This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1 ECE 407 Spring 2009 Farhan Rana Cornell University Handout 9 Application of LCAO to Energy Bands in Solids and the Tight Binding Method In this lecture you will learn: An approach to energy bands in solids using LCAO and the tight binding method Energy k a a s E ss V 4 ECE 407 Spring 2009 Farhan Rana Cornell University Example: A 1D Crystal with 1 Orbital per Primitive Cell Consider a 1D lattice of atoms: a x Each atom has the energy levels as shown The electrons in the lowest energy level(s) are well localized and do not take part in bonding with neighboring atoms The electrons in the outermost sorbital participate in bonding r ( ) r V a r Energy levels s E The crystal has the Hamiltonian: ( ) + = m m a R r V m H r r h 2 2 2 1 a m R m r r = 1 a r ( ) r s r x ( ) r V r Potential in a crystal a 2 ECE 407 Spring 2009 Farhan Rana Cornell University Tight Binding Approach for a 1D Crystal ( ) + = m m a R r V m H r r h 2 2 2 a x 1 a r We assume that the solution is of the LCAO form: ( ) ( ) = m m s m R r c r r r r If we have N atoms in the lattice, then our solution is made up of N different s orbitals that are sitting on the N atoms In principle one can take the assumed solution, as written above, plug it in the Schrodinger equation, get an N x N matrix and solve it (just as we did in the case of molecules). But one can do better .. Periodic potential We know from Blochs theorem that the solution must satisfy the following: ( ) ( ) r e R r R k i r r r r r . = + ( ) ( ) 2 2 r R r r r r = + 1 a m R m r r = ECE 407 Spring 2009 Farhan Rana Cornell University a x 1 a r For the solution: to satisfy: one must have the same value of for all m (i.e. all coefficients must have the same weight). So we can write without loosing generality: 2 m c N e c m i m = Consideration 1: Consideration 2: ( ) ( ) = m m s i R r N e r m r r r For the solution: to satisfy: ( ) 1 3 2 = r d r r r one must have the phase value equal to: m m R k r r . = Tight Binding Approach for a 1D Crystal 1 a m R m r r = ( ) ( ) = m m s m R r c r...
View
Full
Document
This note was uploaded on 04/04/2009 for the course ECE 4070 taught by Professor Rana during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 RANA

Click to edit the document details