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  S-Matrix Interpretation and Predictions

+ 6 like - 0 dislike
1540 views

How does one distinguish between the second-loop contribution of a known particle, and the first-loop contribution of a more massive-and as yet undiscovered-particle in the S-matrix and/or differential cross section?

Imagine: I calculate a one-loop correction to some scattering process in QED. My photon propagator needs to be corrected for all three lepton generations. I then calculate the differential cross section. The experimentalist then shows me a plot of his precise measurements and tells me that my calculations need further corrections.

Question: How do I know that the necessary correction is from two-loop diagrams, and not from a fourth lepton in the first-loop diagram that I haven't considered?


This post imported from StackExchange Physics at 2016-11-08 14:00 (UTC), posted by SE-user Optimus Prime

asked Nov 1, 2016 in Theoretical Physics by Optimus Prime (105 points) [ revision history ]
edited Nov 8, 2016 by Optimus Prime
Huh, because they are quite different beasts?

This post imported from StackExchange Physics at 2016-11-08 14:00 (UTC), posted by SE-user OON
Please elaborate in what context. If experimentally - you calculate for both models and look what fit observations better.

This post imported from StackExchange Physics at 2016-11-08 14:00 (UTC), posted by SE-user OON
I modified my question. Does it help?

This post imported from StackExchange Physics at 2016-11-08 14:00 (UTC), posted by SE-user Optimus Prime

2 Answers

+ 4 like - 0 dislike

Made up example:

enter image description here

where the solid line represents the one-loop calculation, and the dashed one the two-loop one.

On the other hand,

enter image description here

where the solid line represents the one-loop calculation with three generations, and the dashed one represents the one-loop calculation with four generations (the new particle has a mass close to $s=4$ in this scale).

In other words: more loops slightly change the overall look of the cross section. More particles change its behaviour close to the mass of such particles.

This post imported from StackExchange Physics at 2016-11-08 14:00 (UTC), posted by SE-user AccidentalFourierTransform
answered Nov 1, 2016 by AccidentalFourierTransform (480 points) [ no revision ]
Interesting. So what these plots say is that higher order loops change the diagram as a whole, whereas a new particle would merely show up as a bump. The only question I have is how come the three original leptons didn't show up as bumps near their respective masses?

This post imported from StackExchange Physics at 2016-11-08 14:00 (UTC), posted by SE-user Optimus Prime
They actually do show up in their respective masses! My diagram only shows the fourth bump for simplicity.

This post imported from StackExchange Physics at 2016-11-08 14:00 (UTC), posted by SE-user AccidentalFourierTransform
What are their masses in this scale?

This post imported from StackExchange Physics at 2016-11-08 14:00 (UTC), posted by SE-user Optimus Prime
This is a made up diagram, it doesn't correspond to anything physical. It's only meant to illustrate the general trend of loop corrections and resonances. In this case, you can take the three first masses to be, say, very close to the origin (so that they can be treated as massless).

This post imported from StackExchange Physics at 2016-11-08 14:00 (UTC), posted by SE-user AccidentalFourierTransform
Great. Thanks for the effort.

This post imported from StackExchange Physics at 2016-11-08 14:00 (UTC), posted by SE-user Optimus Prime
@OptimusPrime no problem. If you have any more questions, feel free to ask.

This post imported from StackExchange Physics at 2016-11-08 14:00 (UTC), posted by SE-user AccidentalFourierTransform
+ 0 like - 0 dislike

The question is rather strange. If you can, you make one- and two-loop calculations within an old model, and if possible, estimate contributions of higher order corrections. Then you compare your results and estimations with the experimental data. If the old model fails, you need a new model, for example, with a new and heavier particle, why not?

If you cannot make a two loop calculation and you have a significant uncertainty in your estimations of the old model, you stay with this uncertainty and you cannot decide.

answered Nov 9, 2016 by Vladimir Kalitvianski (102 points) [ no revision ]

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