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

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


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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – Danu
    Commented Aug 2, 2015 at 11:32
  • $\begingroup$ 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. $\endgroup$
    – Qmechanic
    Commented Aug 2, 2015 at 21:04
  • $\begingroup$ Related: physics.stackexchange.com/q/94532/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Aug 2, 2015 at 21:43
  • $\begingroup$ 404: Links now dead. $\endgroup$
    – Qmechanic
    Commented Jan 15, 2022 at 19:48

1 Answer 1


Comments to the question (v4):

  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.


  1. M.D. Schwartz, QFT and the Standard Model; Section 6.1, p.72, footnote 2.


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

  • $\begingroup$ I plan to update this answer. $\endgroup$
    – Qmechanic
    Commented Jul 17, 2016 at 18:43

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