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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. The energy eigenstates of the unperturbed Hamiltonian, $ H_0$ , are denoted

$\displaystyle H_0\, \vert n\rangle = E_n\, \vert n\rangle,$ (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:

$\displaystyle (H_0 + H_1) \,\vert E\rangle = E\,\vert E\rangle.$ (602)

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

$\displaystyle \vert E\rangle = \sum_k \langle k \vert E\rangle \vert k\rangle,$ (603)

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

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

where

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

Let us now develop our perturbation expansion. We assume that

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

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

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

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

$\displaystyle E= E_n + O(\epsilon),$ (608)

and
$\displaystyle \langle n\vert E\rangle$ $\displaystyle =$ $\displaystyle 1,$ (609)
$\displaystyle \langle m\vert E\rangle$ $\displaystyle \sim$ $\displaystyle O(\epsilon),$ (610)

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

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

giving

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

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

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

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

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

and a eigenket

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

Note that

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

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


next up previous
Next: Quadratic Stark Effect Up: Time-Independent Perturbation Theory Previous: Two-State System
Richard Fitzpatrick 2013-04-08