# Computing Slater determinants

I need to compute Slater determinants. I'm wondering if I would benefit from assigning each of my functions to a variable prior to computation. I'm working with Slater determinants, but my question goes beyond them and applies to the computation of any determinant.

Say I have functions $f_n[x]$ which are sums of complex exponentials. If I want to compute the determinant of an $N\times N$ matrix, should I use the expanded form of $f_n[x]$ or variables ($a_1,a_2,b_1,b_2$, etc.), or else? Does Mathematica benefit from prior "simplification" of the matrix or not?

At the moment, I'm not concerned with speed because the systems I work with contain few electrons, but that will change.

• You should be more descriptive in your question... give an example for the functions and the matrix constructed from your wavefunctions. It is not clear what $f_n$ refers to, what $a_i$ and $b_i$ are...
– rm -rf
Commented Jun 26, 2012 at 16:33
• I've been RM'ited. @R.M Well, my functions can be any functions, really. What I'm wondering is if Mathematica's algorithm for computing the determinant benefits from working with variables (a1,b2) rather than the whole functions.
– CHM
Commented Jun 26, 2012 at 16:34
• lol! "RMiting" is a thing now? :P
– rm -rf
Commented Jun 26, 2012 at 16:37
• @R.M I was referring to this ;) But concerning my question, I expressly wanted to make it general, and not specific to Slater determinants. The question has to do with how MMA computes the determinant.
– CHM
Commented Jun 26, 2012 at 16:41
• When dealing with many-particle systems, it's useful to know what observables you're after. E.g., are you planning to calculate expectation values of single-particle operators in the end? For most applications you will end up not needing the whole Slater determinant. Instead it is then computationally much more efficient to work with creation and annihilation operators which are defined as the act of adding and removing single-particle orbitals from the Slater determinant. That way you end up cleverly using the orthogonality of the orbitals.
– Jens
Commented Jun 26, 2012 at 17:16