# 2D reciprocal lattice vectors

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

Last edited:

Gold Member
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'.

Tiantian
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.