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The conditional von Neumann entropy is a concave function: if $\rho$ and $\sigma$ are states of a pair of registers $(\mathsf{X},\mathsf{Y})$ and $\lambda\in[0,1]$ is a real number, then $$\mathrm{H}(\mathsf{X}|\mathsf{Y})_{\lambda\rho + (1-\lambda)\sigma} \geq \lambda\, \mathrm{H}(\mathsf{X}|\mathsf{Y})_{\rho} + (1-\lambda)\,\mathrm{H}(\mathsf{X}|\mathsf{Y}... 8 Geometric characterization (as any other characterization) of subsets of the quantum state space in relation with their locality and entanglement properties becomes very complicated as the number of qubits rises. The geometry of the space of negative conditional entropy two qubit states, which are also locally maximally mixed (Weyl states) is known; it is ... 5 Cirq uses numpy's pseudo random number generator to pick measurement results, e.g. here is code from XmonStepper.simulate_measurement: def simulate_measurement(self, index: int) -> bool: [...] prob_one = np.sum(self._pool.map(_one_prob_per_shard, args)) result = bool(np.random.random() <= prob_one) [...] Cirq ... 5 A channel \Phi is said to be degradable if there exists another channel \Xi such that \Xi\Phi is complementary to \Phi. The idea here is as follows. Suppose \Phi is a channel and \Psi is complementary to \Phi. If \Phi is applied to a state \rho, then the output of the channel is \Phi(\rho) (of course), while \Psi(\rho) represents ... 4 That seems to restrict the output probability distributions of all quantum circuits to rather high entropy distributions. The output of a typical randomly chosen quantum circuit is rather high entropy. That doesn't mean you can't construct circuits that have low entropy outputs (you can), it just means that picking random gates is a bad strategy for ... 4 My favourite way of proving that the Shannon entropy is minimized for a measurement in the qubit basis is through the notion of majorizaion (see Nielsen and Chuang or the book on Matrix Analyis by Bhatia for a formal definition). Specifically p and (1-p) is related to p' and (1-p') with the following relation \left(\begin{array}{c} ... 3 As @NorbertSchuch said in a comment, matlab has a function for taking the logarithm of a matrix: logm. In general, there is a standard method for calculating the function f(\sigma) of a matrix \sigma. You first diagonalise the matrix:$$ \sigma=UDU^\dagger, $$where U is a unitary and D is diagonal. We then say$$ f(\sigma)=Uf(D)U^\dagger, $$where ... 3 I'm not an expert with this sort of thing (i.e. there may be imperfections in this argument), but hopefully this will set you in the right direction... Consider \rho_{AB}=\rho_A\otimes |0\rangle\langle 0| and \sigma_{AB}=\sigma_A\otimes |0\rangle\langle 0|. It must be that S(\rho_A\|\sigma_A)=S(\rho_{AB}\|\sigma_{AB}). Now, your superoperator can be ... 3 You can always do that. Subspaces \text{Im}(V) and A\otimes |0\rangle have the same dimension, so there must be some unitary that translates one subspace to another. That is, \exists W \in \text{Unitary}(A\otimes B), W(\text{Im}(V)) = A\otimes |0\rangle. Now WV translates A to A\otimes |0\rangle. Since V is isometry WV is also isometry, hence ... 3 You don't need any additional conditions beyond those already stated in the question. That is, for any isometry V: A \rightarrow A\otimes B and any unit vector |\psi\rangle_B, there will always be a unitary U satisfying the equation in the question (simultaneously for every choice of \rho_A). One way to see this is to first pick any orthonormal ... 3 These are not really the definitions of classical and quantum capacity, as I will explain. Before doing that, let me adjust the notation being used slightly: let \Phi:\text{L}(\mathcal{X}) \rightarrow \text{L}(\mathcal{Y}) be the channel whose capacities we are interested in and let \Psi:\text{L}(\mathcal{X}) \rightarrow \text{L}(\mathcal{Z}) be a ... 3 The mutual information can be written in terms of the relative entropy, please see Nielsen and Chuang (the entropy Venn diagram figure 11.2). I am writing the equation in the question's notation:$$I(\rho^{AB}) = S(\rho^{AB}|\rho^{A} \otimes \rho^{B})$$The relative entropy can be estimated without full tomography. The procedure is described in Bengtsson ... 