Computational Physics

6.7 The pseudopotential method 147

d1,2=± 1 / 8 (a 1 +a 2 +a 3 )relative to the lattice points:

Vps(tot)(K)=Vps(at)(K)(eiK·d^1 +eiK·d^2 ). (6.54)

The sum of the exponentials on the right hand side is known as thestructure factor.
Therefore, we find for a vectorK=

∑ 3

i= 1 nibithat

Vps(tot)(K)=cos[π(n 1 +n 2 +n 3 )/ 4 ]Vps(at)(K). (6.55)

It follows immediately that the pseudopotential components vanish if the sum of
theniis an odd multiple of 2: the structure factor causes extinction of certain wave
vectors. Furthermore, we can choose the componentVps(tot)(K= 000 )to be equal to
zero, as this induces a mere shift in the energy offset. Collecting all bits and pieces,
we are left with the following values of|K|=Kfor which the pseudopotential
does not vanish (apart from a factor 4π^2 /a^2 ):

K^2 =3, 8, 11,... (6.56)

and only these first three components are taken into account in the pseudopotential.
This means that only three numbers are to be fitted and the whole band structure
follows from them.
We shall not carry out the fitting procedure but quote the resulting values for the
potential from literature, resulting from a fit to optical transition energies[22]–
they read (in atomic units):

Vps(tot)(

√

3 )=−0.1121;Vps(tot)(

√

8 )=0.0276;Vps(tot)(

√

11 )=0.0362. (6.57)

The matrix element of the pseudopotential Hamiltonian for plane wavesk+K
andk+K′is given by (K=K−K′)

HK,K′=

1

2

|k+K|^2 δK,K′+V(|K|)cos

[

(K 1 +K 2 +K 3 )

π

4

]

. (6.58)

The diagonalisation of the resulting matrix is straightforward: the plane waves are
orthogonal and hence no overlap matrix has to be taken into account. You can use
basis sets of size 9, 15, 27 etc., just as in the APW case. In Figure 6.13 the band
structure is represented for a basis with 113 states.
The band structure in Figure 6.13 matches the results of calculations using more
sophisticated methods very well, which is remarkable if you note that only three free
parameters enter into the potential. The fact that the band gap comes out well is not
surprising since it has been used in the fitting procedure. It turns out to be 1.17 eV,
and you might compare this withkBTat room temperature in order to estimate
the fraction of electrons excited into the conduction band using the Fermi–Dirac
distribution.