- #1

jonesj314

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## Homework Statement

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].

## Homework 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

## 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

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