How can we define Quantum entanglement (in QFT)?
What are the known mathematical settings and special physical (or logical) conditions of QE applied to Quantum computing?
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$\begingroup$ That depends what do you mean by random and organised? Generally quantum entanglement is when particles interact in such a way that they affecting ones quantum state affects all the others states $\endgroup$– user43087Commented May 2, 2016 at 6:38
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2$\begingroup$ Possible duplicate of How does QFT help with entanglement? $\endgroup$– AccidentalFourierTransformCommented May 2, 2016 at 9:24
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$\begingroup$ Why do you think entanglement in QFT is defined any differently than in QM? Why do you think you can regard entanglement as "information exchange" at all, etiher organized or random? It's not clear what you're asking. $\endgroup$– ACuriousMind ♦Commented May 2, 2016 at 10:40
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$\begingroup$ @ACuriousMind I'm not saying that entanglement is defined differently. The reason I mentioned QFT is that it answers the question of nonlocality in a clearer way. As for the "information exchange" part It appeared that it's not the way physics works. I was asking for a mathematical answer for the spooky action at a distance... $\endgroup$– user115519Commented May 2, 2016 at 11:06
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$\begingroup$ @AccidentalFourierTransform I think you noticed that my question is threefold :) and I mentioned QFT only because I know it makes more sense $\endgroup$– user115519Commented May 2, 2016 at 11:10
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2 Answers
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Quantum entanglement is the property of two objects $A,B$ – more precisely two subsystems – or a relationship between these two objects whose quantities or observables aren't independent of each other. It means that there exist some quantities $a_j$ and $b_k$ describing $A,B$, respectively, such that the probability distribution for these observations doesn't factorize, as expected for "independent propositions": $$ P(a_j=\lambda_c, b_k=\mu_d) \neq P(a_j=\lambda_c) \times P(b_k=\mu_d) $$ In other words, there exists at least one measurement that may be done on $A$ and one measurement done on $B$ such that the results of the two measurements are predicted to be correlated.
In quantum mechanics, such state (situation of the two objects) almost always results from the interaction of the systems $A,B$ in the past – when they were in contact or close enough to influence each other – and the mathematical description of the pure (maximally known) state of $A,B$ in quantum mechanics is in terms of superpositions: $$ |\psi\rangle = \sum_{m=1}^N c_m |\alpha_m\rangle \otimes |\beta_m\rangle $$ Whenever at least $N\geq 2$ terms on the right hand side are needed to express the state $|\psi\rangle$, we say that this state $|\psi\rangle$ is entangled. As I said, it's almost always the case when the two objects interacted in the past but weren't observed separately so far.
Quantum entanglement is nothing else than the correlation of the two objects $A,B$ in the "quantum regime" i.e. when the description in terms of state vectors is needed because the quantum coherence (information about the relative phases of the probability amplitudes) is preserved.
So quantum entanglement may be perhaps said to be a particular feature of the "organization of information", although the definition of the entanglement is in no way given by the words "organization of information". While "organization of information" is at least slightly correct, the phrase "random data exchange" isn't appropriate for the quantum entanglement in any way.
The entanglement is a correlation that resulted from some interactions in the past and doesn't imply any exchange in the present. The correlations between the two measurements are consequences of the entanglement which is a consequence of the contact in the past; the correlations are not a consequence of any information exchange at the present.
answered May 2, 2016 at 7:11
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2$\begingroup$ Do some people get confused about causality and entanglement because they're confused about classical probability? As you say $p(A|B) \neq p(A)$ for entangled observations (or any dependent events). But the connection between $A$ and $B$ is logical not causal, so the fact that $A$ and $B$ could be causally disconnected is irrelevant. Is that fair? $\endgroup$ Commented May 2, 2016 at 10:40
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2$\begingroup$ @user115519 I think you're going to keep being confused if you keep thinking about "collapsing the wave-function". It's kind of a left-over from some early quantum theories, such as the Copenhagen interpretation. Just think about the physical reality of the configurations - things affect other things, and if those things are entangled to each other, their configuration space is limited. Interaction necessitates entanglement; what you may be thinking about is something more like "exclusive entanglement", e.g. the two particles are not entangled with anything else, useful in computing and comms. $\endgroup$– LuaanCommented May 2, 2016 at 12:24
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1$\begingroup$ @user115519 Based on a quick skim of wikipedia, it might be. We always called it configuration space - it's all about the joint configurations of particles, and reality is the sum over all the paths through the configuration space. As for the entanglement, what we (historically) call entanglement is actually a special case where there's hardly any entanglement (in an ideal scenario, just the two entangled particles) - so "quantum independence" might be a better name. The default state is "entangled to tons of stuff" - not very useful for quantum computers etc., since there's too much "noise". $\endgroup$– LuaanCommented May 2, 2016 at 14:34
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1$\begingroup$ @user115519 But really, whatever I say in comments is probably going to be even more confusing and kind of wrong. There's tons of books on the subjects, and I don't know of a simple way to teach it in 500 characters :) Just note that we definitely aren't stuck in the old Copenhagen view - there's plenty of alternate interpretations that seem to work better (Luboš is having loads of fun with strings, for example :P). $\endgroup$– LuaanCommented May 2, 2016 at 14:36
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1$\begingroup$ Dear @user115519 - the logic is exactly the other way around than you suggest. The "noise" (decoherence) that ruins (or makes harder, and requiring fixes or repetitions) a quantum calculation isn't due to "too much entanglement". It's due to the quantum computer's being measured - and therefore disturbed - by the environment which reduces the entanglement. These reductions of the entanglement is something we need to prevent. $\endgroup$ Commented May 4, 2016 at 5:44
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Quantum entanglement is a physical phenomenon that occurs when pairs or groups of particles are generated or interact in ways such that the quantum state of each particle cannot be described independently — instead, a quantum state must be described for the system as a whole.
Measurements of physical properties such as position, momentum, spin, polarization, etc., performed on entangled particles are found to be appropriately correlated. For example, if a pair of particles are generated in such a way that their total spin is known to be zero, and one particle is found to have clockwise spin on a certain axis, then the spin of the other particle, measured on the same axis, will be found to be counterclockwise, as to be expected due to their entanglement. However, this behavior gives rise to paradoxical effects: any measurement of a property of a particle can be seen as acting on that particle (e.g., by collapsing a number of superposed states) and will change the original quantum property by some unknown amount; and in the case of entangled particles, such a measurement will be on the entangled system as a whole. It thus appears that one particle of an entangled pair "knows" what measurement has been performed on the other, and with what outcome, even though there is no known means for such information to be communicated between the particles, which at the time of measurement may be separated by arbitrarily large distances.
