I About time-independent non-degenerate perturbation expansion

[Ishika_96_sparkles]

TL;DR

Confusion regarding the state vector expansion over a set of basis functions and the perturbative expansion in the powers of $lambda$.

We know that any state vector in Hilbert space can be expanded as
$$ \Psi = \sum_{n=1}^{\infty} \alpha_i \psi_i $$

However, when we start the perturbation expansion of the eigenvalues and the eigenfunctions as
$$E_a = \lambda^0 E_a^0+\lambda^1 E_a^1+\lambda^2 E_a^2+\lambda^3 E_a^3+...+\lambda^n E_a^n+...$$
and
$$\psi_a = \lambda^0 \psi_a^0+\lambda^1 \psi_a^1+\lambda^2 \psi_a^2+\lambda^3 \psi_a^3+...+\lambda^n \Psi_a^n+...$$
respectively, is it the single state being expanded over its basis set or what? What exactly is this set ${\psi_a^i}_{i=1}^n$?
How should I differentiate between the two?

PS: If my question is not clear, then please help me refine it as I really want to understand the basics of perturbation theory.

 

[DrClaude]

To reduce confusion, don't reuse the same symbols. For instance, consider
$$ \Psi = \sum_{n=1}^{\infty} \alpha_i \phi_i $$
where the $\{ \phi_i \}$ form some basis.

Then, in

$$\psi_a = \lambda^0 \psi_a^0+\lambda^1 \psi_a^1+\lambda^2 \psi_a^2+\lambda^3 \psi_a^3+...+\lambda^n \Psi_a^n+...$$

$\psi_a^0$ would usually be an eigenstate of the unperturbed Hamiltonian $\hat{H}_0$, and the other $\psi_a^i$ more generic functions (such that $\psi_a$ looks more and more like an eigenstate of the full Hamiltonian).

All the $\psi_a^i$ could be expressed in terms of basis states $\{ \phi_i \}$ (which would typically be the eigenstates of $\hat{H}_0$, but could also be the eigenstates of the full Hamiltonian, or any other basis functions).

 

[Ishika_96_sparkles]

Thank you @DrClaude! for a clear writeup and reply.

ψai more generic functions (such that ψa looks more and more like an eigenstate of the full Hamiltonian).

which probably are the "first-order... second-order...correction" terms that modify the unperturbed eigenfunction to match the eigenfunction of the full Hamiltonian. did i get that right?

Furthermore, all the $\psi_a^i$ are expandable in the original basis set $\{\phi_i\}$.

 

[DrClaude]

Thank you @DrClaude! for a clear writeup and reply.

which probably are the "first-order... second-order...correction" terms that modify the unperturbed eigenfunction to match the eigenfunction of the full Hamiltonian. did i get that right?

Furthermore, all the $\psi_a^i$ are expandable in the original basis set $\{\phi_i\}$.

Correct.