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From Schwartz's QFT textbook, under Ch.18, Mass renormalization, Schwartz introduces a new LSZ formula with renormalized Green's function. He states that the new LSZ formula for QED, with pole mass $m_p$, is $$ \langle f|S|i\rangle = (\not\!{p_f} - m_p)\dots (\not\!{p_i} - m_p) \langle\psi^R\dots\psi^R\rangle_{\text{amputated}} $$ where amputated diagrams signify external lines chopped off until they begin interacting with the other fields.

But don't we get amputated diagram after we apply the $(\not\!{p_j}-m_p)$ onto the Green's function to project it to $S$-matrix? I don't understand why external lines are already 'amputated' even before applying $(\not\!{p_j}-m_p)$.

Also, another side question, is pole mass defined to be physical mass by definition? I don't understand what forces pole mass to be the actual observable physical mass.

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  • $\begingroup$ Yeah, that's why I wrote "(sorry I don't know how to type Dirac slash.)"... Hopefully someone can edit it for me. $\endgroup$
    – Quantization
    Commented Jun 11, 2015 at 6:36

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Your questions are closely related :)

The overall factors $(\not{p} - m)$ are indeed responsible for amputating the external legs. I don't have a copy of Schwartz on hand so I can't comment on what he might have meant, but the correlation functions do have propagators on the external legs before putting them on shell by applying LSZ.

The mass that goes into the LSZ formula is the physical, observable mass (so long as you normalize everything properly)--the mass of the asymptotic state. You can see this if you go through the LSZ derivation. The reason that the pole of the propagator corresponds to the physical mass is precisely because when the LSZ formula puts external particles on shell, it "picks out" the poles in the propagators (since you have $(\not{p} - m)\times 1 / (\not{p} - m)$), at the physical mass.

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  • $\begingroup$ I agree. Schwartz shouldn't write "amputated" on that Green's function, should he? $\endgroup$
    – innisfree
    Commented Jun 11, 2015 at 13:47
  • $\begingroup$ According to Schwartz derivation, renormalized green function has pole at physical mass even before being projected by LSZ formula. Also in the derivation of LSZ, the mass in the LSZ formula is the Lagrangian mass, but when he modifies it, he says including 1PI in the external lines shifts the pole to physical mass. Is this physical mass because it's the mass of 'asymptotic' state? $\endgroup$
    – Quantization
    Commented Jun 11, 2015 at 14:11
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    $\begingroup$ Ah I see. Ok the bottom line is that there's an ambiguity when you do renormalization, and the details of where you choose to put the "physical mass" that you actually observe depend on the scheme you choose. Personally I prefer setting things up so the mass in the lsz formula is the physical one, but you don't necessarily have to do that, Schwarz uses a different convention. The bottom line is that the location of the pole in the propagator is the physical mass, because this tells you how things propagate at large distances. $\endgroup$
    – Andrew
    Commented Jun 11, 2015 at 16:45

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