1
$\begingroup$

In this pdf it says

Because the number of electron states in a Brillouin zone is twice the number of primitive cells in a Bravais lattice (the factor two comes from spin), the zone can be completely occupied only if there are an odd number electrons per primitive cell in the lattice. Thus, all materials, with an odd number of electrons in a primitive cell, are metals. [...]

I have often read the result that (in a simplified model) materials with an odd number of electrons per primitive cells do conduct while those with an even number do not. This is, however, the first time I've read an explanation. And I don't understand it. Why does it take an odd number of electrons to fill an even number of states, and why does this determine the conductivity?

Alternatively, if anyone can provide another explanation, I'll be glad to hear it.

$\endgroup$
2
$\begingroup$

There is clearly a mistake in the pdf.

Quoting from Ashcroft-Mermin (Chapter 8),

Because the number of levels in a band is equal to the number of primitive cells in the crystal and because each level can accommodate two electrons (one of each spin), a configuration with a band gap can arise (though it need not) only if the number of electrons per primitive cell is even.

In short, materials with a band gap (insulators/semiconductors) are guaranteed to have even number of electrons per primitive cell. But the converse is not true, a solid with even number of atoms per unit cell may not be an insulator (as in the case where bands overlap and some electrons go into the upper band). A solid with odd number of electrons per unit cell is a metal as it will definitely have a partially filled band.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you very much for your answer! It seems my confusion was legitimate. But now I think to understand. The author of the pdf has probably just confused even and odd in the first sentence. $\endgroup$ – Fred Aug 7 '18 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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