Chemistry Reference
In-Depth Information
2
the wavefunction
ψ
and its derivative d
ψ/
d
x
must also be continuous
at all
x
.
This holds automatically within the well from eq. (1.45) and within the
barrier from eq. (1.47);
in order to be satisfied everywhere, we then
require
ψ
and d
ψ/
d
x
to be continuous at the well/barrier interfaces, at
x
2. This gives rise to four linear equations involving the four
unknown parameters
A
,
B
,
D
, and
F
:
=±
a
/
D
e
−
κ
a
/
2
ψ(
a
/
2
)
:
A
sin
(
ka
/
2
)
+
B
cos
(
ka
/
2
)
=
(1.48a)
ψ
(
D
e
−
κ
a
/
2
a
/
2
)
:
Ak
cos
(
ka
/
2
)
−
Bk
sin
(
ka
/
2
)
=−
κ
(1.48b)
F
e
−
κ
a
/
2
ψ(
−
a
/
2
)
:
−
A
sin
(
ka
/
2
)
+
B
cos
(
ka
/
2
)
=
(1.48c)
ψ
(
−
F
e
−
κ
a
/
2
a
/
2
)
:
Ak
cos
(
ka
/
2
)
+
Bk
sin
(
ka
/
2
)
=
κ
(1.48d)
These four equations can be solved directly, as we do below, to find
the allowed energy levels,
E
n
, for states confined within the quantum
well. However, because the potential is symmetric, it is easier to calculate
separately the even and odd allowed states.
For the even states (fig. 1.8),
D
=
F
,and
ψ(
x
)
=
B
cos
(
kx
)
within the
well, giving as boundary conditions at
x
=
a
/
2
D
e
−
κ
a
/
2
B
cos
(
ka
/
2
)
=
(1.49a)
e
−
κ
a
/
2
−
Bk
sin
(
ka
/
2
)
=−
D
κ
(1.49b)
with two identical boundary conditions obtained at
x
2. Dividing
(1.49b) by (1.49a), we obtain for the even states within the quantum
well that
k
tan
=−
a
/
(
ka
/
2
)
=
κ
or
k
sin
(
ka
/
2
)
−
κ
cos
(
ka
/
2
)
=
0
(1.50)
We obtain the odd states by letting
D
=−
F
and
ψ(
x
)
=
A
sin
(
kx
)
within
the well, so that
k
cot
(
ka
/
2
)
=−
κ
or
k
cos
(
ka
/
2
)
+
κ
sin
(
ka
/
2
)
=
0
(1.51)
