# Do we know the limits of the quantum Tsallis entropy?

From the two main generalizations of the von Neumann entropy:

$$$$S(\rho)=-\operatorname{Tr}(\rho \log \rho)$$$$

meaning Rényi:

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

and Tsallis:

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

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

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

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

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

and

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

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

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 $$-\infty$$.
• 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. Commented Dec 16, 2020 at 12:58