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# THE LCAO METHOD OF CALCULATING ENERGY LEVELS

We now come to piece together the knowledge that we have previously obtained and see how metals, semiconductors, and insulators arise from common principles. This will be achieved by the linear combination of atomic orbitals method (abbreviated LCAO). This

FIGURE 2.9 The schematic formation of Bloch waves in the linear chain of hydrogen atoms according to the LCAO method.

method can be used as both a first principles method or a semiempirical method, a distinction which will become evident later.

As usual, we will begin with a simplified model. Consider the linear chain of atoms described in section 2.1. We redraw it here in figure 2.9, showing the Is orbital of each hydrogen atom when the atoms are near enough to have formed a solid. Our aim is to see how the Bloch functions are formed out of the original orbitals of the isolated atoms. We assume that the linear chain has been stabilized at an interatomic distance found in most solids, i.e. that of 2-3A0. As a consequence, the overlapping between a Is orbital and a far distant one must be negligible but that between nearest neighbour atoms is significant. Futhermore, as a simplification we will assume that with second and further nearest neighbours overlapping is zero, as shown in figure 2.9. This is only approximately true—overlapping of atomic orbitals may occur up to third nearest neighbours, but their overlapping (defined mathematically below) decays rapidly with distance, so our approximation will not qualitatively change our results. Mathematically our approximation can be written as

if n is not the same or a nearest neighbour of m

where Ф(х-К„) is a Is orbital centred on the nth site and as usual * denotes complex conjugate.

Looking at figure 2.9, it appears that a linear combination of atomic orbitals, i.e. of the form

where the Ckn, are constants to be determined, might reproduce the wavefunctions of the hypothetical crystal in question, that is the wavefunctions that have a propagating wave character. However, since these must obey Bloch’s theorem, the Ckn must be of the form

To prove this, we substitute 2.21 in 2.20 and observe that the 1-dimensional form of Bloch’s theorem is verified. We get, putting x+Rm instead of x in 2.20,

The wavefunctions as given by 2.20 and 2.21 are not normalized. A normalization constant is required such that

It is easily verified that the normalization constant required is 1 /yj{N) where N is the number of atoms or unit cells in the chain. Hence

From the Schroedinger equation НТ=£АЧ'А., we can obtain the energy Ek by multiplying by ЧК* and integrating. Then

Now in 2.25 we choose to ignore all integrals for which the integers n and m do not denote either the same atom or nearest neighbour atoms. This approximation is essentially the approximation given by equation 2.19. The validity of this approximation derives from the fact that at every point in space either one or both factors of the integrand (the orbitals) are zero. We note that operating by H does not change the extent of the orbital and therefore we can ignore all integrals in which n, m denote atoms that are further apart than the first nearest neighbour distance. Then we can write

where Rj=R„ —Rm denotes first nearest neighbours or zero (if n=m). In the linear chain there are only two neighbours of a given atom, one located at a distance a to its right and another at a to its left so R/ = ±a or 0. Hence

where fnn means first nearest neighbours, and

This A factor comes from the n=m terms and В from the first nearest neighbours. Note also that by symmetry the + or - terms will give the same В in 2.27c.

There is one more step before we reach our desired conclusion: we write the crystal potential Vcr as the sum of the atomic potentials of each atom plus a perturbation that resulted from the bonding of atoms and the redistribution of electron charge that results from it. Then

or

Obviously very near the nucleus of the ith atom the total Vcr will be equal to Val(x-R,). Hence

where £(x)HlllO(x)dx = the atomic energy level of the Ф(х) orbital (i.e. the Is orbital) in the isolated atom and

so that equation 2.27a can be written

For each value of к given by 2.12, a corresponding value of energy E(k) will be defined. There will be a maximum value of E(k) equal to Eal+C+2B and a minimum equal to £Я,+С-2В. What does equation 2.31 tell us? Exactly what we described in section 2.1 on intuitive grounds: every atomic level Eat in an isolated atom splits in a crystal in a band (of width 4B) centred around a shifted Eal (see figures 2.10a and 2.10b). Of course, this simplified model does not hold for most crystals. The nearest solid to which this model can be applied is lithium (Li) which has 3 electrons, 1 electron in the outer 2s orbital and 2 in the inner Is orbitals, the potential of which can be grouped together with that of its nucleus. Therefore, it looks very much like it is hydrogenic. Like hydrogen, if N is the number of atoms, there will be 2N states (counting spin also) and N electrons. The band will be half-filled and therefore Li, having a half filled band, will be a metal, as we will show later in this chapter.

