Solid state physics, exercise about reciprocal lattice

[fluidistic]

Homework Statement

Hi guys, I don't reach the correct answer to an exercise. I'm following Ashcroft's book.
I must find that the reciprocal of the bcc Bravais lattice is a fcc one and the reciprocal of the fcc Bravais lattice is a bcc one.

Homework Equations

If a_1, a_2 and a_3 are vectors spaning the Bravais lattice, then the reciprocal lattice is spanned by the vectors $b_1= 2 \pi \frac{a_2 \times a_3}{a_1 \cdot (a_2 \times a_3)}$, $b_2= 2 \pi \frac{a_3 \times a_1}{a_1 \cdot (a_2 \times a_3)}$ and $b_3= 2 \pi \frac{a_1 \times a_2}{a_1 \cdot (a_2 \times a_3)}$

The Attempt at a Solution

For a bcc lattice: $a_1=\frac{a}{2}(\hat y +\hat z-\hat x )$, $a_2=\frac{a}{2}(\hat z +\hat x-\hat y )$ and $a_3=\frac{a}{2}(\hat x +\hat y-\hat z )$.
While for a fcc lattice, $a_1=\frac{a}{2}(\hat y +\hat z)$, $a_2=\frac{a}{2}(\hat z +\hat x )$ and $a_3=\frac{a}{2}(\hat y +\hat x )$.
I first tried to show that the reciprocal of the fcc lattice is a bcc lattice. So I just used the formulae given but didn't reach what I should have.
$b_1=2\pi \frac{[ \frac{a}{2}(\hat z + \hat x )] \times [ \frac{a}{2} (\hat x + \hat y ) ]}{[\frac{a}{2} (\hat y + \hat x )] \cdot [ \frac{a}{2} (\hat z + \hat x ) \times \frac{a}{2} (\hat x + \hat y ) ] }$.
I calculated $(\hat z + \hat x ) \times (\hat x + \hat y )$ to be worth $\hat y + \hat z$. So that I reached that $b_1=\frac{4 \pi }{a} (\hat y + \hat z )$. This is already wrong, according to the book I should have reached $b_1=\frac{4 \pi }{a} \cdot \frac{1}{2} (\hat y + \hat z -\hat x )$.
I have no clue on what I've done wrong.

 

[M Quack]

z times x = y
z times y = -x
x times x = 0
x times y = z

 

[fluidistic]

z times x = y
z times y = -x
x times x = 0
x times y = z

Ok thanks. I see that I made a mistake in doing that cross product (I did it via a determinant and did the arithmetics too fast, forgetting a term).
I now reach that $(\hat z +\hat x) \times (\hat x +\hat y)=\hat y +\hat z -\hat x$. But now the denominator is worth $(\hat y +\hat x )\cdot (\hat y + \hat z -\hat x )=1-1=0$ which can't be right.

 

[ehild]

I now reach that $(\hat z +\hat x) \times (\hat x +\hat y)=\hat y +\hat z -\hat x$. But now the denominator is worth $(\hat y +\hat x )\cdot (\hat y + \hat z -\hat x )=1-1=0$ which can't be right.

The denominator should be

$(\hat y +\hat z )\cdot (\hat y + \hat z -\hat x )$

ehild

 

[fluidistic]

The denominator should be

$(\hat y +\hat z )\cdot (\hat y + \hat z -\hat x )$

ehild

Ok thanks! That made it, I now reach what I should.