I'm currently learning the Hubbard model. Under the assumption of contact potential, the interaction Hamiltonian written in the second quantization is \begin{equation} H_{int} = U\sum_{i}\sum_{\lambda}\,c^\dagger_{i,\lambda}c_{i,\lambda}c^\dagger_{i,-\lambda}c_{i,-\lambda} \end{equation} Now I transfer it into $\pmb{k}$-space \begin{equation} H_{int} = \frac{U}{\Omega}\sum_{\pmb{k},\pmb{p},\pmb{q}}\sum_{\lambda}\,c^\dagger_{k+q,\lambda}c^\dagger_{p-q,-\lambda}c_{p,-\lambda}c_{k,\lambda} \end{equation} and try to compute the first order perturbation due to this interaction $\langle F|H_{int}|F\rangle$, where $|F\rangle$ is the ground state of the system. My questions are:

  1. Can I argue that the only way to have non-zero matrix element is to let $\pmb{q} = 0$ (otherwise the matrix element becomes an inner product of two orthonormal states)?
  2. If so, what if I insist the electrons interact via Coulomb interaction? Would the requirement $\pmb{q}\neq 0$ always makes $\langle F|H_{int}|F\rangle = 0$ (due to the orthogonality of the state after annihilating two electrons with different spins)?

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