User:Espen180/Quantum Mechanics/Time-Dependent Perturbation Theory

As the title suggests, this chapter deals with the situation where we have a solvable system with a small, time-dependent perturbation.

Let $H_{0}$ be the unperturbed Hamiltonian, with known energy levels and eigenstates $H_{0}|n\rangle =E_{n}|n\rangle$. We are interested in the system

$i\hbar \partial _{t}|\Psi (t)\rangle =H|\Psi (t)\rangle =(H_{0}+H_{1}(t))|\Psi (t)\rangle$,

where $H_{1}(t)$ is a small perturbation. Of course, we can forget about finding stationary states of the system, since it is time-dependent. However, since the $|n\rangle$'s are an orthonormal basis set, we can write

$|\Psi (t)\rangle =\sum _{n=1}^{\infty }a_{n}(t)e^{-iE_{n}t/\hbar }|n\rangle$.

Inserting this into the Schrödinger equation and multiplying by $\langle n|e^{iE_{n}t/\hbar }$, we obtain

$i\hbar {\frac {\partial a_{n}(t)}{\partial t}}+a_{n}(t)E_{n}=a_{n}(t)\langle n|H_{0}|n\rangle +\sum _{m=1}^{\infty }a_{m}(t)e^{-i(E_{m}-E_{n})t/\hbar }\langle n|H_{1}(t)|m\rangle$.

We see that $a(t)\langle n|H_{0}|n\rangle =a(t)E_{n}$ cancels with the $a(t)E_{n}$ term on the right side, and we are left with

$i\hbar {\frac {\partial a_{n}(t)}{\partial t}}=\sum _{m=1}^{\infty }a_{m}(t)e^{i(E_{n}-E_{m})t/\hbar }\langle n|H_{1}(t)|m\rangle$.

So far we have made no approximations, so this equation is exact and equivalent to the initial Schrödinger equation.

Now, if the perturbation is small, we can assume that the $a_{i}(t)$'s vary slowly with time. Therefore we can, to first order, neglect the time dependence of the $a_{m}(t)$'s on the right side of the equation. The resulting differential equation is trivial, with solution

$a_{n}(t)=a_{n}(t_{0})+{\frac {1}{i\hbar }}\sum _{m=1}^{\infty }a_{k}(t_{0})\int _{t_{0}}^{t}e^{i(E_{n}-E_{m})t^{\prime }/\hbar }\langle n|H_{1}(t^{\prime })|m\rangle \mathrm {d} t^{\prime }$.