Hot answers tagged

4

Yes, particles become entangled without our intervention all the time. However, it isn't relevant because of the monogamy of entanglement. What monogamy of entanglement means (intuitively) is that a particle can have a large amount of entanglement with one, or maybe two, other particles, but it can only have a little bit of entanglement with each of many ...


3

Let's write the full Hamiltonian of the problem as $$ H = H_A + H_P + H_C + V,$$ where $H_A + H_P + H_C$ are the local Hamiltonians of $A$, $P$ and $C$, and their interaction is $V$. If you want the interaction to take place in arm $a$ of the interferometer, the interaction should be of the form $$ V = H_{AP} \otimes \lvert C_a \rangle \langle C_a\rvert.$$ ...


2

A three-party GHZ state can't be used for teleportation because extra copies of a qubit's value act like de-facto measurements. Charlie's copy of the entanglement's 'secret value' prevents some crucial interference from happening when Alice and Bob interact. You can still transfer information from Alice to Bob using a GHZ state and classical ...


2

Most importantly, the pattern detected at S2 will not depend on the detector D1. Otherwise, this could be used for faster-than-light signalling, which is impossible within quantum mechanics. Beyond that, what you will see at S2 will depend on which degree of freedom has been entangled. Generally, one would expect to see an interference pattern, but if you ...


2

When entangled photons are created in Bell-type experiments, it is their polarizations which are entangled. It is not clear to me that their positions or momenta are entangled in the same way. But anyway, engtanglement does not violate "no action at a distance" and in particular there is no way to see at location S1 what measurement is being made at location ...


2

If B released energy when A was measured, you could use it for communication. And you know the answer to "Can entanglement be used for communication?" is NO. You mentioned it in your question! Therefore... No, entangled particles don't release energy when their partner is measured. They don't change in any locally determinable way when their partner is ...


2

The physical description of what Bell's inequality is about is as follows. Suppose that the outcomes of measurements are described by stochastic variables. That is, each quantity has a value that is some number picked at random in some way. And suppose that for each system the quantities influencing how the numbers are picked are determined locally. That is, ...


2

tl;dr: Your points 1 to 5 are misunderstandings. The answer to your question follows from a better understanding: Entanglement is no active link, hence there is no need for instantaneous action. However, if you want a certain kind of ontology, then you must accept a certain kind of instantaneous action/non-locality (which doesn't necessarily violate ...


1

How by citing a statistical correlation, we can say "instantaneous action at a distance" is necessary. Suppose that there is a challenge which a simple argument establishes can only be solved by a classical probability distribution with 75% effectiveness, but which a quantum team of players (i.e. a team with access to a shared quantum state) can solve ...


1

If the particles are nontrivially entangled, then particle $A$ cannot be in an eigenstate of the energy operator (or any other operator that acts just on $A$'s state space) in the first place. If the initial entangled state is, say, $X\otimes Y+Z\otimes W$, where (for example) $X$ and $Z$ are energy eigenstates, then an observation of particle $B$ will ...


1

No-cloning theorem says that you cannot "clone" the state. As in there is no way to make to make perfect copies of a quantum state. In the simple teleportation case with two parties the Alice's state is transferred to Bob by using the EPR pair. There is no copying. But if Alice were to try and send her state to two people(Bob and Dina) then that would mean ...



Only top voted, non community-wiki answers of a minimum length are eligible