I'm considering one paper about electron recombination and there is an expression for S-matrix that confuses me $${S_{fi}} = i\mathop {\lim }\limits_{t' \to \infty \atop t \to - \infty } \left\langle {{\Phi _{out}}(t')\left| {G(t',t)} \right.\left| {{\Phi _{in}}(t)} \right\rangle } \right.$$

Source: https://doi.org/10.1103/PhysRevA.95.063401

But I'm familiar with the following expression for the matrix element in interaction picture $${A_{fi}}(t) = \langle {E_f}|U(t;{t_0})|{E_i}\rangle,$$

Source: http://www.nyu.edu/classes/tuckerman/stat.mechII/lectures/lecture_21/node3.html

where the initial and final states have different denotations but actually they are the same states as preceding ones. And $U(t;{t_0})$ is equal to $${U_I}(t;{t_0}) = T\left[ {\exp \left( { - \frac{i}{\hbar }\int_{{t_0}}^t \; dt'\;{H_I}(t')} \right)} \right]{\rm{ }},$$ where $T$ is time-order operator.

In this respect, ${U_I}(t;{t_0})$ doen't look like Green's function. So, is it the time part of a time-dependent Shrodinger equation propagator? And U and G have the same role? Or do I get it in a wrong way?

  • $\begingroup$ $U(t;t_0)$ is the formal operator which propagates the initial solution at time $t_0$ to the solution at time $t$. Your writing with the time-order operator $T$ is obviously purely formal, and $T$ is just a mathematical rewording of the propagation properties of $U$. The Green's function is the solution of the Schrödinger equation with $\delta$ distribution on the right-hand-side, when the Hamiltonian is supposed to be a linear operator with time dependence only. In that case, $U$ is a formal rewriting of the Green's function. Nothing to do with the Green's function in QFT, though. $\endgroup$
    – FraSchelle
    Commented Sep 27, 2018 at 3:22


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