💼 Case: Consider an unperturbed system described by a time-independent Hamiltonian $\hat H_0$ for which we know the complete set of unperturbed eigenstates $\ket{n}$ and eigenenergy's $E_n$ i.e.
$$ \hat H_0\ket{n} = E_n \ket{n} $$
💥 Perturbation: to this Hamiltonian we add a time dependent effect such that the total Hamiltonian
$$ \hat H=\hat H_0+ \lambda \hat V(t) $$
where $\hat V(t)$ describes the perturbing field and $\lambda$ is a constant which will be use full in expansions
🧽 Assume: $\hat V(t)$ is small so that we can use it as a perturbation (low order influence ~ first order)
⚽ Goal: finding the probability that a system initially in an eigenstate of $\hat H_0$ transitions to a different eigenstate after the perturbation $\hat V(t)$
In the presence of the perturbation the time-dependent Schrodinger equation becomes
$$ i\hbar \frac{\text d}{\text dt}\ket{\psi(t)} =\hat H\ket{\psi(t)} =(\hat H_0 + \lambda \hat V(t)) \ket{\psi(t)} $$
for some system state $\ket{\psi(t)}$
We expand the time-dependent state in terms of the complete set of eigenstates $\ket{n}$ of $\hat H_0$ as
$$ \ket{\psi(t)} = \sum_n c_n (t) e^{-iE_n t/\hbar} \ket{n} $$
where $c_n(t)$ are time-dependent coefficients that we wish to determine/approximate
🗒️ Note: in the unperturbed state $c_n(t)$ are constant and the exponent is $1$
Substituting the expansion into the TDSE we get
$$ \sum_n \left [ i \hbar \frac{\text d c_n(t)}{\text dt}+E_n c_n (t) \right ]e^{-iE_n t/\hbar} \ket{n} = \sum_n \left [ E_n + \lambda \hat V (t) \right ]c_n (t) e^{-iE_n t/\hbar} \ket{n} $$
which after simplifications becomes
$$ \sum_n i \hbar \frac{\text d c_n(t)}{\text dt}e^{-iE_n t/\hbar} \ket{n} = \lambda \sum_n \hat V (t) c_n (t) e^{-iE_n t/\hbar} \ket{n} $$
We multiply both sides by an arbitrary eigenstate $\bra{m}$ such that $\braket{m|n}=\delta_{mn}$
$$ i \hbar \frac{\text d c_n(t)}{\text dt}e^{-iE_m t/\hbar} = \lambda \sum_n c_n (t) e^{-iE_n t/\hbar} \bra{m}\hat V (t)\ket{n} $$
We can now divide by $e^{-iE_m t/\hbar}$ to get
$$ i \hbar \frac{\text d c_n(t)}{\text dt} = \lambda \sum_n c_n (t) e^{i\omega_{mn} t} \bra{m}\hat V (t)\ket{n} $$
where $\omega_{mn}=(E_m-E_n) /\hbar$
⚽ Goal: our goal still hasn't changed
💼 Case: the perturbation $\lambda \hat V(t)$ is switched on at some time $t=t_0$ such that
$$ \hat H= \left \{ \begin{matrix} \hat H_0 & \text{for} & t\le t_0 \\ \hat H_0 + \lambda \hat V(t) & \text{for} & t>t_0 \end{matrix} \right . $$
Initially the system is in $\ket i$ an eigenstate of $\hat H_0$ such that $\ket{\psi(t_0}=\ket{i}$ and $c_n(t_0)=\delta_{ni}$
To start we will expand the coefficients in a power series of $\lambda$
$$ c_n (t)=c_n ^{(0)} (t)+ \lambda c^{(1)} _n (t) + \lambda ^2 c_n^{(2)} (t) + \ldots $$
Subbing in to our expression for $\dot c_n$ we get
$$ \footnotesize{i\hbar \frac{\text d}{\text dt} [c_m ^{(0)}(t) + \lambda c^{(1)}_m (t) + \lambda ^2 c_m ^{(2)} (t)+\ldots ]=\lambda \sum_n [c^{(0)}n (t) + \lambda c_n^{(1)} (t)+\lambda ^2 c_n ^{(2)}(t)+\ldots ]e^{i\omega{mn}t} \bra{m} \hat V(t) \ket{n}} $$