On page 51 Srednicki states, "Note that the operators are in time order…we can insert $T$ without changing anything". This I agree with. But then on the next paragraph he states "The time order operator $T$, moves all…where they annihilate…". How correct is this paragraph? Is equation (5.14) always valid? Is equation (5.14) the reason why the time-ordering symbol $T$ appears all over the place in QFT?
The equations I'm referring to are (braket package available?):
$$\langle f|i\rangle = \langle0|~a_{1'}(+\infty)a_{2'}(+\infty)a_{1}^\dagger(-\infty)a_{2}^\dagger(-\infty)~|0\rangle \tag{5.13} $$
and since time goes from right to left inside the braket we put the $T$-symbol
$$\langle f|i\rangle = \langle 0|T ~a_{1'}(+\infty)a_{2'}(+\infty)a_{1}^\dagger(-\infty)a_{2}^\dagger(-\infty)~|0\rangle \tag{5.14} $$
without changing anything.
After this, the $T$-symbol appears everywhere!
Best Answer
Hi I know this is an old question but this tripped me up too and I thought it might be good to share the answer. This is more of a notational issue, but what he is doing is absolutely correct.
If I understand it correctly, you're asking whether it's valid to insert the time ordering symbol there. In particular it may seem a little strange because if you remove the time ordering symbol from, say, (5.15), the answer obtained will obviously different, therefore it may seem that inserting the time ordering symbol does change the value of the expression. This is not the case.
First, by definition of the time ordering symbol, the following is indeed trivially true. $$\langle0|~a_{1'}(+\infty)a_{2'}(+\infty)a_{1}^\dagger(-\infty)a_{2}^\dagger(-\infty)~|0\rangle =\langle 0|T ~a_{1'}(+\infty)a_{2'}(+\infty)a_{1}^\dagger(-\infty)a_{2}^\dagger(-\infty)~|0\rangle \equiv \mathcal{A} $$
In the next step, Srednicki uses (5.11) and (5.12) to rewrite the ladder operators. In particular, he puts (5.11) and (5.12) into the RIGHT side of the above equality, and conclude that all terms in this expansion containing ladder operators will vanish, because the time ordered ladder operator terms will kill off the ground states. This gives (5.15)
Now try substituting (5.11) and (5.12) into the left side of the above. Because we no longer have time ordering, the terms containing the ladder operators no longer vanish. In particular, the expression you get is (5.15) without the time ordering operators, PLUS a whole bunch of other terms containing ladder operators.
Therefore if we took (5.15) and removed the time ordering symbol, this does NOT give you $\mathcal{A}$. In other words, the time ordering symbol $T$ appears in the LSZ formula because we made the simplification in (5.15) to kill off the ladder operator terms.
In regards to whether this is why the time ordering operator appears everywhere in field theory, I think another (perhaps more fundamental) reason is (6.13) and (6.18) in Srednicki: i.e. when we construct path integrals with functional derivatives, the resulting expression can be written as a bunch of time ordered position operators sandwiched between 2 states.