quantum mechanics - What is the physical interpretation of a field operator

So far in our lecture we defined creation operators $a^{\dagger}_{n}$ in the following way, that we said:

Somebody got you a antisymmetric or symmetric N- particle state and now $a^{\dagger}_{n}$ puts another particle in state n, so that we end up with a symmetric/antisymmetric N+1-particle state. This interpretation is somehow clear to me in the sense that these $a^{\dagger},a$ operators avoid the cumbersome slater determinants and so on. Despite, we are still dealing with well-defined symmetrized/antisymmetrized product states that become extended or reduced by one state, which are hidden behind this notation.

Now, we also defined field-operators in QM by $\psi^{\dagger}(r) = \sum_{i;\text{all states}} \psi_i^*(r) a_i^{\dagger}.$ We said that they create a particle at position $r$. Somehow, it is not clear to me what this means:

To create a particle at an exact position $r_0$ in QM would mean that we now have an additional state $\psi_i(r) = \delta(r-r_0)$ in our slater determinant. I doubt that this is the idea behind this. But, since the $a_i^{\dagger}$ operators act on $N$-particle state and map to $N+1$ particle states, the same must be true for $\psi^{\dagger}(r)$. Nevertheless, I have difficulties interpreting the result.

If anything is unclear, please let me know.