# Why ground-state configuration is consistent ("additive") from one element to the next?

Ground-state configuration ("electron shells") is consistent throughout the periodic table, for example:

(source)

Why is it this way if the Hamiltonian of each atom is different? In other words, why Al is $\mathrm{[Ne]3s^2 3p^1}$ and not something completely different: (1) at the core, and (2) at the valence.

Well, this configurations arise in what is usually called the orbital approximation and the consistent pattern is due to the very nature of this approximation.

In the orbital approximation the overall electronic wave function describing the atom is represented as the product of the wave functions for the individual electrons known as atomic orbitals. This can be sketched as follows. $$\newcommand{\op}[1]{\hat{#1}} \newcommand{\el}[1][]{_{\mathrm{e}}} \newcommand{\nuc}[1][]{_{\mathrm{n}}} \newcommand{\elel}[1][]{_{\mathrm{ee}}} \newcommand{\elnuc}[1][]{_{\mathrm{en}}}$$

## Electronic Schrödinger Equation

The electronic wave function $\psi_{\mathrm{e}}$ is an eigenfunction of the electronic Hamiltonian $\op{H}\el$ $$\op{H}\el \psi\el = E\el \psi \el \, ,$$ $$\op{H}\el = \op{T}\el + \op{V}\elnuc + \op{V}\elel = - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2} - \sum\limits_{i=1}^{n} \frac{Z}{r_{i}} + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,$$ which indeed is different for different atoms. Thus, the electronic wave function can also be expected to be different. However, the exact analytical solutions of the electronic Schrödinger equation are known only for one and two-electron systems, such as $\ce{H}$, $\ce{H-}$, $\ce{He}$, $\ce{Li+}$, etc., while for all other chemical systems only approximate solutions can be found.

## Separation of variables

The electronic Schrödinger equation is partial differential equation and one of the most frequently used analytical methods to solve such equations is technique known as the separation of variables aimed to broke one compound partial differential equations of many variables into a set of simpler partial differential equations of fewer variables. The main disadvantage of the separation of variables is that it 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 partial differential equation in a function $f(x_1, x_2, \dotsc, x_n)$ of $n$ variables $x_{i}$ is writing a solution as a product of $n$ functions $f_{i}(x_{i})$ each of which is a function of one variable only $$f(x_1, x_2, \dotsc, x_n) = f_1(x_1) f_2(x_2) \dotsm f_n(x_n) \, ,$$ 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 then one could solve these independent equations and by plugging all $f_{i}(x_{i})$ back into the product obtain the solution of the starting equation.

## Orbital approximation

So, it is tempting to try to separate the electronic coordinates by representing $\psi\el$ as a product of one-electron wave functions, however, it is fairly well known that this attempt will be successful only if the potential energy can be expressed as the sum of the separate potentials for each particle. This clearly is not the case for the electronic Hamiltonian: the potential energy of the Coulomb repulsions between the electrons $\op{V}\elel$ prevents the separation of electronic coordinates \begin{equation*} \psi\el(\vec{r}_{1}, \vec{r}_{2}, \dotsc, \vec{r}_{n}) \neq \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n})\, . \end{equation*} So, exact separation in this form is not possible, but what about approximate one? Can we somehow approximate $\op{V}\elel$ with some model potential which will allow the separation of electronic coordinates? To some extent, the answer is, yes, we can.

In the orbital approximation it is indeed assumed that the many-electron wave function can be written as a product of one-electron wave functions called orbitals. As it was already said, it will work only if the exact $\op{V}\elel$ potential is replaced by the model potential $\op{V}_{\mathrm{MF}} = \sum\nolimits_{i=1}^{n} \op{v}_{\mathrm{MF}}(r_{i})$ which allows the separation. The approximate Hamiltonian is $$\op{H}_{\mathrm{MF}} = \op{T}\el + \op{V}\elnuc + \op{V}_{\mathrm{MF}} = - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2} - \sum\limits_{i=1}^{n} \frac{Z}{r_{i}} + \sum\limits_{i=1}^{n} \op{v}_{\mathrm{MF}}(r_{i}) \, .$$ Physically it means 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, another name of the approximation, the mean field approximation.

