9
$\begingroup$

I can't seem to articulate what made me ask this question but it sprang up when I asked: What does it mean to say a material is 'anharmonic'?.

I've heard before that the electronic entropy becomes a force to reckon at extreme (near melting) temperatures since electrons acquire tremendous energy at such temperatures. Does the failure of the QHA have something to do with this? I don't as such understand the physical meaning (rather the implication) of 'coupling' but is electron-phonon coupling the demise of QHA? Also, does phonon-phonon coupling have any role to play here?

$\endgroup$
5
$\begingroup$

The starting point of most calculations is the Born-Oppenheimer approximation, which separates the electronic and nuclear degrees of freedom. The electronic structure problem is then solved using a variety of methods (DFT, wave function methods, etc), and the nuclear problem is typically solved using the (quasi)harmonic approximation in solids.

To observe a failure of the (quasi)harmonic approximation, you do not need to involve electrons directly. The (quasi)harmonic approximation assumes that the phonons are non-interacting albeit volume-dependent. This will fail when phonon-phonon interactions need to be taken into account, which can happen in a variety of cases (high temperature, light elements, structural phase transitions).

The other concepts you mention involving electrons are also important, although not directly related to the (quasi)harmonic approximation. For example, electronic entropy is important in situations including warm dense matter where temperatures are really high, and electron-phonon interactions are important in a variety of cases, and they allow one to go beyond the Born-Oppenheimer approximation by coupling the electronic and nuclear subsystems.

$\endgroup$

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.