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For classical expanders, the spectral definition can be expressed in terms of the second-smallest eigenvalue of the graph Laplacian, which can be thought of as the minimum of a quadratic form over all unit vectors orthogonal to the all-ones vector. If we restrict this minimization to vectors of the form (a,a,...,a, b,b, ..b), then this yields the edge ...

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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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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 ) = ... 0 You have a serious serious misconception. The misconception is that you write as if each object has its own spin state. This is wrong. That only happens sometimes, specifically, when they are not entangled. The state could start out like |0\rangle\otimes|0\rangle=|00\rangle. And then each has a spin state, namely |0\rangle. But later they can be in a ... 0 If not (i.e it does not yield to an eigen equation), then we say that the list of possible values on measuring the particular property will be the set of all basis vectors corresponding to that operator, with respective probabilities P^2 where P is the projection of the state vector into the corresponding basis vector. That can be wrong. If the ... 0 When you do a measurement you most certainly do change the state vector (in the Schrödinger picture). The state ends up being projected onto an eigenspace of the corresponding operator. A quantum gate, also changes the state, but unlike a projection it should be a change represented by a unitary operator. 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 ... 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 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 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" 0 Out of the many quantum-mechanically possible states of an oscillator (be it a mechanical one or light waves), the ones we almost exclusively observe are the coherent states. In a way, they are the states where uncertainty is evenly distributed, such that every uncertain quantity scales as \sqrt{N} for N quanta (e.g. photons or energy quanta in an ... 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. 0 The short answer is no, this experiment would not lead to nonlocal signalling, as long as quantum mechanics holds. The reduced density matrix for B, after tracing over A, is completely mixed if you start with a maximally entangled state, irrespective of what Alice is doing (and Alice's density matrix is mixed too, so she won't see interference using her ... 1 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 = ...

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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 ...

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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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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 ...

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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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$\rho_{AB}$ 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, $$\rho_{AB}=\sum_{i,x,y} p_i q_{x} r_y ... 2 Is the basis in which the density matrix diagonal an orthogonal basis? Note that if the density operator is 1/N times the identity, then it is diagonal in every basis. On the other hand, every density operator on a finite dimensional Hilbert space is diagonal in an orthonormal basis: this follows from the spectral theorem, which asserts that this is ... 0 In Bayesian probability there is some objective truth that can be discovered with higher and higher certainty if we learn more information and update our distribution. The distribution does not fully describe the system. Rather it helps us to guess what the system might look like. In the Copenhagen interpretation of quantum mechanics there is no objective ... 1 It is always possible to find a purification of a Gaussian state in terms of a Gaussian state. This is most easily seen in terms of covariance matrices. A convenient way of doing so is to bring your covariance matrix into normal form by sympectic transformations. This decouples the system into a set of decoupled normal modes at some temperature, which can ... 1 Let \vert\psi_i\rangle be the 4 Bell states and \vert\phi_k\rangle the 4 "BB84" states. Then, define a POVM measurement with elements$$ P_{ik\pi} = \tfrac{1}{12}\vert\psi_i\rangle\langle\psi_i\vert\otimes \vert\phi_k\rangle\otimes \langle\phi_k\vert\ , $$where \pi is a permutation of the three qubits (i.e., the three possible choices for the ... 0 Particles A and B are entangled, quantum information of particle A is transmitted to particle B via quantum teleportation, If the particles are already entangled they don't have single particle information, they have a joint state. Your options are: 1) interact with one to produce random results thus forcing the other to have random results correlated ... 2 I doubt you'll be able to get useful bounds without restricting the class of channels. Consider the following quantum channel (for arbitrary dimension d,n):$$ 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 ... 0 The no cloning theorem (which follows directly from linearity of time evolution) says you can't make a copy so you can't do that. But that is not the point. The point is that you can make a state and even of you made a thousands states just like, the state itself will not give good results for position and good results for velocity. There are states that ... 4 The fact that the overall phase factor does not matter means that we can choose it to be whatever we like. This gives us an extra constraint (even if is one we choose arbitrary) and so reduces the number of degrees of freedom by 1. For example, given that |a_u|^2 + |a_d|^2 = 1 we can write our coefficients as \begin{align} a_u = &\cos\theta ... 8 An arbitrary normalized quantum state on two dimensions can always be written as |\psi⟩=e^{i\alpha}\left(\cos\theta|\uparrow⟩+e^{i\phi}\sin\theta|\downarrow⟩\right)  without loss of generality. The phase factor $e^{i\alpha}$ has no bearing on experiment, as all measurements will be proportional to $⟨\psi|\propto e^{-i\alpha}$. This means that you can ...

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The paradox is not solved yet because so far Hawking has only given a 1-page proposal (The Information Paradox for Black Holes ) on how to solve the problem. The details are still missing, but they will appear in a future work with M.J. Perry and A. Strominger. Basically the proposal is that the information is conserved, because particles that fall into the ...

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