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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

$\displaystyle H_0\, \vert n\rangle = E_n\, \vert n\rangle,$ (7.19)

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

$\displaystyle (H_0 + H_1) \,\vert E\rangle = E\,\vert E\rangle.$ (7.20)

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

$\displaystyle \vert E\rangle = \sum_k \langle k \vert E\rangle \vert k\rangle,$ (7.21)

where the summation is from $ k=1$ to $ N$ . Substituting the previous equation into Equation (7.20), and right-multiplying by $ \langle m\vert$ , we obtain

$\displaystyle (E_m + e_{mm} - E)\, \langle m\vert E\rangle + \sum_{k\neq m} e_{mk}\, \langle k\vert E\rangle = 0,$ (7.22)


$\displaystyle e_{mk} = \langle m \vert\,H_1\,\vert k\rangle.$ (7.23)

Let us now develop our perturbation expansion. We assume that

$\displaystyle \frac{\vert e_{mk}\vert}{E_m - E_k} \sim {\cal O}(\epsilon),$ (7.24)

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

$\displaystyle \frac{\vert e_{mm}\vert}{E_m} \sim {\cal O}(\epsilon),$ (7.25)

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

$\displaystyle E= E_n + {\cal O}(\epsilon),$ (7.26)


$\displaystyle \langle n\vert E\rangle$ $\displaystyle =1,$ (7.27)
$\displaystyle \langle m\vert E\rangle$ $\displaystyle \sim {\cal O}(\epsilon),$ (7.28)

for $ m\neq n$ . Suppose that we write out Equation (7.22) for $ m\neq n$ , neglecting terms that are $ {\cal O}(\epsilon^{\,2})$ according to our expansion scheme. We find that

$\displaystyle (E_m - E_n) \,\langle m \vert E \rangle + e_{mn} \simeq 0,$ (7.29)


$\displaystyle \langle m\vert E\rangle \simeq -\frac{e_{mn}}{E_m - E_n}.$ (7.30)

Substituting the previous expression into Equation (7.22), evaluated for $ m=n$ , and neglecting $ {\cal O}(\epsilon^{\,3})$ terms, we obtain

$\displaystyle (E_n + e_{nn} - E) - \sum_{k\neq n} \frac{\vert e_{nk}\vert^{\,2}} {E_k-E_n} \simeq 0.$ (7.31)

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

$\displaystyle E_n' = E_n + e_{nn} + \sum_{k\neq n} \frac{\vert e_{nk}\vert^{\,2}} {E_n-E_k} + {\cal O}(\epsilon^{\,3}),$ (7.32)

and the eigenket

$\displaystyle \vert n\rangle' = \vert n\rangle +\sum_{k\neq n}\frac{e_{kn}}{E_n - E_k}\,\vert k\rangle + {\cal O}(\epsilon^{\,2}).$ (7.33)

Note that

$\displaystyle \langle m\vert n\rangle' = \delta_{mn} + \frac{e_{nm}^{\,\ast}}{E...
... {E_n-E_m} + {\cal O}(\epsilon^{\,2}) = \delta_{mn} + {\cal O}(\epsilon^{\,2}).$ (7.34)

Thus, the modified eigenkets remain orthonormal to $ {\cal O}(\epsilon^{\,2})$ .

Note, finally, that if the perturbing Hamiltonian, $ H_1$ , commutes with the unperturbed Hamiltonian, $ H_0$ , then

$\displaystyle e_{ml}= e_{mm}\,\delta_{ml},$ (7.35)


$\displaystyle E_n'$ $\displaystyle =E_n + e_{nn},$ (7.36)
$\displaystyle \vert n\rangle'$ $\displaystyle =\vert n\rangle.$ (7.37)

The previous two equations are exact (i.e., they hold to all orders in $ \epsilon$ ).

next up previous
Next: Quadratic Stark Effect Up: Time-Independent Perturbation Theory Previous: Two-State System
Richard Fitzpatrick 2016-01-22