I'd like to prove the LSZ formula, but there is a specific step that is bugging me a lot. I know there are many subtleties in its derivation, but I'm not worrying about this right now: I'm trying to understand the naive proof, so to speak.
You can find an example of the usual proof here: http://isites.harvard.edu/fs/docs/icb.topic473482.files/06-LSZ.pdf
My question is: how to get eqs. (24-25) from eq. (22-23). In (22-23) the time-ordering symbol is to the left of (∂2+m2)ϕ and in (24-25) it is to the right of the differential operator.
I'm asking how to go from T(∂2+m2)ϕ⋯ to (∂2+m2)Tϕ⋯. I feel that this cannot be done in general, because the symbol T will introduce some Θ(x0−y0) functions, which when differentiated will give rise to some deltas.
This step makes no sense to me... Is my question legitimate? Is there anything from the proof I'm missing?
If I think of specific examples, I find different results depending on whether the T symbol is to the right or to the left of the Klein-Gordon differential operator. This means that the order is important... so, which is the right order? Should the T symbol be placed to the right or to the left of the KG operator? In the begining of the proof the T symbol is always to the left, and in the end, it is always to the right.
This same problem appears on many proofs online, such as
http://isites.harvard.edu/fs/docs/icb.topic473482.files/06-LSZ.pdf (eqs. 22-25)
http://www2.ph.ed.ac.uk/~egardi/MQFT_2013/MQFT_2013_lecture_2.pdf (eqs. 32-33)
Srednicki's book, page 51 (eqs. 5.14-5.15) online: http://web.physics.ucsb.edu/~mark/qft.html
etc.
Edit: I was asked to post a self-contained question, so I'll write the delails here:
Take the in state to be |i⟩∝a†1(−∞)a†2(−∞)|0⟩ and the out state to be |f⟩∝a†3(+∞)a†4(+∞)|0⟩. Then the transition amplitude is ⟨i|f⟩∝⟨0|a2(−∞)a1(−∞)a†3(+∞)a†4(+∞)|0⟩=
Now write a(+∞)−a(−∞)∝∫dx e⋯(∂2+m2)ϕ; all terms with a's or a†'s annihilate the vacuum, so that the only remaining term is ⟨i|f⟩∝⟨0|T∫dx1dx2dx3dx4 e⋯(∂21+m2)(∂22+m2)(∂23+m2)(∂24+m2)ϕ1ϕ2ϕ3ϕ4|0⟩
This is usually writen as ⟨i|f⟩∝∫dx1dx2dx3dx4 e⋯(∂21+m2)(∂22+m2)(∂23+m2)(∂24+m2)⟨0|Tϕ1ϕ2ϕ3ϕ4|0⟩
Answer
Comments to the question (v4):
OP is wondering about the contact terms from commuting time-derivatives and time-ordering symbol T, cf. e.g. this and this Phys.SE posts.
Consider the on-shell S-matrix side of the LSZ reduction formula. The time-differentiation from the boundary terms T[n∏i=1{a#pi(ti=∞)−a#pi(ti=−∞)}]
= T[n∏i=1∫Rdti ddtia#pi(ti)] = [n∏i=1∫Rdti ddti]T[n∏j=1a#pj(tj)]can be moved outside the time-ordering symbol T because the contact terms vanish δ(ti−tj)[a#pi(ti),a#pj(tj)] = 0for generic1 3-momenta pi≠pj. Eq. (B) follows from locality, i.e. spatially separated operators commute. [Here # refers to creation/annihilation operators, i.e. with or without Hermitian conjugate.]In the Hamiltonian formulation with only first-order time-derivatives, the above shows that contact terms vanish.
In the Lagrangian formulation with second-order time-derivatives (i.e. one more time differentiation, which is the case OP is asking about), one may show using similar arguments, that contact terms do not contribute to the S-matrix. See also e.g. Ref. 1.
References:
- M.D. Schwartz, QFT and the Standard Model; Section 6.1, p.72, footnote 2.
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1 Also note that we are usually not interested in disconnected parts of the S-matrix, which implies more momentum conservation laws, and hence more special values of the momenta, such as, e.g., pi=pj.
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