Chapter 3 Quantum Mechanics (PP 73-95)
Chapter 3 Quantum Mechanics (PP 73-95)
Chapter 3 Quantum Mechanics (PP 73-95)
CHAPTER 3
QUANTUM MECHANICS
3-1 EXPECTION VALUES, PROBABILITY,
NORMALIZATION AND ORTHOGONALITY
PROPERTIES
Problem 3-1
The normalized eigenfunction of a particle in a one-
dimensional box is given by
|
|
\
|
=
L
x n
L
x
n
sin
2
) ( for L x < < 0
Calculate the expectation value . > < x
Solution
Now
|
|
\
|
=
L
x n
L
x
n
sin
2
) ( and
|
|
\
|
=
L
x n
L
x
n
sin
2
) (
The expectation value < x> is defined as
>= <
L
n n
dx x x x x
0
) ( ) (
(
=
|
|
\
|
>= <
L L
dx
L x n
x
L
dx
L
x n
x
L
x
0 0
2
2
) / 2 cos( 1 2
sin
2
dx
L
x n
x
L
xdx
L
x
L
o
L
|
|
\
|
>= <
0
2
cos
1 1
( )
( ) ( )
( ) ( ) { }
( ) ( )
L L
L n
L x n
L n
L x n x
L
x
L
0
2
0
2
/ 2
/ 2 cos ) 1 (
/ 2
/ 2 sin 1
2
1
(
( )
( ) ( ) ( ) ( )
(
(
)
`
+
)
`
+
(
=
2 2
2
/ 2
) 0 cos(
/ 2
) 0 sin( ) 0 (
/ 2
) 2 cos(
/ 2
2 sin 1
0
2
1
L n L n L n
n
L n
n L
L
L
L
CHAPTER 03 QUANTUM MECHANICS 74
2 4
0
4
0
1
2
2 2
2
2 2
2
L
n
L
n
L
L
L
=
(
)
`
+
)
`
=
Problem 3-2
The normalized eigenfunction of a particle in a one-
dimensional box is given by
|
\
|
=
L
x n
L
x
n
sin
2
) ( for L x < < 0
Calculate the expectation value . > < p B.U. B.Sc. 2007A
Solution
Now
|
\
|
=
L
x n
L
x
n
sin
2
) (
|
\
|
=
L
x n
L
x
n
sin
2
) (
)
`
\
|
|
|
\
|
=
L
n
L
x n
L dt
d
n
cos
2
The expectation value <p> is defined as
>= <
L
n n
dx x p x p
0
) ( ) (
=
|
\
|
>= <
L
n
n n n
dx
dx
d
i dx x
dx
d
i x p
0
0
) ( ) (
h h
dx
d
i p h Q
dx
L
x n
L
x n
L
n
L
i p
L
|
|
\
|
|
|
\
|
|
\
|
|
|
\
|
>= <
cos sin
2
0
2
h
L
L n
L x n
L
n i
p
0
2
2
) / )( 2 (
) / ( sin 2
(
>= <
h
CHAPTER 03 QUANTUM MECHANICS 75
{ } 0 ) 0 ( sin ) ( sin
2 2
= >= < n
L
i
p
h
Problem 3-3
The eigenfunction of a particle in a one-dimensional box is
given by
|
\
|
=
L
x n
A x
n
L
n n
dx x x
0
1 ) ( ) (
=
|
|
\
|
L
dx
L
x n
A
0
2 2
1 sin
2
0
2 2
1
) / (
sin
L n
d
A where
L
x n
=
2
0
2
1 )} 2 cos( 1 {
2
d
n
L A
1
2
) 2 sin(
2
2
0
2
=
(
n
L A
1
2
) 0 sin(
0
2
) 2 sin(
2
2
=
(
)
`
)
`
n
n
n
L A
1
2
2
=
L A
L
A
2
=
CHAPTER 03 QUANTUM MECHANICS 76
Problem 3-4
The normalized eigenfunction of a particle in n
th
state for a
one-dimensional box is given by
|
\
|
=
L
x n
L
x
n
sin
2
) ( for L x < < 0
Determine the probability of finding the particle within
dimensions x = 0 to x = L / 4.
