next up previous
Next: Quadratic Stark Effect Up: Time-Independent Perturbation Theory Previous: Two-State System


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
\begin{displaymath}
H_0 \psi_n = E_n \psi_n,
\end{displaymath} (894)

where $n$ runs from 1 to $N$. The eigenstates are assumed to be orthonormal, so that
\begin{displaymath}
\langle m\vert n\rangle = \delta_{nm},
\end{displaymath} (895)

and to form a complete set. Let us now try to solve the energy eigenvalue problem for the perturbed Hamiltonian:
\begin{displaymath}
(H_0+H_1) \psi_E = E \psi_E.
\end{displaymath} (896)

If follows that
\begin{displaymath}
\langle m\vert H_0+H_1\vert E\rangle = E \langle m \vert E\rangle,
\end{displaymath} (897)

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:
\begin{displaymath}
\psi_E = \sum_k \langle k\vert E\rangle \psi_k,
\end{displaymath} (898)

where $k$ runs from 1 to $N$. We can combine the above equations to give
\begin{displaymath}
(E_m-E+e_{mm}) \langle m\vert E\rangle + \sum_{k\neq m}
e_{mk} \langle k\vert E\rangle = 0,
\end{displaymath} (899)

where
\begin{displaymath}
e_{mk} =\langle m\vert H_1\vert k\rangle.
\end{displaymath} (900)

Let us now develop our perturbation expansion. We assume that

\begin{displaymath}
\frac{e_{mk}}{E_m-E_k} \sim {\cal O}(\epsilon)
\end{displaymath} (901)

for all $m\neq k$, where $\epsilon\ll 1$ is our expansion parameter. We also assume that
\begin{displaymath}
\frac{e_{mm}}{E_m}\sim {\cal O}(\epsilon)
\end{displaymath} (902)

for all $m$. Let us search for a modified version of the $n$th unperturbed energy eigenstate for which
\begin{displaymath}
E = E_n + {\cal O}(\epsilon),
\end{displaymath} (903)

and
$\displaystyle \langle n\vert E\rangle$ $\textstyle =$ $\displaystyle 1,$ (904)
$\displaystyle \langle m\vert E\rangle$ $\textstyle =$ $\displaystyle {\cal O}(\epsilon)$ (905)

for $m\neq n$. Suppose that we write out Eq. (899) for $m\neq n$, neglecting terms which are ${\cal O}(\epsilon^2)$ 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} (906)

giving
\begin{displaymath}
\langle m\vert E\rangle \simeq - \frac{e_{mn}}{E_m-E_n}.
\end{displaymath} (907)

Substituting the above expression into Eq. (899), evaluated for $m=n$, and neglecting ${\cal O}(\epsilon^3)$ terms, we obtain
\begin{displaymath}
(E_n-E+e_{nn})-\sum_{k\neq n}\frac{\vert e_{nk}\vert^2}{E_k-E_n} \simeq 0.
\end{displaymath} (908)

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}
+ {\cal O}(\epsilon^3),
\end{displaymath} (909)

and a wavefunction
\begin{displaymath}
\psi_n' = \psi_n + \sum_{k\neq n} \frac{e_{kn}}{E_n-E_k} \psi_k + {\cal O}(\epsilon^2).
\end{displaymath} (910)

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


next up previous
Next: Quadratic Stark Effect Up: Time-Independent Perturbation Theory Previous: Two-State System
Richard Fitzpatrick 2010-07-20