I want to prove that the basis vectors of the antisymmetrized, $N$-particle vector space generated by the Slater determinants form a complete basis.
Attempt: I started from the definition of a Slater determinant as $$ \lvert\psi\rangle = \frac{1}{\sqrt{N!}} \textrm{det} \left[ \,\,\lvert u_1\rangle\, |u_2\rangle\, ... |u_N\rangle \,\,\right], $$ and want to prove that $$ \sum_{u_1\, u_2\, ... ,u_N} \frac{1}{N!}|\psi\rangle \langle \psi| = I. $$ Ok, first I think that $N!$ shouldn't be there in the equation immediately above because each Slater determinant already contains it. Am I right? Secondly about solving this problem itself, being expanded each determinant will have $N!$ terms. So that each $|\psi\rangle \langle\psi|$ contains $(N!)^2$ terms, $N!$ of which are "self" terms having the form $$ \sum_{u_1\, u_2\, ... ,u_N} |u_1\rangle\, |u_2\rangle\, ... |u_N\rangle \langle u_1|\, \langle u_2|\, ... \langle u_N| = I $$ Since there are $N!$ terms with such form, they will add up to $N!I$. Now I am concerned with the "cross" terms in the product of the two determinant. For if they are all zero, then it's straightforward to see that we will get something like $N!I/N! = I$ hence proved. But are the cross terms indeed zero?
EDIT: To illustrate what I have been working on, I will take a special case of N=2. $$ \psi_{kl}(x_1,x_2) = \frac{1}{\sqrt{2}}(u_k(x_1)u_l(x_2)-u_k(x_2)u_l(x_1)) $$ Then $$ \sum_{k,l} \psi_{kl}(x_1,x_2) \psi^*_{kl}(x_1,x_2) = 1 $$ (note that here I don't use extra 2!). Inserting the expression for $\psi_{kl}(x_1,x_2)$ we will get four terms. The first two are $$ \frac{1}{2}\left(\sum_{k,l} u_k(x_1)u_l(x_2)u^*_k(x_1)u^*_l(x_2) + \sum_{k,l} u_l(x_1)u_k(x_2) u^*_l(x_1)u^*_k(x_2) \right)= 1 $$ because $\sum_{k,l} u_k(x_1)u_l(x_2)u^*_k(x_1)u^*_l(x_2) = \langle x_1|\sum_{k} |u_k\rangle \langle u_k| |x_1\rangle \langle x_2|\sum_{l} |u_l\rangle \langle u_l| |x_2\rangle = \langle x_1|x_1\rangle \langle x_2|x_2\rangle = 1$. The other two terms are the cross terms $$ \frac{1}{2}\left(\sum_{k,l} u_k(x_1)u^*_k(x_2)u^*_l(x_1)u_l(x_2) + \sum_{k,l} u_l(x_1)u^*_l(x_2) u^*_k(x_1)u_k(x_2) \right)= 0 $$ Because $\sum_{k,l} u_k(x_1)u^*_k(x_2)u^*_l(x_1)u_l(x_2) = \langle x_1|x_2\rangle \langle x_2|x_1\rangle = 0$. In this last step I used the fact that $x_1$ must be different from $x_2$ otherwise all eigenfunctions vanish. Summing all those four terms, I get 1 as required.