17.3. PERTURBATIONTHEORYINMATRIXNOTATION 273
isveryeasy. Theeigenvectorsare
%φ(0) 1 =
1
0
0
.
.
, %φ(0) 2 =
0
1
0
.
.
, φ%(0) 3 =
0
0
1
.
.
... (17.55)
andtheeigenvaluesarethediagonalelementsofthematrix.
Sonowtheproblemistofindtheeigenvectorsofthe(non-diagonal)matrix
H=H 0 +V′=
E 1 (0)+V 11 ′ V 12 ′ V 13 ′ ...
V 21 ′ E 2 (0)+V 22 ′ V 23 ′ ...
V 31 ′ V 32 ′ E 3 (0)+V 33 ′ ...
... ...
... ...
(17.56)
where
Hij=〈φ(0)i |H 0 +V′|φ(0)j >=Ejδij+Vij′ (17.57)Butthisproblemisalreadysolved! Tofirstorder(inV′)fortheeigenvectors,and
secondorderintheeigenvalues,thesolutionis
%φn = φ%(0)n +
∑i(=nVin′
En(0)−E(0)i%φ(0)i +O(V′^2 )En = En^0 +Vnn′ +∑i(=n∣∣
∣Vin′∣∣
∣2En(0)−E(0)i+O(V′^3 ) (17.58)
Youcanseethatwecanapplytheseequationstofindingtheeigenstatesandeigen-
valuesofanymatrixoftheform
M=M 0 +Q (17.59)whereM 0 isadiagonalmatrix(Mij=miδij),andQisa“perturbation”matrixsuch
that
|Qij| 5 |mi−mj| , |Qii| 5 |mi| (17.60)Itsjustamatterofchangingnotation(replaceH 0 byM 0 ,E(0)n bymn,etc.).
Gettingbacktotheoriginalproblem,supposewehavefoundtheeigenstates{%φn}
frombytheperturbativemethod(eq.(17.58)),orbysomeothermethod,andthen
normalizedthestatessothat
φ%n·φ%n= 1 (17.61)