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I've seen that, a possible defining condition for the reciprocal lattice is:

$\vec R_s \cdot \vec G=2 \pi l$, where, $R_s= n_1 \vec a_1+n_2 \vec a_2+n_3 \vec a_3$ is the direct lattice, $\vec G$ is the reciprocal lattice, and $l \in \mathbb{Z}$

Using this definition in the one dimensional case, i find a wrong result, indeed: let $a$ be the distance between atoms, it follows that $R_s=na$ is the direct lattice. The defining equation, for the reciprocal lattice, is $R_sG=2 \pi l$. Substituting it becomes $naG=2 \pi l$ and so $G=\frac {2 \pi l}{a n}$, which is wrong, because it should be $G=\frac {2 \pi l}{a }$. Why do i have that extra integer, $n$, at the denominator?

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In the formula $\vec R_s \cdot \vec G = 2 \pi l$, $\vec R_s$ are the basis vectors of the real lattice, rather than the entire lattice itself. You can work in the basis $\vec R_s = a \hat x$, in which case $G = 2 \pi l/a$, or you can choose a larger basis, like $\vec R_s = 2 a \hat x$, then $G = 2 \pi l/2a$. In this case, the output $\vec G$ is just the basis vector of the reciprocal space. It will of course be different if your real space lattice basis vector is different.

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  • $\begingroup$ Ok, so i can choose whichever vector of the real lattice, but just one $\endgroup$
    – SimoBartz
    Commented Feb 1, 2023 at 12:02
  • $\begingroup$ $R_s$ is not necessarily the basis vectors - see here, for example. But $l$ is not an arbitrary integer that one is free to choose - rather it is a constraint that the product could be only integer. $\endgroup$
    – Roger V.
    Commented Feb 1, 2023 at 12:38

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