# Regarding a small step in the derivation of the LSZ formula

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 $(\partial^2+m^2)\phi$ and in (24-25) it is to the right of the differential operator.

I'm asking how to go from $T(\partial^2+m^2)\phi\cdots$ to $(\partial^2+m^2) T\phi\cdots$. I feel that this cannot be done in general, because the symbol $T$ will introduce some $\Theta(x_0-y_0)$ 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

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 $|i\rangle\propto a^\dagger_1(-\infty)a^\dagger_2(-\infty)|0\rangle$ and the out state to be $|f\rangle\propto a^\dagger_3(+\infty)a^\dagger_4(+\infty)|0\rangle$. Then the transition amplitude is $$\langle i|f\rangle \propto \langle 0|a_2(-\infty)a_1(-\infty)a^\dagger_3(+\infty)a^\dagger_4(+\infty)|0\rangle=$$ $$=\langle 0|Ta_2(-\infty)a_1(-\infty)a^\dagger_3(+\infty)a^\dagger_4(+\infty)|0\rangle$$

Now write $a(+\infty)-a(-\infty)\propto \int \mathrm d x\ \mathrm e^{\cdots}(\partial^2+m^2)\phi$; all terms with $a$'s or $a^\dagger$'s annihilate the vacuum, so that the only remaining term is $$\langle i|f\rangle \propto \langle 0|T \int \mathrm dx_1\mathrm dx_2\mathrm dx_3 \mathrm dx_4\ \mathrm e^{\cdots}(\partial_1^2+m^2)(\partial_2^2+m^2)(\partial_3^2+m^2)(\partial_4^2+m^2) \phi_1\phi_2\phi_3\phi_4 |0\rangle$$

This is usually writen as $$\langle i|f\rangle \propto \int \mathrm dx_1\mathrm dx_2\mathrm dx_3 \mathrm dx_4\ \mathrm e^{\cdots}(\partial_1^2+m^2)(\partial_2^2+m^2)(\partial_3^2+m^2)(\partial_4^2+m^2) \langle 0|T\phi_1\phi_2\phi_3\phi_4 |0\rangle$$ but this step is problematic, because the $T$ symbol and the KG operator don't commute, right?

• Instead of providing links, you should make your post self-contained, so that we don't have to visit other pages just to see if we can/want to help you. – Danu Aug 2 '15 at 11:32
• Echoing what @Danu said, consider typing in eqs. (22)-(25). Also consider to mention explicitly author, title, etc. of links, so it is possible to reconstruct links in case of link rot. – Qmechanic Aug 2 '15 at 21:04
• Related: physics.stackexchange.com/q/94532/2451 and links therein. – Qmechanic Aug 2 '15 at 21:43

1. 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.

2. Consider the on-shell $S$-matrix side of the LSZ reduction formula. The time-differentiation from the boundary terms $$T\left[\prod_{i=1}^n \left\{ a_{{\bf p}_i}^{\#}(t_i\!=\!\infty) -a_{{\bf p}_i}^{\#}(t_i\!=\!-\infty)\right\}\right]$$ $$~=~T\left[\prod_{i=1}^n \int_{\mathbb{R}}\! \mathrm{d}t_i~\frac{d}{dt_i}a_{{\bf p}_i}^{\#}(t_i)\right] ~=~\left[\prod_{i=1}^n \int_{\mathbb{R}}\! \mathrm{d}t_i~\frac{d}{dt_i}\right]T\left[\prod_{j=1}^n a_{{\bf p}_j}^{\#}(t_j)\right]\tag{A}$$ can be moved outside the time-ordering symbol $T$ because the contact terms vanish $$\delta(t_i-t_j) \left[a_{{\bf p}_i}^{\#}(t_i) ,a_{{\bf p}_j}^{\#}(t_j) \right]~=~0\tag{B}$$ for generic$^1$ 3-momenta ${\bf p}_i\neq {\bf p}_j$. Eq. (B) follows from locality, i.e. spatially separated operators commute. [Here $\#$ refers to creation/annihilation operators, i.e. with or without Hermitian conjugate.]

3. In the Hamiltonian formulation with only first-order time-derivatives, the above shows that contact terms vanish.

4. 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:

1. 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., ${\bf p}_i={\bf p}_j$.

• I plan to update this answer. – Qmechanic Jul 17 '16 at 18:43