How do I show that all Brillouin zones have the same volume? I have read in a few books that all Brillouin zones have the same volume, and I can vaguely see how it works, but have not been able to think up a formal proof.
Help?
 A: A Brillouin zone is defined as the range of k's which represent a unique one particle pseudomomentum state in the crystal. If you count the total number of states in the Brillouin zone, you do an integral over k:
$$ \int {d^dk\over (2\pi)^d} = {V\over (2\pi)^d} $$
Where V is equal to the total volume in momentum space. This is equal to the dimension of the Hilbert space, so it is independent of how you define the zone. Another way of saying this is if you have a large crystal of volume V, the dimension of the Hilbert space is $N= V/v$ where V is the total crystal volume and v is the volume of a unit cell, and in Fourier space, you get a dual lattice of equidistributed discrete momentum states, with total number of k states in any Brillouin zone equal to N. In the limit $V\rightarrow\infty$, the total number of states is proportional to the volume of the zone, so it must be the same no matter how you chop it up.
I don't know how formal you want, but one can get as formal as you like.
A: We can define an bijective mapping between the first and nth Brillouin zones. Therefore, the regions are of equal volume. First, some notation:
Let $\vec{k}\in\mathbb{R}^{3}$. I'll call
$\vec{G}_j(\vec{k})$ = jth closest reciprocal lattice vector to $\vec{k}$.
The nth Brillouin zone, $\Gamma^{*}_n$, is the set of points in $\mathbb{R}^3$ which have the origin as their nth nearest point in $\Lambda^{*}$, the reciprocal lattice. That is,
$$\Gamma^{*}_n\equiv\{\vec{k} \in \mathbb{R}^{3}: \vec{G}_n(\vec{k}) = \vec{0}\}$$
Lastly, we denote
$$\Gamma^{*}_n(\vec{G}_j = \vec{q})\equiv\{\vec{k} \in \Gamma^{*}_n: \vec{G}_j(\vec{k}) = \vec{q}\}$$.
Now, to the bijection. The collection of sets $\Gamma_{n}^{*}(\vec{G}_1 = \vec{q})$ for $\vec{q} \in \Lambda^{*}$ form a partition for $\Gamma_n^{*}$, so $\forall \vec{k} \in \Gamma_n^{*}$ there is a single $\vec{q} \in \Lambda^{*}$ for which $\vec{k} \in\Gamma_{n}^{*}(\vec{G}_1 = \vec{q})$. This $\vec{q}$ is the single reciprocal lattice vector closest to $\vec{k}$.
Given a $\vec{k}$ and its corresponding $\vec{q}$, the function
$$f(\vec{k}) = \vec{k} - \vec{q}$$
maps $\Gamma_{n}^{*}(\vec{G}_1 = \vec{q}) \rightarrow \Gamma_{1}^{*}(\vec{G}_n = \vec{-q})$. This is because due to translational invariance of $\Lambda^{*}$
$$
\vec{G}_1(\vec{k} - \vec{q}) = \vec{G}_1(\vec{k}) - \vec{q} = \vec{q} - \vec{q} = \vec{0}
$$
, and
$$
\vec{G}_{n}(\vec{k} - \vec{q}) = \vec{G}_n(\vec{k}) - \vec{q} = \vec{0} - \vec{q} = -\vec{q}
$$
From this we conclude that $f(\vec{k}) \in \Gamma_{1}^{*}(\vec{G}_n = \vec{-q})$.
$f(\vec{k})$ is injective. We can prove this by
$$f(\vec{k}_1) = f(\vec{k}_2) \Rightarrow \vec{k}_1 - \vec{q}_1 = \vec{k}_2 - \vec{q}_2$$
But we also have $f(\vec{k}_1)\in \Gamma_{1}^{*}(\vec{G}_n = -\vec{q}_1)$ and $f(\vec{k}_2)\in \Gamma_{1}^{*}(\vec{G}_n = -\vec{q}_2)$ so $f(\vec{k}_1) = f(\vec{k}_2) \Rightarrow \vec{q}_1 = \vec{q}_2 \Rightarrow \vec{k}_1 = \vec{k}_2$
$f(\vec{k})$ is surjective. We can prove this by assuming $\vec{k} \in \Gamma_{1}^{*}(\vec{G}_n = -\vec{q})$. Again, due to translational invariance of $\Lambda^{*}$,
$$\vec{G}_n(\vec{k} + \vec{q}) = \vec{0}$$
and
$$\vec{G}_1(\vec{k} + \vec{q}) = \vec{q}$$
so we can conclude that for each $\vec{k} \in \Gamma_{1}^{*}(\vec{G}_n = -\vec{q})$ there is a $\vec{k} + \vec{q} \in \Gamma_{n}^{*}(\vec{G}_1 = \vec{q})$ such that $f(\vec{k} + \vec{q}) = \vec{k}$.
$f$ surjective and $f$ injective $\equiv f$ bijective. QED.
