Do orbital models derive from the solution of the stationary Schrödinger equation?

Question

I know that hydrogen-like atomic orbitals can be seen as a solution of the stationary Schrödinger equation - but what about atoms with more electrons? As I am a pupil and quite uneducated about this subject, I would be very grateful for an answer that is easy to understand.

Answer 1

I know that hydrogen-like atomic orbitals can be seen as a solution of the stationary Schrödinger equation - but what about atoms with more electrons?

Orbitals are one-electron wave functions, i.e. they are wave functions that describe the state of only one electron. For a hydrogen-like atom, which is a one-electrons system, if treating its nucleus as a classical positively charged point particle, the wave function is indeed an orbital, since there is just one quantum particle out there and it is the single electron. So, right, orbitals are the solutions of the (non-relativistic) time-independent Schrödinger equation if the nucleus is assumed to be a classical point particle.

For any many-electron atom, though, even when its nucleus is though of as a classical point particle, the wave function describes more than one electron, so by the very definition it is not an orbital. It can be written in terms of the one-electron wave functions (orbitals) as their product, but this can be done only approximately due to interactions between the electrons. It is in this sense we say that orbitals aren't real.

Note though that, strictly speaking, even for a hydrogen-like atom orbitals aren't real since they arise in the non-relativistic treatment which is also an approximation. The reality is even more weird than its non-relativistic description, but for atoms from the first and second rows of the periodic table the non-relativistic description is usually a good one in a sense that it introduces relatively small errors that can be neglected for the most of the chemical purposes. The orbital approximation on the other hand might be the source of much larger errors. For instance, for a $\ce{He}$ atom in its ground electronic state the orbital approximation introduces the error of the $10^{-2} E_{\mathrm{h}}$ order while the relativistic correction is about two orders of the magnitude smaller, $10^{-4} E_{\mathrm{h}}$.

Answer 2

Let's take a look at the Hamiltonian for Hydrogen atom with one electron:

$\hat{H}_{H} = T_e + T_p + V_{ep}$

You need for each particle (electron and proton) an expression for the kinetic energy ($T$) and a expression for their interactions as potential energy ($V$). For an additional electron you have to add additional expressions:

$\hat{H}_{H^-} = T_{e_1} + T_{e_2} + T_p + V_{e_1p} + V_{e_2p} + V_{e_1e_2}$

Each expression increases the difficulty for solving the Schrödinger Equation. Thus there are no other element orbitals that can described analytically exact. But there are a lot of good numerical approaches.