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It seems that most of the literature dealing with q-analogs defines q-numbers according to $$[n]_q\equiv \frac{q^n-1}{q-1}.$$ Even Mathematica uses this definition: with the built-in function QGamma you obtain

QGamma[n+1,q] / QGamma[n,q] = (q^n-1) / (q-1)

However, in some books and papers you find that the authors use a different notion of q-numbers, namely $$[n]_q\equiv \frac{q^n-q^{-n}}{q-q^{-1}}.$$ To me it seems that this alternative definition is exclusively used in the context of physics dealing with quantum groups.

Now my questions:

  1. What is the notion/motivation for the latter definition? Somehow, it seems to be more relevant in (physical) applications.

  2. Are these two approaches equivalent? If so, to what extent?

  3. On the technical side: The QGamma function can be defined in terms of QPochhammer functions (see e.g. ). If I wanted to define a QGamma function such that $$\frac{\Gamma_q(n+1)}{\Gamma_q(n)} = \frac{q^n-q^{-n}}{q-q^{-1}}$$ what would be the (modified) definition of the respective QPochhammer functions?

Thanks in advance to everyone considering my questions!

P.S.: I'd also appreciate, if anyone can suggest some good literature on this topic in general.

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Notice that $[n]_q^{\text{symmetric}} = \frac{q^n - q^{-n}}{q-q^{-1}} = q^{1-n} \frac{1-(q^{2})^n}{1-q^2} = q^{1-n} [n]_{q^2}^\text{conventional}$. – Sasha Jul 20 '12 at 17:03
I already spent some time manipulating the expressions, trying to find a simple transformation $(q,n) \rightarrow (q^2, ? )$ and to me, your prefactor $q^{1-n}$ indicates that the two definitions are indeed inequivalent. Moreover I figured out, that $$\Gamma_q^{\text{symmetric}}(n) = \frac{\Gamma_{q^2}^{\text{conventional}}(n)}{f(n,q^2)}$$ where $f(n,q)\equiv q^{\frac{1}{4}(n-\frac{3}{2})^2}$. – André Jul 20 '12 at 17:56
I found a short paper which deals with both definitions and motivates them by different notions of q-deformed (bosonic) harmonic oscillators: – André Jul 21 '12 at 9:45

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