### 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 as a Slater de­ter­mi­nant of two func­tions and . You would write

A gen­eral func­tion can­not be writ­ten as a com­bi­na­tion of the same two func­tions and at every value of . How­ever well cho­sen the two func­tions are.

In fact, for a gen­eral an­ti­sym­met­ric func­tion , a sin­gle Slater de­ter­mi­nant can get right at only two non­triv­ial val­ues and . (Non­triv­ial here means that func­tions and should not just be mul­ti­ples of each other.) Just take and . You might ob­ject that in gen­eral, you have

where , , , and are some con­stants. (They are or val­ues at or , to be pre­cise). But if you plug these two ex­pres­sions into the Slater de­ter­mi­nant formed with and and mul­ti­ply out, you get the Slater de­ter­mi­nant formed with and within a con­stant, so it makes no dif­fer­ence.

If you add a sec­ond Slater de­ter­mi­nant, you can get right at two more val­ues and . Just take the sec­ond Slater de­ter­mi­nant's func­tions to be and , where is the de­vi­a­tion be­tween the true func­tion and what the first Slater de­ter­mi­nant gives. Keep adding Slater de­ter­mi­nants to get more and more -​val­ues right. Since there are in­fi­nitely many -​val­ues to get right, you will in gen­eral need in­fi­nitely many de­ter­mi­nants.

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