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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!

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  • $\begingroup$ The time ordered operator tells us which fields go into the interaction in the Feynman diagrams. So the paragraph you've cited is true. $\endgroup$ – user28355 Jan 20 '14 at 20:59
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    $\begingroup$ It would be helpful if you could write down the relevant equation. $\endgroup$ – JeffDror Jan 20 '14 at 21:56
  • $\begingroup$ Comment to the question: Echoing JeffDror's comment, it would be good if OP (or somebody else?) could try to make the question formulation self-contained, so one doesn't have to open the link to understand the question. $\endgroup$ – Qmechanic Jul 16 '14 at 7:27
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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.

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    $\begingroup$ This would be helpful to more people if you explicitly included all the equations you reference. $\endgroup$ – Wouter Sep 28 '14 at 12:21
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    $\begingroup$ @bechira I'm pretty confused by your answer. In your second paragraph, you said:" it may seem that inserting the time ordering symbol does change the value of the expression. This is not the case." So I thought you are going to show that nothing would be changed when the ordering is imposed. However, in your second last paragraph, you said:"Therefore if we took (5.15) and removed the time ordering symbol, this does NOT give you A". The two statements are obviously contradictory. $\endgroup$ – M. Zeng Jul 5 '15 at 12:38
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    $\begingroup$ @bechira You also mentioned that "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." How is this so-called "simplification" justified? $\endgroup$ – M. Zeng Jul 5 '15 at 12:41
  • $\begingroup$ @M.Zeng this was a long time ago, later today I'll rewrite this answer completely - the answer is correct but perhaps not phrased in the most pedagogical manner. In the meantime feel free to read it over a couple of more times. $\endgroup$ – zzz Jul 5 '15 at 18:12
  • $\begingroup$ Can you really substitute inside the time irdering symbol though. It seems to me like time ordering depends on the explicit way of how you write the expression it looks at labels, which don't necessarily have any physical meaning. By substituting you change the explicit time dependence and hence the action of the timeordering symbol. I see that, when I don't use time ordering I get a lot of terms which could cancel, but I don't quite see the reason why they should. $\endgroup$ – QuantumStudent Mar 3 '17 at 12:31
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My understanding is that the introduction of time-ordering in this derivation can be justified if the momenta of all outgoing particles are different from the momenta of the incoming particles. This isn't a serious constraint, since we are dealing with a scattering process, after all!

Having said that, here's how you could justify time-ordering. First, note that equation (5.10) essentially comes from $ a_1^\dagger (t+dt) - a_1^\dagger(t) = \partial_0 a_1^\dagger(t) \cdot dt$. Formally we could write $$ a_1^\dagger (t_2) - a_1^\dagger (t_1) = \int_{t_1}^{t_2} dt \partial_0 a_1^\dagger(t) $$

Now, we are going to evaluate $ \left<a_{1'}(t_2) a_{2'}(t_2) a_1^\dagger (t_1) a_2^\dagger (t_1) \right>$. One can take the term $a_{2'}(t_2)$ and use the above formula to bring it back in time to $t_1$: $$\left<a_{1'}(t_2) a_{2'}(t_2) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> = \left<a_{1'}(t_2) \left(a_{2'}(t_1) + \int_{t_1}^{t_2} dt \partial_0 a_{2'}(t)\right) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> $$

Recall that the commutation relations for operators $a(\vec{k},t)$ is such that the only non-zero equal-time commutator is $[a(\vec{k}, t), a^\dagger(\vec{k'}, t)]=(2\pi)^3 2\omega \delta^3(\vec{k}-\vec{k'})$. Note that for non-equal-time commutators, because the Lagrangian now can have interaction terms, time-evolution will mix both raising and lowering operators with different $\vec{k}$, which is why we brought $a_{2'}(t_2)$ back to $t_1$ to check how it would commute with the term behind it, $a_1^\dagger(t_1)$.

