solve the time-independent Schro¨dinger equation for this particle to obtain the
energy levels and the normalized wave functions. (Note that the boundary
conditions are different from those in Section 2.5.)
2.3 Consider a particle of mass m confined to move on a circle of radius a. Express
the Hamiltonian operator in plane polar coordinates and then determine the energy
levels and wave functions.
2.4 Consider a particle of mass m and energy E approaching from the left a potential
barrier of height V0, as shown in Figure 2.3 and discussed in Section 2.6. However,
suppose now that E is greater than V0 (E . V0). Obtain expressions for the
reflection and transmission coefficients for this case. Show that T equals unity
when E ÿ V0  n2ð2&quot;2=2ma2 for n  1, 2, . . . Show that between these periodic
maxima T has minima which lie progressively closer to unity as E increases.
2.5 Find the expression for the transmission coefficient T for Problem 2.4 when the
energy E of the particle is equal to the potential barrier height V0.
64 Schro¨dinger wave mechanics
3
General principles of quantum theory
3.1 Linear operators
The wave mechanics discussed in Chapter 2 is a linear theory. In order to
develop the theory in a more formal manner, we need to discuss the properties
of linear operators. An operator A^ is a mathematical entity that transforms a
function \u142 into another function ö
ö  A^\u142 (3:1)
Throughout this book a circumflex is used to denote operators. For example,
multiplying the function \u142(x) by the variable x to give a new function ö(x)
may be regarded as operating on the function \u142(x) with the operator x^, where x^
means multiply by x: ö(x)  x^\u142(x)  x\u142(x). Generally, when the operation is
simple multiplication, the circumflex on the operator is omitted. The operator
D^x, defined as d=dx, acting on \u142(x) gives the first derivative of \u142(x) with
respect to x, so that in this case
ö  D^x\u142  d\u142
dx
The operator A^ may involve a more complex procedure, such as taking the
integral of \u142 with respect to x either implicitly or between a pair of limits.
The operator A^ is linear if```