# Antisymmetric functions as Slater determinants

Can any antisymmetric function, i.e., a function of $N$ spatial-plus-spin variables $\{x_i\ | \ i= 1, \ldots, N\}$ satisfying $$\psi(x_1,\ldots, x_i, \ldots, x_j, \ldots, x_N) = -\psi(x_1,\ldots, x_j, \ldots, x_i, \ldots, x_N)\ ,$$ always be written as a Slater-Determinant of functions ("orbitals"), not necessarily solutions to Schrödinger's Equation? (I know that there are no orbitals describing interacting electrons, but they may [or may not] be solutions to a "Schrödinger-like equation, but that's beside the point.)

(As context: I have been reading up on Density Functional Theory. I am teaching an optional second course on quantum mechanics and I am explaining some of DFT's basic concepts. In Parr & Yang's "Density-functional theory of atoms and molecules" (Oxford University Press, 1989) it is mentioned that there is a proof that any "reasonable" (electron) density can be derived from an antisymmetric wave function which can always be written as Slater-Determinant of orbitals. The authors cite a couple of articles on some journals to which we, at the university where I work, have no subscription [Phys. Rev. A & B, Int. J. Quantum Chem.] )

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which articles are these? –  Danu Mar 27 at 8:00
The articles are: 1. Gilbert, T.L., "Hohenberg-Kohn theorem for nonlocal external potentials". Phys. Rev. B , 12: 2111-2120, 1975 2. Harriman, J.E., "Orthonormal orbitals for the representation of an arbitrary density". Phys. Rev. A , 24: 680-682, 1980 3. Lieb, E.H., "Density functionals for Coulomb systems". Int. J. Quantum Chem. , 24: 243-277, 1983 –  gildardo Mar 27 at 8:23
I can confirm that Gilbert, in his paper, shows that any non-negative density function $n(\vec{r})$ which satisfies $\int d\vec{r}\ n(\vec{r})=N$ is what Gilbert calls 'N-representable'. The final sentence of his paper reads 'Hence, an obvious corollary is that any N-representable particle density $n(\vec{r})$ can be obtained from a single Slater determinant' –  Danu Mar 27 at 8:44

The short answer is: No, it is not true without other strong hypotheses.

What it is true is that any completely antisymmetric wavefunction $\psi(x_1,\ldots,x_N) \in L^2(\mathbb R^{3N})$ (not necessarily solution of Schroedinger equation) can always be written as a, generally infinite, linear combination of Slater determinants.

Indeed, if $\{\phi_k\}_{k=1,2\ldots,}$ is a Hilbert basis of $L^2(\mathbb R^3)$ and $\phi \in L^2(\mathbb R^{3N})$ then: $$\phi(x_1,\ldots,x_N) = \sum_{i_1,\ldots, i_N} C_{i_1...i_N} \phi_{i_1}(x_1)\cdots \phi_{i_n}(x_N)$$ where the convergence is that in $L^2$. Then consider the orthogonal projector $A$ from $L^2(\mathbb R^{3N})$ onto the subspace of completely antisymmetric wavefunctions. if $\phi$ is generic, $\psi = A\phi$ is the generic completely antisymmetric wavefunction, so we have that any completely antisymmetric wavefunction of $N$ entries can be decomposed as: $$\psi(x_1,\ldots, x_N) = \sum_{i_1,\ldots, i_N} C_{i_1...i_N} A(\phi_{i_1}(x_1)\cdots \phi_{i_n}(x_N)) \:.$$ Above $$A(\phi_{i_1}(x_1)\cdots \phi_{i_n}(x_N))$$ is nothing but the Slater determinant of $\phi_{i_1}(x_1)\:,\ldots\:, \phi_{i_n}(x_N)$.
The generalization to the case where $x_k$ includes spin variables is obvious.

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That is what I suspected... of course, there is a subset of functions which can be represented by a single Slater determinant. The further point I am not completely clear is rather about the density derivable from an antisymmetric wave function $\rho(x) = \int |\psi(x, x_2,\ldots,x_N)|^2 dx_2\ldots dx_N$. Gilbert's paper I mention in the comments (and @Danu has confirmed that it) states that "any" density can be derived from a single slater determinant. It can't be ANY density since it is derived from a subset, correct? And what conditions are imposed on such a density? –  gildardo Mar 27 at 9:27
What I am concerned with is the conceptual and mathematical basis of Density Functional theory, and the Kohn-Sham ansatz concerning the construction of the "non-interacting" reference system –  gildardo Mar 27 at 9:32
My feeling is that the statement concerning densities may be correct for ANY positive density, but I do not know that paper and, sorry, I have no spare tame to try to construct a proof. –  Valter Moretti Mar 27 at 10:56
The point is that for a density $\rho(x)$ you have much many constraints on a possible Slater determinant producing it, since you are given a function of a unique variable $x$ and you have to look for $n$ functions $\psi_j$ each of a corresponding variable $x_1,\ldots, x_N$ to fill the identity $\rho(x) = \int |A(\psi_1(x)\psi_2(x_2)\cdots \psi_n(x_N))|^2 dx_2\cdots dx_N$. –  Valter Moretti Mar 27 at 11:08
Sorry for all my typos above. I hope you understand, I am in a hurry. I wrote "much many constraints" but I actually meant "many fewer constraints". –  Valter Moretti Mar 27 at 11:19