QuantumPhysics.dvi

assuming that the state|E^0 n〉is normalized. The equation determining the first order cor-

rection to the state are obtained by expanding (10.15) to orderλ, and are given as follows,

(H 0 −E^0 n)|En^1 〉= (En^1 −H 1 )|En^0 〉 (10.19)

The operatorH 0 −E^0 nis not invertible, however, since it has a zero eigenvalue with eigenvector

|En^0 〉. To solve this equation, we now contract with an arbitrary state〈Em^0 |, and again use

the self-adjointness ofH 0 ,

(E^0 m−En^0 )〈Em^0 |En^1 〉=〈E^0 m|(En^1 −H 1 )|En^0 〉 (10.20)

Whenm=n, this equation is automatically satisfied, and thus yields no information on the

quantity〈En^0 |E^1 n〉. But all other matrix elements are uniquely determined and we have

〈E^0 m|E^1 n〉=−

〈E^0 m|H 1 |En^0 〉

E^0 m−En^0

m 6 =n (10.21)

This result allows us to write down the solution, using the completeness relation on all the

states|Em^0 〉, and we have

|En^1 〉=c|En^0 〉−

∑

m 6 =n

|Em^0 〉

〈E^0 m|H 1 |En^0 〉

E^0 m−En^0

(10.22)

The coefficientcis, at this time arbitrary and not determined by the equation. The correction

to the energy of second order inλmay be derived from by expanding (10.17) to second order

inλ, and we obtain,

En^2 =〈En^0 |H 1 |En^1 〉−cEn^1 (10.23)

Substituting in here the first order correction to the state, we obtain,

En^2 =−

∑

m 6 =n

|〈E^0 m|H 1 |En^0 〉|^2

Em^0 −En^0

(10.24)

Here, we have used the self-adjointness ofH 1 to relate〈Em^0 |H 1 |E^0 n〉=〈En^0 |H 1 |Em^0 〉∗.

An important consequence of this formula is thatthe second order correction to the

ground state energy is always negative, since for the ground state we haveEm^0 −En^0 >0 for

allm 6 =n, and numerator is manifestly positive.