# General concepts
## Basic problem

Solve the time dependent Schrödinger equation (or in case of a stationary state, the time independent one, as can be seen in the second line) for a system of many electrons (coordinates ${{\mathbf {r} }_{i}}$, spins ${{s}_{i}}$)ː
${\begin{aligned}&i{\frac {\partial }{\partial t}}\psi \left(\left\{{{\mathbf {r} }_{i}},{{s}_{i}}\right\},t\right)=\left(-\sum \limits _{i}{{\frac {1}{2}}{{\Delta }_{{\mathbf {r} }_{i}}}}+{\frac {1}{2}}\sum \limits _{i\neq j}{\frac {1}{\left|{{\mathbf {r} }_{i}}-{{\mathbf {r} }_{j}}\right|}}-\sum \limits _{i,J}{\frac {{Z}_{J}}{\left|{{\mathbf {r} }_{i}}-{{\mathbf {R} }_{J}}\right|}}\right)\psi \left(\left\{{{\mathbf {r} }_{i}},{{s}_{i}}\right\},t\right)\\&E\psi \left(\left\{{{\mathbf {r} }_{i}},{{s}_{i}}\right\}\right)=\left(-\sum \limits _{i}{{\frac {1}{2}}{{\Delta }_{{\mathbf {r} }_{i}}}}+{\frac {1}{2}}\sum \limits _{i\neq j}{\frac {1}{\left|{{\mathbf {r} }_{i}}-{{\mathbf {r} }_{j}}\right|}}-\sum \limits _{i,J}{\frac {{Z}_{J}}{\left|{{\mathbf {r} }_{i}}-{{\mathbf {R} }_{J}}\right|}}\right)\psi \left(\left\{{{\mathbf {r} }_{i}},{{s}_{i}}\right\}\right)\\\end{aligned}}$
To deal with the dimensionality problem, we shall try to reduce the problem to the calculation of one-electron wavefunction(s). In the upcoming chapters, different methods will be shown to manage it.

