How Do Feynman Diagrams Work in Phi^4 Theory?

  • #1
Diracobama2181
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TL;DR Summary
Is there any way of finding what $$\bra{\Omega}T(\partial^{\mu}\Phi\partial^{\nu}\Phi)\ket{\Omega}$$ would be?
In this case, the lagrangian density would be
$$\mathcal{L}=\frac{1}{2}((\partial_{\mu}\Phi)^2-m^2\Phi^2)-\frac{\lambda}{4!}\Phi^4$$
whe $$\Phi$$ is the scalar field in the Heisenburg picture and $$\ket{\Omega}$$ is the interacting ground state. Was just curious if there were ways to do Feynman diagrams in this sitution.
 
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  • #2
Yes, it is possible to do Feynman diagrams in this situation. The Lagrangian density provides the rules for constructing Feynman diagrams in the theory, which are used to calculate physical observables such as scattering amplitudes and correlation functions. In particular, the Feynman rules for a scalar field theory with $$\mathcal{L}$$ given above are: 1) Draw a line for each field, representing a propagator. 2) Associate a vertex with each term in the lagrangian. 3) Calculate the amplitude for each diagram by integrating over all internal momenta, and summing over all possible intermediate states. For example, the following Feynman diagram would represent a $\Phi^4$ interaction: [Insert Diagram Here]This diagram describes the process of two scalar fields ($\Phi$) interacting to produce a pair of scalar fields ($\Phi$). The amplitude for this process is given by\begin{equation}A = \int \frac{d^4p}{(2\pi)^4}\frac{1}{p^2-m^2}\frac{\lambda}{3!}\end{equation}where $p$ is the momentum of the intermediate state.
 

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