7.3: Non-Degenerate Perturbation Theory

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

$H_0\, \vert n\rangle = E_n\, \vert n\rangle,$ $\ref{601}$

where $n$ runs from 1 to $N$ . The eigenkets $\vert n\rangle$ are orthogonal, form a complete set, and have their lengths normalized to unity. Let us now try to solve the energy eigenvalue problem for the perturbed Hamiltonian:

$(H_0 + H_1) \,\vert E\rangle = E\,\vert E\rangle.$ $\ref{602}$

We can express $\vert E\rangle$ as a linear superposition of the unperturbed energy eigenkets,

$\vert E\rangle = \sum_k \langle k \vert E\rangle \vert k\rangle,$ $\ref{603}$

where the summation is from $k=1$ to $N$ . Substituting the above equation into Equation $\ref{602}$, and right-multiplying by $\langle m\vert$ , we obtain

$(E_m + e_{mm} - E)\, \langle m\vert E\rangle + \sum_{k\neq m} e_{mk}\, \langle k\vert E\rangle = 0,$ $\ref{604}$

where

$e_{mk} = \langle m \vert\,H_1\,\vert k\rangle.$ $\ref{605}$

Let us now develop our perturbation expansion. We assume that

$\frac{\vert e_{mk}\vert}{E_m - E_k} \sim O(\epsilon),$ $\ref{606}$

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

$\frac{\vert e_{mm}\vert}{E_m} \sim O(\epsilon),$ $\ref{607}$

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

$E= E_n + O(\epsilon),$ $\ref{608}$

and

$\langle n\vert E\rangle$ $=$ $1,$ $\ref{609}$ $\langle m\vert E\rangle$ $\sim$ $O(\epsilon),$ $\ref{610}$

for $m\neq n$ . Suppose that we write out Equation $\ref{604}$ for $m\neq n$ , neglecting terms that are $O(\epsilon^2)$ according to our expansion scheme. We find that

$(E_m - E_n) \,\langle m \vert E \rangle + e_{mn} \simeq 0,$ $\ref{611}$

giving

$\langle m\vert E\rangle \simeq -\frac{e_{mn}}{E_m - E_n}.$ $\ref{612}$

Substituting the above expression into Equation $\ref{604}$, evaluated for $m=n$ , and neglecting $O(\epsilon^3)$ terms, we obtain

$(E_n + e_{nn} - E) - \sum_{k\neq n} \frac{\vert e_{nk}\vert^{\,2}} {E_k-E_n} \simeq 0.$ $\ref{613}$

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

$E_n' = E_n + e_{nn} + \sum_{k\neq n} \frac{\vert e_{nk}\vert^{\,2}} {E_n-E_k} + O(\epsilon^3),$ $\ref{614}$

and a eigenket

$\vert n\rangle' = \vert n\rangle +\sum_{k\neq n}\frac{e_{kn}}{E_n - E_k}\,\vert k\rangle + O(\epsilon^2).$ $\ref{615}$

Note that

$\langle m\vert n\rangle' = \delta_{mn} + \frac{e_{nm}^{\,\ast}}{E_m-E_n} + \frac{e_{mn}} {E_n-E_m} + O(\epsilon^2) = \delta_{mn} + O(\epsilon^2).$ $\ref{616}$

Thus, the modified eigenkets remain orthogonal and properly normalized to $O(\epsilon^2)$ .

Contributors

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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