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Feynman Factors & Relativistic Scalar Propagator

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Nov26-12, 05:13 PM
P: 209
Hey again,

I have a question on a couple of things related to feynman diagrams but also the relativistic scalar propagator term.

First of all, this interaction:

The cross represents a self-interaction via the mass and characterised by the term: -im^2, is this just some initial state then self-interacting with itself via the mass, with nothing changing and it entering into a final state the same as the initial state? Can anybody explain what exactly is happening in this interaction?

My second question is on the propagator for a relativisitc scalar particle, I believe it has form:


My professor said that this is where this form comes from, he said to imagine summing up all the possible number of self-interaction from 0 self interactions to (presumably) and infinite number of interactions, so :

So the first line has no mass interaction so m=0 and the factor for the first one is :


then the second line has one mass interaction so the factor associated with it is :


and the third line has 2 mass interactions and so the factor is :


And so we sum all these factors up (to the maximum number of self-interactions) and can make factorisation below:

[tex]\frac{i}{E^{2}-\mathbf{p}^{2}}(1+\frac{m^{2}}{E^{2}-\mathbf{p}^{2}}+\frac{m^{4}}{(E^{2}-\mathbf{p}^{2})^2}+\cdots )[/tex]

We use identity:

[tex](1+x)^{-1}=1+x+x^{2}+x^{3}+\cdots\: ,\: x=\frac{m^{2}}{E^{2}-\mathbf{p}^{2}}[/tex]

and thus obtain the relativistic scalar propagator:


How does this work? Is the scalar interaction just the sum of all the self-interactions by mass?

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Nov27-12, 05:53 AM
P: 1,020
It is rather different from what you think.the identity used is rather
the cross sign represents all self interaction summation and writing the modified propagator which is obtained simply by supplying the feynman rules for those interaction and summing them.Also mass change is simply identified by well known relation,
EdE=mdm,you can see page 284 here and forward for more on this

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