The first Brillouin zone boundaries are at the wave vectors $\mathbf{k}= \pm \pi / a$, so that a normal dispersion curve looks something like this: enter image description here

It is common to identify the $\mathbf{k}$ vectors with three coordinates: $(k_x, k_y, k_z)$. These coordinates arise from using reduced wave vector coordinates, as stated in Dove Introduction to Lattice Dynamics, page 23:

It is common practice to define the wave vector as normalized by the first reciprocal lattice vector lying along the direction of the wave vector. This gives hat is called the reduced wave vector. For our one-dimensional example, the reduced wave vector has a value of $1/2$ at the Brillouin zone boundary, obtained by dividing the wave vector $a^*/2$ by the reciprocal lattice vector. Thus, in common with most other workers, we will usually show dispersion curves with reduced wave vectors between 0 and 1/2, noting that for non-primitive unit cells some of the zone boundaries occur with reduced wave vector values of 1.

I have got problems obtaining the reduced wave vector, for example at the first Brillouin zone boundary, $\mathbf{k} = \pi / a$, I do not see why this point in reciprocal space is $a^*/2$.

  • $\begingroup$ What is $a*$ here? $\endgroup$
    – noah
    Commented May 10, 2017 at 20:08
  • $\begingroup$ @Michael If $a$ is the direct space lattice vector, $a^{*}$ is the reciprocal space lattice vector $\endgroup$
    – DavidC.
    Commented May 10, 2017 at 20:44
  • $\begingroup$ Then it should be obvious. The boundary of the Brillouin zone passes trough the middle point of the reciprocal lattice vector with length a*. See again Wigner-Seitz cell. The B.Z. is the W-S cell in the reciprocal space. $\endgroup$
    – nasu
    Commented May 10, 2017 at 20:50
  • 1
    $\begingroup$ Is this question actually about why the first brillouin zone boundary is halfway to the first reciprocal lattice point (in the direction of closest packing anyway)? $\endgroup$
    – noah
    Commented May 10, 2017 at 21:08
  • $\begingroup$ @Michael Yes... (thank you very much for all the effort) $\endgroup$
    – DavidC.
    Commented May 10, 2017 at 21:29

1 Answer 1


This is simply a rescaling of the axes in $k$-space. Since in your 1D-example the first reciprocal lattice point is at $2\pi/a$, dividing the point at the brillouin zone boundary by this value gives $1/2$, as is stated in the text. So the point $a^*/2$ is not, as you assumed, the position of the brillouin zone boundary in reduced units, but the boundary in the not-reduced units.

I assume $a^* \equiv 2\pi/a$ is the length of the reciprocal lattice vector, since it would make sense in this context.

Edit 1

For phonons, the reason why the brillouin zone boundary is halfway to the first reciprocal lattice point is that the shortest wavelength you can have is a sign change from one atom to the other. Picture a chain of atoms with the first up, the second down, the third up again. There is no shorter wavelength than this. We also know that the solutions are plane waves (in the simplest case), which means (1D) $s(x) = \mathfrak{Re}(A\cdot e^{ikx})$, where $s(x)$ is the amplitude of the atom at position $x$ and $A$ is the maximum amplitude of the oscillation. For this to change sign from site to site, $k\overset{!}{=}\pi/a$, which you can easily verify.

As to how to construct the first brillouin zone, have a look at any solid state physics book. You just draw lines from the origin to every reciprocal lattice point and bisect them with a plane perpendicular to the line. Every point you can get to without crossing any of these planes is in the first brillouin zone, and the planes themselves are the boundaries.

Edit 2

The brillouin zone is constructed in such a way that it is sufficient to consider all $k$-points inside it, as it can be shown that they are equivalent to points outside. We know the waves to have bloch form $$s(x) = e^{ika} u(x)$$ where $u(x)$ has the periodicity of the lattice. From this expression we can see, that $ka$ gives you the phase change from one lattice site to the next. If now $ka$ is bigger than $\pi$, say $\pi+\Delta$, this point on the $k$-axis is equivalent to $-\pi+\Delta$, because $e^{i(\pi+\Delta)}=e^{i(\pi+\Delta -2\pi)}=e^{i(-\pi+\Delta)}$. So we see that looking at $k$-points up to $\pi/a$ is sufficient for all properties, because the points outside have an equivalent point inside. And by construction of the reciprocal lattice (its first point in the positive direction is at $2\pi/a$), this is precisely at $a^*/2$.

  • $\begingroup$ In my 1D-example the first reciprocal lattice point is not at $2 \pi /a$, but at $\pi /a$ $\endgroup$
    – DavidC.
    Commented May 10, 2017 at 20:47
  • $\begingroup$ The "vector" itself is $ \frac{2\pi}{a} $. The Brillouin zone ends at half this value. $\endgroup$
    – nasu
    Commented May 10, 2017 at 20:51
  • $\begingroup$ No... In 1D, Given the $a$ vector in direct space, the one in reciprocal space is $a^{*}$ and has length of $2 \pi /a $; because this is how reciprocal vectors are constructed ($2 \pi \cdot \text{inverse of length} $) $\endgroup$
    – DavidC.
    Commented May 10, 2017 at 21:00
  • $\begingroup$ @DavidC. You said it yourself, the brillouin zone boundary is at $\pi/a$ but as brillouin zones are constructed, this is halfway to the first reciprocal lattice point. $\endgroup$
    – noah
    Commented May 10, 2017 at 21:02
  • $\begingroup$ @noah I agree that for a chain of atoms in which one is "up" and one is "down", (picture this as spherical $s$ orbitals whose wavefunction change sign), is satisfied when $k = \pi / a$, as it is very well explained in the Hoffmann paper (onlinelibrary.wiley.com/doi/10.1002/anie.198708461/pdf), Figure 7. However, this does not mean that the zone boundary in reciprocal space is located at $a^{*} / 2$ $\endgroup$
    – DavidC.
    Commented May 11, 2017 at 19:13

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