- #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:
<br /> H=H_{o}+\lambda \hat{W}<br />
<br /> H_{o}|\varphi_{p}^{i}\rangle=E_{p}^{o}|\varphi_{p}^{i}\rangle<br /> (1)
and we want to solve the problem:
<br /> H(\lambda)|\Psi(\lambda)\rangle=E(\lambda)\Psi(\lambda)\rangle (2)
Expanding in \lambda series equation (2) I get after equating each term:
<br /> zeroth: (H_{o}-E_{o})|0\rangle=0<br /> (3)
<br /> first: (H_{o}-E_{o})|1\rangle+(\hat{W}-E_{1})|0\rangle=0<br /> (4)
<br /> second: (H_{o}-E_{o})|3\rangle+(\hat{W}-E_{1})|2\rangle<br /> E_{2}|1\rangle-E_{3}|0\rangle=0<br /> (5)
from normalizing the wave fuction order by order I get:
<br /> zeroth: \langle0|0\rangle=1<br /> (6)
<br /> first: \langle0|1\rangle= \langle1|0\rangle=0<br /> (7)
<br /> second: \langle0|2\rangle=<br /> \langle2|0\rangle=-\frac{1}{2}\langle1|1\rangle<br /> (8)
Solution for the non-degenerated level<br /> H_{o}|\varphi_{n}^{o}\rangle=E_{n}^{o}|\varphi_{n}^{o}\rangle<br />
zeroth order:
<br /> E_{o}=E_{n}^{o}<br />
<br /> |0\rangle=|\varphi_{n}\rangle<br />
first order projecting (4) onto the vector |\varphi_{n}\rangle
<br /> E_{n}(\lambda)=E_{n}^{o}+\langle\varphi_{n}|W|\varphi_{n}\rangle<br />
now this is the part that I don't understand:
when finding the eigenvector |1> the project equation (4) onto |\varphi_{p}^{i}\rangle why the putting the supscript i if it is non-degenerated?
|\Psi_{n}(\lambda)\rangle=|\varphi_{n}\rangle+\sum_{p\neq<br /> n}\sum_{i}\frac{\langle\varphi_{p}^{i}|W|\varphi_{n}\rangle}{E_{n}^{o}-E_{p}^{o}}|\varphi_{p}^{i}\rangle
see the book page 1101
in stationary perturbation theory.
the problem the Hamiltonian is split in the known part an the perturbation:
<br /> H=H_{o}+\lambda \hat{W}<br />
<br /> H_{o}|\varphi_{p}^{i}\rangle=E_{p}^{o}|\varphi_{p}^{i}\rangle<br /> (1)
and we want to solve the problem:
<br /> H(\lambda)|\Psi(\lambda)\rangle=E(\lambda)\Psi(\lambda)\rangle (2)
Expanding in \lambda series equation (2) I get after equating each term:
<br /> zeroth: (H_{o}-E_{o})|0\rangle=0<br /> (3)
<br /> first: (H_{o}-E_{o})|1\rangle+(\hat{W}-E_{1})|0\rangle=0<br /> (4)
<br /> second: (H_{o}-E_{o})|3\rangle+(\hat{W}-E_{1})|2\rangle<br /> E_{2}|1\rangle-E_{3}|0\rangle=0<br /> (5)
from normalizing the wave fuction order by order I get:
<br /> zeroth: \langle0|0\rangle=1<br /> (6)
<br /> first: \langle0|1\rangle= \langle1|0\rangle=0<br /> (7)
<br /> second: \langle0|2\rangle=<br /> \langle2|0\rangle=-\frac{1}{2}\langle1|1\rangle<br /> (8)
Solution for the non-degenerated level<br /> H_{o}|\varphi_{n}^{o}\rangle=E_{n}^{o}|\varphi_{n}^{o}\rangle<br />
zeroth order:
<br /> E_{o}=E_{n}^{o}<br />
<br /> |0\rangle=|\varphi_{n}\rangle<br />
first order projecting (4) onto the vector |\varphi_{n}\rangle
<br /> E_{n}(\lambda)=E_{n}^{o}+\langle\varphi_{n}|W|\varphi_{n}\rangle<br />
now this is the part that I don't understand:
when finding the eigenvector |1> the project equation (4) onto |\varphi_{p}^{i}\rangle why the putting the supscript i if it is non-degenerated?
|\Psi_{n}(\lambda)\rangle=|\varphi_{n}\rangle+\sum_{p\neq<br /> n}\sum_{i}\frac{\langle\varphi_{p}^{i}|W|\varphi_{n}\rangle}{E_{n}^{o}-E_{p}^{o}}|\varphi_{p}^{i}\rangle
see the book page 1101