Solution
The desired probability is given by
4 /
0
) ( ) (
L
n n
dx x x
( )
=
4 /
0
2
/ sin
2
L
dx L x n
L
=
4 /
0
)] / 2 cos( 1 [
1
L
dx L x n
L
4 /
0
) / 2 (
) / 2 sin( 1
L
L n
L x n
x
L
(
)
`
)
`
=
) / 2 (
) 0 sin(
0
) / 2 (
) 2 / sin(
4
1
L n L n
n L
L
n
n
2
) 2 / sin(
4
1
=
n
n
2
) 1 (
4
1
2 / ) 1 (
= for n = 1,3,5,.
4
1
= for n = 2,4,6,..
Problem 3-5
The normalized eigenfunctions of a particle in a one-
dimensional box are given by
|
\
|
=
L
x n
L
x
n
sin
2
) ( for L x < < 0
Show that these functions are orthogonal.
Solution
For proof of orthogonality property we solve the following
integral
CHAPTER 03 QUANTUM MECHANICS 77
L
n m
dx x x
0
) ( ) ( where 0 > m , 0 > n and n m
( ) ( )
=
L
dx L x n L x m
L
0
/ sin / sin
2
dx
L
x n m
L
x n m
L
L
)
`
)
`
=
0
) (
cos
) (
cos
1
L
L n m
L x n m
L n m
L x n m
L
0
/ ) (
} / ) sin{(
/ ) (
} / ) sin{( 1
(
+
+
0 = since 0 ) sin( = k for integral values of k.
As 0 ) ( ) (
0
=
L
n m
dx x x therefore the given eigenfunctions
are orthogonal.
Problem 3-6
An object in one-dimension is described by
3 x = for 0 < x < 1
= 0 elsewhere
(a) What is the probability of finding the object within the
interval (0 , 0.5)?
(b) What is average position of the object?
Solution
(a)
=
(
= =
5 . 0
0
5 . 0
0
3
2
5 . 0
0
125 . 0
3
3 3
x
dx x dx
(b) dx x x x dx x ) 3 )( )( 3 (
1
0
1
0
=
4
3
4
3 3
1
0
4 1
0
3
=
(
= =
x
dx x
CHAPTER 03 QUANTUM MECHANICS 78
Problem 3-7
Show that
|
|
\
|
|
|
\
|
=
2
exp ) (
2 2
4 / 1
2
0
x
x
is a normalized eigenfunction.
Solution
Here we evaluate the following integral
= dx I
o
*
0
|
|
\
|
|
|
\
|
|
|
\
|
|
|
\
|
= dx
x x
I
2
exp
2
exp
2 2
4 / 1
2 2 2
4 / 1
2
=
|
|
\
|
=
0
2 2 2 2
2 / 1
2
) exp(
2
) exp( dx x dx x I
Substitute x y = then dy dx
1
= . It may be noted that the
limits of integration remain same for new variable. Hence
\
|
|
|
\
|
=
0
2
) exp(
1 2
dy y I
) 2 / 1 (
1
) exp( 2
1
0
2
|
|
\
|
=
)
`
dy y I
=
0
2 1 2
) exp( 2 ) ( dy y y n
n
Q
( ) 1
1
=
|
|
\
|
=
I = ) 2 / 1 ( Q
Hence the given function is a normalized eigenfunction.
CHAPTER 03 QUANTUM MECHANICS 79
Problem 3-8
The eigenfunction for a certain object is defined as
) ( cos ) (
2
x A x = for
2 2
< < x
(a) Evaluate the value of normalization constant A.
(b) What is the probability of finding the object between
0 = x and
4
= x ?
