5.

8 EXAMPLE EXERCISES
Exercise 5.5
Find (k) for a particle with state function (x) =1,

2L for x - L and (x) =0 for

V(x). Show that Newtons definition of force acting on the particle can be expressed in
quantum mechanics by Ehrenfests theorem (J,Jt)
x
= (o,ox)V(x). Under what
circumstances does (J,Jt)
x
=(o,ox)V(x)?
Exercise 5.7
Prove that the expectation value of the (Hermitian) momentum operator in Cartesian
coordinates is real. Show that i(o,or) is not a Hermitian operator in radial coordinates.
Show that the radial momentum operator
r
=i(1,r)(o,or)r is Hermitian.
Exercise 5.8
Consider a particle of mass m in a finite, one-dimensional, rectangular potential well for
which V(x) = 0 for L - x - L and V(x) = V
0
elsewhere. The value of V
0
is a finite
positive constant. Calculate the average kinetic energy of the particle ground state, and
show that the contribution from the region outside the quantum well is negative.
Exercise 5.9
Show that a consequence of the hermiticity of the Hamiltonian is that probability
(t)(t) is conserved.
Exercise 5.10
Discuss the similarities and differences between classical electrodynamics and quantum
mechanics.
Solutions
Solution 5.1
To illustrate the L
2
nature of space consider wave functions
1
(r) and
2
(r) that
belong to . In a linear space one may form the linear combination
(r) =\
1

1
(r) +\
2

2
(r)
where \
1
and \
2
are complex numbers. To show that (r) is also square integrable, we
expand:
(r)
2
=\
1

1
(r)
2
+\
2

2
(r)
2
+\

1
\
2

1
(r)
2
(r) +\
1
\

1
(r)

2
(r)
The last two terms have the same modulus with an upper limit:
\
1
\
2
(
1
(r)
2
+
2
(r)
2
)
267
EIGENSTATES AND OPERATORS
(r)
2
is therefore smaller than a function the integral of which converges, since
1
(r)
and
2
(r) are square integrable.
Solution 5.2
Starting with the expression for the one-dimensional density of k states, we have
D
1
(k)Jk =
Jk
2t
D
1
(E)JE =D
1
(k)
Jk
JE
JE =
m

2
k
1
(2t)
JE =
m

2
(2t)
JE
(2mE)
1,2
=
1
4t
_
2m

2
_
1,2
E
1,2
JE
The density of states in two-dimensions is just
D
2
(E)JE =D
2
(k)
Jk
JE
JE =
2tk
(2t)
2
m

2
k
JE
D
2
(E)JE =
m
2t
2
JE
where we have used the fact that wave number k =
_
2mE,
2
, energy E =
2
k
2
,2m, and
the energy increment JE =
2
kJk,m.
Solution 5.3
An electron is in an infinite, one-dimensional, rectangular potential well of width L. The
electron is in the simple superposition state consisting of the ground and first excited state
so that
(x, t) =
1

2
(
1
(x, t) +
2
(x, t))
(a) To find the probability density (x, t)
2
, we must first find expressions for the
wave functions
1
(x, t) and
2
(x, t). The first two lowest-energy wave functions for a
particle of mass m confined to an infinite potential well of width L centered at x =0 are

1
(x, t) =
_
2
L
_
1,2
cos
_
tx
L
_
e
iw
1
t
and

2
(x, t) =
_
2
L
_
1,2
sin
_
2tx
L
_
e
iw
2
t
268
5.8 EXAMPLE EXERCISES
where E
n
=w
n
=

2
2m
n
2
t
2
L
2
and n is a positive nonzero integer. The expression for proba-
bility density

2
=
1
2
(
1

1
+
1

2
+
2

1
+
2

2
)

2
=
1
2
(
1

2
+
2

2
+
1

2
e
i(w
1
w
2
)t
+

2
e
i(w
1
w
2
)t
)
(x, t)
2
=
1
2
(
1
(x)
2
+
2
(x)
2
+2Re(
1
(x)

2
(x)) cos((w
1
w
2
)t))
shows an oscillatory solution in which the average position of the particle moves from
one side of the well to the other. The sinusoidal oscillation frequency is (w
2
w
2
) =
3t
2
,2mL
2
.
(b) The average position of the particle is
x(t) =
L,2
_
L,2

xJx =
1
2
L,2
_
L,2
(
1

2
+
2

2
+2Re(
1

2
) cos((w
1
w
2
)t))xJx
x(t) =
1
2
L,2
_
L,2
(2Re(
1

2
) cos((w
1
w
2
)t))xJx
x(t) =(cos((w
1
w
2
)t))
_
2
L
_
L,2
_
L,2
xcos
_
tx
L
_
sin
_
2tx
L
_
Jx
x(t) =(cos((w
1
w
2
)t))
_
1
L
__
2
2
9
_
L
2
t
2
x(t) =(16L,9t
2
) cos((w
2
w
1
)t)
(c) The momentum probability density (
x
, t)
2
is found using the Fourier transform
(
x
, t) =
1

