Time dependent perturbation theory addresses the transition between one eigenstate to another eigenstate when there is a small perturbation with respect to time such that it doesn't change the eigenfunctions.
The Hamiltonian for these problems of perturbation is, $$ H = H_0 + \lambda {H'(t)}$$ where now the perturbing Hamiltonian is time dependent. Solution for $H_0$ is, $$\psi_n(x,t) = \phi_n(x) e^{i\frac{E_n}{\hbar}t} = \phi_n e^{-i\omega_nt}$$ where $$H_0\phi_n = E^0_n\phi_n$$ and $ \omega_n = E_n/\hbar$ and the general state vector is given by, $$ \Psi = \sum_nc_n\psi_n(x,t)$$
In our problem we start with a particular eigenstate and assume that, after switching on the perturbation the state evolves to a general state always expressed as the linear combination of initial eigenstates where the time dependence is accompanied with the coefficient as $$\Psi (x,t) = \sum_n c_n(t) \phi_n(x) e^{i\omega{t}}$$ Because of this assumption, the particle now can be found in any other eigenstate and needn't to be in the same eigenstate forever. As usual, the modulus square of the coefficient gives the probability of finding the particle in any one of eigenstates, except now it changes with perturbation time. Substituting our wave function in general Schrodinger's equation (to find the time dependence of c(t)), $$i\hbar\frac{\partial\Psi}{\partial{t}} = H\Psi$$ gives (taking the inner product with $\psi_k$)$$ i\hbar\dot{c_k} = \lambda\sum_n \langle\phi_k\vert{H'}\vert\phi_n\rangle e^{i(\omega_k-\omega_n)t} f(t) c_n$$ where f(t) is the time-dependent part of the perturbing Hamiltonian H' (in variable separable form). In the limit $\lambda\rightarrow{0}$ the coefficients are all constant in time and so we seek a solution of the form, $$ c_k(t) = c_k^0 + \lambda{c_k^1(t)} + \lambda^2c_k^2(t) + \cdots$$ with corresponding order of correction in the upper index. With this expansion we get, $$ i\hbar \dot{c_k}^0 = 0 $$ which means $c_k^0 = constant$ whatever the value of $c_n^0$ is at the beginning of the perturbation remains a constant throughout the perturbation. In specific, if we denote the starting time $t_0$ then $c_n^0(t_0) = \delta_{nl}$ where we assumed the state is in a specific eigenstate at the start.
Similarly, the first order equation gives, $$ i\hbar\dot{c_k}^1 = \sum_n H'_{kn} f(t) c_n^0 e^{i(\omega_k-\omega_n)t}$$ substituting for $c_n^0$ we get, $$ i\hbar\dot{c_k}^1 = H'_{kl}f(t)e^{i(\omega_k-\omega_l)t}$$ finally gives the expression for first order correction as,$$c_k^1(t) = \frac{H'_{kl}}{i\hbar} \int_{t_0}^t f(t) e^{i\omega_{kl}t'} \,dt'$$ where $\omega_{kl} = \omega_k-\omega_l$ From this we can transition probability for each eigenstate by taking modulus square of the above coefficient.
The Hamiltonian for these problems of perturbation is, $$ H = H_0 + \lambda {H'(t)}$$ where now the perturbing Hamiltonian is time dependent. Solution for $H_0$ is, $$\psi_n(x,t) = \phi_n(x) e^{i\frac{E_n}{\hbar}t} = \phi_n e^{-i\omega_nt}$$ where $$H_0\phi_n = E^0_n\phi_n$$ and $ \omega_n = E_n/\hbar$ and the general state vector is given by, $$ \Psi = \sum_nc_n\psi_n(x,t)$$
In our problem we start with a particular eigenstate and assume that, after switching on the perturbation the state evolves to a general state always expressed as the linear combination of initial eigenstates where the time dependence is accompanied with the coefficient as $$\Psi (x,t) = \sum_n c_n(t) \phi_n(x) e^{i\omega{t}}$$ Because of this assumption, the particle now can be found in any other eigenstate and needn't to be in the same eigenstate forever. As usual, the modulus square of the coefficient gives the probability of finding the particle in any one of eigenstates, except now it changes with perturbation time. Substituting our wave function in general Schrodinger's equation (to find the time dependence of c(t)), $$i\hbar\frac{\partial\Psi}{\partial{t}} = H\Psi$$ gives (taking the inner product with $\psi_k$)$$ i\hbar\dot{c_k} = \lambda\sum_n \langle\phi_k\vert{H'}\vert\phi_n\rangle e^{i(\omega_k-\omega_n)t} f(t) c_n$$ where f(t) is the time-dependent part of the perturbing Hamiltonian H' (in variable separable form). In the limit $\lambda\rightarrow{0}$ the coefficients are all constant in time and so we seek a solution of the form, $$ c_k(t) = c_k^0 + \lambda{c_k^1(t)} + \lambda^2c_k^2(t) + \cdots$$ with corresponding order of correction in the upper index. With this expansion we get, $$ i\hbar \dot{c_k}^0 = 0 $$ which means $c_k^0 = constant$ whatever the value of $c_n^0$ is at the beginning of the perturbation remains a constant throughout the perturbation. In specific, if we denote the starting time $t_0$ then $c_n^0(t_0) = \delta_{nl}$ where we assumed the state is in a specific eigenstate at the start.
Similarly, the first order equation gives, $$ i\hbar\dot{c_k}^1 = \sum_n H'_{kn} f(t) c_n^0 e^{i(\omega_k-\omega_n)t}$$ substituting for $c_n^0$ we get, $$ i\hbar\dot{c_k}^1 = H'_{kl}f(t)e^{i(\omega_k-\omega_l)t}$$ finally gives the expression for first order correction as,$$c_k^1(t) = \frac{H'_{kl}}{i\hbar} \int_{t_0}^t f(t) e^{i\omega_{kl}t'} \,dt'$$ where $\omega_{kl} = \omega_k-\omega_l$ From this we can transition probability for each eigenstate by taking modulus square of the above coefficient.
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