Handout 10 The Tight Binding Method Contd And Crystal Symmetries and Energy Bands In this lecture you will learn The tight binding method contd The bands in conjugated hydrocarbons The relationship between symmetries and energy bands ECE 407 Spring 2009 Farhan Rana Cornell University Tight Binding for a Square Lattice with a Two Atom Basis Consider a 2D square lattice with a two atom basis a2 a B d2 The primitive vectors and basis vectors are as follows a1 a x a d1 x 2 a1 d1 A a2 a y a d 2 y 2 Each basis atom contributes one s orbital that participates in bonding Each primitive cell contributes two s orbitals that participate in bonding a SA r ESA SB r ESB One can write the trial tight binding solution for wavevector k as k i k Rm r e cSA k e ik d1 SA r Rm d1 cSB k e ik d 2 SB r Rm d 2 N m ECE 407 Spring 2009 Farhan Rana Cornell University 1 Analysis of the Tight Binding Solution e i k Rm k r cSA k e ik d 1 SA r Rm d1 cSB k e ik d 2 SB r Rm d 2 N m 1 Summation over all primitive cells 3 Summation over all orbitals within a primitive cell with undetermined coefficients 2 Common phase factor for each primitive cell a2 a 4 A phase factor for each orbital that is related to the position of the orbital within the primitive cell w r t lattice point B d2 a1 d1 A a ECE 407 Spring 2009 Farhan Rana Cornell University Tight Binding Solution Plug the solution into the Schrodinger equation H k r E k k r And then one by one multiply by from the left by the bra s corresponding to every orbital in one primitive cell to generate as many equations as the number of orbitals per primitive cell a2 Step 1 a Multiply the equation with SA r d1 and keep the energy matrix elements for orbitals that are nearest neighbors and assume that the orbitals on different atoms are orthogonal ESA B d2 a1 d1 A a cSA k 4Vss cos k d1 cos k d 2 cSB k E k cSA k where the following identity has been used e i k d1 d 2 e i k d 1 d 2 e i k d 1 d 2 e i k d 1 d 2 4 cos k d1 cos k d 2 ECE 407 Spring 2009 Farhan Rana Cornell University 2 Tight Binding Solution Step 2 Multiply the equation with SB r d 2 and keep the energy matrix elements for orbitals that are nearest neighbors and assume that the orbitals on different atoms are a orthogonal a2 B d2 ESB cSB k 4Vss cos k d1 cos k d 2 cSA k E k cSB k a1 d1 A a Write the equations obtained in a matrix form ESA 4Vss cos k d1 cos k d 2 cSA k 4Vss cos k d1 cos k d 2 cSA k E k ESB cSB k cSB k ECE 407 Spring 2009 Farhan Rana Cornell University Tight Binding Solution a2 a a1 ESB ESA B d2 d1 A a FBZ ESA 4Vss cos k d1 cos k d 2 cSA k 4Vss cos k d1 cos k d 2 cSA k E k ESB cSB k cSB k ECE 407 Spring 2009 Farhan Rana Cornell University 3 Polyacetylene Polyacetylene is a one dimensional conducting hydrocarbon polymer H H C C C C H H y x Carbon atoms are all sp2 hybridized one 2s orbital together with the 2px and the 2py orbitals generate three sp2 orbitals Two sp2 orbitals form bonds with the sp2 orbitals of the neigboring carbon atoms and one remaining sp2 orbital forms a bond with the 1s orbital of the hydrogen atom The bonding orbital associated with each bond is occupied by two electrons spin up and spin down There is one electron per carbon atom left in the 2pz orbital ECE 407 Spring 2009 Farhan Rana Cornell University Bands in Polyacetylene H H 2pz orbitals x The 2pz orbital stick out of the plane of the chain and form bonds with neigboring 2pz orbitals The p bonding results in energy bands that we will study via tight binding The primitive cell of the 1D chain is as shown below it consists of two carbon atoms and two hydrogen atoms H H C C C C H H 0 a1 a x ECE 407 Spring 2009 Farhan Rana Cornell University 4 Bands in Polyacetylene H H C C B C C H H A a 1 x 0 a Two carbon atoms per primitive cell implies we have a 1D crystal with a two atom basis with basis vectors d1 0 a d 2 x 2 Each basis atom contributes one 2pz orbital that participates in bonding Each primitive cell contributes two 2pz orbitals that participate in bonding pzA r pzB r Ep Ep One can write the trial tight binding solution for wavevector k as k i k Rm r e c pzA k pzA r Rm c pzB k e ik d 2 pzB r Rm d 2 N m ECE 407 Spring 2009 Farhan Rana Cornell University Bands in Polyacetylene H H C C B C C H H A 0 a1 a Plug the solution into the Schrodinger equation H k r E k k r x And then one by one multiply by from the left by the bra s corresponding to every orbital in one primitive cell to generate as many equations as the number of orbitals per primitive cell Step 1 Multiply the equation with pzA r and keep the energy matrix elements for orbitals that are nearest neighbors and assume that the orbitals on different atoms are orthogonal E p c pzA k 2Vpp cos k d 2 c pzB k E k c pzA k ECE 407 Spring 2009 Farhan Rana Cornell University 5 Bands in Polyacetylene H H C C B C C H H A a1 0 a Step 2 Multiply the equation with pzB r and keep the energy matrix elements for orbitals that are nearest neighbors and assume that the orbitals on different atoms are orthogonal x E p c pzB k 2Vpp cos k d 2 c pzA k E k c pzB k Write the equations obtained in a matrix form Ep 2Vpp cos k d 2 2Vpp cos k d 2 Ep c pzA k c pzA k E k c pzB k c pzB k ECE 407 Spring 2009 Farhan Rana Cornell University Bands in Polyacetylene Ep 2Vpp cos k d 2 2Vpp cos k d 2 Ep c pzA k c pzA k E k c pzB k c pzB k Solutions are E k E p 2Vpp cos k d 2 Energy c pzA k 1 1 c k 2 1 pzB c pzA k 1 1 c k 2 1 pzB There is no bandgap between the upper and lower bands Since …