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Operator $\rho$ is not a tensor product, it's a sum of tensor products $$p_1|1\rangle\langle 1| \otimes \rho_1 + p_2|2\rangle\langle 2| \otimes \rho_2 + \dots + p_d|d\rangle\langle d| \otimes \rho_d.$$ This is not the same as $$\big(\sum_ip_i|i\rangle\langle i|\big) \otimes \big(\sum_i\rho_i\big),$$ so your expansion isn't correct. Also in general $S(... 4 That seems to restrict the output probability distributions of all quantum circuits to rather high entropy distributions. The output of a typical randomly chosen quantum circuit is rather high entropy. That doesn't mean you can't construct circuits that have low entropy outputs (you can), it just means that picking random gates is a bad strategy for ... 3 As @NorbertSchuch said in a comment, matlab has a function for taking the logarithm of a matrix: logm. In general, there is a standard method for calculating the function$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$...

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 $\text{tr} (\rho) = \sum \limits_i p_i \, \text{tr}\left(\vert \psi_i \rangle \langle \psi_i \vert \right) = \sum \limits_i ... 2 Short answer You are getting that error because your example does not use a valid (normalized) statevector. If you remove the decimals=3 kwarg where you call result.get_statevector it will work. Long Answer The Von-Neuman entropy function in the qiskit.quantum_info works with either Statevector or DensityMatrix object inputs, or inputs that can be ... 2 You've computed$\rho^{AC},\rho^{BC}$incorrectly$-$they are matrices of size$8\times 8$, also they are not equal to$I/8$. Nevertheless their entropy is equal, i.e.$S(A,C) = S(B,C)$. So$|\psi \rangle$is indeed a counterexample to the statement. The inequality$S(A,B)\geq S(A|C)-S(B|C)$, which is equivalent to $$S(A,B) + S(B,C) \geq S(A,C)$$ is ... 1 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$I_{\max}(\rho_{AB}) = 1$... 1 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=\... 1 Considering your question is not specific, you can take a look at Chapters 10 (classical entropy) and 11 (quantum entropy) of this book. Quantum entanglement entropy as we know it comes from classical Shannon theory. Claude Shannon introduced entropy in order to quantify information. You can read his paper here. 1 The entropy of a system depends on your state of knowledge about it. If you know that a state is pure, then its entropy is zero. If you "tell$A,B,C$that the overall shared state is pure", then they know as much and will, therefore, say that the entropy of the overall state is$0\$ (note that this is essentially a tautological statement). However, for them ...

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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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