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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.


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To be clear: You mean all Brillouin zones for a given crystal. The first Brillouin zone of NaCl has a different volume than the first Brillouin zone of NaF. But the first Brillouin zone of NaCl has the same volume as any other Brillouin zone of NaCl. –  Steve B Jul 14 '12 at 19:10

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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.

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