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As far as I understand, the construction of Brillouin zones stems from the relation$$ 2 \vec{k}\cdot \vec{G} +G^2 = 0 \,,$$where $\vec{k}$ is the wave vector and and $\vec{G}$ is the reciprocal lattice vector. This condition is supposed to be fulfilled when $\vec{k}$ terminates on a line normal to $\vec{G}$ at half of the length of $\vec{G}.$

If so, why does the third Brillouin zone take the form of Figure 1, rather than Figure 2, for a quadratic lattice? The second figure shows the area enclosed by lines situated half of $2b\hat{x}$ and $2b\hat{y}$ from the origin, perpendicular to the $x$- and $y$- directions respectively.

$\hspace{50px} {\def\place#1#2#3{\smash{\rlap{\hskip{#1px}\raise{#2px}{#3}}}}} \place{0}{220}{\begin{array}{c}\textbf{Figure 1} \\ \hspace{250px} \end{array}} $ $\hspace{50px}{\def\place#1#2#3{\smash{\rlap{\hskip{#1px}\raise{#2px}{#3}}}}} \place{0}{220}{\begin{array}{c}\textbf{Figure 2} \\ \hspace{250px} \end{array}}$

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2 Answers 2

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The first BZ should contain one lattice point, the second BZ two, and the third BZ three, etc. Figure 1 is correct because it is exactly the case. The third BZ in Figure 2 contains 4 lattice points, so it is not correct.

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  • $\begingroup$ Why is it that "the first BZ should contain one lattice point, the second BZ two, and the third BZ three, etc"? I thought that all Brillouin zones must contain 1 lattice point. $\endgroup$
    – Electra
    Jun 2, 2020 at 18:37
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One reason is that the volume of your third BZ is twice the volume of your first BZ.

To see it another way, notice that the black dot on the far right (the middle of the three vertical dots) is a BZ center, a $\Gamma$ point. The black dot directly above that one (far upper right corner) is also a $\Gamma$ point, but it must be in the next BZ. Half-way between these two $\Gamma$ points must be a zone boundary, exactly as shown in your Figure 1.

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