The LCAO method can be extended to take into account more complicated solids that have more than one electron in their outer shell. In particular, we are interested in such semiconductors as Si, Ge, and GaAs, etc., which have two atoms in their unit cell and s and p electrons on each atom. On intuitive grounds we expect each atomic energy level to split into a band. The LCAO method can be an accurate method for these and other materials by including in the summation in equation 2.23 all the orbitals of all the atoms in the unit

FIGURE 2.10 Exact model of the splitting of the Is energies of the hydrogen chain in a band (a) and their representation in terms of a quasi-continuous wavevector к (b).

cell of the material in question and further interactions than the first nearest neighbours. So, if the positions of the atoms within a unit cell are denoted by the index 1, the unit cells by the index n, as previously, and the position vectors of equivalent atoms in different unit cells by R„ |, equation 2.23 can be generalized in 3 dimensions for more than 1 atom in the unit cell and more than one kind of orbital as follows:

where are the expansion (or weighting) coefficients of the Ф/ ( orbital which is of type j (i.e. s or p) and is centred on the equivalent atoms at positions R„j.

Equation 2.32 does not lead to a simple expression of the form E(k)= a known quantity. Instead it leads to a set of linear equations. To see this, we substitute the expansion of 2.33 into the Schroedinger equation HXV=EXV to obtain

Now following the standard procedure of linear algebra, we multiply on the left by vF)r (k) and integrate. We get, hiding the к dependence for the moment,

where

and

and V is the volume of the crystal.

For the set of linear equations 2.34 in the unknowns to have a non-trivial (i.e. nonzero) solution we must have

where we have reinstated the к dependence and have treated НщГ and S/(;r as the matrix elements of the corresponding matrices H and S. Equation 2.36 is called the secular equation and the reader would recognize the above equations as a standard procedure for turning a differential equation into an algebraic one. The above equations would only be a mathematical trick with no computational significance if the summation in 2.34 were not over but a small number of values of j and /. Indeed, for most semiconductors (not all) we only need to consider the s and the three p type of orbitals with only two atoms in the unit cell so that the size of the matrices H and S is manageable, although, in principle, one would have to include in 2.32 an infinite sum of orbital types.

The bands that we get from the secular equation are radically different from the simple model of the linear chain with only one s orbital in the unit cell. The main feature of the realistic model that led to the secular equation is that we do not get only one E(k) curve as in figure 2.10 but many E(k) curves as the size of the matrices in 2.36 since the determinant of a matrix gives an algebraic equation of order equal to the size of its matrix. So we should write Em(k) where m is a band index referring to the (j,l)th eigenvalue of the secular equation. The same holds for the wavefunctions. We should write 4/,„ifc(r) and not ВД. Figures 2.11a-c shows the bands in Al, Si, and GaAs respectively. Given that к is a

FIGURE 2.11a The band structure of aluminum, calculations by B. Segall, Phys. Rev. 124, 1797 (1961). Capital letters (English or Greek) denote points in the Brillouin Zone.

FIGURE 2.11b The band structure of silicon, calculations by C. S. Wang and В. M. Klein, Phys. Rev. В 24, 3393 (1981). Capital letters (English or Greek) denote points in the Brillouin Zone.

3-dimensional vector, the 1-dimensional vector к appearing in figure 2.11 portrays the values of к along the axes of high symmetry of the Brillouin zone (BZ) shown in figure 2.8— all 3 materials have the same BZ. So far we have not linked the theory with the properties of the solid in question. We have merely stated without a proper explanation that partially filled bands give metallic behaviour. We are now ready to tackle the questions we posed ourselves at the beginning of this chapter (see section 2.1).