## The case of non-interacting electrons: Slater determinants

Let us assume, we find a way to approximate the many-electron Hamilton operator by a sum of single-electron operators:
$E\psi \left(\left\{{{\mathbf {r} }_{k}},{{s}_{k}}\right\}\right)=\sum \limits _{i}{\left(-{\frac {1}{2}}{{\Delta }_{{\mathbf {r} }_{i}}}+V^{\text{eff}}\left({{\mathbf {r} }_{i}}\right)\right)}\cdot \psi \left(\left\{{{\mathbf {r} }_{k}},{{s}_{k}}\right\}\right)$
Then, we might try to write the wave-function as a product of single electron wave functions which fulfill the single-electron Schrödinger equationsː
${\begin{aligned}&\psi \left(\left\{{{\mathbf {r} }_{k}},{{s}_{k}}\right\}\right)=\prod \limits _{k}{{{\phi }_{k}}\left({{\mathbf {r} }_{k}},{{s}_{k}}\right)}\\&{{E}_{k}}{{\phi }_{k}}\left(\mathbf {r} ,s\right)=\left(-{\frac {1}{2}}{{{\text{ }}\!\!\Delta \!\!{\text{ }}}_{k}}+{{V}^{\text{eff}}}\left(\mathbf {r} \right){{\phi }_{k}}\right){{\phi }_{k}}\left(\mathbf {r} ,s\right)\\\end{aligned}}$
These single-electron wave-functions have the orthogonality and completeness propertiesː
${\begin{aligned}&\left\langle {{\phi }_{k}},{{\phi }_{l}}\right\rangle =\int _{V}{\sum \limits _{s\in \left\{\uparrow ,\downarrow \right\}}{\phi _{k}^{*}\left(\mathbf {r} ,s\right){{\phi }_{l}}\left(\mathbf {r} ,s\right){{d}^{3}}r}}={{\delta }_{kl}}\quad &{\text{orthogonality}}\\&\sum \limits _{j}{{{\phi }_{j}}\left(\mathbf {r} ,s\right)\phi _{j}^{*}\left(\mathbf {r} ',s'\right)}=\delta \left(\mathbf {r} -\mathbf {r} '\right){{\delta }_{s,s'}}\quad &{\text{completeness}}\end{aligned}}$
It is evident that the product $\psi \left(\left\{{{\mathbf {r} }_{k}},{{s}_{k}}\right\}\right)$ is a solution of the many-particle Schrödinger equation with energy $E=\sum \limits _{k}{{E}_{k}}$. However, this solution does not fulfill the symmetry requirement for Fermion systemsː
$\psi \left({{\mathbf {r} }_{1}},{{s}_{1}},...,{{\mathbf {r} }_{i}},{{s}_{i}},...,{{\mathbf {r} }_{k}},{{s}_{k}},...,{{\mathbf {r} }_{N}},{{s}_{N}}\right)=-\psi \left({{\mathbf {r} }_{1}},{{s}_{1}},...,{{\mathbf {r} }_{k}},{{s}_{k}},...,{{\mathbf {r} }_{i}},{{s}_{i}},...,{{\mathbf {r} }_{N}},{{s}_{N}}\right)\quad \forall i,k$
To construct a solution with the correct symmetry, first let's notice that if we take any permutation $P$ of the indices of the single-electron wave function and construct the following wave functions then they will also be eigenfunctions of the many-electron Hamiltonian.
${\begin{aligned}&\psi \left(\left\{{{\mathbf {r} }_{j}},{{s}_{j}}\right\}\right)={{\phi }_{1}}\left({{\mathbf {r} }_{1}},{{s}_{1}}\right)\cdot ,...\cdot {{\phi }_{i}}\left({{\mathbf {r} }_{i}},{{s}_{i}}\right)\cdot ...\cdot {{\phi }_{k}}\left({{\mathbf {r} }_{k}},{{s}_{k}}\right)\cdot ...\cdot {{\phi }_{N}}\left({{\mathbf {r} }_{N}},{{s}_{N}}\right)\\&{{\psi }_{P}}\left(\left\{{{\mathbf {r} }_{j}},{{s}_{j}}\right\}\right):={{\phi }_{P\left(1\right)}}\left({{\mathbf {r} }_{1}},{{s}_{1}}\right)\cdot ...\cdot {{\phi }_{P\left(i\right)}}\left({{\mathbf {r} }_{i}},{{s}_{i}}\right)\cdot ...\cdot {{\phi }_{P\left(k\right)}}\left({{\mathbf {r} }_{k}},{{s}_{k}}\right)\cdot ...\cdot {{\phi }_{P\left(N\right)}}\left({{\mathbf {r} }_{N}},{{s}_{N}}\right)\\\end{aligned}}$
The same holds for any linear combination of such permutations. We may thus try to build an anti-symmetric wave-function by linear combination of product wavefunctions. But before this, let's illustrate this construction procedure.

