2
$\begingroup$

In solid state physics, features of crystals are explained by using the concept of energy bands, the existence of which is shown to be a result of the periodic order of the atoms in a crystal.

When this periodic order breaks, e.g. when the crystal melts, what does happen to energy bands? Do they still exist? If no, why do some features persist? (e.g. molten metals are shiny and conduct electricity like solid metals.) if yes, how can they exist without that periodic order of atom positions?

|cite|improve this question
$\endgroup$
  • $\begingroup$ At the most naive level, I would say that the energy bands gets smeared out in momentum so that they become better defined in position instead. $\endgroup$ – KF Gauss May 25 at 8:24
  • $\begingroup$ @KFGauss, could you explain what you mean a little more? $\endgroup$ – apadana May 25 at 10:58
1
$\begingroup$

Many properties of metals are described quite well with the free-electron model and some scattering: conductivity as well as the Drude peak and plasmons in the optical conductivity.

Crystallinity and periodicity just provide us with a manageable way to calculate the whole solid. But it is not essential. As you noticed, most electronic properties hardly change upon melting.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Interesting. And why don't molten dielectrics usually have so much free electrons? I guess your answer would be: because the ionization energies of their atoms are significantly higher than those of the atoms composing metals. $\endgroup$ – apadana May 26 at 1:10
  • 1
    $\begingroup$ Indeed, see also the answer by @my2cts. And also for most dielectrics, optical properties do not change much upon melting. The electronic structure in the tight-binding approximation does not change much because usually the local atomic structure (coordination, distances) does not change much in the phase transition. $\endgroup$ – Pieter May 26 at 6:15
1
$\begingroup$

Whether electrons conduct depends on whether they are delocalised. The Hubbard model is instructive here. In its simplest form it has two parameters, the hopping energy t and the on-site electron-electron repulsion U. For large t/U the material is a conductor with partially filled bands. For small t/U it will be an insulator with fully occupied bands.

https://en.m.wikipedia.org/wiki/Hubbard_model

|cite|improve this answer
$\endgroup$
0
$\begingroup$

Well, I am not an expert in amorphous material, but I agree with you we cannot define energy band in such cases. However, the defining property of metal is gapless-ness in the energy spectrum, and this can be defined in any quantum many-body system, even in the absence of lattice translational symmetry. Many basic properties of metal which depends only on the gaplessness survives. Some phenomena which require the existence of lattice, for example Umklapp scattering, should change.
Of course, if we have lattice translation symmetries in the Hamiltonian, all the many-body energy eigenstates are also eigenstate of that symmetry operators, so situation is much easier and we can use k-space picture.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks for your answer. How can "gapless-ness" be defined without referring to bands? Are molten dielectrics "gapless"? How do you explain the physical differences between molten metals and molten dielectrics? e.g. optical properties, electrical properties $\endgroup$ – apadana May 25 at 10:38
  • $\begingroup$ Since we have many-body quantum Hamiltonian, we can find many-body energy eigenstate, whether or not there's lattice behind. The energy difference between the ground state and the first excited state vanishes in the proper thermodynamic limit. How to obtain proper thermodynamic limit contains some subtlety but that's not big a problem here. $\endgroup$ – hwang May 26 at 1:25
  • $\begingroup$ Also, about molten metal and molten dielectrics, I don't think there's a very general answer to that unless we are given a specific material. In the case of metal, there were free conducting electrons before, so it's most likely to be a conductor even after the atomic lattice breaks down. If it was an insulator before, I cannot certainly tell what's the fate after the melting. $\endgroup$ – hwang May 26 at 1:33
  • $\begingroup$ In most cases, as also in the answer above, electrons are not released from the atom, which keeps the material non-conducting. On the other extreme, if you totally break down all the structure(atoms can freely move, and release all the electrons from atoms), it is more like a plasma, which is of course conducting. $\endgroup$ – hwang May 26 at 1:33

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.