Now, because $a_{2'}$ is centered around $\vec{k}_{2'}$ in momentum space, while $a^\dagger_{1}$ is centered around $\vec{k}_{1}$, we have $\delta^3(\vec{k}_{2'}-\vec{k}_1) = 0$. Therefore all equal-time pairs of $a(t)$ and $a^\dagger(t)$ commute. You now see that $a_{2'}(t_1)$ can move all the way to the right and annihilate the vacuum. The equation now reads

$$\left<a_{1'}(t_2) a_{2'}(t_2) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> = \left<a_{1'}(t_2) \left(\int_{t_1}^{t_2} dt \partial_0 a_{2'}(t)\right) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> $$

Next, pick a certain term in the integral, and try to evaluate $$\frac{\partial}{\partial t} \left<a_{1'}(t_2) a_{2'}(t) a_1^\dagger (t_1) a_2^\dagger (t_1) \right>$$ You may have guessed that I would bring $a_{1'}(t_2)$ back in time to $t$ to get

$$ \begin{eqnarray} & & \left<\left(a_{1'}(t) + \int_{t}^{t_2} dt' \partial_0 a_{1'}(t')\right) a_{2'}(t) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> \\ & = & \left<a_{2'}(t) a_{1'}(t) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> + \left<\left(\int_{t}^{t_2} dt' \partial_0 a_{1'}(t')\right) a_{2'}(t) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> \\ & = & \left<a_{2'}(t) \left(a_{1'}(t_1)+\int_{t_1}^t dt' \partial_0 a_{1'}(t')\right) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> + \\ & & \left<\left(\int_{t}^{t_2} dt' \partial_0 a_{1'}(t')\right) a_{2'}(t) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> \\ & = & \left<a_{2'}(t) \left(\int_{t_1}^t dt' \partial_0 a_{1'}(t')\right) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> + \\ & & \left<\left(\int_{t}^{t_2} dt' \partial_0 a_{1'}(t')\right) a_{2'}(t) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> \end{eqnarray}$$

Therefore $$ \begin{eqnarray} & & \left<a_{1'}(t_2) a_{2'}(t_2) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> \\ & = & \int_{t_1}^{t_2} dt \frac{\partial}{\partial t} \left<a_{2'}(t) \left(\int_{t_1}^t dt' \partial_0 a_{1'}(t')\right) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> + \\ & & \int_{t_1}^{t_2} dt \frac{\partial}{\partial t} \left<\left(\int_{t}^{t_2} dt' \partial_0 a_{1'}(t')\right) a_{2'}(t) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> \\ & = & \int_{t_1}^{t_2} dt \frac{\partial}{\partial t} \int_{t_1}^{t_2} dt' \frac{\partial}{\partial t'}\left<\left(\mathrm{T} a_{2'}(t) a_{1'}(t')\right) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> \end{eqnarray} $$

After applying the same trick to the $a^\dagger$ terms, we have $$ \left<a_{1'}(+\infty) a_{2'}(+\infty) a_1^\dagger (-\infty) a_2^\dagger (\infty) \right> = \int_{-\infty}^{+\infty} dt_{2'} \frac{\partial}{\partial t_{2'}} \int_{-\infty}^{+\infty} dt_{1'} \frac{\partial}{\partial t_{1'}} \int_{-\infty}^{+\infty} dt_{1} \frac{\partial}{\partial t_{1}} \int_{-\infty}^{+\infty} dt_{2} \frac{\partial}{\partial t_{2}} \left<\mathrm{T} a_{2'}(t_{2'}) a_{1'}(t_{1'}) a_1^\dagger (t_1) a_2^\dagger (t_2) \right> $$

Finally, we invoke the commutability of the time-ordering operator $\mathrm{T}$ and time-derivative $\partial_0$ to get the final form of the LSZ formula.

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I googled the LSZ reduction formula, but many of the are almost the same as Srednicki's except this one http://porthos.ist.utl.pt/~romao/homepage/publications/Lectures/Lectures-TCA-c2.pdf by Jorge Romao. He didn't insert a T directly in the expression of S matrix element like (5.13), so his final result contains a "disconnected term", plus (5.15). He stated that this disconnected term "will vanish if none of the initial momenta coincides with one of the final momenta", and "from now on we will no longer consider the disconnected terms, because in practice we are only interested in the cases where all the particles interact."

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