Question

2. This problem will help you understand how two atoms can form a molecule through the process of chemical bonding. The physics behind the chemical bonding is very much the same as that discussed in energy splitting process to form energy bands in a macroscopic matter state where we have a lot of atoms involved but all atoms are nicely arranged to form a kind of periodic structure. In this problem, let's make things even simpler: we only consider two atoms in action. Each atom has its own Hamiltonian H and H, and their own eigen-energy and eigen-wavefunction as follows: HV (2) = EVD), H2V2+) = E,V2() where V.) and 2) are both normalized and orthogonal to each other. Now we start to move these two atoms from far away (no interaction between the two atoms and thus we have two free atoms with the above independent wave function and energy structure) closer to each other. Due to the Coulomb interactions between the electrons of the two atoms, the interaction between the two nuclei, and between the electrons of one atom and the nucleus of the other atom, you can imagine that the energy structure of the two atoms will have to change. To simplify the matter significantly, we will assume the total interaction between the two atoms are modeled as an operator V and the interactions between the two atoms will cause the wave function of electrons to be the linear combination of the two eingen-wave function of the independent atoms: () = C4 (2) + C242) and this wavefunction will satisfy the total Hamiltonian û = û + H2+, i.e., HV) = EVT). (a) Please derive the equation to determine the eigenenergy E; (b) Solve the equation to determine the eigenenergy E for the molecule, i.e., the entity of two atoms combined; (c) Determine the wavefunction for the ground state of the molecule and the first excited state of the molecule. Tips: You can assume the the integral &41) () = Vo = f (T) 42() and SPV)) = t = SP1V ().

Answer 1

Four Simplicity I use Hi(2) 11) and 42 (2) > 12) Vo = <li Û11) = (21Û12) Now <11012) = {211112=t (a) 14) = Cill) +(212) How Ĥ 147 = E14) How use deffinate of complete set into Hamiltonian = Elixir HŞ 1)231147 = 514) => [ LilH;) ¢j = Edi Hij = Lilni) and a j = Lj14) and ai = 2:14) then 1 Ç Hij Q; = Edi Now Now HII = LIHT) = E + Vo Hic = H21 = H22 {11H12) = <11012) =t {21411) = ľ = <21H12) = Eo tv

Now matrix will be I Eo tvo Eo EVO (so tvo-> ]? - HIP = 0 =) (to tvo – +It! ) (50 + vo-1-1+1) = 0 => [a= to two titl I a Eigen value I Eotvo-ltl da = @ it to the ground state energy will be lower energy that is de = Eo two-it And first excited state have energy di = Eo tvo till Now fon Eigen vector for de is 1+2) = ta (11)+12)) bonding - (Ground state ) fon do is HID = tz (11)-12)) (anti bonding ) anti = R12 separation between two atom sepole bond