# Tag Info

1

Both limits are dealt with in a fair amount of detail in the work that originally defined the sandwiched entropies: On quantum Renyi entropies: a new generalization and some properties. In particular, you'll find the relevant results in section IV.C.

2

There are lots of different situations one can talk about from a cryptography perspective. But here's one that has a huge practical relevance: There are physical realisations of quantum computers which, individually, are limited in the number of qubits they can use. For example, ion traps. For the sake of argument, assume you can hold 10 qubits in each ion ...

1

From a brief skim of the paper, the variance (second moment) is defined at the start of section 3, and worked with extensively throughout that section. Higher moments are defined at the start of section 4.

6

Given $\rho$ and a fixed ensemble $\{ |\psi_i \rangle \}$ it might not be possible to write $\rho$ as $\sum_i p_i |\psi_i \rangle \langle \psi_i |$. For example, let $| + \rangle = \frac{1}{\sqrt{2}} (| 0 \rangle + | 1 \rangle )$. Then the state $|+\rangle \langle + |$ cannot be expressed as a convex combination in the ensemble $\{ | 0 \rangle, |1\rangle \... 2 There's no need to increase the dimension of$\newcommand{\calH}{{\mathcal H}}\calH_E$to$D+1$. All you care about is that$U$must send orthonormal vectors to orthonormal vectors. What the state$|0\rangle_E$actually represents is purely conventional. You could put an arbitrary (pure) state there, and get the same channel (as long as you also change the ... 2 As per N&C, fidelity is "analogous to the probability of doing the decompression correctly" (emphasis added). The goal is to do the operation correctly with 100% probability, which means the probability is 1. This is the desired limit of fidelity, so no error means the fidelity is 1. 3 Source of the problem The purported contradiction arises due to the use of incorrect assumptions for Klein equality $$S(\rho||\sigma) \ge 0.$$ The inequality does not require any particular relationship$^1$between the support of$\rho$and the support of$\sigma$. However, it does require that$\rho$and$\sigma$be states, i.e. unit trace positive ... 1 First, encode the bipartite ensemble$\gamma_{12}$into a CQ state$\omega$$$\omega_{XAB}=\sum_{x}p_{x}(x)|x\rangle\langle x|\otimes\rho_{AB}^{x}$$ now we can take the difference between $$H(A|B)\gamma_{12}-\sum_{x}p_{x}(x)H(A|B)_{\rho^{x}}$$ where$\rho_{x}\$ will be one of the density operators you have in your ensemble. Using the CQ state to rewrite the ...

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