Tag Info

New answers tagged

0

I've found this paper: Cosmological quantum entanglement, of E. Martin-Martinez and N. C. Menicucci. (last revised 19 Oct. 2012) Abstract We review recent literature on the connection between quantum entanglement and cosmology, with an emphasis on the context of expanding universes. We discuss recent theoretical results reporting on the ...


1

You can always write any density matrix in the form $\rho = e^{-H_A}$ in a limiting sense, as long as you allow some of the parameters entering $H_A$ to become infinite. This follows from the identity $$\rho = e^{\ln \rho},$$ but notice that if $\rho$ has some eigenvalues equal to zero, then the eigenvalues of the exponent become infinite (or are not ...


3

I'm not a fan of explicit calculations, so let us answer a bit more general question: Given two two-dimensional Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ with basis $\lvert \uparrow \rangle_i$, $\lvert \downarrow \rangle_i$, when is a state in $\mathcal{H}_1 \otimes \mathcal{H}_2$ entangled? Observe that $\lvert \psi \rangle_i = u_i\lvert \uparrow ...


1

This problem can be solved using Peres–Horodecki criterion (or PPT criterion). But the simple way to do this is to compute directly. Suppose $|\phi⟩ = (A|0⟩ + B|1⟩)_A (C|0⟩ + D|1⟩)_B$. Compare the coeffient, we must have: $AC = α_0β_0, BC=0, AD= \alpha_0 \beta_1, BD= \alpha_1 \beta_1$. If $\alpha_1, \beta_0, \beta_1$ all non-zero, the equations above ...


1

The spin is entangled if $\alpha_1\beta_1\neq0$ because a measurement on the first spin gives you information on the expected measurement values on the second. In particular if you get a 1 in the first spin you know you will get a 1 in the second spin. And knowing that you got a 0 in the first spin will changes the probablility of the measurement of spin 2. ...


3

It is an entangled state, as you concluded yourself it cannot be written as a direct tensor product. To recap again: $$ \left( \begin{array}{c} \alpha_0\\ \alpha_1\\ \end{array} \right) \otimes \left( \begin{array}{c} \beta_0\\ \beta_1\\ \end{array} \right) = \left( \begin{array}{c} \alpha_0 \beta_0\\ \alpha_0 \beta_1\\ \alpha_1 \beta_0\\ \alpha_1 \beta_1\\ ...


3

A system's density matrix can be written in the form you cite if and only if it is a classical mixture of factorisable pure states. A factorisable pure state is, of course, one that can be written as a tensor product $\psi_A\otimes\psi_B$, where $\psi_A$ and $\psi_B$ are pure states in subsystems $A$ and $B$, respectively. Correlations between measurements ...


0

Yes, it's enough! Require $S \le 2$ is the similar as require percent of heads to be always $\le 1/2 $ when flipping a coin. It should be always $1/2$ in theory, but obviously not in practice. Even 100 000 repetitions of coin flipping won't guarantee you $\le$ 50000 heads. Most or results will be 50000 $\pm$ 800 ($5\sigma$). Returning to the S, 100000 ...


1

As opposed to type II phase matching that produces orthogonally polarized photons in parametric down conversion (PDC), the type I PDC process produces identically polarized photons in the output signal and idler modes (labels $s$ and $i$ below). Normally the output state from type I PDC is not entangled: to get the required phase matching in the nonlinear ...


1

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.


1

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 ...


1

Any two particles become entangled after an interaction. Entanglement is truly everywhere and occurs constantly. The reaching of equilibrium of the temperature of a coffee cup and the room has not a lot to do with entanglement and more with the irreversibility of the motion of many particles (entropy).



Top 50 recent answers are included