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?
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What do you mean by "justified mathematically"? – Frederic Brünner Jan 21 '13 at 17:10
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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 Answers
I give a full answer to the time-ordering operator in Srednicki's derivation here
To summarize:
Inserting the time ordering symbol $T$ in (5.14) is completely justified just by the definition of the $T$.
The result (5.15) in Srednicki should be interpreted this way: terms containing ladder operators, such as the one you gave, do not contribute additionally to (5.15). because of the following:
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.
With regards to "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.". Contrary to your suggestion, in quantum mechanics all we can compute is transition amplitudes from initial to final momenta. To do this you do not need to know the final momentum, the transition amplitude to a certain momenta gives you the probability amplitude the the initial state, after some evolution, will become that particular final state.
