QuantumPhysics.dvi

We will now use these projectors to decompose the eigenvalueequation (10.47). Inserting the
identity in front of the state gives,

(Ed,i−Ed^0 −λH 1 )P 0 |Ed,i〉+ (Ed,i−H 0 −λH 1 )P 1 |Ed,i〉= 0 (10.50)

Here, we have used the fact that all states inP 0 |Ed,i〉have eigenvalueEd^0 underH 0. Our next step
is to project this equation withP 0 andP 1 , and we obtain,

P 0

(

Ed,i−E^0 d−λP 0 H 1

)

P 0 |Ed,i〉=λP 0 H 1 P 1 |Ed,i〉

P 1

(

Ed,i−H 0 −λP 1 H 1

)

P 1 |Ed,i〉=λP 1 H 1 P 0 |Ed,i〉 (10.51)

We shall deduce from these the equations that yieldP 0 |Ed,i〉andP 1 |Ed,i〉.

The operatorP 1 (Ed,i−H 0 −λH 1 )P 1 is invertible onH 1 providedλis sufficiently small. The
reason is that forλ= 0, it reduces to the operatorP 1 (Ed^0 −H 0 )P 1 which is by construction invertible
onH 1. Invertibility is anopen condition, which means that in the space of all self-adjoint operators,
any operator in a sufficiently small neighborhood of the invertible operator will also be invertible.
Concretely, a finite-dimensional matrixAis invertible if detA 6 = 0. But this means that if we
consider a family of operatorsA+λBfor sufficiently smallλ, we will still have det(A+λB) 6 = 0.
Thus, perturbatively speaking, the operatorP 1 (Ed,i−H 0 −λH 1 )P 1 will be invertible onH 1. As a
result, we have

P 1 |Ed,i〉=λ

(

1

Ed,i−H 0 −λP 1 H 1

)

H 1

P 1 H 1 P 0 |Ed,i〉 (10.52)

Here the subscriptH 1 stands for the inverse operator restricted to the subspaceH 1. It remains to
determineP 0 |Ed,i〉and the corresponding energiesEd,i. To do so, we substitute the expression for
P 0 |Ed,i〉, found above, into equationP 0 , and we find,

(

Ed,i−Ed^0 −λP 0 H 1

)

P 0 |Ed,i〉=λ^2 P 0 H 1 P 1

(

1

Ed,i−H 0 −λP 1 H 1

)

H 1

P 1 H 1 P 0 |Ed,i〉 (10.53)

To orderO(λ^0 ), this equation is satisfied trivially.

10.8.1 Solution to first order

To orderO(λ), the right hand side may be dropped, and the energy may be approximated by
Ed,i−Ed^0 =λE^1 d,i+O(λ^2 ). The resulting equation becomes,

(

Ed,i^1 −P 0 H 1 P 0

)

|E^0 d;i〉= 0 (10.54)

The energy eigenvalues are determined by the eigenvalues ofthe operatorP 0 H 1 P 0 restricted toH 0 ,
and are solutions of the characteristic equation,

det

(

Ed,i^1 −P 0 H 1 P 0

)

H 0

= 0 (10.55)