# Miller Indices: How to deal with some weird(-ish) cases?

I've recently been doing a spot of self-learning (Crystallography), and some of the examples provided in the Wikipedia article for Miller Indices have stumped me.

I can't for the world figure out how they were obtained!

Quick introduction to Miller Indices (the basic stuff):

Including this here, because it was what I was referring to when learning about this stuff

## Miller Indices (hkl)

The orientation of a surface or a crystal plane may be defined by considering how the plane (or indeed any parallel plane) intersects the main crystallographic axes of the solid. The application of a set of rules leads to the assignment of the Miller Indices , (hkl) ; a set of numbers which quantify the intercepts and thus may be used to uniquely identify the plane or surface.

The following treatment of the procedure used to assign the Miller Indices is a simplified one (it may be best if you simply regard it as a "recipe") and only a cubic crystal system (one having a cubic unit cell with dimensions $a$ x $a$ x $a$ ) will be considered.

The procedure is most easily illustrated using an example so we will first consider the following surface/plane:

• Step 1 : Identify the intercepts on the x- , y- and z- axes.

In this case the intercept on the x-axis is at x = a ( at the point (a,0,0) ), but the surface is parallel to the y- and z-axes - strictly therefore there is no intercept on these two axes but we shall consider the intercept to be at infinity ( ∞ ) for the special case where the plane is parallel to an axis. The intercepts on the x- , y- and z-axes are thus: $a$ , $∞$ , $∞$ (in that order).

• Step 2 : Specify the intercepts in fractional co-ordinates

Co-ordinates are converted to fractional co-ordinates by dividing by the respective cell-dimension - for example, a point (x,y,z) in a unit cell of dimensions $a$ x $b$ x $c$ has fractional co-ordinates of ( x/a , y/b , z/c ). In the case of a cubic unit cell each co-ordinate will simply be divided by the cubic cell constant , $a$.

This gives: $a/a$ , $∞/a$, $∞/a$ ; i.e- $1$ , $∞$ , $∞$.

• Step 3 : Take the reciprocals of the fractional intercepts

This final manipulation generates the Miller Indices which (by convention) should then be specified without being separated by any commas or other symbols. The Miller Indices are also enclosed within standard brackets (…) when one is specifying a unique surface such as that being considered here.

The reciprocals of 1 and ∞ are 1 and 0 respectively, thus yielding the Miller Indices ($100$)

So the surface/plane illustrated is the ($100$) plane of the cubic crystal.

Further on, it is mentioned:

If any of the intercepts are at negative values on the axes then the negative sign will carry through into the Miller indices; in such cases the negative sign is actually denoted by overstriking (i.e- adding a little bar/dash over) the relevant number.

Now I was going to through the Wiki article on Miller Indices (linked earlier on in this post), and the following examples were provided:

The corresponding Miller Indices have been provided in the boxes right under each entry. Also, the little "T"s are actually "1"s with a bar on top... kinda small, so it's easy to see it wrong.

Now the first six entries are pretty straightforward (I count left-right), but it's the others that are confusing.

Take for example, the seventh entry:

Okay, the third axis ($a_3$) lies along orange-colored plane, so the plane has no intercept with the ($a_3$) axis (i.e- intercept here is taken to be $∞$, as the orange plane and the axis in question are parallel). Which would naturally result in the Miller Index $0$ (third digit) which is the reciprocal of $∞$.

Alright, no issues there.

But what's the deal with the intercepts on the the first two axes ( $a_1$ and $a_2$ )?

The "intercept at infinity" logic doesn't work here (since the orange plane isn't parallel to $a_1$ or $a_2$).

According to the (box under the) entry, the orange plane makes an intercept of $1$ (second digit) with the $a_2$ axis. Moreover, it also (apparently) makes an intercept of $-1$ (first digit) with the $a_3$ axis.

This is incredible...I am totally lost.

The eighth and ninth entries, too, are made in the same vein (well, obviously... because the seventh, eighth and ninth entries are essentially the same thing)

My question:

How exactly, are the Miller Indices for the seventh entry (the one with the orange plane), obtained?

I've just asked for the rationale behind the Miller Indices of the seventh entry because I can easily work out the Indices for the eighth and ninth entries after that.

