# Using the optical theorem to calculate the imaginary part of a loop diagram

I'm trying to calculate the imaginary part of this diagram in $\phi^4$ theory, using the optical theorem, and I'm having trouble.

All of the examples I can find use the theorem to relate the imaginary part of the total 2-particle to 2-particle forward scattering amplitude to the total cross-section; that's not what I'm trying to do. I see a lot of equations that look like this: $$2 \operatorname{Im} A = \int d\Pi | B |^2$$ wherein $A$ is the diagram above and $B$ is but nobody really discusses such equations or works any examples. Does that mean that I just take the modulus squared of the tree-level diagram: $|i \lambda|^2 = \lambda^2$? That seems too simple, which brings me to the integration over $d\Pi$. What the heck is $d\Pi$? Peskin and Schroeder don't say, but they make vague mention of the phase space of the intermediate particles, so is $d\Pi$ just the differential phase space of the "two" $\phi$s in the loop? If so, how do I go about setting up and evaluating that integral? If not, what is $d\Pi$, and how do I evaluate the integral over it?

• Hi calavicci - on this site each post should have one question, or at least a few closely related questions, so I've removed the other questions from your post. You can post them separately if you like (although questions that just as "is this right?" tend not to be well received here). – David Z Mar 28 '15 at 18:08
• Well, it really is just "one" question... "how do I go about evaluating this particular diagram using the optical theorem." However, I've already done some research on this, and there seem to be a couple of possible approaches, but are probably the same. Omitting what I've found about them so far, or splitting them up, doesn't seem helpful. I could just remove most of my work to this point and ask "how do I do this thing," but that seems less useful than including what I've found so far. Please revert the change; I can remove the separation into sub-questions if that would be better. – calavicci Mar 28 '15 at 18:12
• That one question is probably too broad. There are chapters of textbooks written about it. (That's also a criterion for a question not to be acceptable here, i.e. if it's too broad.) If you've done prior research, that's great, but if the research leads you to an approach, you should try that approach, and if you run into trouble, then you post a question about the thing you ran into trouble on. Don't just ask if what you found is going to work. So in this case, your question about $\mathrm{d}\Pi$ are a pretty good one for this site on its own. – David Z Mar 28 '15 at 18:16
• I have tried several approaches, and I did run into trouble on each of them, and now I am asking for help at the point I got stuck in any of them. As for too broad, the calculation of a specific diagram doesn't seem broad to me, personally. I'm not going to try to argue with a moderator, so I'll post three more questions, but I respectfully disagree with your evaluation. I'm also mildly offended at your implication that I just found something in a textbook and asked if it was going to work without trying it on my own; I've tried to get these things to work out for hours at this point. – calavicci Mar 28 '15 at 18:21
• I didn't imply that - at least I didn't mean to, but sorry if I did. I was just explaining how the ideal question-asking process works, in general, not describing what I thought you did in this specific case. – David Z Mar 28 '15 at 18:24

d$\Pi$ is, indeed, the differential phase-space. Peskin and Schroeder have an equation of exactly the form above in figure 7.6 on page 235, and although they don't say what $d\Pi$ is there, they do define a similar, but more specific, quantity $d\Pi_n$, the differential phase-space for $n$ particles, in equation 4.80 on page 106: $$d\Pi_n = \left ( \prod _f \frac{d^3 p_f}{(2\pi)^3} \frac{1}{2E_f} \right ) (2\pi)^4 \delta^{(4)}\left (P-\sum_f p_f \right)$$ wherein the $\Pi$ on the left is a variable indicating the $n$-body phase space, the $\prod$ on the right is a product symbol, $P$ is the net external 4-momentum, and the subscript $f$ indicates the final state 4-momenta of the $n$ particles.
Noting that $$\int \frac{d^4 p}{(2\pi)^4} 2\pi\delta(p^2-m^2) = \int\frac{d^3 p}{(2\pi)^3}\frac{1}{2 E_{\vec p}}$$ this prescription for $d\Pi$ yields the same integrals for evaluation of a diagram via the optical theorem as do Cutkosky's cutting rules, confirming that this is the correct $d\Pi$ and not just a coincidence of notation.