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This paradox is an artifact of describing the mirror classically. Quantum mechanics is itself free from paradoxes, and it is easy to see that this paradox vanishes the moment you describe the mirror as a quantum mechanical object. A simple way to see what goes wrong is to apply the uncertainty relation. Suppose that we have a freely floating mirror and we ...


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An entangled state is a joint state for both the particles. For instance if you had electrons then when you write $\left|\uparrow\downarrow\right\rangle -\left|\downarrow\uparrow\right\rangle$ for instance the minus sign in the middle there is a definite phase between the two parts. And the joint state itself has an overall phase. So for instance ...


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The rotation $R(\theta)$ in the Deutsch gate should be a rotation about the $X$ axis, not about the $Z$ axis, i.e., $$ R(\theta)\vert k \rangle = i\cos(\theta) \vert k \rangle + \sin(\theta)\vert 1-k\rangle\ , $$ $k=0,1$, as e.g. described on Wikipedia. Then, setting $\theta=\pi/2$ one immediately obtains the Toffoli gate. Note that if $R(\theta)$ were a ...


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A sum-frequency system with a "hot" mirror could act something like an optical switch: Unlike a switch, the output frequency will be different from either of the inputs. Edit: For an example of sum-frequency generation crystals see: Thorlabs Introduction To Periodically Poled Lithium Niobate (PPLN) (PDF) Thorlabs also sells hot mirrors.


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If you just sent a qubit in some state $|\psi\rangle$ from one place to another, it would be very difficult to protect the information it instantiates from decoherence. The decoherence is a result of the dependence of an outside system on the quantum information in the qubit. But by using entanglement it is possible for a system to instantiate quantum ...


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Entanglement is what cannot be created by local operations and classical communication (LOCC), so it can be used to do otherwise impossible tasks in a scenario where one is restricted to LOCC. Now what is the most general LOCC protocol? A measures, communicates the outcome, B measures (conditional on the first outcome), communicates the outcome, A ...


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It's important to remember that the quantum state is physical, and not merely a description of our knowledge or ignorance of a system. At every point during Alice's experiment, the pair has a single quantum state (insofar as the measurement apparatus can be treated as classical), but Alice, Bob and Charlie have different degrees of knowledge about what that ...


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Any set of commuting stabilizers can be mapped to any other set of commuting stabilizers by applying a unitary from the Clifford group, see http://arxiv.org/abs/quant-ph/9807006. The Clifford group is generated by CNOT, Hadamard, and the phase gate $\left(\begin{smallmatrix}1\\&i\end{smallmatrix}\right)$, so your $U$'s indeed have a special form. ...


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Yes, such a kind of exact decomposition is always possible. This is shown in Barenco et al., Elementary gates for quantum computation (Sec. 8, pg. 27 in the arXiv preprint), based on work by Reck et al. (which gives a corresponding decomposition where each elementary operation is the identity except for a $2\times 2$ submatrix). The construction given ...


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Not with certainty. However, Alice can probabilistically tell them apart by performing a measurement that asks "is the state of the system $(|0\rangle + |1\rangle)/\sqrt{2}$?" For example, if the qubit were a spin $1/2$ particle and $|0\rangle$ and $|1\rangle$ were spin up and spin down, this would correspond to measuring spin along the $\hat{x}$ direction ...


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I define the set of projectors $P_{a} = |a\rangle\langle a|$, what is the physical observable (like 'mopentum') that I need to measure to perform the operation? Even if you have some projections, even if they are each self adjoint, and even if they are mutually orthogonal and their range spans the whole space they still don't linearly combine into a ...


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Yes, you hit the nail right on the head, with respect to the H-H NBP, as well as Vilenkin's related tunnelling mechanism. But, there is a further problem as to why the H-H proposal simply does not work: Conformal modes lead to the Einstein-Hilbert action not being bounded from below, which in turn implies that the sum over all 4-geometries leads to a sum ...


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need to normalize the result? this is allready normalized? Do the frequencies add up to 100%? For the first qubit to be 1 lead us for 2 states 10 and 11 and therfor 4/6+1/6=5/6. Am i right? You can answer this with the same technique as for any measurement. A measurement needs a bunch of orthogonal finals states. You need to orthogonally project ...


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I am not sure I agree with the statement that this is not possible. Here is my scenario. We have two space explorers who both set off from Earth at the same time. Their destination is in opposite directions. The travel at the same speed. When they set-off from earth, at that time, a beam of entangled electrons are generated. One particle is transmitted ...


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Partial transposition flips the sign of $\sigma_y$ on the $B$ system. If you now conjugate the resulting matrix with $I\otimes\sigma_y$, you moreover flip the sign of $\sigma_x$ and $\sigma_z$, and thus of all Pauli matrix. On the other hand, this conjugation preserves the spectrum, as claimed. Note that it is only claimed that the corresponding matrices ...


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Just like an electron, light can have spin angular momentum and orbital angular momentum. To make a beam of light with orbital angular momentum you can make a diffraction pattern where instead of a series of parallel lines you have one line that forks where the one line comes in and two (or more) lines come out. Then if you shin the right kind of light beam ...


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If you want to hard code a specific set of qubits you could measure the spin (pass it through an inhomogeneous magnetic field) of a spin 1/2 system and then if you get the wrong spin, flip it (tune a photon to a wavelength based on a given uniform magnetic field strength). If you want to read an output you could do the same thing, measure the spin of each ...


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It is assumed and proven that before measurement there is no distinct spin up or spin down state unless of course it is somehow prepared. General state of a particle is described as a linear combination of spin up and spin down state. If you measure the z projection you get 50 % upz and 50 % downz. If you then take upz particles and measure the x projection ...


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If you prepare many silver atoms in the exact same state each time you might get 1000 out of 1000 to come out spin up. Or if you prepare many silver atoms in the exact same state (but different than the one above) each time you might get 0 out of 1000 to come out spin up. And when I say might it depends on if you pick the right state to prepare them and ...



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