$$\require{cancel}$$

# 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 $P_{i\rightarrow f}(t)\ll 1.$
