Handout 8 Linear Combination of Atomic Orbitals (LCAO) In this lecture you will learn: H • An approach to energy states in molecules based on the linear combination of atomic orbitals H C H H ECE 407 – Spring 2009 – Farhan Rana – Cornell University Energy Bands and Atomic Potentials in Crystals V r The potential energy of an electron due 0 to a single isolated atom looks like: Potential of an isolated Energy levels In a crystal, the potential energy due to atom all the atoms in the lattice looks like: x V r 0 Vacuum level 0 Energy Potential in bands a crystal x 0 The lowest energy levels and wavefunctions of electrons remain unchanged when going from an isolated atom to a crystal The higher energy levels (usually corresponding to the outermost atomic shell) get modified, and the corresponding wavefunctions are no longer localized at individual atoms but become spread over the entire crystal ECE 407 – Spring 2009 – Farhan Rana – Cornell University 1 Failure of the Nearly-Free-Electron Approach V r Vacuum level 0 3 2 Energy Potential in 1 bands a crystal x 0 • For energy bands that are higher in energy (e.g. 2 & 3 in the figure above) the periodic potential of the atoms can be taken as a small perturbation For higher energy bands, the nearly-free-electron approach works well and gives almost the correct results • For energy bands that are lower in energy (e.g. 1 in the figure above) the periodic potential of the atoms is a strong perturbation For lower energy bands, the nearly-free-electron approach does not usually work very well ECE 407 – Spring 2009 – Farhan Rana – Cornell University Nearly-Free-Electron Approach Vs LCAO for Germanium Energy Energy Empirical Energy NFA (eV) LCAO (eV) Pseudopotential (eV) 1 2 2 1 1 1 3 4 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 FBZ (FCC lattice) • For most semiconductors, the nearly-free- electron approach does not work very well • LCAO (or tight binding) works much better and provides additional insights ECE 407 – Spring 2009 – Farhan Rana – Cornell University 2 LCAO: From Hydrogen Atom to Hydrogen Molecule Consider a Hydrogen atom with one electron in the 1s orbital: V r One can solve the Schrodinger equation: 2 2 r V r r E r 2m Potential of 1s energy level a Hydrogen and find the energy of the 1s orbital and atom its wavefunction ˆ Ho 1s r E1s 1s r 0 r where: r 2 1s Hˆ 2 V r o 2m Angular probability 0 r distribution for the 1 1s orbital r er ao 1s 3 ao Radial amplitude for the 1s orbital ECE 407 – Spring 2009 – Farhan Rana – Cornell University Linear Combination of Atomic Orbitals (LCAO) V r Now consider a Hydrogen molecule made up of two covalently bonded a atom b atom Hydrogen atoms sitting at a distance of 2d from each other, as shown: Hamiltonian for an electron is: -d 0 d x 2 Hˆ 2 V r dxˆ V r dxˆ 2m The basic idea behind LCAO approach is to construct a trial variational solution in which the wavefunction is made up of a linear combination (or superposition) of orbitals of isolated atoms: r ca 1s r d xˆ cb 1s r dxˆ And then plugging the trial solution into the Schrodinger equation to find the coefficients ca and cb and the new eigenenergies: Hˆ r E r ECE 407 – Spring 2009 – Farhan Rana – Cornell University 3 LCAO: From Hydrogen Atom to Hydrogen Molecule Plug the LCAO solution: r ca 1s r d xˆ cb 1s r dxˆ 2 into: Hˆ r E r Hˆ 2 V r dxˆ V r dxˆ 2m STEP 1: take the bra of the equation first with 1 s r d x ˆ to get: ˆ 1s r d xˆ H ca 1s r d xˆ cb 1s r dxˆ E 1s r d xˆ ca 1s r d xˆ cb 1s r dxˆ Note that: ˆ V r 1s r dxˆ H 1s r dxˆ E1s Let: a atom b atom ˆ 1s r dxˆ H 1s r dxˆ Vss -d 0 d x 1s r dxˆ 1s r dxˆ 0 Not exactly zero – but we will assume so for simplicity ECE 407 – Spring 2009 – Farhan Rana – Cornell University LCAO: From Hydrogen Atom to Hydrogen Molecule So we get finally: E1s ca Vss cb E ca STEP 2: take the bra of the equation now with 1 s r d x ˆ to get: E1s cb Vss ca E cb Write the two equations obtained in matrix form: E1s Vss ca ca E Vss E1s cb cb This is now an eigenvalue equation and the two solutions are: E E1s Vss ca 1 1 ca 1 1 cb 2 1 cb 2 1 ECE 407 – Spring 2009 – Farhan Rana – Cornell University 4 Bonding and Anti-Bonding Orbitals V r For the lower energy solution we have: EA E EB E1s Vss B 1 r r d xˆ r dxˆ B 2 1s 1s -d 0 d x B r This is called the “Bonding molecular orbital” For the higher energy solution we have: -d 0 d x r EA E1s Vss A 1 r r d xˆ r dxˆ A 2 1s 1s -d 0 d x This is called the “Anti-bonding molecular orbital” ECE 407 – Spring 2009 – Farhan Rana – Cornell University LCAO: Energy Level Splitting and the Energy Matrix Element V r V r