Perturbation Expansion

Let us recall the analysis of Sect. 13.2. 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 $\langle n\vert m\rangle=\delta_{nm}$. Now, in the presence of a small time-dependent perturbation to the Hamiltonian, $H_1(t)$, the wavefunction of the system takes the form

$$\psi(t)= \sum_n c_n(t) \exp(-{\rm i} \omega_n t) \psi_n,$$ (1057)

where $\omega_n=E_n/\hbar$. The amplitudes $c_n(t)$ satisfy

$${\rm i} \hbar \frac{d c_n}{dt} = \sum_m H_{nm} \exp( {\rm i} \omega_{nm} t) c_m,$$ (1058)

where $H_{nm}(t)=\langle n\vert H_1(t)\vert m\rangle$ 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

$$P_n(t)= \vert c_n(t)\vert^2$$ (1059)

(assuing that, initially, $\sum_n\vert c_n\vert^2=1$).

Suppose that at $t=0$ the system is in some initial energy eigenstate labeled $i$. Equation (1058) is, thus, subject to the initial condition

$$c_n(0) = \delta_{ni}.$$ (1060)

Let us attempt a perturbative solution of Eq. (1058) 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 (1058), the $c_n$ are constant in time in the absence of the perturbation. Hence, the zeroth-order solution is simply

$$c_n^{(0)} (t) = \delta_{ni}.$$ (1061)

The first-order solution is obtained, via iteration, by substituting the zeroth-order solution into the right-hand side of Eq. (1058). Thus, we obtain

$${\rm i} \hbar \frac{dc_n^{(1)}}{dt} = \sum_m H_{nm} \exp(... ...{nm} t) c_m^{(0)} = H_{ni} \exp( {\rm i} \omega_{ni} t),$$ (1062)

subject to the boundary condition $c^{(1)}_n(0)=0$. The solution to the above equation is

$$c_n^{(1)} = -\frac{i}{\hbar}\int_0^t H_{ni}(t') \exp( {\rm i} \omega_{ni} t') dt'.$$ (1063)

It follows that, up to first-order in our perturbation expansion,

$$c_n(t) = \delta_{ni} -\frac{i}{\hbar}\int_0^t H_{ni}(t') \exp( {\rm i} \omega_{ni} t') dt'.$$ (1064)

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

$$P_{i\rightarrow f}(t) =\vert c_f(t)\vert^2 = \left\vert -\fr... ...}(t') \exp( {\rm i} \omega_{fi} t') dt'\right\vert^{ 2}.$$ (1065)

Note, finally, that our perturbative solution is clearly only valid provided

$$P_{i\rightarrow f}(t)\ll 1.$$ (1066)