- #1
cire
I'm reading the Cohen-Tannoudji book and I found somthing I don't understand
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 don't 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
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 don't 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