QuantumPhysics.dvi

the integral for negative but smallλwould be well-defined and finite. So, the fact that the

integral is divergent for any negative value ofλ, no matter how small, forces us to have a

series expansion that must have zero radius of convergence.

On the other hand, changing the sign ofω^2 in the integral may change its value, but not

its convergence properties, no matter how largeω^2 is. This explains why the series expansion

on power ofω^2 can have infinite radius of convergence.

Another way of looking at this problem is that the non-convergenceof the smallλex-

pansion comes from the behavior of the integral atq→ ±∞. There, it is theq^4 term that

dominates, and so expanding in its strength is a singular thing to do. Generally, expanding

in the dominant behavior will lead to non-convergent expansions.

Finally, note that the integralI(ω,λ) is a perfectly fine function. In fact, it is given by

a modified Bessel function as follows,

I(ω,λ) =

ω

√

λ

e

ω 4 λ^4

K (^14)
(

ω^4

4 λ

)

(10.14)

Looking up the asymptotics of this function in a standard reference (Bateman Vol II, page

86, formula (7)), we recover the asymptotic expansion derived above.

10.2 Non-degenerate perturbation theory

The basic assumption is that the energy level which we want to studyof the unperturbed

HamiltonianH 0 is non-degenerate (other energy levels may or may not be degenerate). We

also assume that the energy and the state have an expansion of the type (10.3). Thus, we

need to solve the following problem,

(H 0 +λH 1 )|En〉= (En^0 + ∆n)|En〉 (10.15)

where we use the abbreviation ∆n=En−En^0 =λEn^1 +λ^2 En^2 +O(λ^3 ). Contracting both

sides of the first equation with the unperturbed state〈En^0 |, we find,

〈E^0 n|(H 0 +λH 1 )|En〉=〈En^0 |(En^0 + ∆n)|En〉 (10.16)

Using now the self-adjointness ofH 0 , we see that the first term on the left of the equations

cancels with the first term on the right. The remaining terms give us an expression for ∆n,

∆n=λ

〈En^0 |H 1 |En〉

〈E^0 n|En〉

(10.17)

We learn immediately from this that the calculation of the energy correction to a given order

requires the correction to the state to one lesser order. In fact, it is immediate to find the

first order energy correction,

En^1 =〈En^0 |H 1 |En^0 〉 (10.18)