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I have recently found a reference proving the statement from my question. It is: Universal quantum gates Jean-Luc Brylinski and Ranee Brylinski In Mathematics of Quantum Computation, Chapman & Hall (2002) arXiv:quant-ph/0108062 Theorem 1.4, proven in section 8.

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I do know something about the history of the problem, but I don't know the answer. The first reference is S. Lloyd, Almost any quantum logic gate is universal, Phys. Rev. Lett. 10, 346–349 (1995). which gives a "proof" of this... but it isn't strictly correct. I don't know what exactly was wrong with it, because I never looked at it in detail. It doesn't ...

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No. Local unitary operations do not change the spectrum of the reduced density matrix. Thus, all eigenvalues of the reduced density operator must be the same in both states. On the other hand, if all eigenvalues of the RDM are equal, the states can be transformed into each other by local unitaries, since they have the same Schmidt coefficients. Note that ...

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There are a lot of misconceptions here so let's take it one step at a time. The entropy in classical mechanics is called the Gibbs entropy, $$S = - k_B \sum_i p_i \ln p_i,$$ where $p_i$ is the probability of some microstate $i$. This is essentially the same thing as Shannon entropy for physical systems. With this concept one can view knowing ...

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I think what you say is correct. By saying it should be unitary, he means it should be time-reversible. It is indeed time-reversible if before reaching the detector 2 it is in the same state as the initial state. It starts off with the superposition, and then only vertical polarization in the top part and horizontal in the lower part of the arms, which ...

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First, note that a unitary transformation can not modify the commutation relations.. $$AB-BA=C$$ Use the fact that $U^\dagger U=U U^\dagger=1$ to get, $$AU U^\dagger B-BU U^\dagger A=C$$ and then multiply by the conjugate transpose from the left and $U$ from the right, $$U^\dagger AU U^\dagger B U^\dagger- U^\dagger BU U^\dagger AU^\dagger= U^\dagger C ... 3 If A^\prime and B^\prime commute then there exists a set of mutual eigenvectors of A^\prime and B^\prime. For any eigenbasis of A^\prime there exists a unitary transformation W which takes that basis to the mutual eigenbasis of A^\prime and B^\prime. Consequently if there is a unitary operation such that  |\langle \psi | b \rangle |^2 = ... 2 Correct me if I'm wrong, but your line of thinking goes like this... Since quantum fields do not commute in general one can have finite variances for, e.g., particle number. Since the vacuum states defines a probability distribution we can find the corresponding entropy. However, here we are dealing with quantum physics. The entropy is in general S ... 0 First to start, and to be clear, let's talk about a macroscopic string. Then we can talk about a quantum mechanical system. For the macroscopic string, the state is more than just the position of each infinitesimal piece of the string - it is also the momentum of each infinitesimal unit of the string. So when you are looking at the spot where there has ... 1 I will give you an easier assignment to start with: explain the origin of Newton's laws, using nothing but statistical mechanics of Newtonian systems. Can you do it? No. Statistical mechanics follows from Newton's laws PLUS a few assumptions about phase space averaging. In the same way decoherence does not lead you beyond the framework of quantum ... 5 No, decoherence is not a new fundamental feature of quantum physics. It is a phenomenon which occurs when you couple a system with a few degrees of freedom to one with a lot of degrees of freedom and which you can derive from the postulates of quantum physics. There really is no measurement problem. Once you get a classical probability distribution (up to ... 0 I think that yes, but it's a bit intricate. The problem here is that different notions are mixed up. Usually, what people mean by "quantum superposition" is the fact that you can have a pure state that is partly in one state and partly in another. Now, as Timo points out, pure states in classical physics are just points in phase space. No simplex there, the ... -1 In classical physics, the state of a system is represented by points in configuration space or phase space. However in quantum mechanics, a state is represented by a vector and the state vectors are assumed to come from a vector space (or more precisely Hilbert space). From linear algebra you should know that we can add up two vectors from a vector space and ... 1 Hint: Start by representing \psi and \rho in the basis \{ \vert x\rangle,\vert y\rangle\}. Shouldn't be too difficult to calculate the action of U once you've done that. If you don't know how to take the partial trace, post the intermediate result and ask back. These steps will produce an equation like$$ A_{xy}(\psi,\rho)\vert x\rangle\langle ...

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The identity is true in any dimension. To see this, notice that both the left and right hand sides of the equation you wrote down are linear operators on the tensor product of vector spaces, so to show that they are equal, it suffices to show that they agree on a basis. Since the basis for a tensor product is the set of all tensor products of basis ...

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The charge of an atom is defined by its constituent number of protons/electrons and local fluctuations in their density distribution which cause instantaneous dipoles, unless we are talking about ions which have a permanent charge. Charge is a classical concept that has real meaning in classical physics and can be described in various fields ...

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Best answer I could find - Anyone care to chime in on this? I can say with 100% certainty, it would not be the same person. The amount of information contained in a human body is astronomical. This creates a higher mathematical probability (nearing infinity) that some of that information would be lost. Even if applying a future chaos theory to extract ...

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The density matrix for a pure state $|\psi\rangle$ is $\rho=|\psi\rangle\langle\psi|$. Note that this is a matrix in the sense that it takes some vector $|\phi\rangle$ to the vector $\rho|\phi\rangle=|\psi\rangle\langle\psi|\phi\rangle$. It has components \rho_{ij}=\langle e_i|\rho|e_j\rangle=\langle e_i|\psi\rangle\langle\psi|e_j\rangle=\psi_i ...

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Two types of cloning going on here. Biological cloning is possible now. Take a cell, make it behave like a freshly fertilized egg and grow a copy. Your clone will look the same but be somewhat younger. Random features (like some animal hair patterns) are not clonable. As biological clones grow from a single cell it will not be atomically identical if the ...

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Actually there is a theorem in quantum mechanics called the no cloning theorem which says that you can't clone a quantum state. However teleportation is possible and has been done experimentally, the teleportation here is the sense that you destroy the quantum state in one place and recreate in another place.

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