N.16 A single Slater determinant is not exact

The simplest example that illustrates the problem with representing a general wave function by a single Slater determinant is to try to write a general two-variable function $F(x,y)$ as a Slater determinant of two functions $f_1$ and $f_2$. You would write


A general function $F(x,y)$ cannot be written as a combination of the same two functions $f_1(x)$ and $f_2(x)$ at every value of $y$. However well chosen the two functions are.

In fact, for a general antisymmetric function $F$, a single Slater determinant can get $F$ right at only two nontrivial values $y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $y_1$ and $y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $y_2$. (Nontrivial here means that functions $F(x,y_1)$ and $F(x,y_2)$ should not just be multiples of each other.) Just take $f_1(x)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $F(x,y_1)$ and $f_2(x)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $F(x,y_2)$. You might object that in general, you have

F(x,y_1) = c_{11} f_1(x) + c_{12} f_2(x)
F(x,y_2) = c_{21} f_1(x) + c_{22} f_2(x)

where $c_{11}$, $c_{12}$, $c_{21}$, and $c_{22}$ are some constants. (They are $f_1$ or $f_2$ values at $y_1$ or $y_2$, to be precise). But if you plug these two expressions into the Slater determinant formed with $F(x,y_1)$ and $F(x,y_2)$ and multiply out, you get the Slater determinant formed with $f_1$ and $f_2$ within a constant, so it makes no difference.

If you add a second Slater determinant, you can get $F$ right at two more $y$ values $y_3$ and $y_4$. Just take the second Slater determinant's functions to be $f_1^{(2)}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\Delta{F}(x,y_3)$ and $f_2^{(2)}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\Delta{F}(x,y_4)$, where $\Delta{F}$ is the deviation between the true function and what the first Slater determinant gives. Keep adding Slater determinants to get more and more $y$-​values right. Since there are infinitely many $y$-​values to get right, you will in general need infinitely many determinants.

You might object that maybe the deviation $\Delta{F}$ from the single Slater determinant must be zero for some reason. But you can use the same ideas to explicitly construct functions $F$ that show that this is untrue. Just select two arbitrary but different functions $f_1$ and $f_2$ and form a Slater determinant. Now choose two locations $y_1$ and $y_2$ so that $f_1(y_1),f_2(y_1)$ and $f_1(y_2),f_2(y_2)$ are not in the same ratio to each other. Then add additional Slater determinants whose functions $f_1^{(2)},f_2^{(2)},f_1^{(3)},f_2^{(3)},\ldots$ you choose so that they are zero at $y_1$ and $y_2$. The so constructed function $F$ is different from just the first Slater determinant. However, if you try to describe this $F$ by a single determinant, then it could only be the first determinant since that is the only single determinant that gets $y_1$ and $y_2$ right. So a single determinant cannot get $F$ right.