Understanding the numerical complexity in solving a Schrödinger equation Usually, when I read the numerical complexity of solving a multielectron Schrödinger equation is due to its very big size.
I came across the following explanation:
Say, there are $N$ electrons, and we want to solve it numerically, therefore we discretize (in other words, converting the differential equation to difference equation and the equation is solved at discrete points and no longer continuous) the $3D$ space to $K \times K \times K$ grid. I don’t understand how they arrive at this.
The size of wave function values to store in the grid is given by $K^{3N}$. Which is huge even for $K=2$ and $N=100$. (Source: https://www.diva-portal.org/smash/get/diva2:935561/FULLTEXT01.pdf)
But my thought process goes like this:
If there are 8 grid points, then shouldn’t there be $3N\times 8$ wavefunction values and not $8^N$ as the formula indicates. Am I missing something?
 A: If you have a system with $N=100$ electrons, the wavefunction is a complex function of $3 N$ variables. Three coordinates for each electron. The magnitude squared of the wavefunction then specifies the probability density to find the electrons at the places the current values the variables specify.
So if you have a $K \times K \times K$ space grid you have $(K \times K \times K)^N$ grid points for the domain of the wavefunction. And this is just $K^{3N}$.
By the way: The Density Functional Theory avoids this problem as it reformulates the Schrödinger equation in terms of the electron density which is only a function of three variables.
A: For $f:\mathbb{R} \rightarrow  \mathbb{R}$ discretized on a grid with $K$ points, you have $K$ possible real values. More formally, this means that the discretized version is $f: \{1,..,K\} \rightarrow  \mathbb{R}$.
For $f:\mathbb{R}^{m} \rightarrow  \mathbb{R}$ discretized on a grid with $K^{m}$ points, you have $K^{m}$ possible real values. More formally, this means that the discretized version is $f: \{1,..,K\}\times...\times\{1,..,K\} \rightarrow  \mathbb{R}$, where the cartesian product is taken $m$ times.
For $f:\mathbb{R}^{m} \rightarrow  \mathbb{R}^w$ discretized on a grid with $K^{m}$ points, you have $wK^{m}$ possible real values, because this is just a collection of $w$ functions $\{f_1,...,f_w\}$ of the kind discussed above, namely $f_i:\mathbb{R}^{m} \rightarrow  \mathbb{R}$ for $i=1,...,w$.
Answer: Now, consider the complex function $\psi(\mathbf{x}_1, ..., \mathbf{x}_N)$ in $d$ space dimensions, namely
$\Psi:\mathbb{R}^{dN} \rightarrow  \mathbb{C}$. Since $\mathbb{C}$ is analogous to $\mathbb{R}^2$ (algebraically they are not the same thing, but in this context they are, practically speaking, the same space), we are in a case $f:\mathbb{R}^{n} \rightarrow  \mathbb{R}^w$ with $n=dN$ and $w=2$. Therefore, the discretized version of the wave function allows you to store $2K^{dN}$ real values.
Note: if the wave function can be factorized as  $\psi(\mathbf{x}_1, ..., \mathbf{x}_N) =\psi_1(\mathbf{x}_1)...\psi_N(\mathbf{x}_N) $, then  the answer is $2NK^d$. If it can be fully factorized as $\psi(\mathbf{x}_1, ..., \mathbf{x}_N) =\psi_1(x_{11})...\psi_{dN}(x_{dN}) $, then you have to store only $2NdK$ real numbers.
Note: if the wave function is normalized, then you have $2K^{dN}-1$ independent real values in the general case. Further constraints that may be specific to the problem at hand can also lower the number.
