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The quantum theory of information and computation sheds light on many issues in fundamental physics. First, it explains that any physical system can be simulated by a universal computer. This sheds some light on the issue of whether we can expect to be able to understand the laws of physics. Second, such a computer can be constructed by composing almost ...

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Given that the $\mathcal E_i$ are linear operators, i.e., $\mathcal E_i(\alpha A + B)=\alpha \mathcal E_i(A) + \mathcal E_i(B)$ for all $A$ and $B$, it is true that $\mathcal E_1(\rho)=\mathcal E_2(\rho)$ for all $\rho\ge0$ implies that $\mathcal E_1=\mathcal E_2$. The argument goes as follows: Every operator $A$ can be written as $A=X+iY$, with ...

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Quantum computing research already has improved our understanding of the basic laws of nature in what I think are several important ways. "Quantum" does not mean "microscopic" One of the types of system used to implement quantum bits (qubits) is a superconducting resonant circuit. These circuits are large enough to see with your naked eye, yet they exhibit ...

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John Preskill's lecture notes http://www.theory.caltech.edu/people/preskill/ph229/ It will be better if there is some answers to the exercise.

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In the context of physics, there are "natural" equivalence relations motivated by the following notion: mathematical objects that determine the same physics should be considered equivalent. These equivalence relations lead to partitions of the sets on which they are defined. Equipped with this idea, let's examine the two points you mention: Let $\mathcal ... 2 A quantum state does not have any classical memories or classical output, period: a classical memory refers eventually in every application to the process of data stored by copying. As it has been proven quantum states cannot be copied. However they can be entangled. A quantum memory may store a quantum state, but not for the purpose of copying in any ... 1 There is no precise answer to the cause of entanglement is known (I did not come across any such answer). The superposition principle and the tensor product structure of Hilbert space of combined system leads to the possibility of entangled state. It is well understood in the case of pure state, but for the mixed state case it is quite involved. 4 I.The cause of the quantum entanglement is that the wave-function (w.f.) of the involved particles doesn't have the form of the w.f. of independent particles. So, let's begin by defining what is the w.f. of independent particles. The w.f. of a single particle, let's call it A, can look like $$|A> = \frac {\lvert x_A \rangle + \lvert y_A \rangle.} ... 4 This is not an answer for the obvious reason that this question cannot be answered easily, hence why it is an open area of research. What I will provide though is links to how something like this is done. The idea resides in the dynamics of open quantum systems, which are systems that are constantly interacting with the environment and hence tend to become ... 4 The role of coherence in biological electron transport, e.g. within chromophores, is an open and actively researched problem in quantum optics/quantum chemistry. The two classic theoretical treatments which kick-started the field are by Plenio & Huelga and Mohseni et al.. Since then an enormous literature has emerged on the topic. A basic, generic model ... 7 I actually don't think that this view of light being in a quantum superposition is anything new: what Discover magazine is describing (I believe) is the stock standard picture of how one would describe a system of cells, molecules, chloroplasts, fluorophores, whatever interacting with the quantised electomagnetic field. My simplified account here (answer to ... 2 x is compatible with p_y that means they have the same eigenvectors. That is perhaps not the best way to say it. The fact that x and p_y are compatible tells you that there exists at least one set of basis vectors which are simultaneously eigenvectors of both x and p_y. Let's call this set \{v\}. Similarly, the fact that x and y are ... 1 A density matrix is a matrix that describes a quantum system in a mixed state, a statistical ensemble of several quantum states. This should be contrasted with a single state vector that describes a quantum system in a pure state. The density matrix is the quantum-mechanical analogue to a phase-space probability measure (probability distribution of position ... 2 Here is the erroneous step: "Now, if Alice between the steps 3 and 4 (the ideal time would be as the photons get very close to 4, right before the double slit as expected from their synchronized clocks and known distance),measures the polarisation of her photons, the entanglement is broken before Bob's photons arrive at the double slit at step 4. Thus the ... -1 Please keep in mind Dirac's dictum "a particle interferes only with itself" (I am not sure about the word "particle", it may be that he said "photon"). The polarization of Bob's photon is not relevant here. Be it vertically polarized, or horizontally polarized, the fringes will fall on the same places, because one photon doesn't care of the polarization of ... 0 One of the answers to "How does one produce entanglement?" gives an example of how to obtain pairs of entangled particles. However, it is a complicated example. The simplest procedure, and widely used in experiments, is down-conversion of ultraviolet (UV) photons. Such photons are sent on a type of crystal that we name "non-linear crystal". In short, inside ... 1 Let's say that "decoherence" is that transition from a pure quantum state to a mixed state due to interactions with the environment. (A reasonable definition?) Mixed states are NOT decohered states, they are states where the phases of the wavefunctions are well defined, just not in an eigenstate that will give a unique eigen value at measurement. ... -2 I haven't studied this in any detail, so take what I say with a grain of salt. But from summaries I've read, like the essays on decoherence.de, I think that in the Copenhagen interpretation "collapse" would still have to be understood as conceptually distinct from decoherence. I think a way to see this would be by imagining an idealized Schroedinger's-cat ... 1 Since this seems a homework exercise, here's a sketch: I'm not sure about the \mathcal{N}-part in the formular, but in general: Note that \sigma_i form an orthonormal basis of the Hermitian matrices according to the Hilbert-Schmidt inner-product (\langle A,B\rangle:=\operatorname{tr} (A^{\dagger}B)\rangle). This means that you can write$$ ... 2 This is quite simple. Consider the operator$H$on the Hilbert space$\mathscr{H}$, in your simple example it has a spectral resolution: $$H=\sum_{n}E_n \lvert n\rangle\langle n \rvert\; .$$ Each eigenvalue has multiplicity 1. Now the operators$H_1$and$H_2$on$\mathscr{H}\otimes\mathscr{H}$have the same spectrum of$H$, but each eigenvalue has ... 0 Something is wrong with your scheme - it just can't work at all (we don't even need to invoke no-cloning): The positron travels to Alice, the electron to Bob. If Alice measures the positron in the spin down direction, Bob makes a lot of copies of the electron using a cloning device, and then measures them. If he gets all spin up, he knows Alice made the ... 1 The density operator combines pure quantum states into a mixed quantum state. The basic idea is to take a system composed of many pure states and to represent them as a single object, which evolves in time, as a complete system. In this example, the mixed state is represented as a Block sphere, and the$\vec{\sigma}$is a pauli matrix. The Bloch sphere is ... 2 What makes you think that information can be transferred in this way in the first place? I cannot tell you why your method won't work if I don't know what method you have in mind. In any case, I can take a guess as to what you're referring to. Suppose we have a pair of electrons in the state (up to normalization)$\left| +\right> \left| -\right> ...

