Tag Info

New answers tagged

0

Here are my two cents on this. In quantum mechanics, one, two,...many particles are described by a state function. The state function is gives a probability distribution that includes all the possible measurable values of the one,two...many particles. Let us take two for simplicity and because here is where confusion arises. Because of conservation of ...


1

In case your box was classical, it'd output either vertically polarized A and horizontally polarized B or vise-versa, a 45 degree polarizer placed in front of both would allow both to pass half of the time, and block them half of the time - but the pass\block statistics would not be correlated, while in the entangled case, if A passed, B will be blocked, and ...


5

What you are proposing is called a local hidden variable theory. Bell's theorem proves that any such local hidden variable theory is inconsistent with behavior predicted by quantum mechanics. Bell test experiments have been performed, which show that the predictions made by quantum mechanics are correct, in ways that cannot be explained by a local hidden ...


1

My understanding is that the Aspect experiment shows that your understanding is wrong. http://en.wikipedia.org/wiki/Aspect_experiment


0

Seeing your last question, why would you even consider this scenario with entangled photons? Seems like the same question can be perfectly asked with one single photon. But anyway, ignoring the possible optical issues in the scenario, like the exact color of the film you use, or what Nanite has commented about, either way you will not see a superposition of ...


0

Mathematically you have a superposition of two Eigenstates and you use it because you don't know all parameters of the source. And yes, you use a probabilistic source, otherwise you will not do any experiment with it. Using your special source you get always the next result: Measuring one of the photons you know immediately the Eigenstates of the twisted ...


1

There exists a precise way of calculating the entanglement entropy in a conformal field theory via the Ryu-Takayanagi (RT) prescription in the context of the AdS/CFT correspondence. The RT prescription says that the entanglement entropy of a sub-system $A$ in the CFT$_{d+1}$ that lives on the boundary of AdS$_{d+2}$ is given by the minimal area surface ...


1

All different QKD protocols are well covered in most Quantum Information textbooks, e.g. Quantum Information by Jaeger. Trying to give a complete, understandable coverage of the matter is impossible in a post like this, so I'll just give you an overview: In the E91 scheme, entangled photon pairs are used between Alice & Bob, and unlike the single-photon ...


2

The answer is definitely yes. The ground states (and low-lying eigenstates) of many-body systems are generically entangled. Examples include the ground states of local quantum field theories (which describe the fundamental particles and forces of nature) and ground states of fermionic lattice models (which describe much of the solid matter we see around us). ...


1

In short, entanglement is perfectly normal. I am sure that entanglement is ubiquitous in, say, atoms with more than one incomplete subshell, as well as in some kind of organic molecules, but I am not a quantum chemistry expert and can’t provide an easy-to-realize example. Generally, any decay process produce particle states that are entangled in some way ...


6

The morally correct answer is that the measurement of one spin in the EPR-entangled pair eliminates the entanglement as it picks a particular factorized basis vector for the measured spin, and the total wave function therefore has to factorize to $\psi_\text{just measured}\otimes \psi_{\rm something}$. If you parameterize the multi-fermion states in terms ...



Top 50 recent answers are included