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In the page 91 of Many particle physics by Mahan, why $S(+\infty,t) C(t)S(t,t')C'(t')S(t`,-\infty)$ in the numerator can be written as $C(t)C`(t`)S(\infty,-\infty)$? And why in the first place the wavefunction in the positve infinity is presumed to be the same as in the negative infinity?

Derivation of Green function

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2 Answers 2

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As is written on that page just underneath the simplifying step you are asking about, the crucial ingredient is the time ordering operator $T$. It allows you to interchange the $\hat{C}$ and the $\hat{S}$ operators within the expectation value freely, giving you the desired expression. So in general you can write for some non-commuting operators $\hat{A},\hat{B}$ \begin{equation} \hat{A}(t_1) \hat{B}(t_2) = T \hat{B}(t_2) \hat{A}(t_1), ~~~~~t_1>t_2 \end{equation} which would not be correct if one would skip the time ordering on the right hand side.

Concerning your second question; I would say the made assumption is made to ease the computations but it is also a very plausible one. In practice you will deal with physical systems that will undergo some perturbation which is manually switched on at some specific point in time. Prior to that time you simply have an unperturbed system, described by a free Hamiltonian $H_0$. Then you do some computations and eventually assume that the perturbation switches off such that the behaviour of your system in the very far future is again dictated by $H_0$. The assumption made by Mahan here is that only a phase factor $e^{iL}$ survives, which is given one or two pages before the one you show here.

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  • $\begingroup$ Thanks so much for your prompt answer:) I didn't realise S(+∞,-∞) should be split into three parts in the time order operator, and time order operator is more like a rewording for comvenience. And for the second question, what is the reason <|S(-∞,0) shall be written as <|S(∞,0)/<|S(∞,-∞)|>, as the equation in the page under the sentence "we also replace the left-hand braket by". $\endgroup$
    – Tim Peng
    Sep 10, 2019 at 1:40
  • $\begingroup$ I have edited my answer, note that the expression for $e^{iL}$ that is used to show the step you ask about is given in the previous chapter. $\endgroup$ Sep 10, 2019 at 13:24
  • $\begingroup$ Thanks heaps! I understand it is for the sake of convenience when doing the time ordered integral. Otherwise time order operator can not be used here, and $$S(-∞,t) {\hat C_\lambda(t)}S(t,t`){\hat C^ \dagger_{\lambda}(t')}S(t',-∞)$$ basically is very complicated to be calculated. $\endgroup$
    – Tim Peng
    Sep 11, 2019 at 1:46
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Adding to Formelverleger's answer, the formulation here for Green's functions that relies on the wave function at time -inf being equal to the wave function at +inf is not always true. It is a likely ad hoc assumption for many scattering problems. However, several sections later in this book, Mahan motivates Schwinger's time loop formalism as a way of getting around this assumption at the expense of a more complicated expansion with several time orderings.

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  • $\begingroup$ Thanks! I haven't read Schiwinger's formalism yet, but I can assume that $S(-∞,t) {\hat C_\lambda(t)}S(t,t`){\hat C^ \dagger_{\lambda}(t')}S(t',-∞)$ must be much more complicated to calculate without the using of time ordered operator. $\endgroup$
    – Tim Peng
    Sep 11, 2019 at 1:53

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