2
$\begingroup$

A raman tensor $\gamma_m$ is defined as the derivative of the polarizability tensor $\alpha$ with respect to a raman mode $Q_m$, so $$ \gamma_m =\frac{\partial\alpha}{\partial Q_m} $$ $\gamma_m$ will show certain symmetry properties which depend on the point group of the material (e.g C2v, Td etc.) and irreducible representation of the respective raman mode (e.g. Ag, B2g etc.). I have recognized that the trace of $\gamma_m$ is only non zero if the raman mode has $A_g$ symmetry, so $$ \text{Tr}(\gamma_m) = 0 \quad \text{if} \quad\text{Irrep}(m) \neq A_g $$ Is this always true and is there a rigorous proof for this?

$\endgroup$
2
+100
$\begingroup$

This is a general aspect of representation theory. The polarizability tensor $\alpha$ is rank (1,1), and is acted on by a group of transformations $G$. The class of all possible polarization tensors forms a vector space, that decomposes into mutually orthogonal representations of $G$. One of these representations is the 'trivial' representation, invariant under the group action. Since $\alpha$ is an element of $V\otimes \bar V$ for some vector space $V$ (e.g. the vector space of spherical harmonics, acted on by the group $G=SO(3)$), elements $\alpha$ transform under $G$ as $g(\alpha)=g\alpha g^{-1}$. Since the trace is cyclic, $\text{tr}(g(\alpha))=\text{tr}(g\alpha g^{-1})=\text{tr}(g^{-1}g\alpha)=\text{tr}(\alpha)$, so the trace operation projects $V\otimes \bar V$ to the trivial subrepresentation. By orthogonality, all non-trivial representations therefore have zero trace.

$\endgroup$
2
  • $\begingroup$ Thanks for the very nice answer. Could you explain the effect of the derivative with respect to $Q_m$ and how the symmetry of $Q_m$ comes into play? $\endgroup$ – Jannick Jun 30 '16 at 8:34
  • $\begingroup$ As I read your question, I interpreted the $Q_m$ as coefficients for $\alpha$ in a basis of irreducible tensors. If we write $\alpha = \sum_{m} Q_m \gamma_m$, where $\gamma_m$ are the basis elements, then $\frac{\partial\alpha}{\partial Q_m}=\gamma_m$. The 'symmetry of $Q_m$' comes into play when I discuss the transformation rule of $\alpha$: if $Q_m$ is a coefficient for a copy of the trivial representation, then it is invariant under the group action. If not, then it will transform in some finite dimensional irreducible vector subspace of coefficients for other basis elements. $\endgroup$ – TLDR Jun 30 '16 at 12:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.