# 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 $$\phi_{k}$$?

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.