# Trying to understand the necessity of orbital approximation or the independent electron model

I'm reading the approximations used for solving the Schroedinger's equation for molecules; the first is the the Born-Oppenheimer approximation:

Freeze the molecule with a fixed internuclear separation $R;$ then carry out calculations to obtain the total energy, $U,$ and wavefunctions for that $R$ value.$^1$

Then another approximation is used which is known as the orbital approximation to get rid of the inter-electronic repulsion as said by the same source.

Consider each electron to move in some sort of "average potential" which incorporates the interactions with the two nuclei and an "averaged interaction" with the other electron.

The electronic Hamiltonian can then be separated into two parts: $$\hat{H}_e = \hat{H}_1 + \hat{H}_2 \\ \text{where}:\\ \hat{H}_1 \;\text{is dependent only upon the properties of electron (1) and upon R .}\\ \hat{H}_2 \;\text{is dependent only upon the properties of electron (2) and upon R .}$$

I couldn't understand for what purpose this approximation is meant for. Also, I couldn't able to conceive what average potential actually means.

I then googled the term finally to come before a pdf file where it is written that

$\color{red}{\bullet}$ Electrons repel each other according to Coulomb's law, with the repulsion energy $r^{-1} _{ij} .$

$\color{red}{\bullet}$ Hartree-Fock replaces this instantaneous electron-electron repulsion with the repulsion of each electron with an average electron charge cloud.

Prior to that it is written:

$\color{red}{\bullet}$ It assumes that each electron interacts with an average charge distribution due to the other electrons. This is wrong!

I've spent a hard time understanding this approximation.

Could anyone please explain why this approximation is taken & how it actually solved that problem? The source cites the reason of inter-electronic-repulsion as the problem; how does this term in the Hamiltonian cause problem & how is it solved by the approximation? Also, what is meant by average potential?

An intuitive explanation would be appreciated.

$^1$ http://www.chem.qmul.ac.uk/software/download/mo/3.pdf

• I'm tempted to repeat my advice here to stop switching from one source to another in a hope to find a simple explanation. Pick one (good) book and read it (from cover to cover). It also looks like OP might be trying to jump the gun: it looks just to early for him/her to dive into the quantum chemistry. This question is just another indication of OP lacking a good foundation of quantum theory in his/her head. – Wildcat Oct 19 '15 at 19:56
• The question though is a good one, so I made an attempt to answer is as shortly as possible. :) – Wildcat Oct 19 '15 at 20:06
• You may want to go and check the equations, try to write them down and see what you should put where, instead of fiddling with every single paragraph in your book. Try to understand the actual tools, not the pictures or simple explanations textbooks try to give you. Textbooks often oversimplify, or work ony as a metaphor, math clears things up. – Greg Oct 19 '15 at 22:23

## Setting the stage


## Separation of variables technique

The separation of variables is aimed to broke a partial differential equation of many variables into a set of simpler partial differential equations of fewer variables. Note though, that the separation of variables is by no means a universal approach: for some equations a solution of the form mentioned above make the separation of variables possible, for others it is of no use.

The most common way of separation of variables for a partial differential equation in a function $f(x_1, x_2, \dotsc, x_n)$ of $n$ variables $x_{i}$ is by writing its solution as a product of $n$ functions $f_{i}(x_{i})$ each of which is a function of one variable only $$f(x_1, x_2, \dotsc, x_n) = f_1(x_1) f_2(x_2) \dotsm f_n(x_n) = \prod_{i=1}^{n} f_i(x_i) \, , \tag{2}$$ and then substituting solution of this form into the equation. Using algebraic manipulation one could try to break the resulting equation into a set of $n$ independent ordinary differential equations for each function $f_{i}(x_{i})$ and if succeed, one could solve these independent equations and by plugging all $f_{i}(x_{i})$ back into (2) obtain the solution of the starting equation.

## Separation of particles coordinates

It is fairly well known that an attempt to separate the coordinates of individual particles in the time-independent Schrödinger equation $$\op{H} \psi(\vec{r}) = E \psi(\vec{r}) \, ,$$ by making a substitution $\psi(\vec{r}) = \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i})$ will be successful only if the potential energy function can be expressed as the sum of the separate potentials for each individual particle $$\op{V}(\vec{r}) = \sum_{i=1}^{n} \op{v}(\vec{r}_{i}) \, ,$$ which implies that the particles do not interact with each other. For such a system of non-interacting particles the time-independent Schrödinger equation, partial differential equation of a function of $n$ variables, breaks up into a system of $n$ independent partial differential equations of functions of $1$ variable only, $$\op{h}_{i} \psi_{i}(\vec{r}_{i}) = \varepsilon_{i} \psi_{i}(\vec{r}_{i}) \, ,$$ where $\op{H} = \sum\limits_{i=1}^{n} \op{h}_{i}$ and $E = \sum\limits_{i=1}^{n} \varepsilon_{i}$. All this is relatively easy to establish, and besides, this derivation is done in many textbooks in great details, so I leave at as an educational exercise to OP.

