Transforming Crystal Field Parameters by Rotation of the Coordinate System

There is a special program cf2mcphas for rotating crystal field parameters (Stevens Notation) given in the orientation $abc\vert\vert$xyz to the McPhase notation $abc\vert\vert$yzx. The original crystal field parameters have to be given in a file of the above format of a single ion parameter file. Output is written to stdout. Note, there is also a program mcphas2cf for transforming parameters back. For more general rotations of crystal field parameters you can use the program rotateBlm(see chapter 18).

As an example we consider a tetragonal system, where the crystal field parameters are usually given such that $abc\vert\vert$xyz - we denote these parameters as $B_l^m$. For tetragonal symmetry only $B_2^0, B_4^0, B_4^4, B_6^0$ and $B_6^6$ are not zero. In order to use these parameters in mocule cfield, they have to be transformed to a coordinate system $xyz$ with $abc\vert\vert yzx$. Denoting the new crystal field parameters which are to be used as $Blm$ the transformation is given by


$\displaystyle B20$ $\textstyle =$ $\displaystyle -\frac{1}{2} B_2^0$  
$\displaystyle B22$ $\textstyle =$ $\displaystyle +\frac{3}{2} B_2^0$  
$\displaystyle B40$ $\textstyle =$ $\displaystyle +\frac{3}{8} B_4^0+\frac{1}{8} B_4^4$  
$\displaystyle B42$ $\textstyle =$ $\displaystyle -\frac{5}{2} B_4^0+\frac{1}{2} B_4^4$  
$\displaystyle B44$ $\textstyle =$ $\displaystyle \frac{35}{8} B_4^0+\frac{1}{8} B_4^4$  
$\displaystyle B60$ $\textstyle =$ $\displaystyle -\frac{5}{16} B_6^0-\frac{1}{16} B_6^4$  
$\displaystyle B62$ $\textstyle =$ $\displaystyle +\frac{105}{32} B_6^0+\frac{5}{32} B_6^4$  
$\displaystyle B64$ $\textstyle =$ $\displaystyle -\frac{63}{16} B_6^0+\frac{13}{16} B_6^4$  
$\displaystyle B66$ $\textstyle =$ $\displaystyle \frac{231}{32} B_6^0+\frac{11}{32} B_6^4$  

As a second example we consider an orthorhombic system, where the crystal field parameters are given such that $abc\vert\vert$xyz - we denote these parameters as $B_l^m$. For orthorhombic symmetry only $B_2^0,B_2^2, B_4^0,B_4^2, B_4^4, B_6^0, B_6^2, %
B_6^4$ and $B_6^6$ are not zero. In order to use these parameters in mocule cfield, they have to be transformed to a coordinate system $xyz$ with $abc\vert\vert yzx$. Denoting the new crystal field parameters which are to be used as $Blm$ the transformation is given by


$\displaystyle B20$ $\textstyle =$ $\displaystyle -\frac{1}{2} B_2^0 -\frac{1}{2} B_2^2$  
$\displaystyle B22$ $\textstyle =$ $\displaystyle +\frac{3}{2} B_2^0 -\frac{1}{2} B_2^2$  
$\displaystyle B40$ $\textstyle =$ $\displaystyle +\frac{3}{8} B_4^0 +\frac{1}{8} B_4^2 +\frac{1}{8} B_4^4$  
$\displaystyle B42$ $\textstyle =$ $\displaystyle -\frac{5}{2} B_4^0 -\frac{1}{2} B_4^2 +\frac{1}{2} B_4^4$  
$\displaystyle B44$ $\textstyle =$ $\displaystyle \frac{35}{8} B_4^0 -\frac{7}{8} B_4^2 +\frac{1}{8} B_4^4$  
$\displaystyle B60$ $\textstyle =$ $\displaystyle -\frac{5}{16} B_6^0-\frac{1}{16} B_6^2-\frac{1}{16} B_6^4-\frac{1}{16} B_6^6 <tex2html_comment_mark>$  
$\displaystyle B62$ $\textstyle =$ $\displaystyle +\frac{105}{32} B_6^0+\frac{5}{32} B_6^4+...$  
$\displaystyle B64$ $\textstyle =$ $\displaystyle -\frac{63}{16} B_6^0+\frac{13}{16} B_6^4+...$  
$\displaystyle B66$ $\textstyle =$ $\displaystyle \frac{231}{32} B_6^0+\frac{11}{32} B_6^4+...$  

As a second example we consider an dhcp system with quasi cubic sites, where the crystal field parameters are given such that $abc\vert\vert$xyz - we denote these parameters as $B_l^m$. Only $B_2^0,B_2^{-1},B_2^2, B_4^0,B_4^{-1},B_4^2,B_4^{-3}, B_4^4, %
B_6^0,B_6^{-1}, B_6^2,B_6^{-3}, B_6^4$ and $B_6^6$ are not zero. In order to use these parameters in mocule cfield, they have to be transformed to a coordinate system $xyz$ with $abc\vert\vert yzx$. Denoting the new crystal field parameters which are to be used as $Blm$ the transformation is given by


$\displaystyle B21$ $\textstyle =$ $\displaystyle -2B_2^{-1}$  
$\displaystyle B20$ $\textstyle =$ $\displaystyle -\frac{1}{2} B_2^0 -\frac{1}{2} B_2^2$  
$\displaystyle B22$ $\textstyle =$ $\displaystyle +\frac{3}{2} B_2^0 -\frac{1}{2} B_2^2$  
$\displaystyle B4-3$ $\textstyle =$ $\displaystyle -\frac{7}{4} B_4^{-1} -\frac{3}{4} B_4^{-3}$  
$\displaystyle B4-1$ $\textstyle =$ $\displaystyle +\frac{3}{4} B_4^{-1} -\frac{1}{4} B_4^{-3}$  
$\displaystyle B40$ $\textstyle =$ $\displaystyle +\frac{3}{8} B_4^0 +\frac{1}{8} B_4^2 +\frac{1}{8} B_4^4$  
$\displaystyle B42$ $\textstyle =$ $\displaystyle -\frac{5}{2} B_4^0 -\frac{1}{2} B_4^2 +\frac{1}{2} B_4^4$  
$\displaystyle B44$ $\textstyle =$ $\displaystyle \frac{35}{8} B_4^0 -\frac{7}{8} B_4^2 +\frac{1}{8} B_4^4$  
$\displaystyle B60$ $\textstyle =$ $\displaystyle -\frac{5}{16} B_6^0-\frac{1}{16} B_6^4+...$  
$\displaystyle B62$ $\textstyle =$ $\displaystyle +\frac{105}{32} B_6^0+\frac{5}{32} B_6^4+...$  
$\displaystyle B64$ $\textstyle =$ $\displaystyle -\frac{63}{16} B_6^0+\frac{13}{16} B_6^4+...$  
$\displaystyle B66$ $\textstyle =$ $\displaystyle \frac{231}{32} B_6^0+\frac{11}{32} B_6^4+...$  

The general transformation matrices are given by



Subsections
Martin Rotter 2017-01-10