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I'm working on a magnetic system and in the calculations this function appears

\begin{equation} \gamma(\vec{k})=\frac{1}{Z}\sum_{\vec{\mu}}e^{i \vec{\mu} . \vec{k}} \end{equation}

where $\vec{\mu}$ are the primitive vectors of a Bravais Lattice. In the book Spin waves: theory and applications by Anil Prabhakar and Daniel D. Stancil it is said that

For crystals with a center of symmetry $\gamma(\vec{k}) = \gamma(-\vec{k})$

What does this mean? When can I say that the lattice has this symmetry? By looking at the definition of $\gamma$ it seems to me that this symmetry only happens when for every $\vec{\mu}$ there is a $-\vec{\mu}$ in the base generating the lattice but I'm not sure.

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By looking at the definition of $\gamma$ it seems to me that this symmetry only happens when for every $\vec{\mu}$ there is a $-\vec{\mu}$ in the base generating the lattice but I'm not sure.

It is exactly like that. If the crystal has a center of symmetry $O$, it means that for any vector $\vec{OP}$ connecting the center of symmetry to a lattice point, you will have a corresponding vector $-\vec{OP}$ connecting it to another lattice point.

The picture below, with the center of symmetry in red, should clarify this. Notice that the center of symmetry does not have to correspond to a lattice point.

enter image description here

Since this is valid for every Bravais lattice, it follows that

For a perfect crystal, $\gamma(\vec k) = \gamma(-\vec k)$.

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  • $\begingroup$ Great! So... every Bravais Lattice has this symmetry, right? $\endgroup$ Dec 27, 2017 at 15:09
  • $\begingroup$ Yes. For every perfect crystal, $\gamma(\vec k) = \gamma(- \vec k)$. I added this to the answer. $\endgroup$
    – valerio
    Dec 27, 2017 at 15:16

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