How can I calculate the cross-section of a process with three possible Feynman diagrams? I usually see examples with only the scattering amplitude defined by the $t$ channel, but if the scattering amplitude is defined by something like the sum of the three channels, $s$, $t$ and $u$, how can we calculate the square of the matrix? Isn't it gonna be something huge, or am I who doesn't understand something fundamental?
$\begingroup$
$\endgroup$
5
-
$\begingroup$ You add the contributions from each of the diagrams and then take the absolute value squared. From this point onwards, you can use the usual manipulations. Yes, it is huge and you get a lot of contributions. $\endgroup$– Níckolas AlvesCommented Jun 15 at 21:46
-
$\begingroup$ So, if I have some scattering that have all the three channels, so let's assume, incorrectly but just to be easier to write, M = 1/s + 1/t + 1/u. The square would be (1/s + 1/t + 1/u)(1/s + 1/t + 1/u) = ...?? I know it's simple math, but since it appeared in a exam i need to be sure. AHahaha $\endgroup$– LittleBlueCommented Jun 15 at 21:48
-
$\begingroup$ You would also need to complex conjugate one factor. {Or take the absolute value before taking the square, which is more work, usually...) $\endgroup$– Jos BergervoetCommented Jun 15 at 22:08
-
$\begingroup$ Well Nickolas is being sort of brief (they are still right), you take the matrix element which is multiple diagrams and you multiply by the entire complex conjugate, so you will have interference terms that also arise. $\endgroup$– TriatticusCommented Jun 15 at 22:12
-
$\begingroup$ This question (v4) seems rather broad. $\endgroup$– Qmechanic ♦Commented Jun 25 at 10:42
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
As I mentioned in the comments, you just need to sum over all diagrams and then take the absolute valued squared. Once you do this, you have the expression for $|\mathcal{M}|^2$ and you can use the usual expressions for the cross section.
To illustrate the process, suppose the expression for the $s$-channel diagram is $F_s$ and similarly for the other channels. Then $$\mathcal{M} = F_s + F_u + F_t$$ and then $$|\mathcal{M}|^2 = (F_s + F_u + F_t)(F_s + F_u + F_t)^*.$$ Notice the conjugate in the last expression.
The idea is similar when you have other contributions. For example, maybe your theory has more than one kind of interaction. In this case, you will sum over all possible diagrams with the number of loops you're looking for when computing $\mathcal{M}$. Then you take the absolute value squared and proceed.
Some amplitudes may involve a lot of diagrams, so depending on the process you may need to compute a bunch of diagrams and then consider all of the possible combinations you get when squaring the amplitude. The calculations can get quite messy in more complex processes and they can be huge sometimes.
answered Jun 15 at 22:13
-
$\begingroup$ Ah yeah that's what I meant in my comment, your answer came in just afterwords +1 $\endgroup$ Commented Jun 15 at 22:14
-
$\begingroup$ Yea, I understood in the comment but thanks! So in the case we have a scattering process like A + B -> A + B, in the second order, we will get 6 terms to calculate in the cross section right? $\endgroup$ Commented Jun 15 at 22:29
-
$\begingroup$ @LittleBlue I figured my comment should rather be an answer, so I posted it as an answer to fit the site's standards. As for your process, that depends on the details of the theory: at tree level, $e^- e^+ \to e^- e^+$ has a different number of diagrams depending on whether you are considering it in pure QED or in the full standard model (because electrons can interact weakly and through the Higgs boson, in addition to the electromagnetic interaction) $\endgroup$ Commented Jun 15 at 22:44