

One piece closer to a working electronic quantum computer puzzle kit.






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By Torment on 1/13/2010 4:32:29 PM , Rating: 3

They are entangled *because* they the quantum wave function has not collapsed (which corresponds to collapsing to a state). Say, for example, that an atom emits two electrons simultaneously such that the net spin must be zero. Both electrons are in a superposition of quantum wave states of up and down. When you measure electron1's spin state, you force a collapse of the quantum wave state, thus forcing electron1 to be either up or down. Since the net spin must be zero, this forces electron2 to be the opposite spin--instantaneously, no matter the distance. However, since the spin measurement on electron1 is entirely probabilistic, so is the resulting spin state on electron2. In fact, anyone observing electron2 would not be able to determine if its wave state had been collapsed by its counterpart having already been observed (otherwise, you could still transmit information).

IOW, everything looks completely random at both ends.

By Torment on 1/13/2010 4:38:38 PM , Rating: 2
Beware the errant 'they'

By Spuke on 1/13/2010 4:41:32 PM , Rating: 2
IOW, everything looks completely random at both ends.
Are they in completely random states only or can they be stateless also before observation? Or is that the same thing? Thanks for the explanations.

By Fritzr on 1/13/2010 8:59:34 PM , Rating: 2
In the case of unknown until observation, stateless and random but identical to partner give identical results when observed. If stateless the wave form collapses to one value or the other and that value is observed. Random, the particles have an unknown value until observed and that value is observed. It will require some test of the pre-observation state to distinguish the difference.

By Torment on 1/14/2010 6:38:08 PM , Rating: 2
Your entire post is rather incoherent and nonsensical, but I just want you to think about your last sentence. Do you see the logical fallacy?

By Fritzr on 1/15/2010 5:52:43 PM , Rating: 2
Yep. I did not add a big arrow saying this was a fallacy due to my belief that it is obvious to a reader paying attention.

To clear up the other.
1) If the particles are stateless until caused to collapse by observation, then a discrete value is observed.
2) if the values have a random, but discrete value prior to observation, then a discrete value is observed.

The observed value in case 1 is indistinguishable from the observed value in case 2.

The logical fallacy of the final statement implies that a different test is required to determine if the particles are stateless or possessing a definite state that has a discrete value.

Observation of the discrete value says nothing about the state of the particle before observing it's value.

By Torment on 1/13/2010 8:57:16 PM , Rating: 2
An important distinction is that the electron is not both both up and down at the same. Rather, their wave function is in some admixture of states corresponding to up or down. The electron isn't spin up or spin down until you observe it, the act of which collapses the wave function. This is a probabilistic event, meaning that if the wave functions were in an equal admixture, it's a coin toss whether the electron ends up as spin up or spin down.

So, if I understand your question, the electrons, prior to observation, have no (spin) state. It only has wave functions corresponding to physical states. In the example given, those wave functions have equal probabilities, so that when you observe a stream of electron1's, it will have a random distribution of spin-up and spin-down. Because they are entangled, observing electron1 will force electron2 into opposing state, but to the observer of electron2, it is still purely random, even though it is deterministic at this point.

If, however, you could measure whether a wave-state had been collapsed *prior* to your observing it(and you were careful about timing, as doing so is surely an act of observation and you want observer1 to observe first), observer1 could then send signals in the form of stream segments that were either observed or unobserved on his end. Unfortunately, that is not knowable, so still no action at a distance.

