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Reciprocal lattice

  1. Oct 24, 2009 #1
    [tex]f(\vec{r}) = f(\vec{r}+\vec{T}) [/tex]

    [tex]\vec{T}= u_{1} \vec{a_{1}} + u_{2} \vec{a_{2}}+u_{3} \vec{a_{3}} [/tex]

    [tex]u_{1},u_{2},u_{3}[/tex] are integers.

    [tex]f(\vec{r}+\vec{T})= \sum n_{g} e^{(i\vec{G}.(\vec{r}+\vec{R}) )}= f(\vec{r}) [/tex]

    [tex]e^{i\vec{G}.\vec{R} }= 1 [/tex]
    [tex] \vec{G}.\vec{R} = 2\pi m[/tex]

    we call [tex] \vec{G}=h\vec{g_{1}} + k \vec{g_{2}}+l \vec{g_{3}} [/tex] reciprocal lattice vector.

    but what about the primitive lattice vectors [tex]\vec{g_{1}} , \vec{g_{2}} , \vec{g_{3}}[/tex] ?

    To simplify the discussion consider [tex]\vec{T_{1}} [/tex] in 1D;

    [tex]\vec{T_{1}} = u_{1} \vec{a_{1}} [/tex]

    [tex] \vec{G}.\vec{T} =(h\vec{g_{1}} + k \vec{g_{2}}+l \vec{g_{3}}) . ( u_{1} \vec{a_{1}}) = 2 \pi m [/tex]

    Is there any definiton that indicates direct primitive lattice vectors and reciprocal primitive lattice vectors orthogonalities?

    [tex]\vec{g_{1}} . \vec{a_{1}} = 2 \pi [/tex]

    [tex]\vec{g_{2}} . \vec{a_{1}} = \vec{g_{3}} . \vec{a_{1}} = 0 [/tex]
    Last edited: Oct 24, 2009
  2. jcsd
  3. Oct 24, 2009 #2
    g1,g2,g3 are the primitive vectors for the reciprocal lattice.
    And the orthogonality condition is [tex]\vec{a}_i \cdot \vec{g}_j = 2\pi \delta_{i,j}[/tex]

    A trick for calculating the reciprocal vectors is to form the matrix A where the columns are the direct lattice vectors, and the matrix G where the columns are the reciprocal lattice vectors. Then you have
    [tex]G^T \cdot A = 2\pi I[/tex]
    [tex]G = 2\pi (A^{-1})^T[/tex]
  4. Oct 24, 2009 #3
    I is 3x3 identity martix isn't it?
  5. Oct 24, 2009 #4
    yes it is
  6. Oct 25, 2009 #5
    It's still confusing for me..
    so how can I invers a (1x3) matrix?
  7. Oct 25, 2009 #6
    You can get the vectors [tex]\vec{g_1}, \vec{g_2}, \vec{g_3}[/tex] without using any matrices. To do this use the following formulas:
    [tex]\vec{g_1}=2\pi\frac{[\vec{a_2},\vec{a_3}]}{\vec{a_1}[\vec{a_2},\vec{a_3}]},[/tex] [tex]\vec{g_2}=2\pi\frac{[\vec{a_3},\vec{a_1}]}{\vec{a_1}[\vec{a_2},\vec{a_3}]},[/tex] [tex]\vec{g_3}=2\pi\frac{[\vec{a_1},\vec{a_2}]}{\vec{a_1}[\vec{a_2},\vec{a_3}]}.[/tex]
  8. Oct 25, 2009 #7
    A and G are 3x3 matrices, not 1x3 matrices. The columns of A are the vectors of your lattice:
    [tex]A = \left(
    \left( \! \! \begin{array}{c}\\ \vec{a}_1 \\ \, \end{array} \!\! \right)
    \left( \! \! \begin{array}{c}\\ \vec{a}_2 \\ \, \end{array} \!\! \right)
    \left( \! \! \begin{array}{c}\\ \vec{a}_3 \\ \, \end{array} \!\! \right) \right)[/tex]

    Personally, I think this is easier than manually evaluating three separate cross products. But either way works.
  9. Oct 26, 2009 #8
    Thanks but the discussion is how we get these formulas.

    O.K , as you said ai and gi must be vectors which has three components. The question is what are these components?
    Last edited: Oct 26, 2009
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