2t
_
(x, t)e
i
x
x
Jx =
1

4t
L,2
_
L,2
(
1
(x, t)e
i
x
x
+
2
(x, t)e
i
x
x
)Jx
(
x
, t) =
_
2
4tL
_
_
e
iw
1
t
L,2
_
L,2
cos
_
tx
L
_
e
i
x
x
Jx+e
iw
4
t
L,2
_
L,2
sin
_
2tx
L
_
e
i
x
J
Jx
_
_
(
x
, t)
2
=
1
2tL
_
_
e
iw
1
t
L,2
_
L,2
cos
_
tx
L
_
e
i
x
x
Jx+e
iw
4
t
L,2
_
L,2
sin
_
2tx
L
_
e
i
x
x
Jx
_
_

_
_
e
iw
1
t
L,2
_
L,2
cos
_
tx
L
_
e
i
x
x
Jx+e
iw
4
t
L,2
_
L,2
sin
_
2tx
L
_
e
i
x
x
Jx
_
_
269
EIGENSTATES AND OPERATORS
(
x
, t)
2
=
1
2tL
_
_
_

L,2
_
L,2
cos
_
tx
L
_
e
i
x
x
Jx

2
+

L,2
_
L,2
sin
_
2tx
L
_
e
i
x
x
Jx

2
+ 2Re
_
_
e
i(w
4
w
1
)t
L,2
_
L,2
cos
_
tx
L
_
e
i
x
x
sin
_
2tx
L
_
e
i
x
x
Jx
_
_
_
_
We make use of the standard indefinite integrals
8
_
e
ox
cos(lx)Jx =
e
ox
(ocos(lx) +l sin(lx))
o
2
+l
2
_
e
ox
sin(lx)Jx =
e
ox
(osin(lx) l cos(lx))
o
2
+l
2
(
x
, t)
2
=
1
2tL
_

Lt(e
i
x
L,2
+e
i
x
L,2
)
t
2

2
x
L
2

2
+

2Lt(e
i
x
L,2
e
i
x
L,2
)
4t
2

2
x
L
2

2
+2Re
_
e
i(w
4
w
1
)t
Lt(e
i
x
L,2
+e
i
x
L,2
)
t
2

2
x
L
2
2Lt(e
i
x
L,2
e
i
x
L,2
)
4t
2

2
x
L
2
__
(
x
, t)
2
=Acos
2
_

x
L
2
_
+Bsin
2
_

x
L
2
_
+Csin(
x
L) sin((w
4
w
1
)t)
A =
2Lt
(t
2
L
2

2
x
)
2
, B =
2Lt
((4t)
2
L
2

2
x
)
2
, C =
4Lt
(t
2
L
2

2
x
)((2t)
2
L
2

2
x
)
.
(d) The average momentum of the particle is
x
(t) =(8,(3L)) sin((w
2
w
1
)t)
(e) The current flux is
1
x
(x, t) =
2et
mL
2
_
cos
_
tx
L
_
cos
_
2tx
L
_
+
1
2
sin
_
tx
L
_
sin
_
2tx
L
__
sin((w
2
w
1
)t)
Solution 5.4
The values of x and x
2
for a particle confined by the potential V(x) =0 for 0 -x -L
and V(x) = elsewhere are
x =
L
2
and
x
2
=
L
2
3

L
2
2n
2
t
2
8. Gradshteyn and Ryzhik, Table of Integrals, Series, and Products, p. 196.
270
5.8 EXAMPLE EXERCISES
We can now compare these results with those for a classical particle. The classical
particle moves at constant velocity u and traverses the well in time t = L,u. Hence,
classically,
x
classical
=
t=t
_
t=0
ut
t
Jt =
1
2
ut =
L
2
and
x
2

classical
=
t=t
_
t=0
(ut)
2
t
Jt =
1
3
u
2
t
2
=
L
2
3
Thus, as n the quantum results approach the classical solution.
The average values
x
and
2
x
are

x
=0

2
x
=2mE
n
and since
E
n
=

2
k
2
n
2m
=

2
n
2
t
2
2mL
2
we have

2
x
=

2
n
2
t
2
L
2
Hence,
Ax
2
=
L
2
12
_
1
6
n
2
t
2
_
A
2
x
=

2
t
2
n
2
L
2
AxA
x
=

12
(n
2
t
2
6)
1,2
Solution 5.5
To find (k) for a particle the state function of which is (x) =1,

