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Expand the state in the energy eigenstates of $H_0$ asSubstitution into the Time-dependent Schrodinger equation and projection onto $⟨k|$ gives the exact interaction picture equationsIn first-order time-dependent perturbation theory, put $c_j(t)={\delta }_{j0}$ on the right-hand side and integrate from $0$ to $t$:

For a uniform electric field, the electric-dipole interaction is $\Delta (t)=e{\mathcal{E}}_{0}z{e}^{-t/\tau }$ up to an irrelevant overall sign. The supplied hydrogen atom wavefunctions giveFor $t\gg \tau$, the time integral isThe long-time transition fraction is consequentlyBoth $|100⟩$ and $|200⟩$ have even parity, whereas $z$ is odd, so the electric-dipole selection rule gives $⟨200|z|100⟩=0$. Thus the requested fraction in $|200⟩$ is zero.

In the interaction picture, first-order time-dependent perturbation theory gives the transition amplitudeTaking its squared modulus and using $|e^{i\theta }-1{|}^2=4{\sin }^2(\theta /2)$ gives

Let $H_0|r⟩=E_r|r⟩$. In the interaction picture,and the state obeysThe first term of the Dyson series, or equivalently first-order time-dependent perturbation theory, givesThus, if $|\psi (0)⟩={\sum }_r{a}_r|r⟩$,

First-order time-dependent perturbation theory givesThe lowest nontrivial transition probability is $|c_1(t){|}^2$, with errors of higher order in $V$.

To first order in $F$, time-dependent perturbation theory replaces the state on the right-hand side by its unperturbed value $e^{-i{E}_{0}t/\hslash }|0⟩$. The bound state is even, while the position operator $x$ is odd. The parity selection rule therefore givesso ${\dot{b}}_k=0$ and, with no initial continuum component,

Projection onto the odd state givesWriting ${\Omega }_k=(E_k-E_0-\hslash \omega )/\hslash$ and integrating from zero to $t$ yields