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, applying a symmetry operation to a function means the following:

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

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 $\hat{R}$.

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

  • $\begingroup$ 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. $\endgroup$ – user43021 May 21 '18 at 16:24

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 $\hat{R}$ such that $\hat{R}: D\mapsto D$, and $\forall x\in D$, $f(x) = f(\hat{R}x) = (f\circ \hat{R})x$.

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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