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Showing reciprocol lattice is perpenducular to a plane

  1. Nov 21, 2009 #1

    TFM

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    1. The problem statement, all variables and given/known data

    If [tex] a_1 a_2 a_3[/tex] are the unit vectors of a real space lattice, then the so-called “reciprocal lattice” is defined by the unit vectors [tex]b_1 b_2 b_3[/tex] where:

    [tex] b_1 = \frac{2\pi a_2 \times a_3}{a_1 \cdot a_2 \times a_3} [/tex]

    [tex] b_2 = \frac{2\pi a_3 \times a_1}{a_1 \cdot a_2 \times a_3} [/tex]

    [tex] b_3 = \frac{2\pi a_1 \times a_2}{a_1 \cdot a_2 \times a_3} [/tex]


    Consider a plane hkl in a crystal lattice.

    (a) Prove that the reciprocal lattice vector [tex]G = hb_1 + kb_2 + lb_3[/tex] is perpendicular to this plane.

    (b) Prove that the distance between two adjacent parallel planes of the lattice is

    [tex] d(hkl) = \frac{2\pi}{|G|} [/tex]

    2. Relevant equations

    [tex] a \cdot b = |A||B| cos \theta [/tex]
    [tex] a \times b = |A||B| sin \theta [/tex]

    3. The attempt at a solution

    So far I have subsituted in the values of a1/2/3 into the given equation to give:

    [tex]G = h\frac{2\pi a_2 \times a_3}{a_1 \cdot a_2 \times a_3} + k\frac{2\pi a_3 \times a_1}{a_1 \cdot a_2 \times a_3} + l\frac{2\pi a_1 \times a_2}{a_1 \cdot a_2 \times a_3}[/tex]

    but I am not quite sure how to show that this is perpendicular.

    I have added the dot and cross product as I feel these may be useful?
     
  2. jcsd
  3. Nov 22, 2009 #2

    physicsworks

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    Gold Member

    Think of reciprocal lattice as a set of wave vectors [tex]\mathbf{G}[/tex] such that the plane wave [tex]e^{i \mathbf{k}\mathbf{r}}[/tex] with [tex]\mathbf{k = G}[/tex] has a translation symmetry of Bravais lattice.
    So, for any vector [tex]\mathbf{R}[/tex] of Bravais lattice ([tex]\mathbf{R}= n_1 \mathbf{a_1} + n_2 \mathbf{a_2} + n_3 \mathbf{a_3}[/tex]) you should have:
    [tex]e^{i \mathbf{G} \mathbf{r}} = e^{i \mathbf{G (r + R)}}[/tex]
    and
    [tex]e^{i \mathbf {GR}}=1[/tex]
    This is far more physical definition of the reciprocal lattice (RL). From this, you can easily show that three vectors [tex]\math{b_1}, \math{b_2}, \math{b_3}[/tex] as you defined them form a reciprocal lattice and every vector of RL has a form [tex]G=k_1 \mathbf{b_1} + k_2 \mathbf{b_2} + k_3 \mathbf{b_3}[/tex] where [tex]k_1, k_2, k_3 \in Z[/tex].
    Now you can proof that:
    1) for any set of lattice planes separated by the distance d, there are reciprocal vectors perpendicular to them and the shortest vector has a length [tex]2 \pi/d[/tex].
    2) for any reciprocal lattice vector [tex]\mathbf{G}[/tex] there is a set of planes normal to [tex]\mathbf{G}[/tex] separated by the distance d and the the shortest reciprocal vector parallel to [tex]\mathbf{G}[/tex] has a length [tex]2 \pi/d[/tex].
    To do this just use the fact that [tex]e^{i\mathbf{G r}}[/tex] is constant value in lattice planes perpendicular to [tex]\mathbf{G}[/tex] and separated by the distance [tex]\lambda = 2 \pi/K[/tex], so given
    [tex]\mathbf{K}=\frac{2 \pi \mathbf{s}}{d}[/tex]
    where [tex]\mathbf{s}[/tex] is a unit vector perpendicular to the planes
    you have [tex]\lambda = d[/tex].
     
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