Even the mean-field Hamiltonian still looks different for different atoms, however, it is now a sum of one-electron Hamiltonian operators $$\op{H}_{\mathrm{MF}} = \sum\limits_{i=1}^{n}\op{h}(r_{i}) \, ,$$ where $$\op{h}(r_{i}) = - \frac{1}{2} \nabla_{i}^2 - \frac{Z}{ r_{i} } + \op{v}_{\mathrm{MF}}(r_{i}) \, .$$ And in contrast to an eigenfunction $\psi\el$ of the true electronic Hamiltonian $\op{H}\el$ an eigenfunction $\psi_{\mathrm{MF}}$ of the mean-field Hamiltonian $\op{H}_{\mathrm{MF}}$ can be written as products of one-electron wave functions (orbitals) $$\psi_{\mathrm{MF}} = \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n}) \, ,$$ where $\psi_{i}(\vec{r}_{i})$ are solutions of the corresponding one-electron Schrödinger equations $$\op{h}_{i} \psi_{i}(\vec{r}_{i}) = \varepsilon_{i} \psi_{i}(\vec{r}_{i}) \, .$$ Now this resulting equations that defines orbitals are not that different: there will be just a small difference in one-electron Hamiltonians due to slightly different nuclear charges as you move along a group. Of course, there will be more and more occupied orbitals as you proceed through the group, but equations themselves and consequently orbitals will be quite simillar.

• Thanks! If $\hat h (r_i)$ are added to produce $\hat H_{MF}$, why $\psi_i(r_i)$ are multiplied to produce $\psi_{MF}$? Feb 18, 2016 at 16:51
• I expanded the answer a little bit. The key point is that from the get go we want to write the wave function this particular way, but it does make sense only if the Hamiltonian has the form of a sum of one-electron Hamiltonians. Otherwise, we won't get the whole many-particle equation separated into a bunch of one-particles ones. Feb 18, 2016 at 17:37

Sorry, this is too long for a comment. This is an answer to the OP's comment on

If $\op{h}(r_{i})$ are added to produce $\op{H}_{\mathrm{MF}}$, why $\psi_{i}(\vec{r}_{i})$ are multiplied to produce $\psi_{\mathrm{MF}}$?

The combination of these special forms of $\op{H}_{\mathrm{MF}}$ and $\psi_{\mathrm{MF}}$ is consistent with the time-independent Schroedinger Equation $\op{H}_{\mathrm{MF}} \psi_{\mathrm{MF}} = E_{\mathrm{MF}} \psi_{\mathrm{MF}}$:

\begin{align} \op{H}_{\mathrm{MF}} \psi_{\mathrm{MF}} &= \sum\limits_{i=1}^{n}\op{h}(r_{i}) \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n}) \\ &= \underbrace{\op{h}(r_{1}) \psi_{1}(\vec{r}_{1})}_{= \, e_{1} \psi_{1}(\vec{r}_{1})} \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n}) + \psi_{1}(\vec{r}_{1}) \underbrace{\op{h}(r_{2}) \psi_{2}(\vec{r}_{2})}_{= \, e_{2} \psi_{2}(\vec{r}_{2})} \dotsb \psi_{n}(\vec{r}_{n}) + \ldots \\ &\hphantom{=} \, + \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \underbrace{\op{h}(r_{n}) \psi_{n}(\vec{r}_{n})}_{= \, e_{n} \psi_{n}(\vec{r}_{n})} \\ &= e_{1} \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n}) + e_{2} \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n}) + \ldots \\ &\hphantom{=} \, + e_{n} \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n}) \\ &= \underbrace{(e_{1} + e_{2} + \ldots + e_{n})}_{= \, E_{\mathrm{MF}}} \underbrace{\psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n})}_{= \, \psi_{\mathrm{MF}}} \\ &= E_{\mathrm{MF}} \psi_{\mathrm{MF}} \end{align}