Non-degenerate perturbation theory

Let us now generalize our perturbation analysis to deal with systems possessing more than two energy eigenstates. Consider a system in which the energy eigenstates of the unperturbed Hamiltonian, $H_0$, are denoted

$$H_0\,\psi_n = E_n\,\psi_n,$$ (873)

where $n$ runs from 1 to $N$. The eigenstates are assumed to be orthonormal, so that

$$\langle m\vert n\rangle = \delta_{nm},$$ (874)

and to form a complete set. Let us now try to solve the energy eigenvalue problem for the perturbed Hamiltonian:

$$(H_0+H_1)\,\psi_E = E\,\psi_E.$$ (875)

If follows that

$$\langle m\vert H_0+H_1\vert E\rangle = E\,\langle m \vert E\rangle,$$ (876)

where $m$ can take any value from 1 to $N$. Now, we can express $\psi_E$ as a linear superposition of the unperturbed energy eigenstates:

$$\psi_E = \sum_k \langle k\vert E\rangle\,\psi_k,$$ (877)

where $k$ runs from 1 to $N$. We can combine the above equations to give

$$(E_m-E+e_{mm})\,\langle m\vert E\rangle + \sum_{k\neq m} e_{mk}\,\langle k\vert E\rangle = 0,$$ (878)

where

$$e_{mk} =\langle m\vert H_1\vert k\rangle.$$ (879)

Let us now develop our perturbation expansion. We assume that

$$\frac{e_{mk}}{E_m-E_k} \sim {\cal O}(\epsilon)$$ (880)

for all $m\neq k$, where $\epsilon\ll 1$ is our expansion parameter. We also assume that

$$\frac{e_{mm}}{E_m}\sim {\cal O}(\epsilon)$$ (881)

for all $m$. Let us search for a modified version of the $n$th unperturbed energy eigenstate for which

$$E = E_n + {\cal O}(\epsilon),$$ (882)

and

$$\langle n\vert E\rangle = 1,$$ (883)

$$\langle m\vert E\rangle = {\cal O}(\epsilon)$$ (884)

for $m\neq n$. Suppose that we write out Eq. (878) for $m\neq n$, neglecting terms which are ${\cal O}(\epsilon^2)$ according to our expansion scheme. We find that

$$(E_m-E_n)\,\langle m\vert E\rangle + e_{mn} \simeq 0,$$ (885)

giving

$$\langle m\vert E\rangle \simeq - \frac{e_{mn}}{E_m-E_n}.$$ (886)

Substituting the above expression into Eq. (878), evaluated for $m=n$, and neglecting ${\cal O}(\epsilon^3)$ terms, we obtain

$$(E_n-E+e_{nn})-\sum_{k\neq n}\frac{\vert e_{nk}\vert^2}{E_k-E_n} \simeq 0.$$ (887)

Thus, the modified $n$th energy eigenstate possesses an eigenvalue

$$E_n' = E_n + e_{nn} + \sum_{k\neq n}\frac{\vert e_{nk}\vert^2}{E_n-E_k} + {\cal O}(\epsilon^3),$$ (888)

and a wave-function

$$\psi_n' = \psi_n + \sum_{k\neq n} \frac{e_{kn}}{E_n-E_k}\,\psi_k + {\cal O}(\epsilon^2).$$ (889)

Incidentally, it is easily demonstrated that the modified eigenstates remain orthonormal to ${\cal O}(\epsilon^2)$.