Question

2. Derivation of the Kronig-Penney result: Write the solution for the wave function in the region 0 < r < a as a linear combi nation of sin and cos functions and in the region-b< 0 as a linear combination of the hvperbolic sin and cos functions: ψ(z) Asin Kr +Bcos Kx (0<x <a) = By invoking the continuity and differentiability conditions at a- 0, show that B D and AK - CQ, so essentially two constants, say A and B survive. Now use the Bloch's theorem: d p da to connect the wave functions and their derivatives in the two regions, thus giving us two equations involving two unknowns A and B. From the determinantal condition for the existence of nontrivial solutions for A and B, show Q2 -K2 2KOsin Ka sinh b) + cos Kosh bcos k(a+ b) (Note: If x = eika, then x + x-1-2 cos ka) (10 marks

Answer 1

Above both, the equations have a non-trivial solution.

Energies and wave functions for electrons in crystal (periodic potential V(x))

a) Isolated Potential well

- only discrete energies E allowed

- wave functions are standing waves

b) Free electrons

- Plan wave solutions :

ψ(x) = Ce^jkx

Ψ(x,t) = Ce^j(kx-ωt)

- Any energy allowed : E = h²k²/8pi² m

- Parabolic E versus k