On page 254 of "Introduction to Quantum Mechanics Second Edition", by David J. Griffiths he writes the first order correction $\delta u_n(x)$ to the eigenstates $u_n(x)$ as $$\delta u_n(x)=\sum_{m\ne n}\frac{\int {u_m}^*(x) \hat {H_1}u_n(x)\,\mathrm{d}x}{E_n-E_m}u_m(x)\tag{1}$$
He writes the proof using mostly Dirac notation for which I am not familiar and trying to avoid for the time being as I cannot understand his proof due to it. So, instead, I used my own lecture notes (of Imperial College London, Dept. of Physics) to try to understand the derivation:
There are two questions I have about this proof, firstly I can't understand why $$\delta u_n(x)=\sum_{k\ne n}a_{nk}u_k(x)\tag{2}$$ is not being written as $$\delta u_n(x)=\sum_{k\ne n}a_{kn}u_k(x)\tag{3}?$$ I ask this because it was my understanding that the first index of $a$ must correspond to the index of the summation, $k$. Will using $(3)$ instead of $(2)$ mean that final expression for $a$ will read $a_{mn}$ instead of $a_{nm}$?
The second question I have is regarding the final expression $$a_{nm}=\frac{\int {u_m}^*(x) \hat {H_1}u_n(x)\,\mathrm{d}x}{E_n-E_m}\tag{4}$$
So does equation $(4)$ give the coeffients $a_{nm}$ of the expansion $(2)$? The proof shown above does not show how to reach eqn $(1)$. I'm confused because we cannot simply substitute $(4)$ into $(2)$ as $a_{nm}$ does not feature in $(2)$; instead $a_{nk}$ is present. Does anyone know how to obtain equation $(1)$ given equation $(4)$?