# 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.

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.