Learning Objective To Calculate Band Structure in 1D Using the Kronig-Penney Model and Deduce Effective Mass at a Band Minimum Due Tuesday, October 5 Infinite String of Wells We know that each solution has two quantum numbers, energy E and crystal momentum k We want to get the wavefunction The wavefunction and its slope must be continuous at all the boundaries – This will tell us how E must depend on k The wavefunction can’t blow up – This rules out certain ranges of E values 0 a 2a 3a-2a -a E,k w-z 0 V How NOT to Do This E,k 0 V A1eαx +B1e αx Each period has four unknowns An, Bn, Cn, Dn and there are and infinite number of periods A0eαx +B0e αx C0 cosβx +D0 sinβx Let’s Be Clever and Save Work E,k 0 V 1st period 2nd period y1 y2 1. Write an expression for the wavefunction y1 in the first unit cell that already meets boundary conditions at the internal boundary 2. Use Bloch’s Theorem to get an expression for y2 3. Demand continuity of wavefunction and slope at the boundary between the two unit cells 1st Period Wavefunction E,k 0 V coshαx +Dsinhαx Barrier: y1: cosβx +D α β sinβx Well: 0 w-z α = 2m(V E) β = 2mE y and y’ already continuous at x=0. Only one unknown coefficient, D Bloch’s Theorem Restated E,k 0 V 0 w-z ψ(x) = eikxu(x) u(x) = u(x + a) x x+a ψ(x + a) = eik (x+a)u(x) = eikaψ(x) y(x) y(x+a) Wavefunction in Barrier of 2nd Period E,k 0 V coshαx +Dsinhαx Barrier: y: cosβx +Dα β sinβx Well: 0 w-z x-a x ψ(x + a) = eikaψ(x)Bloch’s Theorem: OR: ψ(x) = eikaψ(x a) eika coshα(x a) +Dsinhα(x a) ” # $ % & ‘ Demand Continuity of y and y’ at x=w E,k 0 V coshαx +Dsinhαx Barrier: y: cosβx +Dα β sinβx Well: 0 w-z = w-a eika coshα(x a) +Dsinhα(x a) ” # $ % & ‘ cosβw+Dα β sinβw = eika coshαz Dsinhαz( ) β sinβw+Dα cosβw = eikaα sinhαz+Dcoshαz( ) Solve Each of These Two Expressions for D and Equate Them: cosβwcoshαz+ 12 ( α β β α )sinβwsinhαz = coska α = 2m(V E) β = 2mE f (E) = coskaAbbreviate left hand side as a function f of energy E: This is the E vs k relation or the “band structure” Plot of f(E) f (E) = coska This equation tells us for each E, what’s the corresponding k. BUT k has to be real, otherwise the wavefunction blows up. If coska is in the range -1 to +1, k is real,and the corresponding E is allowed. Otherwise, it’s forbidden (in a band gap) Energy bands in yellow cosβwcoshαz+ 12 ( α β β α )sinβwsinhαz = coska α = 2m(V E) β = 2mE The above is the equation from which the energy bands E(k) are calculated for a string of rectangular wells Let the lattice constant a be 0.4 nm with equal width wells and barriers, w=0.2 nm and z=0.2 nm. Let the barrier height be 2 eV (a) Using Matlab, Mathematica, or your favorite tool, generate an E vs k plot for the lowest three bands. (What you actually do is plug in a value of E and solve for k) E(k) should look something like what is shown on the next page. Be sure to label axes: E in eV and k in m-1 (b) Look more closely at the band minimum in the red oval. To first lowest order E(k) is parabolic: Plot (kb – k)2 vs E from E = Ec to a slightly higher energy, and deduce the effective mass m* from the slope of the plot. I get 0.123 me. E Ec = !2 2m* (kb k) 2 E vs k 2.677 eV Ec = 3.948 eV 10.315 eV 10.420 eV 21.820 eV kb=p/a
