Environmental Engineering Reference
In-Depth Information
We see that the allowed energies increase as the square of the integer quantum
number
n
, and that the energies increase also quadratically as
L
is decreased,
Ea
(1/
L
)
2
.
The condition for allowed values of
k¼np
/
L
is equivalent to
L¼nl
/2, the same
condition that applies to waves on a violin string.
These formulas are easily extended to the three-dimensional box of side
L
,by
taking a product of wavefunctions (2.16) for
x
,
y
, and
z
, and by adding energies in
Equation 2.15 according to
n
2
¼ n
x
þn
y
þn
z
.
This
exact solution
of this simple problem illustrates typical quantum behavior in
which there are discrete allowed energies and corresponding wavefunctions. The
wavefunctions do not precisely locate a particle, they only provide statements on the
probability of
finding a particle in a given range.
The conversion of the 1D picture to a
spherical
finite potential well V(r)
of radius
a
in
the case of zero angular momentum gives
2
2
2
m ¼ n
2
h
2
8
ma
2
E
n
0
¼
h
ðnp=
aÞ
=
=
;
n ¼
1
;
2
;
3
;
...
ð
2
:
17
Þ
and
n
00
ðrÞ¼ð
2
paÞ
1
=
2
½
sin
ðnpr
=
aÞ=
r
:
ð
2
:
18
Þ
The
spherical polar coordinates
are
x ¼ r
sin
cos
w; y ¼ r
sin
sin
w; z ¼ r
cos
:
Here, we are looking at portions of the solutions that are independent of the angles,
fully
spherically symmetric
solutions.
If the value of the potential at
r ¼ a
is a
finite value
V
o
, then it is found that there are
no bound states for
2
8
ma
2
V
o
<
h
=
:
ð
2
:
19
Þ
Alternatively, we can take this as a statement that
to have a single bound state, the
potential strength V
o
must be at least
h
2
/8ma
2
.
This is not an obvious result to obtain
, but it is simple, similar to the 1D case for its
energy formula, and easily remembered. The solutions for such a
nite
potential well,
in either the 1D or the spherical cases, are more dif
cult mathematically than the
in
nite potential well described above. Here, the allowed quantum states (wavefunc-
tions and energies
E
0) are obtained by requiring that the wavefunctions match in
amplitude and in slope at
r ¼ a
. Inside, we have
<
1
(
r
)
¼A
(sin(
kr
))/
r
, (Equation 2.18)
and outside we have
2
ðrÞ¼D
exp
ð
k
rÞ=r:
ð
2
:
19a
Þ
Here,
k ¼
h
1
(2
m
(
E þ V
o
))
1/2
and
k
¼
h
1
(
2
mE
)
1/2
. The solutions of a
transcendental equation are needed, and must be found numerically or graphically:
cot
z ¼
((
z
o
/
z
)
2
1)
1/2
, where
z
o
¼
h
1
(2
mV
o
)
1/2
a
. There is no solution if
z
o
¼
h
1
(2
mV
o
)
1/2
a
<p
/2, which is equivalent to (2.19). Formore details the readermay refer
to the text by Grif
ths [21].
