Feynman diagrams and momentum conservation

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SUMMARY

Feynman diagrams can be represented in both position and momentum space, with 4-momentum conservation applicable in all interactions. While 4-momentum conservation is straightforward in momentum representation, it is not immediately evident in position representation due to the lack of definite momenta assigned to each line. To verify momentum conservation in position space, a Fourier transform is necessary. Ultimately, the mathematical framework ensures that overall momentum conservation holds true regardless of the representation used.

PREREQUISITES
  • Understanding of Feynman diagrams
  • Knowledge of 4-momentum in particle physics
  • Familiarity with Fourier transforms
  • Concept of conservation laws in physics
NEXT STEPS
  • Study the mathematical formulation of Feynman diagrams in both position and momentum space
  • Learn about the implications of 4-momentum conservation in quantum field theory
  • Explore the process of performing Fourier transforms in the context of particle interactions
  • Investigate conservation laws and their applications in various physical systems
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Physicists, students of quantum mechanics, and anyone interested in the mathematical foundations of particle interactions and conservation laws.

johne1618
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I understand that Feynman diagrams can be expressed either in the position or momentum representation.

Is the 4-momentum conserved at the vertices in each case or only in the momentum representation?
 
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4-momentum is conserved in any interaction, regardless of how you write it down. However, if the Feynman diagrams are expressed in position space, that's not immediately apparent, because you don't assign definite momenta to each line. In momentum space the lines all have momenta assigned to them, so you can add them up at each vertex and see that they always sum to zero, but in position space you have to take a Fourier transform to see that.

So I guess the answer to your question is that four-momentum conservation at vertices doesn't make sense in the position representation, because talking about momentum at all doesn't make sense in the position representation. But the math will work out either way to show that the overall interaction conserves momentum.
 
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