Hi Ryan,

You can prove that if $a,b$ are some elements in an algebra such that $[a,b]=\lambda b$ for $\lambda$ a scalar, then (in a context where this expression makes sense) $q^a b q^{-a}=q^{\lambda} b$: rewrite the relation as $$ab=b(a+\lambda)$$ then $$a^nb=b(a+\lambda)^n$$ therefore $$q^ab=\sum \frac{\log(q)^na^n}{n!}b=b\sum \frac{\log(q)^n(a+\lambda)^n}{n!}= bq^{(a+\lambda)}=q^{\lambda}bq^a$$ since $\lambda$ commutes with everything.

Therefore it's in fact not a deformation but the same relation with $q^H$ instead of $H$.

In my opinion, what is more miraculous is the existence of a non-trivial deformation of the Hopf structure. Although it does not leads to explicit formulas, the conceptual explanation for that comes from the Kohno-Drinfeld theorem, or more generally from the Etingof-Kazhdan quantization functor.

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Hi Ryan,

You can prove that if $a,b$ are some elements in an algebra such that $[a,b]=\lambda b$ for $\lambda$ a scalar, then (in a context where this expression makes sense) $q^a b q^{-a}=q^{\lambda} b$. Therefore it's in fact not a deformation but the same relation with $q^H$ instead of $H$.

In my opinion, what is more miraculous is the existence of a non-trivial deformation of the Hopf structure. Although it does not leads to explicit formulas, the conceptual explanation for that comes from the Kohno-Drinfeld theorem, or more generally from the Etingof-Kazhdan quantization functor.