Why are exact solutions limited to hydrogen-like atoms?

Question

Why can we only find exact solutions to the Schrödinger equation for Hydrogen atoms without estimating. What is the problem with the mathematics of extending the Schrödinger equation to more complicated atoms?

Answer 1

How many equations in mathematics have closed-form solutions? Almost none of them. This is a severe problem when using mathematics to construct models of physics, but we have no alternative.

Of course, textbooks dance around this by carefully choosing the problems presented.

Answer 2

Quantum mechanics applies to everything not just hydrogen atoms. It's just that the differential equations that tell us the evolution of the wavefunctions don't always have nice, exact solutions.

There are other simple systems with exact solutions, like a quantum harmonic oscillator. In addition to the hydrogen atom any atom with one electron has nearly the same exact solution. For example, once-ionized helium has one electron and a nuclear charge of $+2e$. The Schrodinger equation for this (or any other single electron atom) is: $$ \nabla^2 \psi = \frac{2m_e}{\hbar^2} \left[ -\frac{Zke^2}{{r}^2} - E \right] \psi, $$ where $Z$ is the atomic number of the atom, so $Z=2$ for helium.

In some sense the hydrogen atom doesn't really have exact solutions. The traditional solution doesn't account for motion of the proton or the magnetic moment of the electron or a host of other things. These are added back in as perturbations to the base solution.

Looking at the equation above, we are assuming the nucleus behaves like a localized point charge. That's how we can define $r$ the separation between the nucleus and the electron wavefunction. If we treat the nucleus like a wavefunction, we need to account for how the nucleus is spread out through space.

Multi-electron atoms

Multi-electron atoms get pretty messy, pretty fast. Let's look at the Schrodinger equations for two electrons in the same atom:

$$ \nabla^2 \psi_1 = \frac{2m_e}{\hbar^2} \left[ -\frac{Zke^2}{{r_1}^2} + \frac{ke^2}{{r_{12}}^2} - E_1 \right] \psi_1 $$

$$ \nabla^2 \psi_2 = \frac{2m_e}{\hbar^2} \left[ -\frac{Zke^2}{{r_2}^2} + \frac{ke^2}{{r_{12}}^2} - E_2 \right] \psi_2 $$

There are now two equations, one for each electron. The first term in the potential is the attractive contribution from the nucleus. We could again assume the nucleus is a fixed point charge.

The second term is the repulsive contribution from the other electron, where $r_{12}$ is the separation between the electrons. Since each electron is a wavefunction, this distance isn't as clear. We need to know what the wavefunction $\psi_2$ is in order to figure the separation and solve for $\psi_1$. But to get $\psi_2$ we also need to know the separation, and therefore need $\psi_1$. Can you see what the problem is?

This does not mean quantum mechanics is not valid. We still used quantum mechanics to set up our physical model. This just means the math prescribed by quantum mechanics is hard.