0
$\begingroup$

So I have a large confusion with QM as applied to solid state. The following is a summary of what I know, what I think I know, and what I know I don't know. I hope to stir a discussion that will help users of QM, but non-experts, like myself in the future.

Bloch's theorem is statement of symmetry if you're in a perfect lattice (infinite, no defects, zero K). Due to the nature of this symmetry, the wave-function has to have a periodic nature (the exp(ik) part). This is fine, and largely unsurprising (although very elegant).

Here we can talk about $k$ space, which are just the harmonics of the crystal lattice. We can talk about how many electrons/holes they can contain, ect. We call $k$-space "momentum" space because the Fourier transform of position operator is the momentum operator. Physically, $k$ space are modes of lattice vibrations. These modes of the crystal lattice are called phonons.

But the crystal is not infinite, the temperature is not 0, and the lattice is not defect free.

Let' say I have a nanoparticle, of a few thousand atoms more or less placed according to a lattice (which doesn't have to necessarily be the bulk lattice, i.e. it could be a quasi-crystal's lattice). There will be "defects", surface strain, surface energy, possibly a grain boundary, and the shape of the particle will be in constant evolution (non-zero $T$). All of these are "imperfections" that violate the assumptions of Bloch's theorem.

Now Bloch's theorem does not apply, but somehow we want to use it anyway because there is some periodicity within the cluster, at least locally. Simultaneously, we'd like to borrow from our analytic answer to the H atom (s, p, d, f orbitals).

And here I stop understanding anything at all. There is a new concept introduced which is partial/projected density of state (pdos). But what is pdos representing, ie what is the space? What do they mean by "partial" what are they projecting when they speak of "projected"? Why and how do they identify f, d, ect orbitals? How are these charts read?

All this and we haven't begun to speak of spin!

$\endgroup$
3
$\begingroup$

Blochs Theorem

Well, you are perfectly right: pretty much every assumption of the Bloch Theorem is violated in even the simplest physical system - there was and there will never be a crystal which statsfies the assumptions. Nontheless it delivers decent results. Lets take one doubt after another:

  1. Size: Its as simple as that: The bigger the crystal, the better result you get with Bloch. I think its well imaginable that a wavefuntion of an electron located in the middle of the crystal is 99,9999% the wavefuntion which the bloch theorem predicts, whereas one at the surface is not. But with increasing size you get a decrease in surface/volume ratio. This means if you use the Bloch Theorem the percentage of wavefunctions you will get right tends more and more to 100%. Crystals used for non-nano measurements almost always fulfill this condition, so if Blochs Theorem is a good approximation for 99,9% of wavefunctions, why not use it? The border where the theorem is considered to be good or bad cant be quantized very well.

  2. Temperature: Including the movement of the atoms and the movement of electrons in one single step is complicated. That is why one tends to decouple both (see Born-Oppenheimer-Approximation), so that in the first step of calculations, the atoms are often assumed as resting and the electrons as moving (of course only if youre not interested in phonon properties). This allows an application of the Bloch-Theorem to the problem. Experience shows that it works and you can include electron-phonon effects afterwards in most cases. So here the answer why we use Blochs Theorem is: Because it works nonetheless (as an approximation).

  3. Impurities: The number of impurities is often very small compared to the number of properly aligned atoms. Therefore, again, the Bloch Theorem is a good apprxomation.

Partial Density of States

It's a bit hard to follow your question, because the concept of partial DOS doesn't necessarily require the Bloch-Theorem.

Lets assume we look at a certain energy in the complete DOS. The energy has a corresponding wavefunction in the crystal. You now can look at or mathematically analyze the resulting wavefunction and its symmetry. The wavefunction can be approximated by wavefuctions of the s,p,d,f,... -like functions we already know. So if we have e.g.. (in a heuristic notation) a wavefunction wich is well approximated by a function 0.8*s-like + 0.1*p-like + ... you could assign the Energy in the DOS a 80% s-character and a 10% p-character and so on. Since s,p,...-like functions are originally functions around a single atom the s,p,...-like in the bulk function has to be understood as averaged about every contributing atom in the unit cell.

This can be transferred into a s-DOS, a p-DOS, etc. (this is the "projection" onto the s,p,...-states) which all of course add up to the complete DOS.

$\endgroup$
  • $\begingroup$ Continuing about the Bloch's theorem: what if we have a quasicrystal? i.e. an ordered lattice that does not have translation symmetry. Is Bloch's theorem still apply? $\endgroup$ – Lenzuola Jan 20 '14 at 4:38
  • $\begingroup$ About the pDOS, lets see if I get this: pDOS is a density of state, meaning that it's a graph showing how many energy states there are at a particular energy. However, to each energy state there corresponds an actual wavefunction (eigenvalue - eigenfunction pair). This wavefunction can be written as a linear sum of spherical functions (the s, p, d). So I can go back to my DOS and plot the four, five points corresponding to how much of each basis function I have. Doing this for all energies, I get continuous lines. Is this right? $\endgroup$ – Lenzuola Jan 20 '14 at 4:53
  • 1
    $\begingroup$ bloch theorem: no, the normal theorem wouldnt apply to quasicrystals, because - like you said - there is no translation symmetry. However, there exist more complex approaches with generalized bloch theorys. pDOS: exactly. $\endgroup$ – LeFitz Jan 20 '14 at 13:32
0
$\begingroup$

By my reckoning, the best way to understand the Bloch Theorem is to go through the math.

In the first place,we just consider a perfect lattice with some periodicity and can be described by some reciprocal lattice. Then from the periodicity of the lattice (let's just consider a 1D chain with periodicity of L), we have the commutator relation $\left[E, T\right]=0$.

For a finite crystal, by employing the periodic boundary condition, and the translation by L operation, we can have the Bloch Theorem, which tells the form of the energy eigenfunction in this crystal for the electron (note that we also take the single electron approximation).

To understand how Bloch Theorem can be applied to the real situation,just think of perturbation theory - by building some models, and adding those perturbation terms, we can have a good approximation for the situation.

$\endgroup$

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.