For atoms, and even molecules, I can understand how the Hamiltonian would be constructed, but what of solids (as in, for electronic structure calculations)? Thank you for any help.

  • 1
    $\begingroup$ What difference do you expect in the Hamiltonian of solids compared to molecules? Also, tagging this with DFT makes even less clear that what kind of Hamiltonian you are talking about. $\endgroup$
    – Greg
    Dec 11, 2020 at 14:18
  • 2
    $\begingroup$ Look into Bloch functions for crystals - the symmetry of the crystal is explicitly introduced into the Hamiltonian. Perhaps more solid state physics than chemistry... $\endgroup$
    – Jon Custer
    Dec 11, 2020 at 17:02

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.