1
$\begingroup$

One of the properties of the reciprocal lattice vector is shown in the following

Theorem. For any family of lattice planes separated by a distance $d$ there are reciprocal lattice vectors $\mathbf{G}_m$ perpendicular to the planes, the shortest of which $\mathbf{g}_{hkl}$have a length $$ |\mathbf{g}_{hkl}|=\frac {2\pi}{d} $$

If $\mathbf {G} _{m}=m_{1}\mathbf {b} _{1}+m_{2}\mathbf {b} _{2}+m_{3}\mathbf {b} _{3}$ is a reciprocal lattice vector pointing to the fixed direction identified by $m=(m_1, m_2, m_3) \in \mathbb{Z}^3$ and $\mathbf {R} _{n}=n_{1}\mathbf {a} _{1}+n_{2}\mathbf {a} _{2}+n_{3}\mathbf {a}_{3}$ is the generic direct lattice vector, the equation

$$ \mathbf{R}_n \cdot \mathbf{G}_m = 2\pi Z, \qquad Z\in\mathbb{Z} $$

represents a parallel sheaf of planes $\pi^{(m)}_Z$ all perpendicular to the line set by $\mathbf{G}_m$.

I sketched a method

enter image description here

that seems to derive the interplanar distance, namely $\displaystyle d=(\mathbf{R}_{n'}-\mathbf{R}_n)\cdot\frac{\mathbf{G}_m}{|\mathbf{G}_m|}$, where $\mathbf{R}_{n'}$ and $\mathbf{R}_n$ are two arbitrary direct lattice vectors satisfying $\mathbf{R}_{n'} \cdot \mathbf{G}_m = 2\pi (Z+1)$ and $\mathbf{R}_{n} \cdot \mathbf{G}_m = 2\pi Z$. This can't be right, since the theorem requires the shortest vector in the direction of $\mathbf{G}_m$, and not $\mathbf{G}_m$ itself. Where is the mistake?

$\endgroup$

2 Answers 2

2
$\begingroup$

Each reciprocal lattice vector $\mathbf{G}$ defines a set of equally-spaced, parallel planes via $$ \mathbf{G}\cdot\mathbf{R}=2\pi m\,, $$ one for every integer $m$. Nearest-neighbor planes have $|\Delta m| =1$. Your calculation indeed shows that the distance between consecutive planes is exactly $2\pi/G$. For completeness, I'll sketch the proof here.

Let $\mathbf{r}_1$ and $\mathbf{r}_2$ be points in consecutive planes, i.e. they satisfy $$ \mathbf{G}\cdot\mathbf{r}_1=2\pi m\,,~~~~~~~ \mathbf{G}\cdot\mathbf{r}_2=2\pi (m+1)\,. $$ Then, $\mathbf{G}\cdot(\mathbf{r}_2 - \mathbf{r}_1)=2\pi$. Dividing by G, we get $\hat{\mathbf{G}}\cdot(\mathbf{r}_2 - \mathbf{r}_1)=2\pi/G$, where $\hat{\mathbf{G}}$ is a unit vector in the direction of ${\mathbf{G}}$. We can clearly interpret the left-hand side as the component of the vector $\mathbf{r}_2 - \mathbf{r}_1$ that is perpendicular to the planes, and hence its length must be the distance between the planes.


The issue is that this set of planes is not a family of lattice planes except in the special case where $\mathbf{G}$ is the smallest reciprocal lattice vector in its direction. The problem is that if $\mathbf{G}$ is not the shortest, then there are too many planes; that is, some of the planes don't have any lattice points at all. (In fact, one of these planes is either a lattice plane$-$it has a two-dimensional lattice of points of the 3D lattice$-$or is it has exactly zero lattice points.)

We only get a family of lattice planes$-$that is, a set of equally-spaced, parallel planes, each of which contains an infinite number of points of the lattice$-$if $\mathbf{G}$ is the shortest reciprocal lattice vector in the direction $\hat{\mathbf{G}}$.


