In condensed matter, one usually considers Bloch states inside the first Brilliouin zone, which, for 1d system with lattice constant $a$, is $-\pi/a<k<\pi/a$.

But the basis of this, Bloch theorem, is based on periodic boundary condition, which implies $k=\frac{2\pi n}{Na}$ with $n=0,1,\cdots N-1$ and $N$ is the number of unit cells. This interval of $k$ is not the same as the interval given by first Brilliouin zone, which is fine if one just wants to consider the dispersion because dispersion is periodic. However, there may be some problems. For example, if we do a Fourier transformation numerically, we need to use $k=\frac{2\pi n}{Na}$ with $n=0,1,\cdots N-1$ instead of $-\pi/a<k<\pi/a$, otherwise we are not really using periodic boundary condition.

Is this right? And is there any more subtle point on which one may easily make mistake?

  • $\begingroup$ I think you are right. One should perhaps use the discreet Fourier transform (DFT) instead of the continuous. $\endgroup$ – garyp Mar 21 '14 at 2:16
  • $\begingroup$ I don't see why periodic boundary condition is necessary of Bloch theorem. We just start on an infinite lattice and consider discrete translation symmetry we get Bloch theorem. $\endgroup$ – Jia Yiyang Mar 21 '14 at 2:20
  • $\begingroup$ @JiaYiyang: I was talking about a finite lattice $\endgroup$ – Mr. Gentleman Mar 21 '14 at 3:12
  • 3
    $\begingroup$ Why do you "need to" do it that way? You can use $0<k<2\frac{\pi}{a}$ for your BZ just as well as you can use $-\frac{\pi}{a}<k<\frac{\pi}{a}$. Likewise, you can shift the DFT; just multiply everything through by the appropriate phase factor. $\endgroup$ – lnmaurer Mar 21 '14 at 4:14

Your Answer

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

Browse other questions tagged or ask your own question.