# What is the concept behind Miller Indices?

Have a look at the attached image below.

If I look at the image intuitively, my common sense tells me that the Miller Indices for the vector should be [1,1,-2,1] since we're travelling one unit along a1 vector, one unit along a2 vector, one unit along z vector and due to vector addition, negative one unit along a3 direction.

However, using the formulae for conversion of the Hexagonal crystal to a Cubic crystal and then back to the hexagonal crystal, the Miller Indices turn out to be [1,1,-2,3] !!

For those interested in the details, the video can be found here.

So what exactly is the basis behind the concept being used here? Why does it seem so non-intuitive to me?

• With the transformations you mention how did you manage to change the last index from [ , , ,1] to [ , , ,3]? Sep 18, 2017 at 12:09
• see, the miller indices are converted to smallest integer values (no hard and fast rule, just preferred for convenience) the first 3 terms have a denominator of 3 while z is just 1. So we multiply all coordinates by 3 to get the smallest integer values :-) Sep 18, 2017 at 13:09

I think I understand what you have done and the question that you have asked about the result that you have obtained.

You started with [1 1 -2 1] in 4-index notation and then transformed it into [1 1 1] in 3-index notation.

Then you started with [1 1 1] and attempted to convert it back into 4-indev notation.

From the first two indices of the 3-index notation you obtained $+\frac 13$, $+\frac 13$ and $-\frac 23$ as the first three indices in the 4-index notation.

Now how to convert from the third index of the 3-index notation to the 4-index notation?
What you did was to say that the last index in the 3-index notation which is a one transforms to become a one in the 4-index notation?
You multiplied by 3 to get only integers which resulted in a 4-index result of [1 1 -2 3] which has mystified you.

The error that you have made is to say that there is no change in the last index.
You transformation of [1 1 .] yielded [$\frac 13$ $\frac 13$ -$\frac 2 3$ .]

Now [$\frac 13$ $\frac 13$ -$\frac 2 3$ .] is the projection of your vector onto the $z=0$ plane and is only $\frac 13^{\rm rd}$ of the projection needed to reach the edge of the unit cell.

So the height of the vector above the $z=0$ plane is $\frac 13$ of the height of the unit cell.

The correct transformation of [1 1 1] is [$\frac 13$ $\frac 13$ -$\frac 2 3$ $\frac 1 3$] which leads to the result that you were expecting.

• Sep 18, 2017 at 15:45
• the z coordinate of 3 index is same as the z coordinate of 4 index system, i had the z coordinate as 1 in the 3 index, so it remains 1 in the 4 index system too right? Further, when i multiply all the coordinates in 4 index system by 3 to get the smallest integer values, i get the z coordinate as 3 Sep 18, 2017 at 15:46
• "....Now.... is the projection of your vector onto the.... plane and is only ... of the projection needed to reach the edge of the unit cell....." i don't understand this, too :-( Sep 18, 2017 at 15:48
• @SakazukiAkainu I do understand but you have to realise that going from 3 to 4 is not necessarily the same as going from 4 to 3. What formulae you use in going from 3 to 4 have to work correctly. Your transformation of [1 1 1] yielded a result which meant that the projection of your 4 index vector did not reach the edge of the unit cell. If you like the correct projection onto the $z=0$ plane should have been [1 1 -2 .] Sep 18, 2017 at 15:57