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So far I have only actually calculated dimensional regularization and I just know about the idea of cutoff regularization. From what I understand, as the name suggests, you just ignore momentum as from some high value and integrate the virtual momenta until this arbitrary chosen value. Depending on which value you chose, and thereby to what order you consider the Feynman diagrams, you get a (little bit) different result.

Now what I was wondering is, why don't people just cutoff exactly at the value that corresponds to the momentum of the process you want to consider, if there is some specific experiment you want to calculate. E.g. if someone wants to calculate some cross section of an experiment that will be/has been done at CERN, why don't they just consider exactly the momentum that the virtual particle must (maximally) have (due to momentum conservation of the particles that are smashed together)? Intuitively it seems (to me) that this must give the best result. And let us ignore of course, that physicists that calculate cross sections for CERN probably do not use cutoff regularization.

Jack
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    The fact that we are integrating over all loop momentum to begin with means that these momenta are not constrained to a maximum value by momentum conservation. – DJBunk Jul 03 '13 at 14:32
  • No of course not, the question is, in the case of an actual experiment, why don't we constrain it in that way using cutoff regularization? (And momentum conservation is always given by the formalism itself.) – Jack Jul 03 '13 at 14:35
  • I am not quite sure where you are confused. Let me just add this - this is a quantum mechanical system so we need to add all possible 'paths' between the in and out state. One of the things we need to sum over is the loop momentum, which ends up just being the momentum integral. In other words, no experimentalist can say what the loop momentum will be. This is the same way no experimentalist can say which slit the electron goes through in the double slit experiment. Does this clarify? – DJBunk Jul 03 '13 at 14:42
  • No not really. The momentum of the virtual particles cannot exceed the momentum of the two particles that collided. It can be any virtual particle and any number of loops, but the restriction of conservation of energy and momentum is always there. – Jack Jul 03 '13 at 14:45
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    Consider the vacuum polarization diagram (photon to photon with a electron loop). Take the momentum of the incoming photon to be $p$, therefore the outgoing must also be $p$ by momentum conservation. However, the momentum of the electrons in the loop must add to $p$, each aren't constrained by a maximum value. One has $p+k$ and the other has $-k$. We integrate over all $k$. – Will Jul 03 '13 at 14:53
  • Ah, oh gosh, okay! I see, right ;) – Jack Jul 03 '13 at 14:55

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