# Reciprocal lattice help needed.

1. May 15, 2010

### Michaelangelo

here is a question on reciprocal lattices that im stuck on

for a simple cubic lattice, the unit cell is defined by a1=a(1,0,0) a2 = a(0,1,0) a3 = a(0,0,1), demonstrate that the reciprocal lattice of its reciprocal lattice is the original crystal lattice.

From what ive found, i think the reciprocal lattice base vectors b1 b2 b3 of the primitive vectors of the crystal lattice a1 a2 a3 is defined by (π is pi btw)

b1=2π (a2xa3/a1.a2xa3), b2=2π (a3xa1/a1.a2xa3) , b3=2π(a1xa2/a1.a2xa3)

the volume V is defined by a1.a2xa3 so i have to figure that out, but if i were to do a2xa3, would it be:
i j k
0 1 0
0 0 1

= (1-0)i + (0-0)j + (0-0)k = so i take it this would equal a(1,0,0) which is a1?, so is a1.a2xa3 basically a1.a1 or am i horribly confused?

it might be the latter but ill soldier on, if it is a1.a1 then does this not just give us the answer a(1,0,0)? or should i have got rid of the a at some point? or should i just be getting an integer? can someone please give me a gentle push in this question i really think i could do most of it myself im just a bit confused and stressed.

But from here im stuck, i took 2 years off my degree and ive found that my basic vector calculation skills have left me completely. Do i do the cross product first or the dot? any help appreciated.

2. May 16, 2010

### nickjer

Hmm... Towards the end you confused me with all the run on questions. The cross product of a2 and a3 is close. You forgot that each of those vectors have the magnitude "a". So the factor out front should be $a^2$ and not $a$.

Then taking the dot product with a1 gives you $a^3$. This makes sense since the volume of a cube with the side 'a' is just $a^3$.