## Hartree method, from Hartree product to HF If we ignore inter electronic repulsion (i.e., no correlation), then we may develop a Hartree product to deal with multiple electron problems. $\Psi_{HP}=\psi_{1}\psi_{2}\cdots\psi_{N}$ $N$ is number of electrons. $\psi_{i}$ is their wavefunctions, and for separated Hamiltonian, $\hat{H}=\sum_{i=1}^{N}h_{i}$ where $h_{i}=-\frac{1}{2}\nabla_{i}^{2}-\sum_{k=1}^{M} \frac{Z_{k}}{r_{ik}}$, $M$ is number of nuclei. So we have $H\Psi_{HP}=(\sum_{i}\epsilon_{i})\Psi_{HP}$ this $\epsilon_{i}$ is the energy for $\psi_{i}$,, i.e., $h_{i}\psi_{i}=\epsilon_{i}\psi_{i}$. >[!Note] >Remember, Hartree does not have Pauli principle (correlation) in it, while [[#^e9965a|Slater determinant]] does! (We do not include e-e interaction in above $h_{i}$ expression) This result looks very nice, but we must consider electron repulsion if we want it to work properly. We still keep $\psi_{i}$ and $\Psi_{HP}$, and consider $\bra{\Psi_{HP}}\hat{H}\ket{\Psi_{HP}}$, the $\hat{H}$ here is the ***true*** Hamiltonian, and $h_{i}$ becomes $h_{i}=-\frac{1}{2}\nabla_{i}^{2}-\sum_{k=1}^{M} \frac{Z_{k}}{r_{ik}} +V_{i}\{j\}$, and if we consider only the Coulombic repulsion, we have $V_{i}\{j\}= \sum_{j\neq i}\int \frac{\rho_{j}}{r_{ij}} \mathrm{d}r$ since $\rho_{j} = |\psi_{j}|^{2}$, this becomes a self-consistent field problem, total energy becomes $E=\sum_{i}\epsilon_{i} - \frac{1}{2}\sum_{i\neq j} \iint \frac{|\psi_{i}|^{2}|\psi_{j}|^{2}}{r_{ij}} \mathrm{d}r_{i}\mathrm{d}r_{j}$ Only including this repulsion (Coulombic) is not enough, although its a big step forward. We have to take spin into account. The spin quantum WFs are represented by $\alpha$ and $\beta$, and are eigenfunctions of operator $\hat{S_{z}}$. >[!Note] > Fermions get antisymmetric properties: exchange the position and spin, the wavefunction should change sign. This is directly given by [Pauli exclusion principle](https://en.wikipedia.org/wiki/Pauli_exclusion_principle). > This switch position/spin is the switch in the wave function. e.g., $\Psi(x_{1}, x_{2}) \rightarrow \Psi(x_{2}, x_{1})$, then $\Psi(x_{1}, x_{2})=-\Psi(x_{2}, x_{1})$. The $\alpha,\ \beta$ are directly multiplied to $\Psi_{HP}$, like for [Triplet oxygen](https://en.wikipedia.org/wiki/Triplet_oxygen), the state is written as $^{3}\Psi_{HP}=\psi_{a}(1)\alpha(1)\psi_{b}(2)\alpha(2)$ Include antisymmetric property (Pauli principle), we have $^{3}\Psi_{SD}= \frac{1}{\sqrt{2}} \biggl(\psi_{a}(1)\alpha(1)\psi_{b}(2)\alpha(2) - \psi_{a}(2)\alpha(2)\psi_{b}(1)\alpha(1) \biggr)$ These $1, \ 2$ denote for electron (position), $a, \ b$ for MO, SD is slater determinant. One may notice, by computing $\bra{^{3}\Psi_{SD}} \frac{1}{r}\ket{^{3}\Psi_{SD}}$ and $\bra{\Psi_{SD}} \frac{1}{r}\ket{\Psi_{SD}}$ (Here $\Psi_{SD}=\frac{1}{\sqrt{2}} \biggl(\psi_{a}(1)\beta(1)\psi_{b}(2)\alpha(2) + \cdots \biggr)$), the first term gains one more energy lowering term., i.e., $-2 \int \psi_{a}(1)\psi_{b}(1) \frac{1}{r_{12}} \psi_{b}(2)\psi_{a}(2) \mathrm{d}r_{1} \mathrm{d}r_{2}$, and this is the ***[exchange energy](https://en.wikipedia.org/wiki/Exchange_interaction)***! This energy is because the same spin (i.e., no node) makes the total energy become lower. It is less probable for electrons with the same spin to be closer in space. To add this effect, antisymmetric property directly goes into the WF expression by calculating the Slater determinant, which is written as $\Psi_{SD} = \frac{1}{\sqrt{N!}} \begin{vmatrix} \chi_{1}(1) & \chi_{2}(1) & \cdots & \chi_{N}(1) \\ \chi_{1}(2) & \chi_{2}(2) & \cdots & \chi_{N}(2) \\ \vdots & \vdots & \ddots & \vdots \\ \chi_{1}(N) & \chi_{2}(N) & \cdots & \chi_{N}(N) \end{vmatrix}$ ^e9965a >[!Note] >in the expression for $\chi$, like >$\chi_{N}(1)$ >This $\chi_{N}$ means the single electron [[Secular equation#MO by LCAO|MO]] wavefunction, $N$ shows the MO index, the number $1$ means electron (position), may also be written as $x_{1}$. And it is this summation form of all MOs under certain basis give the antisymmetric result. Compact notation: $\Psi_{SD} = \ket{\chi_{1},\chi_{2}, \cdots ,\chi_{N}}$ with $\chi_{1} = \psi_{a}(1)\alpha(1)$, etc. These $\chi$ are all MOs. Here this compact form is only a notation, does not necessary means this ket form is the same as the determinant. >[!Notice] > Slater determinants do not give MO by LCAO, but an antisymmetric form fulfill the Pauli exclusion principle for MOs. We have MOs as input and antisymmetric MO as output. >[!Info] >See more on >- https://en.wikipedia.org/wiki/Slater_determinant >- https://en.wikipedia.org/wiki/Exchange_interaction