Showing reciprocol lattice is perpenducular to a plane

[TFM]

Homework Statement

If a_1 a_2 a_3 are the unit vectors of a real space lattice, then the so-called “reciprocal lattice” is defined by the unit vectors b_1 b_2 b_3 where:

b_1 = \frac{2\pi a_2 \times a_3}{a_1 \cdot a_2 \times a_3}

b_2 = \frac{2\pi a_3 \times a_1}{a_1 \cdot a_2 \times a_3}

b_3 = \frac{2\pi a_1 \times a_2}{a_1 \cdot a_2 \times a_3}


Consider a plane hkl in a crystal lattice.

(a) Prove that the reciprocal lattice vector G = hb_1 + kb_2 + lb_3 is perpendicular to this plane.

(b) Prove that the distance between two adjacent parallel planes of the lattice is

d(hkl) = \frac{2\pi}{|G|}

Homework Equations

a \cdot b = |A||B| cos \theta
a \times b = |A||B| sin \theta

The Attempt at a Solution

So far I have subsituted in the values of a1/2/3 into the given equation to give:

G = h\frac{2\pi a_2 \times a_3}{a_1 \cdot a_2 \times a_3} + k\frac{2\pi a_3 \times a_1}{a_1 \cdot a_2 \times a_3} + l\frac{2\pi a_1 \times a_2}{a_1 \cdot a_2 \times a_3}

but I am not quite sure how to show that this is perpendicular.

I have added the dot and cross product as I feel these may be useful?

 

[physicsworks]

Think of reciprocal lattice as a set of wave vectors \mathbf{G} such that the plane wave e^{i \mathbf{k}\mathbf{r}} with \mathbf{k = G} has a translation symmetry of Bravais lattice.
So, for any vector \mathbf{R} of Bravais lattice (\mathbf{R}= n_1 \mathbf{a_1} + n_2 \mathbf{a_2} + n_3 \mathbf{a_3}) you should have:
e^{i \mathbf{G} \mathbf{r}} = e^{i \mathbf{G (r + R)}}
and
e^{i \mathbf {GR}}=1
This is far more physical definition of the reciprocal lattice (RL). From this, you can easily show that three vectors \math{b_1}, \math{b_2}, \math{b_3} as you defined them form a reciprocal lattice and every vector of RL has a form G=k_1 \mathbf{b_1} + k_2 \mathbf{b_2} + k_3 \mathbf{b_3} where k_1, k_2, k_3 \in Z.
Now you can proof that:
1) for any set of lattice planes separated by the distance d, there are reciprocal vectors perpendicular to them and the shortest vector has a length 2 \pi/d.
2) for any reciprocal lattice vector \mathbf{G} there is a set of planes normal to \mathbf{G} separated by the distance d and the the shortest reciprocal vector parallel to \mathbf{G} has a length 2 \pi/d.
To do this just use the fact that e^{i\mathbf{G r}} is constant value in lattice planes perpendicular to \mathbf{G} and separated by the distance \lambda = 2 \pi/K, so given
\mathbf{K}=\frac{2 \pi \mathbf{s}}{d}
where \mathbf{s} is a unit vector perpendicular to the planes
you have \lambda = d.