Let's say we want to calculate the imaginary part of the following scalar diagram in $\varphi^3$ theory:
This amplitude is given by the expression $$i \mathcal{M} = i^5 \int \frac{d^4 \ell_1}{(2\pi)^4} \int \frac{d^4 \ell_2}{(2\pi)^4} \frac{1}{D_1 D_2 D_3 D_4 D_1},$$ where $D_k$ is the $k$-th denominator as in the figure. To take the imaginary part, we have to perform cuts according to Cutkosky. Here we have three possible cuts - we can cut the following lines : $12$, $234$, and again $12$. Employing the cutting rules, for the cut $k$-th line we put $$\frac{1}{D_k} \to -2\pi i \delta (D_k)$$ and get $$\Im \mathcal{M} = 2 \int \frac{d^4 \ell_1}{(2\pi)^4} \int \frac{d^4 \ell_2}{(2\pi)^4} \frac{(-2\pi i)^2}{D_3 D_4} \frac{\delta(D_1)}{D_1} \delta(D_2) + \int \frac{d^4 \ell_1}{(2\pi)^4} \int \frac{d^4 \ell_2}{(2\pi)^4} \frac{(-2\pi i)^3}{D_1^2} \delta(D_2) \delta(D_3) \delta(D_4).$$ The problem is, of course, the first term. How to interpret the explicitly divergent term $\delta(D_1)/D_1$?
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3$\begingroup$ Isn't the Cauchy's residue theorem the origin of the replacement in Cutkosky's rule $1/D \rightarrow -2\pi i \delta(D)$? If you have $1/D^2$, you ought to replace the whole factor with something else. $\endgroup$– QuantumDotCommented Jan 12, 2017 at 1:05
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1$\begingroup$ @QuantumDot Do you mean using the formula for the residue of a higher-order pole? $\endgroup$– user17116Commented Jan 12, 2017 at 1:13
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$\begingroup$ Well, I don't have a concrete answer for you. But the residue of higher order poles vanish, so it can't be exactly that. The best bet is to return to the derivation of Cutkosky's rule and see what to do with double propagator factors. I just now finished calculating a simplified version of the two loop integral above and found that nothing strange is going on in the imaginary part. $\endgroup$– QuantumDotCommented Jan 12, 2017 at 1:15
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$\begingroup$ Essentially, what I did was to pretend that the masses in propagators 3 and 4 are so heavy, that the inner loop can be approximated as a point, yielding a 2-point vertex. The result is a one loop integral with repeated propagator factors. This diagram exhibits the same issue raised in your question. Then I just did the whole integral using standard methods to see what its imaginary part looks like, and found that it is perfectly regular. But, I'm afraid you'll find it unenlightening because I used a computer to do that integral. $\endgroup$– QuantumDotCommented Jan 12, 2017 at 1:22
2 Answers
Going back to Cutkosky's original paper (http://aip.scitation.org/doi/10.1063/1.1703676), it is clear he derives his result via residue theorem, as QuantumDot pointed out in his comment. Therefore, it seems natural that the generalization of the Cutkosky's cutting rule would have to analogous to the formula for the residue of a pole of order higher than one. Explicitly, if the cut propagator is raised to the $n$-th power, we should substitute $$\frac{1}{D^n} \to (-2\pi i) \frac{(-1)^{n-1}}{(n-1)!} \delta^{(n-1)} (D).$$ In case $n=2$, we would then have $$\frac{1}{D^2} \to (2\pi i) \delta'(D).$$ While I have not yet checked this substitution rule in a real calculation, I suspect that it will hold.
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5$\begingroup$ Also, note that $$ \mathrm{Im}\left[\frac{1}{(x+i\epsilon)^n}\right]=\frac{(-1)^n}{(n-1)!}\frac{\partial^{n-1}}{\partial x^{n-1}}\frac{\epsilon}{x^2+\epsilon^2}=\frac{(-1)^n}{(n-1)!}\frac{\partial^{n-1}}{\partial x^{n-1}}\pi\delta(x)+\mathcal O(\epsilon) $$ in agreement with your formula. $\endgroup$ Commented Jan 16, 2017 at 11:44
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$\begingroup$ This approach seems correct, but how is it possible that no one talks about this problem? I’ve searched everywhere for a discussion about this and found nothing, everyone just assume that cutkosky rules (in their original form) can be applied to every feynman diagram, i’m worried about this lack of rigor. Neither i’ve found examples of application of the rules to a diagram with the same problem of the one above… $\endgroup$– dalllaCommented Mar 15 at 15:01
Suppose you cut the propagator #1 on the left, and #2. Think about the diagram you get on the right side of the cut. The cut propagator appears as an external leg. The nested loop inside the diagram that we didn't cut now appears as the one-loop correction to the external leg. With that in mind, if you look at, say, at the discussion around Eq 4.102 in Peskin, you will find that he talks about exactly the issue the you mention, and foreshadows a discussion of the LSZ theorem, which ultimately instructs you to amputate such things from the legs of a Feynman diagram when computing an S-matrix element. That's all that's going on here.