Solution
(a)According to normalization property
=
2 /
2 /
*
1
dx
1 ) ( cos
2 /
2 /
4 *
=
dx x A A (1)
The successive use of the following reduction formula
+ = dx x
n
n
n
x x
dx x
n
n
n 2
1
cos
1 cos sin
cos
will yield
+ = dx x x x dx x
2 3 4
cos
4
3
cos sin
4
1
cos
(
+ + =
dx x x x x dx x
2
1
cos sin
2
1
4
3
cos sin
4
1
cos
3 4
c
x
x x x x dx x + + + =
8
3
cos sin
8
3
cos sin
4
1
cos
3 4
Eq (1) becomes
1
8
3
cos sin
8
3
cos sin
4
1
2 /
2 /
3 *
=
(
+ +
x
x x x x A A
1
2 8
3
0 0
2 8
3
0 0
2
=
(
)
`
\
|
+ +
)
`
\
|
+ +
A
1
8
3 2
= |
\
|
A
CHAPTER 03 QUANTUM MECHANICS 80
3
8 2
= A or
3
8
= A
(b)The normalized eigenfunction can be written as
) ( cos
3
8
) (
2
x x
= for
2 2
< < x
The desired probability is given by
=
4 /
0
4
4 /
0
*
) ( cos
3
8
dx x dx
4 /
0
3
8
3
cos sin
8
3
cos sin
4
1
3
8
+ + =
x
x x x x
{ }
(
+ +
)
`
+ + = 0 0 0
12
3
16
3
16
1
3
8
2 / 1 ) 4 / cos( ) 4 / sin( = = Q
= + =
4 /
0
*
462 . 0
4
1
3
2
dx
CHAPTER 03 QUANTUM MECHANICS 81
3-2 POTENTIAL STEP
Problem 3-9
A particle of energy E approaches a potential step of height
V
o
. What should be the ratio (E/V
o
) so that the reflection
coefficient is 0.5? B.U. B.Sc. 2009A
Solution
The reflection coefficient is given by
2 1
2 1
k k
k k
R
+
=
0
0
0
0
( 2
2
( 2
2
V E E
V E E
V E m
mE
V E m
mE
R
+
=
=
h h
h h
) / (
1
1
1 ) / ( ) / (
1 ) / ( ) / (
0
0 0
0 0
V E x where
x x
x x
V E V E
V E V E
R =
+
=
+
=
1
1
5 . 0
+
=
x x
x x
since R = 0.5
1 1 5 . 0 5 . 0 = + x x x x
x x 5 . 0 1 5 . 1 = or x x = 1 3
Square both sides
x x = ) 1 ( 9 or 9 8 = x or
8
9
= x or 125 . 1
8
9
0
= =
|
|
\
|
V
E
Problem 3-10
A particle of energy 8.1 eV approaches a potential step of
height 8 eV. Calculate the probability of reflection.
Solution
The coefficient of reflection in terms of energy of incident
particle E and height of potential step can be written
as
CHAPTER 03 QUANTUM MECHANICS 82
) / (
1
1
1 ) / ( ) / (
1 ) / ( ) / (
0
0 0
0 0
V E x where
x x
x x
V E V E
V E V E
R =
+
=
+
=
Now eV V eV E 8 , 1 . 8
0
= = therefore
0125 . 1 ) 8 / 1 . 8 ( ) / (
0
= = = V E x .
Hence 8 . 0
1 0125 . 1 0125 . 1
1 0125 . 1 0125 . 1
=
+
= R
The desired probability of reflection of particle from the
potential step is 0.8 or 80 percent.
CHAPTER 03 QUANTUM MECHANICS 83
3-3 POTENTIAL BARRIER
Problem 3-11
A beam of electrons of energy 5 eV is allowed to fall on one-
dimensional potential barrier of height 10 eV and width 1 .
Calculate the percentage of electrons transmitted.
Solution
For E < V
o
the transmission coefficient is given by
1
0 0
2
2
)] / ( 1 )[ / ( 4
) ( sinh
1
+ =
V E V E
a k
T
Now
34
19 31
0
2
10 055 . 1
) 10 602 . 1 )( 5 10 )( 10 109 . 9 ( 2 ) ( 2
=
h
E V m
k
1 10
2
10 145 . 1
= m k
Hence
33 . 0
)} 2 / 1 ( 1 ){ 2 / 1 ( 4
)] 10 1 )( 10 146 . 1 [( sinh
1
1
10 10 2
=
(
+ =
T 3
i.e. 33.3 % or one third of incident electrons will be transmitted.
Problem 3-12
Assume that an alpha particle has energy 10 MeV and
approaches a potential barrier of height equal to 30 MeV.
Determine the width of the potential barrier if the
transmission coefficient is 0.002.