2L for x - L and
(x) =0 for x > L, we take the Fourier transform:
(k) =
1

2t
L
_
L
1

2L
e
ikx
Jx =
_
1

2tL
1
ik
e
ikx
_
L
L
=
1

tL
1
k
sin(kL)
(k)
2
=
sin
2
(kL)
tLk
2
271
EIGENSTATES AND OPERATORS
The wave functions (x) and (k) are plotted in the following figures.
0.3
0.4
0.5
0.6
0.2
0.1
0.0
1 0 1
0.7
0.8
2 2
0.3
0.4
0.5
0.6
0.2
0.1
0.0
1 0 1
Position, x/L Position, x/L
0.7
0.8
2 2
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.1
0.2
0.3
10 0 10
Wave vector, kL/2 Wave vector, kL/2
W
a
v
e

f
u
n
c
t
i
o
n
,

(
k
)
W
a
v
e

f
u
n
c
t
i
o
n
,

(
x
)
0.15
0.20
0.25
0.30
0.35
0.10
0.05
0.00
10
0
10
W
a
v
e

f
u
n
c
t
i
o
n

s
q
u
a
r
e
d
,

|

(
k
)
|
2
W
a
v
e

f
u
n
c
t
i
o
n

s
q
u
a
r
e
d
,

|

(
x
)
|
2
To show that the uncertainty (standard deviation) in particle momentum A is infinite
it is necessary to calculate
x
and
2
x
:

x
=

(k)k(k)Jk =

Lt

1
k
sin
2
(kL)Jk =0
by symmetry, and

2
x
=

(k)k(k)Jk =

2
Lt

sin
2
(kL)Jk =

2
Lt

_
1
2

1
2
cos(2kL)
_
Jk

2
x
=

2
Lt

_
1
2

1
2
cos(2kL)
_
Jk =

2
Lt
_
k
2

1
4L
sin(2kL)
_

272
5.8 EXAMPLE EXERCISES

2
x
=

2
Lt
_

2
0+0
_
=
A
2
x
=
2
x

x

2
=
One understands this as (k)
2
not decreasing to zero fast enough in the limit k .
The average value of position is
x =
L
_
L
(x)x(x)Jx =
_
1

2L
_
2
1
2
|x
2
]
L
L
=0
and the value of x
2
is
x
2
=
_
(x)x
2
(x)Jx =
1
2L
1
3
|x
3
]
L
L
=
1
3L
L
3
=
L
2
3
giving a measure of the spread in measured values of
Ax
2
=x
2
x
2
=
L
2
x
0
It follows that Ax =L,

3, that is finite nonzero, so that

o
ox
o
ox
o
ox
_
=0
(b) |

V(x),
x
] =i
_
V(x)
o
ox

o
ox
V(x) V(x)
o
ox
_
=i
o
ox
V(x)
(c) Making use of the results in part (a) and (b) we have
J
Jt

x
=
i

H,
x
] =
i

_

2
x
2m
+

V(x),
x
_
=
i

V(x),
x
] =
o
ox
V(x)
which we can write as
J
Jt

x
=m
J
2
Jt
2
x =
o
ox
V(x)
If the uncertainty Ax = ( x x)
2

1,2
is small so that the classical value of x = x
and if the potential V(x) is slowly varying such that
J
Jx
V(x) =
J
Jx
V(x)
then one obtains Newtons classical result
J
x
Jt
=m
J
2
x
Jt
2
=
J
Jx
V(x)
273
EIGENSTATES AND OPERATORS
Solution 5.7
The momentum operator is Hermitian that is, it must satisfy A
i]
=A

]i
in matrix elements
or equivalently (
_

J
3
r)

=
_

J
3
r =(
_

J
3
r)

. For simplicity, consider

x
Jx =
_

Jx or
x
=
x

The operator
x
can be seen to be Hermitian if we integrate by parts,
_
UV

Jx = UV
_
U

V Jx, so
_

x
Jx =i
_

o
ox
Jx =|i

+i

ox
Jx
the term |i

=0 and i

ox
Jx =
_

Jx
We have thus shown that = , provided that the wave function 0
at x . To show that is real is now trivial. If = , then =
_

Jr =
_

Jr =

, which can only be true if is real. Hence, the expectation value

m
_
i
o
or
_

n
Jr =
_
4tr
2

m
_
i
o
or

n
__
r=
r=0

r=
_
r=0
i
n
o
or
(4tr
2

m
)Jr
The first term on the right-hand side is zero, assuming that the eigenfunction vanishes at
r =, so that
r=
_
r=0
4tr
2

m
_
i
o
or
_

n
Jr =
_
i4t
n
_
r
2
o
or

m
2r

m
_
Jr
=
_
4tr
2

n
_
i
_
2r
o
or
_

m
_

Jr
Obviously, this does not satisfy A

i]
=A
]i
for a Hermitian operator.
Solution 5.8
A particle mass m is in a finite, one-dimensional rectangular potential well for which
V(x) = 0 for L - x - L and V(x) = V
0
elsewhere, and for which the value of V
0
is a
finite positive constant. The expectation value of kinetic energy is
T =

(x)

2
2m
(x)Jx =

2
2m

(x)
J
Jx
2
(x)Jx
For even-parity bound-state solutions, including the ground state, the spatial wave
functions in the well are of the form

n
(x) =A
n
cos(k
n
x)
274
5.8 EXAMPLE EXERCISES
and in the barrier they are of the form

n
(x) =C
n
e

n
x
where the index n is an odd positive integer that labels the bound-state eigenvalue. For
eigenenergy E
n
,
k
n
=
_
2mE
n
,
and

n
=
_
2m(V
0
E),
For the ground state n =1, and we have an expectation value for kinetic energy that is
T =

