I have few simple questions about Brillouin zone.

I will take the example of a 1D lattice with a period of "a".

If I have well understood, the reciprocal lattice shows all the point in "k-space" allowed for any function that will respect the periodicity of the lattice.

So, it shows us all the allowed wave vectors when we do a decomposition in Fourier series.

(I know that we also can proove that the vector of the reciprocal lattice are the ones that will allow diffraction but I would like if possible not consider experimental arguments to explain me what is going on).

Now, what is physically the (first) Brillouin zone ? In my example I know that it will be all the vector in the reciprocal space that will be between $ [-\pi/a ; \pi/a] $.

But why do we distinguish these vectors ? Why are they this important ?

I have read explanation that says it is because if we take a wave with wavelength $ a $ and another with wavelength $ 2a $ an observer will not see the difference between them.

But I dont understand this argument because : First, why should the observer only look at what happens on atoms of the lattice and not between them ? Next, when we explain this we draw a wave of wavelength 2a starting with a node at x=0, then there will also be a node at x=a and x=2a. And then we draw a wave of wavelength a and we say "thoose waves has the same nodes at the same place : the observer will not be able to distinguish them". Ok but if we have started our wave at a peak and not at a node the observer would be able to distinguish them so I dont understand this argument. We just drawed a particular case.

So, are they this important in link with bloch theory (I asked a question about this in another topic, I am really a beginner in solid states physics, so don't be too hard with your explanations please :) ).

Thank you a lot for your answers


1 Answer 1


Mathematically the reason why we typically only consider the first Brillouin zone is that it is the only area necessary for completely defining both the wavefunction and energy of the system. From Bloch's theorem it follows that $$ \psi_{n,\vec{k} + \vec{K}}(\vec{r}) = \psi_{n,\vec{k}}(\vec{r}) $$ and $$ E_{n,\vec{k} + \vec{K}} = E_{n,\vec{k}} $$

Thus, each energy and wavefunction with quantum number k outside the Brillouin zone can be related to a quantum number within the Brillouin zone. There is nothing wrong with expanding beyond the first Brillouin zone, however the results will be redundant.

Physically, this value is analogous to momentum in the free particle case. However, the symmetries are different as the free particle has a continuous transnational symmetry while a crystal has a discrete transnational symmetry. Therefore, this often called the $\textit{Crystal Momentum}$.

For more detailed information on this, I recommend $\textit{Solid State Physics}$ by Ashcroft and Mermin, chapter 8.

  • $\begingroup$ So, If I understand well, the main reason of introducing Brillouin Zone in solid states physics is because it is used with Bloch functions (because when we consider such wavefunction we can reason only on the brillouin zone) ? Thank you. $\endgroup$
    – StarBucK
    Commented Jan 28, 2016 at 18:00
  • $\begingroup$ Yes, the Brillouin zone contains all necessary information to completely describe you system. All points outside the zone are redundant. $\endgroup$ Commented Jan 28, 2016 at 19:17

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