Symmetry of solids/Topic 2

From WolfWikis

Abelian groups

The reason that irreps are not always one-dimensional is closely related to the fact that elements do not always commute. For example is we first apply say a three fold axis and then mirror, or the other way around we do not arrive at the same spot.

That is a type of behavior that cannot be represented by a one-dimensional matrix (aka a number). Numbers always commute!

Groups, the elements of which all commute are called Abelian

The irreps of Abelian groups are always one-dimensional. (Because we can always mimic their behavior with mere numbers.)

A very important result for solids is that translation is a form of symmetry that always commutes with itself.

Translation symmetry versus rotation symmetry

Symmetry is best described by looking at what a certain element does to the coordinates of each point of the object that the element operates on. Our mirror plane m is fine example. Suppose the mirror plane is horizontal and the origin (0,0,0) lies in it and the z axis is vertical. (We just make it so: we can always chose a coordinate system). In that case each point (x y z) is transformed into (x y -z). We could even describe this operation with a matrix operation on the vector (x y z).

Translations are a bit different. Crystalline solids have a lot of those, because the structure is periodic. A certain motif, known as the unit cell repeats itself endlessly in three directions. Well, not quite endlessly. Crystals are finite after all. But the repetition does go one for quite a while, so we will worry about the surface of the crystal later.

Let's just consider a one-dimensional case, say in the x direction. If we take the period over which the structure repeats as unit (i.e. 1), the translation symmetry says that whatever we have in point (x y z) we must also have at (x+1 y z). We could define that as a symmetry element T, but it is a little harder to write it as a matrix. The best way is to expand our vector (x y z) to four dimensions and add a ('dummy') one: (x y z 1)

As you see this matrix takes (x y z 1) into (x+1 y z 1) and if we leave out our dummy fourth dimension that is exactly what we want.

Of course one element T does not a group make. If we combine T with itself we should get another element of the group, but if we do that we get (x+2 y z 1) and that is a brand new element T² that we could also write as a matrix:

(And yes if you multiply the matrix for T with itself you get T².)

To comply with the definition of a group we also need an inverse element for each element, e.g. T^-2. That is not too hard to do: we just shift to the left twice:

Symmetry of solids/Topic 2/Excel exercise

Closure

But do all these element form a group? We keep getting new ones each time we combine.

There are two ways out of this problem.

1. One is to assume the number of elements is infinite. Of course this is not entirely realistic, because the crystal cannot be infinite. It would fill the universe and collapse into a black hole under its own gravity..
2. The other solution is a real deus ex machina. We assume that there are N elements (N being large like N_avogadro), but that T^N = E.

This equally unrealistic assumption is known as closure and is paramount to assuming that when you exit the crystal one end you are back at the other end.

In a way that is like saying that the crystal does not have a surface, or: we neglect any effects thereof.

Translation groups

Including closure we now have a finite group. It consists of N elements Tⁱ, where i=0,1,2,...,N-1 and T^N=T⁰= E. What the operator Tⁱ does is add i units translation to e.g. the x position of our object:

All of our elements can be written in matrix form:

We can truly use the exponent i as an exponent because:

This simply follows from the fact that applying T¹ twice gives a shift of 2 units, i.e. the same result as applying T² once.

Reciprocal elements are also in the group because T^-i=E.T^-i= T^N.T^-i=T^N-i

So, suppose N=1000, then shifting i=4 to the right can be undone by shifting a further i=996 to the right, because that is equivalent to shifting i= -4 to the left under closure.

The fact that we can treat our elements in this multiplicative way has far-reaching consequences because it makes the translation group abelian: whether we first shift twice and then three times of the other way around does not matter. In both cases we end up with a shift of 5 units:

The fact that all elements commute means that all irreps of the translation group are one-dimensional