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Can this scenario be considered as erasure of information as referred to in Landauer's principle? No. The reason is that the whole process is reversible and Landauer's principle specifically needs an irreversible step - otherwise no "erasure" can have taken place. Where's the problem? It's here: The state $$\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$$ ...

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ACuriousMind's Answer pretty much summed up the reasons, which are essentially mathematical. If you want to grasp the "physical significance", then I suggest you should work through an example: think of two quantum systems, each with three base states: $\left.\left|1\right.\right>$, $\left.\left|2\right.\right>$ and $\left.\left|3\right.\right>$. ...

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This is more of a comment, but my rep is too low... You say, The state is initially in $|g\rangle$, but undergoes a $\pi/2$ rotation about some axis... and you end up with $\theta=\pi/2$, $\phi=0$, ie you are on the $x$-axis and your rotation was simply about $y$, surely?

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The "Kronecker product", better known as the tensor product, is the natural notion of a product for spaces of states, when these are considered properly: A space of states is not a Hilbert space $\mathcal{H}$, but the projective Hilbert space $\mathbb{P}\mathcal{H}$ associated to it. This is the statement that quantum states are rays in a Hilbert space. ...

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What you need is spherical linear interpolation, which is usually done with quaternions. Well, quaternions can be represented by the pauli matrices. If you don't mind the hand-waviness, a rotation operator about the axis $\vec{s}$ is given by $$R = e^{i \frac{\theta}{2} \vec{s} \cdot \vec{\sigma}}$$. The vector $\vec{s}$ is perpendicular to the great ...

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Yes, as it turns out there are distinct types of multipartite entanglement not witnessed by the spectra of reduced states. The simplest example is probably the one from Bennett et al., "Exact and asymptotic measures of multipartite pure-state entanglement", Phys. Rev. A 63, 012307 (2000). For the case of three 4-dimensional systems, consider the states ...

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State vectors define pure quantum states of a system, and, for an isolated system, evolve with time following the Schrödinger equation (in the Schrödinger picture; in the Heisenberg picture they are constant. Density matrices define classical statistical mixtures of pure quantum states. These can arise in two ways: When we have incomplete information ...

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Density matrix is NECESSARY when the quantum system is not in a PURE state, i.e. it cannot be described by a wave-function. Such a case is when a beam of freely moving electrons is not spin-polarized, i.e. there is no direction in space along which all the electrons have the same spin-projection. (For finding whether the electrons are polarized, and on ...

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The density matrix formalism follows naturally from a state vector formalism when we average over a part of variables. It is not a "more general", but a more common approach in practical applications. Sometimes it is the only possible formulation due to experimental restrictions.

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To answer your question shortly: No, they are not the same phenomenon. First of all, it is much easier to think of quantum states as vectors (in something called the Hilbert space, but simply put they obey linearity), and not as particles or waves. Superposition Let's start with a single particle qubit (since you're talking about quantum information) ...

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First of all the words "exchange of information" are not so good. An entanglement is a CONSTRAINT on two particles or more. For instance the famous PHOTON SINGLET is described by the state (1) |Ψ> = [!/sqrt(2)] {|u>|u> + |v>|v>} where u is whatever direction in space that we want to choose, and v is perpendicular to x in the polarization plane (which is ...

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Let me try to (partially) answer the following question then: Can I buy or build a quantum system, where I can perform certain quantum computations (maybe not universal) at home? Tl;dr: The biggest caveat is the amount of control necessary to perform any controlled computation at all. You simply don't have it, except by studying a system with knew ...

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