Solution
For E < V
o
the transmission coefficient is given by
1
0 0
2
2
)] / ( 1 )[ / ( 4
) ( sinh
1
+ =
V E V E
a k
T
)] / ( 1 )[ / ( 4
) ( sinh
1
1
0 0
2
2
V E V E
a k
T
+ =
)] / ( 1 )[ / ( 4
) ( sinh
1
1
0 0
2
2
V E V E
a k
T
=
CHAPTER 03 QUANTUM MECHANICS 84
] 1 ) / 1 )][( / ( 1 )[ / ( 4 ) ( sinh
0 0 2
2
= T V E V E a k
Now
h
) ( 2
0
2
E V m
k
=
34
19 6 27
2
10 055 . 1
) 10 602 . 1 )( 10 )( 10 30 )( 10 645 . 6 ( 2
= k
1 15
10 956 . 1
= m and
] 1 ) / 1 )][( / ( 1 )[ / ( 4
0 0
T V E V E
56 . 443 1
002 . 0
1
30
10
1
30
10
4 = |
\
|
|
\
|
|
\
|
=
Substitute these values in Eq(1)
56 . 443 ] ) 10 956 . 1 [( sinh
15 2
= a
061 . 21 56 . 443 ] ) 10 956 . 1 sinh[(
15
= = a
741 . 3 ) 061 . 21 ( sinh ] ) 10 956 . 1 [(
1 15
= =
a
m a
15
15
10 913 . 1
10 956 . 1
741 . 3
=
=
The desired thickness of the potential barrier is
fm m a 913 . 1 10 913 . 1
15
= =
.
Problem 3-13
A potential barrier has a height of 8 eV and width
10
10 2
m. Calculate the lowest energy of the incident
electron in order to have 100 percent transmission.
Solution
The transmission coefficient for E > V
o
is given by
1
0 0
3
2
] 1 ) / )[( / ( 4
) ( sin
1
+ =
V E V E
a k
T
It is clear that the transmission will be 100 percent if
,....... 3 , 2 ,
3
= a k
For lowest energy electrons we have
= a k
3
CHAPTER 03 QUANTUM MECHANICS 85
=
a
V E m
h
) ( 2
0
2
2
2
0
) ( 2
=
h
a V E m
2
2 2
0
2ma
V E
h
=
eV J V
ma
E 8
) 10 2 )( 10 109 . 9 ( 2
) 10 066 . 1 (
2
2 10 31
2 34 2
0
2
2 2
+
= + =
h
eV eV E 8
) 10 602 . 1 ( ) 10 2 )( 10 109 . 9 ( 2
) 10 066 . 1 (
19 2 10 31
2 34 2
+
eV eV eV E 41 . 17 8 41 . 9 = + =
Problem 3-14
A proton and a deuteron (which has the same charge but
twice the mass) are incident on a barrier of thickness 10 fm
and height 10 MeV. Each particle has a kinetic energy of 3.0
MeV. Find the transmission probabilities for them.
Solution
The approximate formula for transmission probability is
) 2 exp( 1 16
0 0
kL
V
E
V
E
T
|
|
\
|
|
|
\
|
where
h
) ( 2
0
E V m
k
= . Now
36 . 3
10
3
1
10
3
16 1 16
0 0
= |
\
|
|
\
|
=
|
|
\
|
|
|
\
|
V
E
V
E
For proton
h
) ( 2
0
E V m
k
p
=
34
19 6 23
10 055 . 1
) 10 620 . 1 )( 10 )( 3 10 )( 10 673 . 1 ( 2
= k
CHAPTER 03 QUANTUM MECHANICS 86
1 14
10 806 . 5
= m k
612 . 11 ) 10 10 )( 10 806 . 5 ( 2 2
15 14
= =
kL
The desired probability for proton is
5
10 043 . 3 ) 612 . 11 exp( ) 36 . 3 (
= = T
For deuteron, we have
h
) )( 2 ( 2
0
E V m
k
p
=
1 14 14
10 211 . 8 ) 10 806 . 5 ( 2
= = m k
422 . 16 ) 10 10 )( 10 211 . 8 ( 2 2
15 14
= =
kL
The desired transmission probability for deuteron is
7
10 479 . 2 ) 422 . 16 exp( ) 36 . 3 (
= = T
Problem 3-15
When 1.0 eV electrons are incident on a potential barrier of
8.0 eV (such as the work function of a metal), what fraction
of the electrons penetrate their barrier if it is 5.0 wide?
Solution
Here we can use the approximate formula for transmission
coefficient (or probability)
) 2 exp( 1 16
0 0
kL
V
E
V
E
T
|
|
\
|
|
|
\
|
where
h
) ( 2
0
E V m
k
= . Now
75 . 1
8
1
1
8
1
16 1 16
0 0
= |
\
|
|
\
|
=
|
|
\
|
|
|
\
|
V
E
V
E
and
1 10
34
19 31
10 355 . 1
10 055 . 1
) 10 620 . 1 )( 1 8 )( 10 109 . 9 ( 2
= m k
55 . 13 ) 10 5 )( 10 355 . 1 ( 2 2
10 10
= =
kL
6
10 282 . 2 ) 55 . 13 exp( ) 75 . 1 (
= = T
CHAPTER 03 QUANTUM MECHANICS 87
Problem 3-16
A potential barrier has a height 8 eV and thickness
10
10 2
m. What is the lowest energy incident electron may
have and be 100 percent transmitted?
