In Kittel's solid state text, problem 2.3, he says that the volume of the Brillouin zone is the same as a primitive parallelepiped in Fourier space. Somehow I can't see why this is true. Can someone help me see why this is true? Also, is the same relationship true between Wigner-Seitz cells and primitive parallelepiped in real space?
[Physics] Volume of Brillouin zone is the same as Fourier primitive cell
crystalsfourier transformsolid-state-physicsvolume
Related Solutions
For any crystal, the First Brillouin Zone is found using the Wigner-Seitz construction for the reciprocal lattice. The high-symmetry points are labeled by certain letters mainly as a convention--like you said Gamma for (0,0,0) etc.
The important thing to realize as far as the group theory, is that the group of the wavevector at the Gamma point has the full point group symmetry of the real space lattice. However, certain high symmetry wavevectors, labelled by the different Greek letters, are subgroups of this group. That is, only a certain number of the symmetry operations of the point group at the Gamma point (rotations, mirrors, etc.) will leave the new high symmetry point invariant. Thus the benefit of symmetry is that you only have to consider an even smaller region of the BZ to get all of the reciprocal space information about the crystal. This is called the Irreducible Brillouin Zone, and paths along the high symmetry points of the IBZ are used as the x-axis in band structure diagrams.
Use this website to explore different Brillouin zones: http://www.cryst.ehu.es/
This paper gives a thorough treatment of many Brillouin zones: Setyawan, Wahyu, and Stefano Curtarolo. "High-throughput electronic band structure calculations: Challenges and tools." Computational Materials Science 49.2 (2010): 299-312.
The following Book is a great resource which has tables of the high symmetry points in the FBZ and their point groups: Dresselhaus, Mildred S., Gene Dresselhaus, and Ado Jorio. Group theory: application to the physics of condensed matter. Springer Science & Business Media, 2007.
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 $a^*/2$.
Best Answer
This has to be true by construction, either in the real space or in the reciprocal space. There is one primitive parallelepiped and one Wigner-Seitz cell per lattice point, and both of them tile the whole space.