2
k
2
A
2
m
x=L
_
x=0
cos
2
(k
1
x

)Jx

2
k
2
C
2
m
x=
_
x=L
e
2
1
x

Jx

This shows that the contribution to kinetic energy from the barrier region is negative.
Solution 5.9
To show that (t)(t) is constant we evaluate
J
Jt
(t)(t) =
_
J
Jt
(t)
_
(t) +(t)
_
J
Jt
(t)
_
From the Schrdinger equation
o
ot
(t) =
i

H(t)
and
o
ot
(t) =
i

(t)

=
i

(t)

H
from the hermiticity of the Hamiltonian. Hence,
J
Jt
(t)(t) =
_
i

_
(t)

H(t) =0
and we may conclude that the probability density (t)(t) does not evolve in time.
Solution 5.10
In Chapter 1 we showed that for a source-free, linear, frequency-independent, loss-
less dielectric with :(r) = :
0
(r):
r
(r), and with relative magnetic permeability j
r
= 1,
Maxwells equations lead directly to a wave equation for electromagnetic fields. The
linearity allows us to separate out the time and space dependence into a set of harmonic
solutions of the form E(r, t) =E
0
e
ikr
e
iwt
and H(r, t) =H
0
e
ikr
e
iwt
, where we have used
complex numbers for mathematical convenience (always remembering to take the real part
275
Applied quantum mechanics 1
Applied Quantum Mechanics
Chapter 3 Problems and Solutions
LAST NAME FIRST NAME
Useful constants MKS (SI)
Speed of light in free space
Plancks constant
Electron charge
Electron mass
Neutron mass
Proton mass
Boltzmann constant
Permittivity of free space
Permeability of free space
Speed of light in free space
Avagadros number
Bohr radius
Inverse fine-structure constant
c 2.99792458 10
8
m s
1
=
6.58211889 26 ( ) 10
16
eV s =
1.054571596 82 ( ) 10
34
J s =
e 1.602176462 63 ( ) 10
19
C =
m
0
9.10938188 72 ( ) 10
31
kg =
m
n
1.67492716 13 ( ) 10
27
kg =
m
p
1.67262158 13 ( ) 10
27
kg =
k
B
1.3806503 24 ( ) 10
23
J K
1
=
k
B
8.617342 15 ( ) 10
5
eV K
1
=

0
8.8541878 10
12
F m
1
=

0
4 10
7
H m
1
=
c 1
0

0
=
N
A
6.02214199 79 ( ) 10
23
mol
1
=
a
B
0.52917721 19 ( )
10
10 m =
a
B
4
0

2
m
0
e
2
---------------- - =

1
137.0359976 50 ( ) =

1
4
0
c
e
2
----------------- - =
2
PROBLEM 1
Prove that particle flux (current) is zero if the one-dimensional exponential decaying wave func-
tion in tunnel barrier of energy V
0
and finite thickness L is , where is a real
positive number and particle energy .
PROBLEM 2
(a) Use a Taylor expansion to show that the second derivative of a wavefunction sampled
at positions , where j is an integer and h
0
is a small fixed increment in distance x, may be
approximated as
(b) By keeping additional terms in the expansion, show that a more accurate approximation of
the second derivative is
PROBLEM 3
Using the method outlined in Exercise 7 of Chapter 3 as a starting point, calculate numerically
the first four energy eigenvalues and eigenfunctions for an electron with effective mass
confined to a potential well V(x) = V
0
of width L = 10 nm with periodic boundary
conditions.
Periodic boundary conditions require that the wave function at position x = 0 is connected
(wrapped around) to position x = L. The wave function and its first derivative are continuous and
smooth at this connection.
Your solution should include plots of the eigenfunctions and a listing of the computer program
you used to calculate the eigenfunctions and eigenvalues.
PROBLEM 4
Using the method outlined in Exercise 7 of Chapter 3 as a starting point, calculate numerically
the first six energy eigenvalues and eigenfunctions for an electron with effective mass
confined to a triangular potential well of width L = 20 nm bounded by barriers of
infinite energy at x < 0 and x > L. The triangular potential well as a function of distance x is given
by V(x) = V
0
x / L where V
0
= 1 eV.
Explain the change in shape of the wave function with increasing eigenenergy.
Your solution should include plots of the eigenfunctions and a listing of the computer program
you used to calculate the eigenfunctions and eigenvalues.
PROBLEM 5
Calculate the transmission and reflection coefficient for an electron of energy E, moving from
left to right, impinging normal to the plane of a semiconductor heterojunction potential barrier of
energy , where the effective electron mass on the left-hand side is and the effective electron
mass on the right-hand side is .
If the potential barrier energy is V
0
= 1.5 eV and the ratio of effective electron mass on either
side of the heterointerface is m
1
/ m
2
= 3, at what particle energy is the transmission coefficient
unity? What is the transmission coefficient in the limit E ?
x t , ( ) Be
x i t
=
E V
0
< =
x ( )
x
j
jh
0
=
x
2
2
d
d
x
j
( )
x
j 1
( ) 2 x
j
( ) x
j 1 +
( ) +
h
0
2
------------------------------------------------------------------ =
x
2
2
d
d
x
j
( )
x
j 2
( ) 16 + x
j 1
( ) 30 x
j
( ) 16 x
j 1 +
( ) x
j 2 +
( ) +
12h
0
2
--------------------------------------------------------------------------------------------------------------------------------------- =
m
e
*
0.07 m
0
=
m
e
*
0.07 m
0
=
V
0
m
1
m
2
Applied quantum mechanics 3
SOLUTIONS
Solution 1
To prove that particle flux (current) is zero if the one-dimensional exponential decaying wave
function in tunnel barrier of energy V
0
and finite thickness L is , where is a
real positive number and particle energy , we substitute the wave function into the
current operator
Solution 2
(a) Consider the Taylor expansion for the function f(x).
where j is an integer and h
0
is a small fixed increment in distance x. Keeping terms to order , we
see that the first derivative is
Hence, the first derivative of a wavefunction sampled at positions may be approxi-
mated as
To find the second derivative we keep terms to order so that
and
Adding these two equations gives
or
so that the second derivative of a wavefunction sampled at positions may be
approximated as
This is the three-point approximation to the second derivative accurate to second order, .
(b) By keeping five terms in the expansion instead of three one may obtain a more accurate
approximation of the second derivative accurate to fourth order, . To see how, we use the fol-
lowing Taylor expansions with derivatives up to fourth-order.
x t , ( ) Be
x it
=
E V
0
< =
J
i eh
2m
-----------
*
x t , ( )
x d
d
x t , ( ) x t , ( )
x d
d

