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I don't think you need to know quantum physics to understand quantum computing - similarly to how you don't think about the hardware implementation of the classical computers when you write high-level code for them. The field of quantum computing has grown to the point where one cannot really teach all of it in one course, so different approaches to ...

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If you can only use unitary control gates on the second and third qubits, you cannot change the density matrix on the first and last qubits except by single qubit operations. This density matrix is the completely mixed state: $$\frac{1}{4}\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\otimes \left(\begin{array}{cc}1&0\\0&1\end{array}\right)... 3 I believe it is possible to study Quantum Mechanics by studying Quantum Computing. A qubit is a simplest quantum system showing non-classical behavior (superposition of basis states). It is very logical to start studying Quantum Mechanics from the simplest quantum system, and then move to more complex multiqubit systems. If you need Quantum Mechanics to ... 3 Quantum internet is a theoretical concept of global (or at least wide) quantum net. It would allow to interconnect many quantum processors (or computers) and enable them to communicate each other. How is a entanglement related: Quantum entanglement enables to employ superdense coding and quantum teleportation for a communication. The former is used for ... 2 You could probably reach the same conclusions by identifying that tr(\rho \sigma) is just the expectation value of \rho under the mixed state \sigma, but let's do it explicitly: for \rho = \sum_i p_i |\psi_i\rangle \langle \psi_i| and \sigma = \sum_i q_i |\phi_i\rangle \langle \phi_i|, we have: tr(\rho \sigma) = \sum_{ij} p_i q_j tr(|\psi_i\... 2 You've computed \rho^{AC},\rho^{BC} incorrectly - they are matrices of size 8\times 8, also they are not equal to I/8. Nevertheless their entropy is equal, i.e. S(A,C) = S(B,C). So |\psi \rangle is indeed a counterexample to the statement. The inequality S(A,B)\geq S(A|C)-S(B|C), which is equivalent to$$ S(A,B) + S(B,C) \geq S(A,C) $$is ... 1 If you allow for measurements and communication of measurement outcomes, and joint operations on qubits 2 and 3, parties 2 and 3 can prepare the desired state locally and teleport it to 1 and 4. 1 Okay, this is a rather subtle situation, but I think I've figured it out. The key is to be very careful about which mathematical results about Hermitian operators do and do not hold for generic operators. Let H represent an arbitrary Hermitian matrix, N an arbitrary normal one, D be a generic diagonalizable matrix, and M an arbitrary matrix, all ... 1 Square root is not differentiable at 0, so that cyclic property cannot be applied While \rho\sigma has the same non-negative eigenvalues as \sqrt\rho\sigma\sqrt\rho, it's not self-adjoint. Non self-adjoint matrices are not diagonalizable in general, so the square root \sqrt{\rho\sigma} can be not well-defined (see edit below). Anyway, \text{Tr}(\... 1 Number of shots: Number of shots means how many times an algorithm is run to get a probability distribution of results. Experiment is repeated: This means how many times an experiment is repeated with particular number of shots. Imagine, you repeated an experiment 100 times and you have 1,024 shots. In each experiement repetition some metrics is calculated ... 1 So first, let's define a bit your notations. I guess (correct me if I'm wrong) that you consider Bob honest, and that what you denote by \Psi_{a_k,b_k} is the BB84 qubit in basis \{0,1\} if b_k = 0, and in basis \{+,-\} if b_k = 1, whose "value" bit is a_k, i.e.:$$\Psi_{a_k,b_k} = H^{b_k}X^{a_k}|0\rangle Then, Bob will measure in basis $b'_k$...

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I think your misunderstanding is in the wording of the secret key rate; The PLOB-Bound offers an asymptotic, ultimate-upper bound on the secret key rate per use of a lossy bosonic channel. This bound on the secret key rate is computed as a regularisation, where one considers the infinite limit of $n \rightarrow \infty$ transmissions across the channel in ...

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