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I'm reading the section on the LSZ reduction formula in Schwartz's QFT book and he talks about the action of free fields in the formula. Specifically he says (sec. 6.1.1, p. 73):

The LSZ reduction says that to calculate an S-matrix element, multiply the time-ordered product of fields by some $\square + m^2$ factors and Fourier transform. If the fields $\phi(x)$ were free fields, they would satisfy $(\square + m^2)\phi(x)=0$ and so the $(\square_i + m^2)$ terms would give zero. However, as we will see, when calculating amplitudes, there will be factors of propagators $\frac{1}{\square + m^2}$ for the one-particle stales. These blow up as $(\square + m^2)\rightarrow 0$. The LSZ formula guarantees that the zeros and infinities in these terms cancel, leaving a non-zero result.

He talks about the $\square+m^2$ terms giving zero like that's a bad thing, but isn't that what we want? If the fields are free fields then the $S$-matrix will simply be the identity and so the matrix element will vanish, $$\langle f|S|i\rangle=\langle f|i\rangle=0,$$ for distinct initial and final asymptotic states. So why would we want those terms in the LSZ formula to be non-zero for free fields?

For reference the form of the LSZ formula Schwartz is referring to is $$\langle f|S|i\rangle=\left[i\int\mathrm{d}^4x_1\,e^{-ip_1x_1}(\square_1+m^2)\right]\cdots\left[i\int\mathrm{d}^4x_n\,e^{ip_nx_n}(\square_n+m^2)\right]\langle\Omega|T\{\phi(x_1)\cdots\phi(x_n)\}|\Omega\rangle.$$

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If the S-matrix is zero (or actually $1$, since $S = 1 + iT$, so you should say more correctly say "if the T-matrix is zero") then equivalenlty the corresponding theory is non-interacting. However if you switch on an interaction you expect there to be some scattering, observable in a particle collider say, i.e. you expect the $T$-matrix to have some non-zero matrix element.
To clarify: The LSZ formula gives zero for a free field theory, but once you turn on an interaction this is no more the case, the fields $\phi$ in your Greens function $\langle \Omega |T\phi(x_1)...\phi(x_n) |\Omega \rangle$ are then not free fields. In fact those Greens function turn out to be divergent, through a simple pole, in the physical on-shell limit $p^2 \rightarrow m^2$ (or equivalently $\Box^2 + m^2 \rightarrow 0$ in position space) and these poles get exactly cancelled by in the formula, leaving only the coefficient of those poles, which is you scattering amplitude.

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  • $\begingroup$ Right, I understand why, for an interacting theory, the LSZ formula will give a non-zero matrix element. My confusion is why Schwartz claims that it is still non-zero for a free field (due to the propagator terms that pop up) when it should, like you said, give zero. $\endgroup$ Jan 24, 2021 at 10:46
  • $\begingroup$ Sorry, I do not see where it is stated that it should be non-zero for free fields? He explicitely states in your quote that the matrix will vanish by the equations of motion in the free case, no? $\endgroup$
    – jkb1603
    Jan 24, 2021 at 11:15
  • $\begingroup$ Well if I'm reading it correctly, he's saying that the free fields would ordinarily give $(\square_i+m^2)\phi(x)=0$, however $\frac{1}{\square_i+m^2}$ terms pop up during the calculation that blow up and "cancel out" said zeros, giving a non-zero result in the end. $\endgroup$ Jan 24, 2021 at 11:22
  • $\begingroup$ Oh, I think he means that these terms pop up only for the interacting case. $\endgroup$
    – jkb1603
    Jan 24, 2021 at 11:29
  • $\begingroup$ Well crap haha, if that's the case then I've just wasted a long time banging my head over this. Though why would there be zeros in the interacting case? Would those just correspond to the asymptotic states? $\endgroup$ Jan 24, 2021 at 11:39

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