Quantum entanglement seems to share information between particles instantaneously, as well as characteristics of these particles through changes that occur in each particle. If forces transfer energy between one system and another system, then force can be identified as an interaction rather than a force in the classical sense. what is the interaction for quantum entanglement being potentially non-local at very small scales?
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1$\begingroup$ I mentioned this on your other question as well, but just as a reminder, it's not acceptable to remove the content of a question (or answer) with an edit. $\endgroup$– David ZCommented Apr 5, 2015 at 8:08
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$\begingroup$ I understand and its been resolved. I'll remind myself to not be so unreasonable about the prescribed method associated with the standards of practice. $\endgroup$– nick leeCommented Apr 18, 2015 at 4:16
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Though the entangled states are non-local in some sense, they are not non-local in the sense of allowing superluminal transfer of information, by the no-communication theorem and others.
There is no force "responsible" for entanglement.
In fact, in quantum mechanics, there is no clear notion of such a thing as "force", although you might derive classical forces such as the force on the plates in the Casimir effect by it. Force is a classical concept, and although we like to speak of "forces" and "force carriers" when talking about gauge theories in QFT, these are the quantized versions of theories which classically had force. Quantumly, "force" is not to be identified with the classical concept, but rather with the general idea of interactions.
answered Apr 2, 2015 at 0:00
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$\begingroup$ In which sense are they non-local? $\endgroup$– image357Commented Apr 2, 2015 at 0:02
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2$\begingroup$ @MarcelKöpke: Some like to call the correlations entanglement produces "superluminal" or "non-local", since they (sometimes) correlate events that do not lie on each others light cones. Whether that is a useful or sensible thing to say is debatable, especially since also non-entangled states (such as the wavefunction of a photon that went through a beam splitter) exhibit such correlations (the reflected part "collapses instantaneously" when the photon is detected to have been transmitted), so it is not a defining feature of entanglement. $\endgroup$– ACuriousMind ♦Commented Apr 2, 2015 at 0:07
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Yes. Any interaction! Entenglement is only a quantum version of correlation. Let supose that you have two $\frac{1}{2}$ spins that interact one to another via magnetic interaction. This interaction can produce an entenglement. But this is not the only one. Some Amount of electrons are entangled in a metal due electrodynamic interaction.
You can produce entenglement by apply measurements to. In the case of two spins, the entenglement cam be produce by measuring the total angular momentum of the system. In this case, the correlation is made because if you know the state of one particle, this yields the knowledge of the other spin. In classical mechanics this is trivial because if you know the state of the whole system you can determine the state of the subsystems.
In quantum mechanics this is not always true. The total angular momentum $J^2$ don't commute with any projection of the spins $\vec{S}^{(1)}$ and $\vec{S}^{(2)}$. This imply that we can't divide the state of the system in the state of the parts when we have peculiar states (entangled states). Only the sum of the projections commute with the total angular momentum. In this case, the entangled states are $|1,0\rangle$ and $|0,0\rangle$ and the non-entangled states are $|1,-1\rangle$ and $|1,+1\rangle$, with has a trivial correlation as the classical ones.
We have:
$$ |0,0\rangle =\frac{|+,-\rangle -|-,+\rangle}{\sqrt 2} $$ $$ |1,0\rangle =\frac{|+,-\rangle + |-,+\rangle}{\sqrt 2} $$ $$ |1,-1\rangle =|-,-\rangle $$ $$ |1,+1\rangle =|+,+\rangle $$
Note that the non-trivial states (entangled) are superpositions of the $Sz^{(1)}$ and $Sz^{(2)}$ eigenvectors. This superposition tells us that we can't define either the system are in $|+,-\rangle$ or in $|-,+\rangle$. We can only determine the probabilities of this possible measurement.
Now, if you apply some interaction in one spin, the other spin don't feel nothing! For example: if you measure the spin of one particle in the state $|0,0\rangle$, YOU know the spin of the other. Only this. Because the measurement, $J^2$ is not defined anymore and you can define a spin for each particle now. Using the correlation, if you measure +, then the other is -. But nothing actually happens in the other side! A quantum state is only a matemathical object that describe your state of knowledge about the system. And quantum mechanics is not realistc, so this state of knowledge is actually the final word. We don't get any thing "deep" that accounts this features. Quantum mechanics is subjective in this sense.
If we put another observer in the other side, in any way he can determine if you measeure or not by local operations. He needs to act on both spins for determine if you measure or not. As any other correlation. But a weird one, since is combine with some quantum features. This features is only feelled when we are doing things in both spins. You may find in some self studing that this quantum correlations violate some inequalities obeyed by classical correlations.
Entanglement is only a suitable combination of usual correlation and complementarity principle.
