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The optical theorem, which results from the unitarity of the $S$-matrix, relates the imaginary part of the forward scattering amplitude to the total cross section. When using this theorem in practice, one often invokes the fact that in perturbation theory, $S$-matrix elements come out purely real, unless there is a Feynman diagram contribution where an intermediate particle goes on shell. According to Peskin and Schroeder, this is "easily checked" in QED.

This is true, but I had a hard time seeing why it held in more general theories. For example, consider $\phi^n$ theory. Keeping track of only phases mod $\pi/2$, each vertex comes with a factor of $i$, and simultaneously each vertex yields $n/2$ propagators, each with a factor of $i$, so a vertex gives a factor of $i^{n/2+1}$. When $n$ is odd, vertices have to come in pairs, so different orders in perturbation theory are related by a factor of $i^{n+1}$, which is real. But when $n$ is $0 \, (\text{mod} \, 4)$, different orders in perturbation theory contribute with relative factors of $i$, so it looks like the statement is false.

The only way out that I can see is to assign a factor of $i$ to every loop integral $\int d^4 k$. If such a factor exists, then it's simple to establish the result using Euler's formula.

Indeed, this is precisely what happens in dimensional regularization, where the master formula is $$\int \frac{d^dp}{(2\pi)^d} \frac{p^{2a}}{(p^2-\Delta)^b} = i(-1)^{a-b} \frac{1}{(4\pi)^{d/2}} \frac{1}{\Delta^{b-a-d/2}} \frac{ \Gamma(a+\frac{d}{2}) \Gamma(b-a-\frac{d}{2})}{\Gamma(b) \Gamma(\frac{d}{2})}$$ and the right hand side has the factor of $i$ from the Wick rotation. On one hand, this is very strange: a real integral is being regulated to an imaginary number! But on the other hand, dimensional regularization just is strange, e.g. it sets massless integrals to zero.

What I find more disturbing is the apparent requirement that loops each contribute a factor of $i$. This doesn't appear to be true of any regularization scheme I know of besides dimensional regularization. Pauli-Villars, a Wilsonian hard cutoff, and the lattice work by modifying the integrand in the purely real loop integral on the left at high energies, and so can't possibly turn it pure imaginary. That would seem to imply that all these regularization schemes violate unitarity, and in fact violate it maximally. But I've never seen anybody say that, and moreover a lattice theory in a finite box is finite-dimensional, and in this case unitarity is trivial to establish.

What's going on here?

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  • $\begingroup$ Please correct me if I'm wrong. The $i$ from Wick rotation is there regardless of how you regulate, no? The integrand was never purely real - there's an understood $i\epsilon$ there that's always in the denominator of the loop integrals, and the step of Wick rotation generically gives a factor of $i$. Pauli-Villars will give you an $i$ all the same - see for example Srednicki lines 9.22 and 9.23. $\endgroup$
    – user196574
    Mar 29 '20 at 5:07
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    $\begingroup$ @user196574 Yeah, exactly! I realized this right after posting, but I figure I should leave this question up as a lightpost for future people. I forgot that stuff like Pauli-Villars and hard cutoff give the $i$ in the very first step. $\endgroup$
    – knzhou
    Mar 29 '20 at 5:17
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On Peskin and Schroeder's statement:

each diagram contributing to an S-matrix element $\mathcal{M}$ is purely real unless some denominators vanish, so that $i\epsilon$ prescription for the treating the poles becomes relevant.

Doesn't the fact that you have a loop already violate that requirement? For each loop integral, you have to integrate over a free four-momentum, the denominator vanishes at the poles and the $i\epsilon$ prescription matters.

In the loop integral: $$ \int \frac{d^dp}{(2\pi)^d} \frac{p^{2a}}{(p^2-\Delta)^b} = i(-1)^{a-b} \frac{1}{(4\pi)^{d/2}} \frac{1}{\Delta^{b-a-d/2}} \frac{ \Gamma(a+\frac{d}{2}) \Gamma(b-a-\frac{d}{2})}{\Gamma(b) \Gamma(\frac{d}{2})} $$

$\Delta$ contains the $i\epsilon$ prescription (so I wouldn't say it is a real integral). Only because of the $i\epsilon$ prescription can we perform a Wick's rotation to evaluate the integral. The $i$ factor arises naturally from Wick's rotation. Aside from that, the $i\epsilon$ prescription in $\Delta$ on the right hand side is still relevant for some cases. So, in general, I wouldn't call the RSH pure imaginary either.

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  • $\begingroup$ Is it not generally true that we can only get an imaginary part to $\mathcal{M}$ with something like an S-channel loop diagram (where the imaginary part comes from a function with a branch cut and the branch ambiguity is resolved via optical theorem)? E.g. $\phi^4$ $\endgroup$ Mar 28 '20 at 2:24

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