Chemistry Reference
In-Depth Information
origin, where we choose to write the general solution for a state at energy
E
in the form
A
e
i
kx
B
e
−
i
kx
ψ(
x
)
=
+
0
<
x
<
a
(3.10)
with
k
2
2
, as before. The general solution to Schrödinger's
equation in the first barrier to the left of the origin is given by
=
2
mE
/
C
e
κ
x
D
e
−
κ
x
ψ(
x
)
=
+
−
b
<
x
<
0
(3.11)
2
2
, and we assume
E
where
V
0
.
We now introduce four boundary conditions to determine the allowed
solutions of Schrödinger's equation. The first two are easily chosen, requir-
ing that the wavefunction
κ
=
2
m
(
V
0
−
E
)/
<
ψ
and its derivative d
ψ/
d
x
are continuous at the
=
well/barrier interface at
x
0:
ψ(
0
)
:
A
+
B
=
C
+
D
(3.12a)
ψ
(
0
)
:
i
kA
−
i
kB
=
κ
C
−
κ
D
(3.12b)
The two remaining boundary conditions are derived using Bloch's theo-
rem. We apply eq. (3.2b) at
x
e
i
q
(
a
+
b
)
and
=
a
,sothat
ψ
(
a
)
=
ψ
(
−
b
)
w
b
ψ
w
(
)
=
ψ
b
(
−
e
i
q
(
a
+
b
)
:
likewise for the derivatives
a
b
)
:
A
e
i
ka
B
e
−
i
ka
C
e
−
κ
b
D
e
κ
b
e
i
q
(
a
+
b
)
ψ
(
a
)
+
=
(
+
)
(3.12c)
w
ψ
w
(
i
kA
e
i
ka
i
kB
e
−
i
ka
C
e
−
κ
b
D
e
κ
b
e
i
q
(
a
+
b
)
a
)
:
−
=
(κ
−
κ
)
(3.12d)
We can then derive the condition for allowed energy levels by solving the
4
4 determinant involving the four unknowns
A
,
B
,
C
, and
D
,aswas
done with eq. (1.52) in Chapter 1. The energy levels associated with Bloch
wavenumber
q
are then found to be solutions of the equation
×
k
−
sin
1
2
k
κ
(
(
+
))
=
(
)
(κ
)
+
(
)
(κ
)
cos
q
a
b
cos
ka
cosh
b
ka
sinh
b
(3.13)
We can deduce one particularly important result from this equation,
namely, that there will always be energy gaps between the allowed energy
bands in the K-P model. To see this, we first note that the left-hand side of
eq. (3.13), cos
(
q
(
a
+
b
))
, can only take values between
−
1 and 1. If we choose
=
π
±
(κ
)
(
π)
=
ka
n
, then the right-hand side reduces to
cosh
b
, as sin
n
0
and cos
is always greater than 1 for
states which are bound within the well. Hence, there can be no solutions
to eq. (3.13) for
k
(
n
π)
=±
1. The magnitude of cosh
(κ
b
)
=
π/
n
a
, implying no solutions are possible with energy
2
, and likewise for neighbouring values of the energy.
This is illustrated in fig. 3.4(a), where we plot an example of the variation
2
E
=
(
/
2
m
)(
n
π/
a
)
