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Crystallography - (hkl) planes & reciprocal lattices (university level)

This is as much a maths question as it is a chemistry one, but I thought I'd post it here anyway.


"A three dimensional crystal lattice has basis vectors a, b, and c, relative to a common origin at one lattice site. The (hkl)-plane is defined to be the plane that intersects the basis vectors at a/h, b/k, c/l. Show that the equation of the (hkl)-plane can be written in vector form as r.g=1, where g=ha*+kb*+lc*, with the reciprocal lattice vectors given by:

a*=(b x c) / a.(b x c) , b* = (c x a) / b.(c x a), c* = (a x b) / c.(a x b)."


I'm actually a maths student, so if I could have a lucid answer it would be appreciated more than one rich in technical terms :smile:.. Reps will be dished out accordingly.
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Dream Eater
This is as much a maths question as it is a chemistry one, but I thought I'd post it here anyway.


"A three dimensional crystal lattice has basis vectors a, b, and c, relative to a common origin at one lattice site. The (hkl)-plane is defined to be the plane that intersects the basis vectors at a/h, b/k, c/l. Show that the equation of the (hkl)-plane can be written in vector form as r.g=1, where g=ha*+kb*+lc*, with the reciprocal lattice vectors given by:

a*=(b x c) / a.(b x c) , b* = (c x a) / b.(c x a), c* = (a x b) / c.(a x b)."


I'm actually a maths student, so if I could have a lucid answer it would be appreciated more than one rich in technical terms :smile:.. Reps will be dished out accordingly.


OK, now, I'm not a maths student, so this will probably lack the rigour you want, but the way I've convinced myself that the equation given is true is simply by evaluating it.

So, firstly calculate what the reciprocal lattice vectors are:
Unparseable latex formula:

\displaystyle{\vec{a}^\star = \frac{\vec{b}\times \vec{c}}{\vec{a}\cdot\left( \vec{b}\times\vec{c} \right)} = \frac{\left(\begin{array}{c}[br]bc\\[br]0\\[br]0\end{array}\right)}{abc} = \left(\begin{array}{c}[br]1/a\\[br]0\\[br]0\end{array}\right)}



Similarly,
Unparseable latex formula:

\displaystyle{\vec{b}^\star =\left(\begin{array}{c}[br]0\\[br]1/b\\[br]0\end{array}\right)}

and
Unparseable latex formula:

\displaystyle{\vec{c}^\star = \left(\begin{array}{c}[br]0\\[br]0\\[br]1/c\end{array}\right)}

, where a, b and c are the magnitudes of the respective vectors (so a = {a, 0, 0}T and so on).

Now, evaluate the vector g as given:
Unparseable latex formula:

\displaystyle\vec{g} = h \left(\begin{array}{c}[br]1/a\\[br]0\\[br]0\end{array}\right) + k \left(\begin{array}{c}[br]0\\[br]1/b\\[br]0\end{array}\right) + l \left(\begin{array}{c}[br]0\\[br]0\\[br]1/c\end{array}\right) = \left(\begin{array}{c}[br]h/a\\[br]k/b\\[br]l/c\end{array}\right)}




Then write down the equation of the plane, i.e. the position vector r of any point on the plane. There are several forms of writing this, and it can be shown that they're equivalent to each other. So just use the simplest vector form - i.e. some starting position and two vectors which define the plane. I've chosen (not sure how appropriate that is in terms of generality?) to start from the point given by the Miller index h, and then taken the two vectors pointing towards indices k and l to give the equation of the (hkl) plane, so that the vector r is given by
Unparseable latex formula:

\displaystyle\vec{r} = \left(\begin{array}{c}[br]a/h\\[br]0\\[br]0\end{array}\right) +x \left(\begin{array}{c}[br]a/h\\[br]-b/k\\[br]0\end{array}\right) + y \left(\begin{array}{c}[br]a/h\\[br]0\\[br]-c/l\end{array}\right)}

for any real numbers x and y, or equivalently
Unparseable latex formula:

\displaystyle\vec{r} = \left(\begin{array}{c}[br]a/h+ax/h + ay/h\\[br]-bx/k\\[br]-cy/l\end{array}\right)}

.

Now, since r is known to be a representation of the plane given by Miller indices (hkl), we can show the equivalence of giving the equation in terms of g simply by establishing that the scalar product of r and g is unity (as given). So:
Unparseable latex formula:

\displaystyle{\begin{aligned}\vec{r}\cdot\vec{g} & = \left(\begin{array}{c}[br]a/h+ax/h + ay/h\\[br]-bx/k\\[br]-cy/l\end{array}\right) \cdot \left(\begin{array}{c}[br]h/a\\[br]k/b\\[br]l/c\end{array}\right) = \\ [br]& = - x - y + \frac{h}{a} \left( a/h + (a x)/h + (a y)/h \right) \\ [br]& = - x - y+1+x+y \\ & =1[br]\end{aligned}}


- which is exactly what we wanted to show.

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