Scattering Amplitudes for Phi 4 Theory

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SUMMARY

The discussion focuses on the calculation of scattering amplitudes in Phi 4 Theory, specifically addressing the concept of one-loop order in the scattering matrix element. The scattering matrix element is represented as $$\bra{\vec{p_2}\vec{p_1}}\hat{S}\ket{\vec{k_1}\vec{k_2}}$$, and the participant clarifies that one-loop corresponds to the second order in the coupling constant $$\lambda$$, requiring the expansion to include $$\lambda^2$$. The participant initially considered the first order expansion but confirmed that the correct approach involves the second order term.

PREREQUISITES
  • Understanding of scattering theory in quantum field theory
  • Familiarity with Feynman diagrams and their interpretation
  • Knowledge of perturbation theory and loop expansions
  • Basic grasp of the Phi 4 Theory framework
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  • Study the derivation of scattering amplitudes in Phi 4 Theory
  • Learn about Feynman rules for one-loop calculations
  • Explore the significance of connected and amputated diagrams in quantum field theory
  • Investigate the role of the coupling constant $$\lambda$$ in perturbative expansions
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Researchers and students in theoretical physics, particularly those focusing on quantum field theory and scattering processes in particle physics.

Diracobama2181
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How do if calculate ##i\mathcal{M}(\vec{ k_1}\vec{ k_2}\rightarrow \vec{ p_1}\vec{ p_2})(2\pi)^4\delta^{ (4)}(p_1 +p_2-k_1-k_2)## to 1 loop order?
I know $$ i\mathcal{M}(\vec {k_1}\vec{k_2}\rightarrow \vec{p_1}\vec{p_2})(2\pi)^4\delta^{(4)}(p_1 +p_2-k_1-k_2) $$ =sum of all (all connected and amputated Feynman diagrams), but what is meant by 1 loop order? In other words, when I take the scattering matix element $$\bra{\vec{p_2}\vec{p_1}}\hat{S}\ket{\vec{k_1}\vec{k_2}}$$, would 1 loop just be the first order expansion ($$\frac{\lambda}{4!}\int d^4z\bra{\vec{p_1}\vec{p_2}}\phi\phi\phi\phi\ket{\vec{k_1}\vec{k_2}}$$) in this case?
 
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Never mind. Figured it out. I would have to go to $$\lambda^2$$ in the expansion. Thanks.
 
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