What is the connection between reciprocal lattice vectors $\vec G$ and the Miller indices?

Question

We know that a family of crystal planes with Miller indices $(hk\ell)$ is orthogonal to the reciprocal lattice vector $\vec G = h \vec b_1 + k \vec b_2 + \ell\vec b_3$. My question is the converse of this.

For any given set of three integers $m_1, m_2, m_3$, is $\vec G=m_1\vec b_1+m_2\vec b_2+m_3\vec b_3$ always associated with a family of crystal planes in the direct lattice whose Miller indices are $m_1=h, m_2=k$ and $m_3=\ell$? In other words, for every $\vec G$ (or equivalently, every set $m_1,m_2,m_3$) that we can think of, is there a set of parallel planes in the direct lattice to which $\vec G$ is orthogonal?

Answer 1

You can represent a generic position vector of the direct space unit-cell in the basis of the direct lattice, i.e, $\vec r = r_1 ~\vec a_1 + r_2 ~\vec a_2 + r_3 ~\vec a_3$, where $r_i \in \mathbb{R}$ and the Bravais lattice $V=\text{span}\lbrace\vec a_i\rbrace$ whose dual is the reciprocal lattice $V^* = \text{span}\lbrace\vec{b}^j\rbrace$ satisfying $\vec a_i \cdot \vec b^j = 2\pi ~\delta_{ij}$. Any $\vec G\in V^*$ has the representation $\vec G = m_j~ \vec b^j \equiv \vec G_m$, since $V^*$ is a lattice. Now, consider this equality $\vec G_m \cdot \vec r = m_i r_i =0$. It is the equation of a plane passing through the origin, and intercepts the lattice vectors $\vec a_i$ at $\frac{1}{m_i}$ away from the origin. Therefore, by definition, $m_i$ coincides with Miller indices $h_i$. (Notice that any common factor among $m_i$'s is irrelevant to the intercepts, and you can get rid of them at this point.) It is easy to see that $\vec G_m$ is normal that plane. And, since any two parallel planes have the same normal vector, the same $\vec G_m$ applies for them, hence the same Miller indecis $h_i$. You can prove this explicitly by considering a plane shifted from the origin along some vector $\vec d$ and write its equation of loci.