Anti-bonding 2V 1s atomic ss Energy levels Bonding energy level of the molecule 0 x -d 0 d x Energy level diagram going from two isolated atoms to the molecule: The total energy is 1: EA 2 : E1s lowered when 2Vss Hydrogen atoms form 1: EB a Hydrogen molecule The two 1s orbitals on each Hydrogen atom combine to generate two molecular orbitals – the bonding orbital and the anti-bonding orbital – with energy splitting related to the energy matrix element: ˆ 1s r dxˆ H 1s r dxˆ Vss ECE 407 – Spring 2009 – Farhan Rana – Cornell University 5 Atomic Orbitals • Wavefunction amplitudes of the atomic s s and p orbitals in the angular directions are plotted • The s-orbital is spherically symmetric • The p-orbitals have +ve and –ve lobes and are oriented along x-axis, y-axis, and z-axis pz px py ECE 407 – Spring 2009 – Farhan Rana – Cornell University Orbitals and Bonding There are two main types of co-valent bonds: sigma bonds (or -bonds) and pi-bonds (or -bonds) (1) Sigma bonds (or -bonds): s-s -bond ss (Example: Hydrogen molecule, semiconductors) ˆ s r r1 H s r r2 Vss r1 r2 p-p -bond pp (Example: Semiconductors) ˆ p r r1 H p r r2 Vpp r1 r2 s-p -bond p s (Example: Semiconductors) ˆ p r r1 H s r r2 Vsp r1 r2 p s s-p -bond (Example: Semiconductors) ˆ p r r1 H s r r2 Vsp r1 r2 ECE 407 – Spring 2009 – Farhan Rana – Cornell University 6 Orbitals and Bonding What about this situation? p r r Hˆ r r 0 s p 1 s 2 The Hamiltonian is up-down symmetric The s-orbital is up-down symmetric The p-orbital is up-down anti-symmetric The matrix element is zero! No bonding possible r1 r2 What about this situation? What should be the matrix element? p p s p s = sin + cos s r1 r2 r r 1 2 r1 r2 ˆ p r r1 H s r r2 0 . sin Vsp . cos Vsp cos ECE 407 – Spring 2009 – Farhan Rana – Cornell University Orbitals and Bonding (2) Pi bonds (or -bonds): p-p -bond (Example: graphene, carbon nanotubes, conjugated p p conducting molecules) ˆ p r r1 H p r r2 Vpp r1 r2 What about this situation? What should be the matrix element? p p p p p p sin + cos r r r r r1 r2 1 2 1 2 ˆ p r r1 H p r r2 Vpp . sin 0 . cos Vpp sin ECE 407 – Spring 2009 – Farhan Rana – Cornell University 7 LCAO: Methane Molecule A methane molecule consists of one carbon atom covalently bonded to 4 hydrogen atoms in a tetrahedral configuration: H Carbon: Atomic number: 6 Electron Configuration: 1s2 2s2 2p2 Number of electrons in the outermost shell: 4 C H H Hydrogen: Atomic number: 1 H Electron Configuration: 1s1 Number of electrons in the outermost shell: 1 Methane: H The four electrons from the outermost shell of the carbon atom and the four electrons from the four H C H hydrogen atoms take part in covalent bonding H ECE 407 – Spring 2009 – Farhan Rana – Cornell University LCAO: Methane Molecule • The carbon atom is sitting at the origin r1 r2 • The position vectors of the four hydrogen atoms are: d d d r1 1,1,1 r2 1,1,1 r3 1,1,1 r 3 3 3 r4 3 d r 1,1,1 Tetrahedral configuration 2 3 • The carbon atom has one 2s orbital and three 2p orbitals • Each hydrogen atom has one 1s orbital • One can write the solution for the methane molecule as a linear combination of all available orbitals 4 r c j 1s r r j c5 2s r c6 2px r c7 2py r c8 2pz r j 1 But we will pursue a different, and simpler, approach ECE 407 – Spring 2009 – Farhan Rana – Cornell University 8 LCAO: Methane Molecule – sp3 Hybridization For the carbon atom, do a change of basis and define 4 new sp3 atomic orbitals from the 4 existing (one 2s and three 2p) atomic orbitals r 1 1 r r r r r 1 2 2s 2px 2py 2pz 1 r r r r r 2 2 2s 2px 2py 2pz 1 r r r r r 3 2 2s 2px 2py 2pz 1 r r r r r 3 2 2s 2px 2py 2pz r 2r 3r 4 ECE 407 – Spring 2009 – Farhan Rana – Cornell University LCAO: Methane Molecule – sp3 Hybridization • Each carbon sp3 orbital forms a s-sp3 -bond with the 1s orbital of the hydrogen atoms towards which it is pointing: H s-sp3 Average energy of the sp3 orbital: -bond E 3E r Hˆ r E 2s 2p j m sp3 j m 4 j m H C Important matrix element for the s-sp3 bond: H ˆ H 1s r r1 H 1r ˆ 1 1s r r1 H 2s r 2px r 2py r 2pz r 2 z 1 1V 1V 1V 1s sp sp sp 2pz Vss 2 2 3 2 3 2 3 x r1 Vss 3Vsp 2px 2 2py y ECE 407 – Spring 2009 – Farhan Rana – Cornell University 9 LCAO: Methane Molecule – sp3 Hybridization • Matrix elements for all s-sp3 -bonds are the same H s-sp3 Important matrix elements: -bond V 3V r r Hˆ r ss sp 1s 1 1 2 H C V 3V r r Hˆ r ss sp H 1s 2 2 2 H V 3V r r Hˆ r ss sp 1s 3 3 2 V 3V r r Hˆ r ss sp 1s 4 4 2 ECE 407 – Spring 2009 – Farhan Rana – Cornell University LCAO: Methane Molecule – sp3 Hybridization Write the solution for the methane molecule using