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Solid state physics, exercise about reciprocal lattice

  1. Jul 6, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations

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

    3. The attempt at a solution
    For a bcc lattice: [itex]a_1=\frac{a}{2}(\hat y +\hat z-\hat x )[/itex], [itex]a_2=\frac{a}{2}(\hat z +\hat x-\hat y )[/itex] and [itex]a_3=\frac{a}{2}(\hat x +\hat y-\hat z )[/itex].
    While for a fcc lattice, [itex]a_1=\frac{a}{2}(\hat y +\hat z)[/itex], [itex]a_2=\frac{a}{2}(\hat z +\hat x )[/itex] and [itex]a_3=\frac{a}{2}(\hat y +\hat x )[/itex].
    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.
    [itex]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 ) ] }[/itex].
    I calculated [itex](\hat z + \hat x ) \times (\hat x + \hat y )[/itex] to be worth [itex]\hat y + \hat z[/itex]. So that I reached that [itex]b_1=\frac{4 \pi }{a} (\hat y + \hat z )[/itex]. This is already wrong, according to the book I should have reached [itex]b_1=\frac{4 \pi }{a} \cdot \frac{1}{2} (\hat y + \hat z -\hat x )[/itex].
    I have no clue on what I've done wrong.
     
  2. jcsd
  3. Jul 6, 2012 #2
    z times x = y
    z times y = -x
    x times x = 0
    x times y = z
     
  4. Jul 6, 2012 #3

    fluidistic

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    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 [itex](\hat z +\hat x) \times (\hat x +\hat y)=\hat y +\hat z -\hat x[/itex]. But now the denominator is worth [itex](\hat y +\hat x )\cdot (\hat y + \hat z -\hat x )=1-1=0[/itex] which can't be right.
     
  5. Jul 6, 2012 #4

    ehild

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    The denominator should be

    [itex](\hat y +\hat z )\cdot (\hat y + \hat z -\hat x )[/itex]

    ehild
     
  6. Jul 7, 2012 #5

    fluidistic

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