11.4: Non-Degenerate Perturbation Theory

Last updated
Save as PDF

Page ID: 15794

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

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 \[H_0\,\psi_n = E_n\,\psi_n,\] where $n$ runs from 1 to $N$. The eigenstates are assumed to be orthonormal, so that \[\langle m|n\rangle = \delta_{nm},\] and to form a complete set. Let us now try to solve the energy eigenvalue problem for the perturbed Hamiltonian: \[(H_0+H_1)\,\psi_E = E\,\psi_E.\] It follows that \[\langle m|H_0+H_1|E\rangle = E\,\langle m |E\rangle,\] 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: \[\psi_E = \sum_k \langle k|E\rangle\,\psi_k,\] where $k$ runs from 1 to $N$. We can combine the previous equations to give

\[\label{e12.45} (E_m-E+e_{mm})\,\langle m|E\rangle + \sum_{k\neq m} e_{mk}\,\langle k|E\rangle = 0,\] where \[e_{mk} =\langle m|H_1| k\rangle.\]

Let us now develop our perturbation expansion. We assume that \[\frac{e_{mk}}{E_m-E_k} \sim {\cal O}(\epsilon)\] for all $m\neq k$, where $\epsilon\ll 1$ is our expansion parameter. We also assume that \[\frac{e_{mm}}{E_m}\sim {\cal O}(\epsilon)\] for all $m$. Let us search for a modified version of the $n$th unperturbed energy eigenstate for which \[E = E_n + {\cal O}(\epsilon),\] and \[\begin{aligned} \langle n|E\rangle &= 1,\\[0.5ex] \langle m|E\rangle&={\cal O}(\epsilon)\end{aligned}\] for $m\neq n$. Suppose that we write out Equation ([e12.45]) for $m\neq n$, neglecting terms that are ${\cal O}(\epsilon^{\,2})$ according to our expansion scheme. We find that \[(E_m-E_n)\,\langle m|E\rangle + e_{mn} \simeq 0,\] giving \[\langle m|E\rangle \simeq - \frac{e_{mn}}{E_m-E_n}.\] Substituting the previous expression into Equation ([e12.45]), evaluated for $m=n$, and neglecting ${\cal O}(\epsilon^{\,3})$ terms, we obtain \[(E_n-E+e_{nn})-\sum_{k\neq n}\frac{|e_{nk}|^{\,2}}{E_k-E_n} \simeq 0.\] Thus, the modified $n$th energy eigenstate possesses an eigenvalue

\begin{equation}E_{n}^{\prime}=E_{n}+e_{n n}+\sum_{k \neq n} \frac{\left|e_{n k}\right|^{2}}{E_{n}-E_{k}}+\mathcal{O}\left(\epsilon^{3}\right)\end{equation} and a wavefunction

\[\label{e12.57} \psi_n' = \psi_n + \sum_{k\neq n} \frac{e_{kn}}{E_n-E_k}\,\psi_k + {\cal O}(\epsilon^{\,2}).\] Incidentally, it is easily demonstrated that the modified eigenstates remain orthonormal to ${\cal O}(\epsilon^{\,2})$.

Contributors and Attributions

Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

$ \newcommand {\ltapp} {\stackrel {_{\normalsize<}}{_{\normalsize \sim}}}$ $\newcommand {\gtapp} {\stackrel {_{\normalsize>}}{_{\normalsize \sim}}}$ $\newcommand {\btau}{\mbox{\boldmath$\tau$}}$ $\newcommand {\bmu}{\mbox{\boldmath$\mu$}}$ $\newcommand {\bsigma}{\mbox{\boldmath$\sigma$}}$ $\newcommand {\bOmega}{\mbox{\boldmath$\Omega$}}$ $\newcommand {\bomega}{\mbox{\boldmath$\omega$}}$ $\newcommand {\bepsilon}{\mbox{\boldmath$\epsilon$}}$