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Homework Help: 2D reciprocal lattice vectors

  1. Oct 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Si(001) has the following lattice vectors in a (2x1) reconstruction [tex] \vec{a'_1} = \vec{a_1} + \vec{a_2} [/tex] [tex] \vec{a'_2} = -0.5 \vec{a_1} + 0.5 \vec{a_2} [/tex]

    Calculate the reciprocal lattice vectors of the reconstructed unit cell, [itex] \vec{b'_1} [/itex] and [itex] \vec{b'_2} [/itex] in terms of [itex] \vec{a_1} [/itex] and [itex] \vec{a_2} [/itex].

    2. Relevant equations

    I have been using the formulae for finding reciprocal lattice vectors in 3D, i.e

    [tex] \vec{b'_1} = 2 π \frac{(\vec{a'_2} ×\vec{a'_3})}{\vec{a'_1}. (\vec{a'_2} × \vec{a'_3})}[/tex]

    and the usual permutations for the other 2 reciprocal vectors

    3. The attempt at a solution

    Since I'm trying to do this for a 2D lattice I'm running into problems. If I treat [itex] \vec{a'_3} [/itex] as simply being the z unit vector, then i find the numerator to be [itex] \vec{b'_1} = 2π (0.5 \vec{a_1} - 0.5 \vec{a_2}) [/itex] is this correct for the numerator?? (it's orthogonal to [itex] \vec{a'_2} [/itex] as I was expecting)

    however, using this method I find the denominator to be zero since,

    [tex] \vec{a'_1}. (\vec{a'_2} × \vec{a'_3}) = (\vec{a_1} + \vec{a_2}) . (0.5\vec{a_1} - 0.5\vec{a_2} ) [/tex]

    and this dot product equals zero.

    What am I doing wrong? Any help appreciated
    Last edited: Oct 21, 2013
  2. jcsd
  3. Oct 21, 2013 #2


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    I get [itex] \vec{b'_1} = 2π (0.5 \vec{a_1} + 0.5 \vec{a_2}) [/itex]

    And permuting the terms in the triple product try (a1' x a2')°a3'; but the cross product is parallel to a3' (which is OK, 'cause it is a dot product), and a3' is a unit vector so the volume is just |a1' x a2'|= area of the parallelogram with sides a1', a2'.
  4. Oct 21, 2013 #3
    Hi, thanks for the reply. You're right, I evaluated the numerator incorrectly.

    I still don't understand the significance of the denominator. Why do I get zero? Would permuting the triple product to the form you suggest give a different answer?
    Quite confused as to what this should be.
  5. Oct 21, 2013 #4


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    It gives the same answer: I just shifted the form to make the result obvious, and simple to compute. Your product a2' x a3' is incorrect ... same error as with b1.

    The magnitude is 1.
  6. Oct 21, 2013 #5
    oh of course :) thank you! All makes sense now
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