# Find the reciprocal lattice vectors for a triangular lattice with primitive lattice vectors $\vec a_1=(d, 0)$ and $\vec a_2= (d/2, \sqrt{3}d/2)$

Find the reciprocal lattice vectors for a triangular lattice with primitive lattice vectors $$\vec a_1=(d, 0)$$, $$\vec a_2= (d/2, \sqrt{3}d/2)$$

Using the condition that the reciprocal lattice vectors, $$b_j$$ satisfy $$\vec a_i \cdot \vec b_j = 2\pi \delta_{ij}\tag{1}$$

Let $$\vec b_1=\begin{bmatrix}b_x\\ b_y\\ \end{bmatrix}$$

So using $$(1)$$ $$\vec a_1 \cdot \vec b_1=\begin{bmatrix}d\\ 0 \end{bmatrix}\cdot\begin{bmatrix}b_x\\ b_y\\ \end{bmatrix}=2\pi$$

So $$b_x=\frac{2\pi}{d}$$ & $$b_y=0$$, then $$\vec b_1=\begin{bmatrix}\frac{2\pi}{d}\\ 0 \end{bmatrix}$$

The problem is that this reciprocal lattice vector $$\vec b_1$$ is incorrect.

This question can be solved without evaluating the formal expressions for the reciprocal lattice vectors: $$\vec b_1=2\pi\frac{\vec a_2 \times \vec a_3}{\vec a_1 \cdot (\vec a_2 \times \vec a_3)}$$ $$\vec b_2=2\pi\frac{\vec a_3 \times \vec a_1}{\vec a_1 \cdot (\vec a_2 \times \vec a_3)}\tag{2}$$ $$\vec b_3=2\pi\frac{\vec a_1 \times \vec a_2}{\vec a_1 \cdot (\vec a_2 \times \vec a_3)}$$

and setting $$\vec b_3 =\hat k$$

This is the answer given by the author:

The author states that

A choice of the primitive lattice vectors for the reciprocal lattice is $$\vec b_1 =(2\pi/d)\left(1, -1/\sqrt{3}\right)$$ and $$\vec b_2 =\left(0, 4\pi/\sqrt{3}d\right)$$

So does this mean that the primitive lattice vectors are $$\left(1, -1/\sqrt{3}\right)$$ and $$\left(0, 4\pi/\sqrt{3}d\right)$$ or are they $$\left(1, -\sqrt{3}\right)$$ and $$\left(0, \sqrt{3}d/4\pi\right)$$?

You can always use (1). But you've made a mistake in your calculations. $$\vec a_1\vec b_1=2\pi$$ Does not imply $$b_{1y}=0$$. $$\vec a_1\vec b_1=db_{1x}+0b_{1y}=db_{1x}=2\pi$$. From here you can only find out $$b_{1x}=2\pi/d$$. You need the second equation to find $$b_{1y}$$. You get that from $$\vec a_2\vec b_1=0$$, or $$b_{1x}d/2+b_{1y}\sqrt 3d/2=\pi+b_{1y}\sqrt 3/2=0$$, which yields $$b_{1y}=-\frac{2\pi}{d \sqrt 3}$$. You can proceed similarly with $$\vec b_2$$.