How far can an $N$-fermion wavefunction be from the nearest Slater determinant? Consider a system of $N$ non-relativistic spin-$0$ fermions in $D$-dimensional space, all of the same species. The wavefunction of such a system can be represented by a function $\psi(\mathbf{x}_1,...,\mathbf{x}_N)$, where each boldface argument is a point in $D$-dimensional space, and the function is antisymmetric with respect to permutations of those $N$ points. The inner product is the usual one:
$$
\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}
 \la \psi|\phi\ra 
 \equiv \int_{\mathbf{x}_1,...,\mathbf{x}_N}
   \psi^*(\mathbf{x}_1,...,\mathbf{x}_N)
  \phi(\mathbf{x}_1,...,\mathbf{x}_N).
$$
Let $S$ be the set of non-zero wavefunctions of the form
$$
 \sum_\pi (-1)^\pi\prod_{k=1}^N \psi_k(\mathbf{x}_{\pi(k)})
$$
where the sum is over permutations of the $N$ arguments. Such a wavefunction is sometimes called a Slater determinant.
For $N\geq 2$, does the quantity
$$
\max_{|s\ra\in S}
\frac{\la \psi|s\ra\,\la s|\psi\ra}{\la \psi|\psi\ra\,\la s|s\ra}
\tag{1}
$$
have a non-zero minimum value among all non-zero vectors $|\psi\ra$ in the Hilbert space? If so, what is that minimum value as a function of $N$ and $D$?
 A: Curiously I am working on this subject in my research in Theoretical Chemistry. An analytical expression to this is unknown, except for $N=2$ (see 10.1103/PhysRevA.64.022303 or 10.1103/PhysRevA.89.012504). For the general case an analytical expression is unlikely to be possible, and an optimisation over S is required, such as discussed in 10.1103/PhysRevA.89.012504 (although they consider a more general set $S$, you question is a particular case). The set you named as $S$ is actually close related to the Grassmannian (as you probably know, considering the tags you added): If you consider the equivalence classes upon the relation $\psi \sim \lambda \psi$, $\lambda$ a non zero scalar you have an manifold isomorphic to the Grassmannian.
When (hopefully soon) my research on this subject gets published, I will be happy to share it here.
A: There is no minimum, at least for $N=2$.
Without loss of generality, we can assume that all wavefunctions in the problem have norm $1$.
Restricting to $D=1$, we perform the change of coordinates $u=(x_1+x_2)/\sqrt2,\quad v=(x_1-x_2)/\sqrt2$ and consider the wavefunction
$$\psi(u,v)=\sqrt{\frac{2\pi}{\varepsilon}} v\: e^{-\frac{\pi}{2}(v^2/\varepsilon+u^2\varepsilon)}$$
for $\varepsilon>0$. This function is even in $u$ and odd in $v$, so it is antisymmetric in the original coordinates. The integral of its square over all space is
$$\iint \psi(u,v)\:du\:dv= \iint \frac{2\pi}{\varepsilon} v\: e^{-\pi(v^2/\varepsilon+u^2\varepsilon)} = \frac{2\pi}{\varepsilon} \frac{\varepsilon^{3/2}}{2\pi}\varepsilon^{-1/2}=1.$$
On the other hand, we have
$$\psi(u,v)= -2 \sqrt{\frac{2}{\pi}}\varepsilon\: e^{-\frac{\pi}{2}u^2\varepsilon}\:\frac{d}{dv} \left( \frac{1}{\sqrt{2\varepsilon}} e^{-\frac{\pi}{2}v^2/\varepsilon}\right).$$
As $\varepsilon\to 0$, he term inside the derivative is a nascent delta, while the exponential term outside is $1+O(\varepsilon)$.
Now we return to the original coordinates, and consider any (normalized) wavefunction of the form
$$s(x_1,x_2)=f(x_1)g(x_2)-f(x_2)g(x_1),$$
as in the problem. The inner product $\langle s | \psi \rangle$ can be computed by using the properties of the Dirac delta derivative to perform the integration on $x_2$. The result is
$$K \varepsilon \int (f'(x)g(x)-f(x)g'(x)) \:dx + O(\varepsilon^2),$$
for some unimportant constant $K$. This can be made as small as we want by decreasing $\varepsilon$.
If I'm not mistaken, this can be generalized to $D>1$ by taking $\psi$ to be the product of the former function in the first pair of coordinates by nascent Dirac deltas in the other coordinates.
