# An assumption in the non-degenerate perturbation theory

The perturbed Hamiltonian is

$$H = H_0 + g V ,$$

where $g$ is the coupling parameter. The perturbed eigenvalue and eigenstate are of the form

$$E(g) = \sum_{r~=~0}^\infty g^r E_r ,\quad \left|\psi (g)\right\rangle = \sum_{r~=~0}^\infty g^r \left|\psi^{(r)} \right\rangle .$$

It is often assumed that

$$\left\langle \psi^{(0)}\bigg| \psi^{(r\geq 1) } \right\rangle = 0 .$$

This is always possible, because the equation determining $\left|\psi^{(r)} \right\rangle$ is

$$(H_0 - E_0 ) \left|\psi^{(r)}\right\rangle = \sum_{s~=~0}^{r-1} E_{(r-s)} \left|\psi^{(s)}\right\rangle - V \left|\psi^{(r-1)} \right\rangle .$$

Hence it is determined up to a multiplier of $\left|\psi^{(0)}\right \rangle$.

The problem is, is this assumption really necessary? It of course simplifies things. But what if we get rid of it?

Let's assume that, $$\psi=\sum_{k=0}^{+\infty}g^k|\psi^{(k)}\rangle,\quad \forall k>0: \langle\psi^{(k)}|\psi^{(0)}\rangle=n_k$$ Then let's multiply it on $g$-dependent constant $(1+f_mg^m)$ and expand, $$(1+f_m g^m)\psi=\sum_{k=0}^{+\infty}|\tilde{\psi}^{(k)}\rangle=\sum_{k=0}^{m-1}g^k|\psi^{(k)}\rangle+\sum_{k=m}^{+\infty}g^k\Big(|\psi^{(k)}\rangle+f_m|\psi^{(k-m)}\rangle\Big)$$ In particular that means that, $$\langle\tilde{\psi}^{(m)}|\psi^{(0)}\rangle=n_m+f_m$$

And that where that freedom comes from - it's just the possibility to multiply the full state on some $g$-dependent constant. When we expand that constant too, the resulting series will mix different orders with each other.

The usual choice is however not the one you wrote, but the one that conserves the norm, $$\langle\psi|\psi\rangle=\langle\psi^{(0)}|\psi^{(0)}\rangle$$ which looks somewhat more complicated, $$\sum_{k=0}^{m}\langle\psi^{(k)}|\psi^{(m-k)}\rangle=0$$