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Next: The quadratic Stark effect Up: Approximation methods Previous: The two-state system


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
\begin{displaymath}
H_0  \vert n\rangle = E_n  \vert n\rangle,
\end{displaymath} (605)

where $n$ runs from 1 to $N$. The eigenkets $\vert n\rangle$ are orthonormal, 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:
\begin{displaymath}
(H_0 + H_1) \vert E\rangle = E \vert E\rangle.
\end{displaymath} (606)

We can express $\vert E\rangle$ as a linear superposition of the unperturbed energy eigenkets,
\begin{displaymath}
\vert E\rangle = \sum_k \langle k \vert E\rangle \vert k\rangle,
\end{displaymath} (607)

where the summation is from $k=1$ to $N$. Substituting the above equation into Eq. (606), and right-multiplying by $\langle m\vert$, we obtain
\begin{displaymath}
(E_m + e_{mm} - E) \langle m\vert E\rangle + \sum_{k\neq m} e_{mk} \langle k\vert E\rangle = 0,
\end{displaymath} (608)

where
\begin{displaymath}
e_{mk} = \langle m \vert H_1\vert k\rangle.
\end{displaymath} (609)

Let us now develop our perturbation expansion. We assume that

\begin{displaymath}
\frac{\vert e_{mk}\vert}{E_m - E_k} \sim O(\epsilon),
\end{displaymath} (610)

for all $m\neq k$, where $\epsilon\ll 1$ is our expansion parameter. We also assume that
\begin{displaymath}
\frac{\vert e_{mm}\vert}{E_m} \sim O(\epsilon),
\end{displaymath} (611)

for all $m$. Let us search for a modified version of the $n$th unperturbed energy eigenstate, for which
\begin{displaymath}
E= E_n + O(\epsilon),
\end{displaymath} (612)

and
$\displaystyle \langle n\vert E\rangle$ $\textstyle =$ $\displaystyle 1,$ (613)
$\displaystyle \langle m\vert E\rangle$ $\textstyle \sim$ $\displaystyle O(\epsilon),$ (614)

for $m\neq n$. Suppose that we write out Eq. (608) for $m\neq n$, neglecting terms which are $O(\epsilon^2)$ according to our expansion scheme. We find that
\begin{displaymath}
(E_m - E_n) \langle m \vert E \rangle + e_{mn} \simeq 0,
\end{displaymath} (615)

giving
\begin{displaymath}
\langle m\vert E\rangle \simeq -\frac{e_{mn}}{E_m - E_n}.
\end{displaymath} (616)

Substituting the above expression into Eq. (608), evaluated for $m=n$, and neglecting $O(\epsilon^3)$ terms, we obtain
\begin{displaymath}
(E_n + e_{nn} - E) - \sum_{k\neq n} \frac{\vert e_{nk}\vert^2}
{E_k-E_n} = 0.
\end{displaymath} (617)

Thus, the modified $n$th energy eigenstate possesses an eigenvalue
\begin{displaymath}
E_n' = E_n + e_{nn} + \sum_{k\neq n} \frac{\vert e_{nk}\vert^2}
{E_n-E_k} + O(\epsilon^3),
\end{displaymath} (618)

and a eigenket
\begin{displaymath}
\vert n\rangle' = \vert n\rangle +\sum_{k\neq n}\frac{e_{kn}}{E_n - E_k} \vert k\rangle + O(\epsilon^2).
\end{displaymath} (619)

Note that
\begin{displaymath}
\langle m\vert n\rangle' = \delta_{mn} + \frac{e_{nm}^\ast}{...
...{mn}}
{E_n-E_m} + O(\epsilon^2) = \delta_{mn} + O(\epsilon^2).
\end{displaymath} (620)

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


next up previous
Next: The quadratic Stark effect Up: Approximation methods Previous: The two-state system
Richard Fitzpatrick 2006-02-16