What is the form of the Hamiltonian for solids?
There are multiple possibilities. First of all, you can construct them like the ones for molecules, since a solid is also made from nuclei and electrons. So the Hamiltonian looks identical, but with much more terms.
On the practical side, you would do something different. If, for example, your system is periodic (think of a crystal), then you would choose a single elementary cell from the crystal and build the Hamiltonian for that (ignoring all the other cells), so you arrive at a small Hamiltonian.
Then to get the wavefunction for the whole system you do not change the definition of the Hamiltonian but the one of the wavefunction. While for molecules the wavefunction has to vanish at infinity, you now want the wavefunction to have special properties, called periodic boundary conditions:
For example, consider a one-dimensional crystal with unit cell from 0 to L. You now want the wavefunction $\psi$ at 0 and L to be equal, as well as the derivatives of $\psi$ to be equal at these points:
$\psi\left(0\right)=\psi\left(L\right), \frac{\partial\psi}{\partial x}|_{x=0}=\frac{\partial\psi}{\partial x}|_{x=L}$
With this information you can determine the wavefunction in this cell with any method you want. Now you could put this exact wavefunction into the cell from L to 2L and and the given wavefunction would be a solution for this second cell, since the cells are identical. This way you can get a wavefunction $\Psi$ for the whole system, by periodically putting the wavefunctions for the unit cells $\psi$ together:
$\Psi\left(x\right)=\psi\left(x-n\cdot L\right), x\in\left]n\cdot L,(n+1)\cdot L\right], n\in \mathbb{Z}$
The interaction of the wavefunction with the other nuclei in the crystal is taken care of by the periodic boundary conditions, so the wavefunction will be the real solution to the Problem.