The free electron gas: Sommerfeld model Valence electrons: delocalized gas of free electrons (selectrons of alkalies) which is not perturbed by the ion cores (ionic radius of Na + equals 0.97 Å, internuclear separation in solid Na equals 3.66 Å). Constant potential: the variation of the potential due to the ion cores and the rest electrons is negligible in the interstitial regions. It is assumed that the net charge associated with the ion cores and the rest valence electrons is smoothly distributed throughout the volume of the solid ( jellium model ) giving rise to a constant potential everywhere except at the surface of the solid. There a potential barrier prevents the escape of electrons out of the solid. The potential model is that of a 3D square-well. The electrons are treated quantum-mechanically in a meanfield approach The electrons obey Pauli s exclusion principle and Fermi- Dirac statistics 1
References [1] S. Elliott The Physics and Chemistry of Solids, J. Wiley, Chichester, 1998. 2
For a time-independent potential the stationary-state solutions of the time-independent Schrödinger equation are the eigenfunctions and eigenenergies: Hψ(r) = εψ(r) h2 2 ψ(r) + V ψ(r) = εψ(r) The allowed solutions depend upon the boundary conditions, to which the electrons are subject, and are quantized; i.e. the energies of the allowed states are functions of discrete quantum numbers. Independent-electron approximation: it works because of the screening by all other electrons of the interaction between any two electrons. The free particles (electrons) are in reality quasiparticles, comprising an electron and its associated orthogonalization hole. H(r 1, r 2 ) = H(r 1 ) + H(r 2 ) + ψ(r 1, r 2 ) = ψ 1 (r 1 ) ψ 2 (r 2 ) W is the potential well depth or the potential barrier height at the box boundaries. ɛ is the total energy of an electron; φ = W ɛ is the extra energy needed for an electron to escape from the solid; work function. 3
Figure 1: V is the potential energy, assumed constant, set V = 0, in the interior of the box. It involves the ion-core and averaged electronelectron potential energies. 4
The wavefunctions for the free electron gas A general solution is: ψ(r) = Nexp(ik r) = Nexp[i(k x x + k y y + k z z)] 1 = ψ(r) ψ(r)dr. The boundary condition for an electron trapped in a potential box of uniform V is that the wavefunction ψ(r = 0) and ψ(r = L) goes to zero, i.e. it has a node, at the boundaries of the box (stricktly valid only for V (0, L) ). This boundary condition leads to the standing-wave solutions (cf. Appendix 1): ψ(x, y, z) = 8 L 3 1/2 πx sin(n 1 L ) sin(n πy 2 L ) sin(n πz 3 L ) where k x,y,z are the wavevectors of the standing waves k x,y,z = n 1,2,3 π L n 1,2,3 = 1, 2... Each allowed k-state can accommodate two electrons with magnetic spin quantum numbers m s = ± 1 2. 5
Figure 2: Dispersion relation for the free electron gas Substituting the wavefunction in Schrödinger s equation: h2 d 2 Hψ(x, y, z) = εψ(x, y, z) dx + d2 2 dy + d2 2 dz 2 ψ(x, y, z) = εψ(x, y, z) gives for the energy levels of an electron in a 3D potential box: ε(k) = h2 ( k 2 x + k 2 y + k 2 z ) h 2 k 2 The function ε(k) is called the dispersion relation. For the free electron gas ε is a parabolic function of the wavevector k. The highest occupied states at T = 0K have the Fermi energy ε = ε F = h2 k 2 F and their wavevectors terminate on the surface of the so called Fermi sphere with radius equal to the Fermi wavevector k = k F. 6
Figure 3: Figure 4: 7
Density of states of the free electron gas: number of allowed electron states per energy interval For macroscopic samples with large size, L is a large number, therefore the separation in k-space between points corresponding to allowed states becomes infinitesimal. The allowed physically distinct standing-wave solutions of the Schrödinger equation for a 3D potential box occupy the positive octant in k-space (since n i > 0 and hence k i > 0). The number of allowed electron states with k-values between k and k + dk is equal to the volume of the spherical shell with radius k and thickness dk in the positive octant divided by the volume in k-space, which corresponds to a single k-point, multiplied by the spin degeneracy g s = 2. The volume corresponding to a single k-point equals ( π ) 3 L. g(k)dk = 2 1 8 π 3 L 3 4πk 2 dk = V k2 π dk 2 g(k)dk = V π 2 h 2 ε(k)dk where V = L 3 is the volume of the box. The density of electron states as a function of energy g(ε) is given by: g(ε) = g(k) dk dε 8
Figure 5: dk dε = 1 h m e 2 ε 1/2 g(ε) = V π 2 h 2 ε 1 h m e 2 ε 1/2 g(ε) = V 2π 2 h 2 3/2 ε 1/2 At T 0K N electrons fill the N lowest states up to energy ε = ε F : N = ε F 0 g(ε)dε = V 3π 2 h 2 N = 2 3 ε Fg(ε F ) 3/2 ε 3/2 F 9
The Fermi energy is determined by the electron density N/V : ε F = h2 3π 2 N V 2/3. The Fermi wavevector also depends on the electron density N/V : k F = h 2 ε F 1/2 = 3π 2 N V 1/3. 10
Fermi-Dirac distribution function The average value of the occupancy of an electron state of energy ε: n(ε) at temperature T is given by the Fermi-Dirac distribution function f(ε): f(ε) n(ε) = 1 exp[(ε µ)/k B T ] + 1 µ is the chemical potential of an electron. At temperature T 0K it coincides with the Fermi energy: µ(t = 0K) ε F The chemical potential corresponds to the energy at which f(ε) = 1 2. At temperature T 0K f(ε) is a step function. For ε below µ: e (ε µ)/k BT = For ε above µ: 1 exp[(µ ε)/k B T ] 0, hence f(ε) 1 e (ε µ)/k BT, hence f(ε) 0 For the case where the total number of electrons in a 3D solid is constant, the chemical potential must decrease with increasing temperature in order to compensate for the broadening of the distribution. 11
Figure 6: Appendix 1: Normalization of the wavefunction of an electron in a 3D box: Boundary conditions: Hψ(x, y, z) = εψ(x, y, z) H = ˆp2 ˆp x = i h d dx h2 2 ψ(x, y, z) = εψ(x, y, z) ψ(0, 0, 0) = 0 ψ(l, 0, 0) = 0 ψ(0, L, 0) = 0 ψ(0, 0, L) = 0, etc. Trial wavefunction: ψ(x, y, z) = A 3 sin(k x x) sin(k y y) sin(k z z) k x L = n 1 π n 1 = 1, 2,... k y L = n 2 π n 2 = 1, 2,... 12
k z L = n 3 π n 3 = 1, 2,... k x,y,z = n 1,2,3 π L Normalization of the wavefunction: ψ (x, y, z)ψ(x, y, z)dxdydz = 1 ψ (x, y, z)ψ(x, y, z)dxdydz = = L π 3 ψ (x, y, z)ψ(x, y, z)dxdydz = because A 6 ( π 0 sin2 θdθ ) 3 L π 3 ( L 0 A2 sin 2 ( π L x)dx) 3 A 6 ( π ) 3 L = 2 2 sin 2 xdx = 1 2 sinx cosx + x 2 + C 3 A 6 = 1 The normalization constant equals: A 3 = 8 L 3 1/2 13