This formula may be rendered more explicit by introducing a complete set of orthonormalized
eigenstates ofH 0 belonging toH 1 , which we denote by|Ek^0 〉, and we have
Ed,i^2 =−
∑
k
|〈E^0 d;i|H 1 |Ek^0 〉|^2
Ek−Ed^0
(10.62)
OnceE^2 d,iis known, the equation forP 0 |Ed,i^1 〉may be written as,
(
Ed,i^1 −P 0 H 1
)
P 0 |Ed,i^1 〉=Ed,i^2 |Ed^0 ;i〉+P 0 H 1 P 1
(
1
E^0 d−H 0
)
H 1
P 1 H 1 P 0 |Ed^0 ;i〉 (10.63)
By construction, this equation is now orthogonal to〈Ed,i^1 |.
If the eigenvalueEd,i^2 isnon-degeneratewith any of the other eigenvaluesE^1 d,j, forj 6 =i, then
the operator on the left hand side may be inverted on the subspace ofH 0 which is orthogonal to
|Ed,i^1 〉, and we get
P 0 |Ed,i^1 〉=
∑N
j 6 =i
|Ed,j^1 〉
1
E^1 d,i−Ed,j^1
〈E^1 d,j|P 0 H 1 P 1
(
1
Ed^0 −H 0
)
H 1
P 1 H 1 P 0 |E^0 d;i〉 (10.64)
Introducing again a complete set of states|Ek〉forH 1 , we may recast this formula as follows,
P 0 |E^1 d,i〉=
∑N
j 6 =i
|Ed,j^1 〉
1
E^1 d,i−Ed,j^1
∑
k
〈E^1 d,j|H 1 |Ek〉〈Ek|H 1 |E^0 d;i〉
E^0 d−Ek
(10.65)
If, on the other hand, the energy levelE^1 d,iremains degenerate, then the above formula cannot be
applied, since some of the denominators inEd,i^1 −E^1 d,j. When this happens, one needs to rearrange
the corresponding degenerateorderλ^0 states so as to be eigenstates of the second order perturbation
P 0 H 1 P 1
(
1
Ed^0 −H 0
)
H 1
P 1 H 1 P 0 (10.66)
as well. We shall not work out this case explicitly here.
10.9 Periodic potentials and the formation of band structure
The electronic structure in crystals is organized in bands. Empty or completely filled bands are
electronically inactive, while partially filled band allowfor electronic conductivity, providing a
quantitative explanation for the fundamental distinctionbetween insulators and conductors. Here,
we shall derive the basic mechanism responsible for the formation of bands.
For simplicity, we concentrate on one-dimensional systems. The crystalline structure translates
into a periodic potential for the electron, giving rise to a Hamiltonian of the form,
H=
p^2
2 m
+V(x) V(x+a) =V(x) (10.67)