### Two-electron system

The procedure and its implications for a two-electron system looks like the following. The single electron wave functions are ${{\phi }_{1}}\left(\mathbf {r} ,s\right)$ and ${{\phi }_{2}}\left(\mathbf {r} ,s\right)$, therefore a (non-symmetry-correct wave function for the two electron system) $\psi \left(\left\{{{\mathbf {r} }_{k}},{{s}_{k}}\right\}\right)$ is $\psi \left({{\mathbf {r} }_{1}},{{s}_{1}},{{\mathbf {r} }_{2}},{{s}_{2}}\right)={{\phi }_{1}}\left({{\mathbf {r} }_{1}},{{s}_{1}}\right)\cdot {{\phi }_{2}}\left({{\mathbf {r} }_{2}},{{s}_{2}}\right)$.
Now, a still non-symmetry-correct wave function can obtained as ${{\psi }_{P}}\left({{\mathbf {r} }_{1}},{{s}_{1}},{{\mathbf {r} }_{2}},{{s}_{2}}\right)={{\phi }_{P\left(1\right)}}\left({{\mathbf {r} }_{1}},{{s}_{1}}\right)\cdot {{\phi }_{P\left(2\right)}}\left({{\mathbf {r} }_{2}},{{s}_{2}}\right)$, where $P\left(1\right)=2$ and $P\left(2\right)=1$. For this case, the symmetry-correct wave function can be obtained as ${{\psi }_{SC}}=C\left(\psi -{{\psi }_{P}}\right)$, where $C$ is a normalization factor, i.e.
${{\psi }_{SC}}\left({{\mathbf {r} }_{1}},{{s}_{1}},{{\mathbf {r} }_{2}},{{s}_{2}}\right)=\left({{\phi }_{1}}\left({{\mathbf {r} }_{1}},{{s}_{1}}\right)\cdot {{\phi }_{2}}\left({{\mathbf {r} }_{2}},{{s}_{2}}\right)-{{\phi }_{2}}\left({{\mathbf {r} }_{1}},{{s}_{1}}\right)\cdot {{\phi }_{1}}\left({{\mathbf {r} }_{2}},{{s}_{2}}\right)\right)/{\sqrt {2}}$
This function changes sign upon interchange of the two electrons. It has the following properties:
* The wave-function is zero when both electrons are in the same state, i.e. ${{\phi }_{1}}={{\phi }_{2}}$
* In all other cases, the wave-function is zero if both electrons have the same value of $s$ and are at the same position $\mathbf {r}$
* The wave-function is normalized (if it is not zero).
* The electron density is given by
${\begin{aligned}{{\rho }_{1}}\left(\mathbf {r} ,s\right)=&2\int _{V}{\sum \limits _{s'\in \left\{\uparrow ,\downarrow \right\}}{\psi _{SC}^{*}\left(\mathbf {r} ,s,\mathbf {r} ',s'\right)\cdot {{\psi }_{SC}}\left(\mathbf {r} ,s,\mathbf {r} ',s'\right){{d}^{3}}r'}}\\=&2\int _{V}{\sum \limits _{s'\in \left\{\uparrow ,\downarrow \right\}}{\left(\underbrace {\phi _{1}^{*}\left(\mathbf {r} ,s\right)\cdot \phi _{2}^{*}\left(\mathbf {r} ',s'\right)} _{I}-\underbrace {\phi _{2}^{*}\left(\mathbf {r} ,s\right)\cdot \phi _{1}^{*}\left(\mathbf {r} ',s'\right)} _{II}\right)\left(\underbrace {{{\phi }_{1}}\left(\mathbf {r} ,s\right)\cdot {{\phi }_{2}}\left(\mathbf {r} ',s'\right)} _{III}-\underbrace {{{\phi }_{2}}\left(\mathbf {r} ,s\right)\cdot {{\phi }_{1}}\left(\mathbf {r} ',s'\right)} _{IV}\right){{d}^{3}}r'}}/2\\=&\left(\underbrace {\phi _{1}^{*}\left(\mathbf {r} ,s\right){{\phi }_{1}}\left(\mathbf {r} ,s\right)} _{{\text{got from }}I\wedge III}+\underbrace {\phi _{2}^{*}\left(\mathbf {r} ,s\right){{\phi }_{2}}\left(\mathbf {r} ,s\right)} _{{\text{got from }}II\wedge IV}\right)={{\left|{{\phi }_{1}}\left(\mathbf {r} ,s\right)\right|}^{2}}+{{\left|{{\phi }_{2}}\left(\mathbf {r} ,s\right)\right|}^{2}}\end{aligned}}$
We note that the terms got from $I\wedge IV$ and from $II\wedge III$ are both 0 due to the orthogonality of the single-electron wave functions.

