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 $\left(0, \sqrt{3}d/4\pi\right)$?
Best Answer
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 $\vec b_2$.