# Miller indices for hexagonal crystal systems

To find the draw the direction for a given Miller index say, [1234] we first convert this miller index consisting of 4 indices into one containing 3 indices. To do so, we have a set of formulae prescribed in almost every book. Sadly I haven't been able to come across a single book the gives the derivation of those formulae!

I thought that I could use vector-component method to get the results but that gives totally weird formulae not even close to the ones I see in my textbooks. Here is an example, just to be clearer. (and have a look at the attached image)

So, can anyone suggest me a textbook, a link or anything that can help me understand the derivation? I'm not finding the enthusiasm for rote-memorising the formulae if i don't know where they come from...

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From the diagram it is obvious that the Miller-Bravais index has redundant information, as the indices point along three directions that are 120° apart. That means that you can do a simple geometrical derivation using this diagram as your guide:

The third equation follows immediately from the vector addition of $\vec u$ and $\vec v$ - you can see that $\vec t$ points in the opposite direction.

It is equally obvious that, in the way I drew the picture, $\vec u = \vec{v'} + \frac12 \vec{u'}$ and $\vec v = \frac12 \vec{u'} - \vec v'$

Simple manipulation of these equations gets you to the expressions you quote.

• according to the formula i attached, it says u = (2u' - v')/3 though, that's where i'm stuck – Sakazuki Akainu Aug 9 '17 at 14:01
• i can't get that forumla, i in fact reached the same point you did – Sakazuki Akainu Aug 9 '17 at 14:01

In a hexagonal crystal system, just like in any other three dimensional system, every vector can be represented in a basis consisting of 3 linearly independent vectors. Thus, 3 such vectors would be sufficient to describe any direction we want in a crystal.

The image below shows a hexagonal unit cell with 4 axes (represented by vectors): $$\vec{a_1}$$, $$\vec{a_2}$$, $$\vec{a_3}$$ and $$\vec{z}$$, used to index directions in a crystal. Vectors $$\vec{a_1}$$, $$\vec{a_2}$$ and $$\vec{a_3}$$ are $$\mathit{not}$$ linearly independent. Conventionally, we choose $$\vec{a_3}$$ to be the "extra" vector, and $$\vec{a_1}$$, $$\vec{a_2}$$, and $$\vec{z}$$ the "main" vectors, common to both 4 and 3 index systems.

The vector $$\vec{a_3}$$ is defined as $$\vec{a_3} = -\left(\vec{a_1} + \vec{a_2}\right)$$, which gives us some intuition for requiring $$t = -(u+v)$$ to be obeyed, since u,v and t are components along $$\vec{a_1}, \vec{a_2}$$ and $$\vec{a_3}$$, respectively. We could have used a different relation between $$u$$, $$v$$ and $$t$$, but this one is the most straightforward (we need an extra equation for the vector components to be unique since $$\vec{a_1}$$, $$\vec{a_2}$$ and $$\vec{a_3}$$ are not linearly independent).

The difference between the 3 index (denoted by [u'v'w']) and 4 index representations (denoted by [uvtw]) is that when using 3 indices, we ignore the vector $$\vec{a_3}$$, and only use $$\vec{a_1}$$, $$\vec{a_2}$$ and $$\vec{z}$$ (the $$\mathit{same}$$ $$\vec{a_1}$$, $$\vec{a_2}$$ and $$\vec{z}$$ in $$\mathit{both}$$ representations - the only difference is having or not having $$\vec{a_3}$$). Notice that $$\vec{a_1}$$ and $$\vec{a_2}$$ are $$\mathbf{not}$$ orthogonal. Instead, they make a $$120^\circ$$ angle, so they aren't the usual $$\hat{x}$$ and $$\hat{y}$$ vectors, but rather ones that respect the symmetry of the crsystal.

Now, any vector $$\vec{v}$$ can be written in both representations (again, note that the only difference between the two is $$\vec{a_3}$$ being present in 4-index notation and missing in 3-index notation):

$$\vec{v} = u'\vec{a_1} + v'\vec{a_2} + w'\vec{z} \tag{1} \label{1}$$ and $$\vec{v} = u\vec{a_1} + v\vec{a_2} + t\vec{a_3} + w\vec{z} \tag{2} \label{2}$$

As $$\vec{a_3} = -\left(\vec{a_1} + \vec{a_2}\right)$$, inserting this into equation $$\eqref{2}$$ we get:

$$\vec{v} = (u-t)\vec{a_1} + (v-t)\vec{a_2} + w\vec{z}$$

By substituting $$t = -(u+v)$$ we further get

$$\vec{v} = (2u+v)\vec{a_1} + (2v+u)\vec{a_2} + w\vec{z} \tag{3} \label{3}$$

Equating the components along the same vectors from $$\eqref{3}$$ and $$\eqref{1}$$ we get

\begin{aligned} u' &= 2u+v\\ v' &= 2v+u\\ w' &= w \end{aligned}

Assuming $$u'$$, $$v'$$ and $$w'$$ are known, we can solve the system for $$u$$, $$v$$ and $$w$$ and obtain

\begin{aligned} u &= \frac{1}{3} \left(2u'-v'\right)\\ v &= \frac{1}{3} \left(2v'-u'\right)\\ t &= -\left(u+v\right)\\ w &= w' \end{aligned} as desired.