What is the non-relativistic differential equation for the helium atom?

Question

I understand that the Schrodinger Equation for the hydrogen atom is

$$E\Psi=-\frac{\hbar^2}{2m}\nabla^2\Psi+V\Psi$$

with $$-\frac{\hbar^2}{2m}\nabla^2\Psi+V\Psi$$ being the Hamiltonian

and I understand that the Hamiltonian of the Helium Atom is

$$-\frac{\hbar^2}{2m}\left(\nabla_1^2\Psi+\nabla_2^2\Psi\right)+V_1\Psi+V_2\Psi+V_{12}\Psi$$

although I haven't seen what's on the other side of the equation

As I understand it each electron can also be in a different energy level from the other

So is the non relativistic differential equation for the Helium atom

$$E_1\Psi+E_2\Psi=-\frac{\hbar^2}{2m}\left(\nabla_1^2\Psi+\nabla_2^2\Psi\right)+V_1\Psi+V_2\Psi+V_{12}\Psi$$

or is it something else?

Answer 1

Not sure what $E_1 + E_2$ are supposed to be - probably there is no such separation into a sum (and the last equation is missing a wavefunction) - but generally, well, you're right, the time-independent Schrödinger equation for a system (including the Helium atom) is the eigenvalue equation $$ E \Psi = \widehat{H}\Psi $$ for the energy $E$ of the system described by the wavefunction $\Psi$ and with the Hamiltonian operator $\widehat{H}$.