-1
$\begingroup$

Is it a mathematical certainty demonstrated through some proof or is this simply what observation shows? How sensitive are the measurements that determines the "communication" (I know... It is not "communication" however, there is some sort of interaction happening... maybe.) is instant? What I mean is, the instrumentation has some sort of limit. Maybe it says that that the "action at a distance" happens in 0.00000 seconds; however, if the instrumentation were to measure down to 0.000000 seconds, it would say a delay of 0.000001 seconds. Now, those numbers were pulled from thin-air, but my basic premise is How do we know that the action is instantaneous and not something like 100c? I know, general relativity, but what if the mythical negative mass particles are what carries the interaction between the entangled particles. I have no idea how that would be possible, but then again, is instantaneous vs. superluminal really all that more or less mind-blowing?

$\endgroup$
1
$\begingroup$

Quantum entanglement of two particles is a way of stating that there exists a unique quantum mechanical solution ( a mathematical function) that describes the probability of finding the particles with specific attributes at specific spacetime points with specific energy and momentum. The probability is what is described/known/fitted. The shorthand of "entanglement" is what confuses people and leads to all sorts of unphysical expectations.

In the quantum mechanical framework, the underlying framework of matter, conservation laws also hold for these probability functions, energy, momentum, angular momentum, and conservation of quantum numbers as has been documented by innumerable experiments in particle physics. For simplicity let us take two entangled/(statefunction) particles, A and B, with their quantum numbers: this means that we have one function that will give us the crossection ( for example) the probability of decay. To get at the decay probability we have to accumulate a distribution of many similar particles with the same entanglement/state. That is all we can measure, accumulated individual instances.

Take a pi0 decaying into two photons/gamma-rays. We can measure the gamma rays with a photon detector for each individual event and accumulate a decay distribution curve. In principle one can measure the spin of the photons . As the pi0 has zero spin, if we measure the spin of one photon to be +1, we immediately know from conservation of angular momentum that the spin of the other photon is -1. There is no time to be measured or calculated for this knowledge. It comes inevitably from the conservation laws and the velocity with which the neurons in our head make the correlation.

This "instantaneous" business is true in every day life in situations where the bounds are known. If you do not find your keys in your pocket you instantaneously know you forgot them on the table, the velocity of acquiring this knowledge is the velocity of neural connections in thought processes.

$\endgroup$
-1
$\begingroup$

There is no action at a distance and the question makes no sense. A pair of particles must, at any moment occupy some quantum state. That quantum state dictates the probabilities of the outcomes of any measurement you can make. What would it mean for there to be a "delay" in this, and how would that delay differ from a "delay" in any other measurement of any other particle, entangled or not?

Specifically: I have a particle in front of me. I measure, say, its spin in a particular direction. Does "delay" mean that there's a time lapse between triggering the device that does the measuring and being able to read the measurement? Or what? And if that's what it means, why would entanglement have any effect on the so-called "delay"?

$\endgroup$
-2
$\begingroup$

The "instantaneous" collapse of the quantum entanglement is a direct consequence of the axioms of quantum mechanics. Also, in experiment, it has been proved that the speed of "action at a distance" is larger than 10000c. And there is no way to test the absolute instantaneous in experiment.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.