Perturbation Theory - First Order Approximation

[Sunnyocean]

If:

$\hat{H} \psi (x) = E \psi (x)$

where E is the eigenvalue of the *disturbed* eigenfunction $\psi (x)$

and $E_n$ are the eigenvalues of the *undisturbed* Hamiltonian $\hat{H_0}$

and the *disturbed* Hamiltonian is of the form:

$\hat{H} = \hat{H_0} +{\epsilon} \hat{V}$

where $\hat{V}$ is the potential and

then we can expand, using a Taylor expansion, $\psi (x)$ and $\epsilon$ is a very small (positive?) number

$\psi (x) = {\psi}_0 (x) + {\epsilon}{\psi}_1 (x) + {\epsilon}^2 {\psi}_2 (x) + ...$ (1)

where ${\psi}_k (x)$ is the eigenfunction of $\hat{H}$ corresponding to the eigenvalue $E_n$ (eigenvalues are assumed to be non-degenerate)

We also get, using a Taylor expansion:

$E = E_0 +{\epsilon}{E_1} + {\epsilon}^2 {E_2} + ...$ (2)

MY QUESTIONS:

I tried deriving formulae (1) and (2) - i.e. the expansions - on my own but I was unable to do it. Could anyone please show me *in detail* how to obtain those expansions?

Another question is:
How do we know that, when using the Taylor expansion, the coefficient ${\epsilon}$ applies to ${\psi}_1 (x)$ and not to some other eigenfuncion, the coefficient ${\epsilon}^2$ applies to ${\psi}_2 (x)$ and not to some other eigenfuncion, and so on? The same question for eigenvalues.

 

[The_Duck]

Another question is:
How do we know that, when using the Taylor expansion, the coefficient ${\epsilon}$ applies to ${\psi}_1 (x)$ and not to some other eigenfuncion, the coefficient ${\epsilon}^2$ applies to ${\psi}_2 (x)$ and not to some other eigenfuncion, and so on? The same question for eigenvalues.

$E^{(n)}$ is *defined* as the coefficient of $\epsilon^n$ in the power series expansion of $E$. Same for $\psi^{(n)}$.

Note that the $E^{(n)}$'s and $\psi^{(n)}$'s are *not* eigenvalues and eigenfunctions, except for $E^{(0)}$ and $\psi^{(0)}$ which are an eigenvalue and eigenfunction of $H_0$.

 

[Sunnyocean]

Thank you The_Duck, that clarifies it. I was obviously misled by the confusing notation. You made it clear, so thank you again :)