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Physics LibreTexts

12.4: Perturbation Expansion

  • Page ID
    15802
  • Let us recall the analysis of Section 1.2. The \(\psi_n\) are the stationary orthonormal eigenstates of the time-independent unperturbed Hamiltonian, \(H_0\). Thus, \(H_0\,\psi_n=E_n\,\psi_n\), where the \(E_n\) are the unperturbed energy levels, and \(\langle n|m\rangle=\delta_{nm}\). Now, in the presence of a small time-dependent perturbation to the Hamiltonian, \(H_1(t)\), the wavefunction of the system takes the form \[\psi(t)= \sum_n c_n(t)\,\exp(-{\rm i}\,\omega_n\,t)\,\psi_n,\] where \(\omega_n=E_n/\hbar\). The amplitudes \(c_n(t)\) satisfy

    \[\label{e13.42} {\rm i}\,\hbar\,\frac{d c_n}{dt} = \sum_m H_{nm}\,\exp(\,{\rm i}\,\omega_{nm}\,t)\,c_m,\] where \(H_{nm}(t)=\langle n|H_1(t)|m\rangle\) and \(\omega_{nm}=(E_n-E_m)/\hbar\). Finally, the probability of finding the system in the \(n\)th eigenstate at time \(t\) is simply \[P_n(t)= |c_n(t)|^{\,2}\] (assuming that, initially, \(\sum_n|c_n|^{\,2}=1\)).

    Suppose that at \(t=0\) the system is in some initial energy eigenstate labeled \(i\). Equation ([e13.42]) is, thus, subject to the initial condition \[c_n(0) = \delta_{ni}.\] Let us attempt a perturbative solution of Equation ([e13.42]) using the ratio of \(H_1\) to \(H_0\) (or \(H_{nm}\) to \(\hbar\,\omega_{nm}\), to be more exact) as our expansion parameter. Now, according to Equation ([e13.42]), the \(c_n\) are constant in time in the absence of the perturbation. Hence, the zeroth-order solution is simply \[c_n^{(0)} (t) = \delta_{ni}.\] The first-order solution is obtained, via iteration, by substituting the zeroth-order solution into the right-hand side of Equation ([e13.42]). Thus, we obtain \[{\rm i}\,\hbar\,\frac{dc_n^{(1)}}{dt} = \sum_m H_{nm}\,\exp(\,{\rm i}\,\omega_{nm}\,t)\,c_m^{(0)} = H_{ni}\,\exp(\,{\rm i}\,\omega_{ni}\,t),\] subject to the boundary condition \(c^{(1)}_n(0)=0\). The solution to the previous equation is \[c_n^{(1)} = -\frac{\rm i}{\hbar}\int_0^t H_{ni}(t')\,\exp(\,{\rm i}\,\omega_{ni}\,t')\,dt'.\] It follows that, up to first-order in our perturbation expansion,

    \[\label{e13.48} c_n(t) = \delta_{ni} -\frac{\rm i}{\hbar}\int_0^t H_{ni}(t')\,\exp(\,{\rm i}\,\omega_{ni}\,t')\,dt'.\] Hence, the probability of finding the system in some final energy eigenstate labeled \(f\) at time \(t\), given that it is definitely in a different initial energy eigenstate labeled \(i\) at time \(t=0\), is \[P_{i\rightarrow f}(t) =|c_f(t)|^{\,2} = \left| -\frac{\rm i}{\hbar}\int_0^t H_{fi}(t')\,\exp(\,{\rm i}\,\omega_{fi}\,t')\,dt'\right|^{\,2}.\] Note, finally, that our perturbative solution is clearly only valid provided \[P_{i\rightarrow f}(t)\ll 1.\]

    Contributors

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

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