Do we know the limits of the quantum Tsallis entropy?

Question

From the two main generalizations of the von Neumann entropy:

\begin{equation} S(\rho)=-\operatorname{Tr}(\rho \log \rho) \end{equation}

meaning Rényi:

\begin{equation} R_{\alpha}(\rho)=\frac{1}{1-\alpha} \log \operatorname{Tr}\left(\rho^{\alpha}\right), \alpha \in(0,1) \cup(1, \infty) \end{equation}

and Tsallis:

\begin{equation} T_{q}(\rho)=\frac{1}{1-q}\left(\operatorname{Tr}\left(\rho^{q}\right)-1\right), q \in(0,1) \cup(1, \infty) \end{equation}

we know that in both cases the limit to 1 for the entropic parameters gives us back the von Neumann entropy:

\begin{equation} S(\rho)=\lim _{q \rightarrow 1} T_{q}(\rho)=\lim _{\alpha \rightarrow 1} R_{\alpha}(\rho). \end{equation}

In the case of the quantum Rényi entropy we also know its two limits:

\begin{equation} R_{0}(\rho)=\lim _{\alpha \rightarrow 0} R_{\alpha}(\rho)=\log \operatorname{rank}(\rho) \end{equation}

and

\begin{equation} R_{\infty}(\rho)=\lim _{\alpha \rightarrow \infty} R_{\alpha}(\rho)=-\log \|\rho\|. \end{equation}

I couldn't find anything similar for the quantum Tsallis case? Is there something that is considered trivial?

Answer 1

Well for $q \to 0$ we have $$ \lim_{q \to 0} T_q(\rho) = \mathrm{rank}(\rho) - 1. $$ For $q \to \infty$ it's not really interesting as $$ \lim_{q \to \infty} T_{q}(\rho) = 0. $$ For the second result note $\lim_{q\to\infty} \mathrm{Tr}[\rho^q] \leq \lim_{q \to \infty} \mathrm{rank}(\rho)\lambda_{\max}(\rho)^q \leq 1$ as $\rho$ is a quantum state. And so the numerator is finite but the denominator blows up to $-\infty$.