Chemistry Reference
In-Depth Information
It is not immediately obvious why the solutions of eq. (3.43) should
be NFE-like, requiring only a small number of OPW states to determine
the valence energy levels. When we expand the Schrödinger equation,
by substituting eq. (3.40) into (3.43), we find two different types of large
terms in the expansion, the first type associated with the full potential
acting on the plane wave part of the OPWs,
V
e
i
k
·
r
, and the second type
associated with the full Hamiltonian acting on the core state part of the
(
r
)
OPWs,
H
a
i
k
φ
c
i
(
)
=
ε
c
i
a
i
k
φ
c
i
(
)
.
Fortunately, the effect of the two types of terms largely cancel each
other. If we consider a nucleus centred at the origin
r
r
(
r
=
0
)
, then the
e
i
k
·
r
, is large and negative near
r
first term,
V
(
r
)
=
0, while the second
term,
ε
c
i
a
ik
φ
c
i
(
can be shown to be large and positive. We can re-write
eq. (3.43) to take advantage of this cancellation, effectively replacing the
true potential
V
r
)
(
r
)
by a much weaker pseudopotential
V
ps
(
r
)
and the true
by plane-wave pseudobasis states, e
i
k
(
m
)
x
. This gives
a new, so-called pseudoHamiltonian, whose eigenvalues are identical to
the valence eigenvalues of the true Hamiltonian, and with the pseudo-
Hamiltonian basis states given by linear combinations of a small number
of plane wave pseudofunctions.
To illustrate how the two types of term cancel, leaving a weaker
pseudopotential with a smooth pseudowavefunction, we take an exam-
ple based on the K-P model (Weaire and Kermode 1985). We consider a
series of square wells and barriers of equal width
k
m
v
basis states
φ
(
r
)
, with
the zero of energy at the bottom of the well and the barrier of height
V
0
(
a
=
b
=
6Å
)
2.0 eV (fig. 3.16a). The solid lines in fig. 3.16(b)-(d) show the exact
wavefunctions for the three lowest zone centre
=
(
q
=
0
)
states, which have
energies
E
1
2.50 eV.
The dashed line in fig. 3.16(d) shows that the exact wavefunction for the
third state can be well approximated by a single plane wave with
k
=
0.47 eV,
E
2
=
1.82 eV and
E
3
=
=
0
(fig. 3.16e) orthogonalised to the lowest energy level:
a
1
√
2
a
−
ψ
1
√
2
a
ψ
OPW
3
d
x
x
)
φ
(
x
)
=
(
x
)
(
1
1
−
a
1
2
=
1
/(
2
a
)
−
α
ψ
(
x
)
(3.45)
1
1
OPW
3
φ
(
)
We see from fig. 3.16(d) that
is then a good variational estimate
of the true wavefunction. We can, therefore, substitute the orthogonalised
state (eq. (3.45)) into the exact Schrödinger equation,
H
x
ψ
=
E
ψ
, to estimate
OPW
3
can be re-arranged to give a new equation with the same eigenvalue,
E
3
,
but with a weaker pseudopotential and smoother wavefunction. To do so,
OPW
3
φ
=
E
3
φ
the energy,
E
3
, of the third state. We show below how
H
