Reconciling Hartree-Fock approximation and second quantization The Hartree-Fock approximation in solid-state physics, if I understand correctly, is the assumption that the many-particle wavefunction can be written as a Slater determinant of single-particle wavefunctions. This makes sense to me in the contexts in which Hartree-Fock approximations are discussed. However, I am having difficulty reconciling this with my notion of second quantization. The ways in which I've seen second quantization introduced (see here, for example) starts with the initial assumption of a single-particle Hilbert space. A Hilbert space with $n$ particles is obtained by $n$ tensor products of the single-particle Hilbert space, boson/fermions are defined as subspaces of these spaces which have the appropriate symmetrization/anti-symmetrization, and creation/annihilation operators take you from the $n$ particle Hilbert spaces to $n \pm 1$ particle Hilbert spaces (or alternatively, they move you around in a much larger Fock space). In this definition, the wavefunction of $n$ fermions appears to be defined with a Slater determinant in the exact same way that the Hartree-Fock approximation does. This makes sense to some degree - with second quantization, one often talks about creating/annihilating individual particles, so there had better be a notion of a "single particle." But I always had the impression that the second-quantization way of thinking about things was more fundamental - it was just how we were defining systems with multiple particles. Does this mean that every time we use second-quantized notation, we're implicitly making a non-trivial Hartree-Fock-like assumption about our system? This seems to morally conflict with how I think about QFT, where we have no issues tossing around creation/annihilation operators and talking about particles without ever worrying about such an approximation.
 A: There's no contradiction here. The Fock construction is just telling us that the entire space can be spanned by states that are Slater determinants. It doesn't imply that a given state has that form; that's exactly the additional assumption that the Hartree-Fock approximation makes.
As an example, consider the state 
$$|\psi_1 \rangle \otimes_A |\psi_2 \rangle + |\psi_3 \rangle \otimes_A |\psi_4 \rangle$$
where the $|\psi_i \rangle$ are orthogonal single-particle states and $\otimes_A$ is the antisymmetric tensor product. Each of the individual terms is a Slater determinant, but it's easy to check that the superposition isn't, i.e. that there aren't any coefficients so that $(\sum_i a_i |\psi_i \rangle) \otimes_A (\sum_i b_i |\psi_i \rangle)$ yields that state.
A: Your mistake is thinking that any state in second quantization is a Slater determinant. This is not true, you can form linear combinations of such states. More generally, obviously, any state in the Fock space can be written in "second quantization formalism". I say obviously because in a sense this is how the Fock space is built. 
