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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 ${\cal H}_I$.

feynman diagram

PS: Pardon me, a field theorists, may not like my choice of words.

About interacting Hamiltonian: refer Bardeen-Pines.

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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 $k'-k$.

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  • $\begingroup$ What about vertex? $\endgroup$
    – Kartik Chhajed
    Commented May 19, 2020 at 15:30
  • $\begingroup$ @ 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 $\mathcal{H}_I$? $\endgroup$
    – vin92
    Commented May 20, 2020 at 8:22
  • $\begingroup$ Can we do legendre transformation of hamiltonian to get Lagrangian? Or should I think about some Lagrangian whose legendre transformation will give back asforthmention Hamiltonian? I was going through Anderson's paper on RPA in theory of superconductivity where I found the Hamiltonian given in question. And in past I have done course on QFT-1, a introductory course enough to say I know nothing about field theory, but familiar with few terms and definitions. I thought of making Feynman diagram for short note. $\endgroup$
    – Kartik Chhajed
    Commented May 21, 2020 at 0:38
  • $\begingroup$ 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. $\endgroup$
    – vin92
    Commented May 21, 2020 at 9:21
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    $\begingroup$ 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.) $\endgroup$
    – vin92
    Commented May 21, 2020 at 9:21

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