# Making feynman diagram [closed]

I have interaction hamiltonian as $${\cal H}_I = -\frac{1}{2}\sum\limits_{k\neq k',q}\sum\limits_{\sigma,\sigma'}\frac{\omega_{k-k'}M_{k-k'}^2}{\omega_{k-k'}^2-(\epsilon_k-\epsilon_{k'})^2}\hat c_{k'\sigma'}^\dagger\hat c_{-k'+q,\sigma}^\dagger\hat c_{-k+q,\sigma}\hat c_{k\sigma'}$$ and total hamiltonian is $${\cal H} = {\cal H}_K + {\cal H}_I$$ where $${\cal H}_K = \sum\limits_{k,\sigma}\epsilon_k\hat c_{k\sigma}^\dagger\hat c_{k\sigma}$$ What would be feynman diagram corresponding for intercation $${\cal H}_I$$. PS: Pardon me, a field theorists, may not like my choice of words.

Diagram seems fine for me. I would just write the momentum of the propagator on it and be sure that momentum conservation on the vertices occurs, which seems to be the case if the propagator momentum is $$k'-k$$.
• @ Kartik Well, in order to derive the vertex I would like to see the corresponding Lagrangian. But it seems that you have a trivial vertex. Can I ask why do you want a Feynman graph for this? It seems to me that you are working in 2nd order perturbation theory QM? Furthermore, shouldn't be a squared coupling constant in $\mathcal{H}_I$? May 20, 2020 at 8:22
• In principle you could Legendre transfrom it, by identifying the canonical momenta and so on, but you have $\mathcal{H}$ in terms of creation and anihilation operators so it may be a little tricky. In my opinion you don't need to go to the full formalism of QFT to understand what is going on in the system described by the Hamiltonian (althought it can help). First note that the free Hamiltonian is the number operator so it has well defined eigenstates and eigenvalues. Then the interaction Hamiltonian "perturbatively" modifies eigenstates and energies. May 21, 2020 at 9:21
• If u are interested in the QFT approach then you can use a scalar field theory with interaction term $\lambda \phi^3$ and at second order in the perturbative expansion you get this scattering process. Note that in the second order perturbative expansion you would have a squared coupling and six fields (two of them are contracted to give the propagator and the remaining four are the ones needed to anihilate and create the particles involved.) May 21, 2020 at 9:21