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I'm writing my physics bachelor on the Raman scattering effect in solids. I'm trying to evaluate the scattering intensity response to varying polarization angle. This is the well known linear polarized Raman measurement. The intensity is given by:

\begin{align} I_s \propto |e_s\mathfrak{R}(X)e_i|^2+|e_s\mathfrak{R}(Y)e_i|^2+|e_s\mathfrak{R}(Z)e_i|^2 \end{align} $\mathfrak{R}(X),(Y),(Z)$ are the components of the Raman tensor and $e_s$ and $e_i$ are the scattered and incident polarization vector respectively. Since my setup is configured to a parallel polarization configuration, then my polarization vectors are given by: \begin{align} e_i=e_s= \begin{pmatrix} \cos(\alpha)\\ \sin(\alpha)\\ 0\\ \end{pmatrix} \end{align}

Here, I assume that the light is propagating in the Z-direction. I'm studying diamond structure where the raman tensor is given by:

\begin{align} \mathfrak{R}(X)= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & d \\ 0 & d & 0 \\ \end{pmatrix},\\ & \mathfrak{R}(Y)= \begin{pmatrix} 0 & 0 & d \\ 0 & 0 & 0 \\ d & 0 & 0 \\ \end{pmatrix},\\ & \mathfrak{R}(Z)= \begin{pmatrix} 0 & d & 0 \\ d & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} \end{align}

My problem arise when I consider entering the sample form $[111]$ direction. Following the derivation given in (https://pubs.aip.org/aip/jap/article-abstract/120/5/055701/344821/Polarized-Raman-backscattering-selection-rules-for?redirectedFrom=fulltext), I had to transform my coordinates system such that Z-axis is now pointing in [111]. I did this in the following way:

\begin{align} \mathfrak{R}'= T\mathfrak{R}T^{-1} \end{align}

where T is given by two rotation matrices. First, I applied a 45 degree rotation about the Z-axis followed by 54.74 degree rotation about X-axis. This is given by the matrix T:

\begin{align} T= \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\sqrt{\frac{2}{3}}\\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \end{pmatrix} \end{align}

such that the new basis (X,Y,Z) are ([1,-1,0],[1,1,-2],[1,1,1]). If you apply these matrix to the raman matrices you would get:

\begin{align} \begin{split} &\mathfrak{R}'(X')= \begin{pmatrix} 0 & \frac{2\sqrt{3}}{6} d & -\frac{\sqrt{6}}{6}d \\ \frac{2\sqrt{3}}{6}d & -\frac{4}{6}d & -\frac{\sqrt{2}}{6}d \\ -\frac{\sqrt{6}}{6}d & -\frac{\sqrt{2}}{6}d & \frac{4}{6}d \\ \end{pmatrix},\\ &\mathfrak{R}'(Y')= \begin{pmatrix} 0 & -\frac{2\sqrt{3}}{6} d & \frac{\sqrt{6}}{6}d \\ -\frac{2\sqrt{3}}{6}d & -\frac{4}{6}d & -\frac{\sqrt{2}}{6}d \\ \frac{\sqrt{6}}{6}d & -\frac{\sqrt{2}}{6}d & \frac{4}{6}d \\ \end{pmatrix},\\ &\mathfrak{R}'(Z')= \begin{pmatrix} -d& 0 & 0 \\ 0 & \frac{1}{3}d & \frac{\sqrt{2}}{3}d \\ 0 & \frac{\sqrt{2}}{3}d & \frac{2}{3}d \\ \end{pmatrix} \end{split} \end{align}

However, this result doesn't match the experiment data. In the paper previously mentioned, They talked about a transformation of something called phonon polarization such that the new Raman tensor is given by: \begin{align} \mathfrak{R}_{{X},{Y},{Z}}=\mathfrak{R}'_j Q'_j \end{align}

Thus, $\mathfrak{R}_{{X},{Y},{Z}}$ is given by:

\begin{align} \begin{split} &\mathfrak{R}(X')= \begin{pmatrix} 0 & \sqrt{\frac{2}{3}} d & -\frac{1}{\sqrt{3}}d \\ \sqrt{\frac{2}{3}} d & 0 & 0 \\ -\frac{1}{\sqrt{3}}d & 0 & 0 \\ \end{pmatrix},\\ &\mathfrak{R}(Y')= \begin{pmatrix} \sqrt{\frac{2}{3}} d & 0 & 0 \\ 0 & \sqrt{\frac{2}{3}} d & -\frac{1}{\sqrt{3}}d \\ 0 & -\frac{1}{\sqrt{3}}d & 0 \\ \end{pmatrix},\\ &\mathfrak{R}(Z')= \begin{pmatrix} - \frac{1}{\sqrt{3}}d & 0 & 0 \\ 0 & - \frac{1}{\sqrt{3}}d & 0 \\ 0 & 0 & \frac{2}{\sqrt{3}}d \\ \end{pmatrix} \end{split} \end{align}

My question is: How do I find the transformation of phonon polarization $Q'_j$? In this paper there was no mention of what is the phonon polarization transformation and how it's applied. Tracing the refrences in that paper further led me to find the same matrices with out explicitly mentioning the transformation $Q'_j$?

