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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,
, are denoted
![\begin{displaymath}
H_0 \psi_n = E_n \psi_n,
\end{displaymath}](img2148.png) |
(894) |
where
runs from 1 to
. The eigenstates are assumed to
be orthonormal, so that
![\begin{displaymath}
\langle m\vert n\rangle = \delta_{nm},
\end{displaymath}](img2149.png) |
(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}](img2123.png) |
(896) |
If follows that
![\begin{displaymath}
\langle m\vert H_0+H_1\vert E\rangle = E \langle m \vert E\rangle,
\end{displaymath}](img2150.png) |
(897) |
where
can take any value from 1 to
. Now, we can express
as a linear superposition of the unperturbed energy eigenstates:
![\begin{displaymath}
\psi_E = \sum_k \langle k\vert E\rangle \psi_k,
\end{displaymath}](img2151.png) |
(898) |
where
runs from 1 to
. 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}](img2152.png) |
(899) |
where
![\begin{displaymath}
e_{mk} =\langle m\vert H_1\vert k\rangle.
\end{displaymath}](img2153.png) |
(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}](img2154.png) |
(901) |
for all
, where
is our expansion parameter.
We also assume that
![\begin{displaymath}
\frac{e_{mm}}{E_m}\sim {\cal O}(\epsilon)
\end{displaymath}](img2156.png) |
(902) |
for all
. Let us search for a modified version of the
th unperturbed
energy eigenstate for which
![\begin{displaymath}
E = E_n + {\cal O}(\epsilon),
\end{displaymath}](img2157.png) |
(903) |
and
for
. Suppose that we write out Eq. (899) for
,
neglecting terms which are
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}](img2164.png) |
(906) |
giving
![\begin{displaymath}
\langle m\vert E\rangle \simeq - \frac{e_{mn}}{E_m-E_n}.
\end{displaymath}](img2165.png) |
(907) |
Substituting the above expression into Eq. (899),
evaluated for
, and neglecting
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}](img2168.png) |
(908) |
Thus, the modified
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}](img2169.png) |
(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}](img2170.png) |
(910) |
Incidentally, it is easily demonstrated that the modified eigenstates remain orthonormal
to
.
Next: Quadratic Stark Effect
Up: Time-Independent Perturbation Theory
Previous: Two-State System
Richard Fitzpatrick
2010-07-20