2 This is not much of an answer, but is probably too long for a comment... I don't believe that there's a canonical way of doing this. You'd be best off understanding why you're asking the question, and what you want to get out of it. From there, you tailor how you're going to measure it. But multipartite entanglement is a really messy problem, even just for ... 2 After some further consideration I think it's quite clear that the only probability mass function evaluated in the computation of \mathcal{F}_{\text{XEB}} is that of the classically computed ideal distribution, denoted P(x_i) in the main paper. This leads me to the conclusion that the phrasing of the following excerpt from section IV.C of the ... 2 As an initial matter, I think the Supplementary Information (linked in some other answers on this sight) has a significant amount of discussion on \mathcal{F}_{XEB}. However, as I understand it (misunderstandings are my own): There is indeed a concentration of outputs from a random quantum circuit, away from a state wherein the square of the coefficients ... 2 It turns out to be a novice mistake. I was using matlab and this log is elementwise, as @Ahusain pointed out. We must take the matrix logarithm in Matlab which is denoted by logm. Then the calculation becomes:$$-\text{trace}(\rho \log m (\rho)) = \text{NaN}.$$The reason is, we have to define 0 \times \log (0) as 0 instead of \text{NaN} which is ... 2 The entanglement entropy (what you call "von Neumann entropy") is a good measure for entanglement of pure states in the asymptotic setting, i.e. when one is dealing with many copies. However, it is not a good measure for mixed states. Distillable entanglement and entanglement cost are entanglement measures which apply to both pure and mixed states. ... 2 Let's write$$ p'=\frac12+m\cdot n\frac{2p-1}{2}, $$and assume without loss of generality that p>\frac12, which also means that p'>\frac12. Note the binary entropy function h(p') is symmetric about p'=\frac12, and is monotonically decreasing for p'>\frac12, meaning that we want to make p' as large as possible in order to minimise h(p')... 2 Suppose that \mathsf{X} is a register that can store each possible choice for x, as a classical state, while \mathsf{Y} is a register that can store each possible state \rho_x. It is then natural to associate the classical-quantum state$$ \rho = \sum_x p_x |x\rangle \langle x| \otimes \rho_x $$with the ensemble \{(p_x,\rho_x)\}. Now try taking ... 2 Posting an answer because I realised what my issue was: What I didn't realise then: When a density matrix is written in any basis, the diagonal elements correspond to the probabilities of the density matrix landing on the basis states of that basis. So, if in some basis formed by vectors |x_1\rangle, |x_2\rangle, |x_3\rangle, |x_4 \rangle, my density ... 2 It means that if you lose information from your system, that information must have been transferred to the system's surroundings. This shows up as an increase in the entropy in the surroundings. This is directly related to the 2nd law of thermodynamics which says the entropy of an isolated system is always increasing. See Wikipedia: Entropy in thermodynamics ... 2 Yes, unitarity preserves eigenvalues. This is because the definition of eigenvalues is that any Hermitian matrix H can be brought into diagonal form by a unitary V,$$ VHV^\dagger=D, $$and the diagonal elements of D are the eigenvalues. So, now consider UHU^\dagger. This can be diagonalised by a unitary VU^\dagger into the same matrix D:$$ (VU^\...

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You are running into problems because $\rho$ is not a density operator. A mixed state density operator has $\text{tr}(\rho^2) < 1$, but even a mixed state density operator must have $\text{tr}(\rho)=1$. This is necessary because \$\text{tr} (\rho) = \sum \limits_i p_i \, \text{tr}\left(\vert \psi_i \rangle \langle \psi_i \vert \right) = \sum \limits_i ...

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We have, \begin{aligned} S &= - \operatorname { Tr } \left( \varrho \log _ { 2 } \varrho \right) = \log _ { 2 } \left( \frac { \left| \gamma _ { B } \right| ^ { \left( 2 \left| \gamma _ { B } \right| ^ { 2 } \right) / \left( \left| \gamma _ { B } \right| ^ { 2 } - 1 \right) } } { 1 - \left| \gamma _ { B } \right| ^ { 2 } } \right) = \...

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