Time-Dependent Perturbation Theory

Suppose we have a separable Hamiltonian

$$H=H_0+V(t)$$

where $H_0$ is assumed to represent a time-independent Hamiltonian whose energy eigenstates and eigenvalues are known.

$$H_0|n⟩=E_n|n⟩$$

We no longer expect that a state initially populated in one of these eigenstates to remain there for all times. There will generally be transitions to other states with some finite probability due to $V(t)$.

Suppose we have an arbitrary state at time $t=0$ represented as

$$|\alpha ⟩=\sum _n{c}_n(0)|n⟩$$

At a later time, the state will be

$$|\alpha ,t_0=0;t⟩=\sum _n{c}_n(t)e^{-i{E}_{n}t/\hslash }|n⟩$$

where the notation indicated that at time $t_0=0$ the state was in $|\alpha ⟩$. The time dependence that would have been there in a time-independent Hamiltonian is factored out.

Interaction (Dirac) Picture

We define

$$|\alpha ,t_0;t{⟩}_I=e^{i{H}_{0}t/\hslash }|\alpha ,t_0;t{⟩}_S$$

where $|\alpha ,t_0;t{⟩}_S$ is the state ket in the Schrodinger picture at some later time. We can similarly define observables in the interaction picture

$$V_I=e^{i{H}_{0}t/\hslash }V{e}^{-i{H}_{0}t/\hslash }$$

where $V$ is understood as the time dependent potential in the Schrodinger picture. There is a similarity to the Heisenberg picture, the difference being that there is only $H_0$ occurring in the exponential. We see since

$$i\hslash \frac{\partial }{\partial t}|\alpha ,t_0;t{⟩}_S=(H_0+V)|\alpha ,t_0;t{⟩}_S$$ $$i\hslash \frac{\partial }{\partial t}|\alpha ,t_0;t{⟩}_I=e^{i{H}_{0}t/\hslash }V{e}^{-i{H}_{0}t/\hslash }{e}^{i{H}_{0}t/\hslash }|\alpha ,t_0;t{⟩}_S$$

therefore

$$i\hslash \frac{\partial }{\partial t}|\alpha ,t_0;t{⟩}_I=V_I|\alpha ,t_0;t{⟩}_I$$

which is a Schrodinger equation in the interaction picture. Expanding the interaction ket in terms of the original base kets,

$$|\alpha ,t_0;t{⟩}_I=\sum _n{c}_n(t)|n⟩$$

and acting from the left with $⟨n|$ on the interaction Schrodinger equation we get

$$i\hslash \frac{\partial }{\partial t}⟨n|\alpha ,t_0;t{⟩}_I=\sum _m⟨n|V_I|m⟩⟨m|\alpha ,t_0;t⟩$$ $$\to i\hslash \frac{d}{dt}{c}_n(t)=\sum _m{V}_{nm}{e}^{i{\omega }_{nm}t}{c}_m(t)$$

These coefficients provide the exact solution for the time evolution of the state ket.

Rabii Oscillations

Most problems are unsolvable exactly. However a sinusoidally varying potential is exactly solvable. The problem is defined by

$$H_0=E_1|1⟩⟨1|+E_2|2⟩⟨2|$$ $$V(t)=\gamma e^{i\omega t}|1⟩⟨2|+\gamma e^{-i\omega t}|2⟩⟨1|$$

Using the interaction picture,

$$\begin{array}{rl}{\dot{c}}_1& =-i\frac{\gamma }{\hslash }{e}^{i(\omega -{\omega }_{21})t}{c}_2\\ \dot{c_2}& =-i\frac{\gamma }{\hslash }{e}^{-i(\omega -{\omega }_{21})t}{c}_1\end{array}$$

To solve for $c_2$ we can isolate $c_1$ in the second equation and differentiate,

$${\dot{c}}_1=\frac{i\hslash }{\gamma }{e}^{i(\omega -{\omega }_{21})t}(i(\omega -{\omega }_{21})\dot{c_2}+\ddot{c_2})$$

Plugging into the first equation, and rearranging, we get

$${\ddot{c}}_2+i(\omega -{\omega }_{21})\dot{c_2}+\frac{{\gamma }^2}{{\hslash }^2}{c}_2=0$$

To solve this homogeneous second order ODE, we use the solution form $c_2=e^{rt}$

$$r^2+i(\omega -{\omega }_{21})r+\frac{{\gamma }^2}{{\hslash }^2}=0$$ $$r=i\left(-\frac{\omega -{\omega }_{21}}{2}\pm \sqrt{\frac{(\omega -{\omega }_{21}{)}^2}{4}+\frac{{\gamma }^2}{{\hslash }^2}}\right)$$ $$\begin{array}{rl}{c}_2(t)=& A_{+}\exp \left[i\left(-\frac{\omega -{\omega }_{21}}{2}+\sqrt{\frac{(\omega -{\omega }_{21}{)}^2}{4}+\frac{{\gamma }^2}{{\hslash }^2}}\right)t\right]\\ +& A_{-}\exp \left[i\left(-\frac{\omega -{\omega }_{21}}{2}-\sqrt{\frac{(\omega -{\omega }_{21}{)}^2}{4}+\frac{{\gamma }^2}{{\hslash }^2}}\right)t\right]\end{array}$$

We must define some initial condi