5 The Band Structure of the Electrons 5.1 Introduction In the Fourier representation, i.e., in the basis of plane waves 1 i(k+Gn)r χn (k)= e , (5.1) Vg the Hamilton operator of the Kohn-Sham! equation has the form: !2 h = (k + G )2δ + veff (G G ). (5.2) n,m 2m n n,m n − m veff (G G ) decreases with increasing length G G ,andthush is basically n − m | n − m| n,m diagonal for large Gn Gm . The corresponding eigenvalue equation enables the calcu- lation of the energies| and− eigenfunctions| for given k. Figure 5.1: Fourier representation of the atomic potentialsofAl,SiandAg. In Fig. 5.1 we show the function veff (G) for three examples: Al, Si, and Ag. In fact, the figure shows the potential1 of atoms and therefore, veff is defined for continuous values of G .Forasolidwehaveroughly | | M eff eff atom v (r)= v − (r R ) , − I I=1 " 1To be precise, Fig. 5.1 shows the effective potential for atomic pseudopotentials. The definition of pseudopotentials is given in Section 5.5.1 below. 118 eff atom and then, in the Fourier representation v − is only needed at discrete reciprocal lat- 2π 1 tice vectors. These are determined by the lattice structure, and we have: a = 1.55 Å− 1 1 (Al), 1.16 Å− (Si), and 1.54 Å− (Ag). eff If all Fourier components v (Gl) are small, we obtain the dispersion of free electrons: !2 ϵ (k)= (k + G )2. (5.3) n 2m n ϵn(k) as a function of k is called the energy band n. For the one-dimensional case we have shown this dispersion before (cf. Fig. 4.13). However, the one-dimensional case is not typical, because in this case at most two-fold degeneracy can be present. Therefore, we will now discuss a two-dimensional example, which shows all the important characteristics of a band structure – even those of three-dimensional structures. We will investigate a hexagonal lattice. The 1st Brillouin zone and the labels of special k-points are shown in Fig. 5.2. If we now evaluate Eq. (5.3), we obtain the band structure of free electrons, 2π 2π as shown in Fig. 5.3, for G0 =(0, 0) = Γ, G1 = a (0, 1), G2 = a (cos 30◦, sin 30◦)= 2π √3 1 2π 2π √3 1 a ( 2 , 2 ), G3 = a (cos 30◦, sin 30◦)= a ( 2 , 2 ), etc. Here, we restrict ourselves to the boundary of the so-called irreducible− wedge, which− is hatched in Fig. 5.2. By reflection and rotation of this wedge the full 1st Brillouin zone can be obtained. Figure 5.2: Brillouin zone of a hexagonal cell. Special k-points are shown. The hatched area is the irreducible part. The point K is at 1 (G +G )= 2π ( 1 , 0),andalongtheΓK-direction we have k =(k , 0). 3 2 3 a √3 x We obtain the results: !2 2 n =0, i.e., G0 : ϵ0(k)=2m kx ϵ0(Γ)=0 2 ϵ (K) = ! ( 2π 1 )2 0 2m a √3 !2 2 2π 2 n =1, i.e., G1 : ϵ1(k)=2m (kx +(a ) ) !2 2π 2 ϵ1(Γ)=2m ( a ) 2 ϵ (K) = ! ( 1 +1) ( 2π )2 1 2m 3 · a 119 Figure 5.3: Band structure of free electrons in a hexagonal lattice. The numbers in brackets give the degeneracy. Figure 5.4: The band structure of nearly free electrons in a hexagonal lattice for two 1 2 different systems and two different Fermi energies: ϵF for 1 electron per cell and ϵF for 2 electrons per cell. 2 2 n =2, i.e., G : ϵ (k)= ! k + 2π √3 +(2π )2 1 2 2 2m x a 2 a · 4 !2 #2π$ 2 3 1 % & ϵ2(Γ)=2m ( a ) ( 4 + 4 ) !2 2π 2 = 2m ( a ) = ϵ1(Γ) 2 ϵ (K) = ! ( 2π )2 7 2 2m a · 3 etc. The result for ϵn(k) is shown in Fig. 5.3. Even for free electrons the band structure – just because of the reduction to the 1st Brillouin zone – looks rather complicated. Which modifications can we expect if the potential – which has been zero so far – slightly differs from zero? This will be investigated in more detail now. Plane waves with different Gn are coupled by the crystal potential (Bragg-condition, 120 reciprocal lattice ϵ(k) 1st Brillouin 2π zone a π Γ + π − a a Γ Γ Γ Γ Γ Figure 5.5: Three possible representations for ϵn(k) (one-dimensional example). Top: reduced zone scheme. Middle: repeated zone scheme. Bottom: extended zone scheme. hybridization of states), i.e., wave functions and eigenvalues are different: 1 i(k+Gn)r i(k+Gl)r e ϕn,k(r)= l cGn (l)e , √Vg −→ ϵ (k) ϵ (k)+∆ (k) . n −→ 'n n eff 2 For non-degenerate states, the change of the energy levels issmall( v (Gn) ), but it is