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I have some difficulties when calculating the amplitude of a S-matrix using Wick's theorem. The evolution of the $U$-matrix is \begin{align} U(t, t_0) = T \exp(-i\int_{t_0}^{t}H_I(t') dt')=1 - i\int_{t_0}^{t}dt'H_I(t') + \frac{-i^2}{2}[\int_{t_0}^{t}dt'\int_{t'}^{t}dt''H_I(t'')H_I(t')+\int_{t_0}^{t}dt'\int_{t_0}^{t'}dt''H_I(t')H_I(t'')]+...\ . \end{align} That's all fine and clear. However, In QFT Book by Peskin and Schroeder (Eq. 4,92) \begin{align} \langle p_{1}p_{2}| T\{ \frac{-i\lambda}{4!}\int d^{4}x \ \phi_{I}^{4}(x) \}| p_{A}p_{A} \rangle, \end{align} when calculating second order expansion from Dyson's formula, he (they) (take)s the time ordered product again. Why does the Time ordered product $T\{..\} $ appear in the second order expansion? The whole point of Dyson's formula is that we 'Taylor expand' the time ordered exponential to get rid of the $T\{..\} $ in the serie terms.

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  • $\begingroup$ All the $\phi(x)$ fields are computed at the same point (and time). The time ordering in this case is doing nothing. $\endgroup$
    – FrodCube
    May 11, 2021 at 13:50
  • $\begingroup$ If that is the case, how does one get Feynman propagators? Time ordering give rise to normal ordering and contracted terms. If there are no contracted terms, how do we get the Feynman diagrams? $\endgroup$
    – M91
    May 11, 2021 at 14:02
  • $\begingroup$ The last formula you write is the first order term for a $phi^4$ theory. There are no propagators there. The diagram is a simple vertes. You start to get propagators when you go to the next order where you have two insertions of the Hamiltonian. $\endgroup$
    – FrodCube
    May 11, 2021 at 14:04
  • $\begingroup$ But then how do you explain the vacuum bubbles where there are vertices? I really appriciete your help. My POINT is, in order to get anything (propagators, vertices..etc) you NEED field contractions, and to get get field contractions, you need time ordering. So it does make a difference wether the $T{..}$ is present or not $\endgroup$
    – M91
    May 11, 2021 at 15:11
  • $\begingroup$ The T is always present. It's just that at the first order it does nothing. In the vacuum bubble case fields are contracted among each other at the same point, so you get a propagator computed at the origin. I think I don't understand what is your confusion. $\endgroup$
    – FrodCube
    May 11, 2021 at 15:56

1 Answer 1

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It is the other way around. The time-ordering is an succinct way to write the evolution operator, in particular the integral boundary conditions. Even if you expand the exponential, it has to stay there.

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