Note that the eigenstates that are being determined by this eigenvalue equation are the ones to
lowest order. They must be eigenvectors of the HamiltonianP 0 H 1 P 0. This is precisely the setting
that we had encountered already when dealing with the Stark effect to this order.
Next, we derive the corrections to the states to orderλ. The simplest part is given by (10.52),
and to orderO(λ) results in,
P 1
(
|E^0 d;i〉+λ|E^1 d,i〉)
=λ(
1
Ed^0 −H 0)H 1P 1 H 1 P 0 |E^0 d;i〉 (10.56)By construction,P 1 |Ed^0 ;i〉= 0, so that the remaining equation gives,
P 1 |Ed,i^1 〉=(
1
E^0 d−H 0)H 1P 1 H 1 P 0 |Ed^0 ;i〉 (10.57)The solution forP 0 |Ed,i^1 〉needs to be handled at the same time as the second order correction to
the energy, which we proceed to do now.
10.8.2 Solution to second order
DeterminingP 0 |Ed,i^1 〉proceeds analogously from (10.58), which we approximate upto orderλ^2
included. Retaining onlyO(λ^2 ) contributions on the right hand side gives,
(
Ed,i−Ed^0 −λP 0 H 1)
P 0 |Ed,i〉=λ^2 P 0 H 1 P 1(
1
E^0 d−H 0)H 1P 1 H 1 P 0 |Ed^0 ;i〉 (10.58)Expanding the left hand side to this order gives,
(
Ed,i−Ed^0 −λP 0 H 1
)
P 0 |Ed,i〉 = λ(Ed,i^1 −P 0 H 1 )P 0 |Ed^0 ;i〉+λ^2 E^2 d,i|Ed^0 ;i〉+λ^2(
Ed,i^1 −P 0 H 1)
P 0 |E^1 d,i〉 (10.59)The first term on the right hand side vanishes in view of our results forEd,i^1 and|E^0 d;i〉. The other
terms are now all of the same order,λ^2 , as were the terms on the right hand side of (10.58). This
gives the final equation,
Ed,i^2 |E^0 d;i〉+(
Ed,i^1 −P 0 H 1)
P 0 |E^1 d,i〉=P 0 H 1 P 1(
1
E^0 d−H 0)H 1P 1 H 1 P 0 |Ed^0 ;i〉 (10.60)We first determineE^2 d,iby taking the product of the above equation with〈Ed^0 ;i|. The second term
on the right hand side cancels out in this process in view of (10.54), and we have
Ed,i^2 =〈Ed^0 ;i|P 0 H 1 P 1(
1
Ed^0 −H 0)H 1P 1 H 1 P 0 |E^0 d;i〉 (10.61)