# Perturbation theory using Cohen-Tannoudji

1. Sep 22, 2005

### cire

I'm reading the Cohen-Tannoudji book and I found somthing I dont understand
in stationary perturbation theory.
the problem the Hamiltonian is split in the known part an the perturbation:
$$H=H_{o}+\lambda \hat{W}$$
$$H_{o}|\varphi_{p}^{i}\rangle=E_{p}^{o}|\varphi_{p}^{i}\rangle$$ (1)
and we want to solve the problem:
$$H(\lambda)|\Psi(\lambda)\rangle=E(\lambda)\Psi(\lambda)\rangle$$ (2)
Expanding in $$\lambda$$ series equation (2) I get after equating each term:
$$zeroth: (H_{o}-E_{o})|0\rangle=0$$ (3)
$$first: (H_{o}-E_{o})|1\rangle+(\hat{W}-E_{1})|0\rangle=0$$ (4)
$$second: (H_{o}-E_{o})|3\rangle+(\hat{W}-E_{1})|2\rangle E_{2}|1\rangle-E_{3}|0\rangle=0$$ (5)

from normalizing the wave fuction order by order I get:
$$zeroth: \langle0|0\rangle=1$$ (6)
$$first: \langle0|1\rangle= \langle1|0\rangle=0$$ (7)
$$second: \langle0|2\rangle= \langle2|0\rangle=-\frac{1}{2}\langle1|1\rangle$$ (8)

Solution for the non-degenerated level$$H_{o}|\varphi_{n}^{o}\rangle=E_{n}^{o}|\varphi_{n}^{o}\rangle$$
zeroth order:
$$E_{o}=E_{n}^{o}$$
$$|0\rangle=|\varphi_{n}\rangle$$
first order projecting (4) onto the vector $$|\varphi_{n}\rangle$$
$$E_{n}(\lambda)=E_{n}^{o}+\langle\varphi_{n}|W|\varphi_{n}\rangle$$
now this is the part that I dont 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??????
$$|\Psi_{n}(\lambda)\rangle=|\varphi_{n}\rangle+\sum_{p\neq n}\sum_{i}\frac{\langle\varphi_{p}^{i}|W|\varphi_{n}\rangle}{E_{n}^{o}-E_{p}^{o}}|\varphi_{p}^{i}\rangle$$
see the book page 1101

2. Sep 22, 2005

### George Jones

Staff Emeritus
The eigenvalues are not necessarily degenerate. Read the line after equation (B-6) on page 1101.

Regards,
George

3. Sep 22, 2005

### cire

the book is talking about the non-degenerated case
$$H_{o}|\varphi_{n}\rangle=E_{n}^{o}|\varphi_{n} \rangle$$
why projecting in the degenerate subspace $$\{|\varphi_{p}^{i} \rangle\}$$eq B-6 ?????
degenerace refers to the new perturbed energy????

4. Sep 22, 2005

### George Jones

Staff Emeritus
The idea is to expand the first-order eigenvector correction |1> in terms of a complete set of states that consists of unperturbed energy eigenvectors, i.e., to arrive at equation (B-10). Since the only eigenvalue that is known to be non-degenerate is $E_{0}^{n}$, the index $i$ has to used in the labelling of this complete set of states.

Regards,
George

5. Sep 22, 2005

### cire

I got it, thanks but the book is misleading why not to project directly in the entire basis (degenerated in general) and get a matrix in B-5 and B-10 and B-11 all make sense, then the non-degenare case the matrix shrink to one element ...

6. Sep 22, 2005

### vanesch

Staff Emeritus
Of course the non-degenerate case is a special case of the degenerate case. But the degenerate case is slightly more subtle: you cannot just take ANY eigenvector of the unperturbed system and "perturbe it": the perturbation could lift the degeneracy (partly or entirely). So you first have to find out WHICH eigenvector you can perturbe in the first place (in fact, the original, say, 5-dimensional, set of eigenvectors with identical E0 will split in, say, a 2 dimensional set with one Ea, and 3 non-degenerate vectors with Eb, Ec and Ed respectively in first order ; the Ea can still potentially split at higher order).