I'm reading the Cohen-Tannoudji book and I found somthing I dont understand(adsbygoogle = window.adsbygoogle || []).push({});

in stationary perturbation theory.

the problem the Hamiltonian is split in the known part an the perturbation:

[tex]

H=H_{o}+\lambda \hat{W}

[/tex]

[tex]

H_{o}|\varphi_{p}^{i}\rangle=E_{p}^{o}|\varphi_{p}^{i}\rangle

[/tex] (1)

and we want to solve the problem:

[tex]

H(\lambda)|\Psi(\lambda)\rangle=E(\lambda)\Psi(\lambda)\rangle[/tex] (2)

Expanding in [tex]\lambda[/tex] series equation (2) I get after equating each term:

[tex]

zeroth: (H_{o}-E_{o})|0\rangle=0

[/tex] (3)

[tex]

first: (H_{o}-E_{o})|1\rangle+(\hat{W}-E_{1})|0\rangle=0

[/tex] (4)

[tex]

second: (H_{o}-E_{o})|3\rangle+(\hat{W}-E_{1})|2\rangle

E_{2}|1\rangle-E_{3}|0\rangle=0

[/tex] (5)

from normalizing the wave fuction order by order I get:

[tex]

zeroth: \langle0|0\rangle=1

[/tex] (6)

[tex]

first: \langle0|1\rangle= \langle1|0\rangle=0

[/tex] (7)

[tex]

second: \langle0|2\rangle=

\langle2|0\rangle=-\frac{1}{2}\langle1|1\rangle

[/tex] (8)

Solution for the non-degenerated level[tex]

H_{o}|\varphi_{n}^{o}\rangle=E_{n}^{o}|\varphi_{n}^{o}\rangle

[/tex]

zeroth order:

[tex]

E_{o}=E_{n}^{o}

[/tex]

[tex]

|0\rangle=|\varphi_{n}\rangle

[/tex]

first order projecting (4) onto the vector [tex]|\varphi_{n}\rangle[/tex]

[tex]

E_{n}(\lambda)=E_{n}^{o}+\langle\varphi_{n}|W|\varphi_{n}\rangle

[/tex]

now this is the part that I dont understand:

when finding the eigenvector |1> the project equation (4) onto [tex]|\varphi_{p}^{i}\rangle[/tex] why the putting the supscript i if it is non-degenerated??????

[tex]|\Psi_{n}(\lambda)\rangle=|\varphi_{n}\rangle+\sum_{p\neq

n}\sum_{i}\frac{\langle\varphi_{p}^{i}|W|\varphi_{n}\rangle}{E_{n}^{o}-E_{p}^{o}}|\varphi_{p}^{i}\rangle[/tex]

see the book page 1101

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# Perturbation theory using Cohen-Tannoudji

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