*
x t , ( )


=
J
i eh
2m
----------- B
*
e
x i t +
B e
x i t
( ) Be
x i t
B
*
e
x i t +
( ) ( ) =
J
i eh
2m
----------- B
2
e
2x
e
2x
( ) 0 = =
f x
j 1 +
( ) f x
j
( ) f x
j
( ) h
0
f x
j
( )
2!
-------------- h
0
2
f x
j
( )
3!
---------------- h
3
0 + + + +
1
n!
----- f
n ( )
x
j
( ) h
0
n

n

= =
h
0
f x
j
( )
f x
j 1 +
( ) f x
j
( )
h
0
-------------------------------- - =
x ( ) x
j
j h
0
=
x d
d
x
j
( )
x
j 1 +
( ) x
j
( )
h
0
------------------------------------- - =
h
0
2
f x
j 1 +
( ) f x
j
( ) f x
j
( ) h
0
f x
j
( )
2
-------------- h
0
2
+ + =
f x
j 1
( ) f x
j
( ) f x
j
( ) h
0
f x
j
( )
2
-------------- h
0
2
+ =
f x
j 1
( ) f x
j 1 +
( ) + 2f x
j
( ) f x
j
( ) h
0
2
+ =
f x
j
( )
f x
j 1
( ) 2f x
j
( ) f x
j 1 +
( ) +
h
0
2
---------------------------------------------------------- - =
x ( ) x
j
j h
0
=
x
2
2
d
d
x
j
( )
x
j 1
( ) 2 x
j
( ) x
j 1 +
( ) +
h
0
2
------------------------------------------------------------------ =
0 h
0
2
( )
0 h
0
4
( )
4
To eliminate the terms in the first, third, and fourth derivative we multiply each equation by coeffi-
cients a, b, c, d, respectively, to obtain
The coefficients that eliminate terms in the first, third, and fourth derivative must satisfy
The solution to this set of linear equations is
Setting we obtain
so that
so that the second derivative of a wavefunction sampled at positions may be
approximated as
Solution 3
To find numerically the first four energy eigenvalues and eigenfunctions for an electron with
effective mass confined to a potential well V(x) = V
0
of width L = 10 nm with
periodic boundary conditions we descretize the wavefunction and potential appearing in the time-
independent Schrdinger equation
using a discrete set of equally-spaced points such that position where the index
, and is the interval between adjacent sampling points so one may define
. The region in which we wish to solve the Schrdinger equation is of length
. At each sampling point the wave function has value and the potential is
f x
j 2 +
( ) f x
j
( ) 2f x
j
( ) h
0
2f x
j
( ) h
0
2 4
3
--- f x
j
( ) h
0
3 2
3
--- f x
j
( ) h
0
4
+ + + + =
f x
j 1 +
( ) f x
j
( ) f x
j
( ) h
0
1
2
--- f x
j
( )h
0
2 1
6
--- f x
j
( )h
0
3 1
24
------ f x
j
( )h
0
4
+ + + + =
f x
j 1
( ) f x
j
( ) f x
j
( ) h
0
1
2
--- f x
j
( ) h
0
2 1
6
--- f x
j
( )h
0
3