### Generalization to N particles

In the case of $N$ particles, we have to make a linear combination of all possible permutation of indices, the coefficient of each $N$-particle wave function has to be the parity of the permutation $P$ , which is 1 or -1 depending on how many swaps lead to that specific permutation. If the number of interchanges are even, this number is 1, and if they are odd, this number is -1. (Just as in the case of the two-electron system.) The following form is equivalent to this description and called Slater determinantː
$\psi \left(\left\{{{\mathbf {r} }_{k}},{{s}_{k}}\right\}\right)={\frac {1}{\sqrt {N!}}}\sum \limits _{P\in {{S}_{N}}}{{\text{sign}}\left({{P}_{n}}\right)\prod \limits _{i=1}^{N}{{{\phi }_{{{P}_{n}}\left(i\right)}}\left({{\mathbf {r} }_{i}},{{s}_{i}}\right)}}$
* Here ${{\phi }_{k}}\left({{\mathbf {r} }_{k}},{{s}_{k}}\right)$ are still the orthogonal and normalized functions, which forms a complete basis in its space.
* $\sum \limits _{P\in {{S}_{N}}}{\bullet }$ means the summation of all possible permutation of the numbers $1,2,...,N$, i.e. ${{S}_{N}}=\left\{\left(1,2,3,...\right),\left(2,1,3,...\right),\left(1,3,2,...\right),...\right\}$ which is called the symmetric group of order $N$.
* ${\text{sign}}\left({{P}_{n}}\right)={{\left(-1\right)}^{{\text{number of swaps}}\left({{P}_{n}}\right)}}$is 1, if the number of swaps which lead to that specific permutation from the "original" $1,2,...,N$ permutation is even, and -1 otherwise. E.g.ː
${\text{sign}}\left(1,2,3,...,N\right)=1$, because 0 swaps are needed. But ${\text{sign}}\left(2,1,3,...,N\right)=-1$, because one has to make a swap between the first two indices to get this permutation.
* ${{P}_{n}}\left(i\right)$ is the $i$th number in that specific permutation
* Let's practice this a bit and get the two-electron wave function based on this equation!
* ${\begin{aligned}\psi \left({{\mathbf {r} }_{1}},{{s}_{1}},{{\mathbf {r} }_{2}},{{s}_{2}}\right)=&{\frac {1}{\sqrt {2!}}}\left({\text{sign}}\left(1,2\right)\prod \limits _{i=1}^{2}{{{\phi }_{\left(1,2\right)\left(i\right)}}\left({{\mathbf {r} }_{i}},{{s}_{i}}\right)+{\text{sign}}\left(2,1\right)\prod \limits _{i=1}^{2}{{{\phi }_{\left(2,1\right)\left(i\right)}}\left({{\mathbf {r} }_{i}},{{s}_{i}}\right)+}}\right)\\=&\left({{\phi }_{\left(1,2\right)\left(1\right)}}\left({{\mathbf {r} }_{1}},{{s}_{1}}\right){{\phi }_{\left(1,2\right)\left(2\right)}}\left({{\mathbf {r} }_{2}},{{s}_{2}}\right)-{{\phi }_{\left(2,1\right)\left(1\right)}}\left({{\mathbf {r} }_{1}},{{s}_{1}}\right){{\phi }_{\left(2,1\right)\left(2\right)}}\left({{\mathbf {r} }_{2}},{{s}_{2}}\right)\right)/{\sqrt {2}}\\=&\left({{\phi }_{1}}\left({{\mathbf {r} }_{1}},{{s}_{1}}\right){{\phi }_{2}}\left({{\mathbf {r} }_{2}},{{s}_{2}}\right)-{{\phi }_{2}}\left({{\mathbf {r} }_{1}},{{s}_{1}}\right){{\phi }_{1}}\left({{\mathbf {r} }_{2}},{{s}_{2}}\right)\right)/{\sqrt {2}}\end{aligned}}$
The following notation gives reason to name this wave function a determinant:
$\psi \left(\left\{{{\mathbf {r} }_{k}},{{s}_{k}}\right\}\right)={\frac {1}{\sqrt {N!}}}\left|{\begin{matrix}{{\phi }_{1}}\left({{\mathbf {r} }_{1}},{{s}_{1}}\right)&{{\phi }_{1}}\left({{\mathbf {r} }_{2}},{{s}_{2}}\right)&\cdots &{{\phi }_{1}}\left({{\mathbf {r} }_{N}},{{s}_{N}}\right)\\{{\phi }_{2}}\left({{\mathbf {r} }_{1}},{{s}_{1}}\right)&{{\phi }_{2}}\left({{\mathbf {r} }_{2}},{{s}_{2}}\right)&\cdots &{{\phi }_{2}}\left({{\mathbf {r} }_{N}},{{s}_{N}}\right)\\\vdots &\vdots &\ddots &\vdots \\{{\phi }_{N}}\left({{\mathbf {r} }_{1}},{{s}_{1}}\right)&{{\phi }_{N}}\left({{\mathbf {r} }_{2}},{{s}_{2}}\right)&\cdots &{{\phi }_{N}}\left({{\mathbf {r} }_{N}},{{s}_{N}}\right)\\\end{matrix}}\right|$
Each element is a single-electron wave function in this matrix, but the definition of its determinant is the same as for "regular", ${{\mathbb {R} }^{N}}\times {{\mathbb {R} }^{N}}$ matrices.We also note that this notation implies the followings:
* The Slater determinant is zero whenever two electrons are in the same quantum state (same single-electron wave functions) – two rows of the determinant are equal
* The determinant is also zero when two electrons of the same spin are in the same place (two columns are equal)
* Finally, interchanging two particles amounts to interchanging two columns, which results in multiplying the determinant with -1

