Chemistry Reference
In-Depth Information
We require for allowed solutions of Schrödinger's equation that the
wavefunction
ψ
and its derivative d
ψ/
d
x
be continuous at thewell/barrier
interfaces; namely,
e
κ
b
/
2
e
−
κ
b
/
2
ψ(
b
/
2
)
:
A
(
+
)
=
B
cos
(
ka
/
2
)
−
C
sin
(
ka
/
2
)
(2.6a)
ψ
(
e
κ
b
/
2
e
−
κ
b
/
2
/
)
κ
(
−
)
=
(
/
)
+
(
/
)
b
2
:
A
kB
sin
ka
2
kC
cos
ka
2
(2.6b)
ψ(
a
+
b
/
2
)
:
D
=
B
cos
(
ka
/
2
)
+
C
sin
(
ka
/
2
)
(2.6c)
ψ
(
a
+
b
/
2
)
:
−
κ
D
=−
kB
sin
(
ka
/
2
)
+
kC
cos
(
ka
/
2
)
(2.6d)
We could find the conditions for allowed energy levels by solving the 4
4
determinant involving the four unknowns
A
,
B
,
C
,and
D
, as was done
with eq. (1.52) in Chapter 1. Alternatively (or equivalently) we can first
use (2.6c) and (2.6d) to determine
B
and
C
in terms of
D
:
×
D
sin
k
cos
C
=
(
ka
/
2
)
−
(
ka
/
2
)
(2.7a)
D
cos
k
sin
B
=
(
ka
/
2
)
+
(
ka
/
2
)
(2.7b)
and then by substituting the values for
B
and
C
in (2.6a) and (2.6b), and
using the double angle identities, we obtain
e
κ
b
/
2
e
−
κ
b
/
2
A
(
+
)
=
D
((κ/
k
)
sin
(
ka
)
+
cos
(
ka
))
(2.8a)
e
κ
b
/
2
e
−
κ
b
/
2
A
(
−
)
=
D
((
k
/κ)
sin
(
ka
)
−
cos
(
ka
))
(2.8b)
Dividing (2.8b) by (2.8a) we find
tanh
κ
b
2
=
(
k
/κ)
sin
(
ka
)
−
cos
(
ka
)
(2.9a)
(κ/
k
)
sin
(
ka
)
+
cos
(
ka
)
which can be rewritten as
tanh
κ
k
2
sin
tanh
κ
1
cos
b
2
b
2
2
κ
−
(
ka
)
+
k
κ
+
(
ka
)
=
0
(2.9b)
A similar analysis shows that the energy levels for the double well states
of odd parity are found as solutions of
2
coth
κ
k
2
sin
coth
κ
1
cos
b
2
b
2
κ
−
(
ka
)
+
k
κ
+
(
ka
)
=
0
(2.10)
