[NOTES/QM-09008] Perturbation Expansion in Interaction picture

Time evolution in Dirac picture
Let \(H=H_0+ H'\) be the Hamiltonian of system of interest, \(H_0\) and \(H'\), respectively, being the free part and interaction part of the total Hamiltonian. In the interaction picture the time dependence of an operator is given by \begin{equation}\label{eq01} i\hbar \dd[X_I(t)]{t} = [H_0, X_I(t)]. \end{equation} The solution of this equation can be written down explicitly and we have \begin{equation}\label{eq02} X_I(t) = e^{iH_0 t/\hbar}X_I(0)e^{-iH_0 t/\hbar}. \end{equation} The time development of the state vector in the interaction picture is given by the Schrodinger equation \begin{equation}\label{eq03} i\hbar\dd{t} \ket{\psi t}_I = H'_I(t)\ket{\psi t}. \end{equation}

Turning equation of motion into an integral equation
It must be noted that the interaction Hamiltonian in the interaction picture, \(H'_I(t)\), is always time dependent whether the Schrodinger picture operator depends on time or not. This makes it impossible to write an explicit solution to \eqref{eq02} impossible. Here we will be interested in deriving a perturbative expansion in powers of \(H'_I(t)\). This will be achieved by converting \eqref{eq03} into an integral equation. Let \(U(t,t_0)\) be unitary operator which connects the interaction picture states at times \(t, t_0)\): \begin{equation}\label{eq04} \ket{\psi t}_I. = U(t,t_0)\ket{\psi t_0}_I. \end{equation} Obviously we must have \begin{equation}\label{eq05} U(t_0,t_0) = \hat{I}. \end{equation} Substituting \eqref{eq04} in both sides of \eqref{eq03}, we get the following equation for the time evolution operator \(U(t,t_0)\). \begin{equation}\label{eq06} i\hbar \dd{t} U(t,t_0) \ket{\psi t_0}_I = H'_I(t) U(t,t_0)\ket{\psi t_0}_I. \end{equation} Since the initial state \(\ket{\psi,t_0}\) is arbitrary, we get \begin{equation}\label{eq07} \boxed{i\hbar\dd{t}U(t,t_0) = H'_I(t) U(t,t_0).} \end{equation} Integrating this equation w.r.t. time and using the initial condition \eqref{eq05} we get \begin{equation}\label{eq08} \boxed{U(t,t_0) = \hat{I} + \frac{1}{i\hbar} \int_{t_0}^ t H'(t) U(t,t_0)\, dt.} \end{equation} This integral equation is starting point for a perturbative expansion of the time evolution operator \(U(t,t_0)\). \subsection{Perturbative solution} For book keeping purpose we rewrite equation \eqref{eq08} as \begin{equation}\label{eq09} U(t,t_0) = \hat{I} + \frac{\lambda}{i\hbar} \int_{t_0}^ t H'(t)U(t,t_0)\, dt \end{equation} and the parameter \(\lambda\) will be set equal to unity in the end. As zeroth order approximation, we may write \begin{equation}\label{eq10} U^{(0)}(t,t_0) = \hat{I} \end{equation} and obtain the next approximation by inserting the above expression in the right hand side of \eqref{eq09}. This gives \begin{equation}\label{eq11} U^{(1)}(t,t_0) = \hat{I} + \frac{\lambda}{i\hbar} \int_{t_0}^ t H'(t_1)\, dt_1. \end{equation} Repeating this process, by inserting \(U^{(1)}(t,t_0)\) for \(U(t,t_0)\) in the right hand side of \eqref{eq09}, we get the second order approximation as \begin{equation}\label{eq12} U^{(2)}(t,t_0) = \hat{I} + \frac{\lambda}{i\hbar} \int_{t_0}^{t_1}\, dt_1 + \Big(\frac{\lambda}{i\hbar}\Big)^2 \int_{t_0}^{t} \int_{t_0}^{t_1}H'(t_1) H'(t_2)\, dt_2 dt_1. \end{equation} The double integral in the right hand side can be written as \MkGbx{Verify}{01} \begin{equation}\label{eq13} \int_{t_0}^{t} \int_{t_0}^{t_1}H'(t_1) H'(t_2)\, dt_2 dt_1 =\frac{1}{2}\int_{t_0}^{t} \int_{t_0}^{t} T\big(H'(t_1) H'(t_2)\big)\, dt_2 dt_1. \end{equation} Here the symbol \(T\) stands for time ordering, defined by, \begin{equation}\label{eq14} T\big(H'(t_1) H'(t_2)\big) =\begin{cases} H(t_1)H(t_2), & \text{if} \quad t_1 > t_2,\\ H(t_2)H(t_1), & \text{if} \quad t_2 > t_1.\\ \end{cases} \end{equation} The definition time ordered product of several operators is an obvious generalization of the above definition. \noindent {\sf Proof of \eqref{eq13} is left as an exercise in double integration.} Iterating the above process gives an infinite series in powers of \(\lambda\). Later terms of the series can be found and turn out to be multiple integrals of time ordered product of more factors of \(H'(t)\). Skipping some technical details the final answer turns out to be \begin{equation}\label{eq15} U(t,t_0) = I + \sum_{n=1}^\infty \Big(\frac{1}{i\hbar}\Big)^2 \int_{t_0}^ t\, dt_1 \ldots \int_{t_0}^ t\, dt_n T\big\{H(t_1)H(t_2)\ldots H(t_n)\big\}. \end{equation} Retaining first few terms in the above series gives useful approximation for several applications. The series \eqref{eq15} is symbolically written as \begin{equation}\label{eq16} U(t,t_0) = T \exp\left(\frac{-i}{\hbar}\int_{t_0}^t H(t) dt \right). \end{equation} and the right hand side is known as time ordered exponential of the argument.