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

So we have $$\vec a_1 = a\hat i$$ and $$\vec a_2 = 2a\hat j$$ and $$\vec a_3=\hat k$$ since I am going to be using the 3-dimensional formulue for the reciprocal lattice vectors:

$$\vec b_1 =2\pi\frac{\vec a_2 \times \vec a_3}{\vec a_1\cdot(\vec a_2 \times \vec a_3)},\quad\vec b_2 =2\pi\frac{\vec a_3 \times \vec a_1}{\vec a_1\cdot(\vec a_2 \times \vec a_3)},\quad\vec b_3 =2\pi\frac{\vec a_1 \times \vec a_2}{\vec a_1\cdot(\vec a_2 \times \vec a_3)}$$

For a 2D lattice, I was told that

If you want to use the 3D definition with the cross products to deduce the $\vec b$ vectors, choose $\vec a_3 = \hat k$.

This is something my lecturer told me and is in my lecture notes.

So I start by computing $$\vec a_2 \times \vec a_3= \begin{vmatrix} \hat i & \hat j & \hat k \\ 0 & 2a & 0 \\ 0 & 0 & 1 \\ \end{vmatrix}=2a\hat i,\,\,\,\,\,\vec a_3 \times \vec a_1= \begin{vmatrix} \hat i & \hat j & \hat k \\ 0 & 0 & 1 \\ a & 0 & 0 \\ \end{vmatrix}=a\hat j,\,\,\,\,\,\vec a_1 \times \vec a_2= \begin{vmatrix} \hat i & \hat j & \hat k \\ a & 0 & 0 \\ 0 & 2a & 0 \\ \end{vmatrix}=2a^2\hat k $$

Substituting these results into the formulue for the reciprocal lattice vectors gives $$\vec b_1 =2\pi\frac{2a\hat i}{a\hat i\cdot\Big(2a\hat i\Big)},\quad\vec b_2 =2\pi\frac{a\hat j}{a\hat i\cdot\Big(2a\hat i\Big)},\quad\vec b_3 =2\pi\frac{2a^2\hat k}{a\hat i\cdot\Big(2a\hat i\Big)}$$ Since $\hat i^2=$ unity, and all the denominators are identical; on simplification this gives $$\vec b_1 =2\pi\frac{2a\hat i}{2a^2},\quad\vec b_2 =2\pi\frac{a\hat j}{2a^2},\quad\vec b_3 =2\pi\frac{2a^2\hat k}{2a^2}$$ and so

$$\vec b_1 =\frac{2\pi}{a}\hat i,\quad\vec b_2 =\frac{\pi}{a}\hat j,\quad\vec b_3 =2\pi\,\hat k$$

I know $\vec b_3 =2\pi\,\hat k$ cannot be valid as the choices in the question do not contain a $z$ component. But that's the first problem, the other thing is that the only combination of $\vec b_1$ and $\vec b_2$ I can make is $$\dfrac{\pi}{a}\left(2,1 \right)$$ which does not correspond to any of the choices in the question.

The two correct answers are (c) and (d).

Could someone please explain to me how to find reciprocal lattice vectors?


You are almost there. Any reciprocal lattice vector can be written as $\vec{v}=m\vec{b}_1+n\vec{b}_2$, where $m$ and $n$ are integers. By plugging in what you obtained for $\vec{b}_1$ and $\vec{b}_2$, you get $\vec{v}=\frac{\pi}{a}(2n,m)$. So the first element is an even integer (answers a,b,e are wrong) and the second element is an integer (answer a is wrong).

Note that the easier way to compute your reciprocal lattice vectors is $\vec{a}_i\cdot\vec{b}_j=2\pi\delta_{ij}$

  • 1
    $\begingroup$ Thank you for your answer, sorry it took a while for me to respond. You mention that "the easier way to compute your reciprocal lattice vectors is $\vec{a}_i\cdot\vec{b}_j=2\pi\delta_{ij}$". So if I chose $\vec a_1 = a\hat i$ and wrote $a\hat i \cdot \vec {b}_1=2\pi\delta_{ij}$. With an unknown $\vec {b}_1$ how would I proceed? Could you please show me in your answer how to compute reciprocal lattice vectors in this manner? Many thanks. $\endgroup$ – BLAZE Jul 7 '17 at 14:18
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    $\begingroup$ $\vec{b}_1=(x,y)$. Then $\vec{a}_1\cdot \vec{b}_1=ax$. This means $x=2\pi/a$. If $\vec{a}_2=(0,2a)$ then $\vec{a}_2\cdot \vec{b}_1=2ay=0$ so $y=0$. You do the same vor $\vec{b}_2$ $\endgroup$ – Andrei Jul 7 '17 at 15:48

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