Understanding Reciprocal Lattice Vectors and Orthogonality in Primitive Lattices

[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

 

[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

 

[torehan]

I is 3x3 identity martix isn't it?

 

[kanato]

yes it is

 

[torehan]

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

 

[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}]}.

 

[kanato]

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

 

[torehan]

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}]}.

Thanks but the discussion is how we get these formulas.

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

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