PHY 472, Fall 2002, Homework solutions

Ch.1. P.3
In terms of the unit vectors, the three vectors are:
a = (1/2)(x+y-z)
b = (1/2)(x-y+z)
c = (1/2)(-x+y+z)
Let us take any two, like a and b. To get the angle, we use ab = abcos Q. The dot products is
ab=(1/4)(x+y-z)(x-y+z)
This is easy to evaluate, and we get ab= (1/3)(1/4). We get cos Q=1/3, and Q=109.5^o.
Ch.1. P.3
We need to calculate the height of a tetraeder. The side of the tetraeder is the lattice spacing a. The base of the tetraeder is an equilateral triangle, and the height of that triangle is (3/4)^1/2a. (Sorry, there is no easy way to make a square root sign on the WEB!). The distance between a corner and the center of this triangle is (2/3)(3/4)^1/2a. This distance, and one of the edges of the tetraeder, make a right triangle. The third side is the height of the tetraeder. Therefore: a₂ = x₂ + (1/3)a₂ , and x= (8/3)^1/2a
Ch 2. P.3 a:
The volume is V= a₁(a₂ x a₃). Expressing the cross product as a determinant yields:
V =
x y z
0.866a 0.5a 0
-0.866a 0.5a 0
cz = z(a²0.433+a²0.433)cz = ca² (3^1/2/2)

b: Do the vector product in
b₁ = 2 p a₂x b₃/V
by the determinant method as above,
a₂x b₃ =
x y z
-0.866a 0.5a 0
0 0 c
= x (1/2) ca + y (3^1/2/2) ca
This yields
b₁= (2p/a3^1/2) x + (2p/a)y
Repeat the same for the other two vectors.
c:
The BZ is a slab of height 2p/c, centered around the origin. The base of the slab is a hexagon of side (2/3^1/2(2p/a).

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