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