# Time-ordering in QFT [duplicate]

In Srednicki QFT page 37. In the derivation of LSZ reduction formula, he introduces the time-order operator $T$, so no time-dependent creation/annihilation operators are left in the transition amplitude. How can this be justified mathematically? And if I understand this correctly, if time ordering is not used, then a term like $$\langle 0\mid a_{1}\left ( -\infty \right )a_{2}\left ( -\infty \right )a_{1^{\tilde{}}}^{\dagger}\left ( \infty \right )a_{2^{\tilde{}}}^{\dagger}\left ( \infty \right )\mid 0\rangle$$ this mean that there's a contribution that depend on the amplitude of transition from the final momenta to the initial momenta but quantum mechanically the final momenta are not known in advance so such a term can't contribute to the process. Does this explain the use of time-ordering physically?

• What do you mean by "justified mathematically"? – Frederic Brünner Jan 21 '13 at 17:10
• He, probably, means "obtained mathematically as a result of mathematical transformations". It might be there is an implicit deception there. – Vladimir Kalitvianski Jan 21 '13 at 17:23

1. Inserting the time ordering symbol $T$ in (5.14) is completely justified just by the definition of the $T$.
The inserting of T in the term on the last line of (5.15): $$\langle0|\varphi (x_1)... \varphi (x_1)|0\rangle \rightarrow \langle0|T \varphi (x_1)... \varphi (x_1)|0\rangle$$ Neatly cancels out many terms containing ladder operators, this corresponds to writing the operator product time ordered product as a combination of time ordered and normal ordered products, in the canonical derivation of Wick's theorem.