Interpretation of creation and annihilation operators acting in the state of a interacting system If I have a system of $N$ non-interacting fermions, I can write the wave function of the ground-state of the system using a Slater determinant
$$ \Phi_{0}(\textbf{r}_{1}, ..., \textbf{r}_{N}) = \frac{1}{\sqrt{N!}} 
\begin{vmatrix}
\phi_{k_{1}}(\textbf{r}_{1}) & \cdots & \phi_{k_{N}}(\textbf{r}_{1})\\ 
\vdots  & \ddots  & \vdots \\ 
\phi_{k_{1}}(\textbf{r}_{N}) & \cdots & \phi_{k_{N}}(\textbf{r}_{N})
\end{vmatrix}, $$
where $\phi_{k}$ is the single-particle wave function of one fermion. I can define operators $a_{k}$ and $a^{\dagger}_{k}$ that create and annihilate fermions in the state $\phi_{k}$ when they act in the state of the non-interacting system.
But, if now I am interested in the state of N interacting fermions, I can write the total wave function as a linear combination of all possible Slater determinants
$$ \Psi_{0}(\textbf{r}_{1}, ..., \textbf{r}_{N}) = \sum_{n} C_{n} \Phi_{n} (\textbf{r}_{1}, ..., \textbf{r}_{N}).$$
I can interpret $a^{\dagger}\Phi_{0}$ as the state of a system of $N$ fermions with an add fermion in state $\phi_{k}$. My question is: Can I do the same interpretation when I apply one of the creation or annihilation operators to $\Psi$? Can I say that $a^{\dagger}_{k}\Psi_{0}$ is a state of $N$ interacting fermions with an add fermion in state $\phi_{k}$? 
 A: In a strictly non-relativistic model, one in which the canonical annihilation operators actually do annihilate the true vacuum state, the answer is yes: Each application of a canonical creation operator adds another particle to the state, and this is true with or without interactions.
Since we're talking about creation/annihilation operators, a QFT formulation is convenient. In a typical strictly non-relativistic model, the canonical number operator commutes with all observables, even if the Hamiltonian includes interaction terms, like this one:
$$
 H\sim 
 \int d^3x\ 
 \psi^\dagger(x)\frac{-\nabla^2}{2m}\psi(x)
 + 
 \int d^3x\,
 d^3y\ 
 \psi^\dagger(x)\psi^\dagger(y)V(x-y)\psi(y)\psi(x)
$$
with field operators $\psi(x)$ satisfying $\{\psi(x),\psi^\dagger(y)\}\sim\delta^3(x-y)$. The canonical number operator in this case is $\int d^3x\ \psi^\dagger(x)\psi(x)$.
For each $N$, states obtained from the vacuum by applying a product of $N$ creation operators $\psi^\dagger$ are in different superselection sectors: different values of $N$ are not connected to each other by any observables. (This is a stronger statement than merely saying that the number of particles is conserved, which only requires that it commutes with the Hamiltonian.) In this sense, a strictly non-relativistic QFT such as the one illustrated above is like a collection of separate theories, one for each possible number $N$ of particles. The canonical creation/annihilation operators take you back and forth between those separate theories.
Again, this is for strictly non-relativistic models. Relativistic models are a different story.