1
24
------ f x
j
( ) h
0
4
+ + =
f x
j 2
( ) f x
j
( ) 2f x
j
( )h
0
2f x
j
( )h
0
2
+
4
3
--- f x
j
( ) h
0
3 2
3
--- f x
j
( ) h
0
4
+ =
af x
j 2 +
( ) bf x
j 1 +
( ) cf x
j 1
( ) df x
j 2
( ) + + + a b c d + + + ( ) f x
j
( ) 2a b c 2d + ( ) f x
j
( ) h
0
2a
b
2
---
c
2
--- 2d + + +


f x
j
( )h
0
2 4a
3
------
b
6
---
c
6
---
4d
3
------ +


f x
j
( ) h
0
3 2a
3
------
b
24
------
c
24
------
2d
3
------ + + +


f x
j
( )h
0
4
+ +
+ +
=
0 2a b c 2d + ( ) =
0
4a
3
------
b
6
---
c
6
---
4d
3
------ +


=
0
2a
3
------
b
24
------
c
24
------
2d
3
------ + + +


=
a d =
b 16d =
c 16d =
a 1 =
f x
j 2 +
( ) 16f x
j 1 +
( ) 16f x
j 1
( ) f x
j 2
( ) + 30 f x
j
( ) 12f x
j
( )h
0
2
=
f x
j
( )
f x
j 2 +
( ) 16f x
j 1 +
( ) 30f x
j
( ) 16f x
j 1
( ) f x
j 2
( ) + +
12h
0
2
--------------------------------------------------------------------------------------------------------------------------- =
x ( ) x
j
jh
0
=
x
2
2
d
d
x
j
( )
x
j 2
( ) 16 + x
j 1
( ) 30 x
j
( ) 16 x
j 1 +
( ) x
j 2 +
( ) +
12h
0
2
--------------------------------------------------------------------------------------------------------------------------------------- =
m
e
*
0.07 m
0
=
H
n
x ( )

2
2m
-------
x
2
2
d
d
V x ( ) +

n
x ( ) E
n

n
x ( ) = =
N 1 + x
j
j h
0
=
j 0 1 2N , , = h
0
x
j 1 +
x
j
h
0
+
L Nh
0
=
j
x
j
( ) =
Applied quantum mechanics 1
Applied Quantum Mechanics
Chapter 5 Problems and Solutions
LAST NAME FIRST NAME
Useful constants MKS (SI)
Speed of light in free space
Plancks constant
Electron charge
Electron mass
Neutron mass
Proton mass
Boltzmann constant
Permittivity of free space
Permeability of free space
Speed of light in free space
Avagadros number
Bohr radius
Inverse fine-structure constant
c 2.99792458 10
8
m s
1
=
6.58211889 26 ( ) 10
16
eV s =
1.054571596 82 ( ) 10
34
J s =
e 1.602176462 63 ( ) 10
19
C =
m
0
9.10938188 72 ( ) 10
31
kg =
m
n
1.67492716 13 ( ) 10
27
kg =
m
p
1.67262158 13 ( ) 10
27
kg =
k
B
1.3806503 24 ( ) 10
23
J K
1
=
k
B
8.617342 15 ( ) 10
5
eV K
1
=

0
8.8541878 10
12
F m
1
=

0
4 10
7
H m
1
=
c 1
0

0
=
N
A
6.02214199 79 ( ) 10
23
mol
1
=
a
B
0.52917721 19 ( )
10
10 m =
a
B
4
0

2
m
0
e
2
---------------- - =

1
137.0359976 50 ( ) =

1
4
0
c
e
2
----------------- - =
2
PROBLEM 1
An electron in an infinite, one-dimensional, rectangular potential well of width L is in
the simple superposition state consisting of the ground and third excited state so that
Find expressions for:
(a) The probability density, .
(b) The average particle position, .
(c) The momentum probability density,
(d) The average momentum, .
(e) The current flux, .
PROBLEM 2
(a) Show that the density of states for a free-particle of mass m in two-dimensions is
(b) At low temperature, electrons in two electrodes occupy states up to the Fermi
energy, E
F
. The two electrodes are connected by a two dimensional conductance
region. Derive an expression for the conductance of electrons flowing between the two
electrodes as a function of applied voltage V, assuming the transmission coefficient
through the two-dimensional region is unity. Consider the two limiting cases eV >> E
F
and eV << E
F
PROBLEM 3
Derive expressions for the two-dimensional and one-dimensional
density of photon states in a homogeneous dielectric medium characterized by refrac-
tive index, n
r
.
PROBLEM 4
(a) In a particular system the dispersion relation for electrons in one-dimension is
, where t and L are constants and the wave vector in the x-
direction is . This dispersion relation can be derived using a nearest
neighbor tight binding model where t is the overlap integral between atomic orbitals.
Choosing one hundred equally spaced discrete values of k
x
, write a computer program
and plot the electron density of states using
and .
(b) If one includes next nearest neighbor interactions, the dispersion relation in (a) can,
to within a scaling factor, be written . Write a com-
puter program to plot the dispersion relation. Then calculate and plot the electron den-
sity of states using , , and compare with the result you
obtained in (a) including a comparison with the effective electron mass at the band
edges.
x t , ( )
1
2
-------
1
x t , ( )
4
x t , ( ) + ( ) =
x t , ( )
2
x t ( )
p
x
t , ( )
2
p
x
t ( )
J x t , ( )
D
2
E ( )
m
2
2
------------ =
D
2
opt
( ) D
1
opt
( )
E
k
h
k
2t k
x
L ( ) cos = =
0 k
x
L < <
N E ( )