The most trivial question we have so far asked is why A1 is a metal and GaAs is a semiconductor. It is a property of a partly filled band that it can exhibit conductivity, whereas a completely filled band gives a zero conductivity. Why is this so? As an electric field is applied to a solid, the electrons can be accelerated and give rise to electric current, but only if they can accept the energy of the electric field. To do so they must be in states which have empty states just above them (in energy) to which they can jump as they are being accelerated. Electrons in completely filled bands can’t absorb energy from the electric field because there are no such empty states available. Solids in which the highest occupied band is partially filled fulfill the above condition and are metals. We note that in metals

FIGURE 2.11c The band structure of gallium arsenide, calculations by J. P. Walter and M. L. Cohen, Phys. Rev. 183, 763 (1969). Capital letters (English or Greek) denote points in the Brillouin Zone.

the Fermi level is somewhere in the middle of the band and separates the filled from the empty states at T = 0.

Semiconductors are solids in which the highest occupied band, called the valence band, is completely filled at T = 0, but it is separated from the next empty band by a band gap Eg < 2eV. Again at T = 0 these materials can’t exhibit conductivity but as the temperature T rises, electrons can be excited thermally from the valence band to the next empty band called conduction band, thus making both bands conductive. This is the case of GaAs, for example, in which the minimum of the conduction band occurring at the Г point of the BZ is separated by a gap of 1.4eV from the maximum of the valence band occurring also at Г. But how do we know that the highest occupied band at T=0 (i.e. the valence band) is full and the one above it (i.e. the conduction band) is completely empty? This is a mandatory requirement because in either case we will not have a semiconductor but a metal. The answer is that we count the number of states and the number of electrons and then we decide. The procedure is easy and requires only simple arithmetic. We describe this for GaAs immediately below. The arguments are valid only at T=0, but this is the only temperature we need to examine to deduce the character of the material.

Let us draw a figure similar to 2.10a specifically for GaAs. The latter has one Ga and one As atom in its unit cell. We only need to consider the outer shell of each atom. The orbitals of the inner shells of any atom give completely filled bands. The electronic structure of the

FIGURE 2.12 Schematic formation and filling of bands in GaAs as deduced from the atomic structure of the constituent atoms.

outer shell of Ga is 4s24p meaning two electrons in its 4s orbital and one in its 4p orbital, while that of As is 4s24p- The broadening of these atomic levels into bands is shown schematically in figure 2.12. Now according to Bloch’s theorem, each such atomic level gives a band with N such levels in the solid where N is the number of unit cells. We must not forget that each such atomic level has a degeneracy of 2(2/+l) in the atom itself where / is the second quantum number of the specific orbital. We remind the reader (see also equation 1.39) that s-type orbital means 1=0 and p-type orbital means 1=1. We can then construct the following table of bands with electrons and available states in each band.

We can see from figure 2.12 that the valence band of GaAs arises from the splitting of the As atomic levels and the conduction band from the Ga atomic levels. From the table above (table 2.1), it appears as if the valence band contains 5N electrons and the conduction band 3N electrons, but this is misleading because what actually will happen is that the 3N electrons of the Ga states will fall into the empty 3N states of the valence band, filling it completely so that the valence band is full and the conduction band is empty. This type of solid is either a semiconductor or an insulator depending on the value of the

TABLE 2.1 Number of states and electrons in the bands of GaAs

 Atomic Level Number of Electrons Number of States

energy band gap Eg. Solids with Eg<2eV are categorized as semiconductors. Higher values of Eg refer to insulators. We will see in a later section how the value of Eg is related to the number of electrons in a band. Of course the numerical value of Eg or even the existence of a gap cannot be deduced from the simple argument above. Note that we have assumed the existence of a gap in figure 2.12 with no validation—the four bands could have converged into a single band—that is why we need the LCAO (and other methods) of calculation. There is a lot of important information hidden in the Em{k) curves as we will see in a following section.

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