## Separation of electrons coordinates

Electrons, on the contrary, interact with each other, so that a simplification of the electronic Schrödinger equation by means of separation of coordinates of individual electrons won't work. Mathematically, it is the interaction potential $\op{V}\elel$ that will prevent the separation of the electronic coordinates, $$\psi\el(\vec{r}\el) \neq \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i}) \, ,$$

But what if we can approximate $\op{V}\elel$ with some model potential which will allow the separation of electronic coordinates? Obviously, such an approximate potential should have the form of the sum of one-electron potentials each of which depends only on the coordinates of $i$-th electron as we have already seen. Physically it would mean that electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons. Hence the name of the approximation, the mean field approximation. The corresponding approximate potential can be formally written as follows $$\op{V}_{\mathrm{MF}} = \sum\limits_{i=1}^{n} \op{v}_{\mathrm{MF}}(\vec{r}_{i}) \, .$$ The electronic Hamiltonian $\op{H}\el$ is then approximated with the sum of effective one-electron Hamiltonian operators $\op{h}(i)$ $$\op{H}_{\mathrm{MF}} = \sum_{i=1}^{n} \op{h}(\vec{r}_{i}) \, ,$$ where $$\op{h}(\vec{r}_{i}) = - \frac{1}{2} \nabla_{i}^2 - \sum\limits_{α=1}^{ν} \frac{Z_{α}}{ r_{α i} } + \op{v}_{\mathrm{MF}}(\vec{r}_{i}) \, .$$ And in contrast to eigenfunctions $\psi\el(\vec{r}\el)$ of the true electronic Hamiltonian $\op{H}\el$, eigenfunctions $\psi_{\mathrm{MF}}(\vec{r}\el)$ of the mean-field Hamiltonian $\op{H}_{\mathrm{MF}}$ defined by the corresponding eigenvalue equation $$\op{H}_{\mathrm{MF}} \psi_{\mathrm{MF}}(\vec{r}\el) = E_{\mathrm{MF}} \psi_{\mathrm{MF}}(\vec{r}\el) \, ,$$ can be written as products of one-electron wave functions $$\psi_{\mathrm{MF}}(\vec{r}\el) = \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i}) \, ,$$ where $\psi_{i}(\vec{r}_{i})$ are solutions of the corresponding one-electron Schrödinger equation $$\op{h}_{i} \psi_{i}(\vec{r}_{i}) = \varepsilon_{i} \psi_{i}(\vec{r}_{i}) \, .$$

The main problem with the mean field approach described above is that so far we have just assumed that $\op{V}_{\mathrm{MF}}$ exist in one form or another, but we have no idea of how it actually might look like. There is a well-known workaround, but its a different story.

Note that I ignored spin completely in the discussion above which is, of course, fundamentally wrong, but spin won't make a big difference with respect to the key concept discussed, the mean-field approximation.

• First of all, again thanks for answering my question; I've understood that in order to separate the variables, we need to avoid the inter-electronic distance term. But I've not conceived how by not instantaneously acting with each other, we can separate the variable. Physically it would mean that electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons: why should there be no interaction term in the average field? – user5764 Oct 20 '15 at 10:22
• Coulomb interactions between electrons prevents the separation of their coordinates, but if we simply get rid of the corresponding potential, we'll obviously get qualitatively wrong result. So, we proceed more wisely: we not just get rid of Coulomb interactions potential, we also introduce a one-electron additive potential which has to at least approximately recover the energy due to Coulomb interactions between the electrons. – Wildcat Oct 20 '15 at 10:43
• Now, what this one-electron additive potential could mean physically if its aim is to account for energy due to Coulomb interactions? The only sensible meaning that we can give to it is that it is a potential of Coulomb interactions between each and every electron and the average field created by all other electrons in the system. So, the form of the mean-field potential and the role it is supposed to play dictates its physical interpretation. – Wildcat Oct 20 '15 at 10:45
• Sorry, sir, didn't contact as I was in holiday. Okay, I want to ask you that the trial wavefunction $\psi(\vec{r}) = \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i})$ is the Hartree product of $n$ orbitals, but we know that electrons are identical fermions which must be antisymmetrised by using Slater determinant $\psi= |\det(\psi_1\updownarrow_1,\psi_2\updownarrow_2,\ldots,\psi_n\updownarrow_n)\rangle$? So, would the calculation above be different if the later wavefunction is used instead of $\psi(\vec{r}) = \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i})?$ – user5764 Oct 25 '15 at 15:46
• @user36790, antisymmetric product of orbitals is just a linear combination of ordinary products, so the mean-field separation is still possible. – Wildcat Oct 25 '15 at 16:03