When is it necessary to use more than one Slater determinant to write the wave function of a quantum system?

Question

Consider a system of $N$ non-interacting identical fermions of spin $s$, spin quantum number $m_i=m_{s_i}$and position coordinates $\mathbf{r}_i$. Let $\alpha$ denote a state specified by the numbers $n,j_n$, with $n$ the energy level, $j_n \in \{1,2,...,g_n\}$ and where $g_n$ is the degeneracy of the $n$-th energy level. If each particle is in a spin-orbit state $\psi_{\alpha_i}(\mathbf{r}_i,m_i)$, an antisymmetrized state of the system can be constructed with a Slater determinant as follows:

$$ \Psi_{\alpha_{1}, \alpha_{2}, \ldots, \alpha_{N}}\left(\mathbf{r}_{1}, m_{1} ; \ldots ; \mathbf{r}_{N}, m_{N}\right)=\frac{1}{\sqrt{N !}}\left|\begin{array}{cccc} \psi_{\alpha_{1}}\left(\mathbf{r}_{1}, m_{1}\right) & \psi_{\alpha_{1}}\left(\mathbf{r}_{2}, m_{2}\right) & \ldots & \psi_{\alpha_{1}}\left(\mathbf{r}_{N}, m_{N}\right) \\ \psi_{\alpha_{2}}\left(\mathbf{r}_{1}, m_{1}\right) & \psi_{\alpha_{2}}\left(\mathbf{r}_{2}, m_{2}\right) & \ldots & \psi_{\alpha_{2}}\left(\mathbf{r}_{N}, m_{N}\right) \\ \vdots & \vdots & & \vdots \\ \psi_{\alpha_{N}}\left(\mathbf{r}_{1}, m_{1}\right) & \psi_{\alpha_{N}}\left(\mathbf{r}_{2}, m_{2}\right) & \ldots & \psi_{\alpha_{N}}\left(\mathbf{r}_{N}, m_{N}\right) \end{array}\right| $$

However, is it always sufficient to use a single Slater determinant in order to construct the state of the system? Could it be necessary (or even possible) to use more than one determinants?

Answer 1

Only a very restricted class of wavefunctions can be expressed as a single Slater determinant. The Hartree-Fock approximation is only an approximation because of this fact.

Writing an $n$-particle state as single Slater determint is equivalent to decomposing an element $$ {\boldsymbol \omega} = \omega_{i_1,i_2,\ldots, i_n}{\bf e}_{i_1}\wedge {\bf e}_{i_2}\wedge \ldots \wedge {\bf e}_{i_n} $$ of the exterior algebra as a product $$ \omega = {\boldsymbol \eta}_1\wedge {\boldsymbol \eta}_2\wedge\ldots \wedge {\boldsymbol \eta}_n, $$ of $n$ vectors. It is easy to see that there must be restrictions becuase in an $N$ dimensional Hilbert space there are are more free parameters ($N$-choose-$n$) in the antisymmetric tensor $\omega_{i_1,i_2,\ldots, i_n}$ than in the $n$ vectors that only have $nN$ degrees of freedom.

The condition that this can be done are the Plucker relations $$ \omega_{i_1,i_2,\ldots, i_{n-1},[ j_1}\omega_{j_2,j_3 ,\ldots, j_{n+1}]}=0, $$ where the $[\ldots]$ denotes antisymmetrization.

A reference written for physicists is: Calculating the distance from an electronic wave function to the manifold of Slater determinants through the geometry of Grassmannians, Y.Aoto, M. F. da Silva Phys. Rev. A 102, 052803 (2020).