E E
k
( )
2
2 ( )
2
+
---------------------------------------------
k

= t 10 =
t 1 =
E
k
2t k
x
L ( ) 2t 2k
x
L ( ) cos + cos =
t 10 = t 1 = t 0.2 =
Applied quantum mechanics 5
SOLUTIONS
Solution 1
(a) Let the potential in the region be zero and infinity elsewhere. The
eigenfunctions are , where ,
, and . Hence,
and the probability distribution of the particle in the superposition state
is
where we note that .
(b) Expectation value of position is
By symmetry, the first two terms in the integrand are zero, leaving
making use of gives
which, after integrating by parts, results in
(c) The momentum probability density is found using the Fourier transform
L 2 x L 2 < <

n
x t , ( ) 2 L k
n
x L 2 + ( ) ( )e
i
n
t
sin = k
n
n L =

n
k
n
2
2m = n 1 2 3 , , , =

1
x t , ( ) 2 L L ( ) x L 2 + ( ) ( ) e
i
1
t
sin 2 L x L ( ) e
i
1
t
( ) cos = =

4
x t , ( ) 2 L 4 L ( ) x L 2 + ( ) ( ) e
i
4
t
sin 2 L 4x L ( ) e
i
4
t
sin = =
x t , ( ) 1 2
1
x t , ( )
4
x t , ( ) + ( ) =
x t , ( )
2 1
2
---
1
x ( )
2

4
x ( )
2
2
1
x ( )
4
x ( )
4

1
( )t ( ) cos + + ( ) =
x t , ( )
2 1
L
--- cos
2
x
L
------


sin
2
4x
L
---------


2
x
L
------


4x
L
---------



2m
-------
15
2
L
2
-----------t


cos sin cos + +


=

4

1

2m
-------
15
2
L
2
----------- =
x t ( ) x t , ( )
2
x x d

=
x t ( )
1
2
---
1
x ( )
2
x
4
x ( )
2
x 2x
1
x ( )
4
x ( )
4

1
( ) t ( ) cos + + ( )
L 2
L 2

x d =
x t ( )
4

1
( )t ( ) cos
2
L
---
x
L
----- -


4x
L
---------


sin cos
L 2
L 2

x x d =
2 x ( ) y ( ) cos sin x y + ( ) x y ( ) sin + sin =
x t ( )
4

1
( ) t ( ) cos
1
L
---
5x
L
---------


sin
3x
L
---------


sin +


L 2
L 2

x x d =
x t ( )
4

1
( )t ( ) cos
1
L
---
2L
2
25
2
-----------
2L
2
9
2
--------


x x d =
x t ( )
32L
225
2
--------------
4

1
( ) t ( ) cos =
p
x
t , ( )
2
p
x
t , ( )
1
2
-------------- x t , ( )e
ip
x
x
x d

1
4
--------------
1
x t , ( )e
ip
x
x

4
x t , ( ) e
i p
x
x
+ ( ) x d
L 2
L 2

= =
p
x
t , ( )
2
4L
------------- e
i
1
t
x
L
------


cos e
i p
x
x
x e
i
4
t
4x
L
---------


sin e
i p
x
x
x d
L 2
L 2

+ d
L 2
L 2

=
6
We make use of the standard indefinite integrals (I. S. Gradshteyn and I. M. Ryzhik,
Table of integrals, series, and products, Academic Press, San Diego, 1980, p. 196
(ISBN 0 12 294760 6))
, , .
(d) The expectation value of momentum is
The terms and by symmetry, leaving
and
p
x
t , ( )
2 1
2L
------------- e
i
1
t
x
L
------


cos e
ip
x
x
x e
i
4
t
4x
L
---------


sin e
ip
x
x
x d
L 2
L 2

+ d
L 2
L 2

e
i
1
t
x
L
------


cos e
ip
x
x
x e
i
4
t
4x
L
---------


sin e
i p
x
x
x d
L 2
L 2

+ d
L 2
L 2

=
p
x
t , ( )
2 1
2L
-------------
x
L
------


cos e
i p
x
x
x d
L 2
L 2

2
4x
L
---------


sin e
i p
x
x
x d
L 2
L 2

2
2Re e
i
4

1
( )t
x
L
------


cos e
ip
x
x
4x
L
---------


sin e
i p
x
x
x d
L 2
L 2

+ + =
e
ax
bx ( ) cos x d

e
ax
a bx ( ) cos b bx ( ) sin + ( )
a
2
b
2
+
--------------------------------------------------------------- =
e
ax
bx ( ) sin x d

e
ax
a bx ( ) b bx ( ) cos sin ( )
a
2
b
2
+
--------------------------------------------------------------- =
p
x
t , ( )
2 1
2L
-------------
L e
ip
x
L 2
e
i p
x
L 2
+ ( )

