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This is a question motivated by elementary many-body quantum physics, thus I will start by giving some somewhat imprecise descriptions of the objects involved (I will also ignore spin, which is of no concern for this problem). For non-interacting systems, the many-body time-independent wave function in the position representation is of the form $\Phi(x_1,\ldots,x_n)=\text{Det}\left( \phi_j(x_k) \right)_{j,k=1}^{n}$ (here in this notation $\text{Det}\left( A(j,k) \right)_{j,k=1}^{n}$ means the determinant of the $n\times n$ matrix $A$ of elements $A(j,k)$ ), where each $x_j \in \mathbb{R}^3$ and the $\phi_j(x)$ are the so-called one-electron wavefunctions (they are the molecular orbitals). These determinants are known as Slater determinants. However, for interacting systems this is no longer true, although the wave function $\Psi(x_1,\ldots,x_n)$ is always antisymmetrical, i.e $\Psi(x_{\sigma(1)},\ldots,x_{\sigma(n)})=(-1)^{\vert \sigma \vert}\Psi(x_1,\ldots,x_n),$ for any permutation $\sigma \in S_{n},$ because of Pauli's exclusion principle for fermions.

A major topic in many-body physics is how far can one go approximating interacting wavefunctions by Slater determinants (Hartree-Fock or Density Functional Theory are examples of this idea), and in some sophisticated Quantum Chemistry methods, the idea is to employ linear combinations of Slater detrminants. Therefore, the question seems natural: Is the span of the set of Slater determinants "dense" in the subspace of antisymmetric functions? I use quotation marks because of course I haven't specified any particular functional space. Ideally, such a result would hold for a Sobolev space such as $W_{1,2},$ which I believe is the simplest space suitable for a mathematical treatment of the Schrödinger equation.

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When we construct the space of antisymmetric functions, a formal way to do so is taking the so called exterior algebra. In the quantum mechanics case, assuming $\mathscr{H}_j$ is the one-particle Hilbert space of particle number $j$, we could take the $N$-fold wedge product

$$ \wedge_{i=1}^N \mathscr{H}_j$$

as our total Hilbert space. This is actually defined such that slater determinants are dense, because it is defined to be the closure of the following set:

$$ \mathrm{span} \{ f_1 \wedge f_2 \wedge \dots \wedge f_N | f_j \in \mathscr{H}_j \} $$

and this is just another way to write down the Slater determinant built out of $f_1,f_2,...f_N$.

So briefly: The antisymmetric wave functions are exactly the closure of the span of Slater determinants.

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  • $\begingroup$ Hi, Luke, thanks for answering. In your answer, I don't understand what is $\mathcal{H}_j.$ Is it a Sobolev space? Which one? Also, what you say can not be true. Is the span of the set of determinants, and not the set of determinats, what must be dense. Appart from that, you have onlye reformulated the problem: you would still have to prove that the closure of the span of that set you describe is the set of antisymmetric function. Btw, the result is easy for $L^p$ spaces, because it holds for tensor products and the whole $L^p$ space, but I am not so sure about the Sobolev space. $\endgroup$
    – Qwertuy
    Commented Sep 20, 2017 at 10:47
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    $\begingroup$ @DiegoSP: Right, I had to add a "span" in the second equation. Now the second point you make is not clear to me. As I said, $\mathscr{H}_j$ is the one-particle Hilbert space. It depends on what you want to do, it might be some $L^2$-space or a Sobolev space. But still it remains a simple fact that the antisymmetric functions are nothing else than the closure of the span of determinants. $\endgroup$
    – Luke
    Commented Sep 20, 2017 at 12:55
  • $\begingroup$ Maybe I am missing something, but I don't think saying that $\mathcal{H}_j$ is the "one-particle Hilbert space" gives any information at all. Perhaps I don't see your point, but I insist: I agree with the fact that this is easy for the $L^p$ case, but I believe the proof is more involved in the Sobolev case. $\endgroup$
    – Qwertuy
    Commented Sep 21, 2017 at 9:08

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