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I had a hard time understanding what it means to apply a symmetry operator to a function, so i wondered if there is a formal way to define this ?
As far as i understand it, applying a symmetry operation to a function means the followling:

$ \hat Rf(x) \rightarrow f(\hat Rx) $

The result has then to be the same function with a coefficient in front of it. Otherwise it would not be symmetric in regard to the Symmetry Operator R.

Is this correct and can it be shown that this is how to apply symmetry operators to functions ?

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  • $\begingroup$ Well, yes, that's pretty much the size of it. $\endgroup$ – Ivan Neretin May 21 '18 at 10:17
  • $\begingroup$ Do you now any source that formulates it explicitly like this ? $\endgroup$ – Hans Wurst May 21 '18 at 10:19
  • $\begingroup$ Try to work out the functions on the rhs of any point group. Only if the operation say C2(x) on function f, results in 1 then this gives symmetry species, say B1. For squared functions such $(x^2 \pm y^2)$ repeat the operation on x twice and y twice. $\endgroup$ – porphyrin May 21 '18 at 10:40
  • $\begingroup$ Would love to see an answer to this, because the implication in the OP looks odd to me and I haven't seen this notation before. Bishop's discussion of, say, d orbitals seems clear and there is no mention of this property--it seems we could have vectors of functions operating on a matrix operator and the dimensionality could be wrong? $\endgroup$ – daniel May 21 '18 at 12:44
  • $\begingroup$ Symmetry operators produce coordinates transformations, so yes. I have seen it written as you propose in some documents. And as far as I understand that's a property of symmetry operators, not of every operator. You wouldn't write R=partial derivative. $\endgroup$ – santimirandarp May 21 '18 at 16:24
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That's not entirely correct. You're making at least an extra assumption here. You're assuming that $\hat{R}$ is a function from $D\mapsto D$ where $D$ is the domain of $f$, but there's no reason why $f$ must operate as a function on $D\mapsto D$. It might just as well be $f: D\mapsto \mathbb{R}$.

I think what you're really asking for is that a symmetry element $\hat{R}$ is such that $\hat{R}: D\mapsto D$ and some $x\in D$, $f(x) = f(\hat{R}x) = (f\circ \hat{R})x$.

Then for example, for $f(x) = (\cos x, \cos x)$, the transformation $\hat{R}$ mapping $x \rightarrow -x$ would would a valid symmetry element.

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