Here are my constants:
Well width: a = 2b = 1 Angstrom
Well Depth: V0 = 250 eV
Well Spacing (from one center to another): c = 1.75 Angstroms (chosen because it causes essentially no difference in E1, but a very large difference in E3)
E1:
E1 = -226 eV , k = 2.51 Angstroms^(-1) , κ = 7.7 Angstroms^(-1)
D = 1.26 Angstroms^(-1/2), B = 0.39 Angstroms^(-1/2)
E3:
E3 = -50 eV , k = 7.24 Angstroms^(-1) , κ = 3.62 Angstroms^(-1)
D = 1.14 Angstroms^(-1/2), B = -1.01 Angstroms^(-1/2)
E1q=−226+0.46cos(1.75q)
E3q=−50+32cos(1.75q)
Here are the steps to get here and general equations for all states, in case I got anything wrong:
Following from previous discussions we have
Eq=E0+2cos(qc)Im11
Im11=<ϕ0(x)|V(x−c)|ϕ0(x−c)>
<ϕ0(x)|V(x−c)|ϕ0(x−c)>=∫c+bc−bϕ∗0(x)V0(x−c)ϕ0(x−c)dx
=V0∫c+bc−bϕ∗0(x)ϕ0(x−c)dx
ϕ0(x) between c-b and c+b is the exponentially falling tail of the wave function centralized at x = 0, and is the same for even and odd functions.
ϕ0(x−c) between c-b and c+b is the sinusoidal wave function centered at x=c, and varies from even to odd.
=V0∫c+bc−be−κxϕ0(x−c)dx
odd =V0D∫c+bc−be−κxsin(k(x−c))dx
even =V0D∫c+bc−be−κxcos(κ(x−c))dx
odd V0DAe−κck2+κ2[k(e2bκ−1)cos(bk)−κ(e2bκ+1)sin(bk)]
even V0DBe−κck2+κ2[κ(e2bκ−1)cos(bk)+k(e2bκ+1)sin(bk)]
odd Eq=E0+V0DAe−κccos(qc)k2+κ2[k(e2bκ−1)cos(bk)−κ(e2bκ+1)sin(bk)]
even Eq=E0+V0DBe−κccos(qc)k2+κ2[κ(e2bκ−1)cos(bk)+k(e2bκ+1)sin(bk)]
E1q=−226+327,050e−7.7ccos(cq)
E3q=−50+17,885e−3.62ccos(cq)
Does this look reasonable?
Very nice post! Let's discuss and critique these results here in these comments! I would like to see comments from pretty much everyone!
ReplyDeleteThis all looks great.
ReplyDeleteIs the fact that the band width goes to 0 as c goes to infinity something to do with when you have them infinitely far apart it is as if you are simply looking at one by itself? Therefore Eq=E0.
Also, I am still slightly uncertain what q represents. C