2
$\begingroup$

I have read these posts in trying to answer my question: $k$-interval for First Brillouin Zone, Band gaps: are they at the centre or at the edge of the Brillouin zone?. I have also tried to find answers in Ashcroft and Mermin, Ibach and Luth, and Kittle to no avail.

In solid state physics, electron bands are usually introduced by deriving the nearly-free-electron dispersion, using Bloch's theorem. From there the professor folds the dispersion into the First Brillouin Zone (FBZ), and says 'voila: bands and gaps'! I understand from a math perspective that for Bloch wave-functions the k vector only adds a phase, and that it has no impact on physical observables. I am having a hard time understanding this physically. Once could say it is easy to unfold the bands back into the original picture in this case, but what about more complicated bands where the originating BZ is not so obvious?

enter image description here

My question is: why does folding the energy-momentum dispersion into the FBZ not destroy information about the absolute momentum of states? If it does destroy information, but that fact does not matter, why does it not matter?

A state with well defined momentum outside of the FBZ should be measure-able as such, correct? Then why does the FBZ band representation not limit the band diagram?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ I will shamelessly point to this answer to a closely related question. Note the emphsis there on this sentence: "Both schemes show the same information." There is no information lost. $\endgroup$
    – garyp
    Apr 24, 2020 at 20:37
  • $\begingroup$ Thanks for the link. I'll add it to the post so it's more visual. I think the post helps, but it doesn't fully explain why information isn't lost. $\endgroup$
    – NJP
    Apr 26, 2020 at 20:18
  • $\begingroup$ Actually, it does. In the folded scheme all momenta that lie on any vertical line are the same. If they are the same, nothing is lost. It's like counting modulo 5. I can count fifteen consecutive numbers, and the sixteenth will be 1. If you are talking about the sixteenth you are actually talking about the first. $\endgroup$
    – garyp
    Apr 27, 2020 at 3:49

1 Answer 1

Reset to default
1
$\begingroup$

"A state with well defined momentum outside of the FBZ should be measure-able as such, correct?" Well, no.

A Bloch state is labelled by a reciprocal lattice vector $k$, which is often sloppily called "momentum" with or without a factor $\hbar$. The vector space where $k$ lives has symmetry properties that are determined by the structure of the real-space lattice and the space spanned by the lattice basis vectors. Just like translating the wave function by an integer number of lattice sites does not change the wavefunction at all, the same holds true for "translating" the wave function in reciprocal lattice space. This means that the label $k$ cannot be unique, just like a site in an infinite and perfectly periodic lattice cannot be either.

The Wikipedia article on Crystal momentum might help, and the animation in the top right corner is particularly directed at your physical "intuition" for this.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks for the link. It is a big help. So you're saying this phenomenon is a result of aliasing? I guess I have a couple of follow up questions. Can you expand on why the k vector is not a good quantum number? I guess a related question is what are the momentum eigenstates then, and how can they be labeled? $\endgroup$
    – NJP
    Apr 26, 2020 at 20:26
  • 1
    $\begingroup$ According to this meta post you should post these follow-up questions in (a) separate question(s), possibly linking back to here :) $\endgroup$
    – Marius Ladegård Meyer
    Apr 26, 2020 at 21:23

Your Answer

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

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