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Reciprocal Lattice Proof

  1. Feb 25, 2009 #1
    Hey folks,

    Here's my problem:

    Knowing that for reciprocal lattice vectors K and real space lattice vectors R:


    and using the Kronecker delta:


    I need to prove b1, b1, b3 as shown http://www.doitpoms.ac.uk/tlplib/brillouin_zones/reciprocal.php" [Broken]:


    I understand that for the first equation above, the exponential needs to equal zero for the expression to equal 1. So I have K.R=0 as one piece of information, but I don't see how this leads me to the expressions for b1, b1, b3 which I'm trying to find.

    I'm assuming this is part of the proof:


    But how do I use this and where does the 2*pi come from?

    Thanks all!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 25, 2009 #2
    This type of calculation can be found in any book on solid-state physics.

    So we know that [itex]\mb{R}=\sum_i n_i\mb{a}_i, \ n_i\in \mathbb{Z}[/itex], and we want to find a basis [itex]\mb{b}_i[/itex] of our reciprocal lattice such that for [itex]\mb{K}=\sum_i m_i\mb{b}_i, \ m_i\in \mathbb{Z}[/itex] we have [itex]\mb{K}\cdot\mb{R}=2\pi l, \ l\in \mathbb{Z}[/itex], which means [itex]e^{iK\cdot R}=1[/itex]. It is quite easy to see that this is satisfied if [itex]a_i\cdot b_j=2\pi \delta_{ij}[/itex]. Problem is just to find [itex]b_i[/itex] which satisfy this condition. A vector [itex]a_1\times a_2[/itex] will be orthogonal to both [itex]a_1[/itex] and [itex]a_2[/itex] so it makes sense to define [tex]b_3 \propto a_1\times a_2[/tex] since it will naturally give you [itex]a_i\cdot b_3\propto \delta_{i3}[/itex] and so on. The rest is just a matter of finding the correct normalization, which I leave to you.
    Last edited: Feb 26, 2009
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