How Can I Calculate Reciprocal Lattice Vectors for a 2D Lattice?

[jonesj314]

Homework Statement

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

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

Homework Equations

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

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

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 \vec{a'_3} as simply being the z unit vector, then i find the numerator to be \vec{b'_1} = 2π (0.5 \vec{a_1} - 0.5 \vec{a_2}) is this correct for the numerator?? (it's orthogonal to \vec{a'_2} as I was expecting)

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

\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} )

and this dot product equals zero.

What am I doing wrong? Any help appreciated

 

[UltrafastPED]

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


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'.

 

[jonesj314]

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.

 

[UltrafastPED]

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.

 

[jonesj314]

oh of course :) thank you! All makes sense now