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Let $D_{\alpha}(\rho\|\sigma):= \frac{1}{\alpha - 1} \log \mathrm{Tr}[\rho^\alpha \sigma^{1-\alpha}]$ be the Petz-Rényi divergence for $\alpha \in (0,1)\cup(1,\infty)$. Note that for $\alpha \in (0,1)\cup(1,2]$ this quantity satisfies the data processing inequality $$ D_{\alpha}(\rho\|\sigma) \geq D_{\alpha}(\mathcal{E}(\rho) \| \mathcal{E}(\sigma)), $$ ...


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Yes, you can formulate the smooth max-entropy as an SDP. The author of the book you linked notes this when they explain how to derive the SDP for the smooth min-entropy that you reference on page 91. In particular they say that the smoothing constraint $\tilde{\rho}_{AB} \in B^\epsilon(\rho_{AB})$ can be reformulated as the triple of constraints $$ \mathrm{...


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As far as I'm aware there isn't much of a meaningful connection. The corresponding entropy for $D_{\max}$ is the min-entropy (written $H_{\min}$ or $H_{\infty}$). It measures a sort of `worst case' uncertainty whereas the Shannon or von Neumann entropies measure an average uncertainty. To answer your first question: the quantum relative entropies or ...


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No, this is not possible. Consider $\rho_1 = \sigma_2 = \vert 0\rangle\langle 0 \vert$ and $\rho_2 = \sigma_1 = \vert 1\rangle\langle 1 \vert$. Then, $$D_{\max}(\rho_i\|\sigma_i) = \infty\quad \text{for } i = 1,2.$$ Let $p_i = (1/2, 1/2)$ and you see that $D_{\max}(\rho\|\sigma) = 0$.


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Assuming everything is finite dimensional. For $S_0$ we have $$S_0(\rho) = \log \mathrm{rank}(\rho).$$ It's pretty straightforward to see this is not continuous. Take $\rho_{\epsilon} = \epsilon |0\rangle \langle 0 | + (1-\epsilon) |1\rangle \langle 1 |$. Then for all $0 < \epsilon < 1$ we have $S_0(\rho) = \log 2$ but for $\epsilon \in \{0,1\}$ we ...


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There is a problem in the derivation you presented, since $\rho \leq \lambda \sigma$ is only equivalent to $\sigma^{-1/2} \rho \sigma^{-1/2} \leq \lambda I$ when $\sigma$ is invertible (or at least when $\mathrm{supp}(\rho) \subset \mathrm{supp}(\sigma)$, so that you can restrict the space to $\mathrm{supp}(\sigma)$ instead of the whole Hilbert space). The ...


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Unfortunately $D_{\max}$ is not a continuous function and so functions built from it tend not to be continuous. For example consider consider the two states $$ \rho_{AB} = |00 \rangle \langle 00|, $$ and $$ \tau_{AB}(\epsilon) = (1-\epsilon) |00 \rangle \langle 00 | + \epsilon | 11\rangle \langle 11 |. $$ A quick calculation gives $I_{\max}(\rho_{AB}) = 0$ ...


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Can someone provide an example of a state $\rho_{AB}$ for which $\sigma^\star_B \neq \rho_B$? Why not start very easily, with a separable state such as $$ \rho_{AB}=\left(p_0|0\rangle\langle 0|\otimes \tau_0+p_1|1\rangle\langle 1|\otimes \tau_1\right) $$ where $\tau_0$ and $\tau_1$ are different (normalised) single-qubit density matrices. We have that $$ I=\...


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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 ...


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The term max-entropy in quantum information is reserved for the following definition No it's not, many papers like https://arxiv.org/abs/0803.2770 use the term to refer to the quantity $\log \mathrm{rank}(\rho)$. Your first definition comes from the Rényi entropy of order 0, while the second one comes from the Rényi entropy of order $\frac{1}{2}$, and you ...


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