# Homework Help: Desnity matrix method to calculate polarizability of H2 molecule.

1. Oct 8, 2011

### jameson2

1. The problem statement, all variables and given/known data
Have a H2 molecule in a constant E field. This produces a shift in the on site energies of the 2 H atoms so they have energies ε-δ,ε+δ. δ is proportional to field strength.
1)Calculate the tight binding density matrix when the hopping integral is -1. (γ=-1) δ is positive and the molecule contains 2 electrons.
2)Show the trace is 2, independent of δ.
3)Calculate the density matrix for vanishing electrix field, and compare to the previous matrix for δ→0
4)Polarizability is the derivative of the electrical dipole w.r.t. the electric field, at E=0. Given the electrical dipole operator of the H2 molecule P=1/2[|1><1|-|2><2|], where |j><j| is the projector over the 1s atomic orbital centered on the jth atom, and using the previous density matrix, calculate the polarizability of the H2 molecule. (Assume E=cδ, with c just a constant.)

2. Relevant equations
Some included below.

3. The attempt at a solution
I think I've everything right up to the last part. I don't really want to put up the whole density matrix as it's a bit messy and long, but I'm fairly sure I have it right since I get trace 2 and it agrees with the δ=0 case.
For the last part I'm wondering whether for |1> and |2> I use (1,0) and (0,1), or if I use the vectors I got when I solved Hψ=Eψ, with the H matrix entries being ε+δ, -1, -1, ε-δ. I tried calculating the polarizability both ways, by finding P, then finding trace(Pρ), then differentiating this w.r.t. E then setting E to zero. Using the (1,0),(0,1) vectors I get an answer of zero which I presume isn't right. Using the second set of vectors I get a final answer of -4/c. Of course I may just be using a completely wrong method for the last part, I'm not too sure. Does it sound like I'm on the right track?

Oh, and in part 2, what does trace=2 mean? I'm guessing that this is the number of electrons, but I couldn't find why.

Thanks alot for any help.