This post, largely based on Reference 1, is about the Slater determinant. The Slater determinant is related to the Hartree Product. In that post, it was determined that a Hartree Product does not meet the antisymmetry principle. The Slater determinant is one way to fix this problem.

### Two Electrons

It is helpful to first consider the two-electron case [1]. According to the previous post, what spin orbitals should these two electrons occupy? To use the correct notation, one electron–call it electron 1–should be associated with $\vec{x}_1$, and the other electron–call it electron 2–should be associated with $\vec{x}_2$. Furthermore, the spin orbital with argument $\vec{x}_1$ should be $\chi_i$ whereas the spin orbital with argument $\vec{x}_2$ should be $\chi_j$. Hence the corresponding Hartree Product is

$ \Psi_{HP} = \chi_i (\vec{x}_1) \chi_j (\vec{x}_2)$.

This Hartree Product definitely does not obey the antisymmetry principle, because the definition of the Hartree Product does not allow for switching the $\vec{x}$-vectors in the first place (see the previous post for more information).So, the Hartree Product must be abandoned in order to meet the antisymmetry principle.

### Slater Determinant

Slater realized that the antisymmetry principle can be met by a function $\Psi$ that is a* difference* of Hartree products. In particular, for the two-electron system, define:

$ \Psi (\vec{x}_1, \vec{x}_2) \equiv \frac{1}{\sqrt{2}} \big( \chi_i (\vec{x}_1) \chi_j (\vec{x}_2) – \chi_i (\vec{x}_2) \chi_j (\vec{x}_1) \big) $.

With this notation, the order of the arguments in $\Psi$ is critical. One can test whether or not this type of function meets the antisymmetry principle; switching the $\vec{x}$-vectors, one has

$ \Psi (\vec{x}_2, \vec{x}_1) = \frac{1}{\sqrt{2}} \big( \chi_i (\vec{x}_2) \chi_j (\vec{x}_1) – \chi_i (\vec{x}_1) \chi_j (\vec{x}_2) \big) $.

How are these two functions $ \Psi (\vec{x}_1, \vec{x}_2)$ and $ \Psi (\vec{x}_2, \vec{x}_1) $ related? Well, one idea is to multiply $ \Psi (\vec{x}_1, \vec{x}_2)$ by -1 and see what happens:

$ -\Psi (\vec{x}_1, \vec{x}_2) = \frac{1}{\sqrt{2}} \big( -\chi_i (\vec{x}_1) \chi_j (\vec{x}_2) + \chi_i (\vec{x}_2) \chi_j (\vec{x}_1) \big) $

Interesting… that looks like $ \Psi (\vec{x}_2, \vec{x}_1) $! Hence $-\Psi (\vec{x}_1, \vec{x}_2) = \Psi (\vec{x}_2, \vec{x}_1) $. This is the mathematical statement of the antisymmetry principle, so this type of subtraction works–at least for two electrons.

I am going to ignore the factor of $\frac{1}{\sqrt{2}}$, for now.

### More than Two Electrons

Is it possible to extend this idea to the case of more than two electrons? Perhaps. There would be more than one way to exchange the two $\vec{x}$-vectors, since there would be more than two $\vec{x}$-vectors. The resulting function would need to account for all possible ways of exchanging two $\vec{x}$-vectors such that a negative sign results in the desired way. Furthermore, assuming the ordering of the spin orbitals is crucial, the desired function should not change the ordering of the orbitals in any term. Similar to the simplified two-electron case, one should expect a minus sign on any term corresponding to a switch of any two $\vec{x}$-vectors.

This problem is closely related to how many ways it is possible to order the subscripts of the $\vec{x}$-vectors. For instance, if there are three electrons, the numbers 1-3 can be ordered in the following way:

123 (start with 1; remaining two digits ascend) $\rightarrow$ take as first order

132 (start with 1; remaining two digits descend) $\rightarrow$ 2 and 3 are switched, with respect to first order

213 (start with 2; remaining two digits ascend) $\rightarrow$ take as second order

231 (start with 2; remaining two digits descend) $\rightarrow$ 3 and 1 are switched, with respect to second order

312 (start with 3; remaining two digits ascend) $\rightarrow$ take as third order

321 (start with 3; remaining two digits descend) $\rightarrow$ 2 and 1 are switched, with respect to third order

In this three-digit case, once the first digit is fixed, there are only two digits are left, and there are only two ways to order two digits–in an ascending order or in a descending order. Perhaps this idea can be extended to the general case of $N$ numbers or $N$ electrons. The process of listing all unique arrangements–or permutations–of digits can be accomplished with a set of steps.

*Steps for forming all permutations of $N$ digits:*

Step 0: List all digits 1-N in numerical order: $1, … , N$.

Step x: Form a new arrangement by increasing or decreasing the $x^{th}$ digit by 1, until all digits are used up.

### Determinant

Apparently, the desired function is accomplished with a **determinant**, which certainly involves permutations and negative signs. To be continued…