Solution
The transmission coefficient for E > V
0
is given by
1
3
2
2
3 1
2
3
2
1
) ( sin
4
4 1
(
(
|
|
\
|
+ = a k
k k
k k
T
It is clear that the transmission coefficient will be 100 percent if
n a k = = .. ,......... 3 , 2 ,
3
The lowest energy electron will be that for which
= a k
3
=
a
V E m
2
0
) ( 2
h
0 2
2 2
2
V
ma
E + =
h
eV J in E 8
) 10 2 )( 10 109 . 9 ( 2
) 10 055 . 1 (
2 10 31
2 34 2
+
eV eV in E 8
) 10 602 . 1 ( ) 10 2 )( 10 109 . 9 ( 2
) 10 055 . 1 (
19 2 10 31
2 34 2
+
eV eV eV E 41 . 17 8 41 . 9 = + =
CHAPTER 03 QUANTUM MECHANICS 88
3-4 INFINITE POTENTIAL WELL
OR PARTICLE IN A BOX
Problem 3-17
An electron trapped in an infinite potential well of length
100 pm. What are the energies of its three lowest allowed
states? P.U. B.Sc. 2002
Solution
The energy of electron in infinite potential well is given by
2
0
2 2
8 L m
h n
E
n
= in Joules
e L m
h n
E
n
2
0
2 2
8
= in eV
) 10 602 . 1 ( ) 10 100 )( 10 109 . 9 ( 8
) 10 626 . 6 (
19 2 12 31
2 34 2
=
n
E
n
2
61 . 37 n = in eV
Now , 1 = n eV E 61 . 37 ) 1 )( 61 . 37 (
2
1
= =
, 2 = n eV E 44 . 150 ) 2 )( 61 . 37 (
2
2
= =
, 3 = n eV E 49 . 338 ) 3 )( 61 . 37 (
2
3
= =
The desired values of energy are 37.61 eV, 150.44 eV and
338.49 eV.
Problem 3-18
Find the lowest three energies, in MeV, of a proton trapped
in an infinite potential well of width 5 fm.
Solution
The energy of proton in infinite potential well is given by
2
2 2
8 L m
h n
E
p
n
= in Joules
e mpL
h n
E
n 2
2 2
8
= in eV
CHAPTER 03 QUANTUM MECHANICS 89
) 10 602 . 1 ( ) 10 5 )( 10 673 . 1 ( 8
) 10 626 . 6 (
19 2 15 27
2 34 2
=
n
E
n
eV in n
2 6
) 10 19 . 8 ( =
MeV n E
n
2
19 . 8 =
Now , 1 = n MeV E 19 . 8 ) 1 )( 19 . 8 (
2
1
= =
, 2 = n MeV E 76 . 32 ) 2 )( 19 . 8 (
2
2
= =
, 3 = n MeV E 71 . 73 ) 3 )( 19 . 8 (
2
3
= =
The desired values of energy of the trapped proton are
8.19 MeV, 32.76 MeV and 73.71 MeV.
Problem 3-19
Consider an electron trapped in an infinite well whose width
is 98.5 pm. If it is in a state with n = 15, what are (a) its
energy? (b) The uncertainty in its momentum? (c) The
uncertainty in its position?
Solution
(a) The energy of electron in infinite potential well is given by
2
0
2 2
8 L m
h n
E
n
= in Joules
e L m
h n
E
n 2
0
2 2
8
= in eV
Now 15 = n and m pm L
12
10 5 . 98 5 . 98
= = , therefore
) 10 602 . 1 ( ) 10 5 . 98 )( 10 109 . 9 ( 8
) 10 626 . 6 ( ) 15 (
19 2 12 31
2 34 2
= E
keV eV E 72 . 8 10 72 . 8
3
= =
(b)
0
2
0
2 2
0 2
0
2 2 2
1
m
p
m
v m
v m E = = =
E m p
0
2
2 =
E m p
0
2 =
CHAPTER 03 QUANTUM MECHANICS 90
keV E c m pc 4 . 94 ) 72 . 8 )( 511 ( 2 2
2
0
= = =
keV MeV c m 511 511 . 0
2
0
= = Q
The direction of motion of the electron is not known. The
electron bounces back and forth between the walls of given
infinite potential well. Hence the uncertainty in momentum is
given by
L
h n
L
h n
L m
h n
m E m p p
2 4 8
2 2
2
2 2
2
0
2 2
0 0
= =
|
|
\
|
= = =
(c) As electron can be anywhere inside the well, there the
uncertainty in position will be 98.5 pm.
