PH575 Spring 2019 Lecture #10 Density of states Sutton Ch. 3 p 55, Ch.4 pp 88-92 2.0 α = 1 α = 1 β = -0.3 β = -0.3 1.5 N = 10 N = 10 ) ) k k 1.0 E( E( 0.5 0.0 -1 -0.5 0 0.5 1 4 5 6 7 8 9 10 k (units of π/a) DOS MgO - an ionic crystal Density of states Dispersion relation band width band Fermi gap energy band width www.crystal.unito.it/tutojan2004/tutorials/G_properties/ img47.gif&imgrefurl=http://www.crystal.unito.it/tutojan2004/ tutorials/G_properties/properties_tut.html Si - a covalent crystal Density of states Dispersion relation Large DoS No DoS Small DoS www.crystal.unito.it/tutojan2004/tutorials/G_properties/ 3 img47.gif&imgrefurl=http://www.crystal.unito.it/tutojan2004/ tutorials/G_properties/properties_tut.html 2.0 Density of states: α = 1 β = -0.3 Very important concept in 1.5 N = 10 solid state systems. ) Properties of a system k 1.0 heavily influenced by the E( number of states per unit 0.5 energy interval. e.g. absorption of light, 0.0 tunneling current, -1 -0.5 0 0.5 1 k (units of π/a) How densely are states packed along k axis? Spacing is uniform in k! dS = number of states in a small interval Dk (k)dk = dS N L D(k) = density of states in k space – or Dk (k) = = number of states per unit k. It is a constant 2π / a 2π (independent of k) because the states are evenly spaced in k space 2.0 α = 1 β = -0.3 Define the density of states (per unit energy) as 1.5 N = 10 ) D(E), sometimes written as DE(E) to remind us k 1.0 E( that it's a number per unit energy. 0.5 0.0 -1 -0.5 0 0.5 1 k (units of π/a) dS # states in energy interval dE D (E) = = E dE energy interval dE dS dk 1 DE (E) = 2! = 2D(k) ±k dk dE dE dk From dispersion relation E(k) Uniform Na/2π We know density in k-space and we know E(k)! Note – we have not considered spin of electron. E k 2 cos ka 1 ( ) = α + β ( ) D(E)(eV- ) E −α 20 cos(ka) = 2β N =10 β =-0.3 dE = 2βasin(ka) α =1 dk 2 2 = a 4β − (E −α ) E (eV) 0.4 1 1.6 L D (k) = k 2π dk N 1 D (E) = D k = E k ( ) 2 dE π 4β 2 − (E −α ) If we include the spin of the electron, then for every k, there are 2 states - one for each spin, so we would multiply the above by 2. The plot of D(E) is above. SUMMARY – Linear, infinite, homonuclear chain Dispersion relation Density of States N 1 DE,nospin (E) = E(k) = α + 2β cos(ka) 2 2 π 4β − (E −α ) 2.0 α = 1 α = 1 β = -0.3 β = -0.3 1.5 N = 10 N = 10 ) ) k 1.0 k E( E( 0.5 0.0 -1 -0.5 0 0.5 1 4 5 6 7 8 9 10 k (units of π/a) DOS Bandwidth: 4β 2π m 2π k = = λ N a 2-D Density of states: ! dS = Dk (k) × area of k-space area = 2πkdk ! N N N N a2 D k = x y = x y 2 2 2 k ( ) 2π 2π 2 k = kx + ky (2π ) a a E0 E0 + dE dk DE (E)dE = Dk (k)2πkdk kx 2 N x N ya dk DE (E) = 2 2πk (2π ) dE 8 ky 2-D Density of states: 2 N x N ya dk Special case when E depends DE (E) = 2 2πk dE only on the magnitude of k, (2π ) e.g. when band is parabolic or free-particle-like: E E + dE 2 2 * 0 0 k 2m (E −α ) E(k) = α + * k = 2 2m dk d 2m* (E −α ) 1 2m* kx = = dE dE 2 2 2 E ( −α ) m*L2 DE (E) = 2π!2 9 ky 2-D Density of states for parabolic bands (no spin) m*L2 DE (E) = 2 2π! E E + dE 0 0 D(E)is constant! As E increases, the number of states at that energy increases in proportion! If spin is considered, we must kx multiply by 2. (cf Sutton 7.15 – careful of h and hbar) 10 ky 3-D Density of states: Rotate this in your imagination to get 3D picture! You extend 2D->3D (hwk). See Sutton Eqn 7.17 E0 E0 + dE FYI General case of non-circular constant energy surfaces kx 3 " a % dS(k) D(E0 ) = $ ' # 2π & ∫ ∇ E k E = E0 k ( ) 11 ky Si - a covalent crystal Density of states Dispersion relation Large DoS No DoS Small DoS www.crystal.unito.it/tutojan2004/tutorials/G_properties/ 12 img47.gif&imgrefurl=http://www.crystal.unito.it/tutojan2004/ tutorials/G_properties/properties_tut.html .