Second quantisation of interaction potential (Fermions)

Question

If we start with an interaction Hamiltonian for fermions in second quantised form:

$$ H_\text{int} = \frac{1}{2} \int d^3r \int d^3r' V(|r-r'|) \hat{n}(r)\hat{n}(r') $$ where $\hat{n}(r)=c^\dagger(r)c(r)$ is the fermion number operator. With Fourier transformation $c_k = \sum_{k} d^3 k e^{-ikr} c(r)$, it will be transformed into: $$ H_\text{int} = \frac{1}{2} \sum_{k,k',q} \tilde{V}(q) c^\dagger_{k-q} c^\dagger_{k'+q} c_{k'}c_{k} + \frac{V(0)}{2}\sum_k c^\dagger_k c_k $$ where $\tilde{V}(q)$ is the Fourier transformed potential $V(r)$. I am quite worried about the second term $\frac{V(0)}{2}\sum_k c^\dagger_k c_k$, which should diverge for most potentials $\propto \frac{1}{r}$. How should we understand this divergent term here? It should not dominate the Hamiltonian, but how come?

Note: The $\frac{V(0)}{2}\sum_k c^\dagger_k c_k$ comes out of interchange $c(r)c^\dagger(r')=-c^\dagger(r')c(r) + \delta(r-r')$

Answer 1

The correct definition of the two-body interaction potential is (neglecting spin) $$ \hat{H}_{\rm int} = \frac{1}{2}\int{\rm d}^3 r\int {\rm d}^3 r'\, V(r-r') \hat{c}^\dagger(r) \hat{c}^\dagger(r') \hat{c}(r') \hat{c}(r).$$ This differs from the expression you wrote by precisely the spurious term proportional to $V(0)$. This term corresponds to an extensive energy shift that would appear even if there were only a single particle in the system (exercise: what is the expectation value of $\hat{H}_{\rm int}$ in a single-particle state $\lvert \psi\rangle$?). Indeed, the spurious term basically represents "the Coulomb interaction potential that a charged particle would experience due to its own electric field", which is obviously divergent (and also nonsense). The Coulomb interaction (and any other two-body interaction) is an energy associated with the relative position of pairs of particles.