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$\rho_{AB}$ is called separable if it can be written as $$\rho_{AB}=\sum p_i \rho_{A,i}\otimes \rho_{B,i}\ .$$ You can now further decompose $\rho_{A,i}=\sum_{x} q_{x} \vert\alpha_x\rangle\langle \alpha_x\vert$ and $\rho_{B,i}=\sum_{y} r_{y} \vert\beta_y\rangle\langle \beta_y\vert$; then, $$\rho_{AB}=\sum_{i,x,y} p_i q_{x} r_y ... 3 Naively, the collection of data which describes a model of anyons is not its fusion rules, but rather its modular data - that is, a pair of matrices S and T which generate a representation of the (modular) group SL(2,\mathbb Z). The (diagonal) matrix T encodes the mutual statistics of quasi-particles - that is what happens when you exchange two ... 3 The way memory would be organized is that if there are N qubits, each one would be a piece of physical substance, possibly quite small, and possibly in multiple copies. For example, quantum computation has been done using caffeine molecules, each containing a small number of qubits, but with many many molecules, all in the same superposition. When you ask ... 3 Entanglement witnesses are a way to probe entanglement through measurements. An entanglement witness is a linear operator W which is constructed such that \mathrm{tr}[W\rho_s]\ge0 for any unentangled state \rho_s. Thus, if we find that \mathrm{tr}[W\rho]<0, we know that \rho is entangled. Moreover, for any entangled state \rho we can ... 3 In principle: I don't know. Preparing the states would be hard. In practice: of course not. Consider the fractional interaction rates. We routinely measure atmospheric neutrino that have come right through from the other side of the planet. 2 It's therefore a two-qubit operation. 2 Why the outcome after the controlled-phase and Hadamrad is equal to:$$\alpha |++\rangle + \beta |--\rangle = (|0\rangle \otimes H|\psi\rangle+|1\rangle \otimes XH|\psi\rangle)/\sqrt{2}$$Let us label the qubits A and B for clarity. The initial state for circuit (11) is$$ |\psi_A \rangle \otimes |+_B\rangle = \left( \alpha |0_A\rangle + \beta ...

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Is the basis in which the density matrix diagonal an orthogonal basis? Note that if the density operator is $1/N$ times the identity, then it is diagonal in every basis. On the other hand, every density operator on a finite dimensional Hilbert space is diagonal in an orthonormal basis: this follows from the spectral theorem, which asserts that this is ...

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You store an assembly of qubits by storing a bunch of single-qubit-systems next to each other, just like in a classical computer. The laws of physics take care of storing the entanglement for you. In general QM does not give us direct access to each amplitude stored in the overall wavefunction, but only allows us to define certain observables which mix them ...

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I doubt you'll be able to get useful bounds without restricting the class of channels. Consider the following quantum channel (for arbitrary dimension $d,n$): $$T:\mathbb{C}^{d\times d}\to \mathbb{C}^{n\times n}: \rho \mapsto \operatorname{tr}(\rho) \sigma$$ where $\sigma$ is any quantum state. Using the Choi-Jamiolkowski isomorphism, you can easily see ...

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Let $\vert\psi_i\rangle$ be the 4 Bell states and $\vert\phi_k\rangle$ the 4 "BB84" states. Then, define a POVM measurement with elements $$P_{ik\pi} = \tfrac{1}{12}\vert\psi_i\rangle\langle\psi_i\vert\otimes \vert\phi_k\rangle\otimes \langle\phi_k\vert\ ,$$ where $\pi$ is a permutation of the three qubits (i.e., the three possible choices for the ...

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It is always possible to find a purification of a Gaussian state in terms of a Gaussian state. This is most easily seen in terms of covariance matrices. A convenient way of doing so is to bring your covariance matrix into normal form by sympectic transformations. This decouples the system into a set of decoupled normal modes at some temperature, which can ...

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Yes, it is possible to distinguish between $|\phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle +|11\rangle)$ and $|\phi^-\rangle = \frac{1}{\sqrt{2}}(|00\rangle -|11\rangle)$. All that is needed is a local change of basis, on both sides, to states $$|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \;\; \text{and} \;\; |-\rangle = ... 1 The no cloning theorem just says that if you have a state (and you don't know what it is) you can't duplicate it. Nothing says Alice can't prepare the same state twice and send them both to Bob. 1 For linear operators, the support usually denotes the space which is orthogonal to the kernel (equivalently, the space spanned by the columns of the matrix). Density operators are linear operators, and thus it is used in this sense in the papers you cite. See also this question at math.se or this book from a google search "support of a linear map" 1 Even without sending classical infotmation - the state was completley teleported. Unfortunatley, by teleportation identity - after teleporting state  |\psi\rangle at A you get a state U^{\mu \nu}|\psi\rangle at B (where U^{\mu \nu}  is a unitary matrix which depend on mesurement results at A), i.e. the state was teleported to B and transformed to ... 1 Not quite. Your typical qubit is a spin-\frac12 system, where the spin around any axis in 3D will be measured as either "up" or "down" along that axis. The observable matrices are the Pauli matrices; the "computational basis" |0\rangle,|1\rangle is therefore the z-axis. Now if you wrote i|00\rangle + i|11\rangle, that would be an "overall phase ... 1 I'll cover some of the technical stuff. Let us denote the 1st qubit as A (for Alice) and the 2nd as B (for Bob). The initial state of the 2 qubits is$$ |\psi_0\rangle = |0_A 0_B\rangle $$The Hadamard gate applied to qubit A produces$$ |\psi_1\rangle = H_A \otimes I_B |0_A 0_B \rangle = \frac{1}{\sqrt{2}}(|0_A 0_B\rangle + |1_A 0_B \rangle ) = ...

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Since the question is "can you recommend a book that talks about these topics with minimal math," the answer is no. It would be even more confusing to describe quantum information and quantum computing without math than with the math, as the concepts aren't as intuitive as say general relativity, which can fairly effectively be described with mostly words. ...

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The underlying idea of the construction is that you are given a set of POVM measurement operators $M_1,\dots,M_n$ satisfying $\sum_{i=1}^n M_i^\dagger M_i\le 1$ which you want to complete to a complete POVM by adding an extra element (=outcome) $M_0$. Thus, $M_0$ needs to satisfy $$M_0^\dagger M_0 + \sum_{i=1}^n M_i^\dagger M_i = 1\ ,$$ i.e., $M_0$ must be ...

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