This is a claim. To see this, we proceed by way of contradiction. First form the set of planes corresponding to $\mathbf{G}'=2\mathbf{G}$, where $\mathbf{G}$ is a reciprocal lattice vector. Then, $\mathbf{G}'$ is not the shortest reciprocal lattice vector along this direction. The $m=0$ plane is clearly a lattice plane (because $\mathbf{R}=0$ is in this plane), so consider the plane $m=1$. If there is lattice point $\mathbf{R}$ in this plane, then $$ \mathbf{G}'\cdot\mathbf{R} = 2\pi\,, $$ which means that $$ \mathbf{G}\cdot\mathbf{R} = \pi\,. $$ At the same time, since $\mathbf{G}$ is a reciprocal lattice vector, it must satisfy (by definition of the reciprocal lattice!) $$ e^{i\mathbf{G}\cdot\mathbf{R}}=1 $$ which further implies that $$ \mathbf{G}\cdot\mathbf{R} = 2\pi m $$ for some integer $m$. This contradicts the statement above that $\mathbf{G}\cdot\mathbf{R} = \pi$. Hence, our assumption that there was a lattice point $\mathbf{R}$ in this plane must be false. Hence that plane is empty! There are too many planes if we don't use the shortest reciprocal lattice vector along that direction!


Finally, we should probably show that the set of planes formed using a shortest reciprocal lattice vector really is a family of lattice planes, but the proof is essentially dual to the one that we just did. We pretty much just have to reverse the role of the lattice vectors and reciprocal lattice vectors.

$\endgroup$
2
  • $\begingroup$ can you explain why nearest-neighbor planes have |Δm|=1? $\endgroup$
    – Abe
    Feb 20, 2023 at 6:04
  • $\begingroup$ @Abe That's hidden in the calculations above (or in the OP): Given two direct lattice vectors $\vec{R}_1$ and $\vec{R}_2$ satisfying $\vec{G}\cdot\vec{R}_1=2\pi m$ and $\vec{G}\cdot\vec{R}_2=2\pi n$, the distance between the planes defined by $m$ and $n$ is $2\pi(m-n)/G$, which is minized when $m=n\pm1$. $\endgroup$
    – march
    Feb 21, 2023 at 16:28
1
$\begingroup$

You can define a “maximal” family of regularly spaced family of parallel lattice planes by the equation: $$ R_n\cdot G_m \in 2\pi\mathbb Z $$ for a certain $G_m$ in the dual lattice.

The problem with your derivation is that you are implicitly asserting that the values of $R_n\cdot G_m$ is exactly $2\pi\mathbb Z$. This is not true in general, you only have inclusion, and the actual set of possible values is $2\pi gcd(m)\mathbb Z$. Without knowing, you assumed the components of $m$ to be coprime.

Using translation invariance, lets look at the closest plane to the one defined by $R_n\cdot G_m=0$. You are therefore looking for the smallest positive integer $k$ such that there exists $n\in \mathbb Z^3$ satisfying $n\cdot m=k$. A classic result of arithmetic says that $m_1\mathbb Z+m_2\mathbb Z+m_3\mathbb Z=gcd(m)\mathbb Z$ (induction, the case of two terms is proven using euclidean division), hence the result.

You thus obtain the distance between two consecutive planes to be: $$ d= \frac{2\pi gcd(m)}{|G_m|} $$

which is consistent with your theorem and with your proof when $gcd(m)=1$.

Hope this helps and tell me if something’s not clear.

$\endgroup$
2
  • $\begingroup$ I’m sorry, but I don’t see why $m\cdot\mathbb{Z} = \operatorname{gcd}\{m\} \mathbb{Z}$. Suppose $m=(1,2,3)$, then $6\mathbb{Z}\overset{?}{=}\mathbb{Z}$? $\endgroup$
    – ric.san
    Jun 8, 2022 at 6:51
  • $\begingroup$ In your example, you’d get $\mathbb Z+2\mathbb Z+3\mathbb Z=\mathbb Z$ (in which case you don’t even need the result due to the trivial inclusion), the result is also known as Bézout’s identity. $\endgroup$
    – LPZ
    Jun 8, 2022 at 7:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.