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Suppose you have a standard spin qubit: spin "up" is $|1\rangle$ and spin "down" is $|0\rangle$ . Your qubit is in some storage location which does not couple the two spin directions to each other or have any different energy, you store $|\psi\rangle = a |0\rangle + b |1\rangle$ in this qubit. ` Now some proton flies by, or a minor fluctuation in the Earth's ...

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The co-efficient $\frac{1}{4}$ simply ensures that correct normalization so that the probabilities of the different pure states composing the classical mixture (the "mixed state") in question sum to unity: recall that: $$\rho = \sum\limits_j p_j |\psi_j\rangle\langle \psi_j|$$ and when the pure states are normalized properly, we have ${\rm ... 1 Nielsen and Chuang are referring to a scheme known as measurement-based quantum computation, for which many good learning resources are a short google search away. The idea is that you prepare a large, highly entangled state involving a large number of qubits at the start of the experiment. You then proceed to make measurements on the state, potentially ... 1 Alas, it was very simple! $$\delta U_j =-i\Delta t \delta u_k(j)\mathcal{H}_kU_j \, . \tag{1}$$ $$\frac{\delta U_j}{\delta u_k(j)} =-i\Delta t \mathcal{H}_kU_j \, . \tag{2}$$ Now, looking at$\Phi_o$, we see the only terms that actually have a$u_k(j)$dependence are$U_j$and$U_j^{\dagger}$, hence we have the following via the product rule: ... 1 Normally$H(X|Y)$means conditional entropy. In this case I don't think there is any generally accepted definition of Renyi counterpart. There is, however, a recent Master thesis which lists some possibilities including a new propositions by the author, which seems to be quite reasonable: http://web.math.leidenuniv.nl/scripties/MasterBerens.pdf If you ... 0 Let's start with questions 1, for which the answer is "no". An isometric quantum channel$T_{A\to A^{\prime}}$is a channel$T_{A\to A^{\prime}}:\mathcal{B}(\mathcal{H}_A)\to\mathcal{B}(\mathcal{H}^{\prime}_A)$such that$T_A(\rho)=U\rho U^{\dagger}$with$U^{\dagger}U=1$and$UU^{\dagger}=P$with some projection onto a subspace. The conditions on$U$mean ... 3 Since state$\sigma$is not in thermal equilibrium I don't think one can use your definition of "thermodynamical" entropy. In fact, one should instead use Von Neumann entropy, which is a correct measure of statistical (so not quantum!) uncertainty. There is no other "classical" or "thermodynamical" entropy in quantum systems. As you mentioned, for a thermal ... 0 Looking at simple cases might help. You can have a vector$\vert\Psi\rangle$in a Hilbert Space and represent it as a pure state such as$\vert\Psi\rangle\langle\Psi\vert$and you could do the same for the vector$\vert\Phi\rangle$in a Hilbert Space and represent it as a pure state such as$\vert\Phi\rangle\langle\Phi\vert.$You could also imagine that ... 1 If we consider local operators$M_A$and$N_B$acting on Alice's and Bob's part, respectively, then it holds that$[M_A,N_B]=0$, i.e., we can use this property in proofs involving local operations. Note that conversely, however, commutativity need not imply locality (to start with, there need not even be a tensor product structure). 1 Here is a reference for the continuous versions: http://arxiv.org/abs/1402.2966 -2 This is the main point which disallows information exchange. No one had yet discovered the way to set a state on one side and to get a correlated information on the other side. If it was possible, many protocols would allow communication. edit bis : If you cannot set the information to send, even if the two sides are correlated, it's not sending ... 7 It's not completely clear what you're asking, but I can make one thing clear: quantum mechanics was not developed with the specific aim of correctly describing the precise amounts to which CHSH inequalities can be violated. Quantum mechanics was developed in the 1920s and early 30s. Bell published his first paper on Bell inequalities in 1964. The first ... 1 The following calculations are in the basis of$ (|00\rangle,|01\rangle,|10\rangle,|11\rangle $) CNOT1 - when the first qubit is control -$CNOT1|00\rangle = |00\rangleCNOT1|01\rangle = |01\rangleCNOT1|10\rangle = |11\rangleCNOT1|11\rangle = |10\rangle$thus it's matrix representation is $$CNOT1 = \begin{pmatrix} 1 & 0 & 0 & 0\\ ... 4 http://arxiv.org/abs/quant-ph/9607007 discusses necessary conditions on T (more precisely, on its singular values) for \rho to be positive. They don't seem to derive sufficient conditions, however. The basic idea is that one can perform a rotation U_A and U_B on the two qubits, respectively, which correspondingly transforms r\mapsto O_Ar, ... 2 It is the misapplication of QM to problems that may or may not benefit from such application, but which produces no experimentally testable theory. In many cases not even a mathematical model is produced, but "all is handwaving explanation and no proof". -1 yes, it fulfils both criteria. But , AFIK, it uses a quantum device effectively , not yet an universal quantum computer. It's not sure that all agree the same definition of an idealized quantum computer. A clear and functionnal definition is needed. The first promise of the quantum computer theory was the infinite parallelism. If this goal is improbably ... 1 Integration over pure quantum states usually refers to the Haar measure, i.e., the unitarily invariant measure. Vaguely speaking, you assign the same volume (=weight in the integral) to any two set of states which are related by an arbitrary unitary rotation U. In the case of one qubit, this is equivalent to integrating over the Bloch sphere; i.e., we ... 0 While we know that factoring can be solved in polynomial time on a quantum computer, we do not have a proof that factoring cannot be solved in polynomial time on a classical computer. (Indeed, this would be an extremely strong result as it would in particular show that P\neNP.) See also ... 0 The action of η on the pure states is enough because the density operator is a convex combination of pure state projection operators and because η is linear, so all the contributions mixed states add to the average of <η> end up evening out. I think the arrow → was not meant to express an equality, but rather something about the infinitesimal behavior of ... 3 Unitary dynamics means that the evolution of quantum states are describen by unitary operators. Physically this means that there is no dissipation in the system. This is important, because inherently quantum phenomena needed for quantum computation eg. entanglement and quantum superposition are only observable for a long period if no dissipation or other ... 0 1.Is there anyway whatsoever - directly or indirectly - to learn about the path of the atom before the final screen? Yes, via weak measurement. See the physicsworld article In praise of weakness. The wave nature of matter means it doesn't make much difference whether we're using a photon or an electron or a helium atom. For an analogy, think of normal ... 0 When the mean number of photons is huge, the Heisenberg uncertainty becomes negligible and "disappears" (formally it looks like \hbar\to 0). Thus, such a coherent state becomes quite classical one. 5 Quantum "teleportation" is a really strange phenomenon, but most popular-science descriptions fall a little flat, indeed. Here's the basic setup for every quantum teleportation experiment: Two particles A and B are entangled and separated. Two more particles, A2 and B2, are put into two special states: A2 is put into some complicated wavefunction state, ... 1 It is not impossible to send information using quantum entanglement. What is impossible is to send information faster than light, violating causality. The quantum teleportation protocol includes a classical step (sending information about one's measurament, if I'm not mistaken), where things are restricted by the usual laws of relativity. 2 In general, the integral$$ V := \int \mathrm{d} \mu = \int 1 \mathrm{d}\mu$$is the integration of the identity over the space the measure$\mu$is defined on, and should be intuitively understood as the volume of the space with respect to the measure. (This is usually only finite for compact spaces.) Normalizing the measure means sending$\mu \mapsto ...