• I'm still relatively new to Miller indices and I can't say I have an answer for a general procedure, but the Miller indices define an infinite family of planes. So while the particular plane shown for the $7^\text{th}$ entry doesn't have those intercepts, you can picture shifting that plane up to a location where it doesn't intercept $a_3$, intercepts $a_1$ at $-1$, and intercepts $a_2$ at $1$. The same sort of shifting will give you the correct intercepts for 8 and 9.
– Tyberius
Commented Sep 20, 2017 at 17:52
• a similar question from physics.SE: physics.stackexchange.com/questions/62955/… Commented Sep 20, 2017 at 17:55
• This section explains which plane from a set of planes you consider to determine the indices. It has an image that shows planes beyond the unit cell, which resolves the conceptual hurdle of the images shown by the OP.
– Karsten
Commented Aug 12, 2023 at 8:08

## 3 Answers

IFF I remember correctly...

Sometimes you have to "move" the plane, in order to see where the axis crosses.

Take example 7, if you look at the example plane, the one actually shown, it is (0 ∞ ∞) not very useful. The ones making the examples have "moved the planes" (not really, but it is a useful fiction) in order to get the plane into the unit cell so they can show it. If they hadn't moved it, it wouldn't be visible.

The plane (-110) crosses $$a_2$$ axis at $$a_2$$ = $$1$$ and you see, using your spatial imagination, that it then needs to cross $$a_1$$ at $$-1$$ as well. It will never cross the $$a_3$$ axis, so $$0$$. The 8th and 9th are much harder to visualize but it is basically the same thing over again. The planes won't fit into the unit cell unless you "move" them.

Disclaimer: the (hkl) represents the set of all planes of the same err... "orientation", so it is technically not moving anything at all, it is merely choosing a representation of that said plane. If two planes in a certain crystal are equivalent, they can be represented with curly brackets. The sets (100),(010) and (001) are equivalent in a cubic crystal and can be represented collectively as {100}. Source: Basic Solid State Chemistry, A.R West 2nd ed. (1999)

[OP] How to deal with some weird(-ish) cases?

The cases are not weird(-ish), it is the images that are misleading in some cases, and inconsistent overall. Here is a better depiction that shows more than a unit cell:

The description for the image says to take the plane (line in this case because they show a 2D crystal) that is adjacent to that crossing the origin. The origin is marked with "O" and the line you have to pick is marked with "x" and "y" in the images above. The isolated unit cell is shown in red on the right.

[OP] Take for example, the seventh entry [...] How exactly, are the Miller Indices for the seventh entry (the one with the orange plane), obtained?

Here is another picture of the seventh entry, showing how planes graze the unit cell at two edges.

The difference to the original image is subtle, but the hand-drawn orange lines show where the two planes would just touch the unit cell. You would have to extend these planes beyond the unit cell to figure out the Miller indices. One crosses $$a_1$$ at 1 and $$a_2$$ at -1, the other at -1 and 1, respectively. Neither intersect $$a_3$$ (which is in plane). The corresponding Miller indices are the equivalent (1, -1, 0) and (-1, 1, 0). Both describe the same set of planes (i.e. same orientation and same distance between adjacent planes).

Trevor's answer has a helpful practical hint: instead of drawing more planes, you can also move the origin by a unit cell vector translation. What is missing from the answer is that you should consider a plane that is adjacent to that crossing the origin. If you consider one further away, you get a Miller index smaller than one, which should not happen, or turn a higher order set of planes such as (2 8 4) into a lower order set such as (1 4 2).

## A different approach

Once you know about the reciprocal lattice, you can also do the following. Take a vector perpendicular to the planes, and give it the length of the reciprocal of the lattice spacing. This vector (corresponding to a reciprocal lattice point) will have the Miller indices as its coordinates in reciprocal space. With this definition, it is clear that the distance between planes is important, and the it does not matter which one of the parallel planes you consider for the direction.

With the traditional recipe, considering the distance of the planes comes by choosing the plane that is immediately adjacent to the one going through the origin.

For the seventh entry/diagram: The way I understand it, is you can't have the plane cutting the origin at zero. Therefore you must use a different origin. Assume the origin is now on the front RHS corner. Now the plane cuts at a1=-1, a2=1, a3=infinity. Then take the reciprocal and you get (-1 1 0)