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How can we assign an irreducible representation to a derivative operator like $\displaystyle\frac{\mathrm d}{\mathrm dx}$ or the whole gradient operator $\nabla$ in a given point group?

Since the Hamilton operator can be assigned to the totally symmetric representation, it seems like there is a way to to do so for derivatives or operators in general, but I have no clue how to do so. I am well aware how to determine the representation for normal functions, but I have never seen it done for operators.

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    $\begingroup$ You can get better insights by posting this question in MathOverFlow. You may find mathematical researchers there. mathoverflow.net $\endgroup$ – M. Farooq Jul 9 '19 at 13:30
  • $\begingroup$ Did you post your question on mathoverflow? Google shows this answer as one of the first results. It would be helpful to others if you could link to the question. $\endgroup$ – HerpDerpington Feb 8 '20 at 19:49
  • $\begingroup$ I have not posted it on mathoverflow.net. I don't know enough mathematics to ask properly. I only know the physical chemistry approach and i am pretty sure that in pure mathematics the technical language is different. I would prefer it to be answered here since it is a topic that every chemistry student encounters even if it isn't treated in detail or brushed over very quickly. $\endgroup$ – Hans Wurst Feb 9 '20 at 13:26
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The transformation of a derivative operator under a linear transformation $\hat R $ is given by

$$ \frac{\partial}{\partial x_i} \rightarrow \sum_j R_{ji} \frac{\partial}{\partial x_j} = \sum R^T_{ij}\frac{\partial}{\partial x_j} $$ where $R_{nm}$ are the matrix elements of the matrix representation of the transformation operator in regard to cartesian coordiantes. The matrix is defined as the coordinate transformation matrix that takes the initial coordiantes to the transformed coordinates. $$ \hat R : x \rightarrow Rx $$

The character of an representation with the derivative operators as basis is therefore the same as the character of the corresponding coordinate representation, since the trace of a matrix and a matrix transpose are identical. The derivative operators form thus a basis for the same irreducible representation as their coordinates and belong to the same irreducible representation.

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