## Variational principle for the ground state wave-function

The ground-state (lowest energy) solution of the Schrödinger equation fulfills the condition:
$\left\langle \psi \right|{\hat {H}}\left|\psi \right\rangle =\int {{{\psi }^{*}}{\hat {H}}\psi {{d}^{3N}}r{\overset {!}{\mathop {=} }}\,{\underset {\psi }{\mathop {\min } }}\,}$
with the constraint $\left\langle \psi |\psi \right\rangle =1$. Thus we may determine the ground-state wave-function from the variational equation:
$\delta \left[\left\langle \psi \right|{\hat {H}}\left|\psi \right\rangle -\sum \limits _{i,j=1}^{N}{{\lambda }_{i,j}}\left(\left\langle {{\phi }_{i}}|{{\phi }_{j}}\right\rangle -{{\delta }_{ij}}\right)\right]=0$

### Hartree theory

Idea: approximate the wave-function of the interacting many-electron system by a product of single particle wavefunctions. The result matches with the case Hartree-Fock, but this will not contain the exchange term. More to write here...

### Hartree-Fock theory

Idea: Approximate the wave-function of the interacting many-electron system by a Slater determinant (ie the wave-function of a non-interacting system). This is similar to the Hartree theory but with the correct Fermion symmetry. The variational principle to determine the ground state will be used. The expectation value of the Hamiltonian (described at the basic problem section) in the Slater determinant state is given by a cumbersome but straightforward calculation and the result is:
${\begin{aligned}\left\langle \psi \right|{\hat {H}}\left|\psi \right\rangle =&\sum \limits _{s\in \left\{\uparrow ,\downarrow \right\}}{\sum \limits _{i=1}^{N}{\int _{V}{\phi _{i}^{*}}}\left(\mathbf {r} ,s\right)\left(-{\frac {1}{2}}\Delta +{{V}_{e-N}}\left(\mathbf {r} \right)\right){{\phi }_{i}}\left(\mathbf {r} ,s\right){{d}^{3}}r}\\&+{\frac {1}{2}}\sum \limits _{s,s'\in \left\{\uparrow ,\downarrow \right\}}{\sum \limits _{i,j=1}^{N}{\phi _{i}^{*}\left(\mathbf {r} ,s\right)\phi _{j}^{*}\left(\mathbf {r} ',s'\right)}}{\frac {1}{\left|\mathbf {r} -\mathbf {r} '\right|}}{{\phi }_{j}}\left(\mathbf {r} ',s'\right){{\phi }_{i}}\left(\mathbf {r} ,s\right){{d}^{3}}r{{d}^{3}}r'\quad &{\text{Coulomb potential}}\\&-{\frac {1}{2}}\sum \limits _{s,s'\in \left\{\uparrow ,\downarrow \right\}}{\sum \limits _{i,j=1}^{N}{\phi _{i}^{*}\left(\mathbf {r} ,s\right)\phi _{j}^{*}\left(\mathbf {r} ',s'\right)}}{\frac {1}{\left|\mathbf {r} -\mathbf {r} '\right|}}{{\phi }_{j}}\left(\mathbf {r} ,s\right){{\phi }_{i}}\left(\mathbf {r} ',s'\right){{d}^{3}}r{{d}^{3}}r'\quad &{\text{Exchange interaction}}\end{aligned}}$
(There was a ${{\delta }_{s,s'}}$ in the exchange interaction in the lecture noted, but I think this form is the correct one.)