# Translating fractional coordinates resulting from conventional cell and primitive cell

I have calculated the band structure of some material where the unit cell was the conventional one (i.e. the lattice vectors are orthogonal). I'm trying now to calculate the same band structure only now the unit cell is the primitive one.

My question is how do I convert the fractional k-points i.e. $$x_1\vec{b}_1 + x_2\vec{b}_2 + x_3\vec{b}_3$$ in the first Brillouin from the conventional cell to the primitive cell?

The conventional cell is:

    3.3132998943         0.0000000000         0.0000000000
0.0000000000        10.4729995728         0.0000000000
0.0000000000         0.0000000000         4.3740000725


and the primitive is:

    1.6566492265        -5.2364998440         0.0000000000
1.6566492265         5.2364998440         0.0000000000
0.0000000000        -0.0000000000         4.3740000725


for example, what will be the equivalent of (1/2, 0, 0) in the primitive cell?

• +1 Welcome to our new community and thank you so much for contributing your question here! We hope to see much more of you in the future !!! Jun 13 at 20:47
• I think that if you have a "new" cell, you need to generate the new k-point path for the new Brillouin zone.
– Camps
Jun 14 at 22:06

I've never done this before but here is my guess of how you can do it.

If the numbers you posted are the unit cell axes vectors with a vector for each column then their direct matrices are.

$$\tag{1}\mathbf{A} = \begin{bmatrix}3.31 & 0.00 & 0.00 \\ 0.00 & 10.47 & 0.00 \\ 0.00 & 0.00 & 4.37\end{bmatrix}, \qquad\qquad \mathbf{A}' = \begin{bmatrix}1.65 & -5.24 & 0.00 \\ 1.66 & 5.23 & 0.00 \\ 0.00 & 0.00 & 4.37\end{bmatrix}$$

From this you can get the reciprocal space vectors with

$$\mathbf{B} = 2\pi(\mathbf{A}^{-1})^{T}, \qquad\qquad \mathbf{B}' = 2\pi(\mathbf{A}^{-1})^{T}\tag{2}$$

where I have included the $$2\pi$$ to stick with the physicist definition. Then you can convert your example fractional $$k$$-point,

$$\vec{x} = \begin{bmatrix}0.50 \\ 0.00 \\ 0.00\end{bmatrix}\tag{3}$$

to a point in reciprocal space $$\vec{G}$$ with:

$$\vec{G} = \mathbf{B}\vec{x}\tag{4}$$

then convert that point $$\vec{G}$$ in reciprocal space to the fractional $$k$$-point in your other reciprocal lattice.

$$\vec{x}' = (\mathbf{B}')^{-1}\vec{G}\tag{5}$$

Then you'll need to adjust $$\vec{x}'$$ so it is in the first Brillouin zone. I think you can do something like this.

$$\begin{eqnarray} x^{1\mathrm{st}}_{1} &= x'_{1} - \mathrm{round}(x'_{1})\tag{6}\\ x^{1\mathrm{st}}_{2} &= x'_{2} - \mathrm{round}(x'_{2})\tag{7}\\ x^{1\mathrm{st}}_{3} &= x'_{3} - \mathrm{round}(x'_{3})\tag{8} \end{eqnarray}$$

Hopefully that is right.