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Feynman Diagram setup for 4-fermion interaction loop

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Hepth
#1
Jul15-14, 05:29 AM
PF Gold
Hepth's Avatar
P: 472
If one has a 4-fermion vertex, like in Fermi theory : ##G_f (f_1 \Gamma f_2)(l_1 \Gamma l_2)##

And you are calculating a one-loop diagram where you have the diagram of :

f1-> f2,l1,l2 -> f1



(used for dispersive/unitarity approach)

Where in the end you'll use Cutkosky rules to calculate it, but ignore this for now.

The question is about converting this diagram to an equation, which direction does one go? You can follow fermion flow backward through either of the two forward-arrow propagators, but then what. It matters as there is dirac algebra involved.

I assume it has to do with the order of the fermions in your actual interaction term, and when doing the time ordered product of two of these vertices you have to watch for fierz transforming the currents so that its manageable.

Is that correct, that I should approach this "diagram" without usign the diagram, but rather a product of currents, time-ordered, and do it the long way, fierz-transforming the spinor currents so that things simplify? Or is there a prescription for handing this case straight from the diagram.

It seems a feynman diagram is not enough to be honest for 4-fermion interactions of this sort, as the directions matter, and "going against particle flow" is not enough when you have choices. So you must either sum over all the paths, or do it by hand.

Any insight?
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Einj
#2
Jul15-14, 08:32 AM
P: 328
I'm not sure if I understood your question correctly. However, the Fermi Lagrangian just tells you that, at low energy, the coupling between two fermionic currents is [itex]G_F/\sqrt{2}[/itex]; that's what should appear at every four-fermion vertex. The presence of the Dirac algebra, i.e. of spinors, just tells you that you have to use spinors for the external legs of you diagrams and fermionic propagators for the internal ones.
In your case you need to choose a direction for the loop momenta and then write down the integral correctly (don't forget the additional - sign for every fermionic loop). If, for example, [itex]p[/itex] is the momentum of the incoming/outgoing fermion, you can choose the upper internal line to go from left to right with a momentum [itex]q[/itex], the middle internal line to go from left to right with momentum [itex]p-q+k[/itex] and the lower one to go from right to left with momentum [itex]k[/itex].
In this case I would say that you diagram is given by:

$$
\left(\frac{G_F}{\sqrt{2}}\right)^2\bar u_L(p)\int \frac{d^4q}{(2\pi)^4}\frac{d^4k}{(2\pi)^4}\frac{1}{q^\mu\gamma_\mu-m}\frac{1}{(p-q+k)_\mu \gamma^\mu-m}\frac{1}{k_\mu \gamma^\mu-m} u_L(p),
$$

where by [itex]u_L(p)[/itex] I mean the left-handed spinor.
Hepth
#3
Jul15-14, 08:39 AM
PF Gold
Hepth's Avatar
P: 472
Ah yes. I was thinking unclearly. Of course it solves itself as each loop is considered independently, and you have a degree of freedom when choosing because you technically have 3 close loops to choose your 2-loop variables to travel around.

Thanks!


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