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

$$F(x,y)=\frac{a}{\sqrt{2}}\Big(f_1(x)f_2(y)-f_2(x)f_1(y)\Big)$$

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

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

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.