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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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I join the comments of Julian Fernandez. Just some hints for some of your ideas: Spacetime is relative i.e. observer-depending. It seems that behind relative spacetime there is a system of absolute time (= proper time) and absolute space (=space without considering time dilation and length contraction which are relative) where spacetime is "mixing up" ...


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Properties of the Von Neumann entropy 1. Purity. A pure state ρ | φ 〉〈 φ | has S ( ρ ) = 0. 2. Invariance. The entropy is unchanged by a unitary change of basis S ( U ρ U † ) = S ( ρ ) , because the entropy depends only on the eigenvalues of the density matrix. 3. Maximum. If ρ has D non-vanishing eigen- values, then S ( ρ ) ≤ log D, with equality when all ...


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But... all quantum states are really pure states right? There exists the density matrix formulation for the many body system quantum mechanically. What happens in a many body problem, order of 10^23 molecules per mole that are the appropriate numbers for a thermodynamic formulation, is that the off diagonal elements are so small that they are ...


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The issue is that when a wavefunction collapses it has an inherent randomness to it. Since entropy is fundamentally related to information, I'll start with information to explain why this is significant. Information, Randomness, and Entropy If you think of the information of a state as what is needed to completely define a system something interesting ...


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I think it is a mistake, in this case, to think of entropy as "a description of our ignorance." Rather, I would suggest that you think of entropy as a well-defined, objective property provided that you specify which degrees of freedom in the universe are inside and outside of your system. The content of this statement isn't really different, but it ...


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There is a difference between the 'fine grained' and the 'coarse grained' entropy. If we start with a pure state (zero entropy), and we time-evolve it, the entropy indeed stays zero by unitary of time evolution. The fine grained entropy did not change. The coarse grained entropy is what we usually call the thermal entropy, and is the thing that always ...


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You are wrong in a couple of things: 1) As long as your particle (the system {S}) evolves unperturbed, it evolves unitarily - i.e. its evolution can be described as a unitary transformation. (There is an exception from this rule, but if the internal structure of the particle doesn't change during the particle evolution, the exception is not relevant). By ...


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The short answer to the first question is yes. You need enough power above the noise level of your systems (and space) to even receive any signal from somewhere else. You can look up articles on the Shannon-Hartley theorem, which discusses the relationship between information transfer and power, bandwidth, etc. This is one of the most limiting factors ...


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I'll answer my own question and hope this information is useful to someone. I'll take $\hbar = 1$ and will deal with systems of one degree of freedom (the generalisation should be obvious). The normalisation factors are very confusing, so I'll omit most of the reasons for them to be what they are. First, let's take a look at the good old continuous Wigner ...


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The conjecture is false. A counterexample for a single-qubit state and a single-qubit ancilla is: $$P_1:=\left(\begin{array}{rrrr} \frac{1}{2} & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \frac{1}{2} & 0 & 0 & \frac{1}{2} \end{array}\right), \qquad P_2:=\left(\begin{array}{rrrr} \frac{1}{2} ...


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I don't really know what answer you expect here. As you have found out yourself, the Gottesman-Knill theorem tells you that stabilizer circuits can be efficiently simulated by a classical computer. Teleportation can be implemented that way, hence you can efficiently simulate it on a classical computer. What does that mean? Well, give the computer a random ...



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