# Quantum q less than 1

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

Let $q<1$. The set $\overline{q^{\mathbb{Z}}}=\{0, \ldots, q^{2}, q^{1}, 1, q^{-1}, q^{-2}, \ldots \}$ of quantum numbers is a time scale.

 Generic element $t\in \mathbb{T}$: For some $n \in \mathbb{Z}, t =q^n$ Jump operator: $\sigma(t) = \dfrac{t}{q}$ Graininess operator: $\begin{array}{ll} \mu(t) &= \dfrac{t}{q} - t \\ &= \dfrac{t}{q} (1-q) \end{array}$ $\Delta$-derivative: $f^{\Delta}(t) = \dfrac{f(\frac{t}{q})-f(t)}{\frac{t}{q}(1-q)}$ $\Delta$-integral: $\begin{array}{ll} \displaystyle\int_s^t f(\tau) \Delta \tau &= \displaystyle\sum_{k=\log_q(s)}^{\log_q(t)-1} q^{k-1} (1-q) f(q^k) \\ \end{array}$ Exponential function: $\begin{array}{ll} e_p(t,s) &= \exp \left( \displaystyle\int_{s}^{t} \dfrac{1}{\mu(\tau)} \log( 1 + p(\tau) \mu(\tau) ) \Delta \tau \right) \\ &= \exp \left( \displaystyle\sum_{k=\log_q(s)}^{\log_q(t)-1} \mu(q^k) \dfrac{1}{\mu(q^k)} \log(1 + p(q^k)\mu(q^k)) \right) \\ &= \exp \left( \displaystyle\sum_{k=\log_q(s)}^{\log_q(t)-1} \log(1 + p(q^k)\mu(q^k)) \right) \\ &= \displaystyle\prod_{k=\log_q(s)}^{\log_q(t)-1} 1 + p(q^k)q^{k-1}(1-q) \end{array}$