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In this answer we will stick totally to the Copenhagen interpretation of quantum mechanics. Before the system is measured, there is nothing physical (a consequence of realism not playing a part under this interpretation). There is only our mathematical, non-physical description of the possible measurement outcome called the wavefunction. On measurement ...

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The conditional probability talks about things that have happened, so it's an objective reality (ignoring the questions of the objective reality's existence). It's unimportant if you know it's happened or not. This of course changes a bit in experimental reality, where you (presumably) order your events $a$ based on your knowledge of the preconditions ...

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This paper talks about factoring with 1.5n qubits, so you'd need 3072 qubits to factor a 2048 bit semiprime. It says you might be able to do it faster. I'm willing to bet you can't do it with less than n qubits since you need to store the number. You might be able to do it with one or two fewer qubits, since the first and last bits will always be one, but ...

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I'd say that "device independence" is only interesting for cryptographic purposes (at least at present). These are my reasons: If you want to study a physical experiment, you'll always need to consider the device. You can compare different devices, but you can't do an experiment without specifying the device unless you want to run into philosophical ...

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Entanglement is the name for that what you want have explained. The entanglement is combining two particles like EPR Experiment. Till now nobody can say how it is possible, only the facts are well known - a new level of understanding of quantum relations is pointed up. A quantum computer is not direct compound with these topics. A quantum computer of ...

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No. Consider a unitary acting on the tensor product vector space that leaves that special elementary tensor fixed. Then the unitary has a direct sum decomposition $1\oplus U$ where $U$ is a unitary on a vector space of codimension 1 (i.e. the orthogonal complement of the fixed tensor). For simplicity assume that the same finite dimensional vector spaces, of ...

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