Why can't we choose any other antisymmetric function instead of a Slater determinant for a multi-electron system? Why do we choose our wavefunction for a multi-electron atom as a product of single-electron wavefunctions?

  • 3
    $\begingroup$ Of course we can. Go on, choose any other one, see what happens. $\endgroup$ Mar 15 '19 at 8:32
  • 2
    $\begingroup$ Go for it, mate! $\endgroup$
    – user37142
    Mar 15 '19 at 8:37
  • 1
    $\begingroup$ A product of single-electron wavefunctions is, in general, neither symmetric nor antisymmetric with respect to permutation. The second question here seems to be slightly non sequitur. $\endgroup$
    – orthocresol
    Mar 15 '19 at 11:25
  • $\begingroup$ @orthocresol Sorry , By product i mean linear combination of product wavefunction .What does the taking product signify? $\endgroup$ Mar 16 '19 at 9:44

1. A pragmatic view

As the comments suggested you can use different functions than slater determinants. But this would be more or less as bad as an idea as describing a movement in a circle without using polar coordinates.

If you include the constraints of your system implicitly in the coordinate system or basis of your description, nearly everything gets better. More elegant equations, faster convergence... .

Determinants are a fast and elegant way to include constraints like the Pauli Principle in your description.

2. A bit more mathematical view

Let's go back to the definition of a Determinant in general. From Wikipedia:

[...] Another way to define the determinant is expressed in terms of the columns of the matrix. If we write an $n \times n$ matrix $A$ in terms of its column vectors

$$A = \begin{bmatrix} a_1 & a_2 & \cdots & a_n \end{bmatrix}$$

where the $a_j$ are vectors of size $n$, then the determinant of $A$ is defined so that $$ \begin{align} \det \begin{bmatrix} a_1 & \cdots & b a_j + c v & \cdots & a_n \end{bmatrix} &= b\det(A) + c \det \begin{bmatrix} a_1 & \cdots & v & \cdots & a_n \end{bmatrix} \\ \det \begin{bmatrix} a_1 & \cdots & a_j & a_{j+1} & \cdots & a_n \end{bmatrix} &= -\det \begin{bmatrix} a_1 & \cdots & a_{j+1} & a_j & \cdots & a_n \end{bmatrix} \\ \det(I) &= 1 \end{align} $$ [...]

The last constraint comes from the normalization of the atomic orbitals and the second constraint is just the Pauli-Principle. To the best of my knowledge you don't have to force the linearity in the first constraint, but if you do, you end up with a determinant.


Not the answer you're looking for? Browse other questions tagged or ask your own question.