# Schrödinger equation for multi-electron atoms

Does the Schrödinger equation alter its form when we apply it for an atom having a configuration other than hydrogen-like? I think it should happen because of difference in nuclear attraction and shielding effects.

If there is a change, how significant is that?

• May be useful: chem.libretexts.org/Textbook_Maps/… – Felipe S. S. Schneider Jul 17 '17 at 18:58
• The Schroedinger equation doesn't change. The components of the Hamiltonian do. – Todd Minehardt Jul 17 '17 at 19:37
• @ToddMinehardt That's much better than what I said... :) – Zhe Jul 17 '17 at 21:27

That probably depends on what you consider to be the "form" of the Schrödinger equation.

As Todd succinctly said, the Schrödinger equation[s]

$$\mathrm i\hbar \frac{\mathrm d|\psi\rangle}{\mathrm dt} = \hat{H}|\psi\rangle \quad \text{(time-dependent)} \tag{1}$$

or

$$\hat{H}|\psi\rangle = E|\psi\rangle \quad\text{(time-independent)} \tag{2}$$

don't change. The Hamiltonian will, of course, be different, but the question is how different? The general form of the (non-relativistic) Hamiltonian for an atom has three components

$$\hat{H} = \color{red}{\underbrace{-\frac{\hbar^2}{2m_N}\nabla_N^2}_{\text{KE of nucleus}}} \,\,\,\, \color{blue}{\underbrace{- \sum_i \frac{\hbar^2}{2m_e}\nabla_i^2}_{\text{KE of electrons}}} \,\,\,\, \color{green}{\underbrace{-\sum_i\frac{Ze^2}{4\pi\varepsilon_0r_i}}_{\text{nucleus-electron PE}}}\,\,\,\, \color{purple}{\underbrace{+ \sum_i\sum_{j>i} \frac{e^2}{4\pi\varepsilon_0r_{ij}}}_{\text{electron-electron PE}}} \tag{3}$$

where the nucleus has mass $m_N$ and charge $Ze$; and $r_i$ denotes the distance between electron $i$ and the nucleus, and $r_{ij}$ denotes the distance between electrons $i$ and $j$.

Whether you are dealing with a hydrogen atom or a silver atom, this form remains valid. The only special thing about the hydrogen atom (or in a hydrogenic ion) is that there is only one electron, so the first two sums only have one term, and the last sum has no terms. So, the Hamiltonian for the hydrogen atom can be simplified to

$$\hat{H}_\text{H atom} = \color{red}{-\frac{\hbar^2}{2m_p}\nabla_p^2} \color{blue}{- \frac{\hbar^2}{2m_e}\nabla_e^2} \color{green}{- \frac{e^2}{4\pi\varepsilon_0r}} \tag{4}$$

but I'm of the opinion that the form of the Hamiltonian

$$\hat{H} = \color{red}{\hat{T}_n} + \color{blue}{\hat{T}_e} + \color{green}{\hat{V}_{en}} + \color{purple}{\hat{V}_{ee}} \tag{5}$$

doesn't change, it's just that there are fewer terms in each sum. You might disagree, but I don't really care, as long as you write the correct Hamiltonian out.

There are many terms for a given problem as well as a system of equations for all of the interactions. The answers for even 3 body gravitational problems are chaotic, and gravity is less complication than schrodinger.

However, using numerical methods, symmetry, and observation we can model how many-electron atoms behave.

Here is an example of how the mechanics of a helium atom is determined.

Quantum numbers arise as discrete parameters for solutions to the schrodinger equation as well as being observable. The Pauli exclusion principle is empirically found. We combine concepts together to model a larger atom, and then further to model molecules (Molecular orbital theory).

The Schrodinger equation works well, but it isn't the whole story.