Feynman rules - where do the imaginary numbers come from?

  • Thread starter guest1234
  • Start date
  • #1
41
1

Main Question or Discussion Point

I'm trying to learn how to derive Feynman rules (what else to do during xmas, lol).
The book I'm using is QFT 2nd ed by Mandl&Shaw. On p 428 they're trying to show how to derive a Feynman rule for [itex]W W^\dagger Z^2[/itex] interaction term [itex]g^2 \cos^2\theta_W\left[W_\alpha W_\beta^\dagger Z^\alpha Z^\beta - W_\beta^\dagger W^\beta Z_\alpha Z^\alpha\right][/itex]. The idea goes that a) momenta is assigned to every particle b) all fields are replaced with corresponding polarization vectors.
But for some reason they write out the respective amplitude with an imaginary unit [itex]i[/itex] in front, like:
[itex]\mathcal{M} = ig^2 \cos^2\theta_W\left\{ \varepsilon_\alpha(2')\varepsilon_\beta(1')\left[\varepsilon^\alpha (1)\varepsilon^\beta(2) + \varepsilon^\alpha (2)\varepsilon^\beta(1)\right] - \varepsilon_\beta(2')\varepsilon_\beta(1')\left[\varepsilon^\alpha (1)\varepsilon^\alpha(2) + \varepsilon^\alpha (2)\varepsilon^\alpha(1)\right] \right\}[/itex].
Where does this come from? I don't see this mysterious number occur in the next example.

Thanks
 

Answers and Replies

  • #2
129
30
Hi, I should really write it down before I answer, however just from the top of my head,

This 4 gauge boson term comes the the F_mu nu F^mu nu term. The particular piece for this interaction is the extra non abelian in the field strength being squared.

So is the difference between the other example that it is a 3 gauge interaction? I.e that the i isn't squared in this example.
 
  • #3
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
14,406
5,985
There are many places, where i's enter the game. One is the i in the time-evolution operator of quantum theory, and the S-matrix is a special case of a time-evolution operator (which has to defined very carefully indeed!). Then there are i's from the Fourier transform from time-position space to energy-momentum space, where time derivatives ##\partial_{\mu}## in the action map to ##-\mathrm{i} p_{\mu}## in the Fourier decomposition of the propagators and (proper) vertex functions, which make up the building blocks of the perturbative expansion, which is nicely written in Feynman diagrams, which can be seen as a very clever abbreviating notation for the perturbation series to evaluate the S-matrix elements for a given scattering or decay process as well as (with a large grain of salt!) space-time pictures of such processes.
 
  • #4
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
2019 Award
24,034
6,607
Vanhees is correct, there are many places where i's enter in, but I don't think it answers the question about this particular i. The answer to that is "it's convention". Because the matrix element never appears directly, only as it's square, you'll get the same answer if you leave the i in, take it out, or replace it with some other phase.

Now, if you have two amplitudes, by a similar argument, the absolute phases don't matter - but the relative phases between the two amplitudes does. To keep these straight, we have (somewhat, but not completely arbitrarily) assigned phases to various amplitudes. This one happens to be i. We could make it -1 or 1 or -i or (1+i)/sqrt(2) if we wanted, but only if we changed the conventions for every other amplitude.

As an aside, there are amplitudes where the phase convention isn't completely settled, and there have been mistakes made by people who mixed them together in their calculations, and thus got the wrong result. So it is good to have these conventions.
 

Related Threads on Feynman rules - where do the imaginary numbers come from?

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
9
Views
2K
  • Last Post
2
Replies
31
Views
3K
Replies
4
Views
3K
Replies
9
Views
425
Replies
2
Views
2K
  • Last Post
Replies
19
Views
3K
  • Last Post
Replies
3
Views
1K
Replies
36
Views
2K
Replies
9
Views
2K
Top