# How can I transform my unit cell positions to cartesian coordinates?

I have a structure with the parameters bellow:

ibrav = 0

A = 6.05844

CELL_PARAMETERS {alat}

0.497698628   0.497698628   0.000000000

0.497698628   0.000000000   0.497698628

-0.000000000   0.497698628   0.497698628


Can someone help me to convert them to cartesian coordinates, please?

• As far as I understand from this documentation of Quantum Espresso, they are already in Cartesian form Mar 17, 2023 at 11:13
• Each of the three rows of the cell_parameter shows a lattice vector. The three columns are the Cartesian components. So, Here in angstrom unit, you have a unit cell with $\vec{a_1} = 3.01527728 \hat{x} + 3.01527728 \hat{y}$, $\vec{a_2} = 3.01527728 \hat{x} + 3.01527728 \hat{z}$ and $\vec{a_3} = 3.01527728 \hat{y} + 3.01527728 \hat{z}$ because A=6.05844 angstrom is the scaling factor and 0.497698628*6.05844=3.01527728 Mar 17, 2023 at 12:45

$$\begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$$
and the cell parameters are A, B, and C in Angstroms, then, the coordinates in Angstroms are: $$\begin{bmatrix} x_1=a_1*A & y_1=b_1*B & z_1=c_1*C \\ x_2=a_2*A & y_2=b_2*B & z_2=c_2*C \\ x_3=a_3*A & y_3=b_3*B & z_3=c_3*C \end{bmatrix}$$
If the cell is not cubic, the conversion has to take into account the cell angles ($$\alpha_1$$, $$\alpha_2$$, and $$\alpha_3$$):
$${\begin{pmatrix}x_1\\y_2\\z_3\end{pmatrix}}={\begin{pmatrix}A\sin(\alpha _{2}){\sqrt {1-(\cot(\alpha _{1})\cot(\alpha _{2})-\csc(\alpha _{1})\csc(\alpha _{2})\cos(\alpha _{3}))^{2}}}&0&0\\ A\csc(\alpha _{1})\cos(\alpha _{3})-A\cot(\alpha _{1})\cos(\alpha _{2})&B\sin(\alpha _{1})&0\\ A\cos(\alpha _{2})&B\cos(\alpha _{1})&C\\\end{pmatrix}}{\begin{pmatrix}a_1\\b_2\\c_3\end{pmatrix}}$$