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}


\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?


1 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 $-\infty$.

  • $\begingroup$ Perfect, thanks. Are these considered trivial or there is some paper that proves them as for the Renyi case? $\endgroup$
    – jmstf94
    Dec 16, 2020 at 12:23
  • $\begingroup$ Well $S_0(\rho)$ is essentially the rank of $\rho$ so it will certainly turn up wherever you have the rank of $\rho$. And $S_{\infty}(\rho)$ is definitely something I would refer to as trivial, it doesn't depend on the quantum state. I don't know much about Tsallis entropies though as I've never worked with them before. $\endgroup$
    – Rammus
    Dec 16, 2020 at 12:58

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