12.4: Perturbation Expansion

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 and Attributions

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

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