PHYS40352 Spring 2016 2 ELECTRONS IN CRYSTALS The reciprocal lattice for a cubic-I lattice is a cubic-F lattice (FCC), with cubic lattice constant 4 π a , as seen in figure 1.24 Figure 1.24 2 Electrons In Crystals 2.1 A Quantum Approach The electrons in crystals can be described by wavefunctions which solve the Schrödinger wave equation - ~ 2 2 m ∇ 2 ψ ( r ) + V ( r ) ψ ( r ) = Eψ ( r ) (2.1) where V ( r ) includes the potential due to ions and the averaged effect of the other electrons. This is like the ’self-consistent field’ or ’Hartree potential’ in atoms. Also, V ( r ) has the periodicity of a crystal, such that V ( r ) = V ( r + R ) . A periodic V ( r ) can be expressed as a sum of waves with the same periodicity as the crystal, i.e. the same phase on any lattice point, and these are waves with wave vectors G (of the RL). Recall that e i G · ( r + R ) = e i G · r × e i G · R = e i G · r This suggests that we can write the potential like V ( r ) = X G V G e i G · r (2.2) where the Fourier coefficients, V G , are given by the Fourier transform of the crystal potential over a primitive unit cell: V G = 1 v cell Z cell V ( r ) e - G · r d 3 r (2.3) For the crystal potential, V ( r ) , the complex conjugate of a Fourier coefficient, V G , is V * G = V - G There is a special case: 13
PHYS40352 Spring 2016 2 ELECTRONS IN CRYSTALS 2.2 Free Electrons For free electrons, V ( r ) = 0 , so the SWE has plane wave solutions ψ k = e i k · r with energies E k = ~ 2 k 2 2 m with periodic boundary conditions ψ ( x, y, z ) = ψ ( x + L, y, z ) where L a . Therefore, we have exp[ ik x L ] = 1 , which, combined with the k of the other dimensions, implies that k = 2 π L ( n x , n y , n z ) (2.4) Note that L is large, so the spacing of k values is small. We find that the spacing is Δ k x = Δ k y = Δ k z = 2 π L (2.5) which gives us a volume of k -space per k -value of ( 2 π L ) 3 . At T = 0 , the electrons occupy states of lowest energy, consistent with the Pauli exclusion principle, i.e. a sphere of radius k f , the Fermi wave number. Note that this is the whole sphere rather than one octant, due to periodic boundary conditions. The number of electrons is equal to the number of occupied states N e = 2 × 4 π 3 k 3 f (2 π/L ) 3 = k 3 f L 3 3 π 2 (2.6) which gives k 3 f = 3 π 2 N e L 3 (2.7) 2.3 Crystal Potential: V ( r ) 6 = 0 The solutions of the Schrödinger equation are no longer plane waves in the crystal potential. Try a solution of the form ψ k ( r ) = f k e i k · r , and substitute into the SWE: ˆ Hψ k ( r ) = - ~ 2 2 m ∇ 2 + V ( r ) ψ k ( r ) = - ~ 2 2 m ∇ 2 + V ( r ) f k e i k · r = ~ 2 k 2 2 m f k e i k · r + " X G V G e i G · r # f k e i k · r = ~ 2 k 2 2 m f k e i k · r + X G V G f k e i ( k + G ) · r , 14
PHYS40352 Spring 2016 2 ELECTRONS IN CRYSTALS where the sum represents scattered waves with all possible wave numbers of the form k + G . This suggests that the solution of the Schrödinger wave equation can be expressed as ψ k ( r ) = X G f k + G e i ( k + G ) · r . (2.8) What can we deduce from this?
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