Consider a rectangular lattice in two dimensions with primitive lattice vectors $(a,0)$ and $(0,2a)$.

Which of the following are reciprocal lattice vectors for this lattice?

(a) $\quad\dfrac{\pi}{a}\left(1,\frac12 \right)$

(b) $\quad\dfrac{\pi}{a}\left(1,2 \right)$

(c) $\quad\dfrac{\pi}{a}\left(2,-1 \right)$

(d) $\quad\dfrac{\pi}{a}\left(0,2 \right)$

(e) $\quad\dfrac{\pi}{a}\left(\frac12,-2 \right)$

My attempt is:

Let the reciprocal lattice vectors $\vec b_1$ & $\vec b_2$ be $$\vec b_1=\begin{bmatrix}x_1\\ y_1\\ \end{bmatrix}\qquad \text{and}\qquad\vec b_2=\begin{bmatrix}x_2\\ y_2\\ \end{bmatrix}$$ respectively.

and denote the primitive lattice vectors $\vec a_1$ & $\vec a_2$ as $$\vec a_1=\begin{bmatrix}a\\ 0\\ \end{bmatrix}\qquad \text{and}\qquad\vec a_2=\begin{bmatrix}0\\ 2a\\ \end{bmatrix}$$ respectively

Now using the condition, $\vec a_i \cdot \vec b_j=2\pi\delta_{ij}\tag{1}$

$\vec a_1 \cdot \vec b_1=\begin{bmatrix}a\\ 0\\ \end{bmatrix}\cdot\begin{bmatrix}x_1\\ y_1\\ \end{bmatrix}= 2\pi$

So $x_1 =\frac{2\pi}{a}$

$\vec a_2 \cdot \vec b_1=\begin{bmatrix}0\\ 2a\\ \end{bmatrix}\cdot\begin{bmatrix}\frac{2\pi}{a}\\ y_1\\ \end{bmatrix}= 0\implies$ $y_1=0$ and hence $\vec b_1 = \begin{bmatrix}\frac{2\pi}{a}\\ 0\\ \end{bmatrix}=\color{red}{\frac{\pi}{a}\left(2,0\right)}$

Now to find $\vec b_2$,

$\vec a_2 \cdot \vec b_2=\begin{bmatrix}0\\ 2a\\ \end{bmatrix}\cdot\begin{bmatrix}x_2\\ y_2\\ \end{bmatrix}= 2\pi\implies y_2 = \frac{\pi}{a}$

$\vec a_1 \cdot \vec b_2=\begin{bmatrix}a\\ 0\\ \end{bmatrix}\cdot\begin{bmatrix}x_2\\ \frac{\pi}{a}\\ \end{bmatrix}= 0\implies x_2 = 0$ and hence $\vec b_2= \begin{bmatrix}0\\ \frac{\pi}{a}\\ \end{bmatrix}=\color{red}{\frac{\pi}{a}\left(0,1\right)}$

The two vectors calculated (in red) do not correspond to any of the answers above, from which I can tell you that 2 of them are correct. I thought that I was applying the condition $(1)$ correctly.

Could someone please explain to me what I'm doing wrong?


I have just realised that the vectors calculated $\vec b_1=\color{red}{\frac{\pi}{a}\left(2,0\right)}$ & $\vec b_2=\color{red}{\frac{\pi}{a}\left(0,1\right)}$ are still correct answers but they are also the primitive reciprocal lattice vectors.

Any integer multiple of the primitive reciprocal lattice vectors is itself another reciprocal lattice vector:

$2 \vec b_2=\frac{\pi}{a}\left(0,2\right)$ and $\vec b_1 - \vec b_2 = \frac{\pi}{a}\left(2,-1\right)$

So the correct answers are $(\rm{c})$ and $(\rm{d})$

  • $\begingroup$ Not any linear combination. The coefficients must be integers. $\endgroup$ – Andrei May 5 at 14:55
  • 1
    $\begingroup$ Yes. Everything else is fine. $\endgroup$ – Andrei May 6 at 14:20

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