Crystallography, Ashcroft&Mermin

[Feynmanfan]

Homework Statement

I'm having nightmares with this problem, apparently simple (A&M Chapter 7 Prob 2).

I have to show that a trigonal primitive lattice, depending on its angle, can represent fcc or bcc. This I kind of figured it out intuitively. But the serious problem is number b), where I have to prove that a trigonal lattice of basis a_i with angle 60^0, and a two point basis +-\frac{1}{4}(a_1+a_2+a_3) can represent a simple cubic lattice.

Homework Equations

A trigonal primitive cell set of vector can be
$a_1=\frac{a}{2}(x+y)$
$a_2=\frac{a}{2}(z+y)$
$a_3=\frac{a}{2}(x+z)$

The definition of trigonal is, same angle between vectors and same length.
A simple cubic lattice is clear that it can be represented by a (x,y,z)

The Attempt at a Solution

Well... I added a1+a2+a3 and divided by 4 as the problem said but i can't understand why this represents a simple cubic lattice. In addition I am asked to guess what it represents if I divide by 8 instead of 4.

I guess I don't know what "two point basis means".

Thanks for your help!(I need it so badly...)

 

[genxhis]

Here is an intuitive way of reasoning through the problem. Consider the conventional unit cell of size a³ for the FCC lattice. Then the set a_i you have mentioned are just primitive basis vectors for this lattice. Now -- mentally speaking -- subdivide the conventional cell into eight smaller "cubelets" by cutting the cell in half three times using planes perpendicular to the axes. Notice that each lattice point is at the corner of one of the cubelets. The problem calls on you to replace each lattice point by two new lattice points at positions plus or minus a/4( i + j + k ) from the original ones. Well.. this just sends the old (or FCC) lattice points into the centers of the cubelets. It should now be fairly clear that the new lattice is simple cubic with side length a/2.