# Hubbard model and orthogonality of the ground state

I'm currently learning the Hubbard model. Under the assumption of contact potential, the interaction Hamiltonian written in the second quantization is $$$$H_{int} = U\sum_{i}\sum_{\lambda}\,c^\dagger_{i,\lambda}c_{i,\lambda}c^\dagger_{i,-\lambda}c_{i,-\lambda}$$$$ Now I transfer it into $$\pmb{k}$$-space $$$$H_{int} = \frac{U}{\Omega}\sum_{\pmb{k},\pmb{p},\pmb{q}}\sum_{\lambda}\,c^\dagger_{k+q,\lambda}c^\dagger_{p-q,-\lambda}c_{p,-\lambda}c_{k,\lambda}$$$$ and try to compute the first order perturbation due to this interaction $$\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)?