Problem 3-20
What must be the width of an infinite well such that a
trapped electron in the n = 3 state has an energy of 4.70 eV.
Solution
The energy of the electron in an infinite potential well is given
by
2
2 2
8mL
h n
E
n
=
For n = 3,
2
2
3
8
9
L m
h
E = or
3
2
2
8
9
E m
h
L =
or
) 10 602 . 1 70 . 4 )( 10 109 . 9 ( 8
) 10 626 . 6 ( 3
8
3
19 31
34
3
= =
E m
h
L
m L
19
10 486 . 8
= or 8.486
Problem 3-21
The ground-state energy of an electron in infinite potential
well is 2.6 eV. What will be the ground-state energy if the
width of the well is doubled?
Solution
The energy of the electron in an infinite potential well is given
by
CHAPTER 03 QUANTUM MECHANICS 91
2
2 2
8mL
h n
E
n
= or
2
1
L
E
n
If the width of the well is doubled, then its energy will become
one-fourth of the first value i.e.
eV
E
E 65 . 0
4
6 . 2
4
1
1
= = =
Problem 3-22
An electron, trapped in an infinite well of width 253 pm, is
in the ground (n = 1) state. How much energy must it absorb
to jump up to the third excited (n = 4) state?
Solution
The energy absorbed by the electron will be
2
2
2
2
2
2
1 4
8
15
8 8
16
mL
h
mL
h
mL
h
E E = =
2 12 31
2 34
) 10 253 )( 10 109 . 9 ( 8
) 10 626 . 6 ( 15
=
keV or J 74 . 8 10 399 . 1
15
=
Problem 3-23
(a) Calculate the smallest allowed energy of an electron
confined to an infinitely deep well with a width equal to the
diameter of an atomic nucleus (about
14
10 4 . 1
m).
(b) Repeat for a neutron.
Solution
The energy of the electron in an infinite potential well is given
by
2
2 2
8mL
h n
E
n
= . The value of lowest energy is obtained by
substituting n = 1 in above equation i.e.
m m mL
h
E
40
2 14
2 34
2
2
1
10 800 . 2
) 10 4 . 1 ( 8
) 10 626 . 6 (
8
= =
(a) For electron
e
m m = and
CHAPTER 03 QUANTUM MECHANICS 92
GeV J E 918 . 1 10 074 . 3
10 109 . 9
10 800 . 2
10
31
40
1
= =
Q J eV
19
10 602 . 1 1
=
(b) For proton
p
m m = and
MeV J E 045 . 1 10 674 . 1
10 673 . 1
10 800 . 2
13
27
40
1
= =
Problem 3-24
A proton confined in a one-dimensional box has energy 400
keV in its first excited state. How wide is the box?
Solution
The energy of the particle in one-dimensional box is given by
2
2 2
8mL
h n
E
n
=
For first excited state, we substitute n = 2
2
2
2
2 2
2
2 8
) 2 (
mL
h
mL
h
E = =
2
2
2
2mE
h
L = or
2
2mE
h
L =
Now kg m m s J h
p
27 34
10 673 . 1 , 10 626 . 6
= = = and
J J keV E
14 19 6
10 408 . 6 ) 10 602 . 1 )( 10 )( 400 ( 400
= = =
therefore
m L
14
14 27
34
10 525 . 4
) 10 408 . 6 )( 10 673 . 1 ( 2
10 626 . 6
=
CHAPTER 03 QUANTUM MECHANICS 93
CONCEPTUAL QUESTIONS
(1) The wavefunction associated with a physical system is, in
general, a complex quantity but does contain all relevant
information needed to describe the system. How will you get
any significant result from this wavefunction?
Answer: - By measuring | |
2
=
\
|
=
A (b) ) 2 exp(
2
2
2 / 1
x
|
\
|
(c) 1
(2) (a)
5
4
3
a
(b) 0 (3) (49/48) (4)
7
10 9 . 4
(5)
19
10 056 . 2
(6) 904 . 5 ) / (
1 2
= T T