2
p
x
2
L
2

------------------------------------------------
2
4L e
ip
x
L 2
e
i p
x
L 2
( )
16
2
p
x
2
L
2

---------------------------------------------------
2
2Re e
i
4

1
( )t
L e
ip
x
L 2
e
i p
x
L 2
+ ( )

2
p
x
2
L
2

------------------------------------------------
4L e
i p
x
L 2
e
i p
x
L 2
( )
16
2
p
x
2
L
2

---------------------------------------------------



+ + =
p
x
t , ( )
2
Acos
2
p
x
L
2
--------


Bsin
2
p
x
L
2
--------


C p
x
L ( )
4

1
( ) t ( ) sin sin + + =
A
2L

2
L
2
p
x
2
( )
2
-------------------------------- = B
32L
4 ( )
2
L
2
p
x
2
( )
2
---------------------------------------- = C
8 L

2
L
2
p
x
2
( ) 4 ( )
2
L
2
p
x
2
( )
----------------------------------------------------------------- =
p
x
t ( ) i
*
x t , ( )
x

x t , ( ) x d

i
2
--------
1
*

4
*
+ ( )
x

1

4
+ ( )


x d
L 2
L 2

= =

4
*
x

4
x d
L 2
L 2

0 =
1
*
x

1
x d
L 2
L 2

0 =
p
x
t ( )
i
2
--------
2
L
---
4
L
------ e
i
4

1
( )t
x
L
------


4x
L
---------


cos cos x
i
2
-----
2
L
---

L
--- e
i
4

1
( )t
4x
L
---------


sin
x
L
------


sin x d
L 2
L 2

+ d
L 2
L 2

=
p
x
t ( )
i
2
--------
8
L
2
------ e
i
4

1
( )t
2L
15
---------
i
2
-----
2
L
2
------ e
i
4

1
( )t
8L
15
--------- + =
p
x
t ( )
i 8
15L
----------- e
i
4

1
( )t
e
i
4

1
( ) t
+ ( ) =
Applied quantum mechanics 7
(e) The current operator is
Solution 2
(a) Density of states of particle mass m in two dimensions is
per spin.
(b) Current is assumed proportional to applied voltage, electron velocity , trans-
mission coefficient , and density of states.
If we assume , then
When , conductance per electron spin
For the other situation when we use the binomial expansion
so that . Conductance per electron
spin
Solution 3
The two dimensional and one dimensional photon density of states in an isotropic
medium of refractive index is
p
x
t ( )
16
15L
---------
4

1
( )t ( ) sin =
J x t , ( )
ie
2m
-----------
*
x

=
J x t , ( )
ie
4m
-----------
1
*

4
*
+ ( )
x

1

4
+ ( )
1

4
+ ( )
x

1
*

4
*
+ ( )


=
J x t , ( )
i e
2Lm
-----------
x
L
------


x
4x
L
---------


e
i
1

4
( )t
e
i
1

4
( )t
( ) sin cos
4x
L
---------


sin
x
x
L
------


cos e
i
1

4
( )t
e
i
1

4
( )t
( )
+

=
J x t , ( )
e
Lm
---------
x
L
------


x
4x
L
---------


sin cos
4x
L
---------


sin
x
x
L
------


cos



4

1
( ) t ( ) sin =
J x t , ( )
e
mL
2
------------ 4
x
L
---


4
x
L
---


4
x
L
---



x
L
---


sin sin cos cos



4

1
( )t ( ) sin =
D
2
E ( )
m
2
2
----------- - =
v E ( )
T E ( )
I e v E ( )T E ( )D
2
E ( ) E d
E
F
E
F
e V +

e

m
----
2mE

2
-----------


T E ( )
m
2
2
----------- -


E d
E
F
E
F
eV +

= =
T E ( ) T E
F
( ) 1 = =
I e
2m
2
2
------------ E E d
E
F
E
F
e V +

2m
2
2
------------
2
3
--- E
3 2
[ ]E
F
E
F
eV +

2m
3
2
------------ E
F
3 2
1
eV
E
F
------ +


3 2
1


= = =
E
F
eV
G
I
V
---
2m
3
2
----------- -


V = =
E
F
eV
1 x + ( )
n
1 nx + + = 1
eV
E
F
------ +


3 2
1
3
2
---
eV
E
F
------ + =
G
I
V
---
2m
2
2
------------ E
F
= =
n
r