larger for degenerate states2 ( veff ): ∼ ∼ ∆ (k)= veff (G ) . (5.4) n ±| n | Equation (5.4) describes the band structure close to the boundary of the 1st Brillouin zone. This is illustrated in Fig. 5.4. The representation of ϵn(k) in the domain of the 1st Brillouin zone is called a reduced zone scheme. Due to the equivalence of k and (k + G) we can consider the bands ϵn(k) also as periodic functions in k-space, as shown in Fig. 5.5 for a one-dimensional example. However, the “repeated zone scheme” and the “extended zone scheme” illustrated in Fig. 5.5 are very rarely used. 5.1.1 What Can We Learn from a Band Structure? First we note that in an N electron system at T =0K, the N lowest-energy states are occupied, i.e., each state ϕ k(r) can be filled by two electrons: and . n, ↑ ↓ 2cf. e.g. Ashcroft/Mermin, Chapter 9 or Madelung, Chapters 18–19. 121 One band, i.e., a function ϵn(k) with fixed n but variable k (k 1st BZ), can be occupied by ∈ V 2 g d3k (5.5) (2π)3 1.(BZ electrons. The factor 2 takes into account the spin. We have: V V (2π)3 V 2 g d3k =2 g =2 g =2N.ˆ (5.6) (2π)3 (2π)3 Ω Ω 1.(BZ One band can be filled by 2Nˆ electrons, where Nˆ is the number of unit cells in the base volume Vg,andΩ is the volume of the unit cell. Thus, one band can be filled by 2 electrons per primitive unit cell. The number of electrons per primitive unit cell and the band structure determine im- portant electric and optical properties of a solid, and we will now consider three special systems: System #1: One electron per cell. • This can be, for example, alkali metals or noble metals (Cu, Ag, Au). The lowest band of the band structure is then half filled, i.e., the Fermi energy is in the middle of 1 the band (cf. Fig. 5.4, ϵF ). Thus, directly above the highest occupied state there are unoccupied states. Therefore, the energy required to excite an electron and increase its kinetic energy is arbitrarily small. Such a system is an electric conductor. System #2: Two electrons per cell and a band structure as in Fig. 5.4, • left. 2 If ∆ < 0, like in the left firgure of Fig. 5.4, the Fermi energy is at ϵF .Thus,the second band is partially filled while a fraction of the first band remains unoccupied. Also in this case the Fermi edge cuts bands, and thus also this system is a metal. System #3: Two electrons per cell, and ∆ > 0 (Fig. 5.4, left). • When ∆ > 0, the lowest band is filled, and the Fermi energy is in the band gap above this band. The band occupied at T =0K is called valence band (VB), the unoccupied band above the band gap is called conduction band (CB). The position of the Fermi energy is then set equal to the chemical potentialatT =0K.Inthis KS case, the electrons require an energy of at least Egap = ∆ + ∆xc for the excitation from an occupied to an unoccupied state. Here, ∆KS is given by the Kohn-Sham eigenvalues calculated for the N-particle ground state: ∆KS = ϵN ϵN . LB − VB KS The quantity ∆xc is introduced here because in principle ∆ does not correspond to an excitation. We will come back to this point in Section tba. below. Systems with Egap =0 consequently are not electric conductors but, depending on the size of the band̸ gap KS ∆ + ∆xc, they are called insulators or semiconductors. 122 The size of the band gap determines, for instance, the appearance of the material. The measured band gap is the difference between the ionization energy I (removal of an electron from the highest level of the valence band) and the affinity A (addition of an electron in the lowest level of the conduction band)3: E = I A, (5.7) gap − A = EN EN+1 , (5.8) N −1 N I = E − E . (5.9) − From this we obtain N 1 N N+1 N Egap = E − E + E E N 1 − N+1 N− = E − + E 2E . (5.10) − Thus, for a correct evaluation of the band gap we need information for three different systems: The N 1, N,andtheN +1particle system. The Kohn-Sham eigenvalues are typically only evaluated− for the N-particle system. On the other hand for the Kohn-Sham eigenvalues we have: 1 N N+1 N+ 2 A = E E ϵCB , (5.11) − ≈− 1 N 1 N N+ 2 I = E − E ϵ , (5.12) − ≈− VB N˜ N 1 N where we assumed that E is continous and differentiable for E − < N<E˜ and EN < N<E˜ N+1.