Here are two other examples where the $Q'_j$ transformation was applied in the same paper:

Example 1:

First, a transformation of the basis (X,Y,Z) to ([1,-1,0],[1,1,0],[0,0,1]). This can be done by a rotation of 45 degrees about the Z-axis described by the matrix R:

\begin{align} R= \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \end{align}

applying that to the Raman matrices would lead to:

\begin{align} \begin{split} &\mathfrak{R}'(X')= \begin{pmatrix} 0 & 0 & -\frac{1}{\sqrt{2}}d \\ 0 & 0 & \frac{1}{\sqrt{2}}d \\ -\frac{1}{\sqrt{2}}d & \frac{1}{\sqrt{2}}d & 0 \\ \end{pmatrix},\\ &\mathfrak{R}'(Y')= \begin{pmatrix} 0 & 0 & \frac{1}{\sqrt{2}}d \\ 0 & 0 & \frac{1}{\sqrt{2}}d \\ \frac{1}{\sqrt{2}}d & \frac{1}{\sqrt{2}}d & 0 \\ \end{pmatrix},\\ &\mathfrak{R}'(Z')= \begin{pmatrix} -d & 0 & 0 \\ 0 & d & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} \end{split} \end{align}

After applying $Q'_j$ transformation the Raman tensor becomes:

\begin{align} \mathfrak{R}(X')= \begin{pmatrix} 0 & 0 & d \\ 0 & 0 & 0 \\ -d & 0 & 0 \\ \end{pmatrix},\\ & \mathfrak{R}(Y')= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & d \\ 0 & d & 0 \\ \end{pmatrix},\\ & \mathfrak{R}(Z')= \begin{pmatrix} -d & 0 & 0 \\ 0 & d & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} \end{align}

Example 2:

Second transformation of basis (X,Y,Z) to ([1,-1,0],[0,0,-2],[1,1,0]) is done by the same rotation as example one followed by another rotation 90 degree rotation about the new X-axis. This is given by the following matrix R':

\begin{align} R'= \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & 0 \\ 0 & 0 & -1 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \end{pmatrix} \end{align}

which gives the following Raman tensor:

\begin{align} &\mathfrak{R}'(X')= \begin{pmatrix} 0 &\frac{d}{\sqrt{2}} & 0 \\ \frac{d}{\sqrt{2}} & 0 & -\frac{d}{\sqrt{2}} \\ 0 & -\frac{d}{\sqrt{2}} & 0 \\ \end{pmatrix},\\ &\mathfrak{R}'(Y')= \begin{pmatrix} 0 &-\frac{d}{\sqrt{2}} & 0 \\ -\frac{d}{\sqrt{2}} & 0 & -\frac{d}{\sqrt{2}} \\ 0 & -\frac{d}{\sqrt{2}} & 0 \\ \end{pmatrix},\\ &\mathfrak{R}'(Z')= \begin{pmatrix} -d & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & d \\ \end{pmatrix} \end{align}

After applying $Q'_j$ transformation the Raman tensor becomes:

\begin{align} \mathfrak{R}(X')= \begin{pmatrix} 0 & d & 0 \\ d & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix},\\ & \mathfrak{R}(Y')= \begin{pmatrix} d & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -d \\ \end{pmatrix},\\ & \mathfrak{R}(Z')= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -d \\ 0 & -d & 0 \\ \end{pmatrix} \end{align}

Can you please help me find the transformation $Q'_j$?

Note: $\mathfrak{R}'_j$ has zero determinant thus it can't be inverted or $\mathfrak{R'}_j^{-1}$ doesn't exist. Thus trying to find by:

\begin{align} Q'_j =\mathfrak{R'}_j^{-1} \mathfrak{R}_{{X},{Y},{Z}} \end{align}

wouldn't work! what else can I do?

Further references where the same raman tensors for [111] given here. https://journals.aps.org/prb/abstract/10.1103/PhysRevB.8.717

https://pubs.aip.org/aip/jap/article-abstract/82/9/4493/492324/Strain-effects-on-optical-phonons-in-111-GaAs

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