# Reciprocal lattice

1. Oct 24, 2009

### torehan

$$f(\vec{r}) = f(\vec{r}+\vec{T})$$

$$\vec{T}= u_{1} \vec{a_{1}} + u_{2} \vec{a_{2}}+u_{3} \vec{a_{3}}$$

$$u_{1},u_{2},u_{3}$$ are integers.

$$f(\vec{r}+\vec{T})= \sum n_{g} e^{(i\vec{G}.(\vec{r}+\vec{R}) )}= f(\vec{r})$$

$$e^{i\vec{G}.\vec{R} }= 1$$
$$\vec{G}.\vec{R} = 2\pi m$$

we call $$\vec{G}=h\vec{g_{1}} + k \vec{g_{2}}+l \vec{g_{3}}$$ reciprocal lattice vector.

but what about the primitive lattice vectors $$\vec{g_{1}} , \vec{g_{2}} , \vec{g_{3}}$$ ?

To simplify the discussion consider $$\vec{T_{1}}$$ in 1D;

$$\vec{T_{1}} = u_{1} \vec{a_{1}}$$

$$\vec{G}.\vec{T} =(h\vec{g_{1}} + k \vec{g_{2}}+l \vec{g_{3}}) . ( u_{1} \vec{a_{1}}) = 2 \pi m$$

Is there any definiton that indicates direct primitive lattice vectors and reciprocal primitive lattice vectors orthogonalities?
i.e

$$\vec{g_{1}} . \vec{a_{1}} = 2 \pi$$

$$\vec{g_{2}} . \vec{a_{1}} = \vec{g_{3}} . \vec{a_{1}} = 0$$

Last edited: Oct 24, 2009
2. Oct 24, 2009

### kanato

g1,g2,g3 are the primitive vectors for the reciprocal lattice.
And the orthogonality condition is $$\vec{a}_i \cdot \vec{g}_j = 2\pi \delta_{i,j}$$

A trick for calculating the reciprocal vectors is to form the matrix A where the columns are the direct lattice vectors, and the matrix G where the columns are the reciprocal lattice vectors. Then you have
$$G^T \cdot A = 2\pi I$$
so
$$G = 2\pi (A^{-1})^T$$

3. Oct 24, 2009

### torehan

I is 3x3 identity martix isn't it?

4. Oct 24, 2009

### kanato

yes it is

5. Oct 25, 2009

### torehan

It's still confusing for me..
so how can I invers a (1x3) matrix?

6. Oct 25, 2009

### corydalus

You can get the vectors $$\vec{g_1}, \vec{g_2}, \vec{g_3}$$ without using any matrices. To do this use the following formulas:
$$\vec{g_1}=2\pi\frac{[\vec{a_2},\vec{a_3}]}{\vec{a_1}[\vec{a_2},\vec{a_3}]},$$ $$\vec{g_2}=2\pi\frac{[\vec{a_3},\vec{a_1}]}{\vec{a_1}[\vec{a_2},\vec{a_3}]},$$ $$\vec{g_3}=2\pi\frac{[\vec{a_1},\vec{a_2}]}{\vec{a_1}[\vec{a_2},\vec{a_3}]}.$$

7. Oct 25, 2009

### kanato

A and G are 3x3 matrices, not 1x3 matrices. The columns of A are the vectors of your lattice:
$$A = \left( \left( \! \! \begin{array}{c}\\ \vec{a}_1 \\ \, \end{array} \!\! \right) \left( \! \! \begin{array}{c}\\ \vec{a}_2 \\ \, \end{array} \!\! \right) \left( \! \! \begin{array}{c}\\ \vec{a}_3 \\ \, \end{array} \!\! \right) \right)$$

Personally, I think this is easier than manually evaluating three separate cross products. But either way works.

8. Oct 26, 2009

### torehan

Thanks but the discussion is how we get these formulas.

O.K , as you said ai and gi must be vectors which has three components. The question